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Hadwiger-Nelson problem

2018-06-07T09:26:49Z

Pgibbs: /* 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. (''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. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/29/polymath16-sixth-thread-wrestling-with-infinite-graphs/ Polymath16, sixth thread: Wrestling with infinite graphs], Dustin Mixon, May 29, 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://arxiv.org/abs/1805.12181 <math>G_7</math>]
| '''553'''
| 2722
|
|
| 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/2018/05/10/polymath16-fifth-thread-human-verifiable-proofs/#comment-4423 <math>G_{26}</math>]
| 26
|
|Except for the origin, all vertices lie on one of 3 unit circles.
|
| 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://arxiv.org/abs/1805.12181 553]
| 2722
| 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-4583 462]
|-
| <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
* [EI2018b] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.06055 The Hadwiger-Nelson problem with two forbidden distances], arXiv:1805.06055
* [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.
* [S2009] M. Shinohara, [https://www.sciencedirect.com/science/article/pii/S019566980300218X Classification of three-distance sets in two dimensional Euclidean space], European Journal of Combinatorics Volume 25, Issue 7, October 2004, Pages 1039-1058.
* [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]]

Algebraic formulation of Hadwiger-Nelson problem

2018-06-04T20:40:04Z

Pgibbs:

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with a periodic colouring of periodicity <math>p</math> (P<math>p</math>)
:the ring can be <math>k</math>-colored with non-linear colorings (periodic or non-periodic) only (N), or
:the ring can be <math>k</math>-colored with in more than one of the above ways (e.g. L+N).
This table summarises what is known. Where there is more than one option shown the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| L+N
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| L+P8
| L+N
| N
| L+N
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L or L+N
| N
| N
|-
| <math>\mathbb{Z}[\omega, \omega_3, \omega_4, \omega_{25/9}]</math>
| X
| X
| X
| X or N
| X or N
| L+N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X or N
| X or N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finite number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\overline{R}_1 = \mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\overline{R}_H = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_H = \overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring. Therefore <math>\overline{R}_H</math> has a linear 5-coloring.

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Elements of <math>\overline{R}_2</math> take the form <math>z = \frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k}</math> where <math>a,b,c,d</math> are integers subject to some congruence relations (see above). The elements of unit modulus are determined by <math>|z|^2 = 1</math> which is equivalent to

<math>a^2 + 3b^2 + 11c^2 + 33d^2 + (ad + bc)\sqrt{33} = 16 \cdot 3^{2k}</math>

Since <math>\sqrt{33}</math> is irrational this splits into two integer equations. The solutions of <math>ad + bc = 0</math> can be parametrised over integers by <math>a = wx, b = wy, c = zx, d = -zy</math>. Substituting into the rest of the equation gives

<math>(wx)^2 + 3(wy)^2 + 11(zx)^2 + 33(-zy)^2 = 16 \cdot 3^{2k}</math>

which factorizes to <math>(w^2 + 11z^2)(x^2 + 3y^2) = 16 \cdot 3^{2k}</math>

Also, <math>4 \cdot 3^k z = wx + wy\sqrt{3} + zx\sqrt{11} - zy\sqrt{33} = (w+z\sqrt{11})(x + y\sqrt{3})</math>

Therefore <math>3^k z = u v</math> where <math>u \in \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right], |u|^2 = 3^l</math> and <math>v \in \mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right], |v|^2 = 3^{k-l}</math>

Both <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math> are [https://en.wikipedia.org/wiki/Unique_factorization_domain unique factorization domains]. Prime factorizations of 3 are given by <math>3 = (-i\sqrt{3}) \times i\sqrt{3}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>3 = \frac{1+i\sqrt{11}}{2} \times \frac{1-i\sqrt{11}}{2}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>. Units in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> are <math>\omega^i</math> and <math>\pm 1</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>

By prime factorization this gives all solutions for elements of unit modulus in <math>\overline{R}_2</math> in the form <math>z = \omega^i \eta^j, i,j \in \mathbb{Z}</math> where <math>\eta = \frac{\sqrt{33}+i\sqrt{3}}{6} = \sqrt{\omega_3}</math>

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\overline{R}_4 = \overline{R}_2\left[\frac{1}{2},\frac{1}{7}, \phi \right]</math>. Since <math>\mathbb{Z}[\phi]</math> is another unique factorization domain the same method used to find the elements of unit modulus in <math>\overline{R}_2</math> can be used here multiple times to solve this case too. the details are omitted. The result is that elements <math>z</math> of unit modulus in <math>\overline{R}_4</math> must be of the form

<math>z = \omega^p \eta^q \zeta^r \alpha^s \beta^t \gamma^u, p,q,r,s,t,u \in \mathbb{Z}</math>

where <math>\omega = \frac{1 + i\sqrt{3}}{2}, \eta = \frac{\sqrt{33}+i\sqrt{3}}{6}, \zeta = \frac{1 + i\sqrt{15}}{4}, \alpha = \frac{\sqrt{5}+i\sqrt{11}}{4}, \beta = \frac{1+4i\sqrt{3}}{7}, \gamma = \frac{17+3i\sqrt{55}}{28}</math>

For the elements of unit modulus in <math>\overline{R}_4</math> use just <math>\omega^p \eta^q \zeta^r \alpha^s</math>

[[Category:Polymath16]]

Algebraic formulation of Hadwiger-Nelson problem

2018-06-04T18:57:16Z

Pgibbs:

Algebraic formulation of Hadwiger-Nelson problem

2018-06-04T18:43:59Z

Pgibbs: /* example \overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] */

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with a periodic colouring of periodicity <math>p</math> (P<math>p</math>)
:the ring can be <math>k</math>-colored with non-linear colorings (periodic or non-periodic) only (N), or
:the ring can be <math>k</math>-colored with in more than one of the above ways (e.g. L+N).
This table summarises what is known. Where there is more than one option shown the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| L+N
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| L+P8
| L+N
| N
| L+N
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L or L+N
| N
| N
|-
| <math>\mathbb{Z}[\omega, \omega_3, \omega_4, \omega_{25/9}]</math>
| X
| X
| X
| X or N
| X or N
| L+N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X or N
| X or N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finite number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\overline{R}_1 = \mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\overline{R}_H = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_H = \overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring. Therefore <math>\overline{R}_H</math> has a linear 5-coloring.

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Elements of <math>\overline{R}_2</math> take the form <math>z = \frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k}</math> where <math>a,b,c,d</math> are integers subject to some congruence relations (see above). The elements of unit modulus are determined by <math>|z|^2 = 1</math> which is equivalent to

<math>a^2 + 3b^2 + 11c^2 + 33d^2 + (ad + bc)\sqrt{33} = 16 \cdot 3^{2k}</math>

Since <math>\sqrt{33}</math> is irrational this splits into two integer equations. The solutions of <math>ad + bc = 0</math> can be parametrised over integers by <math>a = wx, b = wy, c = zx, d = -zy</math>. Substituting into the rest of the equation gives

<math>(wx)^2 + 3(wy)^2 + 11(zx)^2 + 33(-zy)^2 = 16 \cdot 3^{2k}</math>

which factorizes to <math>(w^2 + 11z^2)(x^2 + 3y^2) = 16 \cdot 3^{2k}</math>

Also, <math>4 \cdot 3^k z = wx + wy\sqrt{3} + zx\sqrt{11} - zy\sqrt{33} = (w+z\sqrt{11})(x + y\sqrt{3})</math>

Therefore <math>3^k z = u v</math> where <math>u \in \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right], |u|^2 = 3^l</math> and <math>v \in \mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right], |v|^2 = 3^{k-l}</math>

Both <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math> are [https://en.wikipedia.org/wiki/Unique_factorization_domain unique factorization domains]. Prime factorizations of 3 are given by <math>3 = (-i\sqrt{3}) \times i\sqrt{3}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>3 = \frac{1+i\sqrt{11}}{2} \times \frac{1-i\sqrt{11}}{2}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>. Units in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> are <math>\omega^i</math> and <math>\pm 1</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>

By prime factorization this gives all solutions for elements of unit modulus in <math>\overline{R}_2</math> in the form <math>z = \omega^i \eta^j, i,j \in \mathbb{Z}</math> where <math>\eta = \frac{\sqrt{33}+i\sqrt{3}}{6} = \sqrt{\omega_3}</math>

== Applications ==

==== <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

It is known that for 4-colourings in the ring <math>\overline{R}_2</math> there can be [https://arxiv.org/abs/1804.02385 no monochromtaic equilateral triangles] with sides of length <math>\sqrt{3}</math> This has not yet been demonstrated by means of a human verifiable proof, but once established computationally it can be shown that the ring <math>\overline{R}_3</math> is not 4-colorable by hand-checking the ways of colouring graph <math>K</math>. Another route to the conclusion that the CNP is at least 5 is to show that for 4-colorings in the ring <math>\overline{R}_2</math> two vertices at a separation of 8/3 are monochromatic (This is equivalent to a periodicity of 8.) It then follows from an immediate spindling argument that <math>\overline{R}_H</math> cannot be 4-colored. This method requires further computational steps.

