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Probabilistic formulation of Hadwiger-Nelson problem

2018-05-23T10:02:57Z

Nazgand: /* Bounds on p_d for 4-colourings */

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> d\geq 1/n, n>1</math>
|
|
| <math>1-\frac{1}{n}</math>
| (n+1)-gon with one edge length 1 and the rest d, Lemma 34
|
|-
| <math> d\geq 1/(n \sqrt{3}), n>1</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d, Lemma 34
| Not better than the above on intervals <math>\left(\frac{1}{7},\frac{1}{4\sqrt{3}}\right),\left(\frac{1}{4},\frac{1}{2\sqrt{3}}\right)</math>
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/4
| <math>G_{21}</math>
| 1/2
| Spindle
| Lower bound computer verified
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|-
| <math>\frac{\sqrt{6} \pm \sqrt{2}}{2}</math>
| 1/6
| An arrangement of five vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{2}</math>
| 1/14
| A graph of 13 vertices
| 1/2
| Spindle
|
|-
| <math>\frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math>
| 1/28
| A graph of 13 vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{3/2 + \sqrt{33}/6}</math>
| 1/196
| A graph of 9 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>\sqrt{5/3}</math>
| 1/756
| A graph of 33 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>2/\sqrt{3}</math>
| 1/177
| A graph of 103 vertices
| 1/2
| Spindle
| Computer-verified
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>p_{\sqrt 3}/(1-p_2)</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

=== Lemma 1 ===
Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

=== Lemma 2 ===
(Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

=== Lemma 3 ===
(Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

=== Corollary 4 ===
(Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

=== Computer-verified Claim 5 ===
(Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

=== Computer-verified Claim 6 ===
(Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

=== Lemma 7 ===
(Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

=== Corollary 8 ===
One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

=== Lemma 9 ===
(Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

=== Corollary 10 ===
(Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

=== Computer-verified claim 11 ===
One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

=== Lemma 12 ===
For certain special distances d, one can improve the bound in Lemma 2:

If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, <math>\gcd(j,2k+1)=1</math>, and <math>r = \frac{1}{2} \csc\left(\frac{j\pi}{2k+1}\right)</math> then
:<math> p_r \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}}} \leq \frac{1}{3}.</math>

'''Proof''' Observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2}} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges. More generally, if <math>a,b,c,d,e</math> are the diagonal lengths of a pentagon with unit sides, then
:<math> 1 \leq p_a + p_b + p_c + p_d + p_e \leq 2.</math>

=== Lemma 13 ===
We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

=== Corollary 14 ===
We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

=== Lemma 15 ===
One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

=== Corollary 16 ===
We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

=== Lemma 17 ===
If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths a,b,c. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also, thus
:<math>{\bf P}(\mathbf{c}(0) \neq \mathbf{c}(a)) + {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(b)) \geq {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(c))</math>
and the claim follows. <math>\Box</math>

Note that Lemma 2 follows from the a=b, c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 1 - 2^{-k}</math> when <math>a \geq 2^{-k}, k\in\mathbb{Z}^+</math>.

Further, we can generalise the a=b case to one in which the triangle is replaced by a (k+1)-gon of which one edge is 1 and the others are all equal, leading to the stronger result <math>p_a \leq 1 - 1/k</math> when <math>a \geq 1/k, k\in\mathbb{Z}^+ \land k>1</math>. Further strengthening is achieved by using <math>1/\sqrt{3}</math> as the long edge, given Lemma 12.

=== Lemma 18 ===
Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio. Also we have
:<math> p_{d/\sqrt{3}} \leq \frac{1}{3} + p_d, \frac{1}{2} + \frac{1}{2} p_d.</math>

Note that this generalises (9), as well as a special case of Lemma 12.

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the first claim. Similarly, if one considers the colorings of an equilateral triangle of sidelength <math>d</math> together with its center, and counts the numbers <math>a,b \in \{0,1,2,3\}</math> of monochromatic edges of length <math>d</math> and <math>d/\sqrt{3}</math> respectively, one observes that one always has <math>\frac{b}{3} \leq \frac{1}{3} + \frac{2}{3} \frac{a}{3}, \frac{1}{2} + \frac{1}{2} \frac{a}{3}</math>, and on taking expectations one obtains the claim.<math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

=== Lemma 19 ===
(Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

=== Theorem 20 ===
One has <math>p_{H = 1tri} > 0</math>.

'''Proof''' Suppose for contradiction that <math>p_{H = 1tri} = 0</math>. One can then run a version of the de Bruijn-Erdos argument to obtain a coloring in which 1tri hexagons are completely nonexistent (since there are arbitrarily large finite colorings with this property). Consider the triangular lattice <math>{\bf Z}[e^{\pi i/3}]</math>. We 2-color the edges of this lattice by coloring an edge black if it is the short diagonal of a unit rhombus with monochromatic long diagonal, and white otherwise. The four colorings of hexagons lead to four possible colorings at each vertex:

* If H is colored 2tri, then all six edges to the centre of H are black.
* If H is colored 1tri, then two edges to the centre of H at 120 degree angles are white, the other four are black.
* If H is colored axisym, then two opposing edges of the centre of H are black, the other four are white.
* If H is colored centralsym, then all six edges to the centre of H are black.

In particular, as we are assuming no 1tri hexagons, the faces cut out by the black edges have angles 60 degrees, and thus must be equilateral triangles, sectors of angle 60, half-planes, or the entire plane. If there is at least one equilateral triangle, then the rest of the black edges must form an equilateral lattice with that triangle sidelength. This leads to only a small number of possible hexagon colorings in the lattice:

# Case 1: All edges white.
# Case 2: All edges black.
# Case 3.k: For some natural number <math>k \geq 2</math>, the length k edges joining adjacent vertices in some coset of <math>k \cdot {\mathbf Z}[ e^{\pi i/3} ]</math> are all black, and the remaining edges are white.
# Case 4: Each horizontal row consists either entirely of black edges, or entirely of white edges.
# Case 5: Each northwest row consists either entirely of black edges, or entirely of white edges.
# Case 6: Each northeast row consists either entirely of black edges, or entirely of white edges.
# Case 7: Six rays of black edges meeting at a common vertex; all other edges white.

Technically, Case 1 is contained in Cases 4,5,6 as written above, but this will not be an issue. One can view Case 7 as a limiting case <math>k=\infty</math> of Case 3.k; Case 2 is similarly the opposite limiting case <math>k=1</math>.

In the first case, the coloring is periodic with periods <math>2, 2 e^{\pi i/3}</math>. In the second case, it is periodic with periods <math>3, 3 e^{\pi i/3}</math>. In the third case, it is periodic with periods <math>3k, 3k e^{\pi i/3}</math>. Also note that for each k, one can check if Case 3.k holds by inspecting the coloring at a finite number of vertices. Thus the event that Case 3.k holds is "measurable" in the sense that a meaningful probability can be assigned. (But Cases 1,2,4,5,6 are not measurable events, they require an infinite number of points to be inspected, and the probability measure we are using is only finitely additive rather than infinitely additive.) In Case 4, the coloring is periodic with period 2; also, every coset of <math>2 \cdot {\bf Z}[e^{\pi i/3}]</math> is 2-colored. Similarly for Case 5 and 6 (where the periods are <math>2 e^{2\pi i/3}</math> and <math>2 e^{4\pi i/3}</math> respectively.)

Let <math>\alpha_k</math> be the probability that Case 3.k holds for the given value of <math>k</math>. Then <math> \sum_{k=2}^K \alpha_k \leq 1</math> for any k, hence <math>\sum_{k=2}^\infty \alpha_k \leq 1</math>. In particular, we can find <math>K_1</math> such that <math>\sum_{k={K_1}}^\infty \alpha_k \leq 0.1</math> (say). Let <math>P</math> be six times the least common multiple of <math>1,2,\dots,K_1</math>. Then the coloring is P- and <math>P e^{\pi i/3}</math>-periodic for Case 1, Case 2, and all Case 3.k with <math>k \leq K_1</math>. On the other hand, if <math>K_2</math> is sufficiently large depending on <math>P</math>, and Case 3.k holds for some <math>k \geq K_2</math>, then almost all of the hexagons are colored centralsym, which makes the coloring "almost <math>P, P e^{\pi i/3}</math>-periodic" in the sense that
:<math>{\bf c}(z+P e^{\pi i j/3}) = {\bf c}(z) \hbox{ for } j=0,1,2,3,4,5</math>
will hold for at least <math>0.9</math> of the lattice points <math>z \in {\bf Z}[e^{\pi i/3}]</math> with <math>|z| \leq K_2</math>. Similarly for Case 7 (which is sort of a <math>k=\infty</math> limiting case of Case 3.k.) Thus, with the probability <math> \geq 1 - \sum_{k=K_1}^{K_2} \alpha_k \geq 0.9</math>, the coloring of the seven vertices <math>{\bf c}(0), {\bf c}(P e^{\pi ij/3}, j=1,\dots,6</math> is (up to rotation and recoloring) one of the three patterns of the central and linking vertices in Figure 3 of Aubrey's paper, namely

* <math>{\bf c}(0) = {\bf c}(P) = {\bf c}(P e^{\pi i/3}) = {\bf c}(P e^{2\pi i/3}) = {\bf c}(P e^{3\pi i/3}) = {\bf c}(P e^{4\pi i/3}) = {\bf c}(P e^{5\pi i/3}) </math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{3\pi i/3})</math>; <math>{\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{3\pi i/3})</math>; <math> {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>.

Using the spindling argument from Aubrey's paper, we conclude that the third possibility must in fact hold with probability at least 0.8; on the other hand, from Lemma 2 this scenario can only occur with probability at most 1/2, giving the required contradiction.
<math>\Box</math>

One should be able to refine this argument to show that <math>p_{H = 1tri} > c</math> for an absolute constant <math> c>0 </math>.

=== Lemma 21 ===
Providing a tighter bound for Lemma 17 with a more thorough proof: If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

=== Lemma 22 ===

One has <math>3 p_{1/\sqrt{3}} \geq {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) )</math>.

'''Proof''' Let <math>z</math> be a complex number of magnitude <math>1/\sqrt{3}</math> that is a unit distance from 1. If <math>\mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) = c</math> (say), then <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> cannot be colored with <math>c</math>; also, <math>z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> are the vertices of a unit equilateral triangle and thus must take on three different colors. By the pigeonhole principle, one of <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> must then take the same color as the origin, and the claim follows. <math>\Box</math>

=== Lemma 23 ===

One has <math>4 p_{(\sqrt{6} \pm \sqrt{2})/2} + p_{\sqrt{2}} \geq 1</math>. In particular, by Lemma 2, <math>p_{(\sqrt{6}+\sqrt{2})/2} \geq 1/8</math>.

'''Proof''' [ExIs2018b] We just prove the claim for the + sign (the - sign can then be obtained after applying the Galois conjugacy that maps <math>\sqrt{3}</math> to <math>-\sqrt{3}</math>, leaving <math>\sqrt{2}</math> unchanged). Set <math>d := \frac{\sqrt{6}+\sqrt{2}}{2}</math>, and consider the five vertices

:<math>0, e^{5\pi i/4}, e^{5\pi i/4} + d, e^{5\pi i/4} + e^{\pi i/3} d, e^{5\pi i/4} + (e^{\pi i/3}-i)d.</math>

One can check that of the ten edges determined by these five vertices, five have unit length, four have length d, and the remaining distance (from 0 to <math>e^{5\pi i/4}+d</math>) has distance <math>\sqrt{2}</math>. Since <math>\mathbf{c}</math> must make one of the latter five edges monochromatic, the claim follows. <math>\Box</math>

=== Lemma 24 ===

One has <math>p_{\sqrt{2}} \geq \frac{1}{14}</math>.

'''Proof''' In page 7 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 20 unit distance edges and 14 edges of length <math>\sqrt{2}</math> is constructed. The coloring <math>\mathbf{c}</math> must make one of the 14 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 25 ===

Let <math>d = \frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math> and <math>e = \frac{3^{1/4} \sqrt{2} + \sqrt{3} - 1}{2}</math>.
Then one has <math>14 p_d + p_e \geq 1</math>. In particular, by Lemma 2, <math>p_d \geq 1/28</math>.

'''Proof''' In page 9 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 19 unit edges, 14 edges of length d, and one edge of length e is constructed. The coloring <math>\mathbf{c}</math> must make one of the 15 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 26 ===

Let <math>d = \sqrt{3/2 + \sqrt{33}/6}</math>. Then <math>7 p_d \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_d \geq \frac{1}{196}</math>.

'''Proof''' In page 11 of [ExIs2018b], a graph of nine vertices consisting of 12 unit edges and 7 edges of length d is constructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Thus, <math>\mathbf{c}</math> can only make the AB edge monochromatic if one of the seven length d edges is monochromatic. The claim follows.
<math>\Box</math>

=== Lemma 27 ===

One has <math>27 p_{\sqrt{5/3}} \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_{\sqrt{5/3}} \geq \frac{1}{756}</math>.

'''Proof''' In page 13 of [ExIs2018], a graph of 33 vertices with some unit edges and 27 edges of length <math>\sqrt{5/3}</math> is contructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Now repeat the proof of Lemma 26. <math>\Box</math>

=== Computer-verified claim 28 ===

One has <math>p_{2/\sqrt{3}} \geq \frac{1}{177}</math>.

'''Proof''' In page 15 of [ExIs2018], a 5-chromatic graph of 103 vertices, 312 unit edges, and 177 edges of length <math>2/\sqrt{3}</math> is constructed. <math>\mathbf{c}</math> must make one of the latter edges monochromatic, giving the claim. <math>\Box</math>

=== Lemma 29 ===

One has <math>p_{(\sqrt{6} \pm \sqrt{2})/2} \geq 1/6</math> (this improves the bound in Lemma 23).

'''Proof''' Use graphs 505 and 507 from [S2004] and the spindle bound. <math>\Box</math>

=== Lemma 30 ===
For <math>m > n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' <math>n</math> colors and <math>m</math> points necessitates at least 2 having equal color. I.e.

:<math>{\bf P}\left(\bigvee_{k=0}^n \bigvee_{j=k+1}^n\ \mathbf{c}(z_k) = \mathbf{c}(z_j)\right) = 1</math>

The lemma then follows immediately from the fact:
:<math>{\bf P}\left(\bigcup_{k} E_k\right) \leq \sum_{k} {\bf P}\left(E_k\right) \,\Box</math>

=== Corollary 31 ===
Let <math>\lvert z_k\rvert=1</math>. Then for <math>m \geq n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' Use lemma 30 on the set <math>\left\{z_k \bigg\vert 1\leq k\leq m \land k\in\mathbb{Z}\right\}\cup\{0\}</math>. Simplify using <math>{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(0) \right)=0</math>.
<math>\Box</math>

=== Lemma 32 ===
For <math>x\in\mathbb{R}</math> and an <math>n</math>-coloring of the plane, <math>\sum_{k=1}^{n-1}\left(n-k\right){\bf P}\left(\mathbf{c}\left(0\right) = \mathbf{c}\left( 2\sin\left(\frac{kx}{2}\right) \right) \right) \geq 1</math>.

'''Proof''' Use corollary 31 on the set <math>\left\{e^{ikx} \bigg\vert 0\leq k < n \land k\in\mathbb{Z}\right\}</math>. and simplify by grouping lengths.
<math>\Box</math>

=== Corollary 33 ===
Interesting(easy to simplify results of) values for <math>x</math> in Lemma 32 are in <math>\left\{x \bigg\vert \sin\left(\frac{kx}{2}\right)=1 \land 1\leq k < n \land k\in\mathbb{Z}\right\}</math>.
For 4-colorings, this gives
:<math>2p_{\sqrt 3}+p_2 \geq 1</math>
:<math>3p_{(\sqrt 3-1)/\sqrt 2}+p_{\sqrt 2} \geq 1</math>
:<math>3p_{2\sin(\pi/18)}+2p_{2\sin(\pi/9)} \geq 1</math>

=== Lemma 34 ===
Generalizing the note of Lemma 17, <math>\lvert d_1\rvert= d_1 > \lvert d_0\rvert= d_0\Rightarrow (1-p_{d_1})\leq \left\lceil\frac{d_1}{d_0}\right\rceil(1-p_{d_0})</math>

'''Proof''' let <math>\lvert z_{j+1} -z_j\rvert=d_0 > 0, \lvert z_{j+n} -z_0\rvert=d_1>0</math>.

Base case, <math>n=2</math>, by Lemma 17 using the <math>\left(z_n-z_0,z_n-z_{n-1},z_{n-1}-z_0\right)</math> triangle:
:<math>2d_0\geq d_1\Rightarrow 2p_{d_0}\leq 1+p_{d_1}</math>
The inductive step is Lemma 17 using the <math>\left(z_n-z_0,z_n-z_{n-1},z_{n-1}-z_0\right)</math> triangle. After induction:
:<math>[n\geq 2\land nd_0\geq d_1]\Rightarrow np_{d_0}\leq n-1+p_{d_1}</math>
Substitute <math>n=\left\lceil\frac{d_1}{d_0}\right\rceil</math>, simplify, rearrange:
:<math>d_1 > d_0\Rightarrow (1-p_{d_1})\leq \left\lceil\frac{d_1}{d_0}\right\rceil(1-p_{d_0})</math>
<math>\Box</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

== Further questions ==
* For <math>n,m\geq CNP</math>, what consistent relationships exist between <math>{\bf P}\left(\mathbf{c}(0)=\mathbf{c}(d)\bigg\vert n\text{ colors}\right)</math> and <math>{\bf P}\left(\mathbf{c}(0)=\mathbf{c}(d)\bigg\vert m\text{ colors}\right)</math>? How can these relationships be used to sharpen arguments of the probabilistic formulation?

[[Category:Polymath16]]

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-21T05:35:04Z

Nazgand: /* Lemma 34 */ Fixed inefficiency.

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> d\geq 1/n, n>1</math>
|
|
| <math>1-\frac{1}{n}</math>
| (n+1)-gon with one edge length 1 and the rest d
|
|-
| <math> d\geq 1/(n \sqrt{3}), n>1</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d
| Not better than the above on intervals <math>\left(\frac{1}{7},\frac{1}{4\sqrt{3}}\right),\left(\frac{1}{4},\frac{1}{2\sqrt{3}}\right)</math>
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/4
| <math>G_{21}</math>
| 1/2
| Spindle
| Lower bound computer verified
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|-
| <math>\frac{\sqrt{6} \pm \sqrt{2}}{2}</math>
| 1/6
| An arrangement of five vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{2}</math>
| 1/14
| A graph of 13 vertices
| 1/2
| Spindle
|
|-
| <math>\frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math>
| 1/28
| A graph of 13 vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{3/2 + \sqrt{33}/6}</math>
| 1/196
| A graph of 9 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>\sqrt{5/3}</math>
| 1/756
| A graph of 33 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>2/\sqrt{3}</math>
| 1/177
| A graph of 103 vertices
| 1/2
| Spindle
| Computer-verified
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>p_{\sqrt 3}/(1-p_2)</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

=== Lemma 1 ===
Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

=== Lemma 2 ===
(Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

=== Lemma 3 ===
(Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

=== Corollary 4 ===
(Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

=== Computer-verified Claim 5 ===
(Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

=== Computer-verified Claim 6 ===
(Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

=== Lemma 7 ===
(Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

=== Corollary 8 ===
One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

=== Lemma 9 ===
(Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

=== Corollary 10 ===
(Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

=== Computer-verified claim 11 ===
One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

=== Lemma 12 ===
For certain special distances d, one can improve the bound in Lemma 2:

If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, <math>\gcd(j,2k+1)=1</math>, and <math>r = \frac{1}{2} \csc\left(\frac{j\pi}{2k+1}\right)</math> then
:<math> p_r \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}}} \leq \frac{1}{3}.</math>

'''Proof''' Observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2}} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges. More generally, if <math>a,b,c,d,e</math> are the diagonal lengths of a pentagon with unit sides, then
:<math> 1 \leq p_a + p_b + p_c + p_d + p_e \leq 2.</math>

=== Lemma 13 ===
We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

=== Corollary 14 ===
We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

=== Lemma 15 ===
One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

=== Corollary 16 ===
We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

=== Lemma 17 ===
If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths a,b,c. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also, thus
:<math>{\bf P}(\mathbf{c}(0) \neq \mathbf{c}(a)) + {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(b)) \geq {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(c))</math>
and the claim follows. <math>\Box</math>

Note that Lemma 2 follows from the a=b, c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 1 - 2^{-k}</math> when <math>a \geq 2^{-k}, k\in\mathbb{Z}^+</math>.

Further, we can generalise the a=b case to one in which the triangle is replaced by a (k+1)-gon of which one edge is 1 and the others are all equal, leading to the stronger result <math>p_a \leq 1 - 1/k</math> when <math>a \geq 1/k, k\in\mathbb{Z}^+ \land k>1</math>. Further strengthening is achieved by using <math>1/\sqrt{3}</math> as the long edge, given Lemma 12.

=== Lemma 18 ===
Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio. Also we have
:<math> p_{d/\sqrt{3}} \leq \frac{1}{3} + p_d, \frac{1}{2} + \frac{1}{2} p_d.</math>

Note that this generalises (9), as well as a special case of Lemma 12.

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the first claim. Similarly, if one considers the colorings of an equilateral triangle of sidelength <math>d</math> together with its center, and counts the numbers <math>a,b \in \{0,1,2,3\}</math> of monochromatic edges of length <math>d</math> and <math>d/\sqrt{3}</math> respectively, one observes that one always has <math>\frac{b}{3} \leq \frac{1}{3} + \frac{2}{3} \frac{a}{3}, \frac{1}{2} + \frac{1}{2} \frac{a}{3}</math>, and on taking expectations one obtains the claim.<math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

=== Lemma 19 ===
(Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

=== Theorem 20 ===
One has <math>p_{H = 1tri} > 0</math>.

'''Proof''' Suppose for contradiction that <math>p_{H = 1tri} = 0</math>. One can then run a version of the de Bruijn-Erdos argument to obtain a coloring in which 1tri hexagons are completely nonexistent (since there are arbitrarily large finite colorings with this property). Consider the triangular lattice <math>{\bf Z}[e^{\pi i/3}]</math>. We 2-color the edges of this lattice by coloring an edge black if it is the short diagonal of a unit rhombus with monochromatic long diagonal, and white otherwise. The four colorings of hexagons lead to four possible colorings at each vertex:

* If H is colored 2tri, then all six edges to the centre of H are black.
* If H is colored 1tri, then two edges to the centre of H at 120 degree angles are white, the other four are black.
* If H is colored axisym, then two opposing edges of the centre of H are black, the other four are white.
* If H is colored centralsym, then all six edges to the centre of H are black.

In particular, as we are assuming no 1tri hexagons, the faces cut out by the black edges have angles 60 degrees, and thus must be equilateral triangles, sectors of angle 60, half-planes, or the entire plane. If there is at least one equilateral triangle, then the rest of the black edges must form an equilateral lattice with that triangle sidelength. This leads to only a small number of possible hexagon colorings in the lattice:

# Case 1: All edges white.
# Case 2: All edges black.
# Case 3.k: For some natural number <math>k \geq 2</math>, the length k edges joining adjacent vertices in some coset of <math>k \cdot {\mathbf Z}[ e^{\pi i/3} ]</math> are all black, and the remaining edges are white.
# Case 4: Each horizontal row consists either entirely of black edges, or entirely of white edges.
# Case 5: Each northwest row consists either entirely of black edges, or entirely of white edges.
# Case 6: Each northeast row consists either entirely of black edges, or entirely of white edges.
# Case 7: Six rays of black edges meeting at a common vertex; all other edges white.

Technically, Case 1 is contained in Cases 4,5,6 as written above, but this will not be an issue. One can view Case 7 as a limiting case <math>k=\infty</math> of Case 3.k; Case 2 is similarly the opposite limiting case <math>k=1</math>.

In the first case, the coloring is periodic with periods <math>2, 2 e^{\pi i/3}</math>. In the second case, it is periodic with periods <math>3, 3 e^{\pi i/3}</math>. In the third case, it is periodic with periods <math>3k, 3k e^{\pi i/3}</math>. Also note that for each k, one can check if Case 3.k holds by inspecting the coloring at a finite number of vertices. Thus the event that Case 3.k holds is "measurable" in the sense that a meaningful probability can be assigned. (But Cases 1,2,4,5,6 are not measurable events, they require an infinite number of points to be inspected, and the probability measure we are using is only finitely additive rather than infinitely additive.) In Case 4, the coloring is periodic with period 2; also, every coset of <math>2 \cdot {\bf Z}[e^{\pi i/3}]</math> is 2-colored. Similarly for Case 5 and 6 (where the periods are <math>2 e^{2\pi i/3}</math> and <math>2 e^{4\pi i/3}</math> respectively.)

Let <math>\alpha_k</math> be the probability that Case 3.k holds for the given value of <math>k</math>. Then <math> \sum_{k=2}^K \alpha_k \leq 1</math> for any k, hence <math>\sum_{k=2}^\infty \alpha_k \leq 1</math>. In particular, we can find <math>K_1</math> such that <math>\sum_{k={K_1}}^\infty \alpha_k \leq 0.1</math> (say). Let <math>P</math> be six times the least common multiple of <math>1,2,\dots,K_1</math>. Then the coloring is P- and <math>P e^{\pi i/3}</math>-periodic for Case 1, Case 2, and all Case 3.k with <math>k \leq K_1</math>. On the other hand, if <math>K_2</math> is sufficiently large depending on <math>P</math>, and Case 3.k holds for some <math>k \geq K_2</math>, then almost all of the hexagons are colored centralsym, which makes the coloring "almost <math>P, P e^{\pi i/3}</math>-periodic" in the sense that
:<math>{\bf c}(z+P e^{\pi i j/3}) = {\bf c}(z) \hbox{ for } j=0,1,2,3,4,5</math>
will hold for at least <math>0.9</math> of the lattice points <math>z \in {\bf Z}[e^{\pi i/3}]</math> with <math>|z| \leq K_2</math>. Similarly for Case 7 (which is sort of a <math>k=\infty</math> limiting case of Case 3.k.) Thus, with the probability <math> \geq 1 - \sum_{k=K_1}^{K_2} \alpha_k \geq 0.9</math>, the coloring of the seven vertices <math>{\bf c}(0), {\bf c}(P e^{\pi ij/3}, j=1,\dots,6</math> is (up to rotation and recoloring) one of the three patterns of the central and linking vertices in Figure 3 of Aubrey's paper, namely

* <math>{\bf c}(0) = {\bf c}(P) = {\bf c}(P e^{\pi i/3}) = {\bf c}(P e^{2\pi i/3}) = {\bf c}(P e^{3\pi i/3}) = {\bf c}(P e^{4\pi i/3}) = {\bf c}(P e^{5\pi i/3}) </math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{3\pi i/3})</math>; <math>{\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{3\pi i/3})</math>; <math> {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>.

Using the spindling argument from Aubrey's paper, we conclude that the third possibility must in fact hold with probability at least 0.8; on the other hand, from Lemma 2 this scenario can only occur with probability at most 1/2, giving the required contradiction.
<math>\Box</math>

One should be able to refine this argument to show that <math>p_{H = 1tri} > c</math> for an absolute constant <math> c>0 </math>.

=== Lemma 21 ===
Providing a tighter bound for Lemma 17 with a more thorough proof: If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

=== Lemma 22 ===

One has <math>3 p_{1/\sqrt{3}} \geq {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) )</math>.

'''Proof''' Let <math>z</math> be a complex number of magnitude <math>1/\sqrt{3}</math> that is a unit distance from 1. If <math>\mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) = c</math> (say), then <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> cannot be colored with <math>c</math>; also, <math>z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> are the vertices of a unit equilateral triangle and thus must take on three different colors. By the pigeonhole principle, one of <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> must then take the same color as the origin, and the claim follows. <math>\Box</math>

=== Lemma 23 ===

One has <math>4 p_{(\sqrt{6} \pm \sqrt{2})/2} + p_{\sqrt{2}} \geq 1</math>. In particular, by Lemma 2, <math>p_{(\sqrt{6}+\sqrt{2})/2} \geq 1/8</math>.

'''Proof''' [ExIs2018b] We just prove the claim for the + sign (the - sign can then be obtained after applying the Galois conjugacy that maps <math>\sqrt{3}</math> to <math>-\sqrt{3}</math>, leaving <math>\sqrt{2}</math> unchanged). Set <math>d := \frac{\sqrt{6}+\sqrt{2}}{2}</math>, and consider the five vertices

:<math>0, e^{5\pi i/4}, e^{5\pi i/4} + d, e^{5\pi i/4} + e^{\pi i/3} d, e^{5\pi i/4} + (e^{\pi i/3}-i)d.</math>

One can check that of the ten edges determined by these five vertices, five have unit length, four have length d, and the remaining distance (from 0 to <math>e^{5\pi i/4}+d</math>) has distance <math>\sqrt{2}</math>. Since <math>\mathbf{c}</math> must make one of the latter five edges monochromatic, the claim follows. <math>\Box</math>

=== Lemma 24 ===

One has <math>p_{\sqrt{2}} \geq \frac{1}{14}</math>.

'''Proof''' In page 7 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 20 unit distance edges and 14 edges of length <math>\sqrt{2}</math> is constructed. The coloring <math>\mathbf{c}</math> must make one of the 14 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 25 ===

Let <math>d = \frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math> and <math>e = \frac{3^{1/4} \sqrt{2} + \sqrt{3} - 1}{2}</math>.
Then one has <math>14 p_d + p_e \geq 1</math>. In particular, by Lemma 2, <math>p_d \geq 1/28</math>.

'''Proof''' In page 9 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 19 unit edges, 14 edges of length d, and one edge of length e is constructed. The coloring <math>\mathbf{c}</math> must make one of the 15 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 26 ===

Let <math>d = \sqrt{3/2 + \sqrt{33}/6}</math>. Then <math>7 p_d \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_d \geq \frac{1}{196}</math>.

'''Proof''' In page 11 of [ExIs2018b], a graph of nine vertices consisting of 12 unit edges and 7 edges of length d is constructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Thus, <math>\mathbf{c}</math> can only make the AB edge monochromatic if one of the seven length d edges is monochromatic. The claim follows.
<math>\Box</math>

=== Lemma 27 ===

One has <math>27 p_{\sqrt{5/3}} \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_{\sqrt{5/3}} \geq \frac{1}{756}</math>.

'''Proof''' In page 13 of [ExIs2018], a graph of 33 vertices with some unit edges and 27 edges of length <math>\sqrt{5/3}</math> is contructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Now repeat the proof of Lemma 26. <math>\Box</math>

=== Computer-verified claim 28 ===

One has <math>p_{2/\sqrt{3}} \geq \frac{1}{177}</math>.

'''Proof''' In page 15 of [ExIs2018], a 5-chromatic graph of 103 vertices, 312 unit edges, and 177 edges of length <math>2/\sqrt{3}</math> is constructed. <math>\mathbf{c}</math> must make one of the latter edges monochromatic, giving the claim. <math>\Box</math>

=== Lemma 29 ===

One has <math>p_{(\sqrt{6} \pm \sqrt{2})/2} \geq 1/6</math> (this improves the bound in Lemma 23).

'''Proof''' Use graphs 505 and 507 from [S2004] and the spindle bound. <math>\Box</math>

=== Lemma 30 ===
For <math>m > n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' <math>n</math> colors and <math>m</math> points necessitates at least 2 having equal color. I.e.

:<math>{\bf P}\left(\bigvee_{k=0}^n \bigvee_{j=k+1}^n\ \mathbf{c}(z_k) = \mathbf{c}(z_j)\right) = 1</math>

The lemma then follows immediately from the fact:
:<math>{\bf P}\left(\bigcup_{k} E_k\right) \leq \sum_{k} {\bf P}\left(E_k\right) \,\Box</math>

=== Corollary 31 ===
Let <math>\lvert z_k\rvert=1</math>. Then for <math>m \geq n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' Use lemma 30 on the set <math>\left\{z_k \bigg\vert 1\leq k\leq m \land k\in\mathbb{Z}\right\}\cup\{0\}</math>. Simplify using <math>{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(0) \right)=0</math>.
<math>\Box</math>

=== Lemma 32 ===
For <math>x\in\mathbb{R}</math> and an <math>n</math>-coloring of the plane, <math>\sum_{k=1}^{n-1}\left(n-k\right){\bf P}\left(\mathbf{c}\left(0\right) = \mathbf{c}\left( 2\sin\left(\frac{kx}{2}\right) \right) \right) \geq 1</math>.

'''Proof''' Use corollary 31 on the set <math>\left\{e^{ikx} \bigg\vert 0\leq k < n \land k\in\mathbb{Z}\right\}</math>. and simplify by grouping lengths.
<math>\Box</math>

=== Corollary 33 ===
Interesting(easy to simplify results of) values for <math>x</math> in Lemma 32 are in <math>\left\{x \bigg\vert \sin\left(\frac{kx}{2}\right)=1 \land 1\leq k < n \land k\in\mathbb{Z}\right\}</math>.
For 4-colorings, this gives
:<math>2p_{\sqrt 3}+p_2 \geq 1</math>
:<math>3p_{(\sqrt 3-1)/\sqrt 2}+p_{\sqrt 2} \geq 1</math>
:<math>3p_{2\sin(\pi/18)}+2p_{2\sin(\pi/9)} \geq 1</math>

=== Lemma 34 ===
Generalizing the note of Lemma 17, <math>\lvert d_1\rvert= d_1 > \lvert d_0\rvert= d_0\Rightarrow (1-p_{d_1})\leq \left\lceil\frac{d_1}{d_0}\right\rceil(1-p_{d_0})</math>

'''Proof''' let <math>\lvert z_{j+1} -z_j\rvert=d_0 > 0, \lvert z_{j+n} -z_0\rvert=d_1>0</math>.

Base case, <math>n=2</math>, by Lemma 17 using the <math>\left(z_n-z_0,z_n-z_{n-1},z_{n-1}-z_0\right)</math> triangle:
:<math>2d_0\geq d_1\Rightarrow 2p_{d_0}\leq 1+p_{d_1}</math>
The inductive step is Lemma 17 using the <math>\left(z_n-z_0,z_n-z_{n-1},z_{n-1}-z_0\right)</math> triangle. After induction:
:<math>[n\geq 2\land nd_0\geq d_1]\Rightarrow np_{d_0}\leq n-1+p_{d_1}</math>
Substitute <math>n=\left\lceil\frac{d_1}{d_0}\right\rceil</math>, simplify, rearrange:
:<math>d_1 > d_0\Rightarrow (1-p_{d_1})\leq \left\lceil\frac{d_1}{d_0}\right\rceil(1-p_{d_0})</math>
<math>\Box</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

== Further questions ==
* For <math>n,m\geq CNP</math>, what consistent relationships exist between <math>{\bf P}\left(\mathbf{c}(0)=\mathbf{c}(d)\bigg\vert n\text{ colors}\right)</math> and <math>{\bf P}\left(\mathbf{c}(0)=\mathbf{c}(d)\bigg\vert m\text{ colors}\right)</math>? How can these relationships be used to sharpen arguments of the probabilistic formulation?

[[Category:Polymath16]]

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-20T23:11:08Z

Nazgand: Clarify Lemma 17 note with new lemma

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> d\geq 1/n, n>1</math>
|
|
| <math>1-\frac{1}{n}</math>
| (n+1)-gon with one edge length 1 and the rest d
|
|-
| <math> d\geq 1/(n \sqrt{3}), n>1</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d
| Not better than the above on intervals <math>\left(\frac{1}{7},\frac{1}{4\sqrt{3}}\right),\left(\frac{1}{4},\frac{1}{2\sqrt{3}}\right)</math>
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/4
| <math>G_{21}</math>
| 1/2
| Spindle
| Lower bound computer verified
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|-
| <math>\frac{\sqrt{6} \pm \sqrt{2}}{2}</math>
| 1/6
| An arrangement of five vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{2}</math>
| 1/14
| A graph of 13 vertices
| 1/2
| Spindle
|
|-
| <math>\frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math>
| 1/28
| A graph of 13 vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{3/2 + \sqrt{33}/6}</math>
| 1/196
| A graph of 9 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>\sqrt{5/3}</math>
| 1/756
| A graph of 33 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>2/\sqrt{3}</math>
| 1/177
| A graph of 103 vertices
| 1/2
| Spindle
| Computer-verified
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>p_{\sqrt 3}/(1-p_2)</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

=== Lemma 1 ===
Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

=== Lemma 2 ===
(Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

=== Lemma 3 ===
(Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

=== Corollary 4 ===
(Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

=== Computer-verified Claim 5 ===
(Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

=== Computer-verified Claim 6 ===
(Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

=== Lemma 7 ===
(Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

=== Corollary 8 ===
One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

=== Lemma 9 ===
(Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

=== Corollary 10 ===
(Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

=== Computer-verified claim 11 ===
One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

=== Lemma 12 ===
For certain special distances d, one can improve the bound in Lemma 2:

If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, <math>\gcd(j,2k+1)=1</math>, and <math>r = \frac{1}{2} \csc\left(\frac{j\pi}{2k+1}\right)</math> then
:<math> p_r \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}}} \leq \frac{1}{3}.</math>

'''Proof''' Observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2}} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges. More generally, if <math>a,b,c,d,e</math> are the diagonal lengths of a pentagon with unit sides, then
:<math> 1 \leq p_a + p_b + p_c + p_d + p_e \leq 2.</math>

=== Lemma 13 ===
We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

=== Corollary 14 ===
We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

=== Lemma 15 ===
One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

=== Corollary 16 ===
We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

=== Lemma 17 ===
If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths a,b,c. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also, thus
:<math>{\bf P}(\mathbf{c}(0) \neq \mathbf{c}(a)) + {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(b)) \geq {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(c))</math>
and the claim follows. <math>\Box</math>

Note that Lemma 2 follows from the a=b, c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 1 - 2^{-k}</math> when <math>a \geq 2^{-k}, k\in\mathbb{Z}^+</math>.

Further, we can generalise the a=b case to one in which the triangle is replaced by a (k+1)-gon of which one edge is 1 and the others are all equal, leading to the stronger result <math>p_a \leq 1 - 1/k</math> when <math>a \geq 1/k, k\in\mathbb{Z}^+ \land k>1</math>. Further strengthening is achieved by using <math>1/\sqrt{3}</math> as the long edge, given Lemma 12.

=== Lemma 18 ===
Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio. Also we have
:<math> p_{d/\sqrt{3}} \leq \frac{1}{3} + p_d, \frac{1}{2} + \frac{1}{2} p_d.</math>

Note that this generalises (9), as well as a special case of Lemma 12.

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the first claim. Similarly, if one considers the colorings of an equilateral triangle of sidelength <math>d</math> together with its center, and counts the numbers <math>a,b \in \{0,1,2,3\}</math> of monochromatic edges of length <math>d</math> and <math>d/\sqrt{3}</math> respectively, one observes that one always has <math>\frac{b}{3} \leq \frac{1}{3} + \frac{2}{3} \frac{a}{3}, \frac{1}{2} + \frac{1}{2} \frac{a}{3}</math>, and on taking expectations one obtains the claim.<math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

=== Lemma 19 ===
(Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

=== Theorem 20 ===
One has <math>p_{H = 1tri} > 0</math>.

'''Proof''' Suppose for contradiction that <math>p_{H = 1tri} = 0</math>. One can then run a version of the de Bruijn-Erdos argument to obtain a coloring in which 1tri hexagons are completely nonexistent (since there are arbitrarily large finite colorings with this property). Consider the triangular lattice <math>{\bf Z}[e^{\pi i/3}]</math>. We 2-color the edges of this lattice by coloring an edge black if it is the short diagonal of a unit rhombus with monochromatic long diagonal, and white otherwise. The four colorings of hexagons lead to four possible colorings at each vertex:

* If H is colored 2tri, then all six edges to the centre of H are black.
* If H is colored 1tri, then two edges to the centre of H at 120 degree angles are white, the other four are black.
* If H is colored axisym, then two opposing edges of the centre of H are black, the other four are white.
* If H is colored centralsym, then all six edges to the centre of H are black.

In particular, as we are assuming no 1tri hexagons, the faces cut out by the black edges have angles 60 degrees, and thus must be equilateral triangles, sectors of angle 60, half-planes, or the entire plane. If there is at least one equilateral triangle, then the rest of the black edges must form an equilateral lattice with that triangle sidelength. This leads to only a small number of possible hexagon colorings in the lattice:

# Case 1: All edges white.
# Case 2: All edges black.
# Case 3.k: For some natural number <math>k \geq 2</math>, the length k edges joining adjacent vertices in some coset of <math>k \cdot {\mathbf Z}[ e^{\pi i/3} ]</math> are all black, and the remaining edges are white.
# Case 4: Each horizontal row consists either entirely of black edges, or entirely of white edges.
# Case 5: Each northwest row consists either entirely of black edges, or entirely of white edges.
# Case 6: Each northeast row consists either entirely of black edges, or entirely of white edges.
# Case 7: Six rays of black edges meeting at a common vertex; all other edges white.

Technically, Case 1 is contained in Cases 4,5,6 as written above, but this will not be an issue. One can view Case 7 as a limiting case <math>k=\infty</math> of Case 3.k; Case 2 is similarly the opposite limiting case <math>k=1</math>.

In the first case, the coloring is periodic with periods <math>2, 2 e^{\pi i/3}</math>. In the second case, it is periodic with periods <math>3, 3 e^{\pi i/3}</math>. In the third case, it is periodic with periods <math>3k, 3k e^{\pi i/3}</math>. Also note that for each k, one can check if Case 3.k holds by inspecting the coloring at a finite number of vertices. Thus the event that Case 3.k holds is "measurable" in the sense that a meaningful probability can be assigned. (But Cases 1,2,4,5,6 are not measurable events, they require an infinite number of points to be inspected, and the probability measure we are using is only finitely additive rather than infinitely additive.) In Case 4, the coloring is periodic with period 2; also, every coset of <math>2 \cdot {\bf Z}[e^{\pi i/3}]</math> is 2-colored. Similarly for Case 5 and 6 (where the periods are <math>2 e^{2\pi i/3}</math> and <math>2 e^{4\pi i/3}</math> respectively.)

Let <math>\alpha_k</math> be the probability that Case 3.k holds for the given value of <math>k</math>. Then <math> \sum_{k=2}^K \alpha_k \leq 1</math> for any k, hence <math>\sum_{k=2}^\infty \alpha_k \leq 1</math>. In particular, we can find <math>K_1</math> such that <math>\sum_{k={K_1}}^\infty \alpha_k \leq 0.1</math> (say). Let <math>P</math> be six times the least common multiple of <math>1,2,\dots,K_1</math>. Then the coloring is P- and <math>P e^{\pi i/3}</math>-periodic for Case 1, Case 2, and all Case 3.k with <math>k \leq K_1</math>. On the other hand, if <math>K_2</math> is sufficiently large depending on <math>P</math>, and Case 3.k holds for some <math>k \geq K_2</math>, then almost all of the hexagons are colored centralsym, which makes the coloring "almost <math>P, P e^{\pi i/3}</math>-periodic" in the sense that
:<math>{\bf c}(z+P e^{\pi i j/3}) = {\bf c}(z) \hbox{ for } j=0,1,2,3,4,5</math>
will hold for at least <math>0.9</math> of the lattice points <math>z \in {\bf Z}[e^{\pi i/3}]</math> with <math>|z| \leq K_2</math>. Similarly for Case 7 (which is sort of a <math>k=\infty</math> limiting case of Case 3.k.) Thus, with the probability <math> \geq 1 - \sum_{k=K_1}^{K_2} \alpha_k \geq 0.9</math>, the coloring of the seven vertices <math>{\bf c}(0), {\bf c}(P e^{\pi ij/3}, j=1,\dots,6</math> is (up to rotation and recoloring) one of the three patterns of the central and linking vertices in Figure 3 of Aubrey's paper, namely

* <math>{\bf c}(0) = {\bf c}(P) = {\bf c}(P e^{\pi i/3}) = {\bf c}(P e^{2\pi i/3}) = {\bf c}(P e^{3\pi i/3}) = {\bf c}(P e^{4\pi i/3}) = {\bf c}(P e^{5\pi i/3}) </math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{3\pi i/3})</math>; <math>{\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{3\pi i/3})</math>; <math> {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>.

Using the spindling argument from Aubrey's paper, we conclude that the third possibility must in fact hold with probability at least 0.8; on the other hand, from Lemma 2 this scenario can only occur with probability at most 1/2, giving the required contradiction.
<math>\Box</math>

One should be able to refine this argument to show that <math>p_{H = 1tri} > c</math> for an absolute constant <math> c>0 </math>.

=== Lemma 21 ===
Providing a tighter bound for Lemma 17 with a more thorough proof: If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

=== Lemma 22 ===

One has <math>3 p_{1/\sqrt{3}} \geq {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) )</math>.

'''Proof''' Let <math>z</math> be a complex number of magnitude <math>1/\sqrt{3}</math> that is a unit distance from 1. If <math>\mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) = c</math> (say), then <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> cannot be colored with <math>c</math>; also, <math>z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> are the vertices of a unit equilateral triangle and thus must take on three different colors. By the pigeonhole principle, one of <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> must then take the same color as the origin, and the claim follows. <math>\Box</math>

=== Lemma 23 ===

One has <math>4 p_{(\sqrt{6} \pm \sqrt{2})/2} + p_{\sqrt{2}} \geq 1</math>. In particular, by Lemma 2, <math>p_{(\sqrt{6}+\sqrt{2})/2} \geq 1/8</math>.

'''Proof''' [ExIs2018b] We just prove the claim for the + sign (the - sign can then be obtained after applying the Galois conjugacy that maps <math>\sqrt{3}</math> to <math>-\sqrt{3}</math>, leaving <math>\sqrt{2}</math> unchanged). Set <math>d := \frac{\sqrt{6}+\sqrt{2}}{2}</math>, and consider the five vertices

:<math>0, e^{5\pi i/4}, e^{5\pi i/4} + d, e^{5\pi i/4} + e^{\pi i/3} d, e^{5\pi i/4} + (e^{\pi i/3}-i)d.</math>

One can check that of the ten edges determined by these five vertices, five have unit length, four have length d, and the remaining distance (from 0 to <math>e^{5\pi i/4}+d</math>) has distance <math>\sqrt{2}</math>. Since <math>\mathbf{c}</math> must make one of the latter five edges monochromatic, the claim follows. <math>\Box</math>

=== Lemma 24 ===

One has <math>p_{\sqrt{2}} \geq \frac{1}{14}</math>.

'''Proof''' In page 7 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 20 unit distance edges and 14 edges of length <math>\sqrt{2}</math> is constructed. The coloring <math>\mathbf{c}</math> must make one of the 14 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 25 ===

Let <math>d = \frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math> and <math>e = \frac{3^{1/4} \sqrt{2} + \sqrt{3} - 1}{2}</math>.
Then one has <math>14 p_d + p_e \geq 1</math>. In particular, by Lemma 2, <math>p_d \geq 1/28</math>.

'''Proof''' In page 9 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 19 unit edges, 14 edges of length d, and one edge of length e is constructed. The coloring <math>\mathbf{c}</math> must make one of the 15 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 26 ===

Let <math>d = \sqrt{3/2 + \sqrt{33}/6}</math>. Then <math>7 p_d \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_d \geq \frac{1}{196}</math>.

'''Proof''' In page 11 of [ExIs2018b], a graph of nine vertices consisting of 12 unit edges and 7 edges of length d is constructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Thus, <math>\mathbf{c}</math> can only make the AB edge monochromatic if one of the seven length d edges is monochromatic. The claim follows.
<math>\Box</math>

=== Lemma 27 ===

One has <math>27 p_{\sqrt{5/3}} \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_{\sqrt{5/3}} \geq \frac{1}{756}</math>.

'''Proof''' In page 13 of [ExIs2018], a graph of 33 vertices with some unit edges and 27 edges of length <math>\sqrt{5/3}</math> is contructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Now repeat the proof of Lemma 26. <math>\Box</math>

=== Computer-verified claim 28 ===

One has <math>p_{2/\sqrt{3}} \geq \frac{1}{177}</math>.

'''Proof''' In page 15 of [ExIs2018], a 5-chromatic graph of 103 vertices, 312 unit edges, and 177 edges of length <math>2/\sqrt{3}</math> is constructed. <math>\mathbf{c}</math> must make one of the latter edges monochromatic, giving the claim. <math>\Box</math>

=== Lemma 29 ===

One has <math>p_{(\sqrt{6} \pm \sqrt{2})/2} \geq 1/6</math> (this improves the bound in Lemma 23).

'''Proof''' Use graphs 505 and 507 from [S2004] and the spindle bound. <math>\Box</math>

=== Lemma 30 ===
For <math>m > n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' <math>n</math> colors and <math>m</math> points necessitates at least 2 having equal color. I.e.

:<math>{\bf P}\left(\bigvee_{k=0}^n \bigvee_{j=k+1}^n\ \mathbf{c}(z_k) = \mathbf{c}(z_j)\right) = 1</math>

The lemma then follows immediately from the fact:
:<math>{\bf P}\left(\bigcup_{k} E_k\right) \leq \sum_{k} {\bf P}\left(E_k\right) \,\Box</math>

=== Corollary 31 ===
Let <math>\lvert z_k\rvert=1</math>. Then for <math>m \geq n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' Use lemma 30 on the set <math>\left\{z_k \bigg\vert 1\leq k\leq m \land k\in\mathbb{Z}\right\}\cup\{0\}</math>. Simplify using <math>{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(0) \right)=0</math>.
<math>\Box</math>

=== Lemma 32 ===
For <math>x\in\mathbb{R}</math> and an <math>n</math>-coloring of the plane, <math>\sum_{k=1}^{n-1}\left(n-k\right){\bf P}\left(\mathbf{c}\left(0\right) = \mathbf{c}\left( 2\sin\left(\frac{kx}{2}\right) \right) \right) \geq 1</math>.

'''Proof''' Use corollary 31 on the set <math>\left\{e^{ikx} \bigg\vert 0\leq k < n \land k\in\mathbb{Z}\right\}</math>. and simplify by grouping lengths.
<math>\Box</math>

=== Corollary 33 ===
Interesting(easy to simplify results of) values for <math>x</math> in Lemma 32 are in <math>\left\{x \bigg\vert \sin\left(\frac{kx}{2}\right)=1 \land 1\leq k < n \land k\in\mathbb{Z}\right\}</math>.
For 4-colorings, this gives
:<math>2p_{\sqrt 3}+p_2 \geq 1</math>
:<math>3p_{(\sqrt 3-1)/\sqrt 2}+p_{\sqrt 2} \geq 1</math>
:<math>3p_{2\sin(\pi/18)}+2p_{2\sin(\pi/9)} \geq 1</math>

=== Lemma 34 ===
Generalizing the note of Lemma 17, <math>d_1\geq 2d_0\Rightarrow (1-p_{d_1})\leq \left\lceil\frac{d_1}{d_0}\right\rceil(1-p_{d_0})</math>

'''Proof''' let <math>\lvert z_{j+1} -z_j\rvert=d_0 > 0, \lvert z_{j+n} -z_0\rvert=d_1>0</math>.

Base case, <math>n=2</math>, by Lemma 17 using the <math>\left(z_n-z_0,z_n-z_{n-1},z_{n-1}-z_0\right)</math> triangle:
:<math>2d_0\geq d_1\Rightarrow 2p_{d_0}\leq 1+p_{d_1}</math>
The inductive step is Lemma 17 using the <math>\left(z_n-z_0,z_n-z_{n-1},z_{n-1}-z_0\right)</math> triangle. After induction:
:<math>[n\geq 2\land nd_0\geq d_1]\Rightarrow np_{d_0}\leq n-1+p_{d_1}</math>
Substitute <math>n=\left\lceil\frac{d_1}{d_0}\right\rceil</math>, rearrange:
:<math>d_1\geq 2d_0\Rightarrow (1-p_{d_1})\leq \left\lceil\frac{d_1}{d_0}\right\rceil(1-p_{d_0})</math>
<math>\Box</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

== Further questions ==
* For <math>n,m\geq CNP</math>, what consistent relationships exist between <math>{\bf P}\left(\mathbf{c}(0)=\mathbf{c}(d)\bigg\vert n\text{ colors}\right)</math> and <math>{\bf P}\left(\mathbf{c}(0)=\mathbf{c}(d)\bigg\vert m\text{ colors}\right)</math>? How can these relationships be used to sharpen arguments of the probabilistic formulation?

[[Category:Polymath16]]

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-20T22:20:33Z

Nazgand: /* Bounds on p_d for 4-colourings */

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> d\geq 1/n, n>1</math>
|
|
| <math>1-\frac{1}{n}</math>
| (n+1)-gon with one edge length 1 and the rest d
|
|-
| <math> d\geq 1/(n \sqrt{3}), n>1</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d
| Not better than the above on intervals <math>\left(\frac{1}{7},\frac{1}{4\sqrt{3}}\right),\left(\frac{1}{4},\frac{1}{2\sqrt{3}}\right)</math>
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/4
| <math>G_{21}</math>
| 1/2
| Spindle
| Lower bound computer verified
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|-
| <math>\frac{\sqrt{6} \pm \sqrt{2}}{2}</math>
| 1/6
| An arrangement of five vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{2}</math>
| 1/14
| A graph of 13 vertices
| 1/2
| Spindle
|
|-
| <math>\frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math>
| 1/28
| A graph of 13 vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{3/2 + \sqrt{33}/6}</math>
| 1/196
| A graph of 9 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>\sqrt{5/3}</math>
| 1/756
| A graph of 33 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>2/\sqrt{3}</math>
| 1/177
| A graph of 103 vertices
| 1/2
| Spindle
| Computer-verified
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>p_{\sqrt 3}/(1-p_2)</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

=== Lemma 1 ===
Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

=== Lemma 2 ===
(Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

=== Lemma 3 ===
(Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

=== Corollary 4 ===
(Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

=== Computer-verified Claim 5 ===
(Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

=== Computer-verified Claim 6 ===
(Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

=== Lemma 7 ===
(Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

=== Corollary 8 ===
One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

=== Lemma 9 ===
(Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

=== Corollary 10 ===
(Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

=== Computer-verified claim 11 ===
One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

=== Lemma 12 ===
For certain special distances d, one can improve the bound in Lemma 2:

If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, <math>\gcd(j,2k+1)=1</math>, and <math>r = \frac{1}{2} \csc\left(\frac{j\pi}{2k+1}\right)</math> then
:<math> p_r \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}}} \leq \frac{1}{3}.</math>

'''Proof''' Observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2}} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges. More generally, if <math>a,b,c,d,e</math> are the diagonal lengths of a pentagon with unit sides, then
:<math> 1 \leq p_a + p_b + p_c + p_d + p_e \leq 2.</math>

=== Lemma 13 ===
We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

=== Corollary 14 ===
We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

=== Lemma 15 ===
One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

=== Corollary 16 ===
We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

=== Lemma 17 ===
If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths a,b,c. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also, thus
:<math>{\bf P}(\mathbf{c}(0) \neq \mathbf{c}(a)) + {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(b)) \geq {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(c))</math>
and the claim follows. <math>\Box</math>

Note that Lemma 2 follows from the a=b, c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 1 - 2^{-k}</math> when <math>a \geq 2^{-k}, k\in\mathbb{Z}^+</math>.

Further, we can generalise the a=b case to one in which the triangle is replaced by a (k+1)-gon of which one edge is 1 and the others are all equal, leading to the stronger result <math>p_a \leq 1 - 1/k</math> when <math>a \geq 1/k, k\in\mathbb{Z}^+ \land k>1</math>. Further strengthening is achieved by using <math>1/\sqrt{3}</math> as the long edge, given Lemma 12.

=== Lemma 18 ===
Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio. Also we have
:<math> p_{d/\sqrt{3}} \leq \frac{1}{3} + p_d, \frac{1}{2} + \frac{1}{2} p_d.</math>

Note that this generalises (9), as well as a special case of Lemma 12.

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the first claim. Similarly, if one considers the colorings of an equilateral triangle of sidelength <math>d</math> together with its center, and counts the numbers <math>a,b \in \{0,1,2,3\}</math> of monochromatic edges of length <math>d</math> and <math>d/\sqrt{3}</math> respectively, one observes that one always has <math>\frac{b}{3} \leq \frac{1}{3} + \frac{2}{3} \frac{a}{3}, \frac{1}{2} + \frac{1}{2} \frac{a}{3}</math>, and on taking expectations one obtains the claim.<math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

=== Lemma 19 ===
(Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

=== Theorem 20 ===
One has <math>p_{H = 1tri} > 0</math>.

'''Proof''' Suppose for contradiction that <math>p_{H = 1tri} = 0</math>. One can then run a version of the de Bruijn-Erdos argument to obtain a coloring in which 1tri hexagons are completely nonexistent (since there are arbitrarily large finite colorings with this property). Consider the triangular lattice <math>{\bf Z}[e^{\pi i/3}]</math>. We 2-color the edges of this lattice by coloring an edge black if it is the short diagonal of a unit rhombus with monochromatic long diagonal, and white otherwise. The four colorings of hexagons lead to four possible colorings at each vertex:

* If H is colored 2tri, then all six edges to the centre of H are black.
* If H is colored 1tri, then two edges to the centre of H at 120 degree angles are white, the other four are black.
* If H is colored axisym, then two opposing edges of the centre of H are black, the other four are white.
* If H is colored centralsym, then all six edges to the centre of H are black.

In particular, as we are assuming no 1tri hexagons, the faces cut out by the black edges have angles 60 degrees, and thus must be equilateral triangles, sectors of angle 60, half-planes, or the entire plane. If there is at least one equilateral triangle, then the rest of the black edges must form an equilateral lattice with that triangle sidelength. This leads to only a small number of possible hexagon colorings in the lattice:

# Case 1: All edges white.
# Case 2: All edges black.
# Case 3.k: For some natural number <math>k \geq 2</math>, the length k edges joining adjacent vertices in some coset of <math>k \cdot {\mathbf Z}[ e^{\pi i/3} ]</math> are all black, and the remaining edges are white.
# Case 4: Each horizontal row consists either entirely of black edges, or entirely of white edges.
# Case 5: Each northwest row consists either entirely of black edges, or entirely of white edges.
# Case 6: Each northeast row consists either entirely of black edges, or entirely of white edges.
# Case 7: Six rays of black edges meeting at a common vertex; all other edges white.

Technically, Case 1 is contained in Cases 4,5,6 as written above, but this will not be an issue. One can view Case 7 as a limiting case <math>k=\infty</math> of Case 3.k; Case 2 is similarly the opposite limiting case <math>k=1</math>.

In the first case, the coloring is periodic with periods <math>2, 2 e^{\pi i/3}</math>. In the second case, it is periodic with periods <math>3, 3 e^{\pi i/3}</math>. In the third case, it is periodic with periods <math>3k, 3k e^{\pi i/3}</math>. Also note that for each k, one can check if Case 3.k holds by inspecting the coloring at a finite number of vertices. Thus the event that Case 3.k holds is "measurable" in the sense that a meaningful probability can be assigned. (But Cases 1,2,4,5,6 are not measurable events, they require an infinite number of points to be inspected, and the probability measure we are using is only finitely additive rather than infinitely additive.) In Case 4, the coloring is periodic with period 2; also, every coset of <math>2 \cdot {\bf Z}[e^{\pi i/3}]</math> is 2-colored. Similarly for Case 5 and 6 (where the periods are <math>2 e^{2\pi i/3}</math> and <math>2 e^{4\pi i/3}</math> respectively.)

Let <math>\alpha_k</math> be the probability that Case 3.k holds for the given value of <math>k</math>. Then <math> \sum_{k=2}^K \alpha_k \leq 1</math> for any k, hence <math>\sum_{k=2}^\infty \alpha_k \leq 1</math>. In particular, we can find <math>K_1</math> such that <math>\sum_{k={K_1}}^\infty \alpha_k \leq 0.1</math> (say). Let <math>P</math> be six times the least common multiple of <math>1,2,\dots,K_1</math>. Then the coloring is P- and <math>P e^{\pi i/3}</math>-periodic for Case 1, Case 2, and all Case 3.k with <math>k \leq K_1</math>. On the other hand, if <math>K_2</math> is sufficiently large depending on <math>P</math>, and Case 3.k holds for some <math>k \geq K_2</math>, then almost all of the hexagons are colored centralsym, which makes the coloring "almost <math>P, P e^{\pi i/3}</math>-periodic" in the sense that
:<math>{\bf c}(z+P e^{\pi i j/3}) = {\bf c}(z) \hbox{ for } j=0,1,2,3,4,5</math>
will hold for at least <math>0.9</math> of the lattice points <math>z \in {\bf Z}[e^{\pi i/3}]</math> with <math>|z| \leq K_2</math>. Similarly for Case 7 (which is sort of a <math>k=\infty</math> limiting case of Case 3.k.) Thus, with the probability <math> \geq 1 - \sum_{k=K_1}^{K_2} \alpha_k \geq 0.9</math>, the coloring of the seven vertices <math>{\bf c}(0), {\bf c}(P e^{\pi ij/3}, j=1,\dots,6</math> is (up to rotation and recoloring) one of the three patterns of the central and linking vertices in Figure 3 of Aubrey's paper, namely

* <math>{\bf c}(0) = {\bf c}(P) = {\bf c}(P e^{\pi i/3}) = {\bf c}(P e^{2\pi i/3}) = {\bf c}(P e^{3\pi i/3}) = {\bf c}(P e^{4\pi i/3}) = {\bf c}(P e^{5\pi i/3}) </math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{3\pi i/3})</math>; <math>{\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{3\pi i/3})</math>; <math> {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>.

Using the spindling argument from Aubrey's paper, we conclude that the third possibility must in fact hold with probability at least 0.8; on the other hand, from Lemma 2 this scenario can only occur with probability at most 1/2, giving the required contradiction.
<math>\Box</math>

One should be able to refine this argument to show that <math>p_{H = 1tri} > c</math> for an absolute constant <math> c>0 </math>.

=== Lemma 21 ===
Providing a tighter bound for Lemma 17 with a more thorough proof: If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

=== Lemma 22 ===

One has <math>3 p_{1/\sqrt{3}} \geq {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) )</math>.

