Upper and lower bounds

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Upper and lower bounds for cn for small values of n.

cn is the size of the largest subset of [3]n that does not contain a combinatorial line (OEIS A156762. A spreadsheet for all the latest bounds on cn can be found here. In this page we record the proofs justifying these bounds. See also this page on higher-dimensional DHJ numbers.


n 0 1 2 3 4 5 6 7
cn 1 2 6 18 52 150 450 [1302,1348]

Contents

Basic constructions

For all n \geq 1, a basic example of a mostly line-free set is

D_n := \{ (x_1,\ldots,x_n) \in [3]^n: \sum_{i=1}^n x_i \neq 0 \ \operatorname{mod}\ 3 \}. (1)

This has cardinality |D_n| = 2 \times 3^{n-1}. The only lines in Dn are those with

  1. A number of wildcards equal to a multiple of three;
  2. The number of 1s unequal to the number of 2s modulo 3.

One way to construct line-free sets is to start with Dn and remove some additional points. We also have the variants Dn,0 = Dn,Dn,1,Dn,2 defined as

D_{n,j} := \{ (x_1,\ldots,x_n) \in [3]^n: \sum_{i=1}^n x_i \neq j \ \operatorname{mod}\ 3 \}. (1')

When n is not a multiple of 3, then Dn,0,Dn,1,Dn,2 are all cyclic permutations of each other; but when n is a multiple of 3, then Dn,0 plays a special role (though Dn,1,Dn,2 are still interchangeable).

Another useful construction proceeds by using the slices \Gamma_{a,b,c} \subset [3]^n for (a,b,c) in the triangular grid

\Delta_n := \{ (a,b,c) \in {\Bbb Z}_+^3: a+b+c = n \},. (2)

where Γa,b,c is defined as the strings in [3]n with a 1s, b 2s, and c 3s. Note that

|\Gamma_{a,b,c}| = \frac{n!}{a! b! c!}. (3)

Given any set B \subset \Delta_n that avoids equilateral triangles (a + r,b,c),(a,b + r,c),(a,b,c + r), the set

\Gamma_B := \bigcup_{(a,b,c) \in B} \Gamma_{a,b,c} (4)

is line-free and has cardinality

|\Gamma_B| = \sum_{(a,b,c) \in B} \frac{n!}{a! b! c!}, (5)

and thus provides a lower bound for cn:

c_n \geq \sum_{(a,b,c) \in B} \frac{n!}{a! b! c!}. (6)

All lower bounds on cn have proceeded so far by choosing a good set of B and applying (6). Note that Dn is the same as \Gamma_{B_n}, where Bn consists of those triples (a,b,c) \in \Delta_n in which a \neq b\ \operatorname{mod}\ 3.

Note that if one takes a line-free set and permutes the alphabet {1,2,3} in any fashion (e.g. replacing all 1s by 2s and vice versa), one also gets a line-free set. This potentially gives six examples from any given starting example of a line-free set, though in practice there is enough symmetry that the total number of examples produced this way is less than six. (These six examples also correspond to the six symmetries of the triangular grid Δn formed by rotation and reflection.)

Another symmetry comes from permuting the n indices in the strings of [3]n (e.g. replacing every string by its reversal). But the sets ΓB are automatically invariant under such permutations and thus do not produce new line-free sets via this symmetry.

The basic upper bound

Because [3]n + 1 can be expressed as the union of three copies of [3]n, we have the basic upper bound

c_{n+1} \leq 3 c_n. (7)

Note that equality only occurs if one can find an n + 1-dimensional line-free set such that every n-dimensional slice has the maximum possible cardinality of cn.

n=0

c0 = 1:

This is clear.

n=1

c1 = 2:

The three sets D1 = {1,2}, D1,1 = {2,3}, and D1,2 = {1,3} are the only two-element sets which are line-free in [3]1, and there are no three-element sets.

n=2

c2 = 6:

There are four six-element sets in [3]2 which are line-free, which we denote x = D2,2, y = D2,1, z = D2, and w and are displayed graphically as follows.