The goal here is to apply algebraic methods to show that 4-colorings of <math>\overline{R}_2</math> have a periodicity of 8 given only two assumptions:
* That there are no monochromatic <math>\sqrt{3}</math>-triangles.
* That any rhombus of unit sides that is not in a single <math>\overline{R}_1</math> sub-ring is always coloured with either two or four colours (never three)

The first step is to use the assumption that there are such no monochromatic triangles to demonstrate two features of any 4-coloring <math>c_4(z), z \in \overline{R}_2</math> when restricted to a subring <math>\overline{R}_1 \subset \overline{R}_2</math>
* That in at least one of the three direction along edges of the <math>\overline{R}_1</math> sub-ring, the colouring has peridicity 2, i.e. <math>c_4(z+2\omega^a) = c_4(z), z \in \overline{R}_1</math> for <math>a = 0,1</math> or <math>2</math>.
* That no two points of the same color are separated by an odd integer distance in a direction along edges of the graph
This was [https://dustingmixon.wordpress.com/2018/05/29/polymath16-sixth-thread-wrestling-with-infinite-graphs/#comment-4623 shown in the sixth thread]

The assumption that a rhombus in any other plane is either 2-coloured or 4-coloured means that given the colour of three of its vertices, the fourth can be determined by an exclusive-or operation. This in turn means that for any parallelogram made from edges of the graph but with sides of any (integer) length, the same rule applies to the colors at its four corner vertices.

Consider now single vertex corresponding to an element of the ring which by translation symmetry we can assume to be the origin without loss of generality. This vertex is also the origin of lattices <math>L(\eta^a)</math> generated additively by <math>\eta^a</math> and <math>\omega \eta^a</math> for all <math>a \in \mathbb{Z}</math>. As a lattice, group or graph, each <math>L(\eta^a)</math> is isomorphic to <math>\overline{R}_1</math> Therefore by assumption it has a periodicity of two in one of its three directions <math>u(a) = \omega^{b(a)} \eta^a, b(a) \in \{0,1,2\}</math> for any four coloring <math>c_4</math>, i.e. <math>c_4(x+2u(a)) = c_4(x)</math>, for all <math>x \in L(\eta^a)</math>.

Consider in particular the four lattices <math>L(1), L(\eta^2), L(\eta^4), L(\eta^6)</math>. <math>b(a)</math> takes one of three values so by the pigeon-hole principle it must take the same value for at least one pair out of these four lattices. By rotational symmetry of the ring it can be assumed without loss of generality that the two lattices with periodicity two in the same direction are <math>L(\eta^6)</math> and <math>L(\eta^a)</math> for one of <math>a = 0,2,4</math>, and furthermore that the direction with the perioicity two for these two lattices is <math>b(a) = b(6) = 0</math>. Therefore <math>c_4(x) = c_4(x+2\eta^a), x \in L(\eta^a)</math> and <math>c_4(x) = c_4(x+2\eta^6), x \in L(\eta^6)</math>.

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\overline{R}_4 = \overline{R}_2\left[\frac{1}{2},\frac{1}{7}, \phi \right]</math>. Since <math>\mathbb{Z}[\phi]</math> is another unique factorization domain the same method used to find the elements of unit modulus in <math>\overline{R}_2</math> can be used here multiple times to solve this case too. the details are omitted. The result is that elements <math>z</math> of unit modulus in <math>\overline{R}_4</math> must be of the form

<math>z = \omega^p \eta^q \zeta^r \alpha^s \beta^t \gamma^u, p,q,r,s,t,u \in \mathbb{Z}</math>

where <math>\omega = \frac{1 + i\sqrt{3}}{2}, \eta = \frac{\sqrt{33}+i\sqrt{3}}{6}, \zeta = \frac{1 + i\sqrt{15}}{4}, \alpha = \frac{\sqrt{5}+i\sqrt{11}}{4}, \beta = \frac{1+4i\sqrt{3}}{7}, \gamma = \frac{17+3i\sqrt{55}}{28}</math>

For the elements of unit modulus in <math>\overline{R}_4</math> use just <math>\omega^p \eta^q \zeta^r \alpha^s</math>

[[Category:Polymath16]]

Hadwiger-Nelson problem

2018-06-01T12:32:18Z

Pgibbs: /* Best known results for the chromatic number in higher dimensions */

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. (''Inactive research thread'')
* [https://dustingmixon.wordpress.com/2018/05/29/polymath16-sixth-thread-wrestling-with-infinite-graphs/ Polymath16, sixth thread: Wrestling with infinite graphs], Dustin Mixon, May 29, 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/2018/05/10/polymath16-fifth-thread-human-verifiable-proofs/#comment-4423 <math>G_{26}</math>]
| 26
|
|Except for the origin, all vertices lie on one of 3 unit circles.
|
| 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://arxiv.org/abs/1805.12181 553]
| 2722
| 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-4583 462]
|-
| <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
* [EI2018b] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.06055 The Hadwiger-Nelson problem with two forbidden distances], arXiv:1805.06055
* [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.
* [S2009] M. Shinohara, [https://www.sciencedirect.com/science/article/pii/S019566980300218X Classification of three-distance sets in two dimensional Euclidean space], European Journal of Combinatorics Volume 25, Issue 7, October 2004, Pages 1039-1058.
* [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]]

Algebraic formulation of Hadwiger-Nelson problem

2018-05-28T22:28:27Z

Pgibbs: /* example \overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] */

Hadwiger-Nelson problem

2018-05-25T11:11:47Z

Pgibbs: /* Best known results for the chromatic number in higher dimensions */

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-4583 462]
|-
| <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
* [EI2018b] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.06055 The Hadwiger-Nelson problem with two forbidden distances], arXiv:1805.06055
* [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.
* [S2009] M. Shinohara, [https://www.sciencedirect.com/science/article/pii/S019566980300218X Classification of three-distance sets in two dimensional Euclidean space], European Journal of Combinatorics Volume 25, Issue 7, October 2004, Pages 1039-1058.
* [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]]

Hadwiger-Nelson problem

2018-05-19T17:05:39Z

Pgibbs: /* Best known results for the chromatic number in higher dimensions */

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-4525 472]
|-
| <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
* [EI2018b] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.06055 The Hadwiger-Nelson problem with two forbidden distances], arXiv:1805.06055
* [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.
* [S2009] M. Shinohara, [https://www.sciencedirect.com/science/article/pii/S019566980300218X Classification of three-distance sets in two dimensional Euclidean space], European Journal of Combinatorics Volume 25, Issue 7, October 2004, Pages 1039-1058.
* [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]]

Hadwiger-Nelson problem

2018-05-19T07:55:41Z

Pgibbs: /* Best known results for the chromatic number in higher dimensions */

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-4520 483]
|-
| <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
* [EI2018b] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.06055 The Hadwiger-Nelson problem with two forbidden distances], arXiv:1805.06055
* [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.
* [S2009] M. Shinohara, [https://www.sciencedirect.com/science/article/pii/S019566980300218X Classification of three-distance sets in two dimensional Euclidean space], European Journal of Combinatorics Volume 25, Issue 7, October 2004, Pages 1039-1058.
* [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]]

Hadwiger-Nelson problem

2018-05-19T07:15:44Z

Pgibbs: /* Best known results for the chromatic number in higher dimensions */

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-4519 484]
|-
| <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
* [EI2018b] G. Exoo, D. Ismailescu, [https://arxiv.org/abs/1805.06055 The Hadwiger-Nelson problem with two forbidden distances], arXiv:1805.06055
* [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.
* [S2009] M. Shinohara, [https://www.sciencedirect.com/science/article/pii/S019566980300218X Classification of three-distance sets in two dimensional Euclidean space], European Journal of Combinatorics Volume 25, Issue 7, October 2004, Pages 1039-1058.
* [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]]

Hadwiger-Nelson problem

2018-05-16T06:42:39Z

Pgibbs: /* Best known results for the chromatic number in higher dimensions */

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.

== 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]]

Hadwiger-Nelson problem

2018-05-15T20:38:47Z

Pgibbs: /* 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. (''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>
| 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/2018/05/05/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/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4316 633]
| 3166
| 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.

== 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]]

Hadwiger-Nelson problem

2018-05-14T19:15:25Z

Pgibbs: /* Best known results for the chromatic number in higher dimensions */

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>
| 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/2018/05/05/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/2018/04/14/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>
|}

== 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.

=== 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/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4316 633]
| 3166
| 7
|-
| <math>\mathbb{R}^3</math>
| [https://www.sciencedirect.com/science/article/pii/S0012365X00004064 6]
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4059 60]
|
| [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]
|
|
|
|-
| <math>\mathbb{R}^6</math>
| [https://arxiv.org/abs/1408.2002 12]
| 175
|
|
|-
| <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.

== 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]]

Hadwiger-Nelson problem

2018-05-14T18:58:58Z

Pgibbs: /* Best known results for the chromatic number in higher dimensions */

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>
| 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/2018/05/05/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/2018/04/14/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>
|}

== 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.

=== 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/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4316 633]
| 3166
| 7
|-
| <math>\mathbb{R}^3</math>
| [https://www.sciencedirect.com/science/article/pii/S0012365X00004064 6]
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4059 60]
|
| [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]
|
|
|
|-
| <math>\mathbb{R}^6</math>
| [https://arxiv.org/abs/1408.2002 12]
| 175
|
|
|-
| <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 21]
| 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.

== 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]]

Hadwiger-Nelson problem

2018-05-14T18:32:08Z

Pgibbs: /* Best known results for the chromatic number in higher dimensions */

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>
| 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/2018/05/05/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/2018/04/14/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>
|}

== 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.

=== 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/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4316 633]
| 3166
| 7
|-
| <math>\mathbb{R}^3</math>
| [https://www.sciencedirect.com/science/article/pii/S0012365X00004064 6]
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4059 60]
|
| [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]
|
|
|
|-
| <math>\mathbb{R}^6</math>
| [https://arxiv.org/abs/1408.2002 12]
| 175
|
|
|-
| <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]
| 240
|
|
|-
| <math>\mathbb{R}^9</math>
| [https://arxiv.org/abs/1512.03472 21]
| 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.