'''Proof''' Let <math>z</math> be a complex number of magnitude <math>1/\sqrt{3}</math> that is a unit distance from 1. If <math>\mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) = c</math> (say), then <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> cannot be colored with <math>c</math>; also, <math>z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> are the vertices of a unit equilateral triangle and thus must take on three different colors. By the pigeonhole principle, one of <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> must then take the same color as the origin, and the claim follows. <math>\Box</math>

=== Lemma 23 ===

One has <math>4 p_{(\sqrt{6} \pm \sqrt{2})/2} + p_{\sqrt{2}} \geq 1</math>. In particular, by Lemma 2, <math>p_{(\sqrt{6}+\sqrt{2})/2} \geq 1/8</math>.

'''Proof''' [ExIs2018b] We just prove the claim for the + sign (the - sign can then be obtained after applying the Galois conjugacy that maps <math>\sqrt{3}</math> to <math>-\sqrt{3}</math>, leaving <math>\sqrt{2}</math> unchanged). Set <math>d := \frac{\sqrt{6}+\sqrt{2}}{2}</math>, and consider the five vertices

:<math>0, e^{5\pi i/4}, e^{5\pi i/4} + d, e^{5\pi i/4} + e^{\pi i/3} d, e^{5\pi i/4} + (e^{\pi i/3}-i)d.</math>

One can check that of the ten edges determined by these five vertices, five have unit length, four have length d, and the remaining distance (from 0 to <math>e^{5\pi i/4}+d</math>) has distance <math>\sqrt{2}</math>. Since <math>\mathbf{c}</math> must make one of the latter five edges monochromatic, the claim follows. <math>\Box</math>

=== Lemma 24 ===

One has <math>p_{\sqrt{2}} \geq \frac{1}{14}</math>.

'''Proof''' In page 7 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 20 unit distance edges and 14 edges of length <math>\sqrt{2}</math> is constructed. The coloring <math>\mathbf{c}</math> must make one of the 14 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 25 ===

Let <math>d = \frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math> and <math>e = \frac{3^{1/4} \sqrt{2} + \sqrt{3} - 1}{2}</math>.
Then one has <math>14 p_d + p_e \geq 1</math>. In particular, by Lemma 2, <math>p_d \geq 1/28</math>.

'''Proof''' In page 9 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 19 unit edges, 14 edges of length d, and one edge of length e is constructed. The coloring <math>\mathbf{c}</math> must make one of the 15 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 26 ===

Let <math>d = \sqrt{3/2 + \sqrt{33}/6}</math>. Then <math>7 p_d \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_d \geq \frac{1}{196}</math>.

'''Proof''' In page 11 of [ExIs2018b], a graph of nine vertices consisting of 12 unit edges and 7 edges of length d is constructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Thus, <math>\mathbf{c}</math> can only make the AB edge monochromatic if one of the seven length d edges is monochromatic. The claim follows.
<math>\Box</math>

=== Lemma 27 ===

One has <math>27 p_{\sqrt{5/3}} \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_{\sqrt{5/3}} \geq \frac{1}{756}</math>.

'''Proof''' In page 13 of [ExIs2018], a graph of 33 vertices with some unit edges and 27 edges of length <math>\sqrt{5/3}</math> is contructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Now repeat the proof of Lemma 26. <math>\Box</math>

=== Computer-verified claim 28 ===

One has <math>p_{2/\sqrt{3}} \geq \frac{1}{177}</math>.

'''Proof''' In page 15 of [ExIs2018], a 5-chromatic graph of 103 vertices, 312 unit edges, and 177 edges of length <math>2/\sqrt{3}</math> is constructed. <math>\mathbf{c}</math> must make one of the latter edges monochromatic, giving the claim. <math>\Box</math>

=== Lemma 29 ===

One has <math>p_{(\sqrt{6} \pm \sqrt{2})/2} \geq 1/6</math> (this improves the bound in Lemma 23).

'''Proof''' Use graphs 505 and 507 from [S2004] and the spindle bound. <math>\Box</math>

=== Lemma 30 ===
For <math>m > n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' <math>n</math> colors and <math>m</math> points necessitates at least 2 having equal color. I.e.

:<math>{\bf P}\left(\bigvee_{k=0}^n \bigvee_{j=k+1}^n\ \mathbf{c}(z_k) = \mathbf{c}(z_j)\right) = 1</math>

The lemma then follows immediately from the fact:
:<math>{\bf P}\left(\bigcup_{k} E_k\right) \leq \sum_{k} {\bf P}\left(E_k\right) \,\Box</math>

=== Corollary 31 ===
Let <math>\lvert z_k\rvert=1</math>. Then for <math>m \geq n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' Use lemma 30 on the set <math>\left\{z_k \bigg\vert 1\leq k\leq m \land k\in\mathbb{Z}\right\}\cup\{0\}</math>. Simplify using <math>{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(0) \right)=0</math>.
<math>\Box</math>

=== Lemma 32 ===
For <math>x\in\mathbb{R}</math> and an <math>n</math>-coloring of the plane, <math>\sum_{k=1}^{n-1}\left(n-k\right){\bf P}\left(\mathbf{c}\left(0\right) = \mathbf{c}\left( 2\sin\left(\frac{kx}{2}\right) \right) \right) \geq 1</math>.

'''Proof''' Use corollary 31 on the set <math>\left\{e^{ikx} \bigg\vert 0\leq k < n \land k\in\mathbb{Z}\right\}</math>. and simplify by grouping lengths.
<math>\Box</math>

=== Corollary 33 ===
Interesting(easy to simplify results of) values for <math>x</math> in Lemma 32 are in <math>\left\{x \bigg\vert \sin\left(\frac{kx}{2}\right)=1 \land 1\leq k < n \land k\in\mathbb{Z}\right\}</math>.
For 4-colorings, this gives
:<math>2p_{\sqrt 3}+p_2 \geq 1</math>
:<math>3p_{(\sqrt 3-1)/\sqrt 2}+p_{\sqrt 2} \geq 1</math>
:<math>3p_{2\sin(\pi/18)}+2p_{2\sin(\pi/9)} \geq 1</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

== Further questions ==
* For <math>n,m\geq CNP</math>, what consistent relationships exist between <math>{\bf P}\left(\mathbf{c}(0)=\mathbf{c}(d)\bigg\vert n\text{ colors}\right)</math> and <math>{\bf P}\left(\mathbf{c}(0)=\mathbf{c}(d)\bigg\vert m\text{ colors}\right)</math>? How can these relationships be used to sharpen arguments of the probabilistic formulation?

[[Category:Polymath16]]

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-20T22:18:43Z

Nazgand: /* Bounds on p_d for 4-colourings */

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> \geq 1/n</math>
|
|
| <math>1-\frac{1}{n}</math>
| (n+1)-gon with one edge length 1 and the rest d
|
|-
| <math> \geq 1/(n \sqrt{3})</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d
| Not better than the above on intervals <math>\left(\frac{1}{7},\frac{1}{4\sqrt{3}}\right),\left(\frac{1}{4},\frac{1}{2\sqrt{3}}\right)</math>
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/4
| <math>G_{21}</math>
| 1/2
| Spindle
| Lower bound computer verified
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|-
| <math>\frac{\sqrt{6} \pm \sqrt{2}}{2}</math>
| 1/6
| An arrangement of five vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{2}</math>
| 1/14
| A graph of 13 vertices
| 1/2
| Spindle
|
|-
| <math>\frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math>
| 1/28
| A graph of 13 vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{3/2 + \sqrt{33}/6}</math>
| 1/196
| A graph of 9 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>\sqrt{5/3}</math>
| 1/756
| A graph of 33 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>2/\sqrt{3}</math>
| 1/177
| A graph of 103 vertices
| 1/2
| Spindle
| Computer-verified
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>p_{\sqrt 3}/(1-p_2)</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

=== Lemma 1 ===
Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

=== Lemma 2 ===
(Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

=== Lemma 3 ===
(Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

=== Corollary 4 ===
(Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

=== Computer-verified Claim 5 ===
(Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

=== Computer-verified Claim 6 ===
(Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

=== Lemma 7 ===
(Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

=== Corollary 8 ===
One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

=== Lemma 9 ===
(Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

=== Corollary 10 ===
(Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

=== Computer-verified claim 11 ===
One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

=== Lemma 12 ===
For certain special distances d, one can improve the bound in Lemma 2:

If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, <math>\gcd(j,2k+1)=1</math>, and <math>r = \frac{1}{2} \csc\left(\frac{j\pi}{2k+1}\right)</math> then
:<math> p_r \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}}} \leq \frac{1}{3}.</math>

'''Proof''' Observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2}} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges. More generally, if <math>a,b,c,d,e</math> are the diagonal lengths of a pentagon with unit sides, then
:<math> 1 \leq p_a + p_b + p_c + p_d + p_e \leq 2.</math>

=== Lemma 13 ===
We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

=== Corollary 14 ===
We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

=== Lemma 15 ===
One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

=== Corollary 16 ===
We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

=== Lemma 17 ===
If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths a,b,c. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also, thus
:<math>{\bf P}(\mathbf{c}(0) \neq \mathbf{c}(a)) + {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(b)) \geq {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(c))</math>
and the claim follows. <math>\Box</math>

Note that Lemma 2 follows from the a=b, c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 1 - 2^{-k}</math> when <math>a \geq 2^{-k}, k\in\mathbb{Z}^+</math>.

Further, we can generalise the a=b case to one in which the triangle is replaced by a (k+1)-gon of which one edge is 1 and the others are all equal, leading to the stronger result <math>p_a \leq 1 - 1/k</math> when <math>a \geq 1/k, k\in\mathbb{Z}^+ \land k>1</math>. Further strengthening is achieved by using <math>1/\sqrt{3}</math> as the long edge, given Lemma 12.

=== Lemma 18 ===
Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio. Also we have
:<math> p_{d/\sqrt{3}} \leq \frac{1}{3} + p_d, \frac{1}{2} + \frac{1}{2} p_d.</math>

Note that this generalises (9), as well as a special case of Lemma 12.

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the first claim. Similarly, if one considers the colorings of an equilateral triangle of sidelength <math>d</math> together with its center, and counts the numbers <math>a,b \in \{0,1,2,3\}</math> of monochromatic edges of length <math>d</math> and <math>d/\sqrt{3}</math> respectively, one observes that one always has <math>\frac{b}{3} \leq \frac{1}{3} + \frac{2}{3} \frac{a}{3}, \frac{1}{2} + \frac{1}{2} \frac{a}{3}</math>, and on taking expectations one obtains the claim.<math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

=== Lemma 19 ===
(Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

=== Theorem 20 ===
One has <math>p_{H = 1tri} > 0</math>.

'''Proof''' Suppose for contradiction that <math>p_{H = 1tri} = 0</math>. One can then run a version of the de Bruijn-Erdos argument to obtain a coloring in which 1tri hexagons are completely nonexistent (since there are arbitrarily large finite colorings with this property). Consider the triangular lattice <math>{\bf Z}[e^{\pi i/3}]</math>. We 2-color the edges of this lattice by coloring an edge black if it is the short diagonal of a unit rhombus with monochromatic long diagonal, and white otherwise. The four colorings of hexagons lead to four possible colorings at each vertex:

* If H is colored 2tri, then all six edges to the centre of H are black.
* If H is colored 1tri, then two edges to the centre of H at 120 degree angles are white, the other four are black.
* If H is colored axisym, then two opposing edges of the centre of H are black, the other four are white.
* If H is colored centralsym, then all six edges to the centre of H are black.

In particular, as we are assuming no 1tri hexagons, the faces cut out by the black edges have angles 60 degrees, and thus must be equilateral triangles, sectors of angle 60, half-planes, or the entire plane. If there is at least one equilateral triangle, then the rest of the black edges must form an equilateral lattice with that triangle sidelength. This leads to only a small number of possible hexagon colorings in the lattice:

# Case 1: All edges white.
# Case 2: All edges black.
# Case 3.k: For some natural number <math>k \geq 2</math>, the length k edges joining adjacent vertices in some coset of <math>k \cdot {\mathbf Z}[ e^{\pi i/3} ]</math> are all black, and the remaining edges are white.
# Case 4: Each horizontal row consists either entirely of black edges, or entirely of white edges.
# Case 5: Each northwest row consists either entirely of black edges, or entirely of white edges.
# Case 6: Each northeast row consists either entirely of black edges, or entirely of white edges.
# Case 7: Six rays of black edges meeting at a common vertex; all other edges white.

Technically, Case 1 is contained in Cases 4,5,6 as written above, but this will not be an issue. One can view Case 7 as a limiting case <math>k=\infty</math> of Case 3.k; Case 2 is similarly the opposite limiting case <math>k=1</math>.

In the first case, the coloring is periodic with periods <math>2, 2 e^{\pi i/3}</math>. In the second case, it is periodic with periods <math>3, 3 e^{\pi i/3}</math>. In the third case, it is periodic with periods <math>3k, 3k e^{\pi i/3}</math>. Also note that for each k, one can check if Case 3.k holds by inspecting the coloring at a finite number of vertices. Thus the event that Case 3.k holds is "measurable" in the sense that a meaningful probability can be assigned. (But Cases 1,2,4,5,6 are not measurable events, they require an infinite number of points to be inspected, and the probability measure we are using is only finitely additive rather than infinitely additive.) In Case 4, the coloring is periodic with period 2; also, every coset of <math>2 \cdot {\bf Z}[e^{\pi i/3}]</math> is 2-colored. Similarly for Case 5 and 6 (where the periods are <math>2 e^{2\pi i/3}</math> and <math>2 e^{4\pi i/3}</math> respectively.)

Let <math>\alpha_k</math> be the probability that Case 3.k holds for the given value of <math>k</math>. Then <math> \sum_{k=2}^K \alpha_k \leq 1</math> for any k, hence <math>\sum_{k=2}^\infty \alpha_k \leq 1</math>. In particular, we can find <math>K_1</math> such that <math>\sum_{k={K_1}}^\infty \alpha_k \leq 0.1</math> (say). Let <math>P</math> be six times the least common multiple of <math>1,2,\dots,K_1</math>. Then the coloring is P- and <math>P e^{\pi i/3}</math>-periodic for Case 1, Case 2, and all Case 3.k with <math>k \leq K_1</math>. On the other hand, if <math>K_2</math> is sufficiently large depending on <math>P</math>, and Case 3.k holds for some <math>k \geq K_2</math>, then almost all of the hexagons are colored centralsym, which makes the coloring "almost <math>P, P e^{\pi i/3}</math>-periodic" in the sense that
:<math>{\bf c}(z+P e^{\pi i j/3}) = {\bf c}(z) \hbox{ for } j=0,1,2,3,4,5</math>
will hold for at least <math>0.9</math> of the lattice points <math>z \in {\bf Z}[e^{\pi i/3}]</math> with <math>|z| \leq K_2</math>. Similarly for Case 7 (which is sort of a <math>k=\infty</math> limiting case of Case 3.k.) Thus, with the probability <math> \geq 1 - \sum_{k=K_1}^{K_2} \alpha_k \geq 0.9</math>, the coloring of the seven vertices <math>{\bf c}(0), {\bf c}(P e^{\pi ij/3}, j=1,\dots,6</math> is (up to rotation and recoloring) one of the three patterns of the central and linking vertices in Figure 3 of Aubrey's paper, namely

* <math>{\bf c}(0) = {\bf c}(P) = {\bf c}(P e^{\pi i/3}) = {\bf c}(P e^{2\pi i/3}) = {\bf c}(P e^{3\pi i/3}) = {\bf c}(P e^{4\pi i/3}) = {\bf c}(P e^{5\pi i/3}) </math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{3\pi i/3})</math>; <math>{\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{3\pi i/3})</math>; <math> {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>.

Using the spindling argument from Aubrey's paper, we conclude that the third possibility must in fact hold with probability at least 0.8; on the other hand, from Lemma 2 this scenario can only occur with probability at most 1/2, giving the required contradiction.
<math>\Box</math>

One should be able to refine this argument to show that <math>p_{H = 1tri} > c</math> for an absolute constant <math> c>0 </math>.

=== Lemma 21 ===
Providing a tighter bound for Lemma 17 with a more thorough proof: If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

=== Lemma 22 ===

One has <math>3 p_{1/\sqrt{3}} \geq {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) )</math>.

'''Proof''' Let <math>z</math> be a complex number of magnitude <math>1/\sqrt{3}</math> that is a unit distance from 1. If <math>\mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) = c</math> (say), then <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> cannot be colored with <math>c</math>; also, <math>z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> are the vertices of a unit equilateral triangle and thus must take on three different colors. By the pigeonhole principle, one of <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> must then take the same color as the origin, and the claim follows. <math>\Box</math>

=== Lemma 23 ===

One has <math>4 p_{(\sqrt{6} \pm \sqrt{2})/2} + p_{\sqrt{2}} \geq 1</math>. In particular, by Lemma 2, <math>p_{(\sqrt{6}+\sqrt{2})/2} \geq 1/8</math>.

'''Proof''' [ExIs2018b] We just prove the claim for the + sign (the - sign can then be obtained after applying the Galois conjugacy that maps <math>\sqrt{3}</math> to <math>-\sqrt{3}</math>, leaving <math>\sqrt{2}</math> unchanged). Set <math>d := \frac{\sqrt{6}+\sqrt{2}}{2}</math>, and consider the five vertices

:<math>0, e^{5\pi i/4}, e^{5\pi i/4} + d, e^{5\pi i/4} + e^{\pi i/3} d, e^{5\pi i/4} + (e^{\pi i/3}-i)d.</math>

One can check that of the ten edges determined by these five vertices, five have unit length, four have length d, and the remaining distance (from 0 to <math>e^{5\pi i/4}+d</math>) has distance <math>\sqrt{2}</math>. Since <math>\mathbf{c}</math> must make one of the latter five edges monochromatic, the claim follows. <math>\Box</math>

=== Lemma 24 ===

One has <math>p_{\sqrt{2}} \geq \frac{1}{14}</math>.

'''Proof''' In page 7 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 20 unit distance edges and 14 edges of length <math>\sqrt{2}</math> is constructed. The coloring <math>\mathbf{c}</math> must make one of the 14 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 25 ===

Let <math>d = \frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math> and <math>e = \frac{3^{1/4} \sqrt{2} + \sqrt{3} - 1}{2}</math>.
Then one has <math>14 p_d + p_e \geq 1</math>. In particular, by Lemma 2, <math>p_d \geq 1/28</math>.

'''Proof''' In page 9 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 19 unit edges, 14 edges of length d, and one edge of length e is constructed. The coloring <math>\mathbf{c}</math> must make one of the 15 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 26 ===

Let <math>d = \sqrt{3/2 + \sqrt{33}/6}</math>. Then <math>7 p_d \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_d \geq \frac{1}{196}</math>.

'''Proof''' In page 11 of [ExIs2018b], a graph of nine vertices consisting of 12 unit edges and 7 edges of length d is constructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Thus, <math>\mathbf{c}</math> can only make the AB edge monochromatic if one of the seven length d edges is monochromatic. The claim follows.
<math>\Box</math>

=== Lemma 27 ===

One has <math>27 p_{\sqrt{5/3}} \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_{\sqrt{5/3}} \geq \frac{1}{756}</math>.

'''Proof''' In page 13 of [ExIs2018], a graph of 33 vertices with some unit edges and 27 edges of length <math>\sqrt{5/3}</math> is contructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Now repeat the proof of Lemma 26. <math>\Box</math>

=== Computer-verified claim 28 ===

One has <math>p_{2/\sqrt{3}} \geq \frac{1}{177}</math>.

'''Proof''' In page 15 of [ExIs2018], a 5-chromatic graph of 103 vertices, 312 unit edges, and 177 edges of length <math>2/\sqrt{3}</math> is constructed. <math>\mathbf{c}</math> must make one of the latter edges monochromatic, giving the claim. <math>\Box</math>

=== Lemma 29 ===

One has <math>p_{(\sqrt{6} \pm \sqrt{2})/2} \geq 1/6</math> (this improves the bound in Lemma 23).

'''Proof''' Use graphs 505 and 507 from [S2004] and the spindle bound. <math>\Box</math>

=== Lemma 30 ===
For <math>m > n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' <math>n</math> colors and <math>m</math> points necessitates at least 2 having equal color. I.e.

:<math>{\bf P}\left(\bigvee_{k=0}^n \bigvee_{j=k+1}^n\ \mathbf{c}(z_k) = \mathbf{c}(z_j)\right) = 1</math>

The lemma then follows immediately from the fact:
:<math>{\bf P}\left(\bigcup_{k} E_k\right) \leq \sum_{k} {\bf P}\left(E_k\right) \,\Box</math>

=== Corollary 31 ===
Let <math>\lvert z_k\rvert=1</math>. Then for <math>m \geq n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' Use lemma 30 on the set <math>\left\{z_k \bigg\vert 1\leq k\leq m \land k\in\mathbb{Z}\right\}\cup\{0\}</math>. Simplify using <math>{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(0) \right)=0</math>.
<math>\Box</math>

=== Lemma 32 ===
For <math>x\in\mathbb{R}</math> and an <math>n</math>-coloring of the plane, <math>\sum_{k=1}^{n-1}\left(n-k\right){\bf P}\left(\mathbf{c}\left(0\right) = \mathbf{c}\left( 2\sin\left(\frac{kx}{2}\right) \right) \right) \geq 1</math>.

'''Proof''' Use corollary 31 on the set <math>\left\{e^{ikx} \bigg\vert 0\leq k < n \land k\in\mathbb{Z}\right\}</math>. and simplify by grouping lengths.
<math>\Box</math>

=== Corollary 33 ===
Interesting(easy to simplify results of) values for <math>x</math> in Lemma 32 are in <math>\left\{x \bigg\vert \sin\left(\frac{kx}{2}\right)=1 \land 1\leq k < n \land k\in\mathbb{Z}\right\}</math>.
For 4-colorings, this gives
:<math>2p_{\sqrt 3}+p_2 \geq 1</math>
:<math>3p_{(\sqrt 3-1)/\sqrt 2}+p_{\sqrt 2} \geq 1</math>
:<math>3p_{2\sin(\pi/18)}+2p_{2\sin(\pi/9)} \geq 1</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

== Further questions ==
* For <math>n,m\geq CNP</math>, what consistent relationships exist between <math>{\bf P}\left(\mathbf{c}(0)=\mathbf{c}(d)\bigg\vert n\text{ colors}\right)</math> and <math>{\bf P}\left(\mathbf{c}(0)=\mathbf{c}(d)\bigg\vert m\text{ colors}\right)</math>? How can these relationships be used to sharpen arguments of the probabilistic formulation?

[[Category:Polymath16]]

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-20T19:37:01Z

Nazgand: /* Further questions */

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> \geq 1/(n \sqrt{3})</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d
|
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/4
| <math>G_{21}</math>
| 1/2
| Spindle
| Lower bound computer verified
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|-
| <math>\frac{\sqrt{6} \pm \sqrt{2}}{2}</math>
| 1/6
| An arrangement of five vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{2}</math>
| 1/14
| A graph of 13 vertices
| 1/2
| Spindle
|
|-
| <math>\frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math>
| 1/28
| A graph of 13 vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{3/2 + \sqrt{33}/6}</math>
| 1/196
| A graph of 9 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>\sqrt{5/3}</math>
| 1/756
| A graph of 33 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>2/\sqrt{3}</math>
| 1/177
| A graph of 103 vertices
| 1/2
| Spindle
| Computer-verified
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>p_{\sqrt 3}/(1-p_2)</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

=== Lemma 1 ===
Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

=== Lemma 2 ===
(Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

=== Lemma 3 ===
(Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

=== Corollary 4 ===
(Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

=== Computer-verified Claim 5 ===
(Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

=== Computer-verified Claim 6 ===
(Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

=== Lemma 7 ===
(Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

=== Corollary 8 ===
One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

=== Lemma 9 ===
(Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

=== Corollary 10 ===
(Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

=== Computer-verified claim 11 ===
One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

=== Lemma 12 ===
For certain special distances d, one can improve the bound in Lemma 2:

If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, <math>\gcd(j,2k+1)=1</math>, and <math>r = \frac{1}{2} \csc\left(\frac{j\pi}{2k+1}\right)</math> then
:<math> p_r \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}}} \leq \frac{1}{3}.</math>

'''Proof''' Observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2}} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges. More generally, if <math>a,b,c,d,e</math> are the diagonal lengths of a pentagon with unit sides, then
:<math> 1 \leq p_a + p_b + p_c + p_d + p_e \leq 2.</math>

=== Lemma 13 ===
We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

=== Corollary 14 ===
We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

=== Lemma 15 ===
One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

=== Corollary 16 ===
We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

=== Lemma 17 ===
If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths a,b,c. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also, thus
:<math>{\bf P}(\mathbf{c}(0) \neq \mathbf{c}(a)) + {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(b)) \geq {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(c))</math>
and the claim follows. <math>\Box</math>

Note that Lemma 2 follows from the a=b, c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 1 - 2^{-k}</math> when <math>a \geq 2^{-k}, k\in\mathbb{Z}^+</math>.

Further, we can generalise the a=b case to one in which the triangle is replaced by a (k+1)-gon of which one edge is 1 and the others are all equal, leading to the stronger result <math>p_a \leq 1 - 1/k</math> when <math>a \geq 1/k, k\in\mathbb{Z}^+ \land k>1</math>. Further strengthening is achieved by using <math>1/\sqrt{3}</math> as the long edge, given Lemma 12.

=== Lemma 18 ===
Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio. Also we have
:<math> p_{d/\sqrt{3}} \leq \frac{1}{3} + p_d, \frac{1}{2} + \frac{1}{2} p_d.</math>

Note that this generalises (9), as well as a special case of Lemma 12.

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the first claim. Similarly, if one considers the colorings of an equilateral triangle of sidelength <math>d</math> together with its center, and counts the numbers <math>a,b \in \{0,1,2,3\}</math> of monochromatic edges of length <math>d</math> and <math>d/\sqrt{3}</math> respectively, one observes that one always has <math>\frac{b}{3} \leq \frac{1}{3} + \frac{2}{3} \frac{a}{3}, \frac{1}{2} + \frac{1}{2} \frac{a}{3}</math>, and on taking expectations one obtains the claim.<math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

=== Lemma 19 ===
(Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

=== Theorem 20 ===
One has <math>p_{H = 1tri} > 0</math>.

'''Proof''' Suppose for contradiction that <math>p_{H = 1tri} = 0</math>. One can then run a version of the de Bruijn-Erdos argument to obtain a coloring in which 1tri hexagons are completely nonexistent (since there are arbitrarily large finite colorings with this property). Consider the triangular lattice <math>{\bf Z}[e^{\pi i/3}]</math>. We 2-color the edges of this lattice by coloring an edge black if it is the short diagonal of a unit rhombus with monochromatic long diagonal, and white otherwise. The four colorings of hexagons lead to four possible colorings at each vertex:

* If H is colored 2tri, then all six edges to the centre of H are black.
* If H is colored 1tri, then two edges to the centre of H at 120 degree angles are white, the other four are black.
* If H is colored axisym, then two opposing edges of the centre of H are black, the other four are white.
* If H is colored centralsym, then all six edges to the centre of H are black.

In particular, as we are assuming no 1tri hexagons, the faces cut out by the black edges have angles 60 degrees, and thus must be equilateral triangles, sectors of angle 60, half-planes, or the entire plane. If there is at least one equilateral triangle, then the rest of the black edges must form an equilateral lattice with that triangle sidelength. This leads to only a small number of possible hexagon colorings in the lattice:

# Case 1: All edges white.
# Case 2: All edges black.
# Case 3.k: For some natural number <math>k \geq 2</math>, the length k edges joining adjacent vertices in some coset of <math>k \cdot {\mathbf Z}[ e^{\pi i/3} ]</math> are all black, and the remaining edges are white.
# Case 4: Each horizontal row consists either entirely of black edges, or entirely of white edges.
# Case 5: Each northwest row consists either entirely of black edges, or entirely of white edges.
# Case 6: Each northeast row consists either entirely of black edges, or entirely of white edges.
# Case 7: Six rays of black edges meeting at a common vertex; all other edges white.

Technically, Case 1 is contained in Cases 4,5,6 as written above, but this will not be an issue. One can view Case 7 as a limiting case <math>k=\infty</math> of Case 3.k; Case 2 is similarly the opposite limiting case <math>k=1</math>.

In the first case, the coloring is periodic with periods <math>2, 2 e^{\pi i/3}</math>. In the second case, it is periodic with periods <math>3, 3 e^{\pi i/3}</math>. In the third case, it is periodic with periods <math>3k, 3k e^{\pi i/3}</math>. Also note that for each k, one can check if Case 3.k holds by inspecting the coloring at a finite number of vertices. Thus the event that Case 3.k holds is "measurable" in the sense that a meaningful probability can be assigned. (But Cases 1,2,4,5,6 are not measurable events, they require an infinite number of points to be inspected, and the probability measure we are using is only finitely additive rather than infinitely additive.) In Case 4, the coloring is periodic with period 2; also, every coset of <math>2 \cdot {\bf Z}[e^{\pi i/3}]</math> is 2-colored. Similarly for Case 5 and 6 (where the periods are <math>2 e^{2\pi i/3}</math> and <math>2 e^{4\pi i/3}</math> respectively.)

Let <math>\alpha_k</math> be the probability that Case 3.k holds for the given value of <math>k</math>. Then <math> \sum_{k=2}^K \alpha_k \leq 1</math> for any k, hence <math>\sum_{k=2}^\infty \alpha_k \leq 1</math>. In particular, we can find <math>K_1</math> such that <math>\sum_{k={K_1}}^\infty \alpha_k \leq 0.1</math> (say). Let <math>P</math> be six times the least common multiple of <math>1,2,\dots,K_1</math>. Then the coloring is P- and <math>P e^{\pi i/3}</math>-periodic for Case 1, Case 2, and all Case 3.k with <math>k \leq K_1</math>. On the other hand, if <math>K_2</math> is sufficiently large depending on <math>P</math>, and Case 3.k holds for some <math>k \geq K_2</math>, then almost all of the hexagons are colored centralsym, which makes the coloring "almost <math>P, P e^{\pi i/3}</math>-periodic" in the sense that
:<math>{\bf c}(z+P e^{\pi i j/3}) = {\bf c}(z) \hbox{ for } j=0,1,2,3,4,5</math>
will hold for at least <math>0.9</math> of the lattice points <math>z \in {\bf Z}[e^{\pi i/3}]</math> with <math>|z| \leq K_2</math>. Similarly for Case 7 (which is sort of a <math>k=\infty</math> limiting case of Case 3.k.) Thus, with the probability <math> \geq 1 - \sum_{k=K_1}^{K_2} \alpha_k \geq 0.9</math>, the coloring of the seven vertices <math>{\bf c}(0), {\bf c}(P e^{\pi ij/3}, j=1,\dots,6</math> is (up to rotation and recoloring) one of the three patterns of the central and linking vertices in Figure 3 of Aubrey's paper, namely

* <math>{\bf c}(0) = {\bf c}(P) = {\bf c}(P e^{\pi i/3}) = {\bf c}(P e^{2\pi i/3}) = {\bf c}(P e^{3\pi i/3}) = {\bf c}(P e^{4\pi i/3}) = {\bf c}(P e^{5\pi i/3}) </math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{3\pi i/3})</math>; <math>{\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{3\pi i/3})</math>; <math> {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>.

Using the spindling argument from Aubrey's paper, we conclude that the third possibility must in fact hold with probability at least 0.8; on the other hand, from Lemma 2 this scenario can only occur with probability at most 1/2, giving the required contradiction.
<math>\Box</math>

One should be able to refine this argument to show that <math>p_{H = 1tri} > c</math> for an absolute constant <math> c>0 </math>.

=== Lemma 21 ===
Providing a tighter bound for Lemma 17 with a more thorough proof: If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

=== Lemma 22 ===

One has <math>3 p_{1/\sqrt{3}} \geq {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) )</math>.

'''Proof''' Let <math>z</math> be a complex number of magnitude <math>1/\sqrt{3}</math> that is a unit distance from 1. If <math>\mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) = c</math> (say), then <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> cannot be colored with <math>c</math>; also, <math>z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> are the vertices of a unit equilateral triangle and thus must take on three different colors. By the pigeonhole principle, one of <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> must then take the same color as the origin, and the claim follows. <math>\Box</math>

=== Lemma 23 ===

One has <math>4 p_{(\sqrt{6} \pm \sqrt{2})/2} + p_{\sqrt{2}} \geq 1</math>. In particular, by Lemma 2, <math>p_{(\sqrt{6}+\sqrt{2})/2} \geq 1/8</math>.

'''Proof''' [ExIs2018b] We just prove the claim for the + sign (the - sign can then be obtained after applying the Galois conjugacy that maps <math>\sqrt{3}</math> to <math>-\sqrt{3}</math>, leaving <math>\sqrt{2}</math> unchanged). Set <math>d := \frac{\sqrt{6}+\sqrt{2}}{2}</math>, and consider the five vertices

:<math>0, e^{5\pi i/4}, e^{5\pi i/4} + d, e^{5\pi i/4} + e^{\pi i/3} d, e^{5\pi i/4} + (e^{\pi i/3}-i)d.</math>

One can check that of the ten edges determined by these five vertices, five have unit length, four have length d, and the remaining distance (from 0 to <math>e^{5\pi i/4}+d</math>) has distance <math>\sqrt{2}</math>. Since <math>\mathbf{c}</math> must make one of the latter five edges monochromatic, the claim follows. <math>\Box</math>

=== Lemma 24 ===

One has <math>p_{\sqrt{2}} \geq \frac{1}{14}</math>.

'''Proof''' In page 7 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 20 unit distance edges and 14 edges of length <math>\sqrt{2}</math> is constructed. The coloring <math>\mathbf{c}</math> must make one of the 14 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 25 ===

Let <math>d = \frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math> and <math>e = \frac{3^{1/4} \sqrt{2} + \sqrt{3} - 1}{2}</math>.
Then one has <math>14 p_d + p_e \geq 1</math>. In particular, by Lemma 2, <math>p_d \geq 1/28</math>.

'''Proof''' In page 9 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 19 unit edges, 14 edges of length d, and one edge of length e is constructed. The coloring <math>\mathbf{c}</math> must make one of the 15 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 26 ===

Let <math>d = \sqrt{3/2 + \sqrt{33}/6}</math>. Then <math>7 p_d \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_d \geq \frac{1}{196}</math>.

'''Proof''' In page 11 of [ExIs2018b], a graph of nine vertices consisting of 12 unit edges and 7 edges of length d is constructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Thus, <math>\mathbf{c}</math> can only make the AB edge monochromatic if one of the seven length d edges is monochromatic. The claim follows.
<math>\Box</math>

=== Lemma 27 ===

One has <math>27 p_{\sqrt{5/3}} \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_{\sqrt{5/3}} \geq \frac{1}{756}</math>.

'''Proof''' In page 13 of [ExIs2018], a graph of 33 vertices with some unit edges and 27 edges of length <math>\sqrt{5/3}</math> is contructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Now repeat the proof of Lemma 26. <math>\Box</math>

=== Computer-verified claim 28 ===

One has <math>p_{2/\sqrt{3}} \geq \frac{1}{177}</math>.

'''Proof''' In page 15 of [ExIs2018], a 5-chromatic graph of 103 vertices, 312 unit edges, and 177 edges of length <math>2/\sqrt{3}</math> is constructed. <math>\mathbf{c}</math> must make one of the latter edges monochromatic, giving the claim. <math>\Box</math>

=== Lemma 29 ===

One has <math>p_{(\sqrt{6} \pm \sqrt{2})/2} \geq 1/6</math> (this improves the bound in Lemma 23).

'''Proof''' Use graphs 505 and 507 from [S2004] and the spindle bound. <math>\Box</math>

=== Lemma 30 ===
For <math>m > n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' <math>n</math> colors and <math>m</math> points necessitates at least 2 having equal color. I.e.

:<math>{\bf P}\left(\bigvee_{k=0}^n \bigvee_{j=k+1}^n\ \mathbf{c}(z_k) = \mathbf{c}(z_j)\right) = 1</math>

The lemma then follows immediately from the fact:
:<math>{\bf P}\left(\bigcup_{k} E_k\right) \leq \sum_{k} {\bf P}\left(E_k\right) \,\Box</math>

=== Corollary 31 ===
Let <math>\lvert z_k\rvert=1</math>. Then for <math>m \geq n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' Use lemma 30 on the set <math>\left\{z_k \bigg\vert 1\leq k\leq m \land k\in\mathbb{Z}\right\}\cup\{0\}</math>. Simplify using <math>{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(0) \right)=0</math>.
<math>\Box</math>

=== Lemma 32 ===
For <math>x\in\mathbb{R}</math> and an <math>n</math>-coloring of the plane, <math>\sum_{k=1}^{n-1}\left(n-k\right){\bf P}\left(\mathbf{c}\left(0\right) = \mathbf{c}\left( 2\sin\left(\frac{kx}{2}\right) \right) \right) \geq 1</math>.

'''Proof''' Use corollary 31 on the set <math>\left\{e^{ikx} \bigg\vert 0\leq k < n \land k\in\mathbb{Z}\right\}</math>. and simplify by grouping lengths.
<math>\Box</math>

=== Corollary 33 ===
Interesting(easy to simplify results of) values for <math>x</math> in Lemma 32 are in <math>\left\{x \bigg\vert \sin\left(\frac{kx}{2}\right)=1 \land 1\leq k < n \land k\in\mathbb{Z}\right\}</math>.
For 4-colorings, this gives
:<math>2p_{\sqrt 3}+p_2 \geq 1</math>
:<math>3p_{(\sqrt 3-1)/\sqrt 2}+p_{\sqrt 2} \geq 1</math>
:<math>3p_{2\sin(\pi/18)}+2p_{2\sin(\pi/9)} \geq 1</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

== Further questions ==
* For <math>n,m\geq CNP</math>, what consistent relationships exist between <math>{\bf P}\left(\mathbf{c}(0)=\mathbf{c}(d)\bigg\vert n\text{ colors}\right)</math> and <math>{\bf P}\left(\mathbf{c}(0)=\mathbf{c}(d)\bigg\vert m\text{ colors}\right)</math>? How can these relationships be used to sharpen arguments of the probabilistic formulation?

[[Category:Polymath16]]

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-20T19:35:02Z

Nazgand: Added questions section

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> \geq 1/(n \sqrt{3})</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d
|
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/4
| <math>G_{21}</math>
| 1/2
| Spindle
| Lower bound computer verified
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|-
| <math>\frac{\sqrt{6} \pm \sqrt{2}}{2}</math>
| 1/6
| An arrangement of five vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{2}</math>
| 1/14
| A graph of 13 vertices
| 1/2
| Spindle
|
|-
| <math>\frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math>
| 1/28
| A graph of 13 vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{3/2 + \sqrt{33}/6}</math>
| 1/196
| A graph of 9 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>\sqrt{5/3}</math>
| 1/756
| A graph of 33 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>2/\sqrt{3}</math>
| 1/177
| A graph of 103 vertices
| 1/2
| Spindle
| Computer-verified
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>p_{\sqrt 3}/(1-p_2)</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

=== Lemma 1 ===
Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

=== Lemma 2 ===
(Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

=== Lemma 3 ===
(Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

=== Corollary 4 ===
(Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

=== Computer-verified Claim 5 ===
(Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

=== Computer-verified Claim 6 ===
(Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

=== Lemma 7 ===
(Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

=== Corollary 8 ===
One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

=== Lemma 9 ===
(Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

=== Corollary 10 ===
(Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

=== Computer-verified claim 11 ===
One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

=== Lemma 12 ===
For certain special distances d, one can improve the bound in Lemma 2:

If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, <math>\gcd(j,2k+1)=1</math>, and <math>r = \frac{1}{2} \csc\left(\frac{j\pi}{2k+1}\right)</math> then
:<math> p_r \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}}} \leq \frac{1}{3}.</math>

'''Proof''' Observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2}} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges. More generally, if <math>a,b,c,d,e</math> are the diagonal lengths of a pentagon with unit sides, then
:<math> 1 \leq p_a + p_b + p_c + p_d + p_e \leq 2.</math>

=== Lemma 13 ===
We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

=== Corollary 14 ===
We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

=== Lemma 15 ===
One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

=== Corollary 16 ===
We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

=== Lemma 17 ===
If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths a,b,c. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also, thus
:<math>{\bf P}(\mathbf{c}(0) \neq \mathbf{c}(a)) + {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(b)) \geq {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(c))</math>
and the claim follows. <math>\Box</math>

Note that Lemma 2 follows from the a=b, c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 1 - 2^{-k}</math> when <math>a \geq 2^{-k}, k\in\mathbb{Z}^+</math>.

Further, we can generalise the a=b case to one in which the triangle is replaced by a (k+1)-gon of which one edge is 1 and the others are all equal, leading to the stronger result <math>p_a \leq 1 - 1/k</math> when <math>a \geq 1/k, k\in\mathbb{Z}^+ \land k>1</math>. Further strengthening is achieved by using <math>1/\sqrt{3}</math> as the long edge, given Lemma 12.

=== Lemma 18 ===
Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio. Also we have
:<math> p_{d/\sqrt{3}} \leq \frac{1}{3} + p_d, \frac{1}{2} + \frac{1}{2} p_d.</math>

Note that this generalises (9), as well as a special case of Lemma 12.

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the first claim. Similarly, if one considers the colorings of an equilateral triangle of sidelength <math>d</math> together with its center, and counts the numbers <math>a,b \in \{0,1,2,3\}</math> of monochromatic edges of length <math>d</math> and <math>d/\sqrt{3}</math> respectively, one observes that one always has <math>\frac{b}{3} \leq \frac{1}{3} + \frac{2}{3} \frac{a}{3}, \frac{1}{2} + \frac{1}{2} \frac{a}{3}</math>, and on taking expectations one obtains the claim.<math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

=== Lemma 19 ===
(Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

=== Theorem 20 ===
One has <math>p_{H = 1tri} > 0</math>.

'''Proof''' Suppose for contradiction that <math>p_{H = 1tri} = 0</math>. One can then run a version of the de Bruijn-Erdos argument to obtain a coloring in which 1tri hexagons are completely nonexistent (since there are arbitrarily large finite colorings with this property). Consider the triangular lattice <math>{\bf Z}[e^{\pi i/3}]</math>. We 2-color the edges of this lattice by coloring an edge black if it is the short diagonal of a unit rhombus with monochromatic long diagonal, and white otherwise. The four colorings of hexagons lead to four possible colorings at each vertex:

* If H is colored 2tri, then all six edges to the centre of H are black.
* If H is colored 1tri, then two edges to the centre of H at 120 degree angles are white, the other four are black.
* If H is colored axisym, then two opposing edges of the centre of H are black, the other four are white.
* If H is colored centralsym, then all six edges to the centre of H are black.

In particular, as we are assuming no 1tri hexagons, the faces cut out by the black edges have angles 60 degrees, and thus must be equilateral triangles, sectors of angle 60, half-planes, or the entire plane. If there is at least one equilateral triangle, then the rest of the black edges must form an equilateral lattice with that triangle sidelength. This leads to only a small number of possible hexagon colorings in the lattice:

# Case 1: All edges white.
# Case 2: All edges black.
# Case 3.k: For some natural number <math>k \geq 2</math>, the length k edges joining adjacent vertices in some coset of <math>k \cdot {\mathbf Z}[ e^{\pi i/3} ]</math> are all black, and the remaining edges are white.
# Case 4: Each horizontal row consists either entirely of black edges, or entirely of white edges.
# Case 5: Each northwest row consists either entirely of black edges, or entirely of white edges.
# Case 6: Each northeast row consists either entirely of black edges, or entirely of white edges.
# Case 7: Six rays of black edges meeting at a common vertex; all other edges white.

Technically, Case 1 is contained in Cases 4,5,6 as written above, but this will not be an issue. One can view Case 7 as a limiting case <math>k=\infty</math> of Case 3.k; Case 2 is similarly the opposite limiting case <math>k=1</math>.

In the first case, the coloring is periodic with periods <math>2, 2 e^{\pi i/3}</math>. In the second case, it is periodic with periods <math>3, 3 e^{\pi i/3}</math>. In the third case, it is periodic with periods <math>3k, 3k e^{\pi i/3}</math>. Also note that for each k, one can check if Case 3.k holds by inspecting the coloring at a finite number of vertices. Thus the event that Case 3.k holds is "measurable" in the sense that a meaningful probability can be assigned. (But Cases 1,2,4,5,6 are not measurable events, they require an infinite number of points to be inspected, and the probability measure we are using is only finitely additive rather than infinitely additive.) In Case 4, the coloring is periodic with period 2; also, every coset of <math>2 \cdot {\bf Z}[e^{\pi i/3}]</math> is 2-colored. Similarly for Case 5 and 6 (where the periods are <math>2 e^{2\pi i/3}</math> and <math>2 e^{4\pi i/3}</math> respectively.)

Let <math>\alpha_k</math> be the probability that Case 3.k holds for the given value of <math>k</math>. Then <math> \sum_{k=2}^K \alpha_k \leq 1</math> for any k, hence <math>\sum_{k=2}^\infty \alpha_k \leq 1</math>. In particular, we can find <math>K_1</math> such that <math>\sum_{k={K_1}}^\infty \alpha_k \leq 0.1</math> (say). Let <math>P</math> be six times the least common multiple of <math>1,2,\dots,K_1</math>. Then the coloring is P- and <math>P e^{\pi i/3}</math>-periodic for Case 1, Case 2, and all Case 3.k with <math>k \leq K_1</math>. On the other hand, if <math>K_2</math> is sufficiently large depending on <math>P</math>, and Case 3.k holds for some <math>k \geq K_2</math>, then almost all of the hexagons are colored centralsym, which makes the coloring "almost <math>P, P e^{\pi i/3}</math>-periodic" in the sense that
:<math>{\bf c}(z+P e^{\pi i j/3}) = {\bf c}(z) \hbox{ for } j=0,1,2,3,4,5</math>
will hold for at least <math>0.9</math> of the lattice points <math>z \in {\bf Z}[e^{\pi i/3}]</math> with <math>|z| \leq K_2</math>. Similarly for Case 7 (which is sort of a <math>k=\infty</math> limiting case of Case 3.k.) Thus, with the probability <math> \geq 1 - \sum_{k=K_1}^{K_2} \alpha_k \geq 0.9</math>, the coloring of the seven vertices <math>{\bf c}(0), {\bf c}(P e^{\pi ij/3}, j=1,\dots,6</math> is (up to rotation and recoloring) one of the three patterns of the central and linking vertices in Figure 3 of Aubrey's paper, namely

* <math>{\bf c}(0) = {\bf c}(P) = {\bf c}(P e^{\pi i/3}) = {\bf c}(P e^{2\pi i/3}) = {\bf c}(P e^{3\pi i/3}) = {\bf c}(P e^{4\pi i/3}) = {\bf c}(P e^{5\pi i/3}) </math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{3\pi i/3})</math>; <math>{\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{3\pi i/3})</math>; <math> {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>.

Using the spindling argument from Aubrey's paper, we conclude that the third possibility must in fact hold with probability at least 0.8; on the other hand, from Lemma 2 this scenario can only occur with probability at most 1/2, giving the required contradiction.
<math>\Box</math>

One should be able to refine this argument to show that <math>p_{H = 1tri} > c</math> for an absolute constant <math> c>0 </math>.

=== Lemma 21 ===
Providing a tighter bound for Lemma 17 with a more thorough proof: If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

=== Lemma 22 ===

One has <math>3 p_{1/\sqrt{3}} \geq {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) )</math>.

'''Proof''' Let <math>z</math> be a complex number of magnitude <math>1/\sqrt{3}</math> that is a unit distance from 1. If <math>\mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) = c</math> (say), then <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> cannot be colored with <math>c</math>; also, <math>z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> are the vertices of a unit equilateral triangle and thus must take on three different colors. By the pigeonhole principle, one of <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> must then take the same color as the origin, and the claim follows. <math>\Box</math>

=== Lemma 23 ===

One has <math>4 p_{(\sqrt{6} \pm \sqrt{2})/2} + p_{\sqrt{2}} \geq 1</math>. In particular, by Lemma 2, <math>p_{(\sqrt{6}+\sqrt{2})/2} \geq 1/8</math>.

'''Proof''' [ExIs2018b] We just prove the claim for the + sign (the - sign can then be obtained after applying the Galois conjugacy that maps <math>\sqrt{3}</math> to <math>-\sqrt{3}</math>, leaving <math>\sqrt{2}</math> unchanged). Set <math>d := \frac{\sqrt{6}+\sqrt{2}}{2}</math>, and consider the five vertices

:<math>0, e^{5\pi i/4}, e^{5\pi i/4} + d, e^{5\pi i/4} + e^{\pi i/3} d, e^{5\pi i/4} + (e^{\pi i/3}-i)d.</math>

One can check that of the ten edges determined by these five vertices, five have unit length, four have length d, and the remaining distance (from 0 to <math>e^{5\pi i/4}+d</math>) has distance <math>\sqrt{2}</math>. Since <math>\mathbf{c}</math> must make one of the latter five edges monochromatic, the claim follows. <math>\Box</math>

=== Lemma 24 ===

One has <math>p_{\sqrt{2}} \geq \frac{1}{14}</math>.

'''Proof''' In page 7 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 20 unit distance edges and 14 edges of length <math>\sqrt{2}</math> is constructed. The coloring <math>\mathbf{c}</math> must make one of the 14 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 25 ===

Let <math>d = \frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math> and <math>e = \frac{3^{1/4} \sqrt{2} + \sqrt{3} - 1}{2}</math>.
Then one has <math>14 p_d + p_e \geq 1</math>. In particular, by Lemma 2, <math>p_d \geq 1/28</math>.

'''Proof''' In page 9 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 19 unit edges, 14 edges of length d, and one edge of length e is constructed. The coloring <math>\mathbf{c}</math> must make one of the 15 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 26 ===

Let <math>d = \sqrt{3/2 + \sqrt{33}/6}</math>. Then <math>7 p_d \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_d \geq \frac{1}{196}</math>.

'''Proof''' In page 11 of [ExIs2018b], a graph of nine vertices consisting of 12 unit edges and 7 edges of length d is constructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Thus, <math>\mathbf{c}</math> can only make the AB edge monochromatic if one of the seven length d edges is monochromatic. The claim follows.
<math>\Box</math>

=== Lemma 27 ===

One has <math>27 p_{\sqrt{5/3}} \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_{\sqrt{5/3}} \geq \frac{1}{756}</math>.

'''Proof''' In page 13 of [ExIs2018], a graph of 33 vertices with some unit edges and 27 edges of length <math>\sqrt{5/3}</math> is contructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Now repeat the proof of Lemma 26. <math>\Box</math>

=== Computer-verified claim 28 ===

One has <math>p_{2/\sqrt{3}} \geq \frac{1}{177}</math>.

'''Proof''' In page 15 of [ExIs2018], a 5-chromatic graph of 103 vertices, 312 unit edges, and 177 edges of length <math>2/\sqrt{3}</math> is constructed. <math>\mathbf{c}</math> must make one of the latter edges monochromatic, giving the claim. <math>\Box</math>

=== Lemma 29 ===

One has <math>p_{(\sqrt{6} \pm \sqrt{2})/2} \geq 1/6</math> (this improves the bound in Lemma 23).

'''Proof''' Use graphs 505 and 507 from [S2004] and the spindle bound. <math>\Box</math>

=== Lemma 30 ===
For <math>m > n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' <math>n</math> colors and <math>m</math> points necessitates at least 2 having equal color. I.e.

:<math>{\bf P}\left(\bigvee_{k=0}^n \bigvee_{j=k+1}^n\ \mathbf{c}(z_k) = \mathbf{c}(z_j)\right) = 1</math>

The lemma then follows immediately from the fact:
:<math>{\bf P}\left(\bigcup_{k} E_k\right) \leq \sum_{k} {\bf P}\left(E_k\right) \,\Box</math>

=== Corollary 31 ===
Let <math>\lvert z_k\rvert=1</math>. Then for <math>m \geq n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' Use lemma 30 on the set <math>\left\{z_k \bigg\vert 1\leq k\leq m \land k\in\mathbb{Z}\right\}\cup\{0\}</math>. Simplify using <math>{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(0) \right)=0</math>.
<math>\Box</math>

=== Lemma 32 ===
For <math>x\in\mathbb{R}</math> and an <math>n</math>-coloring of the plane, <math>\sum_{k=1}^{n-1}\left(n-k\right){\bf P}\left(\mathbf{c}\left(0\right) = \mathbf{c}\left( 2\sin\left(\frac{kx}{2}\right) \right) \right) \geq 1</math>.

'''Proof''' Use corollary 31 on the set <math>\left\{e^{ikx} \bigg\vert 0\leq k < n \land k\in\mathbb{Z}\right\}</math>. and simplify by grouping lengths.
<math>\Box</math>

=== Corollary 33 ===
Interesting(easy to simplify results of) values for <math>x</math> in Lemma 32 are in <math>\left\{x \bigg\vert \sin\left(\frac{kx}{2}\right)=1 \land 1\leq k < n \land k\in\mathbb{Z}\right\}</math>.
For 4-colorings, this gives
:<math>2p_{\sqrt 3}+p_2 \geq 1</math>
:<math>3p_{(\sqrt 3-1)/\sqrt 2}+p_{\sqrt 2} \geq 1</math>
:<math>3p_{2\sin(\pi/18)}+2p_{2\sin(\pi/9)} \geq 1</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

== Further questions ==
* For <math>n,m\geq CNP</math>, what are consistent relationships between <math>{\bf P}\left(\mathbf{c}(0)=\mathbf{c}(d)\bigg\vert n\text{ colors}\right)</math> and <math>{\bf P}\left(\mathbf{c}(0)=\mathbf{c}(d)\bigg\vert m\text{ colors}\right)</math>? How can these relationships be used to sharpen arguments of the probabilistic formulation?

[[Category:Polymath16]]

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-20T18:07:23Z

Nazgand: /* Lemma 17 */ Corrected claim in case of k=1, typo

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> \geq 1/(n \sqrt{3})</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d
|
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/4
| <math>G_{21}</math>
| 1/2
| Spindle
| Lower bound computer verified
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|-
| <math>\frac{\sqrt{6} \pm \sqrt{2}}{2}</math>
| 1/6
| An arrangement of five vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{2}</math>
| 1/14
| A graph of 13 vertices
| 1/2
| Spindle
|
|-
| <math>\frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math>
| 1/28
| A graph of 13 vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{3/2 + \sqrt{33}/6}</math>
| 1/196
| A graph of 9 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>\sqrt{5/3}</math>
| 1/756
| A graph of 33 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>2/\sqrt{3}</math>
| 1/177
| A graph of 103 vertices
| 1/2
| Spindle
| Computer-verified
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>p_{\sqrt 3}/(1-p_2)</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

=== Lemma 1 ===
Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

=== Lemma 2 ===
(Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

=== Lemma 3 ===
(Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

=== Corollary 4 ===
(Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

=== Computer-verified Claim 5 ===
(Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

=== Computer-verified Claim 6 ===
(Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

=== Lemma 7 ===
(Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

=== Corollary 8 ===
One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

=== Lemma 9 ===
(Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

=== Corollary 10 ===
(Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

=== Computer-verified claim 11 ===
One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

=== Lemma 12 ===
For certain special distances d, one can improve the bound in Lemma 2:

If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, <math>\gcd(j,2k+1)=1</math>, and <math>r = \frac{1}{2} \csc\left(\frac{j\pi}{2k+1}\right)</math> then
:<math> p_r \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}}} \leq \frac{1}{3}.</math>

'''Proof''' Observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2}} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges. More generally, if <math>a,b,c,d,e</math> are the diagonal lengths of a pentagon with unit sides, then
:<math> 1 \leq p_a + p_b + p_c + p_d + p_e \leq 2.</math>

=== Lemma 13 ===
We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

=== Corollary 14 ===
We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

=== Lemma 15 ===
One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

=== Corollary 16 ===
We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

=== Lemma 17 ===
If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths a,b,c. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also, thus
:<math>{\bf P}(\mathbf{c}(0) \neq \mathbf{c}(a)) + {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(b)) \geq {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(c))</math>
and the claim follows. <math>\Box</math>

Note that Lemma 2 follows from the a=b, c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 1 - 2^{-k}</math> when <math>a \geq 2^{-k}, k\in\mathbb{Z}^+</math>.

Further, we can generalise the a=b case to one in which the triangle is replaced by a (k+1)-gon of which one edge is 1 and the others are all equal, leading to the stronger result <math>p_a \leq 1 - 1/k</math> when <math>a \geq 1/k, k\in\mathbb{Z}^+ \land k>1</math>. Further strengthening is achieved by using <math>1/\sqrt{3}</math> as the long edge, given Lemma 12.

=== Lemma 18 ===
Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio. Also we have
:<math> p_{d/\sqrt{3}} \leq \frac{1}{3} + p_d, \frac{1}{2} + \frac{1}{2} p_d.</math>

Note that this generalises (9), as well as a special case of Lemma 12.

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the first claim. Similarly, if one considers the colorings of an equilateral triangle of sidelength <math>d</math> together with its center, and counts the numbers <math>a,b \in \{0,1,2,3\}</math> of monochromatic edges of length <math>d</math> and <math>d/\sqrt{3}</math> respectively, one observes that one always has <math>\frac{b}{3} \leq \frac{1}{3} + \frac{2}{3} \frac{a}{3}, \frac{1}{2} + \frac{1}{2} \frac{a}{3}</math>, and on taking expectations one obtains the claim.<math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

=== Lemma 19 ===
(Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

=== Theorem 20 ===
One has <math>p_{H = 1tri} > 0</math>.

'''Proof''' Suppose for contradiction that <math>p_{H = 1tri} = 0</math>. One can then run a version of the de Bruijn-Erdos argument to obtain a coloring in which 1tri hexagons are completely nonexistent (since there are arbitrarily large finite colorings with this property). Consider the triangular lattice <math>{\bf Z}[e^{\pi i/3}]</math>. We 2-color the edges of this lattice by coloring an edge black if it is the short diagonal of a unit rhombus with monochromatic long diagonal, and white otherwise. The four colorings of hexagons lead to four possible colorings at each vertex:

* If H is colored 2tri, then all six edges to the centre of H are black.
* If H is colored 1tri, then two edges to the centre of H at 120 degree angles are white, the other four are black.
* If H is colored axisym, then two opposing edges of the centre of H are black, the other four are white.
* If H is colored centralsym, then all six edges to the centre of H are black.

In particular, as we are assuming no 1tri hexagons, the faces cut out by the black edges have angles 60 degrees, and thus must be equilateral triangles, sectors of angle 60, half-planes, or the entire plane. If there is at least one equilateral triangle, then the rest of the black edges must form an equilateral lattice with that triangle sidelength. This leads to only a small number of possible hexagon colorings in the lattice:

# Case 1: All edges white.
# Case 2: All edges black.
# Case 3.k: For some natural number <math>k \geq 2</math>, the length k edges joining adjacent vertices in some coset of <math>k \cdot {\mathbf Z}[ e^{\pi i/3} ]</math> are all black, and the remaining edges are white.
# Case 4: Each horizontal row consists either entirely of black edges, or entirely of white edges.
# Case 5: Each northwest row consists either entirely of black edges, or entirely of white edges.
# Case 6: Each northeast row consists either entirely of black edges, or entirely of white edges.
# Case 7: Six rays of black edges meeting at a common vertex; all other edges white.

Technically, Case 1 is contained in Cases 4,5,6 as written above, but this will not be an issue. One can view Case 7 as a limiting case <math>k=\infty</math> of Case 3.k; Case 2 is similarly the opposite limiting case <math>k=1</math>.

In the first case, the coloring is periodic with periods <math>2, 2 e^{\pi i/3}</math>. In the second case, it is periodic with periods <math>3, 3 e^{\pi i/3}</math>. In the third case, it is periodic with periods <math>3k, 3k e^{\pi i/3}</math>. Also note that for each k, one can check if Case 3.k holds by inspecting the coloring at a finite number of vertices. Thus the event that Case 3.k holds is "measurable" in the sense that a meaningful probability can be assigned. (But Cases 1,2,4,5,6 are not measurable events, they require an infinite number of points to be inspected, and the probability measure we are using is only finitely additive rather than infinitely additive.) In Case 4, the coloring is periodic with period 2; also, every coset of <math>2 \cdot {\bf Z}[e^{\pi i/3}]</math> is 2-colored. Similarly for Case 5 and 6 (where the periods are <math>2 e^{2\pi i/3}</math> and <math>2 e^{4\pi i/3}</math> respectively.)

Let <math>\alpha_k</math> be the probability that Case 3.k holds for the given value of <math>k</math>. Then <math> \sum_{k=2}^K \alpha_k \leq 1</math> for any k, hence <math>\sum_{k=2}^\infty \alpha_k \leq 1</math>. In particular, we can find <math>K_1</math> such that <math>\sum_{k={K_1}}^\infty \alpha_k \leq 0.1</math> (say). Let <math>P</math> be six times the least common multiple of <math>1,2,\dots,K_1</math>. Then the coloring is P- and <math>P e^{\pi i/3}</math>-periodic for Case 1, Case 2, and all Case 3.k with <math>k \leq K_1</math>. On the other hand, if <math>K_2</math> is sufficiently large depending on <math>P</math>, and Case 3.k holds for some <math>k \geq K_2</math>, then almost all of the hexagons are colored centralsym, which makes the coloring "almost <math>P, P e^{\pi i/3}</math>-periodic" in the sense that
:<math>{\bf c}(z+P e^{\pi i j/3}) = {\bf c}(z) \hbox{ for } j=0,1,2,3,4,5</math>
will hold for at least <math>0.9</math> of the lattice points <math>z \in {\bf Z}[e^{\pi i/3}]</math> with <math>|z| \leq K_2</math>. Similarly for Case 7 (which is sort of a <math>k=\infty</math> limiting case of Case 3.k.) Thus, with the probability <math> \geq 1 - \sum_{k=K_1}^{K_2} \alpha_k \geq 0.9</math>, the coloring of the seven vertices <math>{\bf c}(0), {\bf c}(P e^{\pi ij/3}, j=1,\dots,6</math> is (up to rotation and recoloring) one of the three patterns of the central and linking vertices in Figure 3 of Aubrey's paper, namely

* <math>{\bf c}(0) = {\bf c}(P) = {\bf c}(P e^{\pi i/3}) = {\bf c}(P e^{2\pi i/3}) = {\bf c}(P e^{3\pi i/3}) = {\bf c}(P e^{4\pi i/3}) = {\bf c}(P e^{5\pi i/3}) </math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{3\pi i/3})</math>; <math>{\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{3\pi i/3})</math>; <math> {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>.