    13 .. 33       .. 23 33       13 23 ..       13 23 ..
x = 12 22 ..   y = 12 .. 32   z = .. 22 32   w = 12 .. 32
    .. 21 31       11 21 ..       11 .. 31       .. 21 31

Combining this with the basic upper bound (7) we see that c2 = 6.

n=3

c3 = 18:

We describe a subset A of [3]3 as a string abc, where a, b, c \subset [3]^2 correspond to strings of the form 1 * * , 2 * * , 3 * * in [3]3 respectively. Thus for instance D3 = xyz, and so from (7) we have c3 = 18.

Lemma 1.

  • The only 18-element line-free subset of [3]3 is D3 = xyz.
  • The only 17-element line-free subsets of [3]3 are formed by removing a point from D3 = xyz, or by removing either 111, 222, or 333 from D3,2 = yzx or D3,3 = zxy.

Proof. We prove the second claim. As 17 = 6 + 6 + 5, and c2 = 6, at least two of the slices of a 17-element line-free set must be from x, y, z, w, with the third slice having 5 points. If two of the slices are identical, the last slice can have only 3 points, a contradiction. If one of the slices is a w, then the 5-point slice will contain a diagonal, contradiction. By symmetry we may now assume that two of the slices are x and y, which force the last slice to be z with one point removed. Now one sees that the slices must be in the order xyz, yzx, or zxy, because any other combination has too many lines that need to be removed. The sets yzx, zxy contain the diagonal {111,222,333} and so one additional point needs to be removed.

The first claim follows by a similar argument to the second. \Box

n=4

c4 = 52:

Indeed, divide a line-free set in [3]4 into three blocks 1 * * * ,2 * * * ,3 * * * of [3]3. If two of them are of size 18, then they must both be xyz, and the third block can have at most 6 elements, leading to an inferior bound of 42. So the best one can do is 18 + 17 + 17 = 52 which can be attained by deleting the diagonal {1111,2222,3333} from D_{4,1} = xyz\ yzx\ xzy, D_4 = yzx\ zxy\ xyz, or D_{4,2} = zxy\ xyz\ yzx. In fact,

Lemma 2.

  • The only 52-element line-free sets in [3]4 are formed by removing the diagonal {1111,2222,3333} from D4,j for some j=0,1,2.
  • The only 51-element line-free sets in [3]4 are formed by removing the diagonal and one further point from D4,j for some j=0,1,2.
  • The only 50-element line-free sets in [3]4 are formed by removing the diagonal and two further points from D4,j for some j=0,1,2 OR is equal to one of the three permutations of the set X := \Gamma_{3,1,0} \cup \Gamma_{3,0,1} \cup \Gamma_{2,2,0} \cup \Gamma_{2,0,2} \cup \Gamma_{1,1,2} \cup \Gamma_{1,2,1} \cup \Gamma_{0,2,2}.

Proof It suffices to prove the third claim. In fact it suffices to show that every 50-point line-free set is either contained in the 54-point set D4,j for some j=0,1,2, or is some permutation of the set X. Indeed, if a 50-point line-free set is contained in, say, D4, then it cannot contain 2222, since otherwise it must omit one point from each of the four pairs formed from {2333, 2111} by permuting the indices, and must also omit one of {1111, 1222, 1333}, leading to at most 49 points in all; similarly, it cannot contain 1111, and so omits the entire diagonal {1111,2222,3333}, with two more points to be omitted. Similarly when D4 is replaced by one of the other D4,j

Next, observe that every three-dimensional slice of a line-free set can have at most c3 = 18 points; thus when one partitions a 50-point line-free set into three such slices, it must divide either as 18+16+16, 18+17+15, 17+17+16, or some permutation of these.

Suppose that we can slice the set into two slices of 17 points and one slice of 16 points. By the various symmetries, we may assume that the 1*** slice and 2*** slices have 17 points, and the 3*** slice has 16 points. By Lemma 1, the 1-slice is \{1\} \times D_{3,j} with one point removed, and the 2-slice is \{2\} \times D_{3,k} with one point removed, for some j,k \in \{0,1,2\}.

If j=k, then the 1-slice and 2-slice have at least 15 points in common, so the 3-slice can have at most 27 − 15 = 12 points, a contradiction. If jk = 01, 12, or 20, then observe that from Lemma 1 the *1**, *2**, *3** slices cannot equal a 17-point or 18-point line-free set, so each have at most 16 points, leading to only 48 points in all, a contradiction. Thus we must have jk = 10, 21, or 02.