== 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]]

Hadwiger-Nelson problem

2018-05-14T18:25:11Z

Pgibbs: /* Best known results for the chromatic number in higher dimensions */

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>
| 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/2018/05/05/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/2018/04/14/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>
|}

== 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.

=== 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/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4316 633]
| 3166
| 7
|-
| <math>\mathbb{R}^3</math>
| [https://www.sciencedirect.com/science/article/pii/S0012365X00004064 6]
| [https://dustingmixon.wordpress.com/2018/04/22/polymath16-second-thread-what-does-it-take-to-be-5-chromatic/#comment-4059 60]
|
| [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
|
| [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]
|
|
|
|-
| <math>\mathbb{R}^6</math>
| [https://arxiv.org/abs/1408.2002 12]
| 175
|
|
|-
| <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]
| 240
|
|
|-
| <math>\mathbb{R}^9</math>
| [https://arxiv.org/abs/1512.03472 21]
| 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.

== 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]]

Hadwiger-Nelson problem

2018-05-14T14:45:57Z

Pgibbs: /* 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. (''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>
| 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/2018/05/05/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/2018/04/14/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>
|}

== 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.

=== 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/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4316 633]
| 3166
| 7
|-
| <math>\mathbb{R}^3</math>
| [https://www.sciencedirect.com/science/article/pii/S0012365X00004064 6]
|
|
| [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
|
| [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]
|
|
|
|-
| <math>\mathbb{R}^6</math>
| [https://arxiv.org/abs/1408.2002 12]
| 175
|
|
|-
| <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]
| 240
|
|
|-
| <math>\mathbb{R}^9</math>
| [https://arxiv.org/abs/1512.03472 21]
| 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.

== 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]]

Algebraic formulation of Hadwiger-Nelson problem

2018-05-12T06:48:57Z

Pgibbs: /* Coloring Rings */

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with a periodic colouring of periodicity <math>p</math> (P<math>p</math>)
:the ring can be <math>k</math>-colored with non-linear colorings (periodic or non-periodic) only (N), or
:the ring can be <math>k</math>-colored with in more than one of the above ways (e.g. L+N).
This table summarises what is known. Where there is more than one option shown the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| L+N
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| L+P8
| L+N
| N
| L+N
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L or L+N
| N
| N
|-
| <math>\mathbb{Z}[\omega, \omega_3, \omega_4, \omega_{25/9}]</math>
| X
| X
| X
| X or N
| X or N
| L+N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X or N
| X or N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finite number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\overline{R}_1 = \mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\overline{R}_H = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_H = \overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring. Therefore <math>\overline{R}_H</math> has a linear 5-coloring.

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Elements of <math>\overline{R}_2</math> take the form <math>z = \frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k}</math> where <math>a,b,c,d</math> are integers subject to some congruence relations (see above). The elements of unit modulus are determined by <math>|z|^2 = 1</math> which is equivalent to

<math>a^2 + 3b^2 + 11c^2 + 33d^2 + (ad + bc)\sqrt{33} = 16 \cdot 3^{2k}</math>

Since <math>\sqrt{33}</math> is irrational this splits into two integer equations. The solutions of <math>ad + bc = 0</math> can be parametrised over integers by <math>a = wx, b = wy, c = zx, d = -zy</math>. Substituting into the rest of the equation gives

<math>(wx)^2 + 3(wy)^2 + 11(zx)^2 + 33(-zy)^2 = 16 \cdot 3^{2k}</math>

which factorizes to <math>(w^2 + 11z^2)(x^2 + 3y^2) = 16 \cdot 3^{2k}</math>

Also, <math>4 \cdot 3^k z = wx + wy\sqrt{3} + zx\sqrt{11} - zy\sqrt{33} = (w+z\sqrt{11})(x + y\sqrt{3})</math>

Therefore <math>3^k z = u v</math> where <math>u \in \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right], |u|^2 = 3^l</math> and <math>v \in \mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right], |v|^2 = 3^{k-l}</math>

Both <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math> are [https://en.wikipedia.org/wiki/Unique_factorization_domain unique factorization domains]. Prime factorizations of 3 are given by <math>3 = (-i\sqrt{3}) \times i\sqrt{3}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>3 = \frac{1+i\sqrt{11}}{2} \times \frac{1-i\sqrt{11}}{2}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>. Units in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> are <math>\omega^i</math> and <math>\pm 1</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>

By prime factorization this gives all solutions for elements of unit modulus in <math>\overline{R}_2</math> in the form <math>z = \omega^i \eta^j, i,j \in \mathbb{Z}</math> where <math>\eta = \frac{\sqrt{33}+i\sqrt{3}}{6} = \sqrt{\omega_3}</math>

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\overline{R}_4 = \overline{R}_2\left[\frac{1}{2},\frac{1}{7}, \phi \right]</math>. Since <math>\mathbb{Z}[\phi]</math> is another unique factorization domain the same method used to find the elements of unit modulus in <math>\overline{R}_2</math> can be used here multiple times to solve this case too. the details are omitted. The result is that elements <math>z</math> of unit modulus in <math>\overline{R}_4</math> must be of the form

<math>z = \omega^p \eta^q \zeta^r \alpha^s \beta^t \gamma^u, p,q,r,s,t,u \in \mathbb{Z}</math>

where <math>\zeta = \frac{1 + i\sqrt{15}}{4}, \alpha = \frac{\sqrt{5}+i\sqrt{11}}{4}, \beta = \frac{1+4i\sqrt{3}}{7}, \gamma = \frac{17+3i\sqrt{55}}{28}</math>

For the elements of unit modulus in <math>\overline{R}_4</math> use just <math>\omega^p \eta^q \zeta^r \alpha^s</math>

[[Category:Polymath16]]

Excluding bichromatic vertices

2018-05-11T18:42:35Z

Pgibbs: missing bracket

'''Theorem 1''' A 4-coloring of the unit plane cannot contain a bichromatic vertex (a vertex that can be colored in either of two colors while keeping the color of all other vertices fixed).

Equivalently, in a 4-coloring of the plane, the color of each vertex is determined uniquely by the colors of all the other vertices. This also gives an alternate proof of Falconer's result that the measurable chromatic number of the plane is at least five.

'''Proof''' Without loss of generality we may assume that the colors are <math>{\bf Z}/4{\bf Z}</math> and that the origin is bichromatic with colors 0,2.

Let <math>\omega := \exp( i \pi/3)</math> be the sixth root of vector, then we have the identities
:<math> \omega^6 = 1; \quad \omega^3 = -1; \quad \omega^2 = \omega - 1. \quad (1)</math>
Let <math>\eta := \exp( \frac{i}{2} \arccos(\frac{5}{6})),</math> then <math>\eta</math> is a unit vector that obeys the identity
:<math> \eta (\omega + \omega^2) - \overline{\eta} (\omega + \omega^2)+ 1 = 0 \quad (2).</math>
Let
:<math>C_6 := \{ \omega^j: j=0,\dots,5\}</math>
denote the multiplicative cyclic group of order 6. We consider the coloring of the following seven cosets of <math>C_6</math>:
:<math> C_6, (\eta - \overline{\eta}) C_6, (\eta - \overline{\eta}\omega ) C_6, \eta C_6, (1+\eta\omega^2 ) C_6, \overline{\eta} C_6, (1 + \overline{\eta \omega^2 }) C_6.</math>
Clearly <math>C_6</math> is conjugation invariant. Since
:<math> \overline{\eta - \overline{\eta}} = (\eta - \overline{\eta}) \omega^3 </math>
:<math> \overline{\eta - \omega \overline{\eta}} = (\eta - \omega \overline{\eta}) \omega^2 </math>
we also see that <math>(\eta - \overline{\eta}) C_6</math>, <math>(\eta - \overline{\eta}\omega) C_6</math> are conjugation invariant, while <math>\eta C_6</math> is conjugate to <math> \overline{\eta} C_6</math> and <math>(1+\eta \omega^2 ) C_6</math> is conjugate to <math>(1 + \overline{\eta \omega^2 }) C_6</math>.