Using the spindling argument from Aubrey's paper, we conclude that the third possibility must in fact hold with probability at least 0.8; on the other hand, from Lemma 2 this scenario can only occur with probability at most 1/2, giving the required contradiction.
<math>\Box</math>

One should be able to refine this argument to show that <math>p_{H = 1tri} > c</math> for an absolute constant <math> c>0 </math>.

=== Lemma 21 ===
Providing a tighter bound for Lemma 17 with a more thorough proof: If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

=== Lemma 22 ===

One has <math>3 p_{1/\sqrt{3}} \geq {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) )</math>.

'''Proof''' Let <math>z</math> be a complex number of magnitude <math>1/\sqrt{3}</math> that is a unit distance from 1. If <math>\mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) = c</math> (say), then <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> cannot be colored with <math>c</math>; also, <math>z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> are the vertices of a unit equilateral triangle and thus must take on three different colors. By the pigeonhole principle, one of <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> must then take the same color as the origin, and the claim follows. <math>\Box</math>

=== Lemma 23 ===

One has <math>4 p_{(\sqrt{6} \pm \sqrt{2})/2} + p_{\sqrt{2}} \geq 1</math>. In particular, by Lemma 2, <math>p_{(\sqrt{6}+\sqrt{2})/2} \geq 1/8</math>.

'''Proof''' [ExIs2018b] We just prove the claim for the + sign (the - sign can then be obtained after applying the Galois conjugacy that maps <math>\sqrt{3}</math> to <math>-\sqrt{3}</math>, leaving <math>\sqrt{2}</math> unchanged). Set <math>d := \frac{\sqrt{6}+\sqrt{2}}{2}</math>, and consider the five vertices

:<math>0, e^{5\pi i/4}, e^{5\pi i/4} + d, e^{5\pi i/4} + e^{\pi i/3} d, e^{5\pi i/4} + (e^{\pi i/3}-i)d.</math>

One can check that of the ten edges determined by these five vertices, five have unit length, four have length d, and the remaining distance (from 0 to <math>e^{5\pi i/4}+d</math>) has distance <math>\sqrt{2}</math>. Since <math>\mathbf{c}</math> must make one of the latter five edges monochromatic, the claim follows. <math>\Box</math>

=== Lemma 24 ===

One has <math>p_{\sqrt{2}} \geq \frac{1}{14}</math>.

'''Proof''' In page 7 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 20 unit distance edges and 14 edges of length <math>\sqrt{2}</math> is constructed. The coloring <math>\mathbf{c}</math> must make one of the 14 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 25 ===

Let <math>d = \frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math> and <math>e = \frac{3^{1/4} \sqrt{2} + \sqrt{3} - 1}{2}</math>.
Then one has <math>14 p_d + p_e \geq 1</math>. In particular, by Lemma 2, <math>p_d \geq 1/28</math>.

'''Proof''' In page 9 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 19 unit edges, 14 edges of length d, and one edge of length e is constructed. The coloring <math>\mathbf{c}</math> must make one of the 15 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 26 ===

Let <math>d = \sqrt{3/2 + \sqrt{33}/6}</math>. Then <math>7 p_d \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_d \geq \frac{1}{196}</math>.

'''Proof''' In page 11 of [ExIs2018b], a graph of nine vertices consisting of 12 unit edges and 7 edges of length d is constructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Thus, <math>\mathbf{c}</math> can only make the AB edge monochromatic if one of the seven length d edges is monochromatic. The claim follows.
<math>\Box</math>

=== Lemma 27 ===

One has <math>27 p_{\sqrt{5/3}} \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_{\sqrt{5/3}} \geq \frac{1}{756}</math>.

'''Proof''' In page 13 of [ExIs2018], a graph of 33 vertices with some unit edges and 27 edges of length <math>\sqrt{5/3}</math> is contructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Now repeat the proof of Lemma 26. <math>\Box</math>

=== Computer-verified claim 28 ===

One has <math>p_{2/\sqrt{3}} \geq \frac{1}{177}</math>.

'''Proof''' In page 15 of [ExIs2018], a 5-chromatic graph of 103 vertices, 312 unit edges, and 177 edges of length <math>2/\sqrt{3}</math> is constructed. <math>\mathbf{c}</math> must make one of the latter edges monochromatic, giving the claim. <math>\Box</math>

=== Lemma 29 ===

One has <math>p_{(\sqrt{6} \pm \sqrt{2})/2} \geq 1/6</math> (this improves the bound in Lemma 23).

'''Proof''' Use graphs 505 and 507 from [S2004] and the spindle bound. <math>\Box</math>

=== Lemma 30 ===
For <math>m > n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' <math>n</math> colors and <math>m</math> points necessitates at least 2 having equal color. I.e.

:<math>{\bf P}\left(\bigvee_{k=0}^n \bigvee_{j=k+1}^n\ \mathbf{c}(z_k) = \mathbf{c}(z_j)\right) = 1</math>

The lemma then follows immediately from the fact:
:<math>{\bf P}\left(\bigcup_{k} E_k\right) \leq \sum_{k} {\bf P}\left(E_k\right) \,\Box</math>

=== Corollary 31 ===
Let <math>\lvert z_k\rvert=1</math>. Then for <math>m \geq n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' Use lemma 30 on the set <math>\left\{z_k \bigg\vert 1\leq k\leq m \land k\in\mathbb{Z}\right\}\cup\{0\}</math>. Simplify using <math>{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(0) \right)=0</math>.
<math>\Box</math>

=== Lemma 32 ===
For <math>x\in\mathbb{R}</math> and an <math>n</math>-coloring of the plane, <math>\sum_{k=1}^{n-1}\left(n-k\right){\bf P}\left(\mathbf{c}\left(0\right) = \mathbf{c}\left( 2\sin\left(\frac{kx}{2}\right) \right) \right) \geq 1</math>.

'''Proof''' Use corollary 31 on the set <math>\left\{e^{ikx} \bigg\vert 0\leq k < n \land k\in\mathbb{Z}\right\}</math>. and simplify by grouping lengths.
<math>\Box</math>

=== Corollary 33 ===
Interesting(easy to simplify results of) values for <math>x</math> in Lemma 32 are in <math>\left\{x \bigg\vert \sin\left(\frac{kx}{2}\right)=1 \land 1\leq k < n \land k\in\mathbb{Z}\right\}</math>.
For 4-colorings, this gives
:<math>2p_{\sqrt 3}+p_2 \geq 1</math>
:<math>3p_{(\sqrt 3-1)/\sqrt 2}+p_{\sqrt 2} \geq 1</math>
:<math>3p_{2\sin(\pi/18)}+2p_{2\sin(\pi/9)} \geq 1</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

[[Category:Polymath16]]

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-20T17:30:34Z

Nazgand: More Lemmas

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> \geq 1/(n \sqrt{3})</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d
|
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/4
| <math>G_{21}</math>
| 1/2
| Spindle
| Lower bound computer verified
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|-
| <math>\frac{\sqrt{6} \pm \sqrt{2}}{2}</math>
| 1/6
| An arrangement of five vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{2}</math>
| 1/14
| A graph of 13 vertices
| 1/2
| Spindle
|
|-
| <math>\frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math>
| 1/28
| A graph of 13 vertices; Lemma 2
| 1/2
| Spindle
|
|-
| <math>\sqrt{3/2 + \sqrt{33}/6}</math>
| 1/196
| A graph of 9 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>\sqrt{5/3}</math>
| 1/756
| A graph of 33 vertices; Corollary 16
| 1/2
| Spindle
|
|-
| <math>2/\sqrt{3}</math>
| 1/177
| A graph of 103 vertices
| 1/2
| Spindle
| Computer-verified
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>p_{\sqrt 3}/(1-p_2)</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

=== Lemma 1 ===
Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

=== Lemma 2 ===
(Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

=== Lemma 3 ===
(Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

=== Corollary 4 ===
(Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

=== Computer-verified Claim 5 ===
(Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

=== Computer-verified Claim 6 ===
(Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

=== Lemma 7 ===
(Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

=== Corollary 8 ===
One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

=== Lemma 9 ===
(Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

=== Corollary 10 ===
(Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

=== Computer-verified claim 11 ===
One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

=== Lemma 12 ===
For certain special distances d, one can improve the bound in Lemma 2:

If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, <math>\gcd(j,2k+1)=1</math>, and <math>r = \frac{1}{2} \csc\left(\frac{j\pi}{2k+1}\right)</math> then
:<math> p_r \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}}} \leq \frac{1}{3}.</math>

'''Proof''' Observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2}} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges. More generally, if <math>a,b,c,d,e</math> are the diagonal lengths of a pentagon with unit sides, then
:<math> 1 \leq p_a + p_b + p_c + p_d + p_e \leq 2.</math>

=== Lemma 13 ===
We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

=== Corollary 14 ===
We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

=== Lemma 15 ===
One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

=== Corollary 16 ===
We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

=== Lemma 17 ===
If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths a,b,c. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also, thus
:<math>{\bf P}(\mathbf{c}(0) \neq \mathbf{c}(a)) + {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(b)) \geq {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(c))</math>
and the claim follows. <math>\Box</math>

Note that Lemma 2 follows from the a=b, c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 1 - 2^{-k}</math> when <math>a \geq 2^{-k}, k\in\mathbb{Z}^+</math>. Further, we can generalise the a=b case to one in which the triangle is replaced by a (k+1)-gon of which one edge is 1 and the others are all equal, leading to the stronger result <math>p_a \leq 1 - 1/k</math> when <math>a \geq 1/k, k\in\mathbb{Z}^+</math>. Further strenthening is achieved by using <math>1/\sqrt{3}</math> as the long edge, given Lemma 12.

=== Lemma 18 ===
Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio. Also we have
:<math> p_{d/\sqrt{3}} \leq \frac{1}{3} + p_d, \frac{1}{2} + \frac{1}{2} p_d.</math>

Note that this generalises (9), as well as a special case of Lemma 12.

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the first claim. Similarly, if one considers the colorings of an equilateral triangle of sidelength <math>d</math> together with its center, and counts the numbers <math>a,b \in \{0,1,2,3\}</math> of monochromatic edges of length <math>d</math> and <math>d/\sqrt{3}</math> respectively, one observes that one always has <math>\frac{b}{3} \leq \frac{1}{3} + \frac{2}{3} \frac{a}{3}, \frac{1}{2} + \frac{1}{2} \frac{a}{3}</math>, and on taking expectations one obtains the claim.<math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

=== Lemma 19 ===
(Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

=== Theorem 20 ===
One has <math>p_{H = 1tri} > 0</math>.

'''Proof''' Suppose for contradiction that <math>p_{H = 1tri} = 0</math>. One can then run a version of the de Bruijn-Erdos argument to obtain a coloring in which 1tri hexagons are completely nonexistent (since there are arbitrarily large finite colorings with this property). Consider the triangular lattice <math>{\bf Z}[e^{\pi i/3}]</math>. We 2-color the edges of this lattice by coloring an edge black if it is the short diagonal of a unit rhombus with monochromatic long diagonal, and white otherwise. The four colorings of hexagons lead to four possible colorings at each vertex:

* If H is colored 2tri, then all six edges to the centre of H are black.
* If H is colored 1tri, then two edges to the centre of H at 120 degree angles are white, the other four are black.
* If H is colored axisym, then two opposing edges of the centre of H are black, the other four are white.
* If H is colored centralsym, then all six edges to the centre of H are black.

In particular, as we are assuming no 1tri hexagons, the faces cut out by the black edges have angles 60 degrees, and thus must be equilateral triangles, sectors of angle 60, half-planes, or the entire plane. If there is at least one equilateral triangle, then the rest of the black edges must form an equilateral lattice with that triangle sidelength. This leads to only a small number of possible hexagon colorings in the lattice:

# Case 1: All edges white.
# Case 2: All edges black.
# Case 3.k: For some natural number <math>k \geq 2</math>, the length k edges joining adjacent vertices in some coset of <math>k \cdot {\mathbf Z}[ e^{\pi i/3} ]</math> are all black, and the remaining edges are white.
# Case 4: Each horizontal row consists either entirely of black edges, or entirely of white edges.
# Case 5: Each northwest row consists either entirely of black edges, or entirely of white edges.
# Case 6: Each northeast row consists either entirely of black edges, or entirely of white edges.
# Case 7: Six rays of black edges meeting at a common vertex; all other edges white.

Technically, Case 1 is contained in Cases 4,5,6 as written above, but this will not be an issue. One can view Case 7 as a limiting case <math>k=\infty</math> of Case 3.k; Case 2 is similarly the opposite limiting case <math>k=1</math>.

In the first case, the coloring is periodic with periods <math>2, 2 e^{\pi i/3}</math>. In the second case, it is periodic with periods <math>3, 3 e^{\pi i/3}</math>. In the third case, it is periodic with periods <math>3k, 3k e^{\pi i/3}</math>. Also note that for each k, one can check if Case 3.k holds by inspecting the coloring at a finite number of vertices. Thus the event that Case 3.k holds is "measurable" in the sense that a meaningful probability can be assigned. (But Cases 1,2,4,5,6 are not measurable events, they require an infinite number of points to be inspected, and the probability measure we are using is only finitely additive rather than infinitely additive.) In Case 4, the coloring is periodic with period 2; also, every coset of <math>2 \cdot {\bf Z}[e^{\pi i/3}]</math> is 2-colored. Similarly for Case 5 and 6 (where the periods are <math>2 e^{2\pi i/3}</math> and <math>2 e^{4\pi i/3}</math> respectively.)

Let <math>\alpha_k</math> be the probability that Case 3.k holds for the given value of <math>k</math>. Then <math> \sum_{k=2}^K \alpha_k \leq 1</math> for any k, hence <math>\sum_{k=2}^\infty \alpha_k \leq 1</math>. In particular, we can find <math>K_1</math> such that <math>\sum_{k={K_1}}^\infty \alpha_k \leq 0.1</math> (say). Let <math>P</math> be six times the least common multiple of <math>1,2,\dots,K_1</math>. Then the coloring is P- and <math>P e^{\pi i/3}</math>-periodic for Case 1, Case 2, and all Case 3.k with <math>k \leq K_1</math>. On the other hand, if <math>K_2</math> is sufficiently large depending on <math>P</math>, and Case 3.k holds for some <math>k \geq K_2</math>, then almost all of the hexagons are colored centralsym, which makes the coloring "almost <math>P, P e^{\pi i/3}</math>-periodic" in the sense that
:<math>{\bf c}(z+P e^{\pi i j/3}) = {\bf c}(z) \hbox{ for } j=0,1,2,3,4,5</math>
will hold for at least <math>0.9</math> of the lattice points <math>z \in {\bf Z}[e^{\pi i/3}]</math> with <math>|z| \leq K_2</math>. Similarly for Case 7 (which is sort of a <math>k=\infty</math> limiting case of Case 3.k.) Thus, with the probability <math> \geq 1 - \sum_{k=K_1}^{K_2} \alpha_k \geq 0.9</math>, the coloring of the seven vertices <math>{\bf c}(0), {\bf c}(P e^{\pi ij/3}, j=1,\dots,6</math> is (up to rotation and recoloring) one of the three patterns of the central and linking vertices in Figure 3 of Aubrey's paper, namely

* <math>{\bf c}(0) = {\bf c}(P) = {\bf c}(P e^{\pi i/3}) = {\bf c}(P e^{2\pi i/3}) = {\bf c}(P e^{3\pi i/3}) = {\bf c}(P e^{4\pi i/3}) = {\bf c}(P e^{5\pi i/3}) </math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{3\pi i/3})</math>; <math>{\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{3\pi i/3})</math>; <math> {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>.

Using the spindling argument from Aubrey's paper, we conclude that the third possibility must in fact hold with probability at least 0.8; on the other hand, from Lemma 2 this scenario can only occur with probability at most 1/2, giving the required contradiction.
<math>\Box</math>

One should be able to refine this argument to show that <math>p_{H = 1tri} > c</math> for an absolute constant <math> c>0 </math>.

=== Lemma 21 ===
Providing a tighter bound for Lemma 17 with a more thorough proof: If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

=== Lemma 22 ===

One has <math>3 p_{1/\sqrt{3}} \geq {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) )</math>.

'''Proof''' Let <math>z</math> be a complex number of magnitude <math>1/\sqrt{3}</math> that is a unit distance from 1. If <math>\mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) = c</math> (say), then <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> cannot be colored with <math>c</math>; also, <math>z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> are the vertices of a unit equilateral triangle and thus must take on three different colors. By the pigeonhole principle, one of <math>0, z, e^{2\pi i/3} z, e^{4\pi i/3} z</math> must then take the same color as the origin, and the claim follows. <math>\Box</math>

=== Lemma 23 ===

One has <math>4 p_{(\sqrt{6} \pm \sqrt{2})/2} + p_{\sqrt{2}} \geq 1</math>. In particular, by Lemma 2, <math>p_{(\sqrt{6}+\sqrt{2})/2} \geq 1/8</math>.

'''Proof''' [ExIs2018b] We just prove the claim for the + sign (the - sign can then be obtained after applying the Galois conjugacy that maps <math>\sqrt{3}</math> to <math>-\sqrt{3}</math>, leaving <math>\sqrt{2}</math> unchanged). Set <math>d := \frac{\sqrt{6}+\sqrt{2}}{2}</math>, and consider the five vertices

:<math>0, e^{5\pi i/4}, e^{5\pi i/4} + d, e^{5\pi i/4} + e^{\pi i/3} d, e^{5\pi i/4} + (e^{\pi i/3}-i)d.</math>

One can check that of the ten edges determined by these five vertices, five have unit length, four have length d, and the remaining distance (from 0 to <math>e^{5\pi i/4}+d</math>) has distance <math>\sqrt{2}</math>. Since <math>\mathbf{c}</math> must make one of the latter five edges monochromatic, the claim follows. <math>\Box</math>

=== Lemma 24 ===

One has <math>p_{\sqrt{2}} \geq \frac{1}{14}</math>.

'''Proof''' In page 7 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 20 unit distance edges and 14 edges of length <math>\sqrt{2}</math> is constructed. The coloring <math>\mathbf{c}</math> must make one of the 14 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 25 ===

Let <math>d = \frac{1}{2} \sqrt{3^{1/4} \cdot 2 \sqrt{2} + 2 \sqrt{3} + 2}</math> and <math>e = \frac{3^{1/4} \sqrt{2} + \sqrt{3} - 1}{2}</math>.
Then one has <math>14 p_d + p_e \geq 1</math>. In particular, by Lemma 2, <math>p_d \geq 1/28</math>.

'''Proof''' In page 9 of [ExIs2018b], a non-4-colorable graph of 13 vertices with 19 unit edges, 14 edges of length d, and one edge of length e is constructed. The coloring <math>\mathbf{c}</math> must make one of the 15 latter edges monochromatic, and the claim follows. <math>\Box</math>

=== Lemma 26 ===

Let <math>d = \sqrt{3/2 + \sqrt{33}/6}</math>. Then <math>7 p_d \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_d \geq \frac{1}{196}</math>.

'''Proof''' In page 11 of [ExIs2018b], a graph of nine vertices consisting of 12 unit edges and 7 edges of length d is constructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Thus, <math>\mathbf{c}</math> can only make the AB edge monochromatic if one of the seven length d edges is monochromatic. The claim follows.
<math>\Box</math>

=== Lemma 27 ===

One has <math>27 p_{\sqrt{5/3}} \geq p_{1/\sqrt{3}}</math>. In particular, by Corollary 16, <math>p_{\sqrt{5/3}} \geq \frac{1}{756}</math>.

'''Proof''' In page 13 of [ExIs2018], a graph of 33 vertices with some unit edges and 27 edges of length <math>\sqrt{5/3}</math> is contructed with the property that any 4-coloring of this graph cannot have two specific vertices A,B (which are distance <math>1/\sqrt{3}</math> apart) monochromatic. Now repeat the proof of Lemma 26. <math>\Box</math>

=== Computer-verified claim 28 ===

One has <math>p_{2/\sqrt{3}} \geq \frac{1}{177}</math>.

'''Proof''' In page 15 of [ExIs2018], a 5-chromatic graph of 103 vertices, 312 unit edges, and 177 edges of length <math>2/\sqrt{3}</math> is constructed. <math>\mathbf{c}</math> must make one of the latter edges monochromatic, giving the claim. <math>\Box</math>

=== Lemma 29 ===

One has <math>p_{(\sqrt{6} \pm \sqrt{2})/2} \geq 1/6</math> (this improves the bound in Lemma 23).

'''Proof''' Use graphs 505 and 507 from [S2004] and the spindle bound. <math>\Box</math>

=== Lemma 30 ===
For <math>m > n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' <math>n</math> colors and <math>m</math> points necessitates at least 2 having equal color. I.e.

:<math>{\bf P}\left(\bigvee_{k=0}^n \bigvee_{j=k+1}^n\ \mathbf{c}(z_k) = \mathbf{c}(z_j)\right) = 1</math>

The lemma then follows immediately from the fact:
:<math>{\bf P}\left(\bigcup_{k} E_k\right) \leq \sum_{k} {\bf P}\left(E_k\right) \,\Box</math>

=== Corollary 31 ===
Let <math>\lvert z_k\rvert=1</math>. Then for <math>m \geq n</math> and an <math>n</math>-coloring of the complex plane, <math>\sum_{k=1}^m\sum_{j=k+1}^m{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(z_j) \right) \geq 1</math>.

'''Proof''' Use lemma 30 on the set <math>\left\{z_k \bigg\vert 1\leq k\leq m \land k\in\mathbb{Z}\right\}\cup\{0\}</math>. Simplify using <math>{\bf P}\left(\mathbf{c}(z_k) = \mathbf{c}(0) \right)=0</math>.
<math>\Box</math>

=== Lemma 32 ===
For <math>x\in\mathbb{R}</math> and an <math>n</math>-coloring of the plane, <math>\sum_{k=1}^{n-1}\left(n-k\right){\bf P}\left(\mathbf{c}\left(0\right) = \mathbf{c}\left( 2\sin\left(\frac{kx}{2}\right) \right) \right) \geq 1</math>.

'''Proof''' Use corollary 31 on the set <math>\left\{e^{ikx} \bigg\vert 0\leq k < n \land k\in\mathbb{Z}\right\}</math>. and simplify by grouping lengths.
<math>\Box</math>

=== Corollary 33 ===
Interesting(easy to simplify results of) values for <math>x</math> in Lemma 32 are in <math>\left\{x \bigg\vert \sin\left(\frac{kx}{2}\right)=1 \land 1\leq k < n \land k\in\mathbb{Z}\right\}</math>.
For 4-colorings, this gives
:<math>2p_{\sqrt 3}+p_2 \geq 1</math>
:<math>3p_{(\sqrt 3-1)/\sqrt 2}+p_{\sqrt 2} \geq 1</math>
:<math>3p_{2\sin(\pi/18)}+2p_{2\sin(\pi/9)} \geq 1</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

[[Category:Polymath16]]

Hadwiger-Nelson problem

2018-05-17T01:04:31Z

Nazgand: /* Further questions */

The Chromatic Number of the Plane (CNP) is the chromatic number of the graph whose vertices are elements of the plane, and two points are connected by an edge if they are a unit distance apart. The Hadwiger-Nelson problem asks to compute CNP. The bounds <math>4 \leq CNP \leq 7</math> are classical; recently [deG2018] it was shown that <math>CNP \geq 5</math>. This is achieved by explicitly locating finite unit distance graphs with chromatic number at least 5.

The '''Polymath16''' project seeks to simplify the graphs used in [deG2018] to establish this lower bound. More precisely, the goals are

* '''Goal 1''': Find progressively smaller 5-chromatic unit-distance graphs.
* '''Goal 2''': Reduce (ideally to zero) the reliance on computer assistance for the proof. Computer assistance was leveraged in [deG2018] to analyze a subgraph of size 397.
* '''Goal 3''': Apply these simpler graphs to inform progress in related areas. For example:
** Find a 6-chromatic unit-distance graph in the plane.
** Improve the corresponding bound in higher dimensions.
** Improve the current record of 383/102 for the fractional chromatic number of the plane.
** Find the smallest unit-distance graph of a given minimum degree (excluding, in some natural way, boring cases like Cartesian products of a graph with a hypercube).

== Polymath threads ==

* [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/ Polymath proposal: finding simpler unit distance graphs of chromatic number 5], Aubrey de Grey, Apr 10 2018. ('''Active discussion thread''')
* [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/ Polymath16, first thread: Simplifying de Grey’s graph], Dustin Mixon, Apr 14, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic Polymath16, second thread: What does it take to be 5-chromatic?], Dustin Mixon, Apr 22, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/ Polymath16, third thread: Is 6-chromatic within reach?], Dustin Mixon, May 1, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/ Polymath16, fourth thread: Applying the probabilistic method], Dustin Mixon, May 5, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/10/polymath16-fifth-thread-human-verifiable-proofs/ Polymath16, fifth thread: Human-verifiable proofs], Dustin Mixon, May 10, 2018. ('''Active research thread''')

== Notable unit distance graphs ==

A '''unit distance graph''' is a graph that can be realised as a collection of vertices in the plane, with two vertices connected by an edge if they are precisely a unit distance apart. The chromatic number of any such graph is a lower bound for <math>CNP</math>; in particular, if one can find a unit distance graph with no 4-colorings, then <math>CNP \geq 5</math>. The boldface number of vertices is the current minimal number of vertices of a unit distance graph that is currently known to not be 4-colorable.

<math>G_1 \oplus G_2</math> denotes the Minkowski sum of two unit distance graphs <math>G_1,G_2</math> (vertices in <math>G_1 \oplus G_2</math> are sums of the vertices of <math>G_1,G_2</math>). <math>G_1 \cup G_2</math> denotes the union. <math>\mathrm{rot}(G, \theta)</math> denotes <math>G</math> rotated counterclockwise by <math>\theta</math>. <math>\mathrm{trim}(G,r)</math> denotes the trimming of <math>G</math> after removing all vertices of distance greater than <math>r</math> from the origin.

Another basic operation is '''spindling''': taking two copies of a graph <math>G</math>, gluing them together at one vertex, and rotating the copies so that the two copies of another vertex are a unit distance apart. For instance, the Moser spindle is the spindling of a rhombus graph. If a graph has two vertices forced to be the same color in a k-coloring, then the spindling of that graph at those two vertices is not k-colorable.

Many of the graphs are embeddable in abelian groups or rings, particularly those generated by the complex numbers of unit modulus <math>\omega_t := \exp( i \arccos( 1 - \tfrac{1}{2t} ))</math> for various natural numbers <math>t</math>, particularly the [https://oeis.org/A003136 Loeschian numbers] <math>1,3,4,7,9,12,\dots</math>. These numbers arise naturally as the apex angle of a <math>\sqrt{t}, \sqrt{t}, 1</math> isosceles triangle, and the distances <math>\sqrt{t}</math> are the distances that arise in the triangular lattice. The rings <math>R_n = {\bf Z}[ \omega_{t_1}, \dots, \omega_{t_n}]</math>, where <math>t_1,t_2,\dots</math> are the Loeschian numbers, seem particularly relevant, thus <math>R_0 = {\bf Z}, R_1 = {\bf Z}[\omega_1], R_2 = {\bf Z}[\omega_1, \omega_3]</math>, etc.. Closely related rings are the rings <math>\overline{R_n}</math> generated by the unit vectors in <math>R_n</math> and their inverses.

Note that the square root <math>\eta = \exp( i \frac{1}{2} \arccos \frac{5}{6} )</math> of <math>\omega_3</math> lies in <math>R_2</math>, thanks to the identity
:<math> \eta = (\omega_1^4 + \omega_1^5) (\omega_3 - 1).</math>

{| border=1
|-
!Name!!Number of vertices!! Number of edges !! Structure !! Group !! Colorings
|-
| [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]
| 7
| 11
| Two 60-120-60-120 rhombi with a common vertex, with one pair of sharp vertices coincident and the other joined
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [http://mathworld.wolfram.com/GolombGraph.html Golomb graph]
| 10
| 18
| Contains the center and vertices of a hexagon and equilateral triangle
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [https://arxiv.org/abs/1804.02385 H]
| 7
| 12
| Vertices and center of a hexagon
| <math>{\bf Z}[\omega_1]</math>
| Has essentially four 4-colorings, two of which contain a monochromatic <math>\sqrt{3}</math>-triangle. Every 5-coloring has a monochrome <math>\sqrt{3}</math>-edge or a monochrome <math>2</math>-edge
|-
| [https://arxiv.org/abs/1804.02385 J]
| 31
| 72
| Contains 13 copies of H
| <math>{\bf Z}[\omega_1]</math>
| Has essentially six 4-colorings in which no H has a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 K]
| 61
| 150
| Contains 2 copies of J
|
| In all 4-colorings lacking an H with a monochromatic <math>\sqrt{3}</math>-triangle, all pairs of vertices at distance 4 are monochromatic
|-
| [https://arxiv.org/abs/1804.02385 L]
| 121
| 301
| Contains two copies of K and 52 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ <math>L_1</math>]
| 97
|
| Has 40 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120295 <math>L_2</math>]
| 120
| 354
|
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 T]
| 9
| 15
| Contains one Moser spindle and useful symmetry; three vertices form an equilateral triangle
|
|
|-
| [https://arxiv.org/abs/1804.02385 U]
| 15
| 33
| Three copies of T at 120-degree rotations: <math>T \cup \mathrm{rot}(T, 2\pi/3) \cup \mathrm{rot}(T, 4\pi/3)</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 V]
| 31
| 30
| <math>\{0\} \cup \{ \omega_1^x \omega_3^{y/2}: x=0,\dots,5; y=0,\dots,4\}</math>
| <math>{\bf Z}[\omega_1,\omega_3]</math>
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>V_1</math>]
| 61
| 60
| Union of V and a rotation of V: <math>V \cup \mathrm{rot}(V, \mathrm{arccos}(7/8))</math>
| <math>{\bf Z}[\omega_1,\omega_3, \omega_4]</math>
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_b</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_x</math>]
| 13
| 12
| Subgraph of <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_z</math>]
|
|
| Subgraph of <math>V_a</math>; shares a line of symmetry with <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_y</math>]
| 13
| 12
| Subgraph of <math>V_b</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-3970 <math>V_A</math>]
| 37
| 36
| Unit vectors with angles <math>i \frac{\pi}{3} + j \mathrm{arccos} \frac{5}{6} + k \mathrm{arccos} \frac{7}{8}</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 W]
| 301
| 1230
| Cartesian product of V with itself, minus vertices at more than <math>\sqrt{3}</math> from the centre (i.e. <math>\mathrm{trim}(V \oplus V, \sqrt{3})</math>)
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>W_1</math>]
|
|
| Trimmed product of V with itself (<math>\mathrm{trim}(V \oplus V, 1.95)</math>)
|
|
|-
| [https://arxiv.org/abs/1804.02385 M]
| 1345
| 8268
| Cartesian product of W and H (<math>W \oplus H</math>)
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| No 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120274 <math>M_1</math>]
| 278
|
| Deleting vertices from M while maintaining its restriction on H
|
| No 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>M_2</math>]
| 7075
|
| Sum of H with a trimmed product of <math>V_1</math> with itself
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 N]
| 20425
| 151311
| Contains 52 copies of M arranged around the H-copies of L
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4, \omega_{16}]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_0</math>]
| 1585
| 7909
| N "shrunk" by stepwise deletions and replacements of vertices
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 G]
| 1581
| 7877
| Deleting 4 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_1</math>]
| 1577
|
| Deleting 8 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120318 <math>G_2</math>]
| 874
| 4461
| Juxtaposing two copies of M and shrinking
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3867 <math>G_3</math>]
| 826
| 4273
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4105 <math>G_4</math>]
| 803
| 4144
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4316 <math>G_5</math>]
| 633
| 3166
| Subgraph of two copies of <math>V \oplus V \oplus V</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/05/10/polymath16-fifth-thread-human-verifiable-proofs/#comment-4465 <math>G_6</math>]
| '''610'''
| 3000
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 R]
|
|
| Union of <math>W_1</math> and a rotated copy of <math>W_1</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>\mathrm{trim}(R \oplus H, 1.67)</math>]
| 2563
|
| Trimmed sum of R and H
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 <math>V \oplus V \oplus H</math>]
|
|
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_{64/9}]</math>
| Has two vertices forced to be the same color in a 4-coloring; also no monochromatic <math>\sqrt{3}</math>-triangles
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 <math>G_{745}</math>]
| 745
|
| Subgraph of <math>V \oplus V \oplus H</math>
|
| Has two vertices forced to be the same color in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3085
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_a \oplus V_z \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3049
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>G_{1951}</math>]
| 1951
|
| Trimmed version of <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic <math>V_A \oplus V_A \oplus V_A</math>]
| 6937
| 44439
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{40}</math>]
| 40
| 82
|
|
| Any 4-coloring that avoids a monochromatic <math>\sqrt{11/3}</math>-edge has two specific vertices forced to be the same color
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{79}</math>]
| 79
| 165
| Spindling of <math>G_{40}</math>
|
| Any 4-coloring has a monochromatic <math>\sqrt{11/3}</math>-edge
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{49}</math>]
| 49
| 180
|
| <math>{\mathbf Q}[\sqrt{3},\sqrt{11}]^2</math>
| Any 4-coloring either has a specific <math>\sqrt{11/3}</math>-edge monochromatic, or a monochromatic <math>1/\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{51}</math>]
| 51
|
|
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring of plane
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{627}</math>]
| 627
| 2982
| Contains <math>G_{51}</math>
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4161 <math>G_{103}</math>]
| 103
|
|
|
| All 4-colorings contain a monochromatic <math>2/\sqrt{3}</math>-edge
|-
| regular pentagon with unit side length
| 5
| 5
|
|
| All 4-colorings contain a monochromatic <math>(\sqrt{5}+1)/2</math>-edge
|-
| regular pentagon with <math>(\sqrt{5}-1)/2</math> side length
| 5
| 5
|
|
| All 4-colorings contain a monochromatic <math>(\sqrt{5}-1)/2</math>-edge
|-
| [http://www.hansparshall.com/txt/n7superCoordinates.txt <math>G_{21}</math>]
| 21
| 49
|
| <math>{\bf Z}[\omega_1,\omega_3]</math>
| Implies <math>p_2 \geq 1/4</math>
|-
| [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4391 <math>G_{43}</math>]
| 43
|
| <math>\{0,1,\eta,\overline{\eta}, \eta -\overline{\eta}, \eta - \overline{\eta}\omega_1, 1+\eta \omega_1^2, 1+\overline{\eta\omega_1^2}\} \cdot \langle \omega_1 \rangle</math>
| <math>{\bf Z}[\omega_1,\omega_3]</math>
| Origin cannot be bichromatic
|-
| [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4398 <math>G_{24}</math>]
| 24
|
|
|
| Origin cannot be bichromatic
|-
| [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4409 <math>G_{34}</math>]
| 34
|
|
| Origin cannot be bichromatic
|-
| [https://dustingmixon.wordpress.com/2018/05/10/polymath16-fifth-thread-human-verifiable-proofs/#comment-4420 <math>G_{30}</math>]
| 30
|
|
| Origin cannot be bichromatic
|}

== Lower bounds ==

In [P1988], Pritikin proved that every graph with at most 12 vertices is 4-colorable, and every graph with at most 6197 vertices is 6-colorable. Pritikin's bounds are obtained by coloring the plane with k colors and an additional “wild” color such that points of unit distance are both allowed to receive the wild color. If the wild color occupies a small fraction p of the plane, then an exercise in the probabilistic method gives that any fixed unit-distance graph with n vertices enjoys an embedding in the plane that avoids the wild color (and is therefore k-colorable) provided n<1/p. Pritikin’s coloring for the k=4 case cannot be improved without improving on the densest known subset of the plane that avoids unit distances (originally due to Croft in 1967). See [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable this MO thread] for additional information. As such, improving the bound in the k=4 case might require a new technique, whereas the k=5 and 6 cases might be amenable to optimization.

[https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4176 Every unit distance graph with at most 16 vertices is 5-colorable].

A "tile-based" colouring of the plane must have at least 6 colours, as shown by Townsend [Tow2005]; the same proof with a minor error was also derived by Woodall [W1973]. In [T1999], Thomassen showed that any tiling-based 6-coloring would have to be be "unscaleable", i.e. the maximum diameter of a tile and the minimum separation of same-coloured tiles must both be exactly 1 (so that the distance 1 is excluded by suitable colouring of the boundary points).

== Virtual edge ==

''Warning: other definitions have been proposed and the exact definition of this notion is currently under discussion. The clamping of graph G in this section may be interpreted as the devirtualization of all bichromatic virtual edges with length <math>d</math> and chromatic number <math>4</math>.''

A '''virtual edge''' of a unit distance graph <math>G</math> is a distance <math>d</math> with the property that every 4-coloring of <math>G</math> contains a monochromatic pair of vertices of distance exactly <math>d</math>. Observe that if a unit distance graph <math>G</math> has a virtual edge at distance <math>d</math>, and if there is another unit distance graph <math>H</math> with a pair of vertices at distance <math>d</math> that cannot be monochromatic in a 4-coloring, then we can create a non-4-colorable unit distance graph by "clamping" a copy of <math>H</math> to every virtual edge in <math>G</math>.

Known examples of virtual edges include:

* <math>G_{40}</math> has a virtual edge at distance <math>\sqrt{11/3}</math>.
* <math>V \oplus V \oplus H</math> has a (single) virtual edge at distance <math>8/3</math>.
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4117 <math>(\sqrt{3}\pm 1)/\sqrt{2}</math> are virtual edges of some graphs].

== Bichromatic virtual edge ==

===Definition===
If a unit distance graph <math>H</math> exists with a specific pair of vertices which are distance <math>d</math> apart and is bichromatic in all proper <math>n</math>-colorings of <math>H</math>, then we say that <math>H</math> virtualizes a '''bichromatic virtual edge''' with length <math>d</math> and chromatic number <math>n</math>.

===Vacuous properties===
Considering the graph of the entire plane, all distances correspond to a virtual edge of chromatic number <math>CNP-1</math>.

If a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n</math> exists, then a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n-1</math> exists.

===Convenience and devirtualization===
Bichromatic virtual edges behave the same as the unit edge(which is a bichromatic virtual edge of every chromatic number), and thus may be used to simplify the construction of graphs.

Recursively replacing bichromatic virtual edges of a graph <math>G</math> with graphs which virtualize the bichromatic virtual edges through '''clamping''' produces a unit distance graph <math>G'</math> where <math>\chi(G')\geq\chi(G)</math>.

Devirtualizing bichromatic virtual edges of a graph <math>G</math> (i.e. clamping) has no benefit unless nontrivial points of <math>G</math> and <math>H_0</math> overlap or nontrivial points of <math>H_1</math> and <math>H_0</math> overlap, where <math>H_0</math> and <math>H_1</math> are graphs virtualizing bichromatic virtual edges of <math>G</math>. Such overlaps may cause <math>\chi(G')>\chi(G)</math>. If no nontrivial overlaps exist, then <math>\chi(G')=\max\left(\chi(G),\chi(H_0),\chi(H_1),\dots\right)</math>.

Because more than 1 graph virtualizes any given bichromatic virtual edge for a fixed length and fixed chromatic number, multiple devirtualizations exist for every finite graph.

===Relation to rings===
A '''devirtualized ring''' of graph <math>G</math> is a ring which contains all the vertices of a devirtualization of <math>G</math>, where the vertices are interpreted as complex numbers.

Use the implied shorthand for arbitrary sets <math>\mathbb{S}[\left\{a,b,c,\dots\right\}]=\mathbb{S}[a,b,c,\dots]</math>.

Let the vertices of graph <math>G</math> generate the ring <math>S_0[T_0]</math>.
Let the bichromatic virtual edges be virtualizable by the graphs <math>H_1,\dots,H_m</math>.
Let <math>S_k[T_k]</math> be the ring generated by the vertices of <math>H_k</math>.
Let <math>S</math> be the ring generated by <math>\bigcup_{k=0}^m S_k</math>.
Let <math>T=\bigcup_{k=0}^m T_k</math>.
Then <math>S[T]</math> is a devirtualized ring of <math>G</math>.

As multiple graphs may virtualize the same virtual edge, careful choice may be required to minimize the set <math>T</math>.

===Multiplicative property===
Let <math>H_0</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_0</math> and chromatic number <math>n</math>.
Let <math>H_1</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_1</math> and chromatic number <math>n</math>.
Create a graph <math>H_2</math> by scaling <math>H_0</math> by a factor of <math>d_1</math>. Graph <math>H_2</math> then virtualizes a bichromatic virtual edge length <math>d_0d_1</math> and chromatic number <math>n</math>.

===Notable bichromatic virtual edges===
{| border=1
|-
! Chromatic number !! Length <math>d</math>!! Proof
|-
| any
| 1
| trivial case
|-
| 2
| <math>0\leq d\leq 3</math>
| moving the flexible parts of the graph (0,0),(0,1),(0,2),(0,3)
|-
| 3
| <math>\left\{\left|\frac{(3+\sqrt{-3})k}{2}+\sqrt{-3}j+(-1)^r\right| \mid k,j\in\mathbb{Z}, r\in\mathbb{R}\right\}</math>
| equilateral triangle tiling
|}

===Continuous ranges of bichromatic virtual edges===
Flexible graphs might produce continuous ranges of lengths of bichromatic virtual edges of chromatic number <math>n</math> without having a specific pair which is monochrome for every <math>n</math>-coloring.
Let the range of lengths be on the interval <math>d_0\leq d\leq d_1</math>.

If <math>1 < d_1</math>, then let <math>k\in\mathbb{Z}^+,d_0^{k+1} \leq d_1^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all lengths larger than <math>d_0^k</math>. A finite area allowed to be the same color, the infinite area of the plane, and a finite upper bound on CNP implies <math>CNP> n</math>.

If <math>d_0 < 1</math>, then let <math>k\in\mathbb{Z}^+,d_1^{k+1} \geq d_0^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all non-zero lengths smaller than <math>d_1^k</math>. Considering a set of <math>CNP+1</math> points all within distance <math>d_1^k</math> of each other implies <math>CNP> n</math>.

Using a flexible bichromatic virtual edge of chromatic number <math>n</math> and a graph <math>H</math> of chromatic number <math>n+1</math>, a flexible a graph <math>H'</math> of chromatic number <math>n+1</math> can be created by scaling <math>H</math> to some arbitrarily large or arbitrarily small size and clamping flexible bichromatic virtual edges of chromatic number <math>n</math> as a replacement of each rigid bichromatic virtual edge of <math>H</math>.

<math>H</math> virtualizing a bichromatic virtual edge of chromatic number <math>n+1</math> does not necessarily mean the graph <math>H'</math> virtualizes a bichromatic virtual edge.

==Best known results for the chromatic number in higher dimensions==

{| border=1
|-
! Space !! lower bound on CN !! Number of vertices !! Number of edges !! Upper bound on CN
|-
| <math>\mathbb{R}^1</math>
| 2
| 2
| 1
| 2
|-
| <math>\mathbb{R}^2</math>
| [https://arxiv.org/abs/1804.02385 5]
| [https://dustingmixon.wordpress.com/2018/05/10/polymath16-fifth-thread-human-verifiable-proofs/#comment-4465 610]
| 3000
| 7
|-
| <math>\mathbb{R}^3</math>
| [https://www.sciencedirect.com/science/article/pii/S0012365X00004064 6]
| [https://dustingmixon.wordpress.com/2018/05/10/polymath16-fifth-thread-human-verifiable-proofs/#comment-4463 59]
| [https://dustingmixon.wordpress.com/2018/05/10/polymath16-fifth-thread-human-verifiable-proofs/#comment-4463 183]
| [http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.145.684&rep=rep1&type=pdf 15]
|-
| <math>\mathbb{R}^4</math>
| [https://s3.amazonaws.com/academia.edu.documents/34568816/r4_march_22_revised.pdf?AWSAccessKeyId=AKIAIWOWYYGZ2Y53UL3A&Expires=1526311039&Signature=%2FrBgIHVOjapKNjsPiizHVO3IpJQ%3D&response-content-disposition=inline%3B%20filename%3DOn_the_Chromatic_Number_of_R.pdf 9]
| 65
| 588
| [https://link.springer.com/chapter/10.1007/978-3-642-55566-4_32 54]
|-
| <math>\mathbb{R}^5</math>
| [https://www.sciencedirect.com/science/article/pii/S0097316596800069 9]
|
|
| [https://dustingmixon.wordpress.com/2018/05/10/polymath16-fifth-thread-human-verifiable-proofs/#comment-4463 156]
|-
| <math>\mathbb{R}^6</math>
| [https://arxiv.org/abs/1408.2002 12]
| 175
|
| [https://dustingmixon.wordpress.com/2018/05/10/polymath16-fifth-thread-human-verifiable-proofs/#comment-4463 564]
|-
| <math>\mathbb{R}^7</math>
| [https://arxiv.org/abs/1408.2002 16]
| 168
| 4396
|
|-
| <math>\mathbb{R}^8</math>
| [https://arxiv.org/abs/1409.1278 19]
| 289
|
|
|-
| <math>\mathbb{R}^9</math>
| [https://arxiv.org/abs/1512.03472 22]
| 672
|
|
|-
| <math>\mathbb{R}^{10}</math>
| [https://arxiv.org/abs/1512.03472 30]
| 960
|
|
|-
| <math>\mathbb{R}^{11}</math>
| [https://arxiv.org/abs/1512.03472 35]
| 1320
|
|
|-
| <math>\mathbb{R}^{12}</math>
| [https://arxiv.org/abs/1512.03472 37]
| 1760
|
|
|}

== Probabilistic formulation ==

See [[Probabilistic formulation of Hadwiger-Nelson problem]].

== Algebraic formulation ==

See [[Algebraic formulation of Hadwiger-Nelson problem]].

== Excluding bichromatic vertices ==

See [[Excluding bichromatic vertices]].

== Further questions ==

* What are the [http://mathworld.wolfram.com/IndependenceRatio.html independence ratios] of the above unit distance graphs?
* What are the [https://en.wikipedia.org/wiki/Fractional_coloring fractional chromatic numbers] of these graphs?
* What are the [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz numbers] of these graphs?
** The Lovasz theta function value of Lovasz number of <math>G_2</math> at the complement [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3844 is in the interval [3.3746, 3.3748]].
* What about the Erdos unit distance graph (<math>n</math> vertices, <math>n^{1+c/\log\log n}</math> edges)?
* Can we use de Grey’s graph to construct unit-distance graphs that are not 5-colorable? To answer this question, we first need to understand how 5-colorings of de Grey’s graph force small collections of vertices to be colored. [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3899 Varga] and [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3940 Nazgand] provide some thoughts along these lines. Even if we can’t stitch together a 6-chromatic unit-distance graph with these ideas, we might be able to apply them to [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3900 prove that the measurable chromatic number of the plane is at least 6].
* It appears as though the coordinates of our smallest 5-chromatic graph lie in <math>\mathbb{Q}[\sqrt{3}, \sqrt{5}, \sqrt{11}]</math> (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3879 this]). If we view the plane as the complex plane, what is the smallest ring that admits a 5-chromatic single-distance graph? Every single-distance graph in the Eistenstein integers and Gaussian integers is 2-chromatic (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3914 this]). [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3934 David Speyer suggests] looking at <math>\mathbb{Z}[\frac{1+\sqrt{-71}}{2}]</math> next.
* What is the smallest cardinality of a subset of the plane which contains at least <math>n</math> colours in every colouring of the plane?
* Can the lower bound for <math>CNP\geq 5</math> be extended to all <math>L^p</math> norms where <math>p>1</math>, similar to how the Moser spindle was generalized?

== Blog, forums, and media ==

* [https://rjlipton.wordpress.com/2011/05/22/more-on-coloring-the-plane/ More On Coloring The Plane], Richard Lipton, May 22, 2011.
* [https://mathoverflow.net/questions/236392/has-there-been-a-computer-search-for-a-5-chromatic-unit-distance-graph Has there been a computer search for a 5-chromatic unit distance graph?], Juno, Apr 16, 2016.
* [https://quomodocumque.wordpress.com/2018/04/09/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Jordan Ellenberg, Apr 9 2018.
* [https://gilkalai.wordpress.com/2018/04/10/aubrey-de-grey-the-chromatic-number-of-the-plane-is-at-least-5/ Aubrey de Grey: The chromatic number of the plane is at least 5], Gil Kalai, Apr 10 2018.
* [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Dustin Mixon, Apr 10, 2018.
* [https://www.scottaaronson.com/blog/?p=3697 Amazing progress on long-standing problems], Scott Aaronson, Apr 11 2018.
* [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ The chromatic number of the plane is at least 5, Part II], Dustin Mixon, Apr 13 2018.
* [http://community.wolfram.com/groups/-/m/t/1320004 A 5-chromatic unit distance graph], Ed Pegg, Apr 13 2018.
* [http://aperiodical.com/2018/04/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Katie Steckles, Apr 17 2018.
* [https://www.quantamagazine.org/decades-old-graph-problem-yields-to-amateur-mathematician-20180417/ Decades-Old Graph Problem Yields to Amateur Mathematician], Evelyn Lamb, Quanta, Apr 17, 2018.
* [https://www.heise.de/newsticker/meldung/Zahlen-bitte-Wie-bunt-ist-die-Ebene-4024574.html Zahlen, bitte! 5 - Wie bunt ist die Ebene?], Harald Bögeholz, Heise, Apr 17, 2018.
* [http://www.sciencemag.org/news/2018/04/amateur-mathematician-cracks-decades-old-math-problem Amateur mathematician cracks decades-old math problem], Katie Langin, Science News, Apr 18, 2018.
* [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable How much of the plane is 4-colorable?], Dustin Mixon, Apr 18, 2018.
* [https://www.sciencealert.com/amateur-solves-decades-old-maths-problem-about-colours-that-can-never-touch-hadwiger-nelson-problem An Amateur Solved a 60-Year-Old Maths Problem About Colours That Can Never Touch], Peter Dockrill, ScienceAlert, Apr 19, 2018.
* [https://www.zmescience.com/science/math/amateur-mathematician-solves-problem-20042018/ Amateur mathematician Aubrey de Grey, known for his work on anti-aging, solves decades-old problem], Mihai Andrei, ZME Science, Apr 20, 2018.
* [https://automaths.blog/2018/04/21/5-nuances-daubrey-de-grey/ 5 nuances d’Aubrey de Grey], Automaths, Apr 21, 2018.
* [https://science.howstuffworks.com/math-concepts/amateur-solves-part-of-decades-old-math-problem.htm Amateur Solves Part of Decades-Old Math Problem], HowStuffWorks, Apr 30, 2018.
* [https://www.nemokennislink.nl/publicaties/het-platte-vlak-heeft-minstens-vijf-kleuren-nodig/ Het platte vlak heeft minstens vijf kleuren nodig], Kennislink, May 3, 2018.
* [https://index.hu/tudomany/2018/05/04/egy_biologus_oldott_meg_egy_olyan_problemat_amire_a_matematikusok_mar_60_eve_keptelenek/ Egy biológus oldott meg egy olyan problémát, amire a matematikusok már 60 éve képtelenek], Index.hu, May 4, 2018.

== Code and data ==

[https://www.dropbox.com/sh/ufknm1v9gtbhad3/AACB2xwaXYx5EGda38_L-0foa?dl=0 This dropbox folder] will contain most of the data and images for the project.

Data:

* [https://www.dropbox.com/s/x6kvu7b3wvdvqsn/graph.dimacs?dl=0 The 1585-vertex graph in DIMACS format]
* [https://www.dropbox.com/s/qk3gbpjvvsjfsc3/sat.dimacs?dl=0 A naive translation of 4-colorability of this graph into a SAT problem in DIMACS format]
* [https://www.dropbox.com/s/nipqikfzcfn9o5a/vertices.sage?dl=0 The vertices of this graph in explicit Sage notation]
* The graph <math>G_2</math>: [http://www.cs.utexas.edu/~marijn/CNP/874.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/874.edge edges (DIMACS)]
* The graph <math>G_3</math>: [http://www.cs.utexas.edu/~marijn/CNP/826.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/826.edge edges (DIMACS)] [http://www.cs.utexas.edu/~marijn/CNP/826.pdf Visualization]
* The [https://www.dropbox.com/sh/ufknm1v9gtbhad3/AADfw9WO1ol2ayQNJEtGf8oBa/Graphs?dl=0 densest unit-distance graphs] on an <math>n\times n</math> grid for <math>n=10,20,\ldots,100</math> (DIMACS format).

Code:

* [https://www.dropbox.com/s/6u1jctbjy38t383/lovaszmoser.m?dl=0 MATLAB script for computing Lovasz number]
* [https://files.jixco.de/pm16/zzvtx/ Python code for converting a list of vertices in Mathematica format into vertices in Z^n]
* [https://files.jixco.de/pm16/edge2cnf.py Python code for converting a DIMACS edge list into a CNF formula (forcing up to 3 suitable vertices to a fixed color, to break some symmetries)]

Software:

* [http://cvxr.com/cvx/ CVX]
* [http://www.labri.fr/perso/lsimon/glucose/ Glucose 4.0]
* [http://minisat.se/ Minisat]

== Wikipedia ==

* [https://en.wikipedia.org/wiki/Graph_coloring#Chromatic_number Chromatic number]
* [https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Erd%C5%91s_theorem_(graph_theory) de Bruijn-Erdos theorem]
* [https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem Hadwiger-Nelson problem]
* [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz number]
* [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]

== Bibliography ==
* [B2008] B. Bukh, [https://doi.org/10.1007/s00039-008-0673-8 Measurable sets with excluded distances], Geometric and Functional Analysis 18 (2008), 668-697.
* [CR2015] D. Cranston, L. Rabern, [https://arxiv.org/abs/1501.01647 The fractional chromatic number of the plane], arXiv:1501.01647
* [deBE1951] N. G. de Bruijn, P. Erdős, [http://www.math-inst.hu/~p_erdos/1951-01.pdf A colour problem for infinite graphs and a problem in the theory of relations], Nederl. Akad. Wetensch. Proc. Ser. A, 54 (1951): 371–373, MR 0046630.
* [deG2018] A. de Grey, [https://arxiv.org/abs/1804.02385 The chromatic number of the plane is at least 5], arXiv:1804.02385
* [EI2018] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.00157 The chromatic number of the plane is at least 5 - a new proof], arXiv:1805.00157
* [F1981] K.J. Falconer, The Realization of distances in measurable subsets covering Rn, J. Combin. Theory Ser. A 31 (1981) 184–189.
* [H1945] H. Hadwiger, Uberdeckung des euklidischen Raum durch kongruente Mengen, Portugaliae Math. 4 (1945), 238–242.
* [MM1961] L. Moser and M. Moser, Solution to Problem 10, Can. Math. Bull. 4 (1961), 187–189.
* [P1998] D. Pritikin, [https://www.sciencedirect.com/science/article/pii/S0095895698918196 All unit-distance graphs of order 6197 are 6-colorable], Journal of Combinatorial Theory, Series B 73.2 (1998): 159-163.
* [S2008] A. Soifer, The Mathematical Coloring Book, Springer, 2008, ISBN-13: 978-0387746401.
* [T1999] C. Thomassen, On the Nelson unit distance coloring problem, Amer. Math. Monthly 106 (1999) 850-853.
* [Tow2005] Townsend, S.P., Colouring the plane with no monochrome unit. Geombinatorics XIV(4) (2005), 181-193.
* [W1973] Woodall, D.R., Distances realized by sets covering the plane. J. Combin. Theory Ser. A, 14 (1973), 187-200.

[[Category:Polymath16]]

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-07T13:47:30Z

Nazgand: /* Lemma 21 */

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> \geq 1/(n \sqrt{3})</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d
|
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/6
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>p_{\sqrt 3}/(1-p_2)</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

=== Lemma 1 ===
Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

=== Lemma 2 ===
(Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

=== Lemma 3 ===
(Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

=== Corollary 4 ===
(Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

=== Computer-verified Claim 5 ===
(Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

=== Computer-verified Claim 6 ===
(Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

=== Lemma 7 ===
(Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

=== Corollary 8 ===
One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

=== Lemma 9 ===
(Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

=== Corollary 10 ===
(Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

=== Computer-verified claim 11 ===
One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

For certain special distances d, one can improve the bound in Lemma 2:

=== Lemma 12 ===
If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, <math>\gcd(j,2k+1)=1</math>, and <math>r = \frac{1}{2} \csc\left(\frac{j\pi}{2k+1}\right)</math> then
:<math> p_r \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}}} \leq \frac{1}{3}.</math>

'''Proof''' Observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2}} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges.

=== Lemma 13 ===
We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

=== Corollary 14 ===
We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

=== Lemma 15 ===
One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

=== Corollary 16 ===
We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

=== Lemma 17 ===
If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths a,b,c. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also, thus
:<math>{\bf P}(\mathbf{c}(0) \neq \mathbf{c}(a)) + {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(b)) \geq {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(c))</math>
and the claim follows. <math>\Box</math>

Note that Lemma 2 follows from the a=b, c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 1 - 2^{-k}</math> when <math>a \geq 2^{-k}, k\in\mathbb{Z}^+</math>. Further, we can generalise the a=b case to one in which the triangle is replaced by a (k+1)-gon of which one edge is 1 and the others are all equal, leading to the stronger result <math>p_a \leq 1 - 1/k</math> when <math>a \geq 1/k, k\in\mathbb{Z}^+</math>. Further strenthening is achieved by using <math>1/\sqrt{3}</math> as the long edge, given Lemma 12.

=== Lemma 18 ===
Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio. Also we have
:<math> p_{d/\sqrt{3}} \leq \frac{1}{3} + p_d, \frac{1}{2} + \frac{1}{2} p_d.</math>

Note that this generalises (9), as well as a special case of Lemma 12.

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the first claim. Similarly, if one considers the colorings of an equilateral triangle of sidelength <math>d</math> together with its center, and counts the numbers <math>a,b \in \{0,1,2,3\}</math> of monochromatic edges of length <math>d</math> and <math>d/\sqrt{3}</math> respectively, one observes that one always has <math>\frac{b}{3} \leq \frac{1}{3} + \frac{2}{3} \frac{a}{3}, \frac{1}{2} + \frac{1}{2} \frac{a}{3}</math>, and on taking expectations one obtains the claim.<math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

=== Lemma 19 ===
(Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

=== Theorem 20 ===
One has <math>p_{H = 1tri} > 0</math>.

'''Proof''' Suppose for contradiction that <math>p_{H = 1tri} = 0</math>. One can then run a version of the de Bruijn-Erdos argument to obtain a coloring in which 1tri hexagons are completely nonexistent (since there are arbitrarily large finite colorings with this property). Consider the triangular lattice <math>{\bf Z}[e^{\pi i/3}]</math>. We consider the dual graph, in which each element of the triangular lattice is the center of a hexagon in the dual graph, which is dual to the copy of H centered at that element. Color an edge <math>e^\perp</math> in the dual lattice black if the original edge <math>e</math> is monochromatic, and white otherwise. Thus, each hexagon has one of three coloring patterns up to rotation, depending on the coloring type of H:

* If H is colored 2tri, then all six edges of the hexagon are black.
* If H is colored axisym, then two opposing edges of the hexagon are black, and the other two are white.
* If H is colored centralsym, then none of the edges of the hexagon are black.

In particular, if one hexagon is colored 2tri, then either all adjacent hexagons are colored 2tri, or else all adjacent hexagons are colored axisym. If one follows an axisym hexagon along its axis of symmetry, the adjacent hexagons are either again axisym, or 2tri; thus, one either gets an infinite chain of axisym hexagons, a ray of axisym hexagons terminating at a 2tri hexagon, or an interval of axisym hexagons terminated at both sides by a 2tri hexagon. In the latter case one can continue the coloring to find that the 2tri hexagons are arranged in a lattice, with consecutive 2tri hexagons joined by chains of axisym hexagons. This leads to only a small number of possible hexagon colorings in the lattice:

# Case 1: Every hexagon centred at an element <math>{\bf Z}[ e^{\pi i/3} ]</math> is colored centralsym.
# Case 2: Every hexagon centred at an element <math>{\bf Z}[ e^{\pi i/3} ]</math> is colored 2tri.
# Case 3.k: For some natural number <math>k \geq 2</math>, every hexagon centred in a coset of <math>k \cdot {\mathbf Z}[ e^{\pi i/3} ]</math> is colored 2tri, the hexagons connecting any two adjacent hexagons in this coset are colored axisym, and all other hexagons are colored centralsym.
# Case 4: Each horizontal row of hexagons either consists entirely of centralsym, or consists entirely of axisym (with the axis of symmetry horizontal).
# Case 5: Each northwest row of hexagons consists either entirely of centralsym, or consists entirely of axisym (with axis of symmetry northwest).
# Case 6: Each northeast row of hexagons consists either entirely of centralsym, or consists entirely of axisym (with axis of symmetry northeast).
# Case 7: A single hexagon colored 2tri, with six rays of hexagons colored axisym emanating from this hexagon; all other triangles are centralsym.

Technically, Case 1 is contained in Cases 4,5,6 as written above, but this will not be an issue.

In the first case, the coloring is periodic with periods <math>2, 2 e^{\pi i/3}</math>. In the second case, it is periodic with periods <math>3, 3 e^{\pi i/3}</math>. In the third case, it is periodic with periods <math>3k, 3k e^{\pi i/3}</math>. Also note that for each k, one can check if Case 3.k holds by inspecting the coloring at a finite number of vertices. Thus the event that Case 3.k holds is "measurable" in the sense that a meaningful probability can be assigned. (But Cases 1,2,4,5,6 are not measurable events, they require an infinite number of points to be inspected, and the probability measure we are using is only finitely additive rather than infinitely additive.) In Case 4, the coloring is periodic with period 2; also, every coset of <math>2 \cdot {\bf Z}[e^{\pi i/3}]</math> is 2-colored. Similarly for Case 5 and 6 (where the periods are <math>2 e^{2\pi i/3}</math> and <math>2 e^{4\pi i/3}</math> respectively.)