Let's first suppose that jk=02. Then by Lemma 1, the 2*** slice contains the nine points formed from {2211, 2322, 2331} and permuting the last three indices, while the 1*** slice contains at least eight of the nine points formed from {1211, 1322, 1311} and permuting the last three indices. Thus the 3*** slice can contain at most one of the nine points formed from {3211, 3322, 3311} and permuting the last three indices. If it does contain one of these points, say 3211, then it must omit one point from each of the four pairs {3222, 3233}, {3212, 3213}, {3221, 3231}, {3111, 3311}, leading to at most 15 points on this slice, a contradiction. So the 3*** slice must omit all nine points, and is therefore contained in \{3\} \times D_{4,1}, and so the 50-point set is contained in D4,1, and we are done by the discussion at the beginning of the proof.

The case jk=10 is similar to the jk=02 case (indeed one can get from one case to the other by swapping the 1 and 2 indices). Now suppose instead that jk=12. Then by Lemma 1, the 1*** slice contains the six points from permuting the last three indices of 1123, and similarly the 2*** slice contains the six points from permuting the last three indices of 2123. Thus the 3*** slice must avoid all six points formed by permuting the last three indices of 3123. Similarly, as 1133 lies in the 1*** slice and 2233 lies in the 2*** slice, 3333 must be avoided in the 3*** slice.

Now we claim that 3111 must be avoided also; for if 3111 was in the set, then one point from each of the six pairs formed from {3311, 3211}, {3331, 3221} and permuting the last three indices must lie outside the 3*** slice, which reduces the size of that slice to at most 27 − 6 − 1 − 6 = 14, which is too small. Similarly, 3222 must be avoided, which puts the 3*** slice inside \{3\} \times D_3 and then places the 50-point set inside D4, and we are done by the discussion at the beginning of the proof.

We have handled the case in which at least one of the slicings of the 50-point set is of the form 50=17+17+16. The only remaining case is when all slicings of the 50-point set are of the form 18+17+15 or 18+16+16 (or a permutation thereof). By the symmetries of the situation, we may assume that the 1*** slice has 18 points, and thus by Lemma 1 takes the form \{1\} \times D_3. Inspecting the *1**, *2**, *3** slices, we then see (from Lemma 1) that only the *1** slice can have 18 points; since we are assuming that this slicing is some permutation of 50=18+17+16, we conclude that the *1** slice must have exactly 18 points, and is thus described precisely by Lemma 1. Similarly for the **1* and ***1 slices. Indeed, by Lemma 1, we see that the 50-point set must agree exactly with D4,1 on any of these slices. In particular, on the remaining portion {2,3}4 of the cube, there are exactly 6 points of the 50-point set in {2,3}4.

Suppose that 3333 was in the set; then since all permutations of 3311, 3331 are known to lie in the set, then 3322, 3332 must lie outside the set. Also, as 1222 lies in the set, at least one of 2222, 3222 lie outside the set. This leaves only 5 points in {2,3}4, a contradiction. Thus 3333 lies outside the set; similarly 2222 lies outside the set.

Let a be the number of points in the 50-point set which are some permutation of 2233, thus 0 \leq a \leq 6. If a=0 then the set lies in D4,1 and we are done. If a=6 then the set is exactly X and we are done. Now suppose a=1,2,3. By symmetry we may assume that 2233 lies in the set. Then (since 2133, 1233 2231, 2213 are known to lie in the set) 2333, 3233, 2223, 2232 lie outside the set, which leaves at most 5 points inside {2,3}4, a contradiction.

The remaining case is when a=4,5. Then one of the three pairs {2233, 3322}, {2323, 3232}, {2332, 3223} lie in the set. By symmetry we may assume that {2233, 3322} lie in the set. Then by arguing as before we see that all eight points formed by permuting 2333 or 3222 lie outside the set, leading to at most 5 points inside {2,3}4, a contradiction. \Box

n=5

c5 = 150:

Lemma 3. Any line-free subset of D5,j can have at most 150 points.