We observe the following unit edges:
# Within <math>C_6</math>, there is a unit edge from any <math>\omega^j \in C_6</math> to the origin, to <math>\omega^{j+1} \in C_6</math>, and <math>\omega^{j-1} \in C_6</math>.
# Within <math> \eta C_6</math>, there is a unit edge from any <math>\eta \omega^j \in C_6</math> to the origin, to <math>\eta \omega^{j+1} \in \eta C_6</math>, and <math>\eta \omega^{j-1} \in \eta C_6</math>. Similarly with <math>\eta</math> replaced by <math>\overline{\eta}</math>.
# Using (2) (which can be rewritten for instance as <math>(\eta - \overline{\eta}) (\omega^2 - 1) = -\omega</math>), we see that within <math>(\eta - \overline{\eta}) C_6</math>, there is a unit edge from any <math>(\eta - \overline{\eta}) \omega^j</math> to <math>(\eta - \overline{\eta}) \omega^{j+2}</math> and <math>(\eta - \overline{\eta}) \omega^{j-2}</math>. In particular, <math>(\eta - \overline{\eta}) C_6</math> contains a triangle and thus cannot be 2-colored.
# Each <math> (\eta - \overline{\eta}) \omega^j \in (\eta - \overline{\eta}) C_6</math> has a unit edge to <math>(\eta - \overline{\eta} \omega) \omega^j \in (\eta - \overline{\eta} \omega) C_6</math>, to <math>(\eta - \overline{\eta} \omega) \omega^{j-1} \in (\eta - \overline{\eta} \omega) C_6</math>, to <math>\eta \omega^j \in \eta C_6</math>, to <math>-\overline{\eta} \omega^j \in \overline{\eta} C_6</math>, to <math>\eta \omega^j - \overline{\eta} (\omega^j + \omega^{j+1}) = (1 + \eta \omega^2) \omega^{j+2} \in (1+\eta \omega^2) C_6</math>, and to <math>\eta (\omega^j + \omega^{j-1}) - \overline{\eta} \omega^j = (1+\overline{\eta \omega^2}) \omega^{j+1} \in (1 + \overline{\eta \omega^2}) C_6</math>. (Here (2) is used to obtain the latter two identities.)
# Each <math> (\eta - \overline{\eta} \omega) \omega^j \in (\eta - \overline{\eta} \omega) C_6</math> has a unit edge to <math> \eta \omega^j \in \eta C_6</math>, to <math>-\overline{\eta} \omega^{j+1} \in \overline{\eta} C_6</math>, to <math> \eta \omega^j - \overline{\eta} (\omega^j + \omega^{j+1}) = (1 + \eta \omega^2) \omega^{j+2} \in (1+\eta \omega^2) C_6</math>, and to <math> \eta (\omega^{j+1} + \omega^j) - \overline{\eta} \omega^{j+1} = (1+\overline{\eta \omega^2}) \omega^{j+2} \in (1 + \overline{\eta \omega^2}) C_6</math>. (Again, the latter two identities come from (2).)
# Each <math> \omega^j \in C_6</math> has a unit edge to <math> \omega^j + \eta \omega^{j+2} \in (1 + \eta \omega^2) C_6</math> and to <math> \omega^j + \overline{\eta} \omega^{j-2} \in (1 + \overline{\eta \omega^2}) C_6.</math>
# Each <math> \eta \omega^j \in C_6</math> has a unit edge to <math>\omega^{j-2} + \eta \omega^j \in (1 + \eta \omega^2) C_6</math>; similarly, each <math>\overline{\eta} \omega^j \in C_6</math> has a unit edge to <math>\omega^{j+2} + \overline{\eta} \omega^j \in (1 + \overline{\eta \omega^2}) C_6</math>.

We will show that <math>(\eta - \overline{\eta}) C_6</math> will be forced to be 2-colorable, contradicting item 3 above.

From item 1 and the fact that the origin is colored 0,2 we see that <math>C_6</math> is colored 1,3, and in fact the coloring map <math>c</math> is given by <math>c(\omega^j) = c_1 + 2j</math> for some <math>c_1 = 1,3</math>. Similarly from item 2 we have <math>c(\eta \omega^j) = c_\eta + 2j</math>, <math>c(\overline{\eta} \omega^j) = c_{\overline{\eta}} + 2j</math> for some <math>c_\eta,c_{\overline{\eta}} = 1,3</math>.

Suppose first that <math>c_\eta = c_{\overline{\eta}}</math>. By item 4, each <math>(\eta - \overline{\eta})\omega^j</math> is connected both to <math> \eta \omega^j</math> that has color <math>c_\eta + 2j</math>, and to <math>-\overline{\eta} \omega^j</math> that has color <math>c_{\overline{\eta}} + 2j + 2</math>. Thus <math>(\eta - \overline{\eta})\omega^j</math> has to be colored 0 or 2, giving the 2-coloring of <math>(\eta - \overline{\eta}) C_6</math>.

Now suppose instead that <math>c_\eta \neq c_{\overline{\eta}}</math>, then <math>c_1</math> is distinct from either <math>c_\eta</math> or <math>c_{\overline{\eta}}</math>. Without loss of generality (conjugation symmetry) assume <math>c_\eta \neq c_1</math>. By item 5, each <math>(\eta - \overline{\eta} \omega) \omega^j</math> is connected both to <math>\eta \omega^j</math>, which has color <math>c_\eta + 2j</math>, and to <math>-\overline{\eta}\omega^{j+1}</math>, which has color <math>c_{\overline{\eta}}+2j</math>. Thus all of <math>(\eta - \overline{\eta} \omega) C_6</math> is 0,2-colored. Similarly, from items 6,7, each <math>\omega^j + \eta \omega^{j+2}</math> is connected to <math>\omega^j</math>, which has color <math>c_1 + 2j</math>, and <math>\eta \omega^{j+2}</math>, which has color <math>c_\eta + 2j</math>, and hence <math>(1 + \eta \omega^2) C_6</math> is also 0,2-colored. Now observe from items 4,5 that each <math> (\eta - \overline{\eta}) \omega^j</math> is the vertex of an equilateral triangle with other two vertices <math>\eta \omega^{j-1} - \overline{\eta} \omega^j</math>, <math> \eta (\omega^j + \omega^{j-1}) - \overline{\eta} \omega^j</math> which are 0,2-colored and hence <math>(\eta - \overline{\eta}) C_6</math> is 1,3-colored, again giving the required contradiction. <math>\Box</math>

[[Category:Polymath16]]

Algebraic formulation of Hadwiger-Nelson problem

2018-05-11T17:06:07Z

Pgibbs: /* Coloring Rings */

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (L+N).
This table summarises what is known. Where there is more than one option shown the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| L+N
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| L+N
| L+N
| N
| L+N
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L or L+N
| N
| N
|-
| <math>\mathbb{Z}[\omega, \omega_3, \omega_4, \omega_{25/9}]</math>
| X
| X
| X
| X or N
| X or N
| L+N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X or N
| X or N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finite number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\overline{R}_1 = \mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\overline{R}_H = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_H = \overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring. Therefore <math>\overline{R}_H</math> has a linear 5-coloring.

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Elements of <math>\overline{R}_2</math> take the form <math>z = \frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k}</math> where <math>a,b,c,d</math> are integers subject to some congruence relations (see above). The elements of unit modulus are determined by <math>|z|^2 = 1</math> which is equivalent to

<math>a^2 + 3b^2 + 11c^2 + 33d^2 + (ad + bc)\sqrt{33} = 16 \cdot 3^{2k}</math>

Since <math>\sqrt{33}</math> is irrational this splits into two integer equations. The solutions of <math>ad + bc = 0</math> can be parametrised over integers by <math>a = wx, b = wy, c = zx, d = -zy</math>. Substituting into the rest of the equation gives

<math>(wx)^2 + 3(wy)^2 + 11(zx)^2 + 33(-zy)^2 = 16 \cdot 3^{2k}</math>

which factorizes to <math>(w^2 + 11z^2)(x^2 + 3y^2) = 16 \cdot 3^{2k}</math>

Also, <math>4 \cdot 3^k z = wx + wy\sqrt{3} + zx\sqrt{11} - zy\sqrt{33} = (w+z\sqrt{11})(x + y\sqrt{3})</math>

Therefore <math>3^k z = u v</math> where <math>u \in \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right], |u|^2 = 3^l</math> and <math>v \in \mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right], |v|^2 = 3^{k-l}</math>

Both <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math> are [https://en.wikipedia.org/wiki/Unique_factorization_domain unique factorization domains]. Prime factorizations of 3 are given by <math>3 = (-i\sqrt{3}) \times i\sqrt{3}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>3 = \frac{1+i\sqrt{11}}{2} \times \frac{1-i\sqrt{11}}{2}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>. Units in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> are <math>\omega^i</math> and <math>\pm 1</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>

By prime factorization this gives all solutions for elements of unit modulus in <math>\overline{R}_2</math> in the form <math>z = \omega^i \eta^j, i,j \in \mathbb{Z}</math> where <math>\eta = \frac{\sqrt{33}+i\sqrt{3}}{6} = \sqrt{\omega_3}</math>

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\overline{R}_4 = \overline{R}_2\left[\frac{1}{2},\frac{1}{7}, \phi \right]</math>. Since <math>\mathbb{Z}[\phi]</math> is another unique factorization domain the same method used to find the elements of unit modulus in <math>\overline{R}_2</math> can be used here multiple times to solve this case too. the details are omitted. The result is that elements <math>z</math> of unit modulus in <math>\overline{R}_4</math> must be of the form

<math>z = \omega^p \eta^q \zeta^r \alpha^s \beta^t \gamma^u, p,q,r,s,t,u \in \mathbb{Z}</math>

where <math>\zeta = \frac{1 + i\sqrt{15}}{4}, \alpha = \frac{\sqrt{5}+i\sqrt{11}}{4}, \beta = \frac{1+4i\sqrt{3}}{7}, \gamma = \frac{17+3i\sqrt{55}}{28}</math>

For the elements of unit modulus in <math>\overline{R}_4</math> use just <math>\omega^p \eta^q \zeta^r \alpha^s</math>

[[Category:Polymath16]]

Algebraic formulation of Hadwiger-Nelson problem

2018-05-11T17:01:40Z

Pgibbs: /* Coloring Rings */

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (L+N).
This table summarises what is known. Where there is more than one option shown the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| L+N
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| L+N
| L+N
| N
| L+N
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L or L+N
| N
| N
|-
| <math>\mathbb{Z}[\omega, \omega_3, \omega_4, \omega_{25/9}]</math>
| X
| X
| X
| X or N
| X or N
| L+N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X or N
| X or N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\overline{R}_1 = \mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\overline{R}_H = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_H = \overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring. Therefore <math>\overline{R}_H</math> has a linear 5-coloring.