Let <math>\alpha_k</math> be the probability that Case 3.k holds for the given value of <math>k</math>. Then <math> \sum_{k=2}^K \alpha_k \leq 1</math> for any k, hence <math>\sum_{k=2}^\infty \alpha_k \leq 1</math>. In particular, we can find <math>K_1</math> such that <math>\sum_{k={K_1}}^\infty \alpha_k \leq 0.1</math> (say). Let <math>P</math> be six times the least common multiple of <math>1,2,\dots,K_1</math>. Then the coloring is P- and <math>P e^{\pi i/3}</math>-periodic for Case 1, Case 2, and all Case 3.k with <math>k \leq K_1</math>. On the other hand, if <math>K_2</math> is sufficiently large depending on <math>P</math>, and Case 3.k holds for some <math>k \geq K_2</math>, then almost all of the hexagons are colored centralsym, which makes the coloring "almost <math>P, P e^{\pi i/3}</math>-periodic" in the sense that
:<math>{\bf c}(z+P e^{\pi i j/3}) = {\bf c}(z) \hbox{ for } j=0,1,2,3,4,5</math>
will hold for at least <math>0.9</math> of the lattice points <math>z \in {\bf Z}[e^{\pi i/3}]</math> with <math>|z| \leq K_2</math>. Similarly for Case 7 (which is sort of a <math>k=\infty</math> limiting case of Case 3.k.) Thus, with the probability <math> \geq 1 - \sum_{k=K_1}^{K_2} \alpha_k \geq 0.9</math>, the coloring of the seven vertices <math>{\bf c}(0), {\bf c}(P e^{\pi ij/3}, j=1,\dots,6</math> is (up to rotation and recoloring) one of the three patterns of the central and linking vertices in Figure 3 of Aubrey's paper, namely

* <math>{\bf c}(0) = {\bf c}(P) = {\bf c}(P e^{\pi i/3}) = {\bf c}(P e^{2\pi i/3}) = {\bf c}(P e^{3\pi i/3}) = {\bf c}(P e^{4\pi i/3}) = {\bf c}(P e^{5\pi i/3}) </math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{3\pi i/3})</math>; <math>{\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>;
* <math>{\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{3\pi i/3})</math>; <math> {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3}) </math> for some <math>P' = P e^{\pi i j/3}</math>, <math>j=1,\dots,6</math>.

Using the spindling argument from Aubrey's paper, we conclude that the third possibility must in fact hold with probability at least 0.8; on the other hand, from Lemma 2 this scenario can only occur with probability at most 1/2, giving the required contradiction.
<math>\Box</math>

One should be able to refine this argument to show that <math>p_{H = 1tri} > c</math> for an absolute constant <math> c>0 </math>.

=== Lemma 21 ===
Providing a tighter bound for Lemma 17 with a more thorough proof: If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-06T16:22:00Z

Nazgand: /* Lemma 12 */ Beautify

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> \geq 1/(n \sqrt{3})</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d
|
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/6
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>p_{\sqrt 3}/(1-p_2)</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

=== Lemma 1 ===
Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

=== Lemma 2 ===
(Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

=== Lemma 3 ===
(Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

=== Corollary 4 ===
(Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

=== Computer-verified Claim 5 ===
(Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

=== Computer-verified Claim 6 ===
(Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

=== Lemma 7 ===
(Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

=== Corollary 8 ===
One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

=== Lemma 9 ===
(Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

=== Corollary 10 ===
(Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

=== Computer-verified claim 11 ===
One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

For certain special distances d, one can improve the bound in Lemma 2:

=== Lemma 12 ===
If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, <math>\gcd(j,2k+1)=1</math>, and <math>r = \frac{1}{2} \csc\left(\frac{j\pi}{2k+1}\right)</math> then
:<math> p_r \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}}} \leq \frac{1}{3}.</math>

'''Proof''' Observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2}} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges.

=== Lemma 13 ===
We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

=== Corollary 14 ===
We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

=== Lemma 15 ===
One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

=== Corollary 16 ===
We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

=== Lemma 17 ===
If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

Note that Lemma 2 follows from the c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 1 - 2^{-k}</math> when <math>a \geq 2^{-k}, k\in\mathbb{Z}^+</math>.

=== Lemma 18 ===
Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio.

Note that this generalises (9).

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the claim. <math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

=== Lemma 19 ===
(Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-06T16:15:47Z

Nazgand: /* Bounds on {\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) ) for 4-colourings */ typo fix

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> \geq 1/(n \sqrt{3})</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d
|
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/6
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>p_{\sqrt 3}/(1-p_2)</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

=== Lemma 1 ===
Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

=== Lemma 2 ===
(Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

=== Lemma 3 ===
(Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

=== Corollary 4 ===
(Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

=== Computer-verified Claim 5 ===
(Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

=== Computer-verified Claim 6 ===
(Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

=== Lemma 7 ===
(Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

=== Corollary 8 ===
One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

=== Lemma 9 ===
(Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

=== Corollary 10 ===
(Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

=== Computer-verified claim 11 ===
One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

For certain special distances d, one can improve the bound in Lemma 2:

=== Lemma 12 ===
If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, and <math>\gcd(j,2k+1)=1</math> then
:<math> p_{\frac{1}{2 \sin \frac{j\pi}{2k+1}}} \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}})} \leq \frac{1}{3}.</math>

'''Proof''' Write <math>r = \frac{1}{2 \sin \frac{j\pi}{2k+1}}</math>, then observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2})} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges.

=== Lemma 13 ===
We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

=== Corollary 14 ===
We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

=== Lemma 15 ===
One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

=== Corollary 16 ===
We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

=== Lemma 17 ===
If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

Note that Lemma 2 follows from the c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 1 - 2^{-k}</math> when <math>a \geq 2^{-k}, k\in\mathbb{Z}^+</math>.

=== Lemma 18 ===
Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio.

Note that this generalises (9).

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the claim. <math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

=== Lemma 19 ===
(Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-06T16:15:07Z

Nazgand: /* Bounds on {\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) ) for 4-colourings */ notation

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> \geq 1/(n \sqrt{3})</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d
|
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/6
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>p_(\sqrt 3)/(1-p_2)</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

=== Lemma 1 ===
Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

=== Lemma 2 ===
(Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

=== Lemma 3 ===
(Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

=== Corollary 4 ===
(Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

=== Computer-verified Claim 5 ===
(Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

=== Computer-verified Claim 6 ===
(Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

=== Lemma 7 ===
(Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

=== Corollary 8 ===
One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

=== Lemma 9 ===
(Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

=== Corollary 10 ===
(Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

=== Computer-verified claim 11 ===
One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

For certain special distances d, one can improve the bound in Lemma 2:

=== Lemma 12 ===
If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, and <math>\gcd(j,2k+1)=1</math> then
:<math> p_{\frac{1}{2 \sin \frac{j\pi}{2k+1}}} \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}})} \leq \frac{1}{3}.</math>

'''Proof''' Write <math>r = \frac{1}{2 \sin \frac{j\pi}{2k+1}}</math>, then observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2})} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges.

=== Lemma 13 ===
We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

=== Corollary 14 ===
We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

=== Lemma 15 ===
One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

=== Corollary 16 ===
We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

=== Lemma 17 ===
If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

Note that Lemma 2 follows from the c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 1 - 2^{-k}</math> when <math>a \geq 2^{-k}, k\in\mathbb{Z}^+</math>.

=== Lemma 18 ===
Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio.

Note that this generalises (9).

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the claim. <math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

=== Lemma 19 ===
(Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-06T16:09:17Z

Nazgand: /* Proofs of bounds */ Put Lemmas and such in subsections

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> \geq 1/(n \sqrt{3})</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d
|
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/6
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(\sqrt 3) )/(1-{\bf P}( \mathbf{c}(0) = \mathbf{c}(2) ))</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

=== Lemma 1 ===
Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

=== Lemma 2 ===
(Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

=== Lemma 3 ===
(Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

=== Corollary 4 ===
(Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

=== Computer-verified Claim 5 ===
(Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

=== Computer-verified Claim 6 ===
(Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

=== Lemma 7 ===
(Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

=== Corollary 8 ===
One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

=== Lemma 9 ===
(Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

=== Corollary 10 ===
(Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

=== Computer-verified claim 11 ===
One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

For certain special distances d, one can improve the bound in Lemma 2:

=== Lemma 12 ===
If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, and <math>\gcd(j,2k+1)=1</math> then
:<math> p_{\frac{1}{2 \sin \frac{j\pi}{2k+1}}} \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}})} \leq \frac{1}{3}.</math>

'''Proof''' Write <math>r = \frac{1}{2 \sin \frac{j\pi}{2k+1}}</math>, then observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2})} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges.

=== Lemma 13 ===
We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

=== Corollary 14 ===
We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

=== Lemma 15 ===
One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

=== Corollary 16 ===
We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

=== Lemma 17 ===
If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

Note that Lemma 2 follows from the c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 1 - 2^{-k}</math> when <math>a \geq 2^{-k}, k\in\mathbb{Z}^+</math>.

=== Lemma 18 ===
Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio.

Note that this generalises (9).

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the claim. <math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

=== Lemma 19 ===
(Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-06T15:56:17Z

Nazgand: /* Proofs of bounds */

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> \geq 1/(n \sqrt{3})</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d
|
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/6
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(\sqrt 3) )/(1-{\bf P}( \mathbf{c}(0) = \mathbf{c}(2) ))</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

'''Lemma 1''' Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

'''Lemma 2''' (Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

'''Lemma 3''' (Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

'''Corollary 4''' (Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

'''Computer-verified Claim 5''' (Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

'''Computer-verified Claim 6''' (Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

'''Lemma 7''' (Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

'''Corollary 8''' One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

'''Lemma 9''' (Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

'''Corollary 10''' (Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

'''Computer-verified claim 11''' One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

For certain special distances d, one can improve the bound in Lemma 2:

'''Lemma 12''' If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, and <math>\gcd(j,2k+1)=1</math> then
:<math> p_{\frac{1}{2 \sin \frac{j\pi}{2k+1}}} \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}})} \leq \frac{1}{3}.</math>

'''Proof''' Write <math>r = \frac{1}{2 \sin \frac{j\pi}{2k+1}}</math>, then observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2})} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges.

'''Lemma 13''' We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

'''Corollary 14''' We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

'''Lemma 15''' One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

'''Corollary 16''' We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

'''Lemma 17''' If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

Note that Lemma 2 follows from the c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 1 - 2^{-k}</math> when <math>a \geq 2^{-k}, k\in\mathbb{Z}^+</math>.

'''Lemma 18''' Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio.

Note that this generalises (9).

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the claim. <math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

'''Lemma 19''' (Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-06T15:43:56Z

Nazgand: /* Proofs of bounds */ Tightened the inequality of Lemma 17

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>
(In probabilistic language, this means that the random coloring is a [https://en.wikipedia.org/wiki/Stationary_process stationary process] with respect to the action of <math>S_4 \times E(2)</math>. The extraction of a stationary process from a deterministic object is an example of the ''Furstenberg correspondence principle''.)

== Bounds on <math>p_d</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>. Accordingly, define
:<math>p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).</math>

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph/method !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math> \geq 1/2</math>
|
|
| 1/2
| Spindle
|
|-
| <math> \geq 1/(n \sqrt{3})</math>
|
|
| <math>(3n-2)/3n</math>
| (n+1)-gon with one edge length <math>1/\sqrt{3}</math> and the rest d
|
|-
| 1
| 0
| Unit edge
| 0
| Unit edge
|
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H, Corollary 16
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/6
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H, Corollary 16
| 1/2
| Spindle
|
|-
| <math>{\frac{\sqrt{5}+1}{2}}</math>
| 1/5
| regular pentagon with unit side length
| 2/5
| regular pentagon with unit side length
|
|-
| <math>{\frac{\sqrt{5}-1}{2}}</math>
| 1/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
| 2/5
| regular pentagon with <math>{\frac{\sqrt{5}-1}{2}}</math> side length
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
| 1/2
| Spindle
| computer-verified
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
| 1/2
| Spindle
| computer-verified; leads to contradiction
|}

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )</math> for 4-colourings ==

{| border=1
|-
! <math>d_1</math> !! <math>d_0</math> !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| <math>1+(-1)^{1/3}</math>
| <math>2</math>
| 1/2
| H
|
|
| Equals <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(\sqrt 3) )/(1-{\bf P}( \mathbf{c}(0) = \mathbf{c}(2) ))</math>
|-
| <math>1+(-1)^{1/3}</math>
| <math>1+(-1)^{-1/3}</math>
|
|
| 1/2
| H
|
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

'''Lemma 1''' Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

'''Lemma 2''' (Spindle argument) Let <math>|d| \geq 1/2</math>. Then
:<math> p_d \leq \frac{1}{2} \quad (7).</math>

'''Proof''' We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

'''Lemma 3''' (Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

'''Corollary 4''' (Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

'''Computer-verified Claim 5''' (Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

'''Computer-verified Claim 6''' (Using the graph <math>V \oplus V \oplus H</math>) One has <math>p_{8/3} = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

'''Lemma 7''' (Using <math>G_{40}</math>) We have
:<math>59 p_{\sqrt{11/3}} + p_{8/3} \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

'''Corollary 8''' One has
:<math> p_{\sqrt{11/3}} \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

'''Lemma 9''' (Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

'''Corollary 10''' (Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

'''Computer-verified claim 11''' One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

For certain special distances d, one can improve the bound in Lemma 2:

'''Lemma 12''' If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, and <math>\gcd(j,2k+1)=1</math> then
:<math> p_{\frac{1}{2 \sin \frac{j\pi}{2k+1}}} \leq \frac{k}{2k+1},</math>
thus for instance
:<math> p_{\frac{1}{\sqrt{3}})} \leq \frac{1}{3}.</math>

'''Proof''' Write <math>r = \frac{1}{2 \sin \frac{j\pi}{2k+1}}</math>, then observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

On the other hand, for <math>k=2,j=1</math> we also know from the regular pentagon of unit sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2})} \leq \frac{2}{5} \quad (9)</math>
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.

Similarly, for <math>k=2,j=2</math> we also know from the regular pentagon of <math>\frac{\sqrt{5}-1}{2}</math> sidelength that
:<math> \frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}</math>
since any 4-coloring of that pentagon has either one or two monochromatic edges.

'''Lemma 13''' We have
:<math> 7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>p_{\sqrt{3}}</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

'''Corollary 14''' We have <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

'''Lemma 15''' One has
:<math>p_{\sqrt{3}} + p_2 \geq \frac{2}{3}</math> and <math>2 p_{\sqrt{3}} + p_2 \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

'''Corollary 16''' We have <math>p_2 \geq \frac{1}{6}</math>, <math>p_{\sqrt{3}} \geq \frac{1}{4} </math> and <math>p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

'''Lemma 17''' If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math>.

'''Proof''' Consider a triangle of side lengths <math>a,b,c</math>. Let <math>\left|z_2\right|=b,\left|a-z_2\right|=c</math>. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: <math>\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]</math>. <math>[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)</math> thus
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c</math>
<math>{\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)</math>, so
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>{\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
:<math>1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))</math>
Using the law of cosines: <math>z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)</math>
:<math>{\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c</math> <math>\Box</math>

Note that Lemma 2 follows from the c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>p_a</math> for small values of a, e.g. <math>p_a \leq 3/4</math> when <math>a \geq 1/4</math>.

'''Lemma 18''' Whenever <math>d>0</math>, one has the inequalities
:<math> |p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1</math>
where <math>\phi := \frac{1+\sqrt{5}}{2}</math> is the golden ratio.

Note that this generalises (9).

'''Proof''' Consider the regular pentagon with sidelength <math>d</math>, so it also has 5 diagonals of length <math>\phi d</math>. Let <math>a \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic edges and let <math>b \in \{0,1,2,3,4,5\}</math> denote the number of monochromatic diagonals. Observe:
* <math>a,b</math> cannot both be zero (pigeonhole principle).
* <math>a</math> cannot be 4. Similarly, <math>b</math> cannot be 4.
* If <math>a=5</math>, then <math>b=5</math>, and conversely.
* If <math>a=0</math>, then <math>b=1,2</math>; similarly, if <math>b=0</math>, then <math>a=1,2</math>.

From this we observe the inequalities
:<math> |\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1</math>
and on taking expectations we obtain the claim. <math>\Box</math>

The hexagon <math>H</math> has essentially four distinct colorings: the coloring <math>\hbox{2tri}</math> with two triangles, the coloring <math>\hbox{1tri}</math> with one triangle, the coloring <math>\hbox{axisym}</math> that is symmetric around an axis, and the coloring <math>\hbox{centralsym}</math> that is symmetric around the central point. This gives four probabilities <math>p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}</math> that sum to 1. By counting the number of monochromatic edges of length <math>\sqrt{3}, 2</math> respectively, one also obtains the identities
:<math>p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}</math>
which reproves Lemma 15. Also
:<math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.</math>
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
:<math> p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}</math>
which reproves Corollary 4.

'''Lemma 19''' (Hubai) One has <math>p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}</math>.

'''Proof''' Consider five copies of H centred at 0,1,2,3,4. With probability at least <math>1 - 5( p_{H = 1tri} + p_{H = axisym} )</math>, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. <math>\Box</math>

== Simplification rules for triplets of points in the complex plane ==
Deduced from the rule <math>{\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)</math>.
:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

:<math>{\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )</math>

== Proofs of bounds for conditional probabilities ==
The trivial case, valid where <math>\left|d\right|\neq 1</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0</math>

Trivial plus Baye's Theorem, valid where <math>d\neq 0</math>:
:<math>x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1</math>
Rearrange:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1</math>

Spindle method: for <math>\left|d\right|=d\geq 1/2</math> and <math>\theta=2\text{arcsin}\left(\frac{1}{2d}\right)</math>:
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1</math>
which is another way to see <math>\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>.

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-06T10:19:31Z

Nazgand: /* Simplification rules for triplets of points in the complex plane */

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-06T10:18:31Z

Nazgand: /* Proofs of bounds for conditional probabilities */

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T21:39:16Z

Nazgand: /* Bounds on p_d for 4-colourings */

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T19:41:02Z

Nazgand: /* Bounds on {\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) ) for 4-colourings */

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T19:21:55Z

Nazgand: /* Bounds on {\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) ) for 4-colourings */

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T17:43:14Z

Nazgand: /* Proofs of bounds for conditional probabilities */ removed logic error

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T17:41:16Z

Nazgand: /* Proofs of bounds for conditional probabilities */ Typo fix

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T17:38:12Z

Nazgand: /* Proofs of bounds for conditional probabilities */ Extended inequality to <math>d\neq 0\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2</math>

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T16:32:45Z

Nazgand: /* Proofs of bounds for conditional probabilities */

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T16:18:43Z

Nazgand: /* Proofs of bounds for conditional probabilities */

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T16:09:54Z

Nazgand: /* Proofs of bounds for conditional probabilities */ Added spindle method

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T14:54:09Z

Nazgand: /* Proofs of bounds for conditional probabilities */

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T14:47:01Z

Nazgand: /* Bounds on {\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) ) for 4-colourings */

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T11:57:08Z

Nazgand: /* Proofs of bounds */ trivial conditional case added.

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T11:25:53Z

Nazgand: Added a conditional probability

Hadwiger-Nelson problem

2018-05-05T10:03:22Z

Nazgand: /* Notable unit distance graphs */

The Chromatic Number of the Plane (CNP) is the chromatic number of the graph whose vertices are elements of the plane, and two points are connected by an edge if they are a unit distance apart. The Hadwiger-Nelson problem asks to compute CNP. The bounds <math>4 \leq CNP \leq 7</math> are classical; recently [deG2018] it was shown that <math>CNP \geq 5</math>. This is achieved by explicitly locating finite unit distance graphs with chromatic number at least 5.

The '''Polymath16''' project seeks to simplify the graphs used in [deG2018] to establish this lower bound. More precisely, the goals are

* '''Goal 1''': Find progressively smaller 5-chromatic unit-distance graphs.
* '''Goal 2''': Reduce (ideally to zero) the reliance on computer assistance for the proof. Computer assistance was leveraged in [deG2018] to analyze a subgraph of size 397.
* '''Goal 3''': Apply these simpler graphs to inform progress in related areas. For example:
** Find a 6-chromatic unit-distance graph in the plane.
** Improve the corresponding bound in higher dimensions.
** Improve the current record of 383/102 for the fractional chromatic number of the plane.
** Find the smallest unit-distance graph of a given minimum degree (excluding, in some natural way, boring cases like Cartesian products of a graph with a hypercube).

== Polymath threads ==

* [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/ Polymath proposal: finding simpler unit distance graphs of chromatic number 5], Aubrey de Grey, Apr 10 2018. ('''Active discussion thread''')
* [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/ Polymath16, first thread: Simplifying de Grey’s graph], Dustin Mixon, Apr 14, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic Polymath16, second thread: What does it take to be 5-chromatic?], Dustin Mixon, Apr 22, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/ Polymath16, third thread: Is 6-chromatic within reach?], Dustin Mixon, May 1, 2018. ('''Active research thread''')

== Notable unit distance graphs ==

A '''unit distance graph''' is a graph that can be realised as a collection of vertices in the plane, with two vertices connected by an edge if they are precisely a unit distance apart. The chromatic number of any such graph is a lower bound for <math>CNP</math>; in particular, if one can find a unit distance graph with no 4-colorings, then <math>CNP \geq 5</math>. The boldface number of vertices is the current minimal number of vertices of a unit distance graph that is currently known to not be 4-colorable.

<math>G_1 \oplus G_2</math> denotes the Minkowski sum of two unit distance graphs <math>G_1,G_2</math> (vertices in <math>G_1 \oplus G_2</math> are sums of the vertices of <math>G_1,G_2</math>). <math>G_1 \cup G_2</math> denotes the union. <math>\mathrm{rot}(G, \theta)</math> denotes <math>G</math> rotated counterclockwise by <math>\theta</math>. <math>\mathrm{trim}(G,r)</math> denotes the trimming of <math>G</math> after removing all vertices of distance greater than <math>r</math> from the origin.

Another basic operation is '''spindling''': taking two copies of a graph <math>G</math>, gluing them together at one vertex, and rotating the copies so that the two copies of another vertex are a unit distance apart. For instance, the Moser spindle is the spindling of a rhombus graph. If a graph has two vertices forced to be the same color in a k-coloring, then the spindling of that graph at those two vertices is not k-colorable.

Many of the graphs are embeddable in abelian groups or rings, particularly those generated by the roots of unity <math>\omega_t := \exp( i \arccos( 1 - \tfrac{1}{2t} ))</math> for various natural numbers t.

{| border=1
|-
!Name!!Number of vertices!! Number of edges !! Structure !! Group !! Colorings
|-
| [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]
| 7
| 11
| Two 60-120-60-120 rhombi with a common vertex, with one pair of sharp vertices coincident and the other joined
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [http://mathworld.wolfram.com/GolombGraph.html Golomb graph]
| 10
| 18
| Contains the center and vertices of a hexagon and equilateral triangle
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [https://arxiv.org/abs/1804.02385 H]
| 7
| 12
| Vertices and center of a hexagon
| <math>{\bf Z}[\omega_1]</math>
| Has essentially four 4-colorings, two of which contain a monochromatic <math>\sqrt{3}</math>-triangle. Every 5-coloring has a monochrome <math>\sqrt{3}</math>-edge or a monochrome <math>2</math>-edge
|-
| [https://arxiv.org/abs/1804.02385 J]
| 31
| 72
| Contains 13 copies of H
| <math>{\bf Z}[\omega_1]</math>
| Has essentially six 4-colorings in which no H has a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 K]
| 61
| 150
| Contains 2 copies of J
|
| In all 4-colorings lacking an H with a monochromatic <math>\sqrt{3}</math>-triangle, all pairs of vertices at distance 4 are monochromatic
|-
| [https://arxiv.org/abs/1804.02385 L]
| 121
| 301
| Contains two copies of K and 52 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ <math>L_1</math>]
| 97
|
| Has 40 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120295 <math>L_2</math>]
| 120
| 354
|
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 T]
| 9
| 15
| Contains one Moser spindle and useful symmetry; three vertices form an equilateral triangle
|
|
|-
| [https://arxiv.org/abs/1804.02385 U]
| 15
| 33
| Three copies of T at 120-degree rotations: <math>T \cup \mathrm{rot}(T, 2\pi/3) \cup \mathrm{rot}(T, 4\pi/3)</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 V]
| 31
| 30
| Unit vectors at angles consistent with three interlocking Moser spindles
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>V_1</math>]
| 61
| 60
| Union of V and a rotation of V: <math>V \cup \mathrm{rot}(V, \mathrm{arccos}(7/8))</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_b</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_x</math>]
| 13
| 12
| Subgraph of <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_z</math>]
|
|
| Subgraph of <math>V_a</math>; shares a line of symmetry with <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_y</math>]
| 13
| 12
| Subgraph of <math>V_b</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-3970 <math>V_A</math>]
| 37
| 36
| Unit vectors with angles <math>i \frac{\pi}{3} + j \mathrm{arccos} \frac{5}{6} + k \mathrm{arccos} \frac{7}{8}</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 W]
| 301
| 1230
| Cartesian product of V with itself, minus vertices at more than <math>\sqrt{3}</math> from the centre (i.e. <math>\mathrm{trim}(V \oplus V, \sqrt{3})</math>)
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>W_1</math>]
|
|
| Trimmed product of V with itself (<math>\mathrm{trim}(V \oplus V, 1.95)</math>)
|
|
|-
| [https://arxiv.org/abs/1804.02385 M]
| 1345
| 8268
| Cartesian product of W and H (<math>W \oplus H</math>)
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120274 <math>M_1</math>]
| 278
|
| Deleting vertices from M while maintaining its restriction on H
|
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>M_2</math>]
| 7075
|
| Sum of H with a trimmed product of <math>V_1</math> with itself
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 N]
| 20425
| 151311
| Contains 52 copies of M arranged around the H-copies of L
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4, \omega_{16}]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_0</math>]
| 1585
| 7909
| N "shrunk" by stepwise deletions and replacements of vertices
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 G]
| 1581
| 7877
| Deleting 4 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_1</math>]
| 1577
|
| Deleting 8 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120318 <math>G_2</math>]
| 874
| 4461
| Juxtaposing two copies of M and shrinking
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3867 <math>G_3</math>]
| 826
| 4273
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4105 <math>G_4</math>]
| '''803'''
| 4144
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 R]
|
|
| Union of <math>W_1</math> and a rotated copy of <math>W_1</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>\mathrm{trim}(R \oplus H, 1.67)</math>]
| 2563
|
| Trimmed sum of R and H
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 <math>V \oplus V \oplus H</math>]
|
|
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_{64/9}]</math>
| Has two vertices forced to be the same color in a 4-coloring; also no monochromatic <math>\sqrt{3}</math>-triangles
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 No name assigned]
| 745
|
| Subgraph of <math>V \oplus V \oplus H</math>
|
| Has two vertices forced to be the same color in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3085
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_a \oplus V_z \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3049
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 No name assigned]
| 1951
|
| Trimmed version of <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic <math>V_A \oplus V_A \oplus V_A</math>]
| 6937
| 44439
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{40}</math>]
| 40
| 82
|
|
| Any 4-coloring that avoids a monochromatic <math>\sqrt{11/3}</math>-edge has two specific vertices forced to be the same color
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{79}</math>]
| 79
| 165
| Spindling of <math>G_{40}</math>
|
| Any 4-coloring has a monochromatic <math>\sqrt{11/3}</math>-edge
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{49}</math>]
| 49
| 180
|
| <math>{\mathbf Q}[\sqrt{3},\sqrt{11}]^2</math>
| Any 4-coloring either has a specific <math>\sqrt{11/3}</math>-edge monochromatic, or a monochromatic <math>1/\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{51}</math>]
| 51
|
|
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring of plane
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{627}</math>]
| 627
| 2982
| Contains <math>G_{51}</math>
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4161 No name assigned]
| 103
|
|
|
| All 4-colorings contain a monochromatic <math>2/\sqrt{3}</math>-edge
|-
| regular pentagon with unit side length
| 5
| 5
|
|
| All 4-colorings contain a monochromatic <math>(\sqrt{5}+1)/2</math>-edge
|-
| regular pentagon with <math>(\sqrt{5}-1)/2</math> side length
| 5
| 5
|
|
| All 4-colorings contain a monochromatic <math>(\sqrt{5}-1)/2</math>-edge
|}

== Lower bounds ==

In [P1988], Pritikin proved that every graph with at most 12 vertices is 4-colorable, and every graph with at most 6197 vertices is 6-colorable. Pritikin's bounds are obtained by coloring the plane with k colors and an additional “wild” color such that points of unit distance are both allowed to receive the wild color. If the wild color occupies a small fraction p of the plane, then an exercise in the probabilistic method gives that any fixed unit-distance graph with n vertices enjoys an embedding in the plane that avoids the wild color (and is therefore k-colorable) provided n<1/p. Pritikin’s coloring for the k=4 case cannot be improved without improving on the densest known subset of the plane that avoids unit distances (originally due to Croft in 1967). See [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable this MO thread] for additional information. As such, improving the bound in the k=4 case might require a new technique, whereas the k=5 and 6 cases might be amenable to optimization.

[https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4176 Every unit distance graph with at most 16 vertices is 5-colorable].

A "tile-based" colouring of the plane must have at least 6 colours, as shown by Townsend [Tow2005]; the same proof with a minor error was also derived by Woodall [W1973]. In [T1999], Thomassen showed that any tiling-based 6-coloring would have to be be "unscaleable", i.e. the maximum diameter of a tile and the minimum separation of same-coloured tiles must both be exactly 1 (so that the distance 1 is excluded by suitable colouring of the boundary points).

== Virtual edge ==

''Warning: other definitions have been proposed and the exact definition of this notion is currently under discussion. The clamping of graph G in this section may be interpreted as the devirtualization of all bichromatic virtual edges with length <math>d</math> and chromatic number <math>4</math>.''

A '''virtual edge''' of a unit distance graph <math>G</math> is a distance <math>d</math> with the property that every 4-coloring of <math>G</math> contains a monochromatic pair of vertices of distance exactly <math>d</math>. Observe that if a unit distance graph <math>G</math> has a virtual edge at distance <math>d</math>, and if there is another unit distance graph <math>H</math> with a pair of vertices at distance <math>d</math> that cannot be monochromatic in a 4-coloring, then we can create a non-4-colorable unit distance graph by "clamping" a copy of <math>H</math> to every virtual edge in <math>G</math>.

Known examples of virtual edges include:

* <math>G_{40}</math> has a virtual edge at distance <math>\sqrt{11/3}</math>.
* <math>V \oplus V \oplus H</math> has a (single) virtual edge at distance <math>8/3</math>.
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4117 <math>(\sqrt{3}\pm 1)/\sqrt{2}</math> are virtual edges of some graphs].

== Bichromatic virtual edge ==

===Definition===
If a unit distance graph <math>H</math> exists with a specific pair of vertices which are distance <math>d</math> apart and is bichromatic in all proper <math>n</math>-colorings of <math>H</math>, then we say that <math>H</math> virtualizes a '''bichromatic virtual edge''' with length <math>d</math> and chromatic number <math>n</math>.

===Vacuous properties===
Considering the graph of the entire plane, all distances correspond to a virtual edge of chromatic number <math>CNP-1</math>.

If a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n</math> exists, then a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n-1</math> exists.

===Convenience and devirtualization===
Bichromatic virtual edges behave the same as the unit edge(which is a bichromatic virtual edge of every chromatic number), and thus may be used to simplify the construction of graphs.

Recursively replacing bichromatic virtual edges of a graph <math>G</math> with graphs which virtualize the bichromatic virtual edges through '''clamping''' produces a unit distance graph <math>G'</math> where <math>\chi(G')\geq\chi(G)</math>.

Devirtualizing bichromatic virtual edges of a graph <math>G</math> (i.e. clamping) has no benefit unless nontrivial points of <math>G</math> and <math>H_0</math> overlap or nontrivial points of <math>H_1</math> and <math>H_0</math> overlap, where <math>H_0</math> and <math>H_1</math> are graphs virtualizing bichromatic virtual edges of <math>G</math>. Such overlaps may cause <math>\chi(G')>\chi(G)</math>. If no nontrivial overlaps exist, then <math>\chi(G')=\max\left(\chi(G),\chi(H_0),\chi(H_1),\dots\right)</math>.

Because more than 1 graph virtualizes any given bichromatic virtual edge for a fixed length and fixed chromatic number, multiple devirtualizations exist for every finite graph.

===Relation to rings===
A '''devirtualized ring''' of graph <math>G</math> is a ring which contains all the vertices of a devirtualization of <math>G</math>, where the vertices are interpreted as complex numbers.

Use the implied shorthand for arbitrary sets <math>\mathbb{S}[\left\{a,b,c,\dots\right\}]=\mathbb{S}[a,b,c,\dots]</math>.

Let the vertices of graph <math>G</math> generate the ring <math>S_0[T_0]</math>.
Let the bichromatic virtual edges be virtualizable by the graphs <math>H_1,\dots,H_m</math>.
Let <math>S_k[T_k]</math> be the ring generated by the vertices of <math>H_k</math>.
Let <math>S</math> be the ring generated by <math>\bigcup_{k=0}^m S_k</math>.
Let <math>T=\bigcup_{k=0}^m T_k</math>.
Then <math>S[T]</math> is a devirtualized ring of <math>G</math>.

As multiple graphs may virtualize the same virtual edge, careful choice may be required to minimize the set <math>T</math>.

===Multiplicative property===
Let <math>H_0</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_0</math> and chromatic number <math>n</math>.
Let <math>H_1</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_1</math> and chromatic number <math>n</math>.
Create a graph <math>H_2</math> by scaling <math>H_0</math> by a factor of <math>d_1</math>. Graph <math>H_2</math> then virtualizes a bichromatic virtual edge length <math>d_0d_1</math> and chromatic number <math>n</math>.

===Notable bichromatic virtual edges===
{| border=1
|-
! Chromatic number !! Length <math>d</math>!! Proof
|-
| any
| 1
| trivial case
|-
| 2
| <math>0\leq d\leq 3</math>
| moving the flexible parts of the graph (0,0),(0,1),(0,2),(0,3)
|-
| 3
| <math>\left\{\left|\frac{(3+\sqrt{-3})k}{2}+\sqrt{-3}j+(-1)^r\right| \mid k,j\in\mathbb{Z}, r\in\mathbb{R}\right\}</math>
| equilateral triangle tiling
|}

===Continuous ranges of bichromatic virtual edges===
Flexible graphs might produce continuous ranges of lengths of bichromatic virtual edges of chromatic number <math>n</math> without having a specific pair which is monochrome for every <math>n</math>-coloring.
Let the range of lengths be on the interval <math>d_0\leq d\leq d_1</math>.

If <math>1 < d_1</math>, then let <math>k\in\mathbb{Z}^+,d_0^{k+1} \leq d_1^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all lengths larger than <math>d_0^k</math>. A finite area allowed to be the same color, the infinite area of the plane, and a finite upper bound on CNP implies <math>CNP> n</math>.

If <math>d_0 < 1</math>, then let <math>k\in\mathbb{Z}^+,d_1^{k+1} \geq d_0^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all non-zero lengths smaller than <math>d_1^k</math>. Considering a set of <math>CNP+1</math> points all within distance <math>d_1^k</math> of each other implies <math>CNP> n</math>.

Using a flexible bichromatic virtual edge of chromatic number <math>n</math> and a graph <math>H</math> of chromatic number <math>n+1</math>, a flexible a graph <math>H'</math> of chromatic number <math>n+1</math> can be created by scaling <math>H</math> to some arbitrarily large or arbitrarily small size and clamping flexible bichromatic virtual edges of chromatic number <math>n</math> as a replacement of each rigid bichromatic virtual edge of <math>H</math>.

<math>H</math> virtualizing a bichromatic virtual edge of chromatic number <math>n+1</math> does not necessarily men the graph <math>H'</math> virtualizes a bichromatic virtual edge.

== Probabilistic formulation ==

See [[Probabilistic formulation of Hadwiger-Nelson problem]].

== Further questions ==

* What are the [http://mathworld.wolfram.com/IndependenceRatio.html independence ratios] of the above unit distance graphs?
* What are the [https://en.wikipedia.org/wiki/Fractional_coloring fractional chromatic numbers] of these graphs?
* What are the [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz numbers] of these graphs?
** The Lovasz theta function value of Lovasz number of <math>G_2</math> at the complement [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3844 is in the interval [3.3746, 3.3748]].
* What about the Erdos unit distance graph (<math>n</math> vertices, <math>n^{1+c/\log\log n}</math> edges)?
* Can we use de Grey’s graph to construct unit-distance graphs that are not 5-colorable? To answer this question, we first need to understand how 5-colorings of de Grey’s graph force small collections of vertices to be colored. [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3899 Varga] and [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3940 Nazgand] provide some thoughts along these lines. Even if we can’t stitch together a 6-chromatic unit-distance graph with these ideas, we might be able to apply them to [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3900 prove that the measurable chromatic number of the plane is at least 6].
* It appears as though the coordinates of our smallest 5-chromatic graph lie in <math>\mathbb{Q}[\sqrt{3}, \sqrt{5}, \sqrt{11}]</math> (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3879 this]). If we view the plane as the complex plane, what is the smallest ring that admits a 5-chromatic single-distance graph? Every single-distance graph in the Eistenstein integers and Gaussian integers is 2-chromatic (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3914 this]). [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3934 David Speyer suggests] looking at <math>\mathbb{Z}[\frac{1+\sqrt{-71}}{2}]</math> next.

== Blog, forums, and media ==

* [https://rjlipton.wordpress.com/2011/05/22/more-on-coloring-the-plane/ More On Coloring The Plane], Richard Lipton, May 22, 2011.
* [https://mathoverflow.net/questions/236392/has-there-been-a-computer-search-for-a-5-chromatic-unit-distance-graph Has there been a computer search for a 5-chromatic unit distance graph?], Juno, Apr 16, 2016.
* [https://quomodocumque.wordpress.com/2018/04/09/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Jordan Ellenberg, Apr 9 2018.
* [https://gilkalai.wordpress.com/2018/04/10/aubrey-de-grey-the-chromatic-number-of-the-plane-is-at-least-5/ Aubrey de Grey: The chromatic number of the plane is at least 5], Gil Kalai, Apr 10 2018.
* [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Dustin Mixon, Apr 10, 2018.
* [https://www.scottaaronson.com/blog/?p=3697 Amazing progress on long-standing problems], Scott Aaronson, Apr 11 2018.
* [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ The chromatic number of the plane is at least 5, Part II], Dustin Mixon, Apr 13 2018.
* [http://community.wolfram.com/groups/-/m/t/1320004 A 5-chromatic unit distance graph], Ed Pegg, Apr 13 2018.
* [http://aperiodical.com/2018/04/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Katie Steckles, Apr 17 2018.
* [https://www.quantamagazine.org/decades-old-graph-problem-yields-to-amateur-mathematician-20180417/ Decades-Old Graph Problem Yields to Amateur Mathematician], Evelyn Lamb, Quanta, Apr 17, 2018.
* [https://www.heise.de/newsticker/meldung/Zahlen-bitte-Wie-bunt-ist-die-Ebene-4024574.html Zahlen, bitte! 5 - Wie bunt ist die Ebene?], Harald Bögeholz, Heise, Apr 17, 2018.
* [http://www.sciencemag.org/news/2018/04/amateur-mathematician-cracks-decades-old-math-problem Amateur mathematician cracks decades-old math problem], Katie Langin, Science News, Apr 18, 2018.
* [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable How much of the plane is 4-colorable?], Dustin Mixon, Apr 18, 2018.
* [https://www.sciencealert.com/amateur-solves-decades-old-maths-problem-about-colours-that-can-never-touch-hadwiger-nelson-problem An Amateur Solved a 60-Year-Old Maths Problem About Colours That Can Never Touch], Peter Dockrill, ScienceAlert, Apr 19, 2018.
* [https://www.zmescience.com/science/math/amateur-mathematician-solves-problem-20042018/ Amateur mathematician Aubrey de Grey, known for his work on anti-aging, solves decades-old problem], Mihai Andrei, ZME Science, Apr 20, 2018.
* [https://automaths.blog/2018/04/21/5-nuances-daubrey-de-grey/ 5 nuances d’Aubrey de Grey], Automaths, Apr 21, 2018.
* [https://science.howstuffworks.com/math-concepts/amateur-solves-part-of-decades-old-math-problem.htm Amateur Solves Part of Decades-Old Math Problem], HowStuffWorks, Apr 30, 2018.
* [https://www.nemokennislink.nl/publicaties/het-platte-vlak-heeft-minstens-vijf-kleuren-nodig/ Het platte vlak heeft minstens vijf kleuren nodig], Kennislink, May 3, 2018.

== Code and data ==

[https://www.dropbox.com/sh/ufknm1v9gtbhad3/AACB2xwaXYx5EGda38_L-0foa?dl=0 This dropbox folder] will contain most of the data and images for the project.

Data:

* [https://www.dropbox.com/s/x6kvu7b3wvdvqsn/graph.dimacs?dl=0 The 1585-vertex graph in DIMACS format]
* [https://www.dropbox.com/s/qk3gbpjvvsjfsc3/sat.dimacs?dl=0 A naive translation of 4-colorability of this graph into a SAT problem in DIMACS format]
* [https://www.dropbox.com/s/nipqikfzcfn9o5a/vertices.sage?dl=0 The vertices of this graph in explicit Sage notation]
* The graph <math>G_2</math>: [http://www.cs.utexas.edu/~marijn/CNP/874.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/874.edge edges (DIMACS)]
* The graph <math>G_3</math>: [http://www.cs.utexas.edu/~marijn/CNP/826.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/826.edge edges (DIMACS)] [http://www.cs.utexas.edu/~marijn/CNP/826.pdf Visualization]
* The [https://www.dropbox.com/sh/ufknm1v9gtbhad3/AADfw9WO1ol2ayQNJEtGf8oBa/Graphs?dl=0 densest unit-distance graphs] on an <math>n\times n</math> grid for <math>n=10,20,\ldots,100</math> (DIMACS format).

Code:

* [https://www.dropbox.com/s/6u1jctbjy38t383/lovaszmoser.m?dl=0 MATLAB script for computing Lovasz number]
* [https://files.jixco.de/pm16/zzvtx/ Python code for converting a list of vertices in Mathematica format into vertices in Z^n]
* [https://files.jixco.de/pm16/edge2cnf.py Python code for converting a DIMACS edge list into a CNF formula (forcing up to 3 suitable vertices to a fixed color, to break some symmetries)]

Software:

* [http://cvxr.com/cvx/ CVX]
* [http://www.labri.fr/perso/lsimon/glucose/ Glucose 4.0]
* [http://minisat.se/ Minisat]

== Wikipedia ==

* [https://en.wikipedia.org/wiki/Graph_coloring#Chromatic_number Chromatic number]
* [https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Erd%C5%91s_theorem_(graph_theory) de Bruijn-Erdos theorem]
* [https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem Hadwiger-Nelson problem]
* [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz number]
* [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]

== Bibliography ==
* [B2008] B. Bukh, [https://doi.org/10.1007/s00039-008-0673-8 Measurable sets with excluded distances], Geometric and Functional Analysis 18 (2008), 668-697.
* [CR2015] D. Cranston, L. Rabern, [https://arxiv.org/abs/1501.01647 The fractional chromatic number of the plane], arXiv:1501.01647
* [deBE1951] N. G. de Bruijn, P. Erdős, [http://www.math-inst.hu/~p_erdos/1951-01.pdf A colour problem for infinite graphs and a problem in the theory of relations], Nederl. Akad. Wetensch. Proc. Ser. A, 54 (1951): 371–373, MR 0046630.
* [deG2018] A. de Grey, [https://arxiv.org/abs/1804.02385 The chromatic number of the plane is at least 5], arXiv:1804.02385
* [EI2018] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.00157 The chromatic number of the plane is at least 5 - a new proof], arXiv:1805.00157
* [F1981] K.J. Falconer, The Realization of distances in measurable subsets covering Rn, J. Combin. Theory Ser. A 31 (1981) 184–189.
* [H1945] H. Hadwiger, Uberdeckung des euklidischen Raum durch kongruente Mengen, Portugaliae Math. 4 (1945), 238–242.
* [MM1961] L. Moser and M. Moser, Solution to Problem 10, Can. Math. Bull. 4 (1961), 187–189.
* [P1998] D. Pritikin, [https://www.sciencedirect.com/science/article/pii/S0095895698918196 All unit-distance graphs of order 6197 are 6-colorable], Journal of Combinatorial Theory, Series B 73.2 (1998): 159-163.
* [S2008] A. Soifer, The Mathematical Coloring Book, Springer, 2008, ISBN-13: 978-0387746401.
* [T1999] C. Thomassen, On the Nelson unit distance coloring problem, Amer. Math. Monthly 106 (1999) 850-853.
* [Tow2005] Townsend, S.P., Colouring the plane with no monochrome unit. Geombinatorics XIV(4) (2005), 181-193.
* [W1973] Woodall, D.R., Distances realized by sets covering the plane. J. Combin. Theory Ser. A, 14 (1973), 187-200.

Hadwiger-Nelson problem

2018-05-05T09:53:06Z

Nazgand: /* Notable unit distance graphs */

The Chromatic Number of the Plane (CNP) is the chromatic number of the graph whose vertices are elements of the plane, and two points are connected by an edge if they are a unit distance apart. The Hadwiger-Nelson problem asks to compute CNP. The bounds <math>4 \leq CNP \leq 7</math> are classical; recently [deG2018] it was shown that <math>CNP \geq 5</math>. This is achieved by explicitly locating finite unit distance graphs with chromatic number at least 5.

The '''Polymath16''' project seeks to simplify the graphs used in [deG2018] to establish this lower bound. More precisely, the goals are

* '''Goal 1''': Find progressively smaller 5-chromatic unit-distance graphs.
* '''Goal 2''': Reduce (ideally to zero) the reliance on computer assistance for the proof. Computer assistance was leveraged in [deG2018] to analyze a subgraph of size 397.
* '''Goal 3''': Apply these simpler graphs to inform progress in related areas. For example:
** Find a 6-chromatic unit-distance graph in the plane.
** Improve the corresponding bound in higher dimensions.
** Improve the current record of 383/102 for the fractional chromatic number of the plane.
** Find the smallest unit-distance graph of a given minimum degree (excluding, in some natural way, boring cases like Cartesian products of a graph with a hypercube).

== Polymath threads ==

* [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/ Polymath proposal: finding simpler unit distance graphs of chromatic number 5], Aubrey de Grey, Apr 10 2018. ('''Active discussion thread''')
* [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/ Polymath16, first thread: Simplifying de Grey’s graph], Dustin Mixon, Apr 14, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic Polymath16, second thread: What does it take to be 5-chromatic?], Dustin Mixon, Apr 22, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/ Polymath16, third thread: Is 6-chromatic within reach?], Dustin Mixon, May 1, 2018. ('''Active research thread''')

== Notable unit distance graphs ==

A '''unit distance graph''' is a graph that can be realised as a collection of vertices in the plane, with two vertices connected by an edge if they are precisely a unit distance apart. The chromatic number of any such graph is a lower bound for <math>CNP</math>; in particular, if one can find a unit distance graph with no 4-colorings, then <math>CNP \geq 5</math>. The boldface number of vertices is the current minimal number of vertices of a unit distance graph that is currently known to not be 4-colorable.

<math>G_1 \oplus G_2</math> denotes the Minkowski sum of two unit distance graphs <math>G_1,G_2</math> (vertices in <math>G_1 \oplus G_2</math> are sums of the vertices of <math>G_1,G_2</math>). <math>G_1 \cup G_2</math> denotes the union. <math>\mathrm{rot}(G, \theta)</math> denotes <math>G</math> rotated counterclockwise by <math>\theta</math>. <math>\mathrm{trim}(G,r)</math> denotes the trimming of <math>G</math> after removing all vertices of distance greater than <math>r</math> from the origin.

Another basic operation is '''spindling''': taking two copies of a graph <math>G</math>, gluing them together at one vertex, and rotating the copies so that the two copies of another vertex are a unit distance apart. For instance, the Moser spindle is the spindling of a rhombus graph. If a graph has two vertices forced to be the same color in a k-coloring, then the spindling of that graph at those two vertices is not k-colorable.

Many of the graphs are embeddable in abelian groups or rings, particularly those generated by the roots of unity <math>\omega_t := \exp( i \arccos( 1 - \tfrac{1}{2t} ))</math> for various natural numbers t.

{| border=1
|-
!Name!!Number of vertices!! Number of edges !! Structure !! Group !! Colorings
|-
| [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]
| 7
| 11
| Two 60-120-60-120 rhombi with a common vertex, with one pair of sharp vertices coincident and the other joined
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [http://mathworld.wolfram.com/GolombGraph.html Golomb graph]
| 10
| 18
| Contains the center and vertices of a hexagon and equilateral triangle
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [https://arxiv.org/abs/1804.02385 H]
| 7
| 12
| Vertices and center of a hexagon
| <math>{\bf Z}[\omega_1]</math>
| Has essentially four 4-colorings, two of which contain a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 J]
| 31
| 72
| Contains 13 copies of H
| <math>{\bf Z}[\omega_1]</math>
| Has essentially six 4-colorings in which no H has a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 K]
| 61
| 150
| Contains 2 copies of J
|
| In all 4-colorings lacking an H with a monochromatic <math>\sqrt{3}</math>-triangle, all pairs of vertices at distance 4 are monochromatic
|-
| [https://arxiv.org/abs/1804.02385 L]
| 121
| 301
| Contains two copies of K and 52 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ <math>L_1</math>]
| 97
|
| Has 40 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120295 <math>L_2</math>]
| 120
| 354
|
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 T]
| 9
| 15
| Contains one Moser spindle and useful symmetry; three vertices form an equilateral triangle
|
|
|-
| [https://arxiv.org/abs/1804.02385 U]
| 15
| 33
| Three copies of T at 120-degree rotations: <math>T \cup \mathrm{rot}(T, 2\pi/3) \cup \mathrm{rot}(T, 4\pi/3)</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 V]
| 31
| 30
| Unit vectors at angles consistent with three interlocking Moser spindles
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>V_1</math>]
| 61
| 60
| Union of V and a rotation of V: <math>V \cup \mathrm{rot}(V, \mathrm{arccos}(7/8))</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_b</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_x</math>]
| 13
| 12
| Subgraph of <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_z</math>]
|
|
| Subgraph of <math>V_a</math>; shares a line of symmetry with <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_y</math>]
| 13
| 12
| Subgraph of <math>V_b</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-3970 <math>V_A</math>]
| 37
| 36
| Unit vectors with angles <math>i \frac{\pi}{3} + j \mathrm{arccos} \frac{5}{6} + k \mathrm{arccos} \frac{7}{8}</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 W]
| 301
| 1230
| Cartesian product of V with itself, minus vertices at more than <math>\sqrt{3}</math> from the centre (i.e. <math>\mathrm{trim}(V \oplus V, \sqrt{3})</math>)
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>W_1</math>]
|
|
| Trimmed product of V with itself (<math>\mathrm{trim}(V \oplus V, 1.95)</math>)
|
|
|-
| [https://arxiv.org/abs/1804.02385 M]
| 1345
| 8268
| Cartesian product of W and H (<math>W \oplus H</math>)
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120274 <math>M_1</math>]
| 278
|
| Deleting vertices from M while maintaining its restriction on H
|
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>M_2</math>]
| 7075
|
| Sum of H with a trimmed product of <math>V_1</math> with itself
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 N]
| 20425
| 151311
| Contains 52 copies of M arranged around the H-copies of L
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4, \omega_{16}]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_0</math>]
| 1585
| 7909
| N "shrunk" by stepwise deletions and replacements of vertices
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 G]
| 1581
| 7877
| Deleting 4 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_1</math>]
| 1577
|
| Deleting 8 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120318 <math>G_2</math>]
| 874
| 4461
| Juxtaposing two copies of M and shrinking
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3867 <math>G_3</math>]
| 826
| 4273
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4105 <math>G_4</math>]
| '''803'''
| 4144
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 R]
|
|
| Union of <math>W_1</math> and a rotated copy of <math>W_1</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>\mathrm{trim}(R \oplus H, 1.67)</math>]
| 2563
|
| Trimmed sum of R and H
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 <math>V \oplus V \oplus H</math>]
|
|
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_{64/9}]</math>
| Has two vertices forced to be the same color in a 4-coloring; also no monochromatic <math>\sqrt{3}</math>-triangles
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 No name assigned]
| 745
|
| Subgraph of <math>V \oplus V \oplus H</math>
|
| Has two vertices forced to be the same color in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3085
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_a \oplus V_z \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3049
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 No name assigned]
| 1951
|
| Trimmed version of <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic <math>V_A \oplus V_A \oplus V_A</math>]
| 6937
| 44439
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{40}</math>]
| 40
| 82
|
|
| Any 4-coloring that avoids a monochromatic <math>\sqrt{11/3}</math>-edge has two specific vertices forced to be the same color
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{79}</math>]
| 79
| 165
| Spindling of <math>G_{40}</math>
|
| Any 4-coloring has a monochromatic <math>\sqrt{11/3}</math>-edge
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{49}</math>]
| 49
| 180
|
| <math>{\mathbf Q}[\sqrt{3},\sqrt{11}]^2</math>
| Any 4-coloring either has a specific <math>\sqrt{11/3}</math>-edge monochromatic, or a monochromatic <math>1/\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{51}</math>]
| 51
|
|
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring of plane
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{627}</math>]
| 627
| 2982
| Contains <math>G_{51}</math>
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4161 No name assigned]
| 103
|
|
|
| All 4-colorings contain a monochromatic <math>2/\sqrt{3}</math>-edge
|-
| regular pentagon with unit side length
| 5
| 5
|
|
| All 4-colorings contain a monochromatic <math>(\sqrt{5}+1)/2</math>-edge
|-
| regular pentagon with <math>(\sqrt{5}-1)/2</math> side length
| 5
| 5
|
|
| All 4-colorings contain a monochromatic <math>(\sqrt{5}-1)/2</math>-edge
|}

== Lower bounds ==

In [P1988], Pritikin proved that every graph with at most 12 vertices is 4-colorable, and every graph with at most 6197 vertices is 6-colorable. Pritikin's bounds are obtained by coloring the plane with k colors and an additional “wild” color such that points of unit distance are both allowed to receive the wild color. If the wild color occupies a small fraction p of the plane, then an exercise in the probabilistic method gives that any fixed unit-distance graph with n vertices enjoys an embedding in the plane that avoids the wild color (and is therefore k-colorable) provided n<1/p. Pritikin’s coloring for the k=4 case cannot be improved without improving on the densest known subset of the plane that avoids unit distances (originally due to Croft in 1967). See [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable this MO thread] for additional information. As such, improving the bound in the k=4 case might require a new technique, whereas the k=5 and 6 cases might be amenable to optimization.

[https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4176 Every unit distance graph with at most 16 vertices is 5-colorable].

A "tile-based" colouring of the plane must have at least 6 colours, as shown by Townsend [Tow2005]; the same proof with a minor error was also derived by Woodall [W1973]. In [T1999], Thomassen showed that any tiling-based 6-coloring would have to be be "unscaleable", i.e. the maximum diameter of a tile and the minimum separation of same-coloured tiles must both be exactly 1 (so that the distance 1 is excluded by suitable colouring of the boundary points).

== Virtual edge ==

''Warning: other definitions have been proposed and the exact definition of this notion is currently under discussion. The clamping of graph G in this section may be interpreted as the devirtualization of all bichromatic virtual edges with length <math>d</math> and chromatic number <math>4</math>.''

A '''virtual edge''' of a unit distance graph <math>G</math> is a distance <math>d</math> with the property that every 4-coloring of <math>G</math> contains a monochromatic pair of vertices of distance exactly <math>d</math>. Observe that if a unit distance graph <math>G</math> has a virtual edge at distance <math>d</math>, and if there is another unit distance graph <math>H</math> with a pair of vertices at distance <math>d</math> that cannot be monochromatic in a 4-coloring, then we can create a non-4-colorable unit distance graph by "clamping" a copy of <math>H</math> to every virtual edge in <math>G</math>.

Known examples of virtual edges include:

* <math>G_{40}</math> has a virtual edge at distance <math>\sqrt{11/3}</math>.
* <math>V \oplus V \oplus H</math> has a (single) virtual edge at distance <math>8/3</math>.
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4117 <math>(\sqrt{3}\pm 1)/\sqrt{2}</math> are virtual edges of some graphs].

== Bichromatic virtual edge ==

===Definition===
If a unit distance graph <math>H</math> exists with a specific pair of vertices which are distance <math>d</math> apart and is bichromatic in all proper <math>n</math>-colorings of <math>H</math>, then we say that <math>H</math> virtualizes a '''bichromatic virtual edge''' with length <math>d</math> and chromatic number <math>n</math>.

===Vacuous properties===
Considering the graph of the entire plane, all distances correspond to a virtual edge of chromatic number <math>CNP-1</math>.

If a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n</math> exists, then a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n-1</math> exists.

===Convenience and devirtualization===
Bichromatic virtual edges behave the same as the unit edge(which is a bichromatic virtual edge of every chromatic number), and thus may be used to simplify the construction of graphs.

Recursively replacing bichromatic virtual edges of a graph <math>G</math> with graphs which virtualize the bichromatic virtual edges through '''clamping''' produces a unit distance graph <math>G'</math> where <math>\chi(G')\geq\chi(G)</math>.

Devirtualizing bichromatic virtual edges of a graph <math>G</math> (i.e. clamping) has no benefit unless nontrivial points of <math>G</math> and <math>H_0</math> overlap or nontrivial points of <math>H_1</math> and <math>H_0</math> overlap, where <math>H_0</math> and <math>H_1</math> are graphs virtualizing bichromatic virtual edges of <math>G</math>. Such overlaps may cause <math>\chi(G')>\chi(G)</math>. If no nontrivial overlaps exist, then <math>\chi(G')=\max\left(\chi(G),\chi(H_0),\chi(H_1),\dots\right)</math>.

Because more than 1 graph virtualizes any given bichromatic virtual edge for a fixed length and fixed chromatic number, multiple devirtualizations exist for every finite graph.

===Relation to rings===
A '''devirtualized ring''' of graph <math>G</math> is a ring which contains all the vertices of a devirtualization of <math>G</math>, where the vertices are interpreted as complex numbers.

Use the implied shorthand for arbitrary sets <math>\mathbb{S}[\left\{a,b,c,\dots\right\}]=\mathbb{S}[a,b,c,\dots]</math>.

Let the vertices of graph <math>G</math> generate the ring <math>S_0[T_0]</math>.
Let the bichromatic virtual edges be virtualizable by the graphs <math>H_1,\dots,H_m</math>.
Let <math>S_k[T_k]</math> be the ring generated by the vertices of <math>H_k</math>.
Let <math>S</math> be the ring generated by <math>\bigcup_{k=0}^m S_k</math>.
Let <math>T=\bigcup_{k=0}^m T_k</math>.
Then <math>S[T]</math> is a devirtualized ring of <math>G</math>.

As multiple graphs may virtualize the same virtual edge, careful choice may be required to minimize the set <math>T</math>.

===Multiplicative property===
Let <math>H_0</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_0</math> and chromatic number <math>n</math>.
Let <math>H_1</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_1</math> and chromatic number <math>n</math>.
Create a graph <math>H_2</math> by scaling <math>H_0</math> by a factor of <math>d_1</math>. Graph <math>H_2</math> then virtualizes a bichromatic virtual edge length <math>d_0d_1</math> and chromatic number <math>n</math>.

===Notable bichromatic virtual edges===
{| border=1
|-
! Chromatic number !! Length <math>d</math>!! Proof
|-
| any
| 1
| trivial case
|-
| 2
| <math>0\leq d\leq 3</math>
| moving the flexible parts of the graph (0,0),(0,1),(0,2),(0,3)
|-
| 3
| <math>\left\{\left|\frac{(3+\sqrt{-3})k}{2}+\sqrt{-3}j+(-1)^r\right| \mid k,j\in\mathbb{Z}, r\in\mathbb{R}\right\}</math>
| equilateral triangle tiling
|}

===Continuous ranges of bichromatic virtual edges===
Flexible graphs might produce continuous ranges of lengths of bichromatic virtual edges of chromatic number <math>n</math> without having a specific pair which is monochrome for every <math>n</math>-coloring.
Let the range of lengths be on the interval <math>d_0\leq d\leq d_1</math>.

If <math>1 < d_1</math>, then let <math>k\in\mathbb{Z}^+,d_0^{k+1} \leq d_1^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all lengths larger than <math>d_0^k</math>. A finite area allowed to be the same color, the infinite area of the plane, and a finite upper bound on CNP implies <math>CNP> n</math>.

If <math>d_0 < 1</math>, then let <math>k\in\mathbb{Z}^+,d_1^{k+1} \geq d_0^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all non-zero lengths smaller than <math>d_1^k</math>. Considering a set of <math>CNP+1</math> points all within distance <math>d_1^k</math> of each other implies <math>CNP> n</math>.

Using a flexible bichromatic virtual edge of chromatic number <math>n</math> and a graph <math>H</math> of chromatic number <math>n+1</math>, a flexible a graph <math>H'</math> of chromatic number <math>n+1</math> can be created by scaling <math>H</math> to some arbitrarily large or arbitrarily small size and clamping flexible bichromatic virtual edges of chromatic number <math>n</math> as a replacement of each rigid bichromatic virtual edge of <math>H</math>.

<math>H</math> virtualizing a bichromatic virtual edge of chromatic number <math>n+1</math> does not necessarily men the graph <math>H'</math> virtualizes a bichromatic virtual edge.

== Probabilistic formulation ==

See [[Probabilistic formulation of Hadwiger-Nelson problem]].

== Further questions ==

* What are the [http://mathworld.wolfram.com/IndependenceRatio.html independence ratios] of the above unit distance graphs?
* What are the [https://en.wikipedia.org/wiki/Fractional_coloring fractional chromatic numbers] of these graphs?
* What are the [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz numbers] of these graphs?
** The Lovasz theta function value of Lovasz number of <math>G_2</math> at the complement [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3844 is in the interval [3.3746, 3.3748]].
* What about the Erdos unit distance graph (<math>n</math> vertices, <math>n^{1+c/\log\log n}</math> edges)?
* Can we use de Grey’s graph to construct unit-distance graphs that are not 5-colorable? To answer this question, we first need to understand how 5-colorings of de Grey’s graph force small collections of vertices to be colored. [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3899 Varga] and [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3940 Nazgand] provide some thoughts along these lines. Even if we can’t stitch together a 6-chromatic unit-distance graph with these ideas, we might be able to apply them to [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3900 prove that the measurable chromatic number of the plane is at least 6].
* It appears as though the coordinates of our smallest 5-chromatic graph lie in <math>\mathbb{Q}[\sqrt{3}, \sqrt{5}, \sqrt{11}]</math> (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3879 this]). If we view the plane as the complex plane, what is the smallest ring that admits a 5-chromatic single-distance graph? Every single-distance graph in the Eistenstein integers and Gaussian integers is 2-chromatic (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3914 this]). [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3934 David Speyer suggests] looking at <math>\mathbb{Z}[\frac{1+\sqrt{-71}}{2}]</math> next.