Proof. By rotation we may work with D5. This set has 162 points. By looking at the triplets {10000, 11110, 12220} and cyclic permutations we must lose 5 points; similarly from the triplets {20000,22220, 21110} and cyclic permutations. Finally from {11000,11111,11222} and {22000,22222,22111} we lose two more points. \Box

Equality can be attained by removing Γ0,4,10,5,04,0,15,0,0 from D5. Thus c_5 \geq 150.

Another pattern of 150 points is this: Take the 450 points in [3]6 which are (1,2,3), (0,2,4) and permutations, then select the 150 whose final coordinate is 1. That gives this many points in each cube:

17 18 17

17 17 18

12 17 17

Lemma 4. A line-free subset of [3]5 with over 150 points cannot have two parallel [3]4 slices, each of which contain at least 51 points.

Proof. Suppose not. By symmetry, we may assume that the 1**** and 2**** slices have at least 51 points, and that the whole set has at least 151 points, which force the third slice to have at least 151 − 2c4 = 47 points.

By Lemma 2, the 1**** slice takes the form \{1\} \times D_{4,j} for some j = 0,1,2 with the diagonal {11111,12222,13333} and possibly one more point removed, and similarly the 2**** slice takes the form \{2\} \times D_{4,k} for some k = 0,1,2 with the diagonal {21111,22222,23333} and possibly one more point removed.

Suppose first that j=k. Then the 1-slice and 2-slice have at least 50 points in common, leaving at most 31 points for the 3-slice, a contradiction. Next, suppose that jk=01. Then observe that the *i*** slice cannot look like any of the configurations in Lemma 2 and so must have at most 50 points for i=1,2,3, leading to 150 points in all, a contradiction. Similarly if jk=12 or 20. Thus we must have jk equal to 10, 21, or 02.

Let's suppose first that jk=10. The first slice then is equal to \{1\} \times D_{4,1} with the diagonal and possibly one more point removed, while the second slice is equal to \{2\} \times D_{4,0} with the diagonal and possibly one more point removed. Superimposing these slices, we thus see that the third slice is contained in \{3\} \times D_{4,2} except possibly for two additional points, together with the one point 32222 of the diagonal that lies outside of \{3\} \times D_{4,2}.

The lines x12xx, x13xx (plus permutations of the last four digits) must each contain one point outside the set. The first two slices can only absorb two of these, and so at least 14 of the 16 points formed by permuting the last four digits of 31233, 31333 must lie outside the set. These points all lie in \{3\} \times D_{4,2}, and so the 3**** slice can have at most | D4,2 | − 14 + 3 = 43 points, a contradiction.

The case jk=02 is similar to the case jk=10 (indeed one can obtain one from the other by swapping 1 and 2). Now we turn to the case jk=21. Arguing as before we see that the third slice is contained in \{3\} \times D_4 except possibly for two points, together with 33333.

If 33333 was in the set, then each of the lines xx333, xxx33 (and permutations of the last four digits) must have a point missing from the first two slices, which cannot be absorbed by the two points we are permitted to remove; thus 33333 is not in the set. For similar reasons, 33331 is not in the set, as can be seen by looking at xxx31 and permutations of the last four digits. Indeed, any string containing four threes does not lie in the set; this means that at least 8 points are missing from \{3\} \times D_4, leaving only at most 46 points inside that set. Furthermore, any point in the 3**** slice outside of \{3\} \times D_4 can only be created by removing a point from the first two slices, so the total cardinality is at most 46 + 52 + 52 = 150, a contradiction.\Box

Corollary. c_5 \leq 152

Proof. By Lemma 4 and the bound c4 = 52, any line-free set with over 150 points can have one slice of cardinality 52, but then the other two slices can have at most 50 points. \Box


Lemma 5 Any solution with 151 or more points has a slice with at most 49 points.

Proof Suppose we have 151 points without a line, and each of three slices has at least 50 points.

Using earlier notation, we split subsets of [3]4 into nine subsets of [3]2. So we think of x,y,z,a,b and c as subsets of a square. Each slice is one of the following.