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Elements of <math>\overline{R}_2</math> take the form <math>z = \frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k}</math> where <math>a,b,c,d</math> are integers subject to some congruence relations (see above). The elements of unit modulus are determined by <math>|z|^2 = 1</math> which is equivalent to

<math>a^2 + 3b^2 + 11c^2 + 33d^2 + (ad + bc)\sqrt{33} = 16 \cdot 3^{2k}</math>

Since <math>\sqrt{33}</math> is irrational this splits into two integer equations. The solutions of <math>ad + bc = 0</math> can be parametrised over integers by <math>a = wx, b = wy, c = zx, d = -zy</math>. Substituting into the rest of the equation gives

<math>(wx)^2 + 3(wy)^2 + 11(zx)^2 + 33(-zy)^2 = 16 \cdot 3^{2k}</math>

which factorizes to <math>(w^2 + 11z^2)(x^2 + 3y^2) = 16 \cdot 3^{2k}</math>

Also, <math>4 \cdot 3^k z = wx + wy\sqrt{3} + zx\sqrt{11} - zy\sqrt{33} = (w+z\sqrt{11})(x + y\sqrt{3})</math>

Therefore <math>3^k z = u v</math> where <math>u \in \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right], |u|^2 = 3^l</math> and <math>v \in \mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right], |v|^2 = 3^{k-l}</math>

Both <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math> are [https://en.wikipedia.org/wiki/Unique_factorization_domain unique factorization domains]. Prime factorizations of 3 are given by <math>3 = (-i\sqrt{3}) \times i\sqrt{3}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>3 = \frac{1+i\sqrt{11}}{2} \times \frac{1-i\sqrt{11}}{2}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>. Units in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> are <math>\omega^i</math> and <math>\pm 1</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>

By prime factorization this gives all solutions for elements of unit modulus in <math>\overline{R}_2</math> in the form <math>z = \omega^i \eta^j, i,j \in \mathbb{Z}</math> where <math>\eta = \frac{\sqrt{33}+i\sqrt{3}}{6} = \sqrt{\omega_3}</math>

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\overline{R}_4 = \overline{R}_2\left[\frac{1}{2},\frac{1}{7}, \phi \right]</math>. Since <math>\mathbb{Z}[\phi]</math> is another unique factorization domain the same method used to find the elements of unit modulus in <math>\overline{R}_2</math> can be used here multiple times to solve this case too. the details are omitted. The result is that elements <math>z</math> of unit modulus in <math>\overline{R}_4</math> must be of the form

<math>z = \omega^p \eta^q \zeta^r \alpha^s \beta^t \gamma^u, p,q,r,s,t,u \in \mathbb{Z}</math>

where <math>\zeta = \frac{1 + i\sqrt{15}}{4}, \alpha = \frac{\sqrt{5}+i\sqrt{11}}{4}, \beta = \frac{1+4i\sqrt{3}}{7}, \gamma = \frac{17+3i\sqrt{55}}{28}</math>

For the elements of unit modulus in <math>\overline{R}_4</math> use just <math>\omega^p \eta^q \zeta^r \alpha^s</math>

[[Category:Polymath16]]

Algebraic formulation of Hadwiger-Nelson problem

2018-05-10T15:11:08Z

Pgibbs: /* Construction of ring colorings */

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (L+N).
This table summarises what is known. Where there is more than one option shown the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| L+N
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| L+N
| L+N
| N
| L+N
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L or L+N
| N
| N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X or N
| X or N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\overline{R}_1 = \mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\overline{R}_H = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_H = \overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring. Therefore <math>\overline{R}_H</math> has a linear 5-coloring.

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Elements of <math>\overline{R}_2</math> take the form <math>z = \frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k}</math> where <math>a,b,c,d</math> are integers subject to some congruence relations (see above). The elements of unit modulus are determined by <math>|z|^2 = 1</math> which is equivalent to

<math>a^2 + 3b^2 + 11c^2 + 33d^2 + (ad + bc)\sqrt{33} = 16 \cdot 3^{2k}</math>

Since <math>\sqrt{33}</math> is irrational this splits into two integer equations. The solutions of <math>ad + bc = 0</math> can be parametrised over integers by <math>a = wx, b = wy, c = zx, d = -zy</math>. Substituting into the rest of the equation gives

<math>(wx)^2 + 3(wy)^2 + 11(zx)^2 + 33(-zy)^2 = 16 \cdot 3^{2k}</math>

which factorizes to <math>(w^2 + 11z^2)(x^2 + 3y^2) = 16 \cdot 3^{2k}</math>

Also, <math>4 \cdot 3^k z = wx + wy\sqrt{3} + zx\sqrt{11} - zy\sqrt{33} = (w+z\sqrt{11})(x + y\sqrt{3})</math>

Therefore <math>3^k z = u v</math> where <math>u \in \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right], |u|^2 = 3^l</math> and <math>v \in \mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right], |v|^2 = 3^{k-l}</math>

Both <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math> are [https://en.wikipedia.org/wiki/Unique_factorization_domain unique factorization domains]. Prime factorizations of 3 are given by <math>3 = (-i\sqrt{3}) \times i\sqrt{3}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>3 = \frac{1+i\sqrt{11}}{2} \times \frac{1-i\sqrt{11}}{2}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>. Units in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> are <math>\omega^i</math> and <math>\pm 1</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>

By prime factorization this gives all solutions for elements of unit modulus in <math>\overline{R}_2</math> in the form <math>z = \omega^i \eta^j, i,j \in \mathbb{Z}</math> where <math>\eta = \frac{\sqrt{33}+i\sqrt{3}}{6} = \sqrt{\omega_3}</math>

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\overline{R}_4 = \overline{R}_2\left[\frac{1}{2},\frac{1}{7}, \phi \right]</math>. Since <math>\mathbb{Z}[\phi]</math> is another unique factorization domain the same method used to find the elements of unit modulus in <math>\overline{R}_2</math> can be used here multiple times to solve this case too. the details are omitted. The result is that elements <math>z</math> of unit modulus in <math>\overline{R}_4</math> must be of the form

<math>z = \omega^p \eta^q \zeta^r \alpha^s \beta^t \gamma^u, p,q,r,s,t,u \in \mathbb{Z}</math>

where <math>\zeta = \frac{1 + i\sqrt{15}}{4}, \alpha = \frac{\sqrt{5}+i\sqrt{11}}{4}, \beta = \frac{1+4i\sqrt{3}}{7}, \gamma = \frac{17+3i\sqrt{55}}{28}</math>

For the elements of unit modulus in <math>\overline{R}_4</math> use just <math>\omega^p \eta^q \zeta^r \alpha^s</math>

Algebraic formulation of Hadwiger-Nelson problem

2018-05-10T15:10:04Z

Pgibbs: /* example \overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] */

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (L+N).
This table summarises what is known. Where there is more than one option shown the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| L+N
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| L+N
| L+N
| N
| L+N
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L or L+N
| N
| N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X or N
| X or N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

=== Construction of ring colorings ===

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\overline{R}_1 = \mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\overline{R}_H = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_H = \overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring. Therefore <math>\overline{R}_H</math> has a linear 5-coloring.

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Elements of <math>\overline{R}_2</math> take the form <math>z = \frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k}</math> where <math>a,b,c,d</math> are integers subject to some congruence relations (see above). The elements of unit modulus are determined by <math>|z|^2 = 1</math> which is equivalent to

<math>a^2 + 3b^2 + 11c^2 + 33d^2 + (ad + bc)\sqrt{33} = 16 \cdot 3^{2k}</math>

Since <math>\sqrt{33}</math> is irrational this splits into two integer equations. The solutions of <math>ad + bc = 0</math> can be parametrised over integers by <math>a = wx, b = wy, c = zx, d = -zy</math>. Substituting into the rest of the equation gives

<math>(wx)^2 + 3(wy)^2 + 11(zx)^2 + 33(-zy)^2 = 16 \cdot 3^{2k}</math>

which factorizes to <math>(w^2 + 11z^2)(x^2 + 3y^2) = 16 \cdot 3^{2k}</math>

Also, <math>4 \cdot 3^k z = wx + wy\sqrt{3} + zx\sqrt{11} - zy\sqrt{33} = (w+z\sqrt{11})(x + y\sqrt{3})</math>

Therefore <math>3^k z = u v</math> where <math>u \in \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right], |u|^2 = 3^l</math> and <math>v \in \mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right], |v|^2 = 3^{k-l}</math>

Both <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math> are [https://en.wikipedia.org/wiki/Unique_factorization_domain unique factorization domains]. Prime factorizations of 3 are given by <math>3 = (-i\sqrt{3}) \times i\sqrt{3}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>3 = \frac{1+i\sqrt{11}}{2} \times \frac{1-i\sqrt{11}}{2}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>. Units in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> are <math>\omega^i</math> and <math>\pm 1</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>

By prime factorization this gives all solutions for elements of unit modulus in <math>\overline{R}_2</math> in the form <math>z = \omega^i \eta^j, i,j \in \mathbb{Z}</math> where <math>\eta = \frac{\sqrt{33}+i\sqrt{3}}{6} = \sqrt{\omega_3}</math>

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\overline{R}_4 = \overline{R}_2\left[\frac{1}{2},\frac{1}{7}, \phi \right]</math>. Since <math>\mathbb{Z}[\phi]</math> is another unique factorization domain the same method used to find the elements of unit modulus in <math>\overline{R}_2</math> can be used here multiple times to solve this case too. the details are omitted. The result is that elements <math>z</math> of unit modulus in <math>\overline{R}_4</math> must be of the form

<math>z = \omega^p \eta^q \zeta^r \alpha^s \beta^t \gamma^u, p,q,r,s,t,u \in \mathbb{Z}</math>

where <math>\zeta = \frac{1 + i\sqrt{15}}{4}, \alpha = \frac{\sqrt{5}+i\sqrt{11}}{4}, \beta = \frac{1+4i\sqrt{3}}{7}, \gamma = \frac{17+3i\sqrt{55}}{28}</math>

For the elements of unit modulus in <math>\overline{R}_4</math> use just <math>\omega^p \eta^q \zeta^r \alpha^s</math>

Algebraic formulation of Hadwiger-Nelson problem

2018-05-10T14:17:17Z

Pgibbs: /* example \overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] */

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (L+N).
This table summarises what is known. Where there is more than one option shown the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| L+N
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| L+N
| L+N
| N
| L+N
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L or L+N
| N
| N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X or N
| X or N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

=== Construction of ring colorings ===

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\overline{R}_1 = \mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\overline{R}_H = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_H = \overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring. Therefore <math>\overline{R}_H</math> has a linear 5-coloring.