== Blog, forums, and media ==

* [https://rjlipton.wordpress.com/2011/05/22/more-on-coloring-the-plane/ More On Coloring The Plane], Richard Lipton, May 22, 2011.
* [https://mathoverflow.net/questions/236392/has-there-been-a-computer-search-for-a-5-chromatic-unit-distance-graph Has there been a computer search for a 5-chromatic unit distance graph?], Juno, Apr 16, 2016.
* [https://quomodocumque.wordpress.com/2018/04/09/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Jordan Ellenberg, Apr 9 2018.
* [https://gilkalai.wordpress.com/2018/04/10/aubrey-de-grey-the-chromatic-number-of-the-plane-is-at-least-5/ Aubrey de Grey: The chromatic number of the plane is at least 5], Gil Kalai, Apr 10 2018.
* [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Dustin Mixon, Apr 10, 2018.
* [https://www.scottaaronson.com/blog/?p=3697 Amazing progress on long-standing problems], Scott Aaronson, Apr 11 2018.
* [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ The chromatic number of the plane is at least 5, Part II], Dustin Mixon, Apr 13 2018.
* [http://community.wolfram.com/groups/-/m/t/1320004 A 5-chromatic unit distance graph], Ed Pegg, Apr 13 2018.
* [http://aperiodical.com/2018/04/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Katie Steckles, Apr 17 2018.
* [https://www.quantamagazine.org/decades-old-graph-problem-yields-to-amateur-mathematician-20180417/ Decades-Old Graph Problem Yields to Amateur Mathematician], Evelyn Lamb, Quanta, Apr 17, 2018.
* [https://www.heise.de/newsticker/meldung/Zahlen-bitte-Wie-bunt-ist-die-Ebene-4024574.html Zahlen, bitte! 5 - Wie bunt ist die Ebene?], Harald Bögeholz, Heise, Apr 17, 2018.
* [http://www.sciencemag.org/news/2018/04/amateur-mathematician-cracks-decades-old-math-problem Amateur mathematician cracks decades-old math problem], Katie Langin, Science News, Apr 18, 2018.
* [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable How much of the plane is 4-colorable?], Dustin Mixon, Apr 18, 2018.
* [https://www.sciencealert.com/amateur-solves-decades-old-maths-problem-about-colours-that-can-never-touch-hadwiger-nelson-problem An Amateur Solved a 60-Year-Old Maths Problem About Colours That Can Never Touch], Peter Dockrill, ScienceAlert, Apr 19, 2018.
* [https://www.zmescience.com/science/math/amateur-mathematician-solves-problem-20042018/ Amateur mathematician Aubrey de Grey, known for his work on anti-aging, solves decades-old problem], Mihai Andrei, ZME Science, Apr 20, 2018.
* [https://automaths.blog/2018/04/21/5-nuances-daubrey-de-grey/ 5 nuances d’Aubrey de Grey], Automaths, Apr 21, 2018.
* [https://science.howstuffworks.com/math-concepts/amateur-solves-part-of-decades-old-math-problem.htm Amateur Solves Part of Decades-Old Math Problem], HowStuffWorks, Apr 30, 2018.
* [https://www.nemokennislink.nl/publicaties/het-platte-vlak-heeft-minstens-vijf-kleuren-nodig/ Het platte vlak heeft minstens vijf kleuren nodig], Kennislink, May 3, 2018.

== Code and data ==

[https://www.dropbox.com/sh/ufknm1v9gtbhad3/AACB2xwaXYx5EGda38_L-0foa?dl=0 This dropbox folder] will contain most of the data and images for the project.

Data:

* [https://www.dropbox.com/s/x6kvu7b3wvdvqsn/graph.dimacs?dl=0 The 1585-vertex graph in DIMACS format]
* [https://www.dropbox.com/s/qk3gbpjvvsjfsc3/sat.dimacs?dl=0 A naive translation of 4-colorability of this graph into a SAT problem in DIMACS format]
* [https://www.dropbox.com/s/nipqikfzcfn9o5a/vertices.sage?dl=0 The vertices of this graph in explicit Sage notation]
* The graph <math>G_2</math>: [http://www.cs.utexas.edu/~marijn/CNP/874.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/874.edge edges (DIMACS)]
* The graph <math>G_3</math>: [http://www.cs.utexas.edu/~marijn/CNP/826.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/826.edge edges (DIMACS)] [http://www.cs.utexas.edu/~marijn/CNP/826.pdf Visualization]
* The [https://www.dropbox.com/sh/ufknm1v9gtbhad3/AADfw9WO1ol2ayQNJEtGf8oBa/Graphs?dl=0 densest unit-distance graphs] on an <math>n\times n</math> grid for <math>n=10,20,\ldots,100</math> (DIMACS format).

Code:

* [https://www.dropbox.com/s/6u1jctbjy38t383/lovaszmoser.m?dl=0 MATLAB script for computing Lovasz number]
* [https://files.jixco.de/pm16/zzvtx/ Python code for converting a list of vertices in Mathematica format into vertices in Z^n]
* [https://files.jixco.de/pm16/edge2cnf.py Python code for converting a DIMACS edge list into a CNF formula (forcing up to 3 suitable vertices to a fixed color, to break some symmetries)]

Software:

* [http://cvxr.com/cvx/ CVX]
* [http://www.labri.fr/perso/lsimon/glucose/ Glucose 4.0]
* [http://minisat.se/ Minisat]

== Wikipedia ==

* [https://en.wikipedia.org/wiki/Graph_coloring#Chromatic_number Chromatic number]
* [https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Erd%C5%91s_theorem_(graph_theory) de Bruijn-Erdos theorem]
* [https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem Hadwiger-Nelson problem]
* [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz number]
* [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]

== Bibliography ==
* [B2008] B. Bukh, [https://doi.org/10.1007/s00039-008-0673-8 Measurable sets with excluded distances], Geometric and Functional Analysis 18 (2008), 668-697.
* [CR2015] D. Cranston, L. Rabern, [https://arxiv.org/abs/1501.01647 The fractional chromatic number of the plane], arXiv:1501.01647
* [deBE1951] N. G. de Bruijn, P. Erdős, [http://www.math-inst.hu/~p_erdos/1951-01.pdf A colour problem for infinite graphs and a problem in the theory of relations], Nederl. Akad. Wetensch. Proc. Ser. A, 54 (1951): 371–373, MR 0046630.
* [deG2018] A. de Grey, [https://arxiv.org/abs/1804.02385 The chromatic number of the plane is at least 5], arXiv:1804.02385
* [EI2018] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.00157 The chromatic number of the plane is at least 5 - a new proof], arXiv:1805.00157
* [F1981] K.J. Falconer, The Realization of distances in measurable subsets covering Rn, J. Combin. Theory Ser. A 31 (1981) 184–189.
* [H1945] H. Hadwiger, Uberdeckung des euklidischen Raum durch kongruente Mengen, Portugaliae Math. 4 (1945), 238–242.
* [MM1961] L. Moser and M. Moser, Solution to Problem 10, Can. Math. Bull. 4 (1961), 187–189.
* [P1998] D. Pritikin, [https://www.sciencedirect.com/science/article/pii/S0095895698918196 All unit-distance graphs of order 6197 are 6-colorable], Journal of Combinatorial Theory, Series B 73.2 (1998): 159-163.
* [S2008] A. Soifer, The Mathematical Coloring Book, Springer, 2008, ISBN-13: 978-0387746401.
* [T1999] C. Thomassen, On the Nelson unit distance coloring problem, Amer. Math. Monthly 106 (1999) 850-853.
* [Tow2005] Townsend, S.P., Colouring the plane with no monochrome unit. Geombinatorics XIV(4) (2005), 181-193.
* [W1973] Woodall, D.R., Distances realized by sets covering the plane. J. Combin. Theory Ser. A, 14 (1973), 187-200.

Hadwiger-Nelson problem

2018-05-05T09:47:24Z

Nazgand: /* Continuous ranges of bichromatic virtual edges */

The Chromatic Number of the Plane (CNP) is the chromatic number of the graph whose vertices are elements of the plane, and two points are connected by an edge if they are a unit distance apart. The Hadwiger-Nelson problem asks to compute CNP. The bounds <math>4 \leq CNP \leq 7</math> are classical; recently [deG2018] it was shown that <math>CNP \geq 5</math>. This is achieved by explicitly locating finite unit distance graphs with chromatic number at least 5.

The '''Polymath16''' project seeks to simplify the graphs used in [deG2018] to establish this lower bound. More precisely, the goals are

* '''Goal 1''': Find progressively smaller 5-chromatic unit-distance graphs.
* '''Goal 2''': Reduce (ideally to zero) the reliance on computer assistance for the proof. Computer assistance was leveraged in [deG2018] to analyze a subgraph of size 397.
* '''Goal 3''': Apply these simpler graphs to inform progress in related areas. For example:
** Find a 6-chromatic unit-distance graph in the plane.
** Improve the corresponding bound in higher dimensions.
** Improve the current record of 383/102 for the fractional chromatic number of the plane.
** Find the smallest unit-distance graph of a given minimum degree (excluding, in some natural way, boring cases like Cartesian products of a graph with a hypercube).

== Polymath threads ==

* [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/ Polymath proposal: finding simpler unit distance graphs of chromatic number 5], Aubrey de Grey, Apr 10 2018. ('''Active discussion thread''')
* [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/ Polymath16, first thread: Simplifying de Grey’s graph], Dustin Mixon, Apr 14, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic Polymath16, second thread: What does it take to be 5-chromatic?], Dustin Mixon, Apr 22, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/ Polymath16, third thread: Is 6-chromatic within reach?], Dustin Mixon, May 1, 2018. ('''Active research thread''')

== Notable unit distance graphs ==

A '''unit distance graph''' is a graph that can be realised as a collection of vertices in the plane, with two vertices connected by an edge if they are precisely a unit distance apart. The chromatic number of any such graph is a lower bound for <math>CNP</math>; in particular, if one can find a unit distance graph with no 4-colorings, then <math>CNP \geq 5</math>. The boldface number of vertices is the current minimal number of vertices of a unit distance graph that is currently known to not be 4-colorable.

<math>G_1 \oplus G_2</math> denotes the Minkowski sum of two unit distance graphs <math>G_1,G_2</math> (vertices in <math>G_1 \oplus G_2</math> are sums of the vertices of <math>G_1,G_2</math>). <math>G_1 \cup G_2</math> denotes the union. <math>\mathrm{rot}(G, \theta)</math> denotes <math>G</math> rotated counterclockwise by <math>\theta</math>. <math>\mathrm{trim}(G,r)</math> denotes the trimming of <math>G</math> after removing all vertices of distance greater than <math>r</math> from the origin.

Another basic operation is '''spindling''': taking two copies of a graph <math>G</math>, gluing them together at one vertex, and rotating the copies so that the two copies of another vertex are a unit distance apart. For instance, the Moser spindle is the spindling of a rhombus graph. If a graph has two vertices forced to be the same color in a k-coloring, then the spindling of that graph at those two vertices is not k-colorable.

Many of the graphs are embeddable in abelian groups or rings, particularly those generated by the roots of unity <math>\omega_t := \exp( i \arccos( 1 - \tfrac{1}{2t} ))</math> for various natural numbers t.

{| border=1
|-
!Name!!Number of vertices!! Number of edges !! Structure !! Group !! Colorings
|-
| [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]
| 7
| 11
| Two 60-120-60-120 rhombi with a common vertex, with one pair of sharp vertices coincident and the other joined
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [http://mathworld.wolfram.com/GolombGraph.html Golomb graph]
| 10
| 18
| Contains the center and vertices of a hexagon and equilateral triangle
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [https://arxiv.org/abs/1804.02385 H]
| 7
| 12
| Vertices and center of a hexagon
| <math>{\bf Z}[\omega_1]</math>
| Has essentially four 4-colorings, two of which contain a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 J]
| 31
| 72
| Contains 13 copies of H
| <math>{\bf Z}[\omega_1]</math>
| Has essentially six 4-colorings in which no H has a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 K]
| 61
| 150
| Contains 2 copies of J
|
| In all 4-colorings lacking an H with a monochromatic <math>\sqrt{3}</math>-triangle, all pairs of vertices at distance 4 are monochromatic
|-
| [https://arxiv.org/abs/1804.02385 L]
| 121
| 301
| Contains two copies of K and 52 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ <math>L_1</math>]
| 97
|
| Has 40 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120295 <math>L_2</math>]
| 120
| 354
|
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 T]
| 9
| 15
| Contains one Moser spindle and useful symmetry; three vertices form an equilateral triangle
|
|
|-
| [https://arxiv.org/abs/1804.02385 U]
| 15
| 33
| Three copies of T at 120-degree rotations: <math>T \cup \mathrm{rot}(T, 2\pi/3) \cup \mathrm{rot}(T, 4\pi/3)</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 V]
| 31
| 30
| Unit vectors at angles consistent with three interlocking Moser spindles
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>V_1</math>]
| 61
| 60
| Union of V and a rotation of V: <math>V \cup \mathrm{rot}(V, \mathrm{arccos}(7/8))</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_b</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_x</math>]
| 13
| 12
| Subgraph of <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_z</math>]
|
|
| Subgraph of <math>V_a</math>; shares a line of symmetry with <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_y</math>]
| 13
| 12
| Subgraph of <math>V_b</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-3970 <math>V_A</math>]
| 37
| 36
| Unit vectors with angles <math>i \frac{\pi}{3} + j \mathrm{arccos} \frac{5}{6} + k \mathrm{arccos} \frac{7}{8}</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 W]
| 301
| 1230
| Cartesian product of V with itself, minus vertices at more than <math>\sqrt{3}</math> from the centre (i.e. <math>\mathrm{trim}(V \oplus V, \sqrt{3})</math>)
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>W_1</math>]
|
|
| Trimmed product of V with itself (<math>\mathrm{trim}(V \oplus V, 1.95)</math>)
|
|
|-
| [https://arxiv.org/abs/1804.02385 M]
| 1345
| 8268
| Cartesian product of W and H (<math>W \oplus H</math>)
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120274 <math>M_1</math>]
| 278
|
| Deleting vertices from M while maintaining its restriction on H
|
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>M_2</math>]
| 7075
|
| Sum of H with a trimmed product of <math>V_1</math> with itself
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 N]
| 20425
| 151311
| Contains 52 copies of M arranged around the H-copies of L
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4, \omega_{16}]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_0</math>]
| 1585
| 7909
| N "shrunk" by stepwise deletions and replacements of vertices
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 G]
| 1581
| 7877
| Deleting 4 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_1</math>]
| 1577
|
| Deleting 8 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120318 <math>G_2</math>]
| 874
| 4461
| Juxtaposing two copies of M and shrinking
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3867 <math>G_3</math>]
| 826
| 4273
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4105 <math>G_4</math>]
| '''803'''
| 4144
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 R]
|
|
| Union of <math>W_1</math> and a rotated copy of <math>W_1</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>\mathrm{trim}(R \oplus H, 1.67)</math>]
| 2563
|
| Trimmed sum of R and H
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 <math>V \oplus V \oplus H</math>]
|
|
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_{64/9}]</math>
| Has two vertices forced to be the same color in a 4-coloring; also no monochromatic <math>\sqrt{3}</math>-triangles
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 No name assigned]
| 745
|
| Subgraph of <math>V \oplus V \oplus H</math>
|
| Has two vertices forced to be the same color in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3085
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_a \oplus V_z \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3049
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 No name assigned]
| 1951
|
| Trimmed version of <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic <math>V_A \oplus V_A \oplus V_A</math>]
| 6937
| 44439
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{40}</math>]
| 40
| 82
|
|
| Any 4-coloring that avoids a monochromatic <math>\sqrt{11/3}</math>-edge has two specific vertices forced to be the same color
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{79}</math>]
| 79
| 165
| Spindling of <math>G_{40}</math>
|
| Any 4-coloring has a monochromatic <math>\sqrt{11/3}</math>-edge
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{49}</math>]
| 49
| 180
|
| <math>{\mathbf Q}[\sqrt{3},\sqrt{11}]^2</math>
| Any 4-coloring either has a specific <math>\sqrt{11/3}</math>-edge monochromatic, or a monochromatic <math>1/\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{51}</math>]
| 51
|
|
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring of plane
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{627}</math>]
| 627
| 2982
| Contains <math>G_{51}</math>
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4161 No name assigned]
| 103
|
|
|
| All 4-colorings contain a monochromatic <math>2/\sqrt{3}</math>-edge
|}

== Lower bounds ==

In [P1988], Pritikin proved that every graph with at most 12 vertices is 4-colorable, and every graph with at most 6197 vertices is 6-colorable. Pritikin's bounds are obtained by coloring the plane with k colors and an additional “wild” color such that points of unit distance are both allowed to receive the wild color. If the wild color occupies a small fraction p of the plane, then an exercise in the probabilistic method gives that any fixed unit-distance graph with n vertices enjoys an embedding in the plane that avoids the wild color (and is therefore k-colorable) provided n<1/p. Pritikin’s coloring for the k=4 case cannot be improved without improving on the densest known subset of the plane that avoids unit distances (originally due to Croft in 1967). See [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable this MO thread] for additional information. As such, improving the bound in the k=4 case might require a new technique, whereas the k=5 and 6 cases might be amenable to optimization.

[https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4176 Every unit distance graph with at most 16 vertices is 5-colorable].

A "tile-based" colouring of the plane must have at least 6 colours, as shown by Townsend [Tow2005]; the same proof with a minor error was also derived by Woodall [W1973]. In [T1999], Thomassen showed that any tiling-based 6-coloring would have to be be "unscaleable", i.e. the maximum diameter of a tile and the minimum separation of same-coloured tiles must both be exactly 1 (so that the distance 1 is excluded by suitable colouring of the boundary points).

== Virtual edge ==

''Warning: other definitions have been proposed and the exact definition of this notion is currently under discussion. The clamping of graph G in this section may be interpreted as the devirtualization of all bichromatic virtual edges with length <math>d</math> and chromatic number <math>4</math>.''

A '''virtual edge''' of a unit distance graph <math>G</math> is a distance <math>d</math> with the property that every 4-coloring of <math>G</math> contains a monochromatic pair of vertices of distance exactly <math>d</math>. Observe that if a unit distance graph <math>G</math> has a virtual edge at distance <math>d</math>, and if there is another unit distance graph <math>H</math> with a pair of vertices at distance <math>d</math> that cannot be monochromatic in a 4-coloring, then we can create a non-4-colorable unit distance graph by "clamping" a copy of <math>H</math> to every virtual edge in <math>G</math>.

Known examples of virtual edges include:

* <math>G_{40}</math> has a virtual edge at distance <math>\sqrt{11/3}</math>.
* <math>V \oplus V \oplus H</math> has a (single) virtual edge at distance <math>8/3</math>.
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4117 <math>(\sqrt{3}\pm 1)/\sqrt{2}</math> are virtual edges of some graphs].

== Bichromatic virtual edge ==

===Definition===
If a unit distance graph <math>H</math> exists with a specific pair of vertices which are distance <math>d</math> apart and is bichromatic in all proper <math>n</math>-colorings of <math>H</math>, then we say that <math>H</math> virtualizes a '''bichromatic virtual edge''' with length <math>d</math> and chromatic number <math>n</math>.

===Vacuous properties===
Considering the graph of the entire plane, all distances correspond to a virtual edge of chromatic number <math>CNP-1</math>.

If a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n</math> exists, then a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n-1</math> exists.

===Convenience and devirtualization===
Bichromatic virtual edges behave the same as the unit edge(which is a bichromatic virtual edge of every chromatic number), and thus may be used to simplify the construction of graphs.

Recursively replacing bichromatic virtual edges of a graph <math>G</math> with graphs which virtualize the bichromatic virtual edges through '''clamping''' produces a unit distance graph <math>G'</math> where <math>\chi(G')\geq\chi(G)</math>.

Devirtualizing bichromatic virtual edges of a graph <math>G</math> (i.e. clamping) has no benefit unless nontrivial points of <math>G</math> and <math>H_0</math> overlap or nontrivial points of <math>H_1</math> and <math>H_0</math> overlap, where <math>H_0</math> and <math>H_1</math> are graphs virtualizing bichromatic virtual edges of <math>G</math>. Such overlaps may cause <math>\chi(G')>\chi(G)</math>. If no nontrivial overlaps exist, then <math>\chi(G')=\max\left(\chi(G),\chi(H_0),\chi(H_1),\dots\right)</math>.

Because more than 1 graph virtualizes any given bichromatic virtual edge for a fixed length and fixed chromatic number, multiple devirtualizations exist for every finite graph.

===Relation to rings===
A '''devirtualized ring''' of graph <math>G</math> is a ring which contains all the vertices of a devirtualization of <math>G</math>, where the vertices are interpreted as complex numbers.

Use the implied shorthand for arbitrary sets <math>\mathbb{S}[\left\{a,b,c,\dots\right\}]=\mathbb{S}[a,b,c,\dots]</math>.

Let the vertices of graph <math>G</math> generate the ring <math>S_0[T_0]</math>.
Let the bichromatic virtual edges be virtualizable by the graphs <math>H_1,\dots,H_m</math>.
Let <math>S_k[T_k]</math> be the ring generated by the vertices of <math>H_k</math>.
Let <math>S</math> be the ring generated by <math>\bigcup_{k=0}^m S_k</math>.
Let <math>T=\bigcup_{k=0}^m T_k</math>.
Then <math>S[T]</math> is a devirtualized ring of <math>G</math>.

As multiple graphs may virtualize the same virtual edge, careful choice may be required to minimize the set <math>T</math>.

===Multiplicative property===
Let <math>H_0</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_0</math> and chromatic number <math>n</math>.
Let <math>H_1</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_1</math> and chromatic number <math>n</math>.
Create a graph <math>H_2</math> by scaling <math>H_0</math> by a factor of <math>d_1</math>. Graph <math>H_2</math> then virtualizes a bichromatic virtual edge length <math>d_0d_1</math> and chromatic number <math>n</math>.

===Notable bichromatic virtual edges===
{| border=1
|-
! Chromatic number !! Length <math>d</math>!! Proof
|-
| any
| 1
| trivial case
|-
| 2
| <math>0\leq d\leq 3</math>
| moving the flexible parts of the graph (0,0),(0,1),(0,2),(0,3)
|-
| 3
| <math>\left\{\left|\frac{(3+\sqrt{-3})k}{2}+\sqrt{-3}j+(-1)^r\right| \mid k,j\in\mathbb{Z}, r\in\mathbb{R}\right\}</math>
| equilateral triangle tiling
|}

===Continuous ranges of bichromatic virtual edges===
Flexible graphs might produce continuous ranges of lengths of bichromatic virtual edges of chromatic number <math>n</math> without having a specific pair which is monochrome for every <math>n</math>-coloring.
Let the range of lengths be on the interval <math>d_0\leq d\leq d_1</math>.

If <math>1 < d_1</math>, then let <math>k\in\mathbb{Z}^+,d_0^{k+1} \leq d_1^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all lengths larger than <math>d_0^k</math>. A finite area allowed to be the same color, the infinite area of the plane, and a finite upper bound on CNP implies <math>CNP> n</math>.

If <math>d_0 < 1</math>, then let <math>k\in\mathbb{Z}^+,d_1^{k+1} \geq d_0^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all non-zero lengths smaller than <math>d_1^k</math>. Considering a set of <math>CNP+1</math> points all within distance <math>d_1^k</math> of each other implies <math>CNP> n</math>.

Using a flexible bichromatic virtual edge of chromatic number <math>n</math> and a graph <math>H</math> of chromatic number <math>n+1</math>, a flexible a graph <math>H'</math> of chromatic number <math>n+1</math> can be created by scaling <math>H</math> to some arbitrarily large or arbitrarily small size and clamping flexible bichromatic virtual edges of chromatic number <math>n</math> as a replacement of each rigid bichromatic virtual edge of <math>H</math>.

<math>H</math> virtualizing a bichromatic virtual edge of chromatic number <math>n+1</math> does not necessarily men the graph <math>H'</math> virtualizes a bichromatic virtual edge.

== Probabilistic formulation ==

See [[Probabilistic formulation of Hadwiger-Nelson problem]].

== Further questions ==

* What are the [http://mathworld.wolfram.com/IndependenceRatio.html independence ratios] of the above unit distance graphs?
* What are the [https://en.wikipedia.org/wiki/Fractional_coloring fractional chromatic numbers] of these graphs?
* What are the [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz numbers] of these graphs?
** The Lovasz theta function value of Lovasz number of <math>G_2</math> at the complement [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3844 is in the interval [3.3746, 3.3748]].
* What about the Erdos unit distance graph (<math>n</math> vertices, <math>n^{1+c/\log\log n}</math> edges)?
* Can we use de Grey’s graph to construct unit-distance graphs that are not 5-colorable? To answer this question, we first need to understand how 5-colorings of de Grey’s graph force small collections of vertices to be colored. [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3899 Varga] and [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3940 Nazgand] provide some thoughts along these lines. Even if we can’t stitch together a 6-chromatic unit-distance graph with these ideas, we might be able to apply them to [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3900 prove that the measurable chromatic number of the plane is at least 6].
* It appears as though the coordinates of our smallest 5-chromatic graph lie in <math>\mathbb{Q}[\sqrt{3}, \sqrt{5}, \sqrt{11}]</math> (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3879 this]). If we view the plane as the complex plane, what is the smallest ring that admits a 5-chromatic single-distance graph? Every single-distance graph in the Eistenstein integers and Gaussian integers is 2-chromatic (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3914 this]). [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3934 David Speyer suggests] looking at <math>\mathbb{Z}[\frac{1+\sqrt{-71}}{2}]</math> next.

== Blog, forums, and media ==

* [https://rjlipton.wordpress.com/2011/05/22/more-on-coloring-the-plane/ More On Coloring The Plane], Richard Lipton, May 22, 2011.
* [https://mathoverflow.net/questions/236392/has-there-been-a-computer-search-for-a-5-chromatic-unit-distance-graph Has there been a computer search for a 5-chromatic unit distance graph?], Juno, Apr 16, 2016.
* [https://quomodocumque.wordpress.com/2018/04/09/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Jordan Ellenberg, Apr 9 2018.
* [https://gilkalai.wordpress.com/2018/04/10/aubrey-de-grey-the-chromatic-number-of-the-plane-is-at-least-5/ Aubrey de Grey: The chromatic number of the plane is at least 5], Gil Kalai, Apr 10 2018.
* [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Dustin Mixon, Apr 10, 2018.
* [https://www.scottaaronson.com/blog/?p=3697 Amazing progress on long-standing problems], Scott Aaronson, Apr 11 2018.
* [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ The chromatic number of the plane is at least 5, Part II], Dustin Mixon, Apr 13 2018.
* [http://community.wolfram.com/groups/-/m/t/1320004 A 5-chromatic unit distance graph], Ed Pegg, Apr 13 2018.
* [http://aperiodical.com/2018/04/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Katie Steckles, Apr 17 2018.
* [https://www.quantamagazine.org/decades-old-graph-problem-yields-to-amateur-mathematician-20180417/ Decades-Old Graph Problem Yields to Amateur Mathematician], Evelyn Lamb, Quanta, Apr 17, 2018.
* [https://www.heise.de/newsticker/meldung/Zahlen-bitte-Wie-bunt-ist-die-Ebene-4024574.html Zahlen, bitte! 5 - Wie bunt ist die Ebene?], Harald Bögeholz, Heise, Apr 17, 2018.
* [http://www.sciencemag.org/news/2018/04/amateur-mathematician-cracks-decades-old-math-problem Amateur mathematician cracks decades-old math problem], Katie Langin, Science News, Apr 18, 2018.
* [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable How much of the plane is 4-colorable?], Dustin Mixon, Apr 18, 2018.
* [https://www.sciencealert.com/amateur-solves-decades-old-maths-problem-about-colours-that-can-never-touch-hadwiger-nelson-problem An Amateur Solved a 60-Year-Old Maths Problem About Colours That Can Never Touch], Peter Dockrill, ScienceAlert, Apr 19, 2018.
* [https://www.zmescience.com/science/math/amateur-mathematician-solves-problem-20042018/ Amateur mathematician Aubrey de Grey, known for his work on anti-aging, solves decades-old problem], Mihai Andrei, ZME Science, Apr 20, 2018.
* [https://automaths.blog/2018/04/21/5-nuances-daubrey-de-grey/ 5 nuances d’Aubrey de Grey], Automaths, Apr 21, 2018.
* [https://science.howstuffworks.com/math-concepts/amateur-solves-part-of-decades-old-math-problem.htm Amateur Solves Part of Decades-Old Math Problem], HowStuffWorks, Apr 30, 2018.
* [https://www.nemokennislink.nl/publicaties/het-platte-vlak-heeft-minstens-vijf-kleuren-nodig/ Het platte vlak heeft minstens vijf kleuren nodig], Kennislink, May 3, 2018.

== Code and data ==

[https://www.dropbox.com/sh/ufknm1v9gtbhad3/AACB2xwaXYx5EGda38_L-0foa?dl=0 This dropbox folder] will contain most of the data and images for the project.

Data:

* [https://www.dropbox.com/s/x6kvu7b3wvdvqsn/graph.dimacs?dl=0 The 1585-vertex graph in DIMACS format]
* [https://www.dropbox.com/s/qk3gbpjvvsjfsc3/sat.dimacs?dl=0 A naive translation of 4-colorability of this graph into a SAT problem in DIMACS format]
* [https://www.dropbox.com/s/nipqikfzcfn9o5a/vertices.sage?dl=0 The vertices of this graph in explicit Sage notation]
* The graph <math>G_2</math>: [http://www.cs.utexas.edu/~marijn/CNP/874.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/874.edge edges (DIMACS)]
* The graph <math>G_3</math>: [http://www.cs.utexas.edu/~marijn/CNP/826.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/826.edge edges (DIMACS)] [http://www.cs.utexas.edu/~marijn/CNP/826.pdf Visualization]
* The [https://www.dropbox.com/sh/ufknm1v9gtbhad3/AADfw9WO1ol2ayQNJEtGf8oBa/Graphs?dl=0 densest unit-distance graphs] on an <math>n\times n</math> grid for <math>n=10,20,\ldots,100</math> (DIMACS format).

Code:

* [https://www.dropbox.com/s/6u1jctbjy38t383/lovaszmoser.m?dl=0 MATLAB script for computing Lovasz number]
* [https://files.jixco.de/pm16/zzvtx/ Python code for converting a list of vertices in Mathematica format into vertices in Z^n]
* [https://files.jixco.de/pm16/edge2cnf.py Python code for converting a DIMACS edge list into a CNF formula (forcing up to 3 suitable vertices to a fixed color, to break some symmetries)]

Software:

* [http://cvxr.com/cvx/ CVX]
* [http://www.labri.fr/perso/lsimon/glucose/ Glucose 4.0]
* [http://minisat.se/ Minisat]

== Wikipedia ==

* [https://en.wikipedia.org/wiki/Graph_coloring#Chromatic_number Chromatic number]
* [https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Erd%C5%91s_theorem_(graph_theory) de Bruijn-Erdos theorem]
* [https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem Hadwiger-Nelson problem]
* [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz number]
* [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]

== Bibliography ==
* [B2008] B. Bukh, [https://doi.org/10.1007/s00039-008-0673-8 Measurable sets with excluded distances], Geometric and Functional Analysis 18 (2008), 668-697.
* [CR2015] D. Cranston, L. Rabern, [https://arxiv.org/abs/1501.01647 The fractional chromatic number of the plane], arXiv:1501.01647
* [deBE1951] N. G. de Bruijn, P. Erdős, [http://www.math-inst.hu/~p_erdos/1951-01.pdf A colour problem for infinite graphs and a problem in the theory of relations], Nederl. Akad. Wetensch. Proc. Ser. A, 54 (1951): 371–373, MR 0046630.
* [deG2018] A. de Grey, [https://arxiv.org/abs/1804.02385 The chromatic number of the plane is at least 5], arXiv:1804.02385
* [EI2018] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.00157 The chromatic number of the plane is at least 5 - a new proof], arXiv:1805.00157
* [F1981] K.J. Falconer, The Realization of distances in measurable subsets covering Rn, J. Combin. Theory Ser. A 31 (1981) 184–189.
* [H1945] H. Hadwiger, Uberdeckung des euklidischen Raum durch kongruente Mengen, Portugaliae Math. 4 (1945), 238–242.
* [MM1961] L. Moser and M. Moser, Solution to Problem 10, Can. Math. Bull. 4 (1961), 187–189.
* [P1998] D. Pritikin, [https://www.sciencedirect.com/science/article/pii/S0095895698918196 All unit-distance graphs of order 6197 are 6-colorable], Journal of Combinatorial Theory, Series B 73.2 (1998): 159-163.
* [S2008] A. Soifer, The Mathematical Coloring Book, Springer, 2008, ISBN-13: 978-0387746401.
* [T1999] C. Thomassen, On the Nelson unit distance coloring problem, Amer. Math. Monthly 106 (1999) 850-853.
* [Tow2005] Townsend, S.P., Colouring the plane with no monochrome unit. Geombinatorics XIV(4) (2005), 181-193.
* [W1973] Woodall, D.R., Distances realized by sets covering the plane. J. Combin. Theory Ser. A, 14 (1973), 187-200.

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T09:14:03Z

Nazgand: /* Proofs of bounds */

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T09:07:58Z

Nazgand: /* Bounds on {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ) for 4-colourings */

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-05T08:49:08Z

Nazgand:

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-04T13:30:03Z

Nazgand: /* Bounds on {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ) for 4-colourings */

''This is a part of Polymath16 - for the main page, see [http://michaelnielsen.org/polymath1/index.php?title=Hadwiger-Nelson_problem].''

Suppose for sake of contradiction that we have a 4-coloring <math>c: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with no unit edges monochromatic, thus

:<math>c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)</math>

We can create further such colorings by composing <math>c</math> on the left with a permutation <math>\sigma \in S_4</math> on the left, and with the (inverse of) a Euclidean isometry <math>T \in E(2)</math> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same property. This is an action of the solvable group <math>S_4 \times E(2)</math>.

It is a fact that all solvable groups (viewed as discrete groups) are [https://en.wikipedia.org/wiki/Amenable_group amenable], so in particular <math>S_4 \times E(2)</math> is amenable. This means that there is a finitely additive probability measure <math>\mu</math> on <math>S_4 \times E(2)</math> (with all subsets of this group measurable), which is left-invariant: <math>\mu(gE) = \mu(E)</math> for all <math>g \in S_4 \times E(2)</math> and <math>E \subset S_4 \times E(2)</math>. This gives <math>S_4 \times E(2)</math> the structure of a finitely additive probability space. We can then define a random coloring <math>{\mathbf c}: {\bf C} \to \{1,2,3,4\}</math> by defining <math>{\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}</math>, where <math>({\mathbf \sigma},{\mathbf T})</math> is the element of the sample space <math>S_4 \times E(2)</math>. Thus for any complex number <math>z</math>, the random color <math>{\mathbf c}(z)</math> is a random variable taking values in <math>\{1,2,3,4\}</math>. The left-invariance of the measure implies that for any <math>(\sigma,T) \in S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the color permutation invariance

:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)</math>

for any <math>z_1,\dots,z_k \in {\bf C}</math>, <math>c_1,\dots,c_k \in \{1,2,3,4\}</math>, and <math>\sigma \in S_4</math>, and the Euclidean isometry invariance
:<math> {\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)</math>

== Bounds on <math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(d) )</math> for 4-colourings ==

A class of correlations that is of particular interest is that of vertex pairs at some distance <math>d</math>.

{| border=1
|-
! distance !! Lower bound !! Lower-bounding graph !! Upper bound !! Upper-bounding graph !! Notes
|-
| >=1/2
|
|
| 1/2
| Spindle
|
|-
| >=1/4
|
|
| 3/4
| Spindle plus triangle
|
|-
| <math>{\sqrt{11/3}}</math>
| 1/118
| <math>G_{40}</math>
|
|
| computer-verified
|-
| <math>\frac{1}{\sqrt{3}}</math>
| 1/28
| Unit diamond plus centres of triangles, together with H
| 1/3
| Unit triangle plus its centre
|
|-
| 2
| 1/6
| H
|
|
|
|-
| 8/3
| 1
| <math>V \oplus V \oplus H</math>
|
|
| computer-verified
|-
| <math>{\sqrt{3}}</math>
| 1/4
| H
|
|
| Consider the opposite ends of a double triangle and permutations of colours to obtain the exact value 1/2
|}

== Proofs of bounds ==

One can compute some correlations of the coloring exactly:

'''Lemma 1''' Let <math>z,w \in {\bf C}</math> with <math>|z-w|=1</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)</math>
for all <math>c=1,\dots,4</math>,
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),</math>
and
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)</math>
for any distinct <math>c,c' \in \{1,2,3,4\}</math>. If <math>u</math> is at a unit distance from both z and w, then
:<math> {\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')</math>

'''Proof''' By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').<math>\Box</math>

'''Lemma 2''' (Spindle argument) Let <math>z,w \in {\bf C}</math> with <math>|z-w| \geq 1/2</math>. Then
:<math> {\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) \leq \frac{1}{2} \quad (7).</math>

'''Proof''' Set <math>d = |z-w|</math>. We can find an angle <math>\theta</math> with <math>|de^{i\theta}-d|=1</math>, then <math>\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)</math> almost surely. This means that at least one of the events <math>\mathbf{c}(0) = \mathbf{c}(d)</math>, <math>\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})</math> occurs with probability at most 1/2. The claim now follows from isometry invariance (3). <math>\Box</math>

'''Lemma 3''' (Using the K graph) We have
:<math>52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).</math>

'''Proof''' Consider the 61-vertex graph <math>K</math> from [https://arxiv.org/abs/1804.02385 de Grey's paper]. It has 26 (isometric) copies of H, and thus 52 copies of the triangle <math>(1, e^{2\pi i/3}, e^{4\pi i/3})</math>. With probability at least <math>1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) </math>, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals <math>(-2, +2)</math>, <math>(-2 e^{2\pi i/3}, 2e^{2\pi i/3})</math>, <math>(-2 e^{4\pi i/3}, e^{-4\pi i/3})</math> are monochromatic. This gives the claim. <math>\Box</math>

'''Corollary 4''' (Existence of monochromatic <math>\sqrt{3}</math>-triangles) We have <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}</math>.

'''Proof''' The probability <math>{\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )</math> is at most <math>{\bf P}( \mathbf{c}(-2) = \mathbf{c}(2))</math>, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. <math>\Box</math>

'''Computer-verified Claim 5''' (Using the graph M) One has <math>{\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0</math> (Note this contradicts Corollary 4).

'''Proof''' This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from [https://arxiv.org/abs/1804.02385 de Grey's paper] with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph <math>M_1</math> or <math>V \oplus V \oplus H</math>. <math>\Box</math>

'''Computer-verified Claim 6''' (Using the graph <math>V \oplus V \oplus H</math>) One has <math> {\bf P}( \mathbf{c}(0) = \mathbf{c}(8/3) ) = 1</math> (note this contradicts Lemma 2).

'''Proof''' This reflects the fact that every 4-coloring of <math>V \oplus V \oplus H</math> must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of <math>V \oplus V \oplus H</math> with the same property. <math>\Box</math>

'''Lemma 7''' (Using <math>G_{40}</math>) We have
:<math>59 {\bf P}( \mathbf{c}(0) = \mathbf{c}(\sqrt{11/3}) ) + {\bf P}( \mathbf{c}(0) = \mathbf{c}(8/3) ) \geq 1</math>.

'''Proof''' This reflects the fact that every 4-coloring of the 40-vertex graph <math>G_{40}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which none of the 59 pairs of vertices at distance <math>\sqrt{11/3}</math> apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) <math>\Box</math>

'''Corollary 8''' One has
:<math>{\bf P}( \mathbf{c}(0) = \mathbf{c}(\sqrt{11/3}) ) \geq \frac{1}{118}</math>.

'''Proof''' Combine Lemma 7 and Lemma 2. <math>\Box</math>

'''Lemma 9''' (Using <math>G_{49}</math>) One has
:<math>18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq {\bf P}( \mathbf{c}(0) = \mathbf{c}(\sqrt{11/3}) ).</math>

'''Proof''' This reflects the fact that every 4-coloring of the 49-vertex graph <math>G_{49}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] in which 0 and <math>\sqrt{11/3}</math> have the same color, at least one of the 18 copies of <math>(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)</math> is monochromatic. This is potentially human-verifiable. <math>\Box</math>

'''Corollary 10''' (Existence of monochromatic <math>1/\sqrt{3}</math>-triangles) One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.</math>

'''Proof''' Combine Corollary 8 and Lemma 9. <math>\Box</math>

'''Computer-verified claim 11''' One has <math>{\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0</math>. (This contradicts Corollary 10).

'''Proof''' This reflects the fact that the 627-vertex graph <math>G_{627}</math> from [https://arxiv.org/abs/1805.00157 Exoo-Ismaolescu] does not have any 4-colorings with <math>1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3</math> monochromatic. <math>\Box</math>

For certain special distances d, one can improve the bound in Lemma 2:

'''Lemma 12''' If <math>k \geq 1</math> is a natural number, <math>j\in\mathbb{Z}</math>, and <math>\gcd(j,2k+1)=1</math> then
:<math> {\bf P}( {\mathbf c}( 0) = {\mathbf c}( \frac{1}{2 \sin \frac{j\pi}{2k+1}} ) ) \leq \frac{k}{2k+1},</math>
thus for instance
:<math> {\bf P}( {\mathbf c}(0) = {\mathbf c}(\frac{1}{\sqrt{3}}) ) \leq \frac{1}{3}.</math>

'''Proof''' Write <math>r = \frac{1}{2 \sin \frac{j\pi}{2k+1}}</math>, then observe that the regular 2k+1-polygon <math>r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}</math> has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. <math>\Box</math>

'''Lemma 13''' We have
:<math> 7 {\bf P}( {\mathbf c}(0) = {\mathbf c}(\frac{1}{\sqrt{3}}) ) \geq {\bf P}( {\mathbf c}(0) = {\mathbf c}(\sqrt{3}) )</math>.

'''Proof''' Consider the unit rhombus <math>0, 1, e^{i\pi/3}, e^{-i\pi/3}</math> together with the centers <math>e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}</math>. With probability <math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(\sqrt{3}) )</math>, the two far vertices <math>e^{i\pi/3}, e^{-i\pi/3}</math> are the same color, and then 0,1 will be two other colors. This forces either one of the centers <math>e^{i\pi/6}/\sqrt{3}</math> of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance <math>1/\sqrt{3}</math> is monochromatic, giving the claim. <math>\Box</math>

'''Corollary 14''' We have <math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(\frac{1}{\sqrt{3}}) ) \geq \frac{1}{728}</math>.

This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.

'''Proof''' Combine Corollary 4 and Lemma 13. <math>\Box</math>

'''Lemma 15''' One has
:<math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(\sqrt{3}) ) + {\bf P} ({\mathbf c}(0) = {\mathbf c}(2) ) \geq \frac{2}{3}</math> and <math>2{\bf P}( {\mathbf c}(0) = {\mathbf c}(\sqrt{3}) ) + {\bf P} ({\mathbf c}(0) = {\mathbf c}(2) ) \geq 1</math>.

'''Proof''' As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length <math>\sqrt{3}</math> and three of length <math>2</math>. If we let a denote the number of monochromatic <math>\sqrt{3}</math> edges and b the number of monochromatic <math>2</math>-edges, we see from inspection of all four colorings that <math>(a,b)</math> is either <math>(6, 0), (4,0), (2, 1)</math>, or <math>(0,3)</math>. In particular, one always has <math>\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}</math> and <math>2\frac{a}{6} + \frac{b}{3} \geq 1</math>. Taking expectations, we obtain the claim. <math>\Box</math>

'''Corollary 16''' We have <math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(2) ) \geq \frac{1}{6}</math>, <math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(\sqrt{3} ) ) \geq \frac{1}{4} </math> and <math>{\bf P}( {\mathbf c}(0) = {\mathbf c}(\frac{1}{\sqrt{3}}) ) \geq \frac{1}{28}</math>.

'''Proof''' Combine Lemma 2, Lemma 15, and Lemma 13. <math>\Box</math>

'''Lemma 17''' If <math>a,b,c > 0</math> are the lengths of a triangle, then <math>{\bf P}(\mathbf{c}(0)=\mathbf{c}(a)) + {\bf P}(\mathbf{c}(0)=\mathbf{c}(b)) \leq 1 + {\bf P}(\mathbf{c}(0)=\mathbf{c}(c))</math>.

'''Proof''' Consider a triangle of side lengths a,b,c. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also, thus
:<math>{\bf P}(\mathbf{c}(0) \neq \mathbf{c}(a)) + {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(b)) \geq {\bf P}(\mathbf{c}(0) \neq \mathbf{c}(c))</math>
and the claim follows. <math>\Box</math>

Note that Lemma 2 follows from the c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on <math>{\bf P}(\mathbf{c}(0)=\mathbf{c}(a))</math> for small values of a, e.g. <math>{\bf P}(\mathbf{c}(0)=\mathbf{c}(a)) \leq 3/4</math> when <math>a \geq 1/4</math>.

Hadwiger-Nelson problem

2018-05-04T12:13:21Z

Nazgand: /* Virtual edge */

The Chromatic Number of the Plane (CNP) is the chromatic number of the graph whose vertices are elements of the plane, and two points are connected by an edge if they are a unit distance apart. The Hadwiger-Nelson problem asks to compute CNP. The bounds <math>4 \leq CNP \leq 7</math> are classical; recently [deG2018] it was shown that <math>CNP \geq 5</math>. This is achieved by explicitly locating finite unit distance graphs with chromatic number at least 5.

The '''Polymath16''' project seeks to simplify the graphs used in [deG2018] to establish this lower bound. More precisely, the goals are

* '''Goal 1''': Find progressively smaller 5-chromatic unit-distance graphs.
* '''Goal 2''': Reduce (ideally to zero) the reliance on computer assistance for the proof. Computer assistance was leveraged in [deG2018] to analyze a subgraph of size 397.
* '''Goal 3''': Apply these simpler graphs to inform progress in related areas. For example:
** Find a 6-chromatic unit-distance graph.
** Improve the corresponding bound in higher dimensions.
** Improve the current record of 105/29 for the fractional chromatic number of the plane.
** Find the smallest unit-distance graph of a given minimum degree (excluding, in some natural way, boring cases like Cartesian products of a graph with a hypercube).

== Polymath threads ==

* [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/ Polymath proposal: finding simpler unit distance graphs of chromatic number 5], Aubrey de Grey, Apr 10 2018. ('''Active discussion thread''')
* [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/ Polymath16, first thread: Simplifying de Grey’s graph], Dustin Mixon, Apr 14, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic Polymath16, second thread: What does it take to be 5-chromatic?], Dustin Mixon, Apr 22, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/ Polymath16, third thread: Is 6-chromatic within reach?], Dustin Mixon, May 1, 2018. ('''Active research thread''')

== Notable unit distance graphs ==

A '''unit distance graph''' is a graph that can be realised as a collection of vertices in the plane, with two vertices connected by an edge if they are precisely a unit distance apart. The chromatic number of any such graph is a lower bound for <math>CNP</math>; in particular, if one can find a unit distance graph with no 4-colorings, then <math>CNP \geq 5</math>. The boldface number of vertices is the current minimal number of vertices of a unit distance graph that is currently known to not be 4-colorable.

<math>G_1 \oplus G_2</math> denotes the Minkowski sum of two unit distance graphs <math>G_1,G_2</math> (vertices in <math>G_1 \oplus G_2</math> are sums of the vertices of <math>G_1,G_2</math>). <math>G_1 \cup G_2</math> denotes the union. <math>\mathrm{rot}(G, \theta)</math> denotes <math>G</math> rotated counterclockwise by <math>\theta</math>. <math>\mathrm{trim}(G,r)</math> denotes the trimming of <math>G</math> after removing all vertices of distance greater than <math>r</math> from the origin.

Another basic operation is '''spindling''': taking two copies of a graph <math>G</math>, gluing them together at one vertex, and rotating the copies so that the two copies of another vertex are a unit distance apart. For instance, the Moser spindle is the spindling of a rhombus graph. If a graph has two vertices forced to be the same color in a k-coloring, then the spindling of that graph at those two vertices is not k-colorable.

Many of the graphs are embeddable in abelian groups or rings, particularly those generated by the roots of unity <math>\omega_t := \exp( i \arccos( 1 - \tfrac{1}{2t} ))</math> for various natural numbers t.

{| border=1
|-
!Name!!Number of vertices!! Number of edges !! Structure !! Group !! Colorings
|-
| [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]
| 7
| 11
| Two 60-120-60-120 rhombi with a common vertex, with one pair of sharp vertices coincident and the other joined
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [http://mathworld.wolfram.com/GolombGraph.html Golomb graph]
| 10
| 18
| Contains the center and vertices of a hexagon and equilateral triangle
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [https://arxiv.org/abs/1804.02385 H]
| 7
| 12
| Vertices and center of a hexagon
| <math>{\bf Z}[\omega_1]</math>
| Has essentially four 4-colorings, two of which contain a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 J]
| 31
| 72
| Contains 13 copies of H
| <math>{\bf Z}[\omega_1]</math>
| Has essentially six 4-colorings in which no H has a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 K]
| 61
| 150
| Contains 2 copies of J
|
| In all 4-colorings lacking an H with a monochromatic <math>\sqrt{3}</math>-triangle, all pairs of vertices at distance 4 are monochromatic
|-
| [https://arxiv.org/abs/1804.02385 L]
| 121
| 301
| Contains two copies of K and 52 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ <math>L_1</math>]
| 97
|
| Has 40 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120295 <math>L_2</math>]
| 120
| 354
|
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 T]
| 9
| 15
| Contains one Moser spindle and useful symmetry; three vertices form an equilateral triangle
|
|
|-
| [https://arxiv.org/abs/1804.02385 U]
| 15
| 33
| Three copies of T at 120-degree rotations: <math>T \cup \mathrm{rot}(T, 2\pi/3) \cup \mathrm{rot}(T, 4\pi/3)</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 V]
| 31
| 30
| Unit vectors at angles consistent with three interlocking Moser spindles
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>V_1</math>]
| 61
| 60
| Union of V and a rotation of V: <math>V \cup \mathrm{rot}(V, \mathrm{arccos}(7/8))</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_b</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_x</math>]
| 13
| 12
| Subgraph of <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_z</math>]
|
|
| Subgraph of <math>V_a</math>; shares a line of symmetry with <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_y</math>]
| 13
| 12
| Subgraph of <math>V_b</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-3970 <math>V_A</math>]
| 37
| 36
| Unit vectors with angles <math>i \frac{\pi}{3} + j \mathrm{arccos} \frac{5}{6} + k \mathrm{arccos} \frac{7}{8}</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 W]
| 301
| 1230
| Cartesian product of V with itself, minus vertices at more than <math>\sqrt{3}</math> from the centre (i.e. <math>\mathrm{trim}(V \oplus V, \sqrt{3})</math>)
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>W_1</math>]
|
|
| Trimmed product of V with itself (<math>\mathrm{trim}(V \oplus V, 1.95)</math>)
|
|
|-
| [https://arxiv.org/abs/1804.02385 M]
| 1345
| 8268
| Cartesian product of W and H (<math>W \oplus H</math>)
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120274 <math>M_1</math>]
| 278
|
| Deleting vertices from M while maintaining its restriction on H
|
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>M_2</math>]
| 7075
|
| Sum of H with a trimmed product of <math>V_1</math> with itself
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 N]
| 20425
| 151311
| Contains 52 copies of M arranged around the H-copies of L
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4, \omega_{16}]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_0</math>]
| 1585
| 7909
| N "shrunk" by stepwise deletions and replacements of vertices
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 G]
| 1581
| 7877
| Deleting 4 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_1</math>]
| 1577
|
| Deleting 8 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120318 <math>G_2</math>]
| 874
| 4461
| Juxtaposing two copies of M and shrinking
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3867 <math>G_3</math>]
| 826
| 4273
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4105 <math>G_4</math>]
| '''803'''
| 4144
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 R]
|
|
| Union of <math>W_1</math> and a rotated copy of <math>W_1</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>\mathrm{trim}(R \oplus H, 1.67)</math>]
| 2563
|
| Trimmed sum of R and H
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 <math>V \oplus V \oplus H</math>]
|
|
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_{64/9}]</math>
| Has two vertices forced to be the same color in a 4-coloring; also no monochromatic <math>\sqrt{3}</math>-triangles
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 No name assigned]
| 745
|
| Subgraph of <math>V \oplus V \oplus H</math>
|
| Has two vertices forced to be the same color in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3085
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_a \oplus V_z \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3049
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 No name assigned]
| 1951
|
| Trimmed version of <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic <math>V_A \oplus V_A \oplus V_A</math>]
| 6937
| 44439
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{40}</math>]
| 40
| 82
|
|
| Any 4-coloring that avoids a monochromatic <math>\sqrt{11/3}</math>-edge has two specific vertices forced to be the same color
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{79}</math>]
| 79
| 165
| Spindling of <math>G_{40}</math>
|
| Any 4-coloring has a monochromatic <math>\sqrt{11/3}</math>-edge
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{49}</math>]
| 49
| 180
|
| <math>{\mathbf Q}[\sqrt{3},\sqrt{11}]^2</math>
| Any 4-coloring either has a specific <math>\sqrt{11/3}</math>-edge monochromatic, or a monochromatic <math>1/\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{51}</math>]
| 51
|
|
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring of plane
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{627}</math>]
| 627
| 2982
| Contains <math>G_{51}</math>
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4161 No name assigned]
| 103
|
|
|
| All 4-colorings contain a monochromatic <math>2/\sqrt{3}</math>-edge
|}

== Lower bounds ==

In [P1988], Pritikin proved that every graph with at most 12 vertices is 4-colorable, and every graph with at most 6197 vertices is 6-colorable. Pritikin's bounds are obtained by coloring the plane with k colors and an additional “wild” color such that points of unit distance are both allowed to receive the wild color. If the wild color occupies a small fraction p of the plane, then an exercise in the probabilistic method gives that any fixed unit-distance graph with n vertices enjoys an embedding in the plane that avoids the wild color (and is therefore k-colorable) provided n<1/p. Pritikin’s coloring for the k=4 case cannot be improved without improving on the densest known subset of the plane that avoids unit distances (originally due to Croft in 1967). See [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable this MO thread] for additional information. As such, improving the bound in the k=4 case might require a new technique, whereas the k=5 and 6 cases might be amenable to optimization.

[https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4176 Every unit distance graph with at most 16 vertices is 5-colorable].

A "tile-based" colouring of the plane must have at least 6 colours, as shown by Townsend [Tow2005]; the same proof with a minor error was also derived by Woodall [W1973]. In [T1999], Thomassen showed that any tiling-based 6-coloring would have to be be "unscaleable", i.e. the maximum diameter of a tile and the minimum separation of same-coloured tiles must both be exactly 1 (so that the distance 1 is excluded by suitable colouring of the boundary points).

== Virtual edge ==

''Warning: other definitions have been proposed and the exact definition of this notion is currently under discussion. The clamping of graph G in this section may be interpreted as the devirtualization of all bichromatic virtual edges with length <math>d</math> and chromatic number <math>4</math>.''

A '''virtual edge''' of a unit distance graph <math>G</math> is a distance <math>d</math> with the property that every 4-coloring of <math>G</math> contains a monochromatic pair of vertices of distance exactly <math>d</math>. Observe that if a unit distance graph <math>G</math> has a virtual edge at distance <math>d</math>, and if there is another unit distance graph <math>H</math> with a pair of vertices at distance <math>d</math> that cannot be monochromatic in a 4-coloring, then we can create a non-4-colorable unit distance graph by "clamping" a copy of <math>H</math> to every virtual edge in <math>G</math>.

Known examples of virtual edges include:

* <math>G_{40}</math> has a virtual edge at distance <math>\sqrt{11/3}</math>.
* <math>V \oplus V \oplus H</math> has a (single) virtual edge at distance <math>8/3</math>.
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4117 <math>(\sqrt{3}\pm 1)/\sqrt{2}</math> are virtual edges of some graphs].

== Bichromatic virtual edge ==

===Definition===
If a unit distance graph <math>H</math> exists with a specific pair of vertices which are distance <math>d</math> apart and is bichromatic in all proper <math>n</math>-colorings of <math>H</math>, then we say that <math>H</math> virtualizes a '''bichromatic virtual edge''' with length <math>d</math> and chromatic number <math>n</math>.

===Vacuous properties===
Considering the graph of the entire plane, all distances correspond to a virtual edge of chromatic number <math>CNP-1</math>.

If a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n</math> exists, then a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n-1</math> exists.

===Convenience and devirtualization===
Bichromatic virtual edges behave the same as the unit edge(which is a bichromatic virtual edge of every chromatic number), and thus may be used to simplify the construction of graphs.

Recursively replacing bichromatic virtual edges of a graph <math>G</math> with graphs which virtualize the bichromatic virtual edges through '''clamping''' produces a unit distance graph <math>G'</math> where <math>\chi(G')\geq\chi(G)</math>.

Devirtualizing bichromatic virtual edges of a graph <math>G</math> (i.e. clamping) has no benefit unless nontrivial points of <math>G</math> and <math>H_0</math> overlap or nontrivial points of <math>H_1</math> and <math>H_0</math> overlap, where <math>H_0</math> and <math>H_1</math> are graphs virtualizing bichromatic virtual edges of <math>G</math>. Such overlaps may cause <math>\chi(G')>\chi(G)</math>. If no nontrivial overlaps exist, then <math>\chi(G')=\max\left(\chi(G),\chi(H_0),\chi(H_1),\dots\right)</math>.

Because more than 1 graph virtualizes any given bichromatic virtual edge for a fixed length and fixed chromatic number, multiple devirtualizations exist for every finite graph.

===Relation to rings===
A '''devirtualized ring''' of graph <math>G</math> is a ring which contains all the vertices of a devirtualization of <math>G</math>, where the vertices are interpreted as complex numbers.

Use the implied shorthand for arbitrary sets <math>\mathbb{S}[\left\{a,b,c,\dots\right\}]=\mathbb{S}[a,b,c,\dots]</math>.

Let the vertices of graph <math>G</math> generate the ring <math>S_0[T_0]</math>.
Let the bichromatic virtual edges be virtualizable by the graphs <math>H_1,\dots,H_m</math>.
Let <math>S_k[T_k]</math> be the ring generated by the vertices of <math>H_k</math>.
Let <math>S</math> be the ring generated by <math>\bigcup_{k=0}^m S_k</math>.
Let <math>T=\bigcup_{k=0}^m T_k</math>.
Then <math>S[T]</math> is a devirtualized ring of <math>G</math>.

As multiple graphs may virtualize the same virtual edge, careful choice may be required to minimize the set <math>T</math>.

===Multiplicative property===
Let <math>H_0</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_0</math> and chromatic number <math>n</math>.
Let <math>H_1</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_1</math> and chromatic number <math>n</math>.
Create a graph <math>H_2</math> by scaling <math>H_0</math> by a factor of <math>d_1</math>. Graph <math>H_2</math> then virtualizes a bichromatic virtual edge length <math>d_0d_1</math> and chromatic number <math>n</math>.

===Notable bichromatic virtual edges===
{| border=1
|-
! Chromatic number !! Length <math>d</math>!! Proof
|-
| any
| 1
| trivial case
|-
| 2
| <math>0\leq d\leq 3</math>
| moving the flexible parts of the graph (0,0),(0,1),(0,2),(0,3)
|-
| 3
| <math>\left\{\left|\frac{(3+\sqrt{-3})k}{2}+\sqrt{-3}j+(-1)^r\right| \mid k,j\in\mathbb{Z}, r\in\mathbb{R}\right\}</math>
| equilateral triangle tiling
|}

===Continuous ranges of bichromatic virtual edges===
Flexible graphs might produce continuous ranges of lengths of bichromatic virtual edges of chromatic number <math>n</math> without having a specific pair which is monochrome for every <math>n</math>-coloring.
Let the range of lengths be on the interval <math>d_0\leq d\leq d_1</math>.

If <math>1 < d_1</math>, then let <math>k\in\mathbb{Z}^+,d_0^{k+1} \leq d_1^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all lengths larger than <math>d_0^k</math>. A finite area allowed to be the same color, the infinite area of the plane, and a finite upper bound on CNP implies <math>CNP> n</math>.

If <math>d_0 < 1</math>, then let <math>k\in\mathbb{Z}^+,d_1^{k+1} \geq d_0^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all non-zero lengths smaller than <math>d_1^k</math>. Considering a set of <math>CNP+1</math> all within distance <math>d_1^k</math> of each other implies <math>CNP> n</math>.

Using a flexible bichromatic virtual edge of chromatic number <math>n</math> and a graph <math>H</math> of chromatic number <math>n+1</math>, a flexible a graph <math>H'</math> of chromatic number <math>n+1</math> can be created by scaling <math>H</math> to some arbitrarily large or arbitrarily small size and clamping flexible bichromatic virtual edges of chromatic number <math>n</math> as a replacement of each rigid bichromatic virtual edge of <math>H</math>.

<math>H</math> virtualizing a bichromatic virtual edge of chromatic number <math>n+1</math> does not necessarily men the graph <math>H'</math> virtualizes a bichromatic virtual edge.

== Probabilistic formulation ==

See [[Probabilistic formulation of Hadwiger-Nelson problem]].

== Further questions ==

* What are the [http://mathworld.wolfram.com/IndependenceRatio.html independence ratios] of the above unit distance graphs?
* What are the [https://en.wikipedia.org/wiki/Fractional_coloring fractional chromatic numbers] of these graphs?
* What are the [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz numbers] of these graphs?
** The Lovasz theta function value of Lovasz number of <math>G_2</math> at the complement [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3844 is in the interval [3.3746, 3.3748]].
* What about the Erdos unit distance graph (<math>n</math> vertices, <math>n^{1+c/\log\log n}</math> edges)?
* Can we use de Grey’s graph to construct unit-distance graphs that are not 5-colorable? To answer this question, we first need to understand how 5-colorings of de Grey’s graph force small collections of vertices to be colored. [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3899 Varga] and [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3940 Nazgand] provide some thoughts along these lines. Even if we can’t stitch together a 6-chromatic unit-distance graph with these ideas, we might be able to apply them to [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3900 prove that the measurable chromatic number of the plane is at least 6].
* It appears as though the coordinates of our smallest 5-chromatic graph lie in <math>\mathbb{Q}[\sqrt{3}, \sqrt{5}, \sqrt{11}]</math> (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3879 this]). If we view the plane as the complex plane, what is the smallest ring that admits a 5-chromatic single-distance graph? Every single-distance graph in the Eistenstein integers and Gaussian integers is 2-chromatic (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3914 this]). [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3934 David Speyer suggests] looking at <math>\mathbb{Z}[\frac{1+\sqrt{-71}}{2}]</math> next.