  • D4 = y'zx,zx'y,xyz (with one or two points removed)
  • D4,2 = z'xy,xyz,yzx' (with one or two points removed)
  • D4,1 = xyz,yz'x,zxy' (with one or two points removed)
  • X = xyz,ybw,zwc
  • Y = axw,xyz,wzc
  • Z = awx,wby,xyz

where a, b and c have four points each.

       .. 32 33          31 .. 33         .. .. ..
   a = .. 22 23      b = .. .. ..     c = 21 22 ..
       .. .. ..          11 .. 13         11 12 ..

x', y' and z' are subsets of x, y and z respectively, and have five points each.

Suppose all three slices are subsets of D4,j. We can remove at most five points from the full set of three D_{4,j}. Consider columns 2,3,4,6,7,8. At most two of these columns contain xyz, so one point must be removed from the other four. This uses up all but one of the removals. So the slices must be D4,2,D4,1,D4,0 or a cyclic permutation of that. Then the cube, which contains the first square of slice 1; the fifth square of slice 2; and the ninth square of slice 3, contains three copies of the same square. It takes more than one point removed to remove all lines from that cube. So we can't have all three slices subsets of D4,j.

Suppose one slice is X,Y or Z, and two others are subsets of D4,j. We can remove at most three points from the full D4,j By symmetry, suppose one slice is X. Consider columns 2,3,4 and 7. They must be cyclic permutations of x,y,z, and two of them are not xyz, so must lose a point. Columns 6 and 8 must both lose a point, and we only have 150 points left. So if one slice is X,Y or Z, the full set contains a line.

Suppose two slices are from X,Y and Z, and the other is a subset of D4,j. By symmetry, suppose two slices are X and Y. Columns 3,6,7 and 8 all contain w, and therefore at most 16 points each. Columns 1,5 and 9 contain a,b, or c, and therefore at most 16 points. So the total number of points is at most 7*16+2*18 = 148. This contradicts the assumption of 151 points. \Box

Corollary c_5 \leq 151

Proof By Lemmas 2 and 4, the maximum number of points is 52+50+49=151. \Box

Lemma 5.1 No solution with 151 points contains as a slice the X defined in Lemma 2

Proof Suppose one row is X. Another row is D4,j.

Suppose X is in the first row. Label the other rows with letters from the alphabet.

xyz ybw zwc

mno pqr stu

def ghi jkl

Reslice the array into a left nine, middle nine and right nine. One of these squares contains 52 points, and it can only be the left nine. One of its three columns contains 18 points, and it can only be its left-hand column, xmd. So m=y and d=z. But none of the {math>D_{4,j}</math> begins with y or z, which is a contradiction. So X is not in the first row.

So X is in the second or third row. By symmetry, suppose it is in the second row

def ghi jkl

xyz ybw zwc

mno pqr stu

Again, the left-hand nine must contain 52 points, so it is D4,2. So either the first row is D4,2 or the third row is D4,0. If the first row is D4,2 then the only way to have 50 points in the middle or right-hand nine is if the middle nine is X

z'xy xyz yzx'

xyz ybw zwc

yzx' zwc stu

In the seventh column, s contains 5 points and in the eighth column, t contains 4 points. The final row can now contain at most 48 points, and the whole array contains only 52+50+48 = 150 points.

If the third row is D4,0, then neither the middle nine nor the right-hand nine contains 50 points, by the classification of Lemma 4 and the formulas at the start of Lemma 5. Again, only 52+49+49 = 150 points are possible.

A similar argument is possible if X is in the third row; or if X is replaced by Y or Z.

So when a 151-point set is sliced into three, one slice is D4,j and another slice is 50 points contained in D4,k. \Box

Lemma 5.2 There is no 151-point solution

Proof Assume by symmetry that the first row contains 52 points and the second row contains 50.

If D4,1 is in the first row, then the second row must be contained in D4,0.

xyz yz'x zxy'

y'zx zx'y xyz

def ghi jkl

But then none of the left nine, middle nine or right nine can contain 52 points, which contradicts the corollary to Lemma 5.

Suppose the first row contains D_{4,0}. Then the second row is contained in D4,2, otherwise the cubes formed from the nine columns of the diagram would need to remove too many points.

y'zx zx'y xyz

z'xy xyz yzx'

def ghi jkl

But then neither the left nine, middle nine or right nine contains 52 points.