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Elements of <math>\overline{R}_2</math> take the form <math>z = \frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k}</math> where <math>a,b,c,d</math> are integers subject to some congruence relations (see above). The elements of unit modulus are determined by <math>|z|^2 = 1</math> which is equivalent to

<math>a^2 + 3b^2 + 11c^2 + 33d^2 + (ad + bc)\sqrt{33} = 16 \cdot 3^{2k}</math>

Since <math>\sqrt{33}</math> is irrational this splits into two integer equations. The solutions of <math>ad + bc = 0</math> can be parametrised over integers by <math>a = wx, b = wy, c = zx, d = -zy</math>. Substituting into the rest of the equation gives

<math>(wx)^2 + 3(wy)^2 + 11(zx)^2 + 33(-zy)^2 = 16 \cdot 3^{2k}</math>

which factorizes to <math>(w^2 + 11z^2)(x^2 + 3y^2) = 16 \cdot 3^{2k}</math>

Also, <math>4 \cdot 3^k z = wx + wy\sqrt{3} + zx\sqrt{11} - zy\sqrt{33} = (w+z\sqrt{11})(x + y\sqrt{3})</math>

Therefore <math>3^k z = u v</math> where <math>u \in \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right], |u|^2 = 3^l</math> and <math>v \in \mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right], |v|^2 = 3^{k-l}</math>

Both <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math> are [https://en.wikipedia.org/wiki/Unique_factorization_domain unique factorization domains]. Prime factorizations of 3 are given by <math>3 = (-i\sqrt{3}) \times i\sqrt{3}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> and <math>3 = \frac{1+i\sqrt{11}}{2} \times \frac{1-i\sqrt{11}}{2}</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>. Units in <math>\mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}\right]</math> are <math>\omega^i</math> and <math>\pm 1</math> in <math>\mathbb{Z}\left[\frac{1+i\sqrt{11}}{2}\right]</math>

By prime factorization this gives all solutions for elements of unit modulus in <math>\overline{R}_2</math> in the form <math>z = \omega^i \eta^j, i,j \in \mathbb{Z}</math> where <math>\eta = \frac{\sqrt{33}+i\sqrt{3}}{6} = \sqrt{\omega_3}</math>

Algebraic formulation of Hadwiger-Nelson problem

2018-05-10T13:01:58Z

Pgibbs: /* example \overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] */

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (L+N).
This table summarises what is known. Where there is more than one option shown the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| L+N
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| L+N
| L+N
| N
| L+N
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L or L+N
| N
| N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X or N
| X or N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

=== Construction of ring colorings ===

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\overline{R}_1 = \mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\overline{R}_H = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_H = \overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring. Therefore <math>\overline{R}_H</math> has a linear 5-coloring.

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Elements of <math>\overline{R}_2</math> take the form <math>z = \frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k}</math> where <math>a,b,c,d</math> are integers subject to some congruence relations (see above). The elements of unit modulus are determined by <math>|z|^2 = 1</math> which is equivalent to

<math>a^2 + 3b^2 + 11c^2 + 33d^2 + (ad + bc)\sqrt{33} = 16 \cdot 3^{2k}</math>

Since <math>\sqrt{33}</math> is irrational this splits into two integer equations. The solutions of <math>ad + bc = 0</math> can be parametrised over integers by <math>a = wx, b = wy, c = zx, d = -zy</math>. Substituting into the rest of the equation gives

<math>(wx)^2 + 3(wy)^2 + 11(zx)^2 + 33(-zy)^2 = 16 \cdot 3^{2k}</math>

which factorizes to <math>(w^2 + 11z^2)(x^2 + 3y^2) = 16 \cdot 3^{2k}</math>

Also, <math>4 \cdot 3^k z = wx + wy\sqrt{3} + zx\sqrt{11} - zy\sqrt{33} = (w+z\sqrt{11})(x + y\sqrt{3})</math>

Algebraic formulation of Hadwiger-Nelson problem

2018-05-10T13:01:16Z

Pgibbs: /* example \overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] */

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (L+N).
This table summarises what is known. Where there is more than one option shown the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| L+N
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| L+N
| L+N
| N
| L+N
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L or L+N
| N
| N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X or N
| X or N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

=== Construction of ring colorings ===

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\overline{R}_1 = \mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\overline{R}_H = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_H = \overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring. Therefore <math>\overline{R}_H</math> has a linear 5-coloring.

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Elements of <math>\overline{R}_2</math> take the form <math>z = \frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{4 \cdot 3^k}</math> where <math>a,b,c,d</math> are integers subject to some congruence relations (see above). The elements of unit modulus are determined by <math>|z|^2 = 1</math> which is equivalent to

<math>a^2 + 3b^2 + 11c^2 + 33d^2 + (ad + bc)\sqrt{33} = 16 \cdot 3^{2k}</math>

Since <math>\sqrt{33}</math> is irrational this splits into two integer equations. The solutions of <math>ad + bc = 0</math> can be parametrised over integers by <math>a = wx, b = wy, c = zx, d = -zy</math>. Substituting into the rest of the equation gives

<math>(wx)^2 + 3(wy)^2 + 11(zx)^2 + 33(-zy)^2 = 16 \cdot 3^{2k}</math>

which factorizes to <math>(w^2 + 11z^2)(x^2 + 3y^2) = 16 \cdot 3^{2k}</math>

Also, <math>4 \cdot 3^k z = wx + wy\sqrt{3} + zx\sqrt{11} - zy\sqrt{33} = (w+z\sqrt{11})(x + y\sqrt{3})</math>

Algebraic formulation of Hadwiger-Nelson problem

2018-05-10T12:35:19Z

Pgibbs:

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (L+N).
This table summarises what is known. Where there is more than one option shown the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| L+N
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| L+N
| L+N
| N
| L+N
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L or L+N
| N
| N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X or N
| X or N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

=== Construction of ring colorings ===

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\overline{R}_1 = \mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\overline{R}_H = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_H = \overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring. Therefore <math>\overline{R}_H</math> has a linear 5-coloring.

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Algebraic formulation of Hadwiger-Nelson problem

2018-05-10T12:33:27Z

Pgibbs: /* example \overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] */

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (L+N).
This table summarises what is known. Where there is more than one option shown the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| L+N
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| L+N
| L+N
| N
| L+N
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L or L+N
| N
| N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X or N
| X or N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

=== Construction of ring colorings ===

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\overline{R}_H = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_H = \overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring. Therefore <math>\overline{R}_H</math> has a linear 5-coloring.

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Algebraic formulation of Hadwiger-Nelson problem

2018-05-10T12:32:52Z

Pgibbs: /* example \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] */

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (L+N).
This table summarises what is known. Where there is more than one option shown the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| L+N
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| L+N
| L+N
| N
| L+N
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L or L+N
| N
| N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X or N
| X or N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

=== Construction of ring colorings ===

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' \overline{R}_4 = <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\overline{R}_H = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_H = \overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring. Therefore <math>\overline{R}_H</math> has a linear 5-coloring.

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Algebraic formulation of Hadwiger-Nelson problem

2018-05-10T12:28:03Z

Pgibbs: /* Coloring Rings */

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (L+N).
This table summarises what is known. Where there is more than one option shown the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| L+N
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| L+N
| L+N
| L+N
| L+N
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| L+N
| L+N
| N
| L+N
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L or L+N
| N
| L+N
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L or L+N
| N
| N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X or N
| X or N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

=== Construction of ring colorings ===

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\overline{R}_H = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_H = \overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring. Therefore <math>\overline{R}_H</math> has a linear 5-coloring.

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Algebraic formulation of Hadwiger-Nelson problem

2018-05-10T12:21:20Z

Pgibbs: /* example \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}] */

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (B).
This table summarises what is known. Where there is more than one letter the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| B
| B
| B
| B
| B
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| B
| B
| B
| B
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| B
| B
| N
| B
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L/B
| N
| B
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L/B
| N
| B
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L/B
| N
| N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X/N
| X/N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

=== Construction of ring colorings ===

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\overline{R}_H = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_H = \overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring. Therefore <math>\overline{R}_H</math> has a linear 5-coloring.

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Algebraic formulation of Hadwiger-Nelson problem

2018-05-10T12:10:44Z

Pgibbs:

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\overline{R}_2 = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\overline{R}_2 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\overline{R}_3 = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (B).
This table summarises what is known. Where there is more than one letter the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| B
| B
| B
| B
| B
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| B
| B
| B
| B
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| B
| B
| N
| B
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L/B
| N
| B
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L/B
| N
| B
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L/B
| N
| N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X/N
| X/N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

=== Construction of ring colorings ===

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\overline{R}_2</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\overline{R}_2</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\overline{R}_2</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \overline{R}_2\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring

== Unit vectors ==

A useful application of the algebraic method is to find all edges in a unit distance graph. These are provided by the unit vectors in the additive group of the graph. When the plane is treated as the the Argand diagram these unit vectors are given by complex numbers of unit modulus <math>|z|^2 = 1</math>. When the graph is derived from a subring of the complex numbers, the elements of unit modulus form a multiplicative group which is finitely generated if it is derived from a finite graph. The generators of this group can sometimes be determined by factoring <math>|z|^2 = 1</math> for a given linear presentation of the ring.