== Blog, forums, and media ==

* [https://rjlipton.wordpress.com/2011/05/22/more-on-coloring-the-plane/ More On Coloring The Plane], Richard Lipton, May 22, 2011.
* [https://mathoverflow.net/questions/236392/has-there-been-a-computer-search-for-a-5-chromatic-unit-distance-graph Has there been a computer search for a 5-chromatic unit distance graph?], Juno, Apr 16, 2016.
* [https://quomodocumque.wordpress.com/2018/04/09/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Jordan Ellenberg, Apr 9 2018.
* [https://gilkalai.wordpress.com/2018/04/10/aubrey-de-grey-the-chromatic-number-of-the-plane-is-at-least-5/ Aubrey de Grey: The chromatic number of the plane is at least 5], Gil Kalai, Apr 10 2018.
* [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Dustin Mixon, Apr 10, 2018.
* [https://www.scottaaronson.com/blog/?p=3697 Amazing progress on long-standing problems], Scott Aaronson, Apr 11 2018.
* [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ The chromatic number of the plane is at least 5, Part II], Dustin Mixon, Apr 13 2018.
* [http://community.wolfram.com/groups/-/m/t/1320004 A 5-chromatic unit distance graph], Ed Pegg, Apr 13 2018.
* [http://aperiodical.com/2018/04/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Katie Steckles, Apr 17 2018.
* [https://www.quantamagazine.org/decades-old-graph-problem-yields-to-amateur-mathematician-20180417/ Decades-Old Graph Problem Yields to Amateur Mathematician], Evelyn Lamb, Quanta, Apr 17, 2018.
* [https://www.heise.de/newsticker/meldung/Zahlen-bitte-Wie-bunt-ist-die-Ebene-4024574.html Zahlen, bitte! 5 - Wie bunt ist die Ebene?], Harald Bögeholz, Heise, Apr 17, 2018.
* [http://www.sciencemag.org/news/2018/04/amateur-mathematician-cracks-decades-old-math-problem Amateur mathematician cracks decades-old math problem], Katie Langin, Science News, Apr 18, 2018.
* [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable How much of the plane is 4-colorable?], Dustin Mixon, Apr 18, 2018.
* [https://www.sciencealert.com/amateur-solves-decades-old-maths-problem-about-colours-that-can-never-touch-hadwiger-nelson-problem An Amateur Solved a 60-Year-Old Maths Problem About Colours That Can Never Touch], Peter Dockrill, ScienceAlert, Apr 19, 2018.
* [https://www.zmescience.com/science/math/amateur-mathematician-solves-problem-20042018/ Amateur mathematician Aubrey de Grey, known for his work on anti-aging, solves decades-old problem], Mihai Andrei, ZME Science, Apr 20, 2018.
* [https://automaths.blog/2018/04/21/5-nuances-daubrey-de-grey/ 5 nuances d’Aubrey de Grey], Automaths, Apr 21, 2018.
* [https://science.howstuffworks.com/math-concepts/amateur-solves-part-of-decades-old-math-problem.htm Amateur Solves Part of Decades-Old Math Problem], HowStuffWorks, Apr 30, 2018.
* [https://www.nemokennislink.nl/publicaties/het-platte-vlak-heeft-minstens-vijf-kleuren-nodig/ Het platte vlak heeft minstens vijf kleuren nodig], Kennislink, May 3, 2018.

== Code and data ==

[https://www.dropbox.com/sh/ufknm1v9gtbhad3/AACB2xwaXYx5EGda38_L-0foa?dl=0 This dropbox folder] will contain most of the data and images for the project.

Data:

* [https://www.dropbox.com/s/x6kvu7b3wvdvqsn/graph.dimacs?dl=0 The 1585-vertex graph in DIMACS format]
* [https://www.dropbox.com/s/qk3gbpjvvsjfsc3/sat.dimacs?dl=0 A naive translation of 4-colorability of this graph into a SAT problem in DIMACS format]
* [https://www.dropbox.com/s/nipqikfzcfn9o5a/vertices.sage?dl=0 The vertices of this graph in explicit Sage notation]
* The graph <math>G_2</math>: [http://www.cs.utexas.edu/~marijn/CNP/874.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/874.edge edges (DIMACS)]
* The graph <math>G_3</math>: [http://www.cs.utexas.edu/~marijn/CNP/826.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/826.edge edges (DIMACS)] [http://www.cs.utexas.edu/~marijn/CNP/826.pdf Visualization]
* The [https://www.dropbox.com/sh/ufknm1v9gtbhad3/AADfw9WO1ol2ayQNJEtGf8oBa/Graphs?dl=0 densest unit-distance graphs] on an <math>n\times n</math> grid for <math>n=10,20,\ldots,100</math> (DIMACS format).

Code:

* [https://www.dropbox.com/s/6u1jctbjy38t383/lovaszmoser.m?dl=0 MATLAB script for computing Lovasz number]
* [https://files.jixco.de/pm16/zzvtx/ Python code for converting a list of vertices in Mathematica format into vertices in Z^n]
* [https://files.jixco.de/pm16/edge2cnf.py Python code for converting a DIMACS edge list into a CNF formula (forcing up to 3 suitable vertices to a fixed color, to break some symmetries)]

Software:

* [http://cvxr.com/cvx/ CVX]
* [http://www.labri.fr/perso/lsimon/glucose/ Glucose 4.0]
* [http://minisat.se/ Minisat]

== Wikipedia ==

* [https://en.wikipedia.org/wiki/Graph_coloring#Chromatic_number Chromatic number]
* [https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Erd%C5%91s_theorem_(graph_theory) de Bruijn-Erdos theorem]
* [https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem Hadwiger-Nelson problem]
* [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz number]
* [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]

== Bibliography ==
* [B2008] B. Bukh, [https://doi.org/10.1007/s00039-008-0673-8 Measurable sets with excluded distances], Geometric and Functional Analysis 18 (2008), 668-697.
* [CR2015] D. Cranston, L. Rabern, [https://arxiv.org/abs/1501.01647 The fractional chromatic number of the plane], arXiv:1501.01647
* [deBE1951] N. G. de Bruijn, P. Erdős, [http://www.math-inst.hu/~p_erdos/1951-01.pdf A colour problem for infinite graphs and a problem in the theory of relations], Nederl. Akad. Wetensch. Proc. Ser. A, 54 (1951): 371–373, MR 0046630.
* [deG2018] A. de Grey, [https://arxiv.org/abs/1804.02385 The chromatic number of the plane is at least 5], arXiv:1804.02385
* [EI2018] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.00157 The chromatic number of the plane is at least 5 - a new proof], arXiv:1805.00157
* [F1981] K.J. Falconer, The Realization of distances in measurable subsets covering Rn, J. Combin. Theory Ser. A 31 (1981) 184–189.
* [H1945] H. Hadwiger, Uberdeckung des euklidischen Raum durch kongruente Mengen, Portugaliae Math. 4 (1945), 238–242.
* [MM1961] L. Moser and M. Moser, Solution to Problem 10, Can. Math. Bull. 4 (1961), 187–189.
* [P1998] D. Pritikin, [https://www.sciencedirect.com/science/article/pii/S0095895698918196 All unit-distance graphs of order 6197 are 6-colorable], Journal of Combinatorial Theory, Series B 73.2 (1998): 159-163.
* [S2008] A. Soifer, The Mathematical Coloring Book, Springer, 2008, ISBN-13: 978-0387746401.
* [T1999] C. Thomassen, On the Nelson unit distance coloring problem, Amer. Math. Monthly 106 (1999) 850-853.
* [Tow2005] Townsend, S.P., Colouring the plane with no monochrome unit. Geombinatorics XIV(4) (2005), 181-193.
* [W1973] Woodall, D.R., Distances realized by sets covering the plane. J. Combin. Theory Ser. A, 14 (1973), 187-200.

Hadwiger-Nelson problem

2018-05-04T12:03:57Z

Nazgand: /* Continuous ranges of bichromatic virtual edges */ Typo fix

The Chromatic Number of the Plane (CNP) is the chromatic number of the graph whose vertices are elements of the plane, and two points are connected by an edge if they are a unit distance apart. The Hadwiger-Nelson problem asks to compute CNP. The bounds <math>4 \leq CNP \leq 7</math> are classical; recently [deG2018] it was shown that <math>CNP \geq 5</math>. This is achieved by explicitly locating finite unit distance graphs with chromatic number at least 5.

The '''Polymath16''' project seeks to simplify the graphs used in [deG2018] to establish this lower bound. More precisely, the goals are

* '''Goal 1''': Find progressively smaller 5-chromatic unit-distance graphs.
* '''Goal 2''': Reduce (ideally to zero) the reliance on computer assistance for the proof. Computer assistance was leveraged in [deG2018] to analyze a subgraph of size 397.
* '''Goal 3''': Apply these simpler graphs to inform progress in related areas. For example:
** Find a 6-chromatic unit-distance graph.
** Improve the corresponding bound in higher dimensions.
** Improve the current record of 105/29 for the fractional chromatic number of the plane.
** Find the smallest unit-distance graph of a given minimum degree (excluding, in some natural way, boring cases like Cartesian products of a graph with a hypercube).

== Polymath threads ==

* [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/ Polymath proposal: finding simpler unit distance graphs of chromatic number 5], Aubrey de Grey, Apr 10 2018. ('''Active discussion thread''')
* [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/ Polymath16, first thread: Simplifying de Grey’s graph], Dustin Mixon, Apr 14, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic Polymath16, second thread: What does it take to be 5-chromatic?], Dustin Mixon, Apr 22, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/ Polymath16, third thread: Is 6-chromatic within reach?], Dustin Mixon, May 1, 2018. ('''Active research thread''')

== Notable unit distance graphs ==

A '''unit distance graph''' is a graph that can be realised as a collection of vertices in the plane, with two vertices connected by an edge if they are precisely a unit distance apart. The chromatic number of any such graph is a lower bound for <math>CNP</math>; in particular, if one can find a unit distance graph with no 4-colorings, then <math>CNP \geq 5</math>. The boldface number of vertices is the current minimal number of vertices of a unit distance graph that is currently known to not be 4-colorable.

<math>G_1 \oplus G_2</math> denotes the Minkowski sum of two unit distance graphs <math>G_1,G_2</math> (vertices in <math>G_1 \oplus G_2</math> are sums of the vertices of <math>G_1,G_2</math>). <math>G_1 \cup G_2</math> denotes the union. <math>\mathrm{rot}(G, \theta)</math> denotes <math>G</math> rotated counterclockwise by <math>\theta</math>. <math>\mathrm{trim}(G,r)</math> denotes the trimming of <math>G</math> after removing all vertices of distance greater than <math>r</math> from the origin.

Another basic operation is '''spindling''': taking two copies of a graph <math>G</math>, gluing them together at one vertex, and rotating the copies so that the two copies of another vertex are a unit distance apart. For instance, the Moser spindle is the spindling of a rhombus graph. If a graph has two vertices forced to be the same color in a k-coloring, then the spindling of that graph at those two vertices is not k-colorable.

Many of the graphs are embeddable in abelian groups or rings, particularly those generated by the roots of unity <math>\omega_t := \exp( i \arccos( 1 - \tfrac{1}{2t} ))</math> for various natural numbers t.

{| border=1
|-
!Name!!Number of vertices!! Number of edges !! Structure !! Group !! Colorings
|-
| [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]
| 7
| 11
| Two 60-120-60-120 rhombi with a common vertex, with one pair of sharp vertices coincident and the other joined
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [http://mathworld.wolfram.com/GolombGraph.html Golomb graph]
| 10
| 18
| Contains the center and vertices of a hexagon and equilateral triangle
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [https://arxiv.org/abs/1804.02385 H]
| 7
| 12
| Vertices and center of a hexagon
| <math>{\bf Z}[\omega_1]</math>
| Has essentially four 4-colorings, two of which contain a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 J]
| 31
| 72
| Contains 13 copies of H
| <math>{\bf Z}[\omega_1]</math>
| Has essentially six 4-colorings in which no H has a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 K]
| 61
| 150
| Contains 2 copies of J
|
| In all 4-colorings lacking an H with a monochromatic <math>\sqrt{3}</math>-triangle, all pairs of vertices at distance 4 are monochromatic
|-
| [https://arxiv.org/abs/1804.02385 L]
| 121
| 301
| Contains two copies of K and 52 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ <math>L_1</math>]
| 97
|
| Has 40 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120295 <math>L_2</math>]
| 120
| 354
|
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 T]
| 9
| 15
| Contains one Moser spindle and useful symmetry; three vertices form an equilateral triangle
|
|
|-
| [https://arxiv.org/abs/1804.02385 U]
| 15
| 33
| Three copies of T at 120-degree rotations: <math>T \cup \mathrm{rot}(T, 2\pi/3) \cup \mathrm{rot}(T, 4\pi/3)</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 V]
| 31
| 30
| Unit vectors at angles consistent with three interlocking Moser spindles
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>V_1</math>]
| 61
| 60
| Union of V and a rotation of V: <math>V \cup \mathrm{rot}(V, \mathrm{arccos}(7/8))</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_b</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_x</math>]
| 13
| 12
| Subgraph of <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_z</math>]
|
|
| Subgraph of <math>V_a</math>; shares a line of symmetry with <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_y</math>]
| 13
| 12
| Subgraph of <math>V_b</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-3970 <math>V_A</math>]
| 37
| 36
| Unit vectors with angles <math>i \frac{\pi}{3} + j \mathrm{arccos} \frac{5}{6} + k \mathrm{arccos} \frac{7}{8}</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 W]
| 301
| 1230
| Cartesian product of V with itself, minus vertices at more than <math>\sqrt{3}</math> from the centre (i.e. <math>\mathrm{trim}(V \oplus V, \sqrt{3})</math>)
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>W_1</math>]
|
|
| Trimmed product of V with itself (<math>\mathrm{trim}(V \oplus V, 1.95)</math>)
|
|
|-
| [https://arxiv.org/abs/1804.02385 M]
| 1345
| 8268
| Cartesian product of W and H (<math>W \oplus H</math>)
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120274 <math>M_1</math>]
| 278
|
| Deleting vertices from M while maintaining its restriction on H
|
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>M_2</math>]
| 7075
|
| Sum of H with a trimmed product of <math>V_1</math> with itself
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 N]
| 20425
| 151311
| Contains 52 copies of M arranged around the H-copies of L
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4, \omega_{16}]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_0</math>]
| 1585
| 7909
| N "shrunk" by stepwise deletions and replacements of vertices
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 G]
| 1581
| 7877
| Deleting 4 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_1</math>]
| 1577
|
| Deleting 8 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120318 <math>G_2</math>]
| 874
| 4461
| Juxtaposing two copies of M and shrinking
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3867 <math>G_3</math>]
| 826
| 4273
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4105 <math>G_4</math>]
| '''803'''
| 4144
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 R]
|
|
| Union of <math>W_1</math> and a rotated copy of <math>W_1</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>\mathrm{trim}(R \oplus H, 1.67)</math>]
| 2563
|
| Trimmed sum of R and H
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 <math>V \oplus V \oplus H</math>]
|
|
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_{64/9}]</math>
| Has two vertices forced to be the same color in a 4-coloring; also no monochromatic <math>\sqrt{3}</math>-triangles
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 No name assigned]
| 745
|
| Subgraph of <math>V \oplus V \oplus H</math>
|
| Has two vertices forced to be the same color in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3085
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_a \oplus V_z \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3049
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 No name assigned]
| 1951
|
| Trimmed version of <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic <math>V_A \oplus V_A \oplus V_A</math>]
| 6937
| 44439
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{40}</math>]
| 40
| 82
|
|
| Any 4-coloring that avoids a monochromatic <math>\sqrt{11/3}</math>-edge has two specific vertices forced to be the same color
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{79}</math>]
| 79
| 165
| Spindling of <math>G_{40}</math>
|
| Any 4-coloring has a monochromatic <math>\sqrt{11/3}</math>-edge
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{49}</math>]
| 49
| 180
|
| <math>{\mathbf Q}[\sqrt{3},\sqrt{11}]^2</math>
| Any 4-coloring either has a specific <math>\sqrt{11/3}</math>-edge monochromatic, or a monochromatic <math>1/\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{51}</math>]
| 51
|
|
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring of plane
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{627}</math>]
| 627
| 2982
| Contains <math>G_{51}</math>
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4161 No name assigned]
| 103
|
|
|
| All 4-colorings contain a monochromatic <math>2/\sqrt{3}</math>-edge
|}

== Lower bounds ==

In [P1988], Pritikin proved that every graph with at most 12 vertices is 4-colorable, and every graph with at most 6197 vertices is 6-colorable. Pritikin's bounds are obtained by coloring the plane with k colors and an additional “wild” color such that points of unit distance are both allowed to receive the wild color. If the wild color occupies a small fraction p of the plane, then an exercise in the probabilistic method gives that any fixed unit-distance graph with n vertices enjoys an embedding in the plane that avoids the wild color (and is therefore k-colorable) provided n<1/p. Pritikin’s coloring for the k=4 case cannot be improved without improving on the densest known subset of the plane that avoids unit distances (originally due to Croft in 1967). See [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable this MO thread] for additional information. As such, improving the bound in the k=4 case might require a new technique, whereas the k=5 and 6 cases might be amenable to optimization.

[https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4176 Every unit distance graph with at most 16 vertices is 5-colorable].

A "tile-based" colouring of the plane must have at least 6 colours, as shown by Townsend [Tow2005]; the same proof with a minor error was also derived by Woodall [W1973]. In [T1999], Thomassen showed that any tiling-based 6-coloring would have to be be "unscaleable", i.e. the maximum diameter of a tile and the minimum separation of same-coloured tiles must both be exactly 1 (so that the distance 1 is excluded by suitable colouring of the boundary points).

== Virtual edge ==

''Warning: other definitions have been proposed and the exact definition of this notion is currently under discussion.''

A '''virtual edge''' of a unit distance graph <math>G</math> is a distance <math>d</math> with the property that every 4-coloring of <math>G</math> contains a monochromatic pair of vertices of distance exactly <math>d</math>. Observe that if a unit distance graph <math>G</math> has a virtual edge at distance <math>d</math>, and if there is another unit distance graph <math>H</math> with a pair of vertices at distance <math>d</math> that cannot be monochromatic in a 4-coloring, then we can create a non-4-colorable unit distance graph by "clamping" a copy of <math>H</math> to every virtual edge in <math>G</math>.

Known examples of virtual edges include:

* <math>G_{40}</math> has a virtual edge at distance <math>\sqrt{11/3}</math>.
* <math>V \oplus V \oplus H</math> has a (single) virtual edge at distance <math>8/3</math>.
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4117 <math>(\sqrt{3}\pm 1)/\sqrt{2}</math> are virtual edges of some graphs].

== Bichromatic virtual edge ==

===Definition===
If a unit distance graph <math>H</math> exists with a specific pair of vertices which are distance <math>d</math> apart and is bichromatic in all proper <math>n</math>-colorings of <math>H</math>, then we say that <math>H</math> virtualizes a '''bichromatic virtual edge''' with length <math>d</math> and chromatic number <math>n</math>.

===Vacuous properties===
Considering the graph of the entire plane, all distances correspond to a virtual edge of chromatic number <math>CNP-1</math>.

If a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n</math> exists, then a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n-1</math> exists.

===Convenience and devirtualization===
Bichromatic virtual edges behave the same as the unit edge(which is a bichromatic virtual edge of every chromatic number), and thus may be used to simplify the construction of graphs.

Recursively replacing bichromatic virtual edges of a graph <math>G</math> with graphs which virtualize the bichromatic virtual edges through '''clamping''' produces a unit distance graph <math>G'</math> where <math>\chi(G')\geq\chi(G)</math>.

Devirtualizing bichromatic virtual edges of a graph <math>G</math> (i.e. clamping) has no benefit unless nontrivial points of <math>G</math> and <math>H_0</math> overlap or nontrivial points of <math>H_1</math> and <math>H_0</math> overlap, where <math>H_0</math> and <math>H_1</math> are graphs virtualizing bichromatic virtual edges of <math>G</math>. Such overlaps may cause <math>\chi(G')>\chi(G)</math>. If no nontrivial overlaps exist, then <math>\chi(G')=\max\left(\chi(G),\chi(H_0),\chi(H_1),\dots\right)</math>.

Because more than 1 graph virtualizes any given bichromatic virtual edge for a fixed length and fixed chromatic number, multiple devirtualizations exist for every finite graph.

===Relation to rings===
A '''devirtualized ring''' of graph <math>G</math> is a ring which contains all the vertices of a devirtualization of <math>G</math>, where the vertices are interpreted as complex numbers.

Use the implied shorthand for arbitrary sets <math>\mathbb{S}[\left\{a,b,c,\dots\right\}]=\mathbb{S}[a,b,c,\dots]</math>.

Let the vertices of graph <math>G</math> generate the ring <math>S_0[T_0]</math>.
Let the bichromatic virtual edges be virtualizable by the graphs <math>H_1,\dots,H_m</math>.
Let <math>S_k[T_k]</math> be the ring generated by the vertices of <math>H_k</math>.
Let <math>S</math> be the ring generated by <math>\bigcup_{k=0}^m S_k</math>.
Let <math>T=\bigcup_{k=0}^m T_k</math>.
Then <math>S[T]</math> is a devirtualized ring of <math>G</math>.

As multiple graphs may virtualize the same virtual edge, careful choice may be required to minimize the set <math>T</math>.

===Multiplicative property===
Let <math>H_0</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_0</math> and chromatic number <math>n</math>.
Let <math>H_1</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_1</math> and chromatic number <math>n</math>.
Create a graph <math>H_2</math> by scaling <math>H_0</math> by a factor of <math>d_1</math>. Graph <math>H_2</math> then virtualizes a bichromatic virtual edge length <math>d_0d_1</math> and chromatic number <math>n</math>.

===Notable bichromatic virtual edges===
{| border=1
|-
! Chromatic number !! Length <math>d</math>!! Proof
|-
| any
| 1
| trivial case
|-
| 2
| <math>0\leq d\leq 3</math>
| moving the flexible parts of the graph (0,0),(0,1),(0,2),(0,3)
|-
| 3
| <math>\left\{\left|\frac{(3+\sqrt{-3})k}{2}+\sqrt{-3}j+(-1)^r\right| \mid k,j\in\mathbb{Z}, r\in\mathbb{R}\right\}</math>
| equilateral triangle tiling
|}

===Continuous ranges of bichromatic virtual edges===
Flexible graphs might produce continuous ranges of lengths of bichromatic virtual edges of chromatic number <math>n</math> without having a specific pair which is monochrome for every <math>n</math>-coloring.
Let the range of lengths be on the interval <math>d_0\leq d\leq d_1</math>.

If <math>1 < d_1</math>, then let <math>k\in\mathbb{Z}^+,d_0^{k+1} \leq d_1^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all lengths larger than <math>d_0^k</math>. A finite area allowed to be the same color, the infinite area of the plane, and a finite upper bound on CNP implies <math>CNP> n</math>.

If <math>d_0 < 1</math>, then let <math>k\in\mathbb{Z}^+,d_1^{k+1} \geq d_0^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all non-zero lengths smaller than <math>d_1^k</math>. Considering a set of <math>CNP+1</math> all within distance <math>d_1^k</math> of each other implies <math>CNP> n</math>.

Using a flexible bichromatic virtual edge of chromatic number <math>n</math> and a graph <math>H</math> of chromatic number <math>n+1</math>, a flexible a graph <math>H'</math> of chromatic number <math>n+1</math> can be created by scaling <math>H</math> to some arbitrarily large or arbitrarily small size and clamping flexible bichromatic virtual edges of chromatic number <math>n</math> as a replacement of each rigid bichromatic virtual edge of <math>H</math>.

<math>H</math> virtualizing a bichromatic virtual edge of chromatic number <math>n+1</math> does not necessarily men the graph <math>H'</math> virtualizes a bichromatic virtual edge.

== Probabilistic formulation ==

See [[Probabilistic formulation of Hadwiger-Nelson problem]].

== Further questions ==

* What are the [http://mathworld.wolfram.com/IndependenceRatio.html independence ratios] of the above unit distance graphs?
* What are the [https://en.wikipedia.org/wiki/Fractional_coloring fractional chromatic numbers] of these graphs?
* What are the [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz numbers] of these graphs?
** The Lovasz theta function value of Lovasz number of <math>G_2</math> at the complement [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3844 is in the interval [3.3746, 3.3748]].
* What about the Erdos unit distance graph (<math>n</math> vertices, <math>n^{1+c/\log\log n}</math> edges)?
* Can we use de Grey’s graph to construct unit-distance graphs that are not 5-colorable? To answer this question, we first need to understand how 5-colorings of de Grey’s graph force small collections of vertices to be colored. [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3899 Varga] and [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3940 Nazgand] provide some thoughts along these lines. Even if we can’t stitch together a 6-chromatic unit-distance graph with these ideas, we might be able to apply them to [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3900 prove that the measurable chromatic number of the plane is at least 6].
* It appears as though the coordinates of our smallest 5-chromatic graph lie in <math>\mathbb{Q}[\sqrt{3}, \sqrt{5}, \sqrt{11}]</math> (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3879 this]). If we view the plane as the complex plane, what is the smallest ring that admits a 5-chromatic single-distance graph? Every single-distance graph in the Eistenstein integers and Gaussian integers is 2-chromatic (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3914 this]). [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3934 David Speyer suggests] looking at <math>\mathbb{Z}[\frac{1+\sqrt{-71}}{2}]</math> next.

== Blog, forums, and media ==

* [https://rjlipton.wordpress.com/2011/05/22/more-on-coloring-the-plane/ More On Coloring The Plane], Richard Lipton, May 22, 2011.
* [https://mathoverflow.net/questions/236392/has-there-been-a-computer-search-for-a-5-chromatic-unit-distance-graph Has there been a computer search for a 5-chromatic unit distance graph?], Juno, Apr 16, 2016.
* [https://quomodocumque.wordpress.com/2018/04/09/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Jordan Ellenberg, Apr 9 2018.
* [https://gilkalai.wordpress.com/2018/04/10/aubrey-de-grey-the-chromatic-number-of-the-plane-is-at-least-5/ Aubrey de Grey: The chromatic number of the plane is at least 5], Gil Kalai, Apr 10 2018.
* [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Dustin Mixon, Apr 10, 2018.
* [https://www.scottaaronson.com/blog/?p=3697 Amazing progress on long-standing problems], Scott Aaronson, Apr 11 2018.
* [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ The chromatic number of the plane is at least 5, Part II], Dustin Mixon, Apr 13 2018.
* [http://community.wolfram.com/groups/-/m/t/1320004 A 5-chromatic unit distance graph], Ed Pegg, Apr 13 2018.
* [http://aperiodical.com/2018/04/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Katie Steckles, Apr 17 2018.
* [https://www.quantamagazine.org/decades-old-graph-problem-yields-to-amateur-mathematician-20180417/ Decades-Old Graph Problem Yields to Amateur Mathematician], Evelyn Lamb, Quanta, Apr 17, 2018.
* [https://www.heise.de/newsticker/meldung/Zahlen-bitte-Wie-bunt-ist-die-Ebene-4024574.html Zahlen, bitte! 5 - Wie bunt ist die Ebene?], Harald Bögeholz, Heise, Apr 17, 2018.
* [http://www.sciencemag.org/news/2018/04/amateur-mathematician-cracks-decades-old-math-problem Amateur mathematician cracks decades-old math problem], Katie Langin, Science News, Apr 18, 2018.
* [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable How much of the plane is 4-colorable?], Dustin Mixon, Apr 18, 2018.
* [https://www.sciencealert.com/amateur-solves-decades-old-maths-problem-about-colours-that-can-never-touch-hadwiger-nelson-problem An Amateur Solved a 60-Year-Old Maths Problem About Colours That Can Never Touch], Peter Dockrill, ScienceAlert, Apr 19, 2018.
* [https://www.zmescience.com/science/math/amateur-mathematician-solves-problem-20042018/ Amateur mathematician Aubrey de Grey, known for his work on anti-aging, solves decades-old problem], Mihai Andrei, ZME Science, Apr 20, 2018.
* [https://automaths.blog/2018/04/21/5-nuances-daubrey-de-grey/ 5 nuances d’Aubrey de Grey], Automaths, Apr 21, 2018.
* [https://science.howstuffworks.com/math-concepts/amateur-solves-part-of-decades-old-math-problem.htm Amateur Solves Part of Decades-Old Math Problem], HowStuffWorks, Apr 30, 2018.
* [https://www.nemokennislink.nl/publicaties/het-platte-vlak-heeft-minstens-vijf-kleuren-nodig/ Het platte vlak heeft minstens vijf kleuren nodig], Kennislink, May 3, 2018.

== Code and data ==

[https://www.dropbox.com/sh/ufknm1v9gtbhad3/AACB2xwaXYx5EGda38_L-0foa?dl=0 This dropbox folder] will contain most of the data and images for the project.

Data:

* [https://www.dropbox.com/s/x6kvu7b3wvdvqsn/graph.dimacs?dl=0 The 1585-vertex graph in DIMACS format]
* [https://www.dropbox.com/s/qk3gbpjvvsjfsc3/sat.dimacs?dl=0 A naive translation of 4-colorability of this graph into a SAT problem in DIMACS format]
* [https://www.dropbox.com/s/nipqikfzcfn9o5a/vertices.sage?dl=0 The vertices of this graph in explicit Sage notation]
* The graph <math>G_2</math>: [http://www.cs.utexas.edu/~marijn/CNP/874.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/874.edge edges (DIMACS)]
* The graph <math>G_3</math>: [http://www.cs.utexas.edu/~marijn/CNP/826.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/826.edge edges (DIMACS)] [http://www.cs.utexas.edu/~marijn/CNP/826.pdf Visualization]
* The [https://www.dropbox.com/sh/ufknm1v9gtbhad3/AADfw9WO1ol2ayQNJEtGf8oBa/Graphs?dl=0 densest unit-distance graphs] on an <math>n\times n</math> grid for <math>n=10,20,\ldots,100</math> (DIMACS format).

Code:

* [https://www.dropbox.com/s/6u1jctbjy38t383/lovaszmoser.m?dl=0 MATLAB script for computing Lovasz number]
* [https://files.jixco.de/pm16/zzvtx/ Python code for converting a list of vertices in Mathematica format into vertices in Z^n]
* [https://files.jixco.de/pm16/edge2cnf.py Python code for converting a DIMACS edge list into a CNF formula (forcing up to 3 suitable vertices to a fixed color, to break some symmetries)]

Software:

* [http://cvxr.com/cvx/ CVX]
* [http://www.labri.fr/perso/lsimon/glucose/ Glucose 4.0]
* [http://minisat.se/ Minisat]

== Wikipedia ==

* [https://en.wikipedia.org/wiki/Graph_coloring#Chromatic_number Chromatic number]
* [https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Erd%C5%91s_theorem_(graph_theory) de Bruijn-Erdos theorem]
* [https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem Hadwiger-Nelson problem]
* [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz number]
* [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]

== Bibliography ==
* [B2008] B. Bukh, [https://doi.org/10.1007/s00039-008-0673-8 Measurable sets with excluded distances], Geometric and Functional Analysis 18 (2008), 668-697.
* [CR2015] D. Cranston, L. Rabern, [https://arxiv.org/abs/1501.01647 The fractional chromatic number of the plane], arXiv:1501.01647
* [deBE1951] N. G. de Bruijn, P. Erdős, [http://www.math-inst.hu/~p_erdos/1951-01.pdf A colour problem for infinite graphs and a problem in the theory of relations], Nederl. Akad. Wetensch. Proc. Ser. A, 54 (1951): 371–373, MR 0046630.
* [deG2018] A. de Grey, [https://arxiv.org/abs/1804.02385 The chromatic number of the plane is at least 5], arXiv:1804.02385
* [EI2018] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.00157 The chromatic number of the plane is at least 5 - a new proof], arXiv:1805.00157
* [F1981] K.J. Falconer, The Realization of distances in measurable subsets covering Rn, J. Combin. Theory Ser. A 31 (1981) 184–189.
* [H1945] H. Hadwiger, Uberdeckung des euklidischen Raum durch kongruente Mengen, Portugaliae Math. 4 (1945), 238–242.
* [MM1961] L. Moser and M. Moser, Solution to Problem 10, Can. Math. Bull. 4 (1961), 187–189.
* [P1998] D. Pritikin, [https://www.sciencedirect.com/science/article/pii/S0095895698918196 All unit-distance graphs of order 6197 are 6-colorable], Journal of Combinatorial Theory, Series B 73.2 (1998): 159-163.
* [S2008] A. Soifer, The Mathematical Coloring Book, Springer, 2008, ISBN-13: 978-0387746401.
* [T1999] C. Thomassen, On the Nelson unit distance coloring problem, Amer. Math. Monthly 106 (1999) 850-853.
* [Tow2005] Townsend, S.P., Colouring the plane with no monochrome unit. Geombinatorics XIV(4) (2005), 181-193.
* [W1973] Woodall, D.R., Distances realized by sets covering the plane. J. Combin. Theory Ser. A, 14 (1973), 187-200.

Probabilistic formulation of Hadwiger-Nelson problem

2018-05-04T01:50:41Z

Nazgand: Generalize

Hadwiger-Nelson problem

2018-05-03T23:19:06Z

Nazgand: /* Notable bichromatic virtual edges */

The Chromatic Number of the Plane (CNP) is the chromatic number of the graph whose vertices are elements of the plane, and two points are connected by an edge if they are a unit distance apart. The Hadwiger-Nelson problem asks to compute CNP. The bounds <math>4 \leq CNP \leq 7</math> are classical; recently [deG2018] it was shown that <math>CNP \geq 5</math>. This is achieved by explicitly locating finite unit distance graphs with chromatic number at least 5.

The '''Polymath16''' project seeks to simplify the graphs used in [deG2018] to establish this lower bound. More precisely, the goals are

* '''Goal 1''': Find progressively smaller 5-chromatic unit-distance graphs.
* '''Goal 2''': Reduce (ideally to zero) the reliance on computer assistance for the proof. Computer assistance was leveraged in [deG2018] to analyze a subgraph of size 397.
* '''Goal 3''': Apply these simpler graphs to inform progress in related areas. For example:
** Find a 6-chromatic unit-distance graph.
** Improve the corresponding bound in higher dimensions.
** Improve the current record of 105/29 for the fractional chromatic number of the plane.
** Find the smallest unit-distance graph of a given minimum degree (excluding, in some natural way, boring cases like Cartesian products of a graph with a hypercube).

== Polymath threads ==

* [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/ Polymath proposal: finding simpler unit distance graphs of chromatic number 5], Aubrey de Grey, Apr 10 2018. ('''Active discussion thread''')
* [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/ Polymath16, first thread: Simplifying de Grey’s graph], Dustin Mixon, Apr 14, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic Polymath16, second thread: What does it take to be 5-chromatic?], Dustin Mixon, Apr 22, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/ Polymath16, third thread: Is 6-chromatic within reach?], Dustin Mixon, May 1, 2018. ('''Active research thread''')

== Notable unit distance graphs ==

A '''unit distance graph''' is a graph that can be realised as a collection of vertices in the plane, with two vertices connected by an edge if they are precisely a unit distance apart. The chromatic number of any such graph is a lower bound for <math>CNP</math>; in particular, if one can find a unit distance graph with no 4-colorings, then <math>CNP \geq 5</math>. The boldface number of vertices is the current minimal number of vertices of a unit distance graph that is currently known to not be 4-colorable.

<math>G_1 \oplus G_2</math> denotes the Minkowski sum of two unit distance graphs <math>G_1,G_2</math> (vertices in <math>G_1 \oplus G_2</math> are sums of the vertices of <math>G_1,G_2</math>). <math>G_1 \cup G_2</math> denotes the union. <math>\mathrm{rot}(G, \theta)</math> denotes <math>G</math> rotated counterclockwise by <math>\theta</math>. <math>\mathrm{trim}(G,r)</math> denotes the trimming of <math>G</math> after removing all vertices of distance greater than <math>r</math> from the origin.

Another basic operation is '''spindling''': taking two copies of a graph <math>G</math>, gluing them together at one vertex, and rotating the copies so that the two copies of another vertex are a unit distance apart. For instance, the Moser spindle is the spindling of a rhombus graph. If a graph has two vertices forced to be the same color in a k-coloring, then the spindling of that graph at those two vertices is not k-colorable.

Many of the graphs are embeddable in abelian groups or rings, particularly those generated by the roots of unity <math>\omega_t := \exp( i \arccos( 1 - \tfrac{1}{2t} ))</math> for various natural numbers t.

{| border=1
|-
!Name!!Number of vertices!! Number of edges !! Structure !! Group !! Colorings
|-
| [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]
| 7
| 11
| Two 60-120-60-120 rhombi with a common vertex, with one pair of sharp vertices coincident and the other joined
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [http://mathworld.wolfram.com/GolombGraph.html Golomb graph]
| 10
| 18
| Contains the center and vertices of a hexagon and equilateral triangle
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [https://arxiv.org/abs/1804.02385 H]
| 7
| 12
| Vertices and center of a hexagon
| <math>{\bf Z}[\omega_1]</math>
| Has essentially four 4-colorings, two of which contain a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 J]
| 31
| 72
| Contains 13 copies of H
| <math>{\bf Z}[\omega_1]</math>
| Has essentially six 4-colorings in which no H has a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 K]
| 61
| 150
| Contains 2 copies of J
|
| In all 4-colorings lacking an H with a monochromatic <math>\sqrt{3}</math>-triangle, all pairs of vertices at distance 4 are monochromatic
|-
| [https://arxiv.org/abs/1804.02385 L]
| 121
| 301
| Contains two copies of K and 52 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ <math>L_1</math>]
| 97
|
| Has 40 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120295 <math>L_2</math>]
| 120
| 354
|
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 T]
| 9
| 15
| Contains one Moser spindle and useful symmetry; three vertices form an equilateral triangle
|
|
|-
| [https://arxiv.org/abs/1804.02385 U]
| 15
| 33
| Three copies of T at 120-degree rotations: <math>T \cup \mathrm{rot}(T, 2\pi/3) \cup \mathrm{rot}(T, 4\pi/3)</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 V]
| 31
| 30
| Unit vectors at angles consistent with three interlocking Moser spindles
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>V_1</math>]
| 61
| 60
| Union of V and a rotation of V: <math>V \cup \mathrm{rot}(V, \mathrm{arccos}(7/8))</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_b</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_x</math>]
| 13
| 12
| Subgraph of <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_z</math>]
|
|
| Subgraph of <math>V_a</math>; shares a line of symmetry with <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_y</math>]
| 13
| 12
| Subgraph of <math>V_b</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-3970 <math>V_A</math>]
| 37
| 36
| Unit vectors with angles <math>i \frac{\pi}{3} + j \mathrm{arccos} \frac{5}{6} + k \mathrm{arccos} \frac{7}{8}</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 W]
| 301
| 1230
| Cartesian product of V with itself, minus vertices at more than <math>\sqrt{3}</math> from the centre (i.e. <math>\mathrm{trim}(V \oplus V, \sqrt{3})</math>)
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>W_1</math>]
|
|
| Trimmed product of V with itself (<math>\mathrm{trim}(V \oplus V, 1.95)</math>)
|
|
|-
| [https://arxiv.org/abs/1804.02385 M]
| 1345
| 8268
| Cartesian product of W and H (<math>W \oplus H</math>)
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120274 <math>M_1</math>]
| 278
|
| Deleting vertices from M while maintaining its restriction on H
|
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>M_2</math>]
| 7075
|
| Sum of H with a trimmed product of <math>V_1</math> with itself
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 N]
| 20425
| 151311
| Contains 52 copies of M arranged around the H-copies of L
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4, \omega_{16}]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_0</math>]
| 1585
| 7909
| N "shrunk" by stepwise deletions and replacements of vertices
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 G]
| 1581
| 7877
| Deleting 4 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_1</math>]
| 1577
|
| Deleting 8 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120318 <math>G_2</math>]
| 874
| 4461
| Juxtaposing two copies of M and shrinking
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3867 <math>G_3</math>]
| 826
| 4273
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4105 <math>G_4</math>]
| '''803'''
| 4144
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 R]
|
|
| Union of <math>W_1</math> and a rotated copy of <math>W_1</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>\mathrm{trim}(R \oplus H, 1.67)</math>]
| 2563
|
| Trimmed sum of R and H
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 <math>V \oplus V \oplus H</math>]
|
|
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_{64/9}]</math>
| Has two vertices forced to be the same color in a 4-coloring; also no monochromatic <math>\sqrt{3}</math>-triangles
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 No name assigned]
| 745
|
| Subgraph of <math>V \oplus V \oplus H</math>
|
| Has two vertices forced to be the same color in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3085
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_a \oplus V_z \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3049
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 No name assigned]
| 1951
|
| Trimmed version of <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic <math>V_A \oplus V_A \oplus V_A</math>]
| 6937
| 44439
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{40}</math>]
| 40
| 82
|
|
| Any 4-coloring that avoids a monochromatic <math>\sqrt{11/3}</math>-edge has two specific vertices forced to be the same color
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{79}</math>]
| 79
| 165
| Spindling of <math>G_{40}</math>
|
| Any 4-coloring has a monochromatic <math>\sqrt{11/3}</math>-edge
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{49}</math>]
| 49
| 180
|
| <math>{\mathbf Q}[\sqrt{3},\sqrt{11}]^2</math>
| Any 4-coloring either has a specific <math>\sqrt{11/3}</math>-edge monochromatic, or a monochromatic <math>1/\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{51}</math>]
| 51
|
|
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring of plane
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{627}</math>]
| 627
| 2982
| Contains <math>G_{51}</math>
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4161 No name assigned]
| 103
|
|
|
| All 4-colorings contain a monochromatic <math>2/\sqrt{3}</math>-edge
|}

== Lower bounds ==

In [P1988], Pritikin proved that every graph with at most 12 vertices is 4-colorable, and every graph with at most 6197 vertices is 6-colorable. Pritikin's bounds are obtained by coloring the plane with k colors and an additional “wild” color such that points of unit distance are both allowed to receive the wild color. If the wild color occupies a small fraction p of the plane, then an exercise in the probabilistic method gives that any fixed unit-distance graph with n vertices enjoys an embedding in the plane that avoids the wild color (and is therefore k-colorable) provided n<1/p. Pritikin’s coloring for the k=4 case cannot be improved without improving on the densest known subset of the plane that avoids unit distances (originally due to Croft in 1967). See [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable this MO thread] for additional information. As such, improving the bound in the k=4 case might require a new technique, whereas the k=5 and 6 cases might be amenable to optimization.

[https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4176 Every unit distance graph with at most 16 vertices is 5-colorable].

A "tile-based" colouring of the plane must have at least 6 colours, as shown by Townsend [Tow2005]; the same proof with a minor error was also derived by Woodall [W1973]. In [T1999], Thomassen showed that any tiling-based 6-coloring would have to be be "unscaleable", i.e. the maximum diameter of a tile and the minimum separation of same-coloured tiles must both be exactly 1 (so that the distance 1 is excluded by suitable colouring of the boundary points).

== Virtual edge ==

''Warning: other definitions have been proposed and the exact definition of this notion is currently under discussion.''

A '''virtual edge''' of a unit distance graph <math>G</math> is a distance <math>d</math> with the property that every 4-coloring of <math>G</math> contains a monochromatic pair of vertices of distance exactly <math>d</math>. Observe that if a unit distance graph <math>G</math> has a virtual edge at distance <math>d</math>, and if there is another unit distance graph <math>H</math> with a pair of vertices at distance <math>d</math> that cannot be monochromatic in a 4-coloring, then we can create a non-4-colorable unit distance graph by "clamping" a copy of <math>H</math> to every virtual edge in <math>G</math>.

Known examples of virtual edges include:

* <math>G_{40}</math> has a virtual edge at distance <math>\sqrt{11/3}</math>.
* <math>V \oplus V \oplus H</math> has a (single) virtual edge at distance <math>8/3</math>.
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4117 <math>(\sqrt{3}\pm 1)/\sqrt{2}</math> are virtual edges of some graphs].

== Bichromatic virtual edge ==

===Definition===
If a unit distance graph <math>H</math> exists with a specific pair of vertices which are distance <math>d</math> apart and is bichromatic in all proper <math>n</math>-colorings of <math>H</math>, then we say that <math>H</math> virtualizes a '''bichromatic virtual edge''' with length <math>d</math> and chromatic number <math>n</math>.

===Vacuous properties===
Considering the graph of the entire plane, all distances correspond to a virtual edge of chromatic number <math>CNP-1</math>.

If a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n</math> exists, then a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n-1</math> exists.

===Convenience and devirtualization===
Bichromatic virtual edges behave the same as the unit edge(which is a bichromatic virtual edge of every chromatic number), and thus may be used to simplify the construction of graphs.

Recursively replacing bichromatic virtual edges of a graph <math>G</math> with graphs which virtualize the bichromatic virtual edges through '''clamping''' produces a unit distance graph <math>G'</math> where <math>\chi(G')\geq\chi(G)</math>.

Devirtualizing bichromatic virtual edges of a graph <math>G</math> (i.e. clamping) has no benefit unless nontrivial points of <math>G</math> and <math>H_0</math> overlap or nontrivial points of <math>H_1</math> and <math>H_0</math> overlap, where <math>H_0</math> and <math>H_1</math> are graphs virtualizing bichromatic virtual edges of <math>G</math>. Such overlaps may cause <math>\chi(G')>\chi(G)</math>. If no nontrivial overlaps exist, then <math>\chi(G')=\max\left(\chi(G),\chi(H_0),\chi(H_1),\dots\right)</math>.

Because more than 1 graph virtualizes any given bichromatic virtual edge for a fixed length and fixed chromatic number, multiple devirtualizations exist for every finite graph.

===Relation to rings===
A '''devirtualized ring''' of graph <math>G</math> is a ring which contains all the vertices of a devirtualization of <math>G</math>, where the vertices are interpreted as complex numbers.

Use the implied shorthand for arbitrary sets <math>\mathbb{S}[\left\{a,b,c,\dots\right\}]=\mathbb{S}[a,b,c,\dots]</math>.

Let the vertices of graph <math>G</math> generate the ring <math>S_0[T_0]</math>.
Let the bichromatic virtual edges be virtualizable by the graphs <math>H_1,\dots,H_m</math>.
Let <math>S_k[T_k]</math> be the ring generated by the vertices of <math>H_k</math>.
Let <math>S</math> be the ring generated by <math>\bigcup_{k=0}^m S_k</math>.
Let <math>T=\bigcup_{k=0}^m T_k</math>.
Then <math>S[T]</math> is a devirtualized ring of <math>G</math>.

As multiple graphs may virtualize the same virtual edge, careful choice may be required to minimize the set <math>T</math>.

===Multiplicative property===
Let <math>H_0</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_0</math> and chromatic number <math>n</math>.
Let <math>H_1</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_1</math> and chromatic number <math>n</math>.
Create a graph <math>H_2</math> by scaling <math>H_0</math> by a factor of <math>d_1</math>. Graph <math>H_2</math> then virtualizes a bichromatic virtual edge length <math>d_0d_1</math> and chromatic number <math>n</math>.

===Notable bichromatic virtual edges===
{| border=1
|-
! Chromatic number !! Length <math>d</math>!! Proof
|-
| any
| 1
| trivial case
|-
| 2
| <math>0\leq d\leq 3</math>
| moving the flexible parts of the graph (0,0),(0,1),(0,2),(0,3)
|-
| 3
| <math>\left\{\left|\frac{(3+\sqrt{-3})k}{2}+\sqrt{-3}j+(-1)^r\right| \mid k,j\in\mathbb{Z}, r\in\mathbb{R}\right\}</math>
| equilateral triangle tiling
|}

===Continuous ranges of bichromatic virtual edges===
Flexible graphs might produce continuous ranges of lengths of bichromatic virtual edges of chromatic number <math>n</math> without having a specific pair which is monochrome for every <math>n</math>-coloring.
Let the range of lengths be on the interval <math>d_0\leq d\leq d_1</math>.

If <math>1 > d_1</math>, then let <math>k\in\mathbb{Z}^+,d_0^{k+1} \leq d_1^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all lengths larger than <math>d_0^k</math>. A finite area allowed to be the same color, the infinite area of the plane, and a finite upper bound on CNP implies <math>CNP> n</math>.

If <math>d_0 < 1</math>, then let <math>k\in\mathbb{Z}^+,d_1^{k+1} \geq d_0^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all non-zero lengths smaller than <math>d_1^k</math>. Considering a set of <math>CNP+1</math> all within distance <math>d_1^k</math> of each other implies <math>CNP> n</math>.

Using a flexible bichromatic virtual edge of chromatic number <math>n</math> and a graph <math>H</math> of chromatic number <math>n+1</math>, a flexible a graph <math>H'</math> of chromatic number <math>n+1</math> can be created by scaling <math>H</math> to some arbitrarily large or arbitrarily small size and clamping flexible bichromatic virtual edges of chromatic number <math>n</math> as a replacement of each rigid bichromatic virtual edge of <math>H</math>.

<math>H</math> virtualizing a bichromatic virtual edge of chromatic number <math>n+1</math> does not necessarily men the graph <math>H'</math> virtualizes a bichromatic virtual edge.

== Probabilistic formulation ==

See [[Probabilistic formulation of Hadwiger-Nelson problem]].

== Further questions ==

* What are the [http://mathworld.wolfram.com/IndependenceRatio.html independence ratios] of the above unit distance graphs?
* What are the [https://en.wikipedia.org/wiki/Fractional_coloring fractional chromatic numbers] of these graphs?
* What are the [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz numbers] of these graphs?
** The Lovasz theta function value of Lovasz number of <math>G_2</math> at the complement [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3844 is in the interval [3.3746, 3.3748]].
* What about the Erdos unit distance graph (<math>n</math> vertices, <math>n^{1+c/\log\log n}</math> edges)?
* Can we use de Grey’s graph to construct unit-distance graphs that are not 5-colorable? To answer this question, we first need to understand how 5-colorings of de Grey’s graph force small collections of vertices to be colored. [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3899 Varga] and [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3940 Nazgand] provide some thoughts along these lines. Even if we can’t stitch together a 6-chromatic unit-distance graph with these ideas, we might be able to apply them to [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3900 prove that the measurable chromatic number of the plane is at least 6].
* It appears as though the coordinates of our smallest 5-chromatic graph lie in <math>\mathbb{Q}[\sqrt{3}, \sqrt{5}, \sqrt{11}]</math> (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3879 this]). If we view the plane as the complex plane, what is the smallest ring that admits a 5-chromatic single-distance graph? Every single-distance graph in the Eistenstein integers and Gaussian integers is 2-chromatic (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3914 this]). [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3934 David Speyer suggests] looking at <math>\mathbb{Z}[\frac{1+\sqrt{-71}}{2}]</math> next.

== Blog, forums, and media ==

* [https://rjlipton.wordpress.com/2011/05/22/more-on-coloring-the-plane/ More On Coloring The Plane], Richard Lipton, May 22, 2011.
* [https://mathoverflow.net/questions/236392/has-there-been-a-computer-search-for-a-5-chromatic-unit-distance-graph Has there been a computer search for a 5-chromatic unit distance graph?], Juno, Apr 16, 2016.
* [https://quomodocumque.wordpress.com/2018/04/09/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Jordan Ellenberg, Apr 9 2018.
* [https://gilkalai.wordpress.com/2018/04/10/aubrey-de-grey-the-chromatic-number-of-the-plane-is-at-least-5/ Aubrey de Grey: The chromatic number of the plane is at least 5], Gil Kalai, Apr 10 2018.
* [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Dustin Mixon, Apr 10, 2018.
* [https://www.scottaaronson.com/blog/?p=3697 Amazing progress on long-standing problems], Scott Aaronson, Apr 11 2018.
* [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ The chromatic number of the plane is at least 5, Part II], Dustin Mixon, Apr 13 2018.
* [http://community.wolfram.com/groups/-/m/t/1320004 A 5-chromatic unit distance graph], Ed Pegg, Apr 13 2018.
* [http://aperiodical.com/2018/04/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Katie Steckles, Apr 17 2018.
* [https://www.quantamagazine.org/decades-old-graph-problem-yields-to-amateur-mathematician-20180417/ Decades-Old Graph Problem Yields to Amateur Mathematician], Evelyn Lamb, Quanta, Apr 17, 2018.
* [https://www.heise.de/newsticker/meldung/Zahlen-bitte-Wie-bunt-ist-die-Ebene-4024574.html Zahlen, bitte! 5 - Wie bunt ist die Ebene?], Harald Bögeholz, Heise, Apr 17, 2018.
* [http://www.sciencemag.org/news/2018/04/amateur-mathematician-cracks-decades-old-math-problem Amateur mathematician cracks decades-old math problem], Katie Langin, Science News, Apr 18, 2018.
* [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable How much of the plane is 4-colorable?], Dustin Mixon, Apr 18, 2018.
* [https://www.sciencealert.com/amateur-solves-decades-old-maths-problem-about-colours-that-can-never-touch-hadwiger-nelson-problem An Amateur Solved a 60-Year-Old Maths Problem About Colours That Can Never Touch], Peter Dockrill, ScienceAlert, Apr 19, 2018.
* [https://www.zmescience.com/science/math/amateur-mathematician-solves-problem-20042018/ Amateur mathematician Aubrey de Grey, known for his work on anti-aging, solves decades-old problem], Mihai Andrei, ZME Science, Apr 20, 2018.
* [https://automaths.blog/2018/04/21/5-nuances-daubrey-de-grey/ 5 nuances d’Aubrey de Grey], Automaths, Apr 21, 2018.
* [https://science.howstuffworks.com/math-concepts/amateur-solves-part-of-decades-old-math-problem.htm Amateur Solves Part of Decades-Old Math Problem], HowStuffWorks, Apr 30, 2018.
* [https://www.nemokennislink.nl/publicaties/het-platte-vlak-heeft-minstens-vijf-kleuren-nodig/ Het platte vlak heeft minstens vijf kleuren nodig], Kennislink, May 3, 2018.

== Code and data ==

[https://www.dropbox.com/sh/ufknm1v9gtbhad3/AACB2xwaXYx5EGda38_L-0foa?dl=0 This dropbox folder] will contain most of the data and images for the project.

Data:

* [https://www.dropbox.com/s/x6kvu7b3wvdvqsn/graph.dimacs?dl=0 The 1585-vertex graph in DIMACS format]
* [https://www.dropbox.com/s/qk3gbpjvvsjfsc3/sat.dimacs?dl=0 A naive translation of 4-colorability of this graph into a SAT problem in DIMACS format]
* [https://www.dropbox.com/s/nipqikfzcfn9o5a/vertices.sage?dl=0 The vertices of this graph in explicit Sage notation]
* The graph <math>G_2</math>: [http://www.cs.utexas.edu/~marijn/CNP/874.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/874.edge edges (DIMACS)]
* The graph <math>G_3</math>: [http://www.cs.utexas.edu/~marijn/CNP/826.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/826.edge edges (DIMACS)] [http://www.cs.utexas.edu/~marijn/CNP/826.pdf Visualization]
* The [https://www.dropbox.com/sh/ufknm1v9gtbhad3/AADfw9WO1ol2ayQNJEtGf8oBa/Graphs?dl=0 densest unit-distance graphs] on an <math>n\times n</math> grid for <math>n=10,20,\ldots,100</math> (DIMACS format).

Code:

* [https://www.dropbox.com/s/6u1jctbjy38t383/lovaszmoser.m?dl=0 MATLAB script for computing Lovasz number]
* [https://files.jixco.de/pm16/zzvtx/ Python code for converting a list of vertices in Mathematica format into vertices in Z^n]
* [https://files.jixco.de/pm16/edge2cnf.py Python code for converting a DIMACS edge list into a CNF formula (forcing up to 3 suitable vertices to a fixed color, to break some symmetries)]

Software:

* [http://cvxr.com/cvx/ CVX]
* [http://www.labri.fr/perso/lsimon/glucose/ Glucose 4.0]
* [http://minisat.se/ Minisat]

== Wikipedia ==

* [https://en.wikipedia.org/wiki/Graph_coloring#Chromatic_number Chromatic number]
* [https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Erd%C5%91s_theorem_(graph_theory) de Bruijn-Erdos theorem]
* [https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem Hadwiger-Nelson problem]
* [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz number]
* [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]

== Bibliography ==
* [B2008] B. Bukh, [https://doi.org/10.1007/s00039-008-0673-8 Measurable sets with excluded distances], Geometric and Functional Analysis 18 (2008), 668-697.
* [CR2015] D. Cranston, L. Rabern, [https://arxiv.org/abs/1501.01647 The fractional chromatic number of the plane], arXiv:1501.01647
* [deBE1951] N. G. de Bruijn, P. Erdős, [http://www.math-inst.hu/~p_erdos/1951-01.pdf A colour problem for infinite graphs and a problem in the theory of relations], Nederl. Akad. Wetensch. Proc. Ser. A, 54 (1951): 371–373, MR 0046630.
* [deG2018] A. de Grey, [https://arxiv.org/abs/1804.02385 The chromatic number of the plane is at least 5], arXiv:1804.02385
* [EI2018] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.00157 The chromatic number of the plane is at least 5 - a new proof], arXiv:1805.00157
* [F1981] K.J. Falconer, The Realization of distances in measurable subsets covering Rn, J. Combin. Theory Ser. A 31 (1981) 184–189.
* [H1945] H. Hadwiger, Uberdeckung des euklidischen Raum durch kongruente Mengen, Portugaliae Math. 4 (1945), 238–242.
* [MM1961] L. Moser and M. Moser, Solution to Problem 10, Can. Math. Bull. 4 (1961), 187–189.
* [P1998] D. Pritikin, [https://www.sciencedirect.com/science/article/pii/S0095895698918196 All unit-distance graphs of order 6197 are 6-colorable], Journal of Combinatorial Theory, Series B 73.2 (1998): 159-163.
* [S2008] A. Soifer, The Mathematical Coloring Book, Springer, 2008, ISBN-13: 978-0387746401.
* [T1999] C. Thomassen, On the Nelson unit distance coloring problem, Amer. Math. Monthly 106 (1999) 850-853.
* [Tow2005] Townsend, S.P., Colouring the plane with no monochrome unit. Geombinatorics XIV(4) (2005), 181-193.
* [W1973] Woodall, D.R., Distances realized by sets covering the plane. J. Combin. Theory Ser. A, 14 (1973), 187-200.

Hadwiger-Nelson problem

2018-05-03T23:07:24Z

Nazgand: /* Relation to rings */

The Chromatic Number of the Plane (CNP) is the chromatic number of the graph whose vertices are elements of the plane, and two points are connected by an edge if they are a unit distance apart. The Hadwiger-Nelson problem asks to compute CNP. The bounds <math>4 \leq CNP \leq 7</math> are classical; recently [deG2018] it was shown that <math>CNP \geq 5</math>. This is achieved by explicitly locating finite unit distance graphs with chromatic number at least 5.

The '''Polymath16''' project seeks to simplify the graphs used in [deG2018] to establish this lower bound. More precisely, the goals are

* '''Goal 1''': Find progressively smaller 5-chromatic unit-distance graphs.
* '''Goal 2''': Reduce (ideally to zero) the reliance on computer assistance for the proof. Computer assistance was leveraged in [deG2018] to analyze a subgraph of size 397.
* '''Goal 3''': Apply these simpler graphs to inform progress in related areas. For example:
** Find a 6-chromatic unit-distance graph.
** Improve the corresponding bound in higher dimensions.
** Improve the current record of 105/29 for the fractional chromatic number of the plane.
** Find the smallest unit-distance graph of a given minimum degree (excluding, in some natural way, boring cases like Cartesian products of a graph with a hypercube).

== Polymath threads ==

* [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/ Polymath proposal: finding simpler unit distance graphs of chromatic number 5], Aubrey de Grey, Apr 10 2018. ('''Active discussion thread''')
* [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/ Polymath16, first thread: Simplifying de Grey’s graph], Dustin Mixon, Apr 14, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic Polymath16, second thread: What does it take to be 5-chromatic?], Dustin Mixon, Apr 22, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/ Polymath16, third thread: Is 6-chromatic within reach?], Dustin Mixon, May 1, 2018. ('''Active research thread''')

== Notable unit distance graphs ==

A '''unit distance graph''' is a graph that can be realised as a collection of vertices in the plane, with two vertices connected by an edge if they are precisely a unit distance apart. The chromatic number of any such graph is a lower bound for <math>CNP</math>; in particular, if one can find a unit distance graph with no 4-colorings, then <math>CNP \geq 5</math>. The boldface number of vertices is the current minimal number of vertices of a unit distance graph that is currently known to not be 4-colorable.

<math>G_1 \oplus G_2</math> denotes the Minkowski sum of two unit distance graphs <math>G_1,G_2</math> (vertices in <math>G_1 \oplus G_2</math> are sums of the vertices of <math>G_1,G_2</math>). <math>G_1 \cup G_2</math> denotes the union. <math>\mathrm{rot}(G, \theta)</math> denotes <math>G</math> rotated counterclockwise by <math>\theta</math>. <math>\mathrm{trim}(G,r)</math> denotes the trimming of <math>G</math> after removing all vertices of distance greater than <math>r</math> from the origin.

Another basic operation is '''spindling''': taking two copies of a graph <math>G</math>, gluing them together at one vertex, and rotating the copies so that the two copies of another vertex are a unit distance apart. For instance, the Moser spindle is the spindling of a rhombus graph. If a graph has two vertices forced to be the same color in a k-coloring, then the spindling of that graph at those two vertices is not k-colorable.

Many of the graphs are embeddable in abelian groups or rings, particularly those generated by the roots of unity <math>\omega_t := \exp( i \arccos( 1 - \tfrac{1}{2t} ))</math> for various natural numbers t.