So the first row contains D4,2, and the second row is contained in D4,1. Two points may be removed from the second row of this diagram.

z'xy xyz yzx'

xyz yz'x zxy'

def ghi jkl

Slice it into the left nine, middle nine and right nine. Two of them are contained in D4,j so at least two of def, ghi, and jkl are contained in the corresponding slice of D4,0. Slice along a different axis, and at least two of dgj,ehk,fil are contained in the corresponding slice of D4,0. So eight of the nine squares in the bottom row are contained in the corresponding square of D4,0. Indeed, slice along other axes, and all points except one are contained within D4,0. This point is the intersection of all the 49-point slices.

So, if there is a 151-point solution, then after removal of the specified point, there is a 150-point solution, within D5,j, whose slices in each direction are 52+50+48.

z'xy xyz yzx'

xyz yz'x zxy'

y'zx zx'y xyz

One point must be lost from columns 3, 6, 7 and 8, and four more from the major diagonal z'z'z. That leaves 148 points instead of 150.

So the 150-point solution does not exist with 52+50+48 slices; so the 151 point solution does not exist.\Box


An integer programming method has established the upper bound c_5\leq 150, with 12 extremal solutions.

This file contains the extermisers. One point per line and different extermisers separated by a line with “—”

This is the linear program, readable by Gnu’s glpsol linear programing solver, which also quickly proves that 150 is the optimum.

Each variable corresponds to a point in the cube, numbered according to their lexicografic ordering. If a variable is 1 then the point is in the set, if it is 0 then it is not in the set. There is one linear inequality for each combinatorial line, stating that at least one point must be missing from the line.

n=6

c6 = 450:

The upper bound follows since c_6 \leq 3 c_5. The lower bound can be formed by gluing together all the slices Γa,b,c where (a,b,c) is a permutation of (0,2,4) or (1,2,3).

Computer verification, using the c5 = 150 extremals, has shown that there is exactly one extremiser for c6 = 450.

n=7

1302 \leq c_7 \leq 1348:

To see the upper bound c_7 \leq 3c_6-2, observe that if two parallel six-dimensional slices had c6 points, then by uniqueness they are identical, and the third slice can have at most 36c6 = 279 points, far too few to get anywhere close to 1348. Thus there can be at most one slice with c6 points, and the other two have at most c6 − 1, giving the claim.

The lower bound can be formed by removing 016,106,052,502,151,511,160,610 from D7.

Lemma 6 Any line-free subset of D7 has at most 1302 points.

Proof Start with the 1458 points of D7. You must lose:

  • 42 points from (1,2,4),(1,5,1),(4,2,1)
  • 42 points from (2,1,4),(2,4,1),(5,1,1)
  • 21 points from (0,2,5),(0,5,2),(3,2,2)
  • 21 points from (2,0,5),(2,3,2),(5,0,2)
  • 15 points from (0,1,6),(0,4,3),(3,1,3),(0,7,0),(3,4,0),(6,1,0)
  • 15 points from (1,0,6),(1,3,3),(4,0,3),(7,0,0),(4,3,0),(1,6,0)

where (a,b,c) is shorthand for the slice Γa,b,c. \Box

Larger n

The following construction gives lower bounds for the number of triangle-free points, There are of the order 2.7 \sqrt{log(N)/N}3^N points for large N (N ~ 5000)

It applies when N is a multiple of 3.

  • For N=3M-1, restrict the first digit of a 3M sequence to be 1. So this construction has exactly one-third as many points for N=3M-1 as it has for N=3M.
  • For N=3M-2, restrict the first two digits of a 3M sequence to be 12. This leaves roughly one ninth of the points for N=3M-2 as for N=3M.

The current lower bounds for c3m are built like this, with abc being shorthand for Γa,b,c:

  • c3 from (012) and permutations
  • c6 from (123,024) and perms
  • c9 from (234,135,045) and perms
  • c12 from (345,246,156,02A,057) and perms (A=10)
  • c15 from (456,357,267,13B,168,04B,078) and perms (B=11)

To get the triples in each row, add 1 to the triples in the previous row; then include new triples that have a zero.