==== ''example'' <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

Algebraic formulation of Hadwiger-Nelson problem

2018-05-10T11:38:30Z

Pgibbs: /* Coloring Rings */

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\mathcal{M} = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\mathcal{M} = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\mathcal{M} = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\mathcal{A} = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\mathcal{A} = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (B).
This table summarises what is known. Where there is more than one letter the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| B
| B
| B
| B
| B
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| B
| B
| B
| B
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| B
| B
| N
| B
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L/B
| N
| B
|-
| <math>R_H = \mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L/B
| N
| B
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L/B
| N
| N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X/N
| X/N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

=== Construction of ring colorings ===

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\mathcal{M} = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\mathcal{M}</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\mathcal{M}</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\mathcal{M}</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathcal{M}\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T20:11:48Z

Pgibbs: /* example \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] */

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\mathcal{M} = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\mathcal{M} = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\mathcal{M} = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\mathcal{A} = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\mathcal{A} = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (B).
This table summarises what is known. Where there is more than one letter the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| B
| B
| B
| B
| B
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| B
| B
| B
| B
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| B
| B
| N
| B
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L/B
| N
| B
|-
| <math>\mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L/B
| N
| B
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L/B
| N
| N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X/N
| X/N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

=== Construction of ring colorings ===

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\mathcal{M} = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\mathcal{M}</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\mathcal{M}</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\mathcal{M}</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathcal{M}\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_{64/9}, \overline{\omega_{64/9}}]</math> ====

In the ring <math>\overline{R}_2 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> it is known that the element <math>\frac{8}{3}</math> has the same color as zero. Therefore the ring <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}]</math> formed by spindling this element has no 4-colorings.

<math>\omega_{64/9} = \frac{119+3i\sqrt{247}}{128}</math> so <math>\overline{R}_2[\omega_{64/9}, \overline{\omega_{64/9}}] = \overline{R}_2\left[\sqrt{741}, \frac{1}{2}\right]</math>

741 is a square modulo 5 so this can again by mapped by a ring homomorphism to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{11}]</math> which has a linear 5-coloring

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T19:41:31Z

Pgibbs: /* example \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] */

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\mathcal{M} = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\mathcal{M} = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\mathcal{M} = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\mathcal{A} = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\mathcal{A} = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (B).
This table summarises what is known. Where there is more than one letter the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| B
| B
| B
| B
| B
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| B
| B
| B
| B
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| B
| B
| N
| B
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L/B
| N
| B
|-
| <math>\mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L/B
| N
| B
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L/B
| N
| N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X/N
| X/N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

=== Construction of ring colorings ===

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\mathcal{M} = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\mathcal{M}</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\mathcal{M}</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\mathcal{M}</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathcal{M}\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{7}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

To find the image of the elements of unit modulus in <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{7}]</math> find <math>|x|^2 \equiv a^2 + 3b^2 + c^2 + 3d^2 + (ad + bc)\sqrt{33} \equiv 1 \bmod 5</math> which separates into <math>a^2 + 3b^2 + c^2 + 3d^2 \equiv 1, ad + bc \equiv 0 \bmod 5</math>. Applying this to the 625 elements of <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{7}]</math> finds 24 elements of unit modulus. It can then be verified that the colouring <math>c(x) = a + b + c +d</math> does not give zero for any of these elements and is therefore a linear 5-colouring for <math>\overline{R}_3</math> and <math>\overline{R}_4</math>

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T18:42:36Z

Pgibbs: /* example \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] */

''This is a part of Polymath16 - for the main page, see [[Hadwiger-Nelson problem]].''

Given a [https://en.wikipedia.org/wiki/Unit_distance_graph unit distance graph] <math>G</math>, the edges of the graph are formed by a set of vectors <math>V</math> of unit length. Often a single vector is used many times in the same graph for different edges. The vectors generate an infinite abelian group <math>\mathcal{G}</math> under addition. The [https://en.wikipedia.org/wiki/Cayley_graph Cayley graph] <math>\Gamma(\mathcal{G},S)</math> is a unit distance graph that will contain <math>G</math> as a subgraph. A coloring of the Cayley graph reduces to a coloring of <math>G</math>, and is an efficient way to set an upper bound on its [https://en.wikipedia.org/wiki/Graph_coloring chromatic number].

In two dimensions it is useful to identify the vectors with complex numbers. For example the [https://en.wikipedia.org/wiki/Eisenstein_integer Eisenstein integers] <math>\mathbb{Z}[\omega]</math> are generated by an equilateral triangle or hexagon with unit length sides. These are complex numbers of the form <math>z = a + b\omega, a,b \in \mathbb{Z}</math> where <math>\omega = \frac{1 + i\sqrt{3}}{2}</math> is a cube root of <math>-1</math>. A 3-colouring is given by a mapping to the group <math>\mathbb{Z}_3</math> defined by <math>c(z) = x \in \mathbb{Z}, x \equiv a - b \bmod{3}</math>. To verify that this is valid it is sufficient to check that no element of <math>\mathbb{Z}[\omega]</math> which is of unit modulus is sent to the identity (zero) in <math>\mathbb{Z}_3</math>.

Note that in general a group generated in this way will not be a simple [https://en.wikipedia.org/wiki/Lattice_Group lattice] and in most cases it will be formed from vertices that are dense in the plane.

A <math>k</math>-coloring is '''linear''' if it is given by a group homomorphism onto a finite group of order <math>k</math>. I.e. <math>c(x+y) = c(x)+c(y)</math> for all <math>x,y</math> in the ring.

A <math>k</math>-coloring is '''periodic''' with period <math>p \in \mathbb{Z}</math> if <math>c(x+py) = c(x)</math> for all <math>x,y</math> in the ring.

A linear <math>k</math>-coloring is always periodic with period <math>k</math>.

== Rings ==

The Cayley graph of a group enables us to consider the colorability of graphs formed from all possible repeated '''Minkowski sums''' of a graph with itself. Another operation used to create graphs is '''spindling''' where a graph <math>G</math> is combined with a copy of itself rotated about one of its vertices through an angle <math>\theta</math>. In the complex plane the center of rotation <math>O</math> can be placed at zero and another point <math>A</math> connected by an edge <math>OA</math> can be placed at one in the complex plane. The rotation takes <math>A</math> to the point <math>B</math> at <math>\tau = e^{i\theta}</math>. All other points given by complex numbers <math>z \in G</math> are rotated to <math>\tau z</math>. The combination of all possible spindlings and Minkowski sums is therefore contained in the Cayley graph of a ring <math>\mathcal{R} = \mathbb{Z}[\tau_1,...,\tau_n] \subset \mathbb{C}</math> generated from a set of complex numbers of unit modulus. More specifically <math>\mathcal{R}</math> ia an [https://en.wikipedia.org/wiki/Integral_domain integral domain]

The generators <math>\tau_i</math> are chosen so as to add new edges in the graph in addition to themselves. This will makes them [https://en.wikipedia.org/wiki/Algebraic_number algebraic numbers]. If they are [https://en.wikipedia.org/wiki/Algebraic_integer Algebraic integers] then the additive group <math>(\mathcal{R}, +)</math> will be finitely generated. In most cases they are not algebraic integers, but any algebraic number is equal to an algebraic integer divided by a (rational) integer. The ring <math>\mathcal{R}</math> is therefore a [https://en.wikipedia.org/wiki/Ring_of_integers ring of algebraic integers] <math>\mathcal{O}_{\mathcal{R}}</math> extended with some finite set of [https://en.wikipedia.org/wiki/Unit_fraction unit fractions] <math>\mathcal{R} = \mathcal{O}_{\mathcal{R}}\left[\frac{1}{n_1},...,\frac{1}{n_m}\right]</math>. In the simplest cases <math>\mathcal{O}_{\mathcal{R}}</math> is a ring of [https://en.wikipedia.org/wiki/Quadratic_integer quadratic integers] making its elements easy to describe in terms of sums of square roots.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The squared distances between points on the triangular lattice given by the Eisenstein integers in the complex plane are integers from the sequence of [https://oeis.org/A003136 Loeschian numbers]. Spindling this lattice through angles <math>\omega_t = \exp(i\arccos(1 - \tfrac{1}{2t}))</math> will form new edges of unit length at the base of isosceles triangles with two sides equal to these distances. In particular <math>\omega_3</math> is used to generate a ring <math>\mathcal{M} = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> which contains [https://en.wikipedia.org/wiki/Moser_spindle Moser spindles]. <math>\omega_1 = \omega = \frac{1+i\sqrt{3}}{2}</math> and <math>\omega_3 = \frac{5 + i\sqrt{11}}{6}</math>

This ring can also be given as <math>\mathcal{M} = \mathbb{Z}\left[{\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}}\right]</math> where the first two generators are algebraic integers.

In explicit form it is <math>\mathcal{M} = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}}{3^k} : a,b,c,d,k \in \mathbb{Z}, a \equiv b \equiv c \equiv d \bmod 2, a+b+c-d \equiv 0 \bmod 4, k \ge 0\right\}</math>

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}]</math> ====

This ring includes the Moser spindle and an additional rotation to form a 2,2,1 isosceles triangle.