{| border=1
|-
!Name!!Number of vertices!! Number of edges !! Structure !! Group !! Colorings
|-
| [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]
| 7
| 11
| Two 60-120-60-120 rhombi with a common vertex, with one pair of sharp vertices coincident and the other joined
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [http://mathworld.wolfram.com/GolombGraph.html Golomb graph]
| 10
| 18
| Contains the center and vertices of a hexagon and equilateral triangle
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [https://arxiv.org/abs/1804.02385 H]
| 7
| 12
| Vertices and center of a hexagon
| <math>{\bf Z}[\omega_1]</math>
| Has essentially four 4-colorings, two of which contain a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 J]
| 31
| 72
| Contains 13 copies of H
| <math>{\bf Z}[\omega_1]</math>
| Has essentially six 4-colorings in which no H has a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 K]
| 61
| 150
| Contains 2 copies of J
|
| In all 4-colorings lacking an H with a monochromatic <math>\sqrt{3}</math>-triangle, all pairs of vertices at distance 4 are monochromatic
|-
| [https://arxiv.org/abs/1804.02385 L]
| 121
| 301
| Contains two copies of K and 52 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ <math>L_1</math>]
| 97
|
| Has 40 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120295 <math>L_2</math>]
| 120
| 354
|
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 T]
| 9
| 15
| Contains one Moser spindle and useful symmetry; three vertices form an equilateral triangle
|
|
|-
| [https://arxiv.org/abs/1804.02385 U]
| 15
| 33
| Three copies of T at 120-degree rotations: <math>T \cup \mathrm{rot}(T, 2\pi/3) \cup \mathrm{rot}(T, 4\pi/3)</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 V]
| 31
| 30
| Unit vectors at angles consistent with three interlocking Moser spindles
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>V_1</math>]
| 61
| 60
| Union of V and a rotation of V: <math>V \cup \mathrm{rot}(V, \mathrm{arccos}(7/8))</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_b</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_x</math>]
| 13
| 12
| Subgraph of <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_z</math>]
|
|
| Subgraph of <math>V_a</math>; shares a line of symmetry with <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_y</math>]
| 13
| 12
| Subgraph of <math>V_b</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-3970 <math>V_A</math>]
| 37
| 36
| Unit vectors with angles <math>i \frac{\pi}{3} + j \mathrm{arccos} \frac{5}{6} + k \mathrm{arccos} \frac{7}{8}</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 W]
| 301
| 1230
| Cartesian product of V with itself, minus vertices at more than <math>\sqrt{3}</math> from the centre (i.e. <math>\mathrm{trim}(V \oplus V, \sqrt{3})</math>)
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>W_1</math>]
|
|
| Trimmed product of V with itself (<math>\mathrm{trim}(V \oplus V, 1.95)</math>)
|
|
|-
| [https://arxiv.org/abs/1804.02385 M]
| 1345
| 8268
| Cartesian product of W and H (<math>W \oplus H</math>)
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120274 <math>M_1</math>]
| 278
|
| Deleting vertices from M while maintaining its restriction on H
|
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>M_2</math>]
| 7075
|
| Sum of H with a trimmed product of <math>V_1</math> with itself
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 N]
| 20425
| 151311
| Contains 52 copies of M arranged around the H-copies of L
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4, \omega_{16}]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_0</math>]
| 1585
| 7909
| N "shrunk" by stepwise deletions and replacements of vertices
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 G]
| 1581
| 7877
| Deleting 4 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_1</math>]
| 1577
|
| Deleting 8 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120318 <math>G_2</math>]
| 874
| 4461
| Juxtaposing two copies of M and shrinking
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3867 <math>G_3</math>]
| 826
| 4273
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4105 <math>G_4</math>]
| '''803'''
| 4144
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 R]
|
|
| Union of <math>W_1</math> and a rotated copy of <math>W_1</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>\mathrm{trim}(R \oplus H, 1.67)</math>]
| 2563
|
| Trimmed sum of R and H
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 <math>V \oplus V \oplus H</math>]
|
|
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_{64/9}]</math>
| Has two vertices forced to be the same color in a 4-coloring; also no monochromatic <math>\sqrt{3}</math>-triangles
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 No name assigned]
| 745
|
| Subgraph of <math>V \oplus V \oplus H</math>
|
| Has two vertices forced to be the same color in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3085
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_a \oplus V_z \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3049
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 No name assigned]
| 1951
|
| Trimmed version of <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic <math>V_A \oplus V_A \oplus V_A</math>]
| 6937
| 44439
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{40}</math>]
| 40
| 82
|
|
| Any 4-coloring that avoids a monochromatic <math>\sqrt{11/3}</math>-edge has two specific vertices forced to be the same color
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{79}</math>]
| 79
| 165
| Spindling of <math>G_{40}</math>
|
| Any 4-coloring has a monochromatic <math>\sqrt{11/3}</math>-edge
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{49}</math>]
| 49
| 180
|
| <math>{\mathbf Q}[\sqrt{3},\sqrt{11}]^2</math>
| Any 4-coloring either has a specific <math>\sqrt{11/3}</math>-edge monochromatic, or a monochromatic <math>1/\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{51}</math>]
| 51
|
|
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring of plane
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{627}</math>]
| 627
| 2982
| Contains <math>G_{51}</math>
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4161 No name assigned]
| 103
|
|
|
| All 4-colorings contain a monochromatic <math>2/\sqrt{3}</math>-edge
|}

== Lower bounds ==

In [P1988], Pritikin proved that every graph with at most 12 vertices is 4-colorable, and every graph with at most 6197 vertices is 6-colorable. Pritikin's bounds are obtained by coloring the plane with k colors and an additional “wild” color such that points of unit distance are both allowed to receive the wild color. If the wild color occupies a small fraction p of the plane, then an exercise in the probabilistic method gives that any fixed unit-distance graph with n vertices enjoys an embedding in the plane that avoids the wild color (and is therefore k-colorable) provided n<1/p. Pritikin’s coloring for the k=4 case cannot be improved without improving on the densest known subset of the plane that avoids unit distances (originally due to Croft in 1967). See [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable this MO thread] for additional information. As such, improving the bound in the k=4 case might require a new technique, whereas the k=5 and 6 cases might be amenable to optimization.

[https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4176 Every unit distance graph with at most 16 vertices is 5-colorable].

A "tile-based" colouring of the plane must have at least 6 colours, as shown by Townsend [Tow2005]; the same proof with a minor error was also derived by Woodall [W1973]. In [T1999], Thomassen showed that any tiling-based 6-coloring would have to be be "unscaleable", i.e. the maximum diameter of a tile and the minimum separation of same-coloured tiles must both be exactly 1 (so that the distance 1 is excluded by suitable colouring of the boundary points).

== Virtual edge ==

''Warning: other definitions have been proposed and the exact definition of this notion is currently under discussion.''

A '''virtual edge''' of a unit distance graph <math>G</math> is a distance <math>d</math> with the property that every 4-coloring of <math>G</math> contains a monochromatic pair of vertices of distance exactly <math>d</math>. Observe that if a unit distance graph <math>G</math> has a virtual edge at distance <math>d</math>, and if there is another unit distance graph <math>H</math> with a pair of vertices at distance <math>d</math> that cannot be monochromatic in a 4-coloring, then we can create a non-4-colorable unit distance graph by "clamping" a copy of <math>H</math> to every virtual edge in <math>G</math>.

Known examples of virtual edges include:

* <math>G_{40}</math> has a virtual edge at distance <math>\sqrt{11/3}</math>.
* <math>V \oplus V \oplus H</math> has a (single) virtual edge at distance <math>8/3</math>.
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4117 <math>(\sqrt{3}\pm 1)/\sqrt{2}</math> are virtual edges of some graphs].

== Bichromatic virtual edge ==

===Definition===
If a unit distance graph <math>H</math> exists with a specific pair of vertices which are distance <math>d</math> apart and is bichromatic in all proper <math>n</math>-colorings of <math>H</math>, then we say that <math>H</math> virtualizes a '''bichromatic virtual edge''' with length <math>d</math> and chromatic number <math>n</math>.

===Vacuous properties===
Considering the graph of the entire plane, all distances correspond to a virtual edge of chromatic number <math>CNP-1</math>.

If a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n</math> exists, then a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n-1</math> exists.

===Convenience and devirtualization===
Bichromatic virtual edges behave the same as the unit edge(which is a bichromatic virtual edge of every chromatic number), and thus may be used to simplify the construction of graphs.

Recursively replacing bichromatic virtual edges of a graph <math>G</math> with graphs which virtualize the bichromatic virtual edges through '''clamping''' produces a unit distance graph <math>G'</math> where <math>\chi(G')\geq\chi(G)</math>.

Devirtualizing bichromatic virtual edges of a graph <math>G</math> (i.e. clamping) has no benefit unless nontrivial points of <math>G</math> and <math>H_0</math> overlap or nontrivial points of <math>H_1</math> and <math>H_0</math> overlap, where <math>H_0</math> and <math>H_1</math> are graphs virtualizing bichromatic virtual edges of <math>G</math>. Such overlaps may cause <math>\chi(G')>\chi(G)</math>. If no nontrivial overlaps exist, then <math>\chi(G')=\max\left(\chi(G),\chi(H_0),\chi(H_1),\dots\right)</math>.

Because more than 1 graph virtualizes any given bichromatic virtual edge for a fixed length and fixed chromatic number, multiple devirtualizations exist for every finite graph.

===Relation to rings===
A '''devirtualized ring''' of graph <math>G</math> is a ring which contains all the vertices of a devirtualization of <math>G</math>, where the vertices are interpreted as complex numbers.

Use the implied shorthand for arbitrary sets <math>\mathbb{S}[\left\{a,b,c,\dots\right\}]=\mathbb{S}[a,b,c,\dots]</math>.

Let the vertices of graph <math>G</math> generate the ring <math>S_0[T_0]</math>.
Let the bichromatic virtual edges be virtualizable by the graphs <math>H_1,\dots,H_m</math>.
Let <math>S_k[T_k]</math> be the ring generated by the vertices of <math>H_k</math>.
Let <math>S</math> be the ring generated by <math>\bigcup_{k=0}^m S_k</math>.
Let <math>T=\bigcup_{k=0}^m T_k</math>.
Then <math>S[T]</math> is a devirtualized ring of <math>G</math>.

As multiple graphs may virtualize the same virtual edge, careful choice may be required to minimize the set <math>T</math>.

===Multiplicative property===
Let <math>H_0</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_0</math> and chromatic number <math>n</math>.
Let <math>H_1</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_1</math> and chromatic number <math>n</math>.
Create a graph <math>H_2</math> by scaling <math>H_0</math> by a factor of <math>d_1</math>. Graph <math>H_2</math> then virtualizes a bichromatic virtual edge length <math>d_0d_1</math> and chromatic number <math>n</math>.

===Notable bichromatic virtual edges===
{| border=1
|-
! Chromatic number !! Length <math>d</math>!! Proof
|-
| any
| 1
| trivial case
|-
| 2
| <math>0\leq d\leq 4</math>
| moving the flexible parts of the graph (0,0),(0,1),(0,2),(0,3)
|-
| 3
| <math>\left\{\left|\frac{(3+\sqrt{-3})k}{2}+\sqrt{-3}j+(-1)^r\right| \mid k,j\in\mathbb{Z}, r\in\mathbb{R}\right\}</math>
| equilateral triangle tiling
|}

===Continuous ranges of bichromatic virtual edges===
Flexible graphs might produce continuous ranges of lengths of bichromatic virtual edges of chromatic number <math>n</math> without having a specific pair which is monochrome for every <math>n</math>-coloring.
Let the range of lengths be on the interval <math>d_0\leq d\leq d_1</math>.

If <math>1 > d_1</math>, then let <math>k\in\mathbb{Z}^+,d_0^{k+1} \leq d_1^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all lengths larger than <math>d_0^k</math>. A finite area allowed to be the same color, the infinite area of the plane, and a finite upper bound on CNP implies <math>CNP> n</math>.

If <math>d_0 < 1</math>, then let <math>k\in\mathbb{Z}^+,d_1^{k+1} \geq d_0^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all non-zero lengths smaller than <math>d_1^k</math>. Considering a set of <math>CNP+1</math> all within distance <math>d_1^k</math> of each other implies <math>CNP> n</math>.

Using a flexible bichromatic virtual edge of chromatic number <math>n</math> and a graph <math>H</math> of chromatic number <math>n+1</math>, a flexible a graph <math>H'</math> of chromatic number <math>n+1</math> can be created by scaling <math>H</math> to some arbitrarily large or arbitrarily small size and clamping flexible bichromatic virtual edges of chromatic number <math>n</math> as a replacement of each rigid bichromatic virtual edge of <math>H</math>.

<math>H</math> virtualizing a bichromatic virtual edge of chromatic number <math>n+1</math> does not necessarily men the graph <math>H'</math> virtualizes a bichromatic virtual edge.

== Probabilistic formulation ==

See [[Probabilistic formulation of Hadwiger-Nelson problem]].

== Further questions ==

* What are the [http://mathworld.wolfram.com/IndependenceRatio.html independence ratios] of the above unit distance graphs?
* What are the [https://en.wikipedia.org/wiki/Fractional_coloring fractional chromatic numbers] of these graphs?
* What are the [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz numbers] of these graphs?
** The Lovasz theta function value of Lovasz number of <math>G_2</math> at the complement [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3844 is in the interval [3.3746, 3.3748]].
* What about the Erdos unit distance graph (<math>n</math> vertices, <math>n^{1+c/\log\log n}</math> edges)?
* Can we use de Grey’s graph to construct unit-distance graphs that are not 5-colorable? To answer this question, we first need to understand how 5-colorings of de Grey’s graph force small collections of vertices to be colored. [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3899 Varga] and [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3940 Nazgand] provide some thoughts along these lines. Even if we can’t stitch together a 6-chromatic unit-distance graph with these ideas, we might be able to apply them to [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3900 prove that the measurable chromatic number of the plane is at least 6].
* It appears as though the coordinates of our smallest 5-chromatic graph lie in <math>\mathbb{Q}[\sqrt{3}, \sqrt{5}, \sqrt{11}]</math> (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3879 this]). If we view the plane as the complex plane, what is the smallest ring that admits a 5-chromatic single-distance graph? Every single-distance graph in the Eistenstein integers and Gaussian integers is 2-chromatic (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3914 this]). [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3934 David Speyer suggests] looking at <math>\mathbb{Z}[\frac{1+\sqrt{-71}}{2}]</math> next.

== Blog, forums, and media ==

* [https://rjlipton.wordpress.com/2011/05/22/more-on-coloring-the-plane/ More On Coloring The Plane], Richard Lipton, May 22, 2011.
* [https://mathoverflow.net/questions/236392/has-there-been-a-computer-search-for-a-5-chromatic-unit-distance-graph Has there been a computer search for a 5-chromatic unit distance graph?], Juno, Apr 16, 2016.
* [https://quomodocumque.wordpress.com/2018/04/09/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Jordan Ellenberg, Apr 9 2018.
* [https://gilkalai.wordpress.com/2018/04/10/aubrey-de-grey-the-chromatic-number-of-the-plane-is-at-least-5/ Aubrey de Grey: The chromatic number of the plane is at least 5], Gil Kalai, Apr 10 2018.
* [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Dustin Mixon, Apr 10, 2018.
* [https://www.scottaaronson.com/blog/?p=3697 Amazing progress on long-standing problems], Scott Aaronson, Apr 11 2018.
* [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ The chromatic number of the plane is at least 5, Part II], Dustin Mixon, Apr 13 2018.
* [http://community.wolfram.com/groups/-/m/t/1320004 A 5-chromatic unit distance graph], Ed Pegg, Apr 13 2018.
* [http://aperiodical.com/2018/04/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Katie Steckles, Apr 17 2018.
* [https://www.quantamagazine.org/decades-old-graph-problem-yields-to-amateur-mathematician-20180417/ Decades-Old Graph Problem Yields to Amateur Mathematician], Evelyn Lamb, Quanta, Apr 17, 2018.
* [https://www.heise.de/newsticker/meldung/Zahlen-bitte-Wie-bunt-ist-die-Ebene-4024574.html Zahlen, bitte! 5 - Wie bunt ist die Ebene?], Harald Bögeholz, Heise, Apr 17, 2018.
* [http://www.sciencemag.org/news/2018/04/amateur-mathematician-cracks-decades-old-math-problem Amateur mathematician cracks decades-old math problem], Katie Langin, Science News, Apr 18, 2018.
* [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable How much of the plane is 4-colorable?], Dustin Mixon, Apr 18, 2018.
* [https://www.sciencealert.com/amateur-solves-decades-old-maths-problem-about-colours-that-can-never-touch-hadwiger-nelson-problem An Amateur Solved a 60-Year-Old Maths Problem About Colours That Can Never Touch], Peter Dockrill, ScienceAlert, Apr 19, 2018.
* [https://www.zmescience.com/science/math/amateur-mathematician-solves-problem-20042018/ Amateur mathematician Aubrey de Grey, known for his work on anti-aging, solves decades-old problem], Mihai Andrei, ZME Science, Apr 20, 2018.
* [https://automaths.blog/2018/04/21/5-nuances-daubrey-de-grey/ 5 nuances d’Aubrey de Grey], Automaths, Apr 21, 2018.
* [https://science.howstuffworks.com/math-concepts/amateur-solves-part-of-decades-old-math-problem.htm Amateur Solves Part of Decades-Old Math Problem], HowStuffWorks, Apr 30, 2018.
* [https://www.nemokennislink.nl/publicaties/het-platte-vlak-heeft-minstens-vijf-kleuren-nodig/ Het platte vlak heeft minstens vijf kleuren nodig], Kennislink, May 3, 2018.

== Code and data ==

[https://www.dropbox.com/sh/ufknm1v9gtbhad3/AACB2xwaXYx5EGda38_L-0foa?dl=0 This dropbox folder] will contain most of the data and images for the project.

Data:

* [https://www.dropbox.com/s/x6kvu7b3wvdvqsn/graph.dimacs?dl=0 The 1585-vertex graph in DIMACS format]
* [https://www.dropbox.com/s/qk3gbpjvvsjfsc3/sat.dimacs?dl=0 A naive translation of 4-colorability of this graph into a SAT problem in DIMACS format]
* [https://www.dropbox.com/s/nipqikfzcfn9o5a/vertices.sage?dl=0 The vertices of this graph in explicit Sage notation]
* The graph <math>G_2</math>: [http://www.cs.utexas.edu/~marijn/CNP/874.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/874.edge edges (DIMACS)]
* The graph <math>G_3</math>: [http://www.cs.utexas.edu/~marijn/CNP/826.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/826.edge edges (DIMACS)] [http://www.cs.utexas.edu/~marijn/CNP/826.pdf Visualization]
* The [https://www.dropbox.com/sh/ufknm1v9gtbhad3/AADfw9WO1ol2ayQNJEtGf8oBa/Graphs?dl=0 densest unit-distance graphs] on an <math>n\times n</math> grid for <math>n=10,20,\ldots,100</math> (DIMACS format).

Code:

* [https://www.dropbox.com/s/6u1jctbjy38t383/lovaszmoser.m?dl=0 MATLAB script for computing Lovasz number]
* [https://files.jixco.de/pm16/zzvtx/ Python code for converting a list of vertices in Mathematica format into vertices in Z^n]
* [https://files.jixco.de/pm16/edge2cnf.py Python code for converting a DIMACS edge list into a CNF formula (forcing up to 3 suitable vertices to a fixed color, to break some symmetries)]

Software:

* [http://cvxr.com/cvx/ CVX]
* [http://www.labri.fr/perso/lsimon/glucose/ Glucose 4.0]
* [http://minisat.se/ Minisat]

== Wikipedia ==

* [https://en.wikipedia.org/wiki/Graph_coloring#Chromatic_number Chromatic number]
* [https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Erd%C5%91s_theorem_(graph_theory) de Bruijn-Erdos theorem]
* [https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem Hadwiger-Nelson problem]
* [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz number]
* [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]

== Bibliography ==
* [B2008] B. Bukh, [https://doi.org/10.1007/s00039-008-0673-8 Measurable sets with excluded distances], Geometric and Functional Analysis 18 (2008), 668-697.
* [CR2015] D. Cranston, L. Rabern, [https://arxiv.org/abs/1501.01647 The fractional chromatic number of the plane], arXiv:1501.01647
* [deBE1951] N. G. de Bruijn, P. Erdős, [http://www.math-inst.hu/~p_erdos/1951-01.pdf A colour problem for infinite graphs and a problem in the theory of relations], Nederl. Akad. Wetensch. Proc. Ser. A, 54 (1951): 371–373, MR 0046630.
* [deG2018] A. de Grey, [https://arxiv.org/abs/1804.02385 The chromatic number of the plane is at least 5], arXiv:1804.02385
* [EI2018] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.00157 The chromatic number of the plane is at least 5 - a new proof], arXiv:1805.00157
* [F1981] K.J. Falconer, The Realization of distances in measurable subsets covering Rn, J. Combin. Theory Ser. A 31 (1981) 184–189.
* [H1945] H. Hadwiger, Uberdeckung des euklidischen Raum durch kongruente Mengen, Portugaliae Math. 4 (1945), 238–242.
* [MM1961] L. Moser and M. Moser, Solution to Problem 10, Can. Math. Bull. 4 (1961), 187–189.
* [P1998] D. Pritikin, [https://www.sciencedirect.com/science/article/pii/S0095895698918196 All unit-distance graphs of order 6197 are 6-colorable], Journal of Combinatorial Theory, Series B 73.2 (1998): 159-163.
* [S2008] A. Soifer, The Mathematical Coloring Book, Springer, 2008, ISBN-13: 978-0387746401.
* [T1999] C. Thomassen, On the Nelson unit distance coloring problem, Amer. Math. Monthly 106 (1999) 850-853.
* [Tow2005] Townsend, S.P., Colouring the plane with no monochrome unit. Geombinatorics XIV(4) (2005), 181-193.
* [W1973] Woodall, D.R., Distances realized by sets covering the plane. J. Combin. Theory Ser. A, 14 (1973), 187-200.

Hadwiger-Nelson problem

2018-05-03T22:54:22Z

Nazgand: /* Continuous ranges of bichromatic virtual edges */ Correction

The Chromatic Number of the Plane (CNP) is the chromatic number of the graph whose vertices are elements of the plane, and two points are connected by an edge if they are a unit distance apart. The Hadwiger-Nelson problem asks to compute CNP. The bounds <math>4 \leq CNP \leq 7</math> are classical; recently [deG2018] it was shown that <math>CNP \geq 5</math>. This is achieved by explicitly locating finite unit distance graphs with chromatic number at least 5.

The '''Polymath16''' project seeks to simplify the graphs used in [deG2018] to establish this lower bound. More precisely, the goals are

* '''Goal 1''': Find progressively smaller 5-chromatic unit-distance graphs.
* '''Goal 2''': Reduce (ideally to zero) the reliance on computer assistance for the proof. Computer assistance was leveraged in [deG2018] to analyze a subgraph of size 397.
* '''Goal 3''': Apply these simpler graphs to inform progress in related areas. For example:
** Find a 6-chromatic unit-distance graph.
** Improve the corresponding bound in higher dimensions.
** Improve the current record of 105/29 for the fractional chromatic number of the plane.
** Find the smallest unit-distance graph of a given minimum degree (excluding, in some natural way, boring cases like Cartesian products of a graph with a hypercube).

== Polymath threads ==

* [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/ Polymath proposal: finding simpler unit distance graphs of chromatic number 5], Aubrey de Grey, Apr 10 2018. ('''Active discussion thread''')
* [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/ Polymath16, first thread: Simplifying de Grey’s graph], Dustin Mixon, Apr 14, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic Polymath16, second thread: What does it take to be 5-chromatic?], Dustin Mixon, Apr 22, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/ Polymath16, third thread: Is 6-chromatic within reach?], Dustin Mixon, May 1, 2018. ('''Active research thread''')

== Notable unit distance graphs ==

A '''unit distance graph''' is a graph that can be realised as a collection of vertices in the plane, with two vertices connected by an edge if they are precisely a unit distance apart. The chromatic number of any such graph is a lower bound for <math>CNP</math>; in particular, if one can find a unit distance graph with no 4-colorings, then <math>CNP \geq 5</math>. The boldface number of vertices is the current minimal number of vertices of a unit distance graph that is currently known to not be 4-colorable.

<math>G_1 \oplus G_2</math> denotes the Minkowski sum of two unit distance graphs <math>G_1,G_2</math> (vertices in <math>G_1 \oplus G_2</math> are sums of the vertices of <math>G_1,G_2</math>). <math>G_1 \cup G_2</math> denotes the union. <math>\mathrm{rot}(G, \theta)</math> denotes <math>G</math> rotated counterclockwise by <math>\theta</math>. <math>\mathrm{trim}(G,r)</math> denotes the trimming of <math>G</math> after removing all vertices of distance greater than <math>r</math> from the origin.

Another basic operation is '''spindling''': taking two copies of a graph <math>G</math>, gluing them together at one vertex, and rotating the copies so that the two copies of another vertex are a unit distance apart. For instance, the Moser spindle is the spindling of a rhombus graph. If a graph has two vertices forced to be the same color in a k-coloring, then the spindling of that graph at those two vertices is not k-colorable.

Many of the graphs are embeddable in abelian groups or rings, particularly those generated by the roots of unity <math>\omega_t := \exp( i \arccos( 1 - \tfrac{1}{2t} ))</math> for various natural numbers t.

{| border=1
|-
!Name!!Number of vertices!! Number of edges !! Structure !! Group !! Colorings
|-
| [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]
| 7
| 11
| Two 60-120-60-120 rhombi with a common vertex, with one pair of sharp vertices coincident and the other joined
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [http://mathworld.wolfram.com/GolombGraph.html Golomb graph]
| 10
| 18
| Contains the center and vertices of a hexagon and equilateral triangle
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [https://arxiv.org/abs/1804.02385 H]
| 7
| 12
| Vertices and center of a hexagon
| <math>{\bf Z}[\omega_1]</math>
| Has essentially four 4-colorings, two of which contain a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 J]
| 31
| 72
| Contains 13 copies of H
| <math>{\bf Z}[\omega_1]</math>
| Has essentially six 4-colorings in which no H has a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 K]
| 61
| 150
| Contains 2 copies of J
|
| In all 4-colorings lacking an H with a monochromatic <math>\sqrt{3}</math>-triangle, all pairs of vertices at distance 4 are monochromatic
|-
| [https://arxiv.org/abs/1804.02385 L]
| 121
| 301
| Contains two copies of K and 52 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ <math>L_1</math>]
| 97
|
| Has 40 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120295 <math>L_2</math>]
| 120
| 354
|
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 T]
| 9
| 15
| Contains one Moser spindle and useful symmetry; three vertices form an equilateral triangle
|
|
|-
| [https://arxiv.org/abs/1804.02385 U]
| 15
| 33
| Three copies of T at 120-degree rotations: <math>T \cup \mathrm{rot}(T, 2\pi/3) \cup \mathrm{rot}(T, 4\pi/3)</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 V]
| 31
| 30
| Unit vectors at angles consistent with three interlocking Moser spindles
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>V_1</math>]
| 61
| 60
| Union of V and a rotation of V: <math>V \cup \mathrm{rot}(V, \mathrm{arccos}(7/8))</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_b</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_x</math>]
| 13
| 12
| Subgraph of <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_z</math>]
|
|
| Subgraph of <math>V_a</math>; shares a line of symmetry with <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_y</math>]
| 13
| 12
| Subgraph of <math>V_b</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-3970 <math>V_A</math>]
| 37
| 36
| Unit vectors with angles <math>i \frac{\pi}{3} + j \mathrm{arccos} \frac{5}{6} + k \mathrm{arccos} \frac{7}{8}</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 W]
| 301
| 1230
| Cartesian product of V with itself, minus vertices at more than <math>\sqrt{3}</math> from the centre (i.e. <math>\mathrm{trim}(V \oplus V, \sqrt{3})</math>)
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>W_1</math>]
|
|
| Trimmed product of V with itself (<math>\mathrm{trim}(V \oplus V, 1.95)</math>)
|
|
|-
| [https://arxiv.org/abs/1804.02385 M]
| 1345
| 8268
| Cartesian product of W and H (<math>W \oplus H</math>)
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120274 <math>M_1</math>]
| 278
|
| Deleting vertices from M while maintaining its restriction on H
|
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>M_2</math>]
| 7075
|
| Sum of H with a trimmed product of <math>V_1</math> with itself
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 N]
| 20425
| 151311
| Contains 52 copies of M arranged around the H-copies of L
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4, \omega_{16}]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_0</math>]
| 1585
| 7909
| N "shrunk" by stepwise deletions and replacements of vertices
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 G]
| 1581
| 7877
| Deleting 4 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_1</math>]
| 1577
|
| Deleting 8 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120318 <math>G_2</math>]
| 874
| 4461
| Juxtaposing two copies of M and shrinking
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3867 <math>G_3</math>]
| 826
| 4273
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4105 <math>G_4</math>]
| '''803'''
| 4144
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 R]
|
|
| Union of <math>W_1</math> and a rotated copy of <math>W_1</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>\mathrm{trim}(R \oplus H, 1.67)</math>]
| 2563
|
| Trimmed sum of R and H
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 <math>V \oplus V \oplus H</math>]
|
|
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_{64/9}]</math>
| Has two vertices forced to be the same color in a 4-coloring; also no monochromatic <math>\sqrt{3}</math>-triangles
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 No name assigned]
| 745
|
| Subgraph of <math>V \oplus V \oplus H</math>
|
| Has two vertices forced to be the same color in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3085
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_a \oplus V_z \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3049
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 No name assigned]
| 1951
|
| Trimmed version of <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic <math>V_A \oplus V_A \oplus V_A</math>]
| 6937
| 44439
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{40}</math>]
| 40
| 82
|
|
| Any 4-coloring that avoids a monochromatic <math>\sqrt{11/3}</math>-edge has two specific vertices forced to be the same color
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{79}</math>]
| 79
| 165
| Spindling of <math>G_{40}</math>
|
| Any 4-coloring has a monochromatic <math>\sqrt{11/3}</math>-edge
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{49}</math>]
| 49
| 180
|
| <math>{\mathbf Q}[\sqrt{3},\sqrt{11}]^2</math>
| Any 4-coloring either has a specific <math>\sqrt{11/3}</math>-edge monochromatic, or a monochromatic <math>1/\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{51}</math>]
| 51
|
|
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring of plane
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{627}</math>]
| 627
| 2982
| Contains <math>G_{51}</math>
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4161 No name assigned]
| 103
|
|
|
| All 4-colorings contain a monochromatic <math>2/\sqrt{3}</math>-edge
|}

== Lower bounds ==

In [P1988], Pritikin proved that every graph with at most 12 vertices is 4-colorable, and every graph with at most 6197 vertices is 6-colorable. Pritikin's bounds are obtained by coloring the plane with k colors and an additional “wild” color such that points of unit distance are both allowed to receive the wild color. If the wild color occupies a small fraction p of the plane, then an exercise in the probabilistic method gives that any fixed unit-distance graph with n vertices enjoys an embedding in the plane that avoids the wild color (and is therefore k-colorable) provided n<1/p. Pritikin’s coloring for the k=4 case cannot be improved without improving on the densest known subset of the plane that avoids unit distances (originally due to Croft in 1967). See [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable this MO thread] for additional information. As such, improving the bound in the k=4 case might require a new technique, whereas the k=5 and 6 cases might be amenable to optimization.

[https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4176 Every unit distance graph with at most 16 vertices is 5-colorable].

A "tile-based" colouring of the plane must have at least 6 colours, as shown by Townsend [Tow2005]; the same proof with a minor error was also derived by Woodall [W1973]. In [T1999], Thomassen showed that any tiling-based 6-coloring would have to be be "unscaleable", i.e. the maximum diameter of a tile and the minimum separation of same-coloured tiles must both be exactly 1 (so that the distance 1 is excluded by suitable colouring of the boundary points).

== Virtual edge ==

''Warning: other definitions have been proposed and the exact definition of this notion is currently under discussion.''

A '''virtual edge''' of a unit distance graph <math>G</math> is a distance <math>d</math> with the property that every 4-coloring of <math>G</math> contains a monochromatic pair of vertices of distance exactly <math>d</math>. Observe that if a unit distance graph <math>G</math> has a virtual edge at distance <math>d</math>, and if there is another unit distance graph <math>H</math> with a pair of vertices at distance <math>d</math> that cannot be monochromatic in a 4-coloring, then we can create a non-4-colorable unit distance graph by "clamping" a copy of <math>H</math> to every virtual edge in <math>G</math>.

Known examples of virtual edges include:

* <math>G_{40}</math> has a virtual edge at distance <math>\sqrt{11/3}</math>.
* <math>V \oplus V \oplus H</math> has a (single) virtual edge at distance <math>8/3</math>.
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4117 <math>(\sqrt{3}\pm 1)/\sqrt{2}</math> are virtual edges of some graphs].

== Bichromatic virtual edge ==

===Definition===
If a unit distance graph <math>H</math> exists with a specific pair of vertices which are distance <math>d</math> apart and is bichromatic in all proper <math>n</math>-colorings of <math>H</math>, then we say that <math>H</math> virtualizes a '''bichromatic virtual edge''' with length <math>d</math> and chromatic number <math>n</math>.

===Vacuous properties===
Considering the graph of the entire plane, all distances correspond to a virtual edge of chromatic number <math>CNP-1</math>.

If a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n</math> exists, then a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n-1</math> exists.

===Convenience and devirtualization===
Bichromatic virtual edges behave the same as the unit edge(which is a bichromatic virtual edge of every chromatic number), and thus may be used to simplify the construction of graphs.

Recursively replacing bichromatic virtual edges of a graph <math>G</math> with graphs which virtualize the bichromatic virtual edges through '''clamping''' produces a unit distance graph <math>G'</math> where <math>\chi(G')\geq\chi(G)</math>.

Devirtualizing bichromatic virtual edges of a graph <math>G</math> (i.e. clamping) has no benefit unless nontrivial points of <math>G</math> and <math>H_0</math> overlap or nontrivial points of <math>H_1</math> and <math>H_0</math> overlap, where <math>H_0</math> and <math>H_1</math> are graphs virtualizing bichromatic virtual edges of <math>G</math>. Such overlaps may cause <math>\chi(G')>\chi(G)</math>. If no nontrivial overlaps exist, then <math>\chi(G')=\max\left(\chi(G),\chi(H_0),\chi(H_1),\dots\right)</math>.

Because more than 1 graph virtualizes any given bichromatic virtual edge for a fixed length and fixed chromatic number, multiple devirtualizations exist for every finite graph.

===Relation to rings===
A '''devirtualized ring''' of graph <math>G</math> is a ring which contains all the vertices of a devirtualization of <math>G</math>.

Use the implied shorthand for arbitrary sets <math>\mathbb{S}[\left\{a,b,c,\dots\right\}]=\mathbb{S}[a,b,c,\dots]</math>.

Let the vertices of graph <math>G</math> generate the ring <math>S_0[T_0]</math>.
Let the bichromatic virtual edges be virtualizable by the graphs <math>H_1,\dots,H_m</math>.
Let <math>S_k[T_k]</math> be the ring generated by the vertices of <math>H_k</math>.
Let <math>S</math> be the ring generated by <math>\bigcup_{k=0}^m S_k</math>.
Let <math>T=\bigcup_{k=0}^m T_k</math>.
Then <math>S[T]</math> is a devirtualized ring of <math>G</math>.

As multiple graphs may virtualize the same virtual edge, careful choice may be required to minimize the set <math>T</math>.

===Multiplicative property===
Let <math>H_0</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_0</math> and chromatic number <math>n</math>.
Let <math>H_1</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_1</math> and chromatic number <math>n</math>.
Create a graph <math>H_2</math> by scaling <math>H_0</math> by a factor of <math>d_1</math>. Graph <math>H_2</math> then virtualizes a bichromatic virtual edge length <math>d_0d_1</math> and chromatic number <math>n</math>.

===Notable bichromatic virtual edges===
{| border=1
|-
! Chromatic number !! Length <math>d</math>!! Proof
|-
| any
| 1
| trivial case
|-
| 2
| <math>0\leq d\leq 4</math>
| moving the flexible parts of the graph (0,0),(0,1),(0,2),(0,3)
|-
| 3
| <math>\left\{\left|\frac{(3+\sqrt{-3})k}{2}+\sqrt{-3}j+(-1)^r\right| \mid k,j\in\mathbb{Z}, r\in\mathbb{R}\right\}</math>
| equilateral triangle tiling
|}

===Continuous ranges of bichromatic virtual edges===
Flexible graphs might produce continuous ranges of lengths of bichromatic virtual edges of chromatic number <math>n</math> without having a specific pair which is monochrome for every <math>n</math>-coloring.
Let the range of lengths be on the interval <math>d_0\leq d\leq d_1</math>.

If <math>1 > d_1</math>, then let <math>k\in\mathbb{Z}^+,d_0^{k+1} \leq d_1^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all lengths larger than <math>d_0^k</math>. A finite area allowed to be the same color, the infinite area of the plane, and a finite upper bound on CNP implies <math>CNP> n</math>.

If <math>d_0 < 1</math>, then let <math>k\in\mathbb{Z}^+,d_1^{k+1} \geq d_0^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all non-zero lengths smaller than <math>d_1^k</math>. Considering a set of <math>CNP+1</math> all within distance <math>d_1^k</math> of each other implies <math>CNP> n</math>.

Using a flexible bichromatic virtual edge of chromatic number <math>n</math> and a graph <math>H</math> of chromatic number <math>n+1</math>, a flexible a graph <math>H'</math> of chromatic number <math>n+1</math> can be created by scaling <math>H</math> to some arbitrarily large or arbitrarily small size and clamping flexible bichromatic virtual edges of chromatic number <math>n</math> as a replacement of each rigid bichromatic virtual edge of <math>H</math>.

<math>H</math> virtualizing a bichromatic virtual edge of chromatic number <math>n+1</math> does not necessarily men the graph <math>H'</math> virtualizes a bichromatic virtual edge.

== Probabilistic formulation ==

See [[Probabilistic formulation of Hadwiger-Nelson problem]].

== Further questions ==

* What are the [http://mathworld.wolfram.com/IndependenceRatio.html independence ratios] of the above unit distance graphs?
* What are the [https://en.wikipedia.org/wiki/Fractional_coloring fractional chromatic numbers] of these graphs?
* What are the [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz numbers] of these graphs?
** The Lovasz theta function value of Lovasz number of <math>G_2</math> at the complement [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3844 is in the interval [3.3746, 3.3748]].
* What about the Erdos unit distance graph (<math>n</math> vertices, <math>n^{1+c/\log\log n}</math> edges)?
* Can we use de Grey’s graph to construct unit-distance graphs that are not 5-colorable? To answer this question, we first need to understand how 5-colorings of de Grey’s graph force small collections of vertices to be colored. [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3899 Varga] and [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3940 Nazgand] provide some thoughts along these lines. Even if we can’t stitch together a 6-chromatic unit-distance graph with these ideas, we might be able to apply them to [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3900 prove that the measurable chromatic number of the plane is at least 6].
* It appears as though the coordinates of our smallest 5-chromatic graph lie in <math>\mathbb{Q}[\sqrt{3}, \sqrt{5}, \sqrt{11}]</math> (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3879 this]). If we view the plane as the complex plane, what is the smallest ring that admits a 5-chromatic single-distance graph? Every single-distance graph in the Eistenstein integers and Gaussian integers is 2-chromatic (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3914 this]). [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3934 David Speyer suggests] looking at <math>\mathbb{Z}[\frac{1+\sqrt{-71}}{2}]</math> next.

== Blog, forums, and media ==

* [https://rjlipton.wordpress.com/2011/05/22/more-on-coloring-the-plane/ More On Coloring The Plane], Richard Lipton, May 22, 2011.
* [https://mathoverflow.net/questions/236392/has-there-been-a-computer-search-for-a-5-chromatic-unit-distance-graph Has there been a computer search for a 5-chromatic unit distance graph?], Juno, Apr 16, 2016.
* [https://quomodocumque.wordpress.com/2018/04/09/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Jordan Ellenberg, Apr 9 2018.
* [https://gilkalai.wordpress.com/2018/04/10/aubrey-de-grey-the-chromatic-number-of-the-plane-is-at-least-5/ Aubrey de Grey: The chromatic number of the plane is at least 5], Gil Kalai, Apr 10 2018.
* [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Dustin Mixon, Apr 10, 2018.
* [https://www.scottaaronson.com/blog/?p=3697 Amazing progress on long-standing problems], Scott Aaronson, Apr 11 2018.
* [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ The chromatic number of the plane is at least 5, Part II], Dustin Mixon, Apr 13 2018.
* [http://community.wolfram.com/groups/-/m/t/1320004 A 5-chromatic unit distance graph], Ed Pegg, Apr 13 2018.
* [http://aperiodical.com/2018/04/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Katie Steckles, Apr 17 2018.
* [https://www.quantamagazine.org/decades-old-graph-problem-yields-to-amateur-mathematician-20180417/ Decades-Old Graph Problem Yields to Amateur Mathematician], Evelyn Lamb, Quanta, Apr 17, 2018.
* [https://www.heise.de/newsticker/meldung/Zahlen-bitte-Wie-bunt-ist-die-Ebene-4024574.html Zahlen, bitte! 5 - Wie bunt ist die Ebene?], Harald Bögeholz, Heise, Apr 17, 2018.
* [http://www.sciencemag.org/news/2018/04/amateur-mathematician-cracks-decades-old-math-problem Amateur mathematician cracks decades-old math problem], Katie Langin, Science News, Apr 18, 2018.
* [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable How much of the plane is 4-colorable?], Dustin Mixon, Apr 18, 2018.
* [https://www.sciencealert.com/amateur-solves-decades-old-maths-problem-about-colours-that-can-never-touch-hadwiger-nelson-problem An Amateur Solved a 60-Year-Old Maths Problem About Colours That Can Never Touch], Peter Dockrill, ScienceAlert, Apr 19, 2018.
* [https://www.zmescience.com/science/math/amateur-mathematician-solves-problem-20042018/ Amateur mathematician Aubrey de Grey, known for his work on anti-aging, solves decades-old problem], Mihai Andrei, ZME Science, Apr 20, 2018.
* [https://automaths.blog/2018/04/21/5-nuances-daubrey-de-grey/ 5 nuances d’Aubrey de Grey], Automaths, Apr 21, 2018.
* [https://science.howstuffworks.com/math-concepts/amateur-solves-part-of-decades-old-math-problem.htm Amateur Solves Part of Decades-Old Math Problem], HowStuffWorks, Apr 30, 2018.
* [https://www.nemokennislink.nl/publicaties/het-platte-vlak-heeft-minstens-vijf-kleuren-nodig/ Het platte vlak heeft minstens vijf kleuren nodig], Kennislink, May 3, 2018.

== Code and data ==

[https://www.dropbox.com/sh/ufknm1v9gtbhad3/AACB2xwaXYx5EGda38_L-0foa?dl=0 This dropbox folder] will contain most of the data and images for the project.

Data:

* [https://www.dropbox.com/s/x6kvu7b3wvdvqsn/graph.dimacs?dl=0 The 1585-vertex graph in DIMACS format]
* [https://www.dropbox.com/s/qk3gbpjvvsjfsc3/sat.dimacs?dl=0 A naive translation of 4-colorability of this graph into a SAT problem in DIMACS format]
* [https://www.dropbox.com/s/nipqikfzcfn9o5a/vertices.sage?dl=0 The vertices of this graph in explicit Sage notation]
* The graph <math>G_2</math>: [http://www.cs.utexas.edu/~marijn/CNP/874.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/874.edge edges (DIMACS)]
* The graph <math>G_3</math>: [http://www.cs.utexas.edu/~marijn/CNP/826.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/826.edge edges (DIMACS)] [http://www.cs.utexas.edu/~marijn/CNP/826.pdf Visualization]
* The [https://www.dropbox.com/sh/ufknm1v9gtbhad3/AADfw9WO1ol2ayQNJEtGf8oBa/Graphs?dl=0 densest unit-distance graphs] on an <math>n\times n</math> grid for <math>n=10,20,\ldots,100</math> (DIMACS format).

Code:

* [https://www.dropbox.com/s/6u1jctbjy38t383/lovaszmoser.m?dl=0 MATLAB script for computing Lovasz number]
* [https://files.jixco.de/pm16/zzvtx/ Python code for converting a list of vertices in Mathematica format into vertices in Z^n]
* [https://files.jixco.de/pm16/edge2cnf.py Python code for converting a DIMACS edge list into a CNF formula (forcing up to 3 suitable vertices to a fixed color, to break some symmetries)]

Software:

* [http://cvxr.com/cvx/ CVX]
* [http://www.labri.fr/perso/lsimon/glucose/ Glucose 4.0]
* [http://minisat.se/ Minisat]

== Wikipedia ==

* [https://en.wikipedia.org/wiki/Graph_coloring#Chromatic_number Chromatic number]
* [https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Erd%C5%91s_theorem_(graph_theory) de Bruijn-Erdos theorem]
* [https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem Hadwiger-Nelson problem]
* [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz number]
* [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]

== Bibliography ==
* [B2008] B. Bukh, [https://doi.org/10.1007/s00039-008-0673-8 Measurable sets with excluded distances], Geometric and Functional Analysis 18 (2008), 668-697.
* [CR2015] D. Cranston, L. Rabern, [https://arxiv.org/abs/1501.01647 The fractional chromatic number of the plane], arXiv:1501.01647
* [deBE1951] N. G. de Bruijn, P. Erdős, [http://www.math-inst.hu/~p_erdos/1951-01.pdf A colour problem for infinite graphs and a problem in the theory of relations], Nederl. Akad. Wetensch. Proc. Ser. A, 54 (1951): 371–373, MR 0046630.
* [deG2018] A. de Grey, [https://arxiv.org/abs/1804.02385 The chromatic number of the plane is at least 5], arXiv:1804.02385
* [EI2018] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.00157 The chromatic number of the plane is at least 5 - a new proof], arXiv:1805.00157
* [F1981] K.J. Falconer, The Realization of distances in measurable subsets covering Rn, J. Combin. Theory Ser. A 31 (1981) 184–189.
* [H1945] H. Hadwiger, Uberdeckung des euklidischen Raum durch kongruente Mengen, Portugaliae Math. 4 (1945), 238–242.
* [MM1961] L. Moser and M. Moser, Solution to Problem 10, Can. Math. Bull. 4 (1961), 187–189.
* [P1998] D. Pritikin, [https://www.sciencedirect.com/science/article/pii/S0095895698918196 All unit-distance graphs of order 6197 are 6-colorable], Journal of Combinatorial Theory, Series B 73.2 (1998): 159-163.
* [S2008] A. Soifer, The Mathematical Coloring Book, Springer, 2008, ISBN-13: 978-0387746401.
* [T1999] C. Thomassen, On the Nelson unit distance coloring problem, Amer. Math. Monthly 106 (1999) 850-853.
* [Tow2005] Townsend, S.P., Colouring the plane with no monochrome unit. Geombinatorics XIV(4) (2005), 181-193.
* [W1973] Woodall, D.R., Distances realized by sets covering the plane. J. Combin. Theory Ser. A, 14 (1973), 187-200.

Hadwiger-Nelson problem

2018-05-03T22:33:38Z

Nazgand: /* Continuous ranges of bichromatic virtual edges */

The Chromatic Number of the Plane (CNP) is the chromatic number of the graph whose vertices are elements of the plane, and two points are connected by an edge if they are a unit distance apart. The Hadwiger-Nelson problem asks to compute CNP. The bounds <math>4 \leq CNP \leq 7</math> are classical; recently [deG2018] it was shown that <math>CNP \geq 5</math>. This is achieved by explicitly locating finite unit distance graphs with chromatic number at least 5.

The '''Polymath16''' project seeks to simplify the graphs used in [deG2018] to establish this lower bound. More precisely, the goals are

* '''Goal 1''': Find progressively smaller 5-chromatic unit-distance graphs.
* '''Goal 2''': Reduce (ideally to zero) the reliance on computer assistance for the proof. Computer assistance was leveraged in [deG2018] to analyze a subgraph of size 397.
* '''Goal 3''': Apply these simpler graphs to inform progress in related areas. For example:
** Find a 6-chromatic unit-distance graph.
** Improve the corresponding bound in higher dimensions.
** Improve the current record of 105/29 for the fractional chromatic number of the plane.
** Find the smallest unit-distance graph of a given minimum degree (excluding, in some natural way, boring cases like Cartesian products of a graph with a hypercube).

== Polymath threads ==

* [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/ Polymath proposal: finding simpler unit distance graphs of chromatic number 5], Aubrey de Grey, Apr 10 2018. ('''Active discussion thread''')
* [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/ Polymath16, first thread: Simplifying de Grey’s graph], Dustin Mixon, Apr 14, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic Polymath16, second thread: What does it take to be 5-chromatic?], Dustin Mixon, Apr 22, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/ Polymath16, third thread: Is 6-chromatic within reach?], Dustin Mixon, May 1, 2018. ('''Active research thread''')

== Notable unit distance graphs ==

A '''unit distance graph''' is a graph that can be realised as a collection of vertices in the plane, with two vertices connected by an edge if they are precisely a unit distance apart. The chromatic number of any such graph is a lower bound for <math>CNP</math>; in particular, if one can find a unit distance graph with no 4-colorings, then <math>CNP \geq 5</math>. The boldface number of vertices is the current minimal number of vertices of a unit distance graph that is currently known to not be 4-colorable.

<math>G_1 \oplus G_2</math> denotes the Minkowski sum of two unit distance graphs <math>G_1,G_2</math> (vertices in <math>G_1 \oplus G_2</math> are sums of the vertices of <math>G_1,G_2</math>). <math>G_1 \cup G_2</math> denotes the union. <math>\mathrm{rot}(G, \theta)</math> denotes <math>G</math> rotated counterclockwise by <math>\theta</math>. <math>\mathrm{trim}(G,r)</math> denotes the trimming of <math>G</math> after removing all vertices of distance greater than <math>r</math> from the origin.

Another basic operation is '''spindling''': taking two copies of a graph <math>G</math>, gluing them together at one vertex, and rotating the copies so that the two copies of another vertex are a unit distance apart. For instance, the Moser spindle is the spindling of a rhombus graph. If a graph has two vertices forced to be the same color in a k-coloring, then the spindling of that graph at those two vertices is not k-colorable.

Many of the graphs are embeddable in abelian groups or rings, particularly those generated by the roots of unity <math>\omega_t := \exp( i \arccos( 1 - \tfrac{1}{2t} ))</math> for various natural numbers t.

{| border=1
|-
!Name!!Number of vertices!! Number of edges !! Structure !! Group !! Colorings
|-
| [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]
| 7
| 11
| Two 60-120-60-120 rhombi with a common vertex, with one pair of sharp vertices coincident and the other joined
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [http://mathworld.wolfram.com/GolombGraph.html Golomb graph]
| 10
| 18
| Contains the center and vertices of a hexagon and equilateral triangle
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [https://arxiv.org/abs/1804.02385 H]
| 7
| 12
| Vertices and center of a hexagon
| <math>{\bf Z}[\omega_1]</math>
| Has essentially four 4-colorings, two of which contain a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 J]
| 31
| 72
| Contains 13 copies of H
| <math>{\bf Z}[\omega_1]</math>
| Has essentially six 4-colorings in which no H has a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 K]
| 61
| 150
| Contains 2 copies of J
|
| In all 4-colorings lacking an H with a monochromatic <math>\sqrt{3}</math>-triangle, all pairs of vertices at distance 4 are monochromatic
|-
| [https://arxiv.org/abs/1804.02385 L]
| 121
| 301
| Contains two copies of K and 52 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ <math>L_1</math>]
| 97
|
| Has 40 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120295 <math>L_2</math>]
| 120
| 354
|
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 T]
| 9
| 15
| Contains one Moser spindle and useful symmetry; three vertices form an equilateral triangle
|
|
|-
| [https://arxiv.org/abs/1804.02385 U]
| 15
| 33
| Three copies of T at 120-degree rotations: <math>T \cup \mathrm{rot}(T, 2\pi/3) \cup \mathrm{rot}(T, 4\pi/3)</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 V]
| 31
| 30
| Unit vectors at angles consistent with three interlocking Moser spindles
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>V_1</math>]
| 61
| 60
| Union of V and a rotation of V: <math>V \cup \mathrm{rot}(V, \mathrm{arccos}(7/8))</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_b</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_x</math>]
| 13
| 12
| Subgraph of <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_z</math>]
|
|
| Subgraph of <math>V_a</math>; shares a line of symmetry with <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_y</math>]
| 13
| 12
| Subgraph of <math>V_b</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-3970 <math>V_A</math>]
| 37
| 36
| Unit vectors with angles <math>i \frac{\pi}{3} + j \mathrm{arccos} \frac{5}{6} + k \mathrm{arccos} \frac{7}{8}</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 W]
| 301
| 1230
| Cartesian product of V with itself, minus vertices at more than <math>\sqrt{3}</math> from the centre (i.e. <math>\mathrm{trim}(V \oplus V, \sqrt{3})</math>)
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>W_1</math>]
|
|
| Trimmed product of V with itself (<math>\mathrm{trim}(V \oplus V, 1.95)</math>)
|
|
|-
| [https://arxiv.org/abs/1804.02385 M]
| 1345
| 8268
| Cartesian product of W and H (<math>W \oplus H</math>)
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120274 <math>M_1</math>]
| 278
|
| Deleting vertices from M while maintaining its restriction on H
|
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>M_2</math>]
| 7075
|
| Sum of H with a trimmed product of <math>V_1</math> with itself
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 N]
| 20425
| 151311
| Contains 52 copies of M arranged around the H-copies of L
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4, \omega_{16}]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_0</math>]
| 1585
| 7909
| N "shrunk" by stepwise deletions and replacements of vertices
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 G]
| 1581
| 7877
| Deleting 4 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_1</math>]
| 1577
|
| Deleting 8 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120318 <math>G_2</math>]
| 874
| 4461
| Juxtaposing two copies of M and shrinking
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3867 <math>G_3</math>]
| 826
| 4273
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4105 <math>G_4</math>]
| '''803'''
| 4144
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 R]
|
|
| Union of <math>W_1</math> and a rotated copy of <math>W_1</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>\mathrm{trim}(R \oplus H, 1.67)</math>]
| 2563
|
| Trimmed sum of R and H
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 <math>V \oplus V \oplus H</math>]
|
|
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_{64/9}]</math>
| Has two vertices forced to be the same color in a 4-coloring; also no monochromatic <math>\sqrt{3}</math>-triangles
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 No name assigned]
| 745
|
| Subgraph of <math>V \oplus V \oplus H</math>
|
| Has two vertices forced to be the same color in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3085
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_a \oplus V_z \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3049
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 No name assigned]
| 1951
|
| Trimmed version of <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic <math>V_A \oplus V_A \oplus V_A</math>]
| 6937
| 44439
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{40}</math>]
| 40
| 82
|
|
| Any 4-coloring that avoids a monochromatic <math>\sqrt{11/3}</math>-edge has two specific vertices forced to be the same color
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{79}</math>]
| 79
| 165
| Spindling of <math>G_{40}</math>
|
| Any 4-coloring has a monochromatic <math>\sqrt{11/3}</math>-edge
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{49}</math>]
| 49
| 180
|
| <math>{\mathbf Q}[\sqrt{3},\sqrt{11}]^2</math>
| Any 4-coloring either has a specific <math>\sqrt{11/3}</math>-edge monochromatic, or a monochromatic <math>1/\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{51}</math>]
| 51
|
|
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring of plane
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{627}</math>]
| 627
| 2982
| Contains <math>G_{51}</math>
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4161 No name assigned]
| 103
|
|
|
| All 4-colorings contain a monochromatic <math>2/\sqrt{3}</math>-edge
|}

== Lower bounds ==

In [P1988], Pritikin proved that every graph with at most 12 vertices is 4-colorable, and every graph with at most 6197 vertices is 6-colorable. Pritikin's bounds are obtained by coloring the plane with k colors and an additional “wild” color such that points of unit distance are both allowed to receive the wild color. If the wild color occupies a small fraction p of the plane, then an exercise in the probabilistic method gives that any fixed unit-distance graph with n vertices enjoys an embedding in the plane that avoids the wild color (and is therefore k-colorable) provided n<1/p. Pritikin’s coloring for the k=4 case cannot be improved without improving on the densest known subset of the plane that avoids unit distances (originally due to Croft in 1967). See [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable this MO thread] for additional information. As such, improving the bound in the k=4 case might require a new technique, whereas the k=5 and 6 cases might be amenable to optimization.

[https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4176 Every unit distance graph with at most 16 vertices is 5-colorable].

A "tile-based" colouring of the plane must have at least 6 colours, as shown by Townsend [Tow2005]; the same proof with a minor error was also derived by Woodall [W1973]. In [T1999], Thomassen showed that any tiling-based 6-coloring would have to be be "unscaleable", i.e. the maximum diameter of a tile and the minimum separation of same-coloured tiles must both be exactly 1 (so that the distance 1 is excluded by suitable colouring of the boundary points).

== Virtual edge ==

''Warning: other definitions have been proposed and the exact definition of this notion is currently under discussion.''

A '''virtual edge''' of a unit distance graph <math>G</math> is a distance <math>d</math> with the property that every 4-coloring of <math>G</math> contains a monochromatic pair of vertices of distance exactly <math>d</math>. Observe that if a unit distance graph <math>G</math> has a virtual edge at distance <math>d</math>, and if there is another unit distance graph <math>H</math> with a pair of vertices at distance <math>d</math> that cannot be monochromatic in a 4-coloring, then we can create a non-4-colorable unit distance graph by "clamping" a copy of <math>H</math> to every virtual edge in <math>G</math>.

Known examples of virtual edges include:

* <math>G_{40}</math> has a virtual edge at distance <math>\sqrt{11/3}</math>.
* <math>V \oplus V \oplus H</math> has a (single) virtual edge at distance <math>8/3</math>.
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4117 <math>(\sqrt{3}\pm 1)/\sqrt{2}</math> are virtual edges of some graphs].

== Bichromatic virtual edge ==

===Definition===
If a unit distance graph <math>H</math> exists with a specific pair of vertices which are distance <math>d</math> apart and is bichromatic in all proper <math>n</math>-colorings of <math>H</math>, then we say that <math>H</math> virtualizes a '''bichromatic virtual edge''' with length <math>d</math> and chromatic number <math>n</math>.

===Vacuous properties===
Considering the graph of the entire plane, all distances correspond to a virtual edge of chromatic number <math>CNP-1</math>.

If a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n</math> exists, then a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n-1</math> exists.

===Convenience and devirtualization===
Bichromatic virtual edges behave the same as the unit edge(which is a bichromatic virtual edge of every chromatic number), and thus may be used to simplify the construction of graphs.

Recursively replacing bichromatic virtual edges of a graph <math>G</math> with graphs which virtualize the bichromatic virtual edges through '''clamping''' produces a unit distance graph <math>G'</math> where <math>\chi(G')\geq\chi(G)</math>.

Devirtualizing bichromatic virtual edges of a graph <math>G</math> (i.e. clamping) has no benefit unless nontrivial points of <math>G</math> and <math>H_0</math> overlap or nontrivial points of <math>H_1</math> and <math>H_0</math> overlap, where <math>H_0</math> and <math>H_1</math> are graphs virtualizing bichromatic virtual edges of <math>G</math>. Such overlaps may cause <math>\chi(G')>\chi(G)</math>. If no nontrivial overlaps exist, then <math>\chi(G')=\max\left(\chi(G),\chi(H_0),\chi(H_1),\dots\right)</math>.

Because more than 1 graph virtualizes any given bichromatic virtual edge for a fixed length and fixed chromatic number, multiple devirtualizations exist for every finite graph.

===Relation to rings===
A '''devirtualized ring''' of graph <math>G</math> is a ring which contains all the vertices of a devirtualization of <math>G</math>.

Use the implied shorthand for arbitrary sets <math>\mathbb{S}[\left\{a,b,c,\dots\right\}]=\mathbb{S}[a,b,c,\dots]</math>.

Let the vertices of graph <math>G</math> generate the ring <math>S_0[T_0]</math>.
Let the bichromatic virtual edges be virtualizable by the graphs <math>H_1,\dots,H_m</math>.
Let <math>S_k[T_k]</math> be the ring generated by the vertices of <math>H_k</math>.
Let <math>S</math> be the ring generated by <math>\bigcup_{k=0}^m S_k</math>.
Let <math>T=\bigcup_{k=0}^m T_k</math>.
Then <math>S[T]</math> is a devirtualized ring of <math>G</math>.

As multiple graphs may virtualize the same virtual edge, careful choice may be required to minimize the set <math>T</math>.

===Multiplicative property===
Let <math>H_0</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_0</math> and chromatic number <math>n</math>.
Let <math>H_1</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_1</math> and chromatic number <math>n</math>.
Create a graph <math>H_2</math> by scaling <math>H_0</math> by a factor of <math>d_1</math>. Graph <math>H_2</math> then virtualizes a bichromatic virtual edge length <math>d_0d_1</math> and chromatic number <math>n</math>.

===Notable bichromatic virtual edges===
{| border=1
|-
! Chromatic number !! Length <math>d</math>!! Proof
|-
| any
| 1
| trivial case
|-
| 2
| <math>0\leq d\leq 4</math>
| moving the flexible parts of the graph (0,0),(0,1),(0,2),(0,3)
|-
| 3
| <math>\left\{\left|\frac{(3+\sqrt{-3})k}{2}+\sqrt{-3}j+(-1)^r\right| \mid k,j\in\mathbb{Z}, r\in\mathbb{R}\right\}</math>
| equilateral triangle tiling
|}

===Continuous ranges of bichromatic virtual edges===
Flexible graphs might produce continuous ranges of lengths of bichromatic virtual edges of chromatic number <math>n</math> without having a specific pair which is monochrome for every <math>n</math>-coloring.
Let the range of lengths be on the interval <math>d_0\leq d\leq d_1</math>.

If <math>1 > d_1</math>, then let <math>k\in\mathbb{Z}^+,d_0^{k+1} \leq d_1^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all lengths larger than <math>d_0^k</math>. A finite area allowed to be the same color, the infinite area of the plane, and a finite upper bound on CNP implies <math>CNP> n</math>.

If <math>d_0 < 1</math>, then let <math>k\in\mathbb{Z}^+,d_1^{k+1} \geq d_0^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all non-zero lengths smaller than <math>d_1^k</math>. Considering a set of <math>CNP+1</math> all within distance <math>d_1^k</math> of each other implies <math>CNP> n</math>.