A general formula for these points is given below. I think that they are triangle-free. (For N<21, ignore any triple with a negative entry.)

  • There are thirteen groups of points in the centre, formed from adding one of the following points, or its permutation, to (M,M,M), when N=3M:
    • (-7,-3,+10), (-7, 0,+7),(-7,+3,+4),(-6,-4,+10),(-6,-1,+7),(-6,+2,+4),(-5,-1,+6),(-5,+2,+3),(-4,-2,+6),(-4,+1,+3),(-3,+1,+2),(-2,0,+2),(-1,0,+1)
  • There are also eight string of points, stretching to the edges of the (abc) triangle:
    • For N = 3M
      • M+(-8-2x,-6-2x,14+4x),M+(-8-2x,-3-2x,11+4x),M+(-8-2x,x,8+x),M+(-8-2x,3+x,5+x) and permutations (x>=0, M-8-2x>=0)
      • M+(-9-2x,-5-2x,14+4x),M+(-9-2x,-2-2x,11+4x),M+(-9-2x,1+x,8+x),M+(-9-2x,4+x,5+x) and permutations (x>=0, M-9-2x>=0)


An alternate construction:

First define a sequence, of all positive numbers which, in base 3, do not contain a 1. Add 1 to all multiples of 3 in this sequence. This sequence does not contain a length-3 arithmetic progression.

It starts 1,2,7,8,19,20,25,26,55, …

Second, list all the (abc) triples for which the larger two differ by a number from the sequence, excluding the case when the smaller two differ by 1, but then including the case when (a,b,c) is a permutation of N/3+(-1,0,1)

Asymptotics

DHJ(3) is equivalent to the upper bound

c_n \leq o(3^n)

In the opposite direction, observe that if we take a set S \subset [3n] that contains no 3-term arithmetic progressions, then the set \bigcup_{(a,b,c) \in \Delta_n: a+2b \in S} \Gamma_{a,b,c} is line-free. From this and the Behrend construction it appears that we have the lower bound

c_n \geq 3^{n-O(\sqrt{\log n})}.

More precisely, we have

c_n > C 3^{n - 4\sqrt{\log 2}\sqrt{\log n}+\frac 12 \log \log n}

for some absolute constant C, and where all logarithms are base-3.

Proof For convenience, let n be a multiple of 3. Elkin’s bound gives r_3(\sqrt{n}) > C \sqrt{n} (\log n)^{1/4} \exp_2(-2 \sqrt{\log_2 n}), and let R be a subset of (-3\sqrt{n}/2,3\sqrt{n}/2) without 3-term APs and with size r_3(\sqrt{n}), and with all elements being integer multiples of 3 (again as a matter of convenience). For each r,s\in R, let a = (nrs) / 3. The set A is the union of all Γa,a + r,a + s. Since all of a,a + r,a + s are between n/3-2\sqrt{n} and n/3+2\sqrt{n}, the size of Γa,a + r,a + s is at least C3n / n. Since there are r_3(\sqrt{n})^2 choices for r and s, we have a set with size at least

C (\sqrt{n} (\log n)^{1/4} \exp_2(-2 \sqrt{\log_2 n}))^2 3^n / n.

This simplifies to C \sqrt{\log n} \exp_3(n-\alpha \sqrt{\log_3(n)}), where \alpha=4 \sqrt{\log_3(2)}.

Now suppose that x_i\in \Gamma_{a_i,a_i+r_i,a_i+s_i} is a combinatorial line in the set A. Then (ai + si) − (ai) = si is a 3-term AP contained in R, so the si are all the same. Similarly, all of the ri are the same, and therefore all of the ai are the same, too. But this implies that the xi sequence is constant, which means the line is degenerate. \Box

Numerics suggest that the first large n construction given above above give a lower bound of roughly 2.7 \sqrt{\log(n)/n} \times 3^n, which would asymptotically be inferior to the Behrend bound.

The second large n construction had numerical asymptotics for log(cn / 3n) close to 1.2-\sqrt{\log(n)} between n=1000 and n=10000, consistent with the Behrend bound.

Numerical methods

A greedy algorithm was implemented here. The results were sharp for n \leq 3 but were slightly inferior to the constructions above for larger n.

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