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>

<math>\mathcal{A} = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1+i\sqrt{15}}{2}, \frac{1}{2}, \frac{1}{3}\right]</math>

In explicit form it is <math>\mathcal{A} = \left\{\frac{a+ib\sqrt{3}+ic\sqrt{11}+d\sqrt{33}+e\sqrt{5}+if\sqrt{15}+ig\sqrt{55}+h\sqrt{165}}{2^l \cdot 3^k}, a,b,c,d,e,f,g,h,k,l \in \mathbb{Z}, k,l \ge 0\right\}</math>

== Coloring Rings ==

The main goal is to establish the chromatic number of rings that contain graphs known to be of interest to the Hadwiger-Nelson problem. For each ring and each number of colors <math>k</math> there are four main possibilities:
:The ring cannot be <math>k</math>-colored (X),
:the ring can be coloured but only with linear colourings (L),
:the ring can be <math>k</math>-colored with non-linear colorings only (N), or
:the ring can be <math>k</math>-colored with both linear and non-linear colorings (B).
This table summarises what is known. Where there is more than one letter the correct answer is unknown.

{| border=1
|-
! Ring !! 2-color !! 3-color !! 4-color !! 5-color !! 6-color !! 7-color
|-
| <math>\mathbb{Z}</math>
| L
| B
| B
| B
| B
| B
|-
| <math>R_1 = \mathbb{Z}[\omega]</math>
| X
| L
| B
| B
| B
| B
|-
| <math>R_2 = \mathbb{Z}[\omega, \omega_3]</math>
| X
| X
| B
| B
| N
| B
|-
| <math>R_3 = \mathbb{Z}[\omega, \omega_3, \omega_4]</math>
| X
| X
| X
| L/B
| N
| B
|-
| <math>\mathbb{Z}[\omega, \omega_3, \omega_{64/9}]</math>
| X
| X
| X
| L/B
| N
| B
|-
| <math>R_4 = \mathbb{Z}[\omega, \omega_3, \omega_4, \omega_7]</math>
| X
| X
| X
| L/B
| N
| N
|-
| <math>\mathbb{C}</math>
| X
| X
| X
| X/N
| X/N
| N
|-
|}

If a ring has a linear <math>k</math>-coloring then it has a non-linear <math>(k+1)</math>-coloring

Every ring <math>\mathcal{R} \subset \mathbb{C}</math> has a non-linear 7-coloring. This follows from the Isbell 7-colouring of the plane.

If a ring <math>\mathcal{R}</math> contains a unit fraction <math>\frac{1}{n}</math> and <math>k \mid n</math> then <math>\mathcal{R}</math> does not have a linear <math>k</math>-colouring. This is because for any <math>z \in \mathcal{R}</math> it follows from linearity that <math>z = k\left(\frac{z}{k}\right)</math> must be colored with zero.

Therefore <math>\mathbb{C}</math> has no linear colorings for any finte number of colors.

If a ring has only linear <math>k</math>-colorings then every element that is <math>k</math> times another element in the ring will be colored the same as the origin. In This case a new ring can be constructed by spindling that has no <math>k</math>-coloring.

=== Construction of ring colorings ===

A coloring for a ring <math>\mathcal{R} \subset \mathbb{C}</math> can sometimes be constructed as follows
: Step 1: find a ring-homomorphism from <math>\mathcal{R}</math> to a [https://en.wikipedia.org/wiki/Finite_ring finite ring] <math>\mathcal{F}</math>
: Step 2: find the image <math>\mathcal{T}</math> in <math>\mathcal{F}</math> of the set of all elements of unit modulus in <math>\mathcal{R}</math>
: Step 3: look for a <math>k</math>-coloring of <math>\mathcal{F}</math> such that no two elements that differ by an element in <math>\mathcal{T}</math> have the same color.

The composition of the ring homomorphism and the coloring of the finite ring <math>\mathcal{F}</math> will then be a <math>k</math>-coloring of <math>\mathcal{R}</math>

An ring of algebraic numbers <math>\mathcal{R}</math> has a ring homomorphism onto a finite ring <math>\mathcal{R}/p\mathcal{R}, p \in \mathbb{Z}</math> with the additive structure <math>{\mathbb{Z}_p}^d</math> where <math>d</math> is the algebraic degree of <math>\mathcal{R}</math>. All periodic colorings with period <math>p</math> can be constructed from a coloring of <math>\mathcal{R}/p\mathcal{R}</math>.

==== ''example'' <math>\mathbb{Z}[\omega]</math> ====

<math>\mathbb{Z}[\omega]</math> consists of elements of the form <math>a + b\omega</math> where <math>\omega</math> is the first cube root of -1. There is a ring homomorphism onto <math>\mathcal{F} = \mathbb{Z}[\omega]/3\mathbb{Z}[\omega] = \mathbb{Z}_3[\omega]</math> by taking <math>a</math> and <math>b</math> modulo 3. The additive structure of <math>\mathcal{F}</math> is <math>\mathbb{Z}_3 \times \mathbb{Z}_3</math>.

The elements of <math>\mathbb{Z}[\omega]</math> which have unit modulus will map onto elements of <math>\mathcal{F}</math> for which <math>(a - b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2 \equiv 1 \bmod 3</math>. A check of all 9 elements of the group confirms that the only solutions are <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math>.

Coloring the ring <math>\mathcal{F}</math> is equivalent to filling out a three by three grid with three numbers such that no row or column or cyclic diagonals in one direction has the same number twice. The solution is a latin square unique up to permutations of colours. This can also be specified by the linear colouring <math>c(a + b\omega) = a - b</math>.

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}]</math> ====

The Moser spindle ring is <math>\mathcal{M} = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+i\sqrt{11}}{2}, \frac{1}{3}\right] = \mathbb{Z}\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}, \frac{1}{3}\right]</math>

<math>\mathcal{M}</math> does not have a 3-coloring because it contains the Moser spindle.

To look for a 4-coloring reduce modulo 4, use <math>\frac{1}{3} \equiv 3 \bmod 4</math> to get a ring homomorphism from <math>\mathcal{M}</math> to the 64 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}, \frac{1+\sqrt{33}}{2}\right]</math>.

<math>\alpha = \frac{1+\sqrt{33}}{2}</math> is a solution of <math>\alpha^2 - \alpha - 8 = 0</math>. This has two solutions modulo 4, <math>\alpha \equiv 0,1 \bmod 4</math> either of which can be used to provide a further homomorphism onto the 16 element finite ring <math>\mathbb{Z}_4\left[\frac{1+i\sqrt{3}}{2}\right] = \mathbb{Z}_4\left[\omega\right]</math>. The elements of unit modulus in this ring are again <math>\{\pm 1, \pm \omega, \pm(\omega - 1)\}</math> and the linear coloring <math>c(a + b\omega) = a - b</math>.

A [https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4366 computer assisted analysis] has confirmed that <math>\mathcal{M}</math> has non-linear 4-colorings and that all 4-colorings are periodic with period 8. See also [https://dustingmixon.wordpress.com/2018/05/01/polymath16-third-thread-is-6-chromatic-within-reach/#comment-4206 this]

==== ''example'' <math>\mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}]</math> ====

<math>\omega_4 = \frac{7+i\sqrt{15}}{8}</math>
<math>\omega_7 = \frac{13+i3\sqrt{3}}{14}</math>

<math>\overline{R}_3 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}] = \mathcal{M}\left[\phi, \frac{1}{2}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}\right], \phi = \frac{1+\sqrt{5}}{2}</math>

<math>\overline{R}_4 = \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] = \overline{R}_3\left[\frac{1}{7}\right] = \mathbb{Z}\left[i\sqrt{3}, i\sqrt{11}, \sqrt{5}, \frac{1}{2}, \frac{1}{3}, \frac{1}{7}\right]</math>

These rings do not have linear 4-colorings because they contain <math>\frac{1}{4}</math>. Furthermore they are known not to be 4-colorable by any means because they contain graphs that are not 4-colorable such as <math>G_2</math> and <math>V_A \oplus V_A \oplus V_A</math>

To seek a 5-coloring the rings are first reduced modulo 5 using real equivalences <math> \sqrt{5} \equiv 0, \frac{1}{2} \equiv 3, \frac{1}{3} \equiv 2, \frac{1}{7} \equiv 3 \bmod 5 </math>

This defines a ring homomorphism from <math>\overline{R}_4</math> to <math>\mathbb{Z}_5[i\sqrt{3}, i\sqrt{7}]</math> whose elements are of the form <math>x = a + bi\sqrt{3} + ci\sqrt{11} + d\sqrt{33}, a,b,c,d \in \mathbb{Z}_5</math>

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T18:34:38Z

Pgibbs: /* example \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}] */

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T17:45:07Z

Pgibbs: /* example \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] */

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T17:44:31Z

Pgibbs: /* example \mathbb{Z}[\omega_1, \omega_3, \overline{\omega_3}, \omega_4, \overline{\omega_4}, \omega_7, \overline{\omega_7}] */

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T17:06:08Z

Pgibbs: /* Coloring Rings */

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T13:35:30Z

Pgibbs: /* Coloring Rings */

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T12:40:24Z

Pgibbs: /* Rings */

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T11:01:41Z

Pgibbs:

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T11:00:33Z

Pgibbs: /* Coloring Rings */

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T10:57:15Z

Pgibbs: /* Construction of ring colorings */

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T10:55:40Z

Pgibbs: /* Construction of ring colorings */

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T10:54:44Z

Pgibbs: /* Construction of ring colorings */

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T10:53:36Z

Pgibbs:

Algebraic formulation of Hadwiger-Nelson problem

2018-05-09T10:47:52Z

Pgibbs: /* Construction of ring colorings */

Algebraic formulation of Hadwiger-Nelson problem

2018-05-08T20:57:18Z

Pgibbs: /* Construction of ring colorings */