Using a flexible bichromatic virtual edge of chromatic number <math>n</math> and a bichromatic virtual edge of chromatic number <math>n+1</math> virtualizable by graph <math>H</math>, a flexible bichromatic virtual edge of chromatic number <math>n+1</math> can be made by scaling <math>H</math> to some arbitrarily large or arbitrarily small size and clamping flexible bichromatic virtual edges of chromatic number <math>n</math> as a replacement of each rigid bichromatic virtual edge of <math>H</math>.

Thus no bichromatic virtual edge of chromatic number <math>CNP</math> exists which is not length 1.

== Probabilistic formulation ==

See [[Probabilistic formulation of Hadwiger-Nelson problem]].

== Further questions ==

* What are the [http://mathworld.wolfram.com/IndependenceRatio.html independence ratios] of the above unit distance graphs?
* What are the [https://en.wikipedia.org/wiki/Fractional_coloring fractional chromatic numbers] of these graphs?
* What are the [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz numbers] of these graphs?
** The Lovasz theta function value of Lovasz number of <math>G_2</math> at the complement [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3844 is in the interval [3.3746, 3.3748]].
* What about the Erdos unit distance graph (<math>n</math> vertices, <math>n^{1+c/\log\log n}</math> edges)?
* Can we use de Grey’s graph to construct unit-distance graphs that are not 5-colorable? To answer this question, we first need to understand how 5-colorings of de Grey’s graph force small collections of vertices to be colored. [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3899 Varga] and [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3940 Nazgand] provide some thoughts along these lines. Even if we can’t stitch together a 6-chromatic unit-distance graph with these ideas, we might be able to apply them to [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3900 prove that the measurable chromatic number of the plane is at least 6].
* It appears as though the coordinates of our smallest 5-chromatic graph lie in <math>\mathbb{Q}[\sqrt{3}, \sqrt{5}, \sqrt{11}]</math> (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3879 this]). If we view the plane as the complex plane, what is the smallest ring that admits a 5-chromatic single-distance graph? Every single-distance graph in the Eistenstein integers and Gaussian integers is 2-chromatic (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3914 this]). [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3934 David Speyer suggests] looking at <math>\mathbb{Z}[\frac{1+\sqrt{-71}}{2}]</math> next.

== Blog, forums, and media ==

* [https://rjlipton.wordpress.com/2011/05/22/more-on-coloring-the-plane/ More On Coloring The Plane], Richard Lipton, May 22, 2011.
* [https://mathoverflow.net/questions/236392/has-there-been-a-computer-search-for-a-5-chromatic-unit-distance-graph Has there been a computer search for a 5-chromatic unit distance graph?], Juno, Apr 16, 2016.
* [https://quomodocumque.wordpress.com/2018/04/09/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Jordan Ellenberg, Apr 9 2018.
* [https://gilkalai.wordpress.com/2018/04/10/aubrey-de-grey-the-chromatic-number-of-the-plane-is-at-least-5/ Aubrey de Grey: The chromatic number of the plane is at least 5], Gil Kalai, Apr 10 2018.
* [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Dustin Mixon, Apr 10, 2018.
* [https://www.scottaaronson.com/blog/?p=3697 Amazing progress on long-standing problems], Scott Aaronson, Apr 11 2018.
* [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ The chromatic number of the plane is at least 5, Part II], Dustin Mixon, Apr 13 2018.
* [http://community.wolfram.com/groups/-/m/t/1320004 A 5-chromatic unit distance graph], Ed Pegg, Apr 13 2018.
* [http://aperiodical.com/2018/04/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Katie Steckles, Apr 17 2018.
* [https://www.quantamagazine.org/decades-old-graph-problem-yields-to-amateur-mathematician-20180417/ Decades-Old Graph Problem Yields to Amateur Mathematician], Evelyn Lamb, Quanta, Apr 17, 2018.
* [https://www.heise.de/newsticker/meldung/Zahlen-bitte-Wie-bunt-ist-die-Ebene-4024574.html Zahlen, bitte! 5 - Wie bunt ist die Ebene?], Harald Bögeholz, Heise, Apr 17, 2018.
* [http://www.sciencemag.org/news/2018/04/amateur-mathematician-cracks-decades-old-math-problem Amateur mathematician cracks decades-old math problem], Katie Langin, Science News, Apr 18, 2018.
* [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable How much of the plane is 4-colorable?], Dustin Mixon, Apr 18, 2018.
* [https://www.sciencealert.com/amateur-solves-decades-old-maths-problem-about-colours-that-can-never-touch-hadwiger-nelson-problem An Amateur Solved a 60-Year-Old Maths Problem About Colours That Can Never Touch], Peter Dockrill, ScienceAlert, Apr 19, 2018.
* [https://www.zmescience.com/science/math/amateur-mathematician-solves-problem-20042018/ Amateur mathematician Aubrey de Grey, known for his work on anti-aging, solves decades-old problem], Mihai Andrei, ZME Science, Apr 20, 2018.
* [https://automaths.blog/2018/04/21/5-nuances-daubrey-de-grey/ 5 nuances d’Aubrey de Grey], Automaths, Apr 21, 2018.
* [https://science.howstuffworks.com/math-concepts/amateur-solves-part-of-decades-old-math-problem.htm Amateur Solves Part of Decades-Old Math Problem], HowStuffWorks, Apr 30, 2018.
* [https://www.nemokennislink.nl/publicaties/het-platte-vlak-heeft-minstens-vijf-kleuren-nodig/ Het platte vlak heeft minstens vijf kleuren nodig], Kennislink, May 3, 2018.

== Code and data ==

[https://www.dropbox.com/sh/ufknm1v9gtbhad3/AACB2xwaXYx5EGda38_L-0foa?dl=0 This dropbox folder] will contain most of the data and images for the project.

Data:

* [https://www.dropbox.com/s/x6kvu7b3wvdvqsn/graph.dimacs?dl=0 The 1585-vertex graph in DIMACS format]
* [https://www.dropbox.com/s/qk3gbpjvvsjfsc3/sat.dimacs?dl=0 A naive translation of 4-colorability of this graph into a SAT problem in DIMACS format]
* [https://www.dropbox.com/s/nipqikfzcfn9o5a/vertices.sage?dl=0 The vertices of this graph in explicit Sage notation]
* The graph <math>G_2</math>: [http://www.cs.utexas.edu/~marijn/CNP/874.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/874.edge edges (DIMACS)]
* The graph <math>G_3</math>: [http://www.cs.utexas.edu/~marijn/CNP/826.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/826.edge edges (DIMACS)] [http://www.cs.utexas.edu/~marijn/CNP/826.pdf Visualization]
* The [https://www.dropbox.com/sh/ufknm1v9gtbhad3/AADfw9WO1ol2ayQNJEtGf8oBa/Graphs?dl=0 densest unit-distance graphs] on an <math>n\times n</math> grid for <math>n=10,20,\ldots,100</math> (DIMACS format).

Code:

* [https://www.dropbox.com/s/6u1jctbjy38t383/lovaszmoser.m?dl=0 MATLAB script for computing Lovasz number]
* [https://files.jixco.de/pm16/zzvtx/ Python code for converting a list of vertices in Mathematica format into vertices in Z^n]
* [https://files.jixco.de/pm16/edge2cnf.py Python code for converting a DIMACS edge list into a CNF formula (forcing up to 3 suitable vertices to a fixed color, to break some symmetries)]

Software:

* [http://cvxr.com/cvx/ CVX]
* [http://www.labri.fr/perso/lsimon/glucose/ Glucose 4.0]
* [http://minisat.se/ Minisat]

== Wikipedia ==

* [https://en.wikipedia.org/wiki/Graph_coloring#Chromatic_number Chromatic number]
* [https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Erd%C5%91s_theorem_(graph_theory) de Bruijn-Erdos theorem]
* [https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem Hadwiger-Nelson problem]
* [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz number]
* [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]

== Bibliography ==
* [B2008] B. Bukh, [https://doi.org/10.1007/s00039-008-0673-8 Measurable sets with excluded distances], Geometric and Functional Analysis 18 (2008), 668-697.
* [CR2015] D. Cranston, L. Rabern, [https://arxiv.org/abs/1501.01647 The fractional chromatic number of the plane], arXiv:1501.01647
* [deBE1951] N. G. de Bruijn, P. Erdős, [http://www.math-inst.hu/~p_erdos/1951-01.pdf A colour problem for infinite graphs and a problem in the theory of relations], Nederl. Akad. Wetensch. Proc. Ser. A, 54 (1951): 371–373, MR 0046630.
* [deG2018] A. de Grey, [https://arxiv.org/abs/1804.02385 The chromatic number of the plane is at least 5], arXiv:1804.02385
* [EI2018] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.00157 The chromatic number of the plane is at least 5 - a new proof], arXiv:1805.00157
* [F1981] K.J. Falconer, The Realization of distances in measurable subsets covering Rn, J. Combin. Theory Ser. A 31 (1981) 184–189.
* [H1945] H. Hadwiger, Uberdeckung des euklidischen Raum durch kongruente Mengen, Portugaliae Math. 4 (1945), 238–242.
* [MM1961] L. Moser and M. Moser, Solution to Problem 10, Can. Math. Bull. 4 (1961), 187–189.
* [P1998] D. Pritikin, [https://www.sciencedirect.com/science/article/pii/S0095895698918196 All unit-distance graphs of order 6197 are 6-colorable], Journal of Combinatorial Theory, Series B 73.2 (1998): 159-163.
* [S2008] A. Soifer, The Mathematical Coloring Book, Springer, 2008, ISBN-13: 978-0387746401.
* [T1999] C. Thomassen, On the Nelson unit distance coloring problem, Amer. Math. Monthly 106 (1999) 850-853.
* [Tow2005] Townsend, S.P., Colouring the plane with no monochrome unit. Geombinatorics XIV(4) (2005), 181-193.
* [W1973] Woodall, D.R., Distances realized by sets covering the plane. J. Combin. Theory Ser. A, 14 (1973), 187-200.

Hadwiger-Nelson problem

2018-05-03T22:32:24Z

Nazgand: /* Bichromatic virtual edge */

The Chromatic Number of the Plane (CNP) is the chromatic number of the graph whose vertices are elements of the plane, and two points are connected by an edge if they are a unit distance apart. The Hadwiger-Nelson problem asks to compute CNP. The bounds <math>4 \leq CNP \leq 7</math> are classical; recently [deG2018] it was shown that <math>CNP \geq 5</math>. This is achieved by explicitly locating finite unit distance graphs with chromatic number at least 5.

The '''Polymath16''' project seeks to simplify the graphs used in [deG2018] to establish this lower bound. More precisely, the goals are

* '''Goal 1''': Find progressively smaller 5-chromatic unit-distance graphs.
* '''Goal 2''': Reduce (ideally to zero) the reliance on computer assistance for the proof. Computer assistance was leveraged in [deG2018] to analyze a subgraph of size 397.
* '''Goal 3''': Apply these simpler graphs to inform progress in related areas. For example:
** Find a 6-chromatic unit-distance graph.
** Improve the corresponding bound in higher dimensions.
** Improve the current record of 105/29 for the fractional chromatic number of the plane.
** Find the smallest unit-distance graph of a given minimum degree (excluding, in some natural way, boring cases like Cartesian products of a graph with a hypercube).

== Polymath threads ==

* [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/ Polymath proposal: finding simpler unit distance graphs of chromatic number 5], Aubrey de Grey, Apr 10 2018. ('''Active discussion thread''')
* [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/ Polymath16, first thread: Simplifying de Grey’s graph], Dustin Mixon, Apr 14, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic Polymath16, second thread: What does it take to be 5-chromatic?], Dustin Mixon, Apr 22, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/ Polymath16, third thread: Is 6-chromatic within reach?], Dustin Mixon, May 1, 2018. ('''Active research thread''')

== Notable unit distance graphs ==

A '''unit distance graph''' is a graph that can be realised as a collection of vertices in the plane, with two vertices connected by an edge if they are precisely a unit distance apart. The chromatic number of any such graph is a lower bound for <math>CNP</math>; in particular, if one can find a unit distance graph with no 4-colorings, then <math>CNP \geq 5</math>. The boldface number of vertices is the current minimal number of vertices of a unit distance graph that is currently known to not be 4-colorable.

<math>G_1 \oplus G_2</math> denotes the Minkowski sum of two unit distance graphs <math>G_1,G_2</math> (vertices in <math>G_1 \oplus G_2</math> are sums of the vertices of <math>G_1,G_2</math>). <math>G_1 \cup G_2</math> denotes the union. <math>\mathrm{rot}(G, \theta)</math> denotes <math>G</math> rotated counterclockwise by <math>\theta</math>. <math>\mathrm{trim}(G,r)</math> denotes the trimming of <math>G</math> after removing all vertices of distance greater than <math>r</math> from the origin.

Another basic operation is '''spindling''': taking two copies of a graph <math>G</math>, gluing them together at one vertex, and rotating the copies so that the two copies of another vertex are a unit distance apart. For instance, the Moser spindle is the spindling of a rhombus graph. If a graph has two vertices forced to be the same color in a k-coloring, then the spindling of that graph at those two vertices is not k-colorable.

Many of the graphs are embeddable in abelian groups or rings, particularly those generated by the roots of unity <math>\omega_t := \exp( i \arccos( 1 - \tfrac{1}{2t} ))</math> for various natural numbers t.

{| border=1
|-
!Name!!Number of vertices!! Number of edges !! Structure !! Group !! Colorings
|-
| [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]
| 7
| 11
| Two 60-120-60-120 rhombi with a common vertex, with one pair of sharp vertices coincident and the other joined
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [http://mathworld.wolfram.com/GolombGraph.html Golomb graph]
| 10
| 18
| Contains the center and vertices of a hexagon and equilateral triangle
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [https://arxiv.org/abs/1804.02385 H]
| 7
| 12
| Vertices and center of a hexagon
| <math>{\bf Z}[\omega_1]</math>
| Has essentially four 4-colorings, two of which contain a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 J]
| 31
| 72
| Contains 13 copies of H
| <math>{\bf Z}[\omega_1]</math>
| Has essentially six 4-colorings in which no H has a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 K]
| 61
| 150
| Contains 2 copies of J
|
| In all 4-colorings lacking an H with a monochromatic <math>\sqrt{3}</math>-triangle, all pairs of vertices at distance 4 are monochromatic
|-
| [https://arxiv.org/abs/1804.02385 L]
| 121
| 301
| Contains two copies of K and 52 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ <math>L_1</math>]
| 97
|
| Has 40 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120295 <math>L_2</math>]
| 120
| 354
|
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 T]
| 9
| 15
| Contains one Moser spindle and useful symmetry; three vertices form an equilateral triangle
|
|
|-
| [https://arxiv.org/abs/1804.02385 U]
| 15
| 33
| Three copies of T at 120-degree rotations: <math>T \cup \mathrm{rot}(T, 2\pi/3) \cup \mathrm{rot}(T, 4\pi/3)</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 V]
| 31
| 30
| Unit vectors at angles consistent with three interlocking Moser spindles
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>V_1</math>]
| 61
| 60
| Union of V and a rotation of V: <math>V \cup \mathrm{rot}(V, \mathrm{arccos}(7/8))</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_b</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_x</math>]
| 13
| 12
| Subgraph of <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_z</math>]
|
|
| Subgraph of <math>V_a</math>; shares a line of symmetry with <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_y</math>]
| 13
| 12
| Subgraph of <math>V_b</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-3970 <math>V_A</math>]
| 37
| 36
| Unit vectors with angles <math>i \frac{\pi}{3} + j \mathrm{arccos} \frac{5}{6} + k \mathrm{arccos} \frac{7}{8}</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 W]
| 301
| 1230
| Cartesian product of V with itself, minus vertices at more than <math>\sqrt{3}</math> from the centre (i.e. <math>\mathrm{trim}(V \oplus V, \sqrt{3})</math>)
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>W_1</math>]
|
|
| Trimmed product of V with itself (<math>\mathrm{trim}(V \oplus V, 1.95)</math>)
|
|
|-
| [https://arxiv.org/abs/1804.02385 M]
| 1345
| 8268
| Cartesian product of W and H (<math>W \oplus H</math>)
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120274 <math>M_1</math>]
| 278
|
| Deleting vertices from M while maintaining its restriction on H
|
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>M_2</math>]
| 7075
|
| Sum of H with a trimmed product of <math>V_1</math> with itself
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 N]
| 20425
| 151311
| Contains 52 copies of M arranged around the H-copies of L
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4, \omega_{16}]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_0</math>]
| 1585
| 7909
| N "shrunk" by stepwise deletions and replacements of vertices
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 G]
| 1581
| 7877
| Deleting 4 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_1</math>]
| 1577
|
| Deleting 8 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120318 <math>G_2</math>]
| 874
| 4461
| Juxtaposing two copies of M and shrinking
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3867 <math>G_3</math>]
| 826
| 4273
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4105 <math>G_4</math>]
| '''803'''
| 4144
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 R]
|
|
| Union of <math>W_1</math> and a rotated copy of <math>W_1</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>\mathrm{trim}(R \oplus H, 1.67)</math>]
| 2563
|
| Trimmed sum of R and H
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 <math>V \oplus V \oplus H</math>]
|
|
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_{64/9}]</math>
| Has two vertices forced to be the same color in a 4-coloring; also no monochromatic <math>\sqrt{3}</math>-triangles
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 No name assigned]
| 745
|
| Subgraph of <math>V \oplus V \oplus H</math>
|
| Has two vertices forced to be the same color in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3085
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_a \oplus V_z \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3049
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 No name assigned]
| 1951
|
| Trimmed version of <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic <math>V_A \oplus V_A \oplus V_A</math>]
| 6937
| 44439
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{40}</math>]
| 40
| 82
|
|
| Any 4-coloring that avoids a monochromatic <math>\sqrt{11/3}</math>-edge has two specific vertices forced to be the same color
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{79}</math>]
| 79
| 165
| Spindling of <math>G_{40}</math>
|
| Any 4-coloring has a monochromatic <math>\sqrt{11/3}</math>-edge
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{49}</math>]
| 49
| 180
|
| <math>{\mathbf Q}[\sqrt{3},\sqrt{11}]^2</math>
| Any 4-coloring either has a specific <math>\sqrt{11/3}</math>-edge monochromatic, or a monochromatic <math>1/\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{51}</math>]
| 51
|
|
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring of plane
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{627}</math>]
| 627
| 2982
| Contains <math>G_{51}</math>
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4161 No name assigned]
| 103
|
|
|
| All 4-colorings contain a monochromatic <math>2/\sqrt{3}</math>-edge
|}

== Lower bounds ==

In [P1988], Pritikin proved that every graph with at most 12 vertices is 4-colorable, and every graph with at most 6197 vertices is 6-colorable. Pritikin's bounds are obtained by coloring the plane with k colors and an additional “wild” color such that points of unit distance are both allowed to receive the wild color. If the wild color occupies a small fraction p of the plane, then an exercise in the probabilistic method gives that any fixed unit-distance graph with n vertices enjoys an embedding in the plane that avoids the wild color (and is therefore k-colorable) provided n<1/p. Pritikin’s coloring for the k=4 case cannot be improved without improving on the densest known subset of the plane that avoids unit distances (originally due to Croft in 1967). See [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable this MO thread] for additional information. As such, improving the bound in the k=4 case might require a new technique, whereas the k=5 and 6 cases might be amenable to optimization.

[https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4176 Every unit distance graph with at most 16 vertices is 5-colorable].

A "tile-based" colouring of the plane must have at least 6 colours, as shown by Townsend [Tow2005]; the same proof with a minor error was also derived by Woodall [W1973]. In [T1999], Thomassen showed that any tiling-based 6-coloring would have to be be "unscaleable", i.e. the maximum diameter of a tile and the minimum separation of same-coloured tiles must both be exactly 1 (so that the distance 1 is excluded by suitable colouring of the boundary points).

== Virtual edge ==

''Warning: other definitions have been proposed and the exact definition of this notion is currently under discussion.''

A '''virtual edge''' of a unit distance graph <math>G</math> is a distance <math>d</math> with the property that every 4-coloring of <math>G</math> contains a monochromatic pair of vertices of distance exactly <math>d</math>. Observe that if a unit distance graph <math>G</math> has a virtual edge at distance <math>d</math>, and if there is another unit distance graph <math>H</math> with a pair of vertices at distance <math>d</math> that cannot be monochromatic in a 4-coloring, then we can create a non-4-colorable unit distance graph by "clamping" a copy of <math>H</math> to every virtual edge in <math>G</math>.

Known examples of virtual edges include:

* <math>G_{40}</math> has a virtual edge at distance <math>\sqrt{11/3}</math>.
* <math>V \oplus V \oplus H</math> has a (single) virtual edge at distance <math>8/3</math>.
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4117 <math>(\sqrt{3}\pm 1)/\sqrt{2}</math> are virtual edges of some graphs].

== Bichromatic virtual edge ==

===Definition===
If a unit distance graph <math>H</math> exists with a specific pair of vertices which are distance <math>d</math> apart and is bichromatic in all proper <math>n</math>-colorings of <math>H</math>, then we say that <math>H</math> virtualizes a '''bichromatic virtual edge''' with length <math>d</math> and chromatic number <math>n</math>.

===Vacuous properties===
Considering the graph of the entire plane, all distances correspond to a virtual edge of chromatic number <math>CNP-1</math>.

If a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n</math> exists, then a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n-1</math> exists.

===Convenience and devirtualization===
Bichromatic virtual edges behave the same as the unit edge(which is a bichromatic virtual edge of every chromatic number), and thus may be used to simplify the construction of graphs.

Recursively replacing bichromatic virtual edges of a graph <math>G</math> with graphs which virtualize the bichromatic virtual edges through '''clamping''' produces a unit distance graph <math>G'</math> where <math>\chi(G')\geq\chi(G)</math>.

Devirtualizing bichromatic virtual edges of a graph <math>G</math> (i.e. clamping) has no benefit unless nontrivial points of <math>G</math> and <math>H_0</math> overlap or nontrivial points of <math>H_1</math> and <math>H_0</math> overlap, where <math>H_0</math> and <math>H_1</math> are graphs virtualizing bichromatic virtual edges of <math>G</math>. Such overlaps may cause <math>\chi(G')>\chi(G)</math>. If no nontrivial overlaps exist, then <math>\chi(G')=\max\left(\chi(G),\chi(H_0),\chi(H_1),\dots\right)</math>.

Because more than 1 graph virtualizes any given bichromatic virtual edge for a fixed length and fixed chromatic number, multiple devirtualizations exist for every finite graph.

===Relation to rings===
A '''devirtualized ring''' of graph <math>G</math> is a ring which contains all the vertices of a devirtualization of <math>G</math>.

Use the implied shorthand for arbitrary sets <math>\mathbb{S}[\left\{a,b,c,\dots\right\}]=\mathbb{S}[a,b,c,\dots]</math>.

Let the vertices of graph <math>G</math> generate the ring <math>S_0[T_0]</math>.
Let the bichromatic virtual edges be virtualizable by the graphs <math>H_1,\dots,H_m</math>.
Let <math>S_k[T_k]</math> be the ring generated by the vertices of <math>H_k</math>.
Let <math>S</math> be the ring generated by <math>\bigcup_{k=0}^m S_k</math>.
Let <math>T=\bigcup_{k=0}^m T_k</math>.
Then <math>S[T]</math> is a devirtualized ring of <math>G</math>.

As multiple graphs may virtualize the same virtual edge, careful choice may be required to minimize the set <math>T</math>.

===Multiplicative property===
Let <math>H_0</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_0</math> and chromatic number <math>n</math>.
Let <math>H_1</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_1</math> and chromatic number <math>n</math>.
Create a graph <math>H_2</math> by scaling <math>H_0</math> by a factor of <math>d_1</math>. Graph <math>H_2</math> then virtualizes a bichromatic virtual edge length <math>d_0d_1</math> and chromatic number <math>n</math>.

===Notable bichromatic virtual edges===
{| border=1
|-
! Chromatic number !! Length <math>d</math>!! Proof
|-
| any
| 1
| trivial case
|-
| 2
| <math>0\leq d\leq 4</math>
| moving the flexible parts of the graph (0,0),(0,1),(0,2),(0,3)
|-
| 3
| <math>\left\{\left|\frac{(3+\sqrt{-3})k}{2}+\sqrt{-3}j+(-1)^r\right| \mid k,j\in\mathbb{Z}, r\in\mathbb{R}\right\}</math>
| equilateral triangle tiling
|}

===Continuous ranges of bichromatic virtual edges===
Flexible graphs might produce continuous ranges of lengths of bichromatic virtual edges of chromatic number <math>n</math> without having a specific pair which is monochrome for every <math>n</math>-coloring.
Let the range of lengths be on the interval <math>d_0\leq d\leq d_1</math>.

If <math>1 > d_1</math>, then let <math>k\in\mathbb{Z}^+,d_0^{k+1} \leq d_1^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all lengths larger than <math>d_0^k</math>. A finite area allowed to be the same color, the infinite area of the plane, and a finite upper bound on CNP implies <math>CNP>n</math>.

If <math>d_0 < 1</math>, then let <math>k\in\mathbb{Z}^+,d_1^{k+1} \geq d_0^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all non-zero lengths smaller than <math>d_1^k</math>. Considering a set of <math>CNP+1</math> all within distance <math>d_1^k</math> of each other implies <math>CNP>n</math>.

Using a flexible bichromatic virtual edge of chromatic number <math>n</math> and a bichromatic virtual edge of chromatic number <math>n+1</math> virtualizable by graph <math>H</math>, a flexible bichromatic virtual edge of chromatic number <math>n+1</math> can be made by scaling <math>H</math> to some arbitrarily large or arbitrarily small size and clamping flexible bichromatic virtual edges of chromatic number <math>n</math> as a replacement of each rigid bichromatic virtual edge of <math>H</math>.

Thus no bichromatic virtual edge of chromatic number <math>CNP</math> exists which is not length 1.

== Probabilistic formulation ==

See [[Probabilistic formulation of Hadwiger-Nelson problem]].

== Further questions ==

* What are the [http://mathworld.wolfram.com/IndependenceRatio.html independence ratios] of the above unit distance graphs?
* What are the [https://en.wikipedia.org/wiki/Fractional_coloring fractional chromatic numbers] of these graphs?
* What are the [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz numbers] of these graphs?
** The Lovasz theta function value of Lovasz number of <math>G_2</math> at the complement [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3844 is in the interval [3.3746, 3.3748]].
* What about the Erdos unit distance graph (<math>n</math> vertices, <math>n^{1+c/\log\log n}</math> edges)?
* Can we use de Grey’s graph to construct unit-distance graphs that are not 5-colorable? To answer this question, we first need to understand how 5-colorings of de Grey’s graph force small collections of vertices to be colored. [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3899 Varga] and [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3940 Nazgand] provide some thoughts along these lines. Even if we can’t stitch together a 6-chromatic unit-distance graph with these ideas, we might be able to apply them to [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3900 prove that the measurable chromatic number of the plane is at least 6].
* It appears as though the coordinates of our smallest 5-chromatic graph lie in <math>\mathbb{Q}[\sqrt{3}, \sqrt{5}, \sqrt{11}]</math> (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3879 this]). If we view the plane as the complex plane, what is the smallest ring that admits a 5-chromatic single-distance graph? Every single-distance graph in the Eistenstein integers and Gaussian integers is 2-chromatic (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3914 this]). [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3934 David Speyer suggests] looking at <math>\mathbb{Z}[\frac{1+\sqrt{-71}}{2}]</math> next.

== Blog, forums, and media ==

* [https://rjlipton.wordpress.com/2011/05/22/more-on-coloring-the-plane/ More On Coloring The Plane], Richard Lipton, May 22, 2011.
* [https://mathoverflow.net/questions/236392/has-there-been-a-computer-search-for-a-5-chromatic-unit-distance-graph Has there been a computer search for a 5-chromatic unit distance graph?], Juno, Apr 16, 2016.
* [https://quomodocumque.wordpress.com/2018/04/09/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Jordan Ellenberg, Apr 9 2018.
* [https://gilkalai.wordpress.com/2018/04/10/aubrey-de-grey-the-chromatic-number-of-the-plane-is-at-least-5/ Aubrey de Grey: The chromatic number of the plane is at least 5], Gil Kalai, Apr 10 2018.
* [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Dustin Mixon, Apr 10, 2018.
* [https://www.scottaaronson.com/blog/?p=3697 Amazing progress on long-standing problems], Scott Aaronson, Apr 11 2018.
* [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ The chromatic number of the plane is at least 5, Part II], Dustin Mixon, Apr 13 2018.
* [http://community.wolfram.com/groups/-/m/t/1320004 A 5-chromatic unit distance graph], Ed Pegg, Apr 13 2018.
* [http://aperiodical.com/2018/04/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Katie Steckles, Apr 17 2018.
* [https://www.quantamagazine.org/decades-old-graph-problem-yields-to-amateur-mathematician-20180417/ Decades-Old Graph Problem Yields to Amateur Mathematician], Evelyn Lamb, Quanta, Apr 17, 2018.
* [https://www.heise.de/newsticker/meldung/Zahlen-bitte-Wie-bunt-ist-die-Ebene-4024574.html Zahlen, bitte! 5 - Wie bunt ist die Ebene?], Harald Bögeholz, Heise, Apr 17, 2018.
* [http://www.sciencemag.org/news/2018/04/amateur-mathematician-cracks-decades-old-math-problem Amateur mathematician cracks decades-old math problem], Katie Langin, Science News, Apr 18, 2018.
* [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable How much of the plane is 4-colorable?], Dustin Mixon, Apr 18, 2018.
* [https://www.sciencealert.com/amateur-solves-decades-old-maths-problem-about-colours-that-can-never-touch-hadwiger-nelson-problem An Amateur Solved a 60-Year-Old Maths Problem About Colours That Can Never Touch], Peter Dockrill, ScienceAlert, Apr 19, 2018.
* [https://www.zmescience.com/science/math/amateur-mathematician-solves-problem-20042018/ Amateur mathematician Aubrey de Grey, known for his work on anti-aging, solves decades-old problem], Mihai Andrei, ZME Science, Apr 20, 2018.
* [https://automaths.blog/2018/04/21/5-nuances-daubrey-de-grey/ 5 nuances d’Aubrey de Grey], Automaths, Apr 21, 2018.
* [https://science.howstuffworks.com/math-concepts/amateur-solves-part-of-decades-old-math-problem.htm Amateur Solves Part of Decades-Old Math Problem], HowStuffWorks, Apr 30, 2018.
* [https://www.nemokennislink.nl/publicaties/het-platte-vlak-heeft-minstens-vijf-kleuren-nodig/ Het platte vlak heeft minstens vijf kleuren nodig], Kennislink, May 3, 2018.

== Code and data ==

[https://www.dropbox.com/sh/ufknm1v9gtbhad3/AACB2xwaXYx5EGda38_L-0foa?dl=0 This dropbox folder] will contain most of the data and images for the project.

Data:

* [https://www.dropbox.com/s/x6kvu7b3wvdvqsn/graph.dimacs?dl=0 The 1585-vertex graph in DIMACS format]
* [https://www.dropbox.com/s/qk3gbpjvvsjfsc3/sat.dimacs?dl=0 A naive translation of 4-colorability of this graph into a SAT problem in DIMACS format]
* [https://www.dropbox.com/s/nipqikfzcfn9o5a/vertices.sage?dl=0 The vertices of this graph in explicit Sage notation]
* The graph <math>G_2</math>: [http://www.cs.utexas.edu/~marijn/CNP/874.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/874.edge edges (DIMACS)]
* The graph <math>G_3</math>: [http://www.cs.utexas.edu/~marijn/CNP/826.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/826.edge edges (DIMACS)] [http://www.cs.utexas.edu/~marijn/CNP/826.pdf Visualization]
* The [https://www.dropbox.com/sh/ufknm1v9gtbhad3/AADfw9WO1ol2ayQNJEtGf8oBa/Graphs?dl=0 densest unit-distance graphs] on an <math>n\times n</math> grid for <math>n=10,20,\ldots,100</math> (DIMACS format).

Code:

* [https://www.dropbox.com/s/6u1jctbjy38t383/lovaszmoser.m?dl=0 MATLAB script for computing Lovasz number]
* [https://files.jixco.de/pm16/zzvtx/ Python code for converting a list of vertices in Mathematica format into vertices in Z^n]
* [https://files.jixco.de/pm16/edge2cnf.py Python code for converting a DIMACS edge list into a CNF formula (forcing up to 3 suitable vertices to a fixed color, to break some symmetries)]

Software:

* [http://cvxr.com/cvx/ CVX]
* [http://www.labri.fr/perso/lsimon/glucose/ Glucose 4.0]
* [http://minisat.se/ Minisat]

== Wikipedia ==

* [https://en.wikipedia.org/wiki/Graph_coloring#Chromatic_number Chromatic number]
* [https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Erd%C5%91s_theorem_(graph_theory) de Bruijn-Erdos theorem]
* [https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem Hadwiger-Nelson problem]
* [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz number]
* [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]

== Bibliography ==
* [B2008] B. Bukh, [https://doi.org/10.1007/s00039-008-0673-8 Measurable sets with excluded distances], Geometric and Functional Analysis 18 (2008), 668-697.
* [CR2015] D. Cranston, L. Rabern, [https://arxiv.org/abs/1501.01647 The fractional chromatic number of the plane], arXiv:1501.01647
* [deBE1951] N. G. de Bruijn, P. Erdős, [http://www.math-inst.hu/~p_erdos/1951-01.pdf A colour problem for infinite graphs and a problem in the theory of relations], Nederl. Akad. Wetensch. Proc. Ser. A, 54 (1951): 371–373, MR 0046630.
* [deG2018] A. de Grey, [https://arxiv.org/abs/1804.02385 The chromatic number of the plane is at least 5], arXiv:1804.02385
* [EI2018] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.00157 The chromatic number of the plane is at least 5 - a new proof], arXiv:1805.00157
* [F1981] K.J. Falconer, The Realization of distances in measurable subsets covering Rn, J. Combin. Theory Ser. A 31 (1981) 184–189.
* [H1945] H. Hadwiger, Uberdeckung des euklidischen Raum durch kongruente Mengen, Portugaliae Math. 4 (1945), 238–242.
* [MM1961] L. Moser and M. Moser, Solution to Problem 10, Can. Math. Bull. 4 (1961), 187–189.
* [P1998] D. Pritikin, [https://www.sciencedirect.com/science/article/pii/S0095895698918196 All unit-distance graphs of order 6197 are 6-colorable], Journal of Combinatorial Theory, Series B 73.2 (1998): 159-163.
* [S2008] A. Soifer, The Mathematical Coloring Book, Springer, 2008, ISBN-13: 978-0387746401.
* [T1999] C. Thomassen, On the Nelson unit distance coloring problem, Amer. Math. Monthly 106 (1999) 850-853.
* [Tow2005] Townsend, S.P., Colouring the plane with no monochrome unit. Geombinatorics XIV(4) (2005), 181-193.
* [W1973] Woodall, D.R., Distances realized by sets covering the plane. J. Combin. Theory Ser. A, 14 (1973), 187-200.

Hadwiger-Nelson problem

2018-05-03T22:31:12Z

Nazgand: /* Virtual edge */ Added bichromatic virtual edge info

The Chromatic Number of the Plane (CNP) is the chromatic number of the graph whose vertices are elements of the plane, and two points are connected by an edge if they are a unit distance apart. The Hadwiger-Nelson problem asks to compute CNP. The bounds <math>4 \leq CNP \leq 7</math> are classical; recently [deG2018] it was shown that <math>CNP \geq 5</math>. This is achieved by explicitly locating finite unit distance graphs with chromatic number at least 5.

The '''Polymath16''' project seeks to simplify the graphs used in [deG2018] to establish this lower bound. More precisely, the goals are

* '''Goal 1''': Find progressively smaller 5-chromatic unit-distance graphs.
* '''Goal 2''': Reduce (ideally to zero) the reliance on computer assistance for the proof. Computer assistance was leveraged in [deG2018] to analyze a subgraph of size 397.
* '''Goal 3''': Apply these simpler graphs to inform progress in related areas. For example:
** Find a 6-chromatic unit-distance graph.
** Improve the corresponding bound in higher dimensions.
** Improve the current record of 105/29 for the fractional chromatic number of the plane.
** Find the smallest unit-distance graph of a given minimum degree (excluding, in some natural way, boring cases like Cartesian products of a graph with a hypercube).

== Polymath threads ==

* [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/ Polymath proposal: finding simpler unit distance graphs of chromatic number 5], Aubrey de Grey, Apr 10 2018. ('''Active discussion thread''')
* [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/ Polymath16, first thread: Simplifying de Grey’s graph], Dustin Mixon, Apr 14, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic Polymath16, second thread: What does it take to be 5-chromatic?], Dustin Mixon, Apr 22, 2018. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/ Polymath16, third thread: Is 6-chromatic within reach?], Dustin Mixon, May 1, 2018. ('''Active research thread''')

== Notable unit distance graphs ==

A '''unit distance graph''' is a graph that can be realised as a collection of vertices in the plane, with two vertices connected by an edge if they are precisely a unit distance apart. The chromatic number of any such graph is a lower bound for <math>CNP</math>; in particular, if one can find a unit distance graph with no 4-colorings, then <math>CNP \geq 5</math>. The boldface number of vertices is the current minimal number of vertices of a unit distance graph that is currently known to not be 4-colorable.

<math>G_1 \oplus G_2</math> denotes the Minkowski sum of two unit distance graphs <math>G_1,G_2</math> (vertices in <math>G_1 \oplus G_2</math> are sums of the vertices of <math>G_1,G_2</math>). <math>G_1 \cup G_2</math> denotes the union. <math>\mathrm{rot}(G, \theta)</math> denotes <math>G</math> rotated counterclockwise by <math>\theta</math>. <math>\mathrm{trim}(G,r)</math> denotes the trimming of <math>G</math> after removing all vertices of distance greater than <math>r</math> from the origin.

Another basic operation is '''spindling''': taking two copies of a graph <math>G</math>, gluing them together at one vertex, and rotating the copies so that the two copies of another vertex are a unit distance apart. For instance, the Moser spindle is the spindling of a rhombus graph. If a graph has two vertices forced to be the same color in a k-coloring, then the spindling of that graph at those two vertices is not k-colorable.

Many of the graphs are embeddable in abelian groups or rings, particularly those generated by the roots of unity <math>\omega_t := \exp( i \arccos( 1 - \tfrac{1}{2t} ))</math> for various natural numbers t.

{| border=1
|-
!Name!!Number of vertices!! Number of edges !! Structure !! Group !! Colorings
|-
| [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]
| 7
| 11
| Two 60-120-60-120 rhombi with a common vertex, with one pair of sharp vertices coincident and the other joined
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [http://mathworld.wolfram.com/GolombGraph.html Golomb graph]
| 10
| 18
| Contains the center and vertices of a hexagon and equilateral triangle
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| Not 3-colorable
|-
| [https://arxiv.org/abs/1804.02385 H]
| 7
| 12
| Vertices and center of a hexagon
| <math>{\bf Z}[\omega_1]</math>
| Has essentially four 4-colorings, two of which contain a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 J]
| 31
| 72
| Contains 13 copies of H
| <math>{\bf Z}[\omega_1]</math>
| Has essentially six 4-colorings in which no H has a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 K]
| 61
| 150
| Contains 2 copies of J
|
| In all 4-colorings lacking an H with a monochromatic <math>\sqrt{3}</math>-triangle, all pairs of vertices at distance 4 are monochromatic
|-
| [https://arxiv.org/abs/1804.02385 L]
| 121
| 301
| Contains two copies of K and 52 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ <math>L_1</math>]
| 97
|
| Has 40 copies of H
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120295 <math>L_2</math>]
| 120
| 354
|
|
| All 4-colorings contain an H with a monochromatic <math>\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1804.02385 T]
| 9
| 15
| Contains one Moser spindle and useful symmetry; three vertices form an equilateral triangle
|
|
|-
| [https://arxiv.org/abs/1804.02385 U]
| 15
| 33
| Three copies of T at 120-degree rotations: <math>T \cup \mathrm{rot}(T, 2\pi/3) \cup \mathrm{rot}(T, 4\pi/3)</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 V]
| 31
| 30
| Unit vectors at angles consistent with three interlocking Moser spindles
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>V_1</math>]
| 61
| 60
| Union of V and a rotation of V: <math>V \cup \mathrm{rot}(V, \mathrm{arccos}(7/8))</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_b</math>]
| 25
| 24
| Star graph
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_x</math>]
| 13
| 12
| Subgraph of <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_z</math>]
|
|
| Subgraph of <math>V_a</math>; shares a line of symmetry with <math>V_a</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_y</math>]
| 13
| 12
| Subgraph of <math>V_b</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-3970 <math>V_A</math>]
| 37
| 36
| Unit vectors with angles <math>i \frac{\pi}{3} + j \mathrm{arccos} \frac{5}{6} + k \mathrm{arccos} \frac{7}{8}</math>
|
|
|-
| [https://arxiv.org/abs/1804.02385 W]
| 301
| 1230
| Cartesian product of V with itself, minus vertices at more than <math>\sqrt{3}</math> from the centre (i.e. <math>\mathrm{trim}(V \oplus V, \sqrt{3})</math>)
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>W_1</math>]
|
|
| Trimmed product of V with itself (<math>\mathrm{trim}(V \oplus V, 1.95)</math>)
|
|
|-
| [https://arxiv.org/abs/1804.02385 M]
| 1345
| 8268
| Cartesian product of W and H (<math>W \oplus H</math>)
| <math>{\bf Z}[\omega_1, \omega_3]</math>
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120274 <math>M_1</math>]
| 278
|
| Deleting vertices from M while maintaining its restriction on H
|
| All 4-colorings have a monochromatic triangle in the central copy of H
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3933 <math>M_2</math>]
| 7075
|
| Sum of H with a trimmed product of <math>V_1</math> with itself
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 N]
| 20425
| 151311
| Contains 52 copies of M arranged around the H-copies of L
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4, \omega_{16}]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_0</math>]
| 1585
| 7909
| N "shrunk" by stepwise deletions and replacements of vertices
|
| Not 4-colorable
|-
| [https://arxiv.org/abs/1804.02385 G]
| 1581
| 7877
| Deleting 4 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ <math>G_1</math>]
| 1577
|
| Deleting 8 vertices from <math>G_0</math>
|
| Not 4-colorable
|-
| [https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/#comment-120318 <math>G_2</math>]
| 874
| 4461
| Juxtaposing two copies of M and shrinking
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3867 <math>G_3</math>]
| 826
| 4273
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4105 <math>G_4</math>]
| '''803'''
| 4144
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 R]
|
|
| Union of <math>W_1</math> and a rotated copy of <math>W_1</math>
|
|
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3946 <math>\mathrm{trim}(R \oplus H, 1.67)</math>]
| 2563
|
| Trimmed sum of R and H
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 <math>V \oplus V \oplus H</math>]
|
|
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_{64/9}]</math>
| Has two vertices forced to be the same color in a 4-coloring; also no monochromatic <math>\sqrt{3}</math>-triangles
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4074 No name assigned]
| 745
|
| Subgraph of <math>V \oplus V \oplus H</math>
|
| Has two vertices forced to be the same color in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3085
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3961 <math>V_a \oplus V_z \oplus H \cup V_b \oplus V_y \oplus H</math>]
| 3049
|
|
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3950 No name assigned]
| 1951
|
| Trimmed version of <math>V_a \oplus V_x \oplus H \cup V_b \oplus V_y \oplus H</math>
|
| Not 4-colorable
|-
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic <math>V_A \oplus V_A \oplus V_A</math>]
| 6937
| 44439
|
| <math>{\bf Z}[\omega_1, \omega_3, \omega_4]</math>
| Not 4-colorable
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{40}</math>]
| 40
| 82
|
|
| Any 4-coloring that avoids a monochromatic <math>\sqrt{11/3}</math>-edge has two specific vertices forced to be the same color
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{79}</math>]
| 79
| 165
| Spindling of <math>G_{40}</math>
|
| Any 4-coloring has a monochromatic <math>\sqrt{11/3}</math>-edge
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{49}</math>]
| 49
| 180
|
| <math>{\mathbf Q}[\sqrt{3},\sqrt{11}]^2</math>
| Any 4-coloring either has a specific <math>\sqrt{11/3}</math>-edge monochromatic, or a monochromatic <math>1/\sqrt{3}</math>-triangle
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{51}</math>]
| 51
|
|
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring of plane
|-
| [https://arxiv.org/abs/1805.00157 <math>G_{627}</math>]
| 627
| 2982
| Contains <math>G_{51}</math>
|
| Has a specific <math>1/\sqrt{3}</math>-triangle which cannot be monochromatic in a 4-coloring
|-
| [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4161 No name assigned]
| 103
|
|
|
| All 4-colorings contain a monochromatic <math>2/\sqrt{3}</math>-edge
|}

== Lower bounds ==

In [P1988], Pritikin proved that every graph with at most 12 vertices is 4-colorable, and every graph with at most 6197 vertices is 6-colorable. Pritikin's bounds are obtained by coloring the plane with k colors and an additional “wild” color such that points of unit distance are both allowed to receive the wild color. If the wild color occupies a small fraction p of the plane, then an exercise in the probabilistic method gives that any fixed unit-distance graph with n vertices enjoys an embedding in the plane that avoids the wild color (and is therefore k-colorable) provided n<1/p. Pritikin’s coloring for the k=4 case cannot be improved without improving on the densest known subset of the plane that avoids unit distances (originally due to Croft in 1967). See [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable this MO thread] for additional information. As such, improving the bound in the k=4 case might require a new technique, whereas the k=5 and 6 cases might be amenable to optimization.

[https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4176 Every unit distance graph with at most 16 vertices is 5-colorable].

A "tile-based" colouring of the plane must have at least 6 colours, as shown by Townsend [Tow2005]; the same proof with a minor error was also derived by Woodall [W1973]. In [T1999], Thomassen showed that any tiling-based 6-coloring would have to be be "unscaleable", i.e. the maximum diameter of a tile and the minimum separation of same-coloured tiles must both be exactly 1 (so that the distance 1 is excluded by suitable colouring of the boundary points).

== Virtual edge ==

''Warning: other definitions have been proposed and the exact definition of this notion is currently under discussion.''

A '''virtual edge''' of a unit distance graph <math>G</math> is a distance <math>d</math> with the property that every 4-coloring of <math>G</math> contains a monochromatic pair of vertices of distance exactly <math>d</math>. Observe that if a unit distance graph <math>G</math> has a virtual edge at distance <math>d</math>, and if there is another unit distance graph <math>H</math> with a pair of vertices at distance <math>d</math> that cannot be monochromatic in a 4-coloring, then we can create a non-4-colorable unit distance graph by "clamping" a copy of <math>H</math> to every virtual edge in <math>G</math>.

Known examples of virtual edges include:

* <math>G_{40}</math> has a virtual edge at distance <math>\sqrt{11/3}</math>.
* <math>V \oplus V \oplus H</math> has a (single) virtual edge at distance <math>8/3</math>.
* [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4117 <math>(\sqrt{3}\pm 1)/\sqrt{2}</math> are virtual edges of some graphs].

== Bichromatic virtual edge ==

===Definition===
If a unit distance graph <math>H</math> exists with a specific pair of vertices which are distance <math>d</math> apart and is bichromatic in all proper <math>n</math>-colorings of <math>H</math>, then we say that <math>H</math> virtualizes a '''bichromatic virtual edge''' with length <math>d</math> and chromatic number <math>n</math>.

===Vacuous properties===
Considering the graph of the entire plane, all distances correspond to a virtual edge of chromatic number <math>CNP-1</math>.

If a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n</math> exists, then a bichromatic virtual edge with length <math>d</math> and chromatic number <math>n-1</math> exists.

===Convenience and devirtualization===
Bichromatic virtual edges behave the same as the unit edge(which is a bichromatic virtual edge of every chromatic number), and thus may be used to simplify the construction of graphs.

Recursively replacing bichromatic virtual edges of a graph <math>G</math> with graphs which virtualize the bichromatic virtual edges through '''clamping''' produces a unit distance graph <math>G'</math> where <math>\chi(G')\geq\chi(G)</math>.

Devirtualizing bichromatic virtual edges of a graph <math>G</math> (i.e. clamping) has no benefit unless nontrivial points of <math>G</math> and <math>H_0</math> overlap or nontrivial points of <math>H_1</math> and <math>H_0</math> overlap, where <math>H_0</math> and <math>H_1</math> are graphs virtualizing bichromatic virtual edges of <math>G</math>. Such overlaps may cause <math>\chi(G')>\chi(G)</math>. If no nontrivial overlaps exist, then <math>\chi(G')=\max\left(\chi(G),\chi(H_0),\chi(H_1),\dots\right)</math>.

Because more than 1 graph virtualizes any given bichromatic virtual edge for a fixed length and fixed chromatic number, multiple devirtualizations exist for every finite graph.

===Relation to rings===
A '''devirtualized ring''' of graph <math>G</math> is a ring which contains all the vertices of a devirtualization of <math>G</math>.

Use the implied shorthand for arbitrary sets <math>\mathbb{S}[\left\{a,b,c,\dots\right\}]=\mathbb{S}[a,b,c,\dots]</math>.

Let the vertices of graph <math>G</math> generate the ring <math>S_0[T_0]</math>.
Let the bichromatic virtual edges be virtualizable by the graphs <math>H_1,\dots,H_m</math>.
Let <math>S_k[T_k]</math> be the ring generated by the vertices of <math>H_k</math>.
Let <math>S</math> be the ring generated by <math>\bigcup_{k=0}^m S_k</math>.
Let <math>T=\bigcup_{k=0}^m T_k</math>.
Then <math>S[T]</math> is a devirtualized ring of <math>G</math>.

As multiple graphs may virtualize the same virtual edge, careful choice may be required to minimize the set <math>T</math>.

===Multiplicative property===
Let <math>H_0</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_0</math> and chromatic number <math>n</math>.
Let <math>H_1</math> be a graph virtualizing a bichromatic virtual edge with length <math>d_1</math> and chromatic number <math>n</math>.
Create a graph <math>H_2</math> by scaling <math>H_0</math> by a factor of <math>d_1</math>. Graph <math>H_2</math> then virtualizes a bichromatic virtual edge length <math>d_0d_1</math> and chromatic number <math>n</math>.

===Notable bichromatic virtual edges===
{| border=1
|-
! Chromatic number !! Length <math>d</math>!! Proof
|-
| any
| 1
| trivial case
|-
| 2
| <math>0\leq d\leq 4</math>
| moving the flexible parts of the graph (0,0),(0,1),(0,2),(0,3)
|-
| 3
| <math>\left\{\left|\frac{(3+\sqrt{-3})k}{2}+\sqrt{-3}j+(-1)^r\right| \mid k,j\in\mathbb{Z}, r\in\mathbb{R}\right\}</math>
| equilateral triangle tiling
|}

===Continuous ranges of bichromatic virtual edges===
Flexible graphs might produce continuous ranges of lengths of bichromatic virtual edges of chromatic number <math>n</math> without having a specific pair which is monochrome for every <math>n</math>-coloring.
Let the range of lengths be on the interval <math>d_0\leq d\leq d_1</math>.

If <math>1 > d_1</math>, then let <math>k\in\mathbb{Z}^+,d_0^{k+1} \leq d_1^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all lengths larger than <math>d_0^k</math>. A finite area allowed to be the same color, the infinite area of the plane, and a finite upper bound on CNP implies <math>CNP>n</math>.

If <math>d_0 < 1</math>, then let <math>k\in\mathbb{Z}^+,d_1^{k+1} \geq d_0^k</math>. The multiplicative property of bichromatic virtual edges implies the existence of bichromatic virtual edges of chromatic number <math>n</math> for all non-zero lengths smaller than <math>d_1^k</math>. Considering a set of <math>CNP+1</math> all within distance <math>d_1^k</math> of each other implies <math>CNP>n</math>.

Using a flexible bichromatic virtual edge of chromatic number <math>n</math> and a bichromatic virtual edge of chromatic number <math>n+1</math> virtualizable by graph <math>H</math>, a flexible bichromatic virtual edge of chromatic number <math>n+1</math> can be made by scaling <math>H</math> to some arbitrarily large or arbitrarily small size and clamping flexible bichromatic virtual edges of chromatic number <math>n</math> as a replacement of each rigid bichromatic virtual edge of <math>H</math>.

Thus no bichromatic virtual edge of chromatic number <math>CNP</math> exists.

== Probabilistic formulation ==

See [[Probabilistic formulation of Hadwiger-Nelson problem]].

== Further questions ==

* What are the [http://mathworld.wolfram.com/IndependenceRatio.html independence ratios] of the above unit distance graphs?
* What are the [https://en.wikipedia.org/wiki/Fractional_coloring fractional chromatic numbers] of these graphs?
* What are the [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz numbers] of these graphs?
** The Lovasz theta function value of Lovasz number of <math>G_2</math> at the complement [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3844 is in the interval [3.3746, 3.3748]].
* What about the Erdos unit distance graph (<math>n</math> vertices, <math>n^{1+c/\log\log n}</math> edges)?
* Can we use de Grey’s graph to construct unit-distance graphs that are not 5-colorable? To answer this question, we first need to understand how 5-colorings of de Grey’s graph force small collections of vertices to be colored. [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3899 Varga] and [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3940 Nazgand] provide some thoughts along these lines. Even if we can’t stitch together a 6-chromatic unit-distance graph with these ideas, we might be able to apply them to [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3900 prove that the measurable chromatic number of the plane is at least 6].
* It appears as though the coordinates of our smallest 5-chromatic graph lie in <math>\mathbb{Q}[\sqrt{3}, \sqrt{5}, \sqrt{11}]</math> (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3879 this]). If we view the plane as the complex plane, what is the smallest ring that admits a 5-chromatic single-distance graph? Every single-distance graph in the Eistenstein integers and Gaussian integers is 2-chromatic (see [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3914 this]). [https://dustingmixon.wordpress.com/2018/04/14/polymath16-first-thread-simplifying-de-greys-graph/#comment-3934 David Speyer suggests] looking at <math>\mathbb{Z}[\frac{1+\sqrt{-71}}{2}]</math> next.

== Blog, forums, and media ==

* [https://rjlipton.wordpress.com/2011/05/22/more-on-coloring-the-plane/ More On Coloring The Plane], Richard Lipton, May 22, 2011.
* [https://mathoverflow.net/questions/236392/has-there-been-a-computer-search-for-a-5-chromatic-unit-distance-graph Has there been a computer search for a 5-chromatic unit distance graph?], Juno, Apr 16, 2016.
* [https://quomodocumque.wordpress.com/2018/04/09/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Jordan Ellenberg, Apr 9 2018.
* [https://gilkalai.wordpress.com/2018/04/10/aubrey-de-grey-the-chromatic-number-of-the-plane-is-at-least-5/ Aubrey de Grey: The chromatic number of the plane is at least 5], Gil Kalai, Apr 10 2018.
* [https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Dustin Mixon, Apr 10, 2018.
* [https://www.scottaaronson.com/blog/?p=3697 Amazing progress on long-standing problems], Scott Aaronson, Apr 11 2018.
* [https://dustingmixon.wordpress.com/2018/04/13/the-chromatic-number-of-the-plane-is-at-least-5-part-ii/ The chromatic number of the plane is at least 5, Part II], Dustin Mixon, Apr 13 2018.
* [http://community.wolfram.com/groups/-/m/t/1320004 A 5-chromatic unit distance graph], Ed Pegg, Apr 13 2018.
* [http://aperiodical.com/2018/04/the-chromatic-number-of-the-plane-is-at-least-5/ The chromatic number of the plane is at least 5], Katie Steckles, Apr 17 2018.
* [https://www.quantamagazine.org/decades-old-graph-problem-yields-to-amateur-mathematician-20180417/ Decades-Old Graph Problem Yields to Amateur Mathematician], Evelyn Lamb, Quanta, Apr 17, 2018.
* [https://www.heise.de/newsticker/meldung/Zahlen-bitte-Wie-bunt-ist-die-Ebene-4024574.html Zahlen, bitte! 5 - Wie bunt ist die Ebene?], Harald Bögeholz, Heise, Apr 17, 2018.
* [http://www.sciencemag.org/news/2018/04/amateur-mathematician-cracks-decades-old-math-problem Amateur mathematician cracks decades-old math problem], Katie Langin, Science News, Apr 18, 2018.
* [https://mathoverflow.net/questions/298198/how-much-of-the-plane-is-4-colorable How much of the plane is 4-colorable?], Dustin Mixon, Apr 18, 2018.
* [https://www.sciencealert.com/amateur-solves-decades-old-maths-problem-about-colours-that-can-never-touch-hadwiger-nelson-problem An Amateur Solved a 60-Year-Old Maths Problem About Colours That Can Never Touch], Peter Dockrill, ScienceAlert, Apr 19, 2018.
* [https://www.zmescience.com/science/math/amateur-mathematician-solves-problem-20042018/ Amateur mathematician Aubrey de Grey, known for his work on anti-aging, solves decades-old problem], Mihai Andrei, ZME Science, Apr 20, 2018.
* [https://automaths.blog/2018/04/21/5-nuances-daubrey-de-grey/ 5 nuances d’Aubrey de Grey], Automaths, Apr 21, 2018.
* [https://science.howstuffworks.com/math-concepts/amateur-solves-part-of-decades-old-math-problem.htm Amateur Solves Part of Decades-Old Math Problem], HowStuffWorks, Apr 30, 2018.
* [https://www.nemokennislink.nl/publicaties/het-platte-vlak-heeft-minstens-vijf-kleuren-nodig/ Het platte vlak heeft minstens vijf kleuren nodig], Kennislink, May 3, 2018.

== Code and data ==

[https://www.dropbox.com/sh/ufknm1v9gtbhad3/AACB2xwaXYx5EGda38_L-0foa?dl=0 This dropbox folder] will contain most of the data and images for the project.

Data:

* [https://www.dropbox.com/s/x6kvu7b3wvdvqsn/graph.dimacs?dl=0 The 1585-vertex graph in DIMACS format]
* [https://www.dropbox.com/s/qk3gbpjvvsjfsc3/sat.dimacs?dl=0 A naive translation of 4-colorability of this graph into a SAT problem in DIMACS format]
* [https://www.dropbox.com/s/nipqikfzcfn9o5a/vertices.sage?dl=0 The vertices of this graph in explicit Sage notation]
* The graph <math>G_2</math>: [http://www.cs.utexas.edu/~marijn/CNP/874.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/874.edge edges (DIMACS)]
* The graph <math>G_3</math>: [http://www.cs.utexas.edu/~marijn/CNP/826.vtx vertices (Mathematica)] [http://www.cs.utexas.edu/~marijn/CNP/826.edge edges (DIMACS)] [http://www.cs.utexas.edu/~marijn/CNP/826.pdf Visualization]
* The [https://www.dropbox.com/sh/ufknm1v9gtbhad3/AADfw9WO1ol2ayQNJEtGf8oBa/Graphs?dl=0 densest unit-distance graphs] on an <math>n\times n</math> grid for <math>n=10,20,\ldots,100</math> (DIMACS format).

Code:

* [https://www.dropbox.com/s/6u1jctbjy38t383/lovaszmoser.m?dl=0 MATLAB script for computing Lovasz number]
* [https://files.jixco.de/pm16/zzvtx/ Python code for converting a list of vertices in Mathematica format into vertices in Z^n]
* [https://files.jixco.de/pm16/edge2cnf.py Python code for converting a DIMACS edge list into a CNF formula (forcing up to 3 suitable vertices to a fixed color, to break some symmetries)]

Software:

* [http://cvxr.com/cvx/ CVX]
* [http://www.labri.fr/perso/lsimon/glucose/ Glucose 4.0]
* [http://minisat.se/ Minisat]

== Wikipedia ==

* [https://en.wikipedia.org/wiki/Graph_coloring#Chromatic_number Chromatic number]
* [https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Erd%C5%91s_theorem_(graph_theory) de Bruijn-Erdos theorem]
* [https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem Hadwiger-Nelson problem]
* [https://en.wikipedia.org/wiki/Lov%C3%A1sz_number Lovasz number]
* [https://en.wikipedia.org/wiki/Moser_spindle Moser spindle]

== Bibliography ==
* [B2008] B. Bukh, [https://doi.org/10.1007/s00039-008-0673-8 Measurable sets with excluded distances], Geometric and Functional Analysis 18 (2008), 668-697.
* [CR2015] D. Cranston, L. Rabern, [https://arxiv.org/abs/1501.01647 The fractional chromatic number of the plane], arXiv:1501.01647
* [deBE1951] N. G. de Bruijn, P. Erdős, [http://www.math-inst.hu/~p_erdos/1951-01.pdf A colour problem for infinite graphs and a problem in the theory of relations], Nederl. Akad. Wetensch. Proc. Ser. A, 54 (1951): 371–373, MR 0046630.
* [deG2018] A. de Grey, [https://arxiv.org/abs/1804.02385 The chromatic number of the plane is at least 5], arXiv:1804.02385
* [EI2018] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.00157 The chromatic number of the plane is at least 5 - a new proof], arXiv:1805.00157
* [F1981] K.J. Falconer, The Realization of distances in measurable subsets covering Rn, J. Combin. Theory Ser. A 31 (1981) 184–189.
* [H1945] H. Hadwiger, Uberdeckung des euklidischen Raum durch kongruente Mengen, Portugaliae Math. 4 (1945), 238–242.
* [MM1961] L. Moser and M. Moser, Solution to Problem 10, Can. Math. Bull. 4 (1961), 187–189.
* [P1998] D. Pritikin, [https://www.sciencedirect.com/science/article/pii/S0095895698918196 All unit-distance graphs of order 6197 are 6-colorable], Journal of Combinatorial Theory, Series B 73.2 (1998): 159-163.
* [S2008] A. Soifer, The Mathematical Coloring Book, Springer, 2008, ISBN-13: 978-0387746401.
* [T1999] C. Thomassen, On the Nelson unit distance coloring problem, Amer. Math. Monthly 106 (1999) 850-853.
* [Tow2005] Townsend, S.P., Colouring the plane with no monochrome unit. Geombinatorics XIV(4) (2005), 181-193.
* [W1973] Woodall, D.R., Distances realized by sets covering the plane. J. Combin. Theory Ser. A, 14 (1973), 187-200.