Moser's cube problem

Define a Moser set to be a subset of [math]\displaystyle{ [3]^n }[/math] which does not contain any geometric line, and let [math]\displaystyle{ c'_n }[/math] denote the size of the largest Moser set in [math]\displaystyle{ [3]^n }[/math]. The first few values are (see OEIS A003142):

[math]\displaystyle{ c'_0 = 1; c'_1 = 2; c'_2 = 6; c'_3 = 16; c'_4 = 43. }[/math]

Beyond this point, we only have some upper and lower bounds, e.g. [math]\displaystyle{ 124 \leq c'_5 \leq 127 }[/math]; see this spreadsheet for the latest bounds.

The best known asymptotic lower bound for [math]\displaystyle{ c'_n }[/math] is

[math]\displaystyle{ c'_n \gg 3^n/\sqrt{n} }[/math],

formed by fixing the number of 2s to a single value near n/3. Is it possible to do any better? Note that we have a significantly better bound for [math]\displaystyle{ c_n }[/math]:

[math]\displaystyle{ c_n \geq 3^{n-O(\sqrt{\log n})} }[/math].

A more precise lower bound is

[math]\displaystyle{ c'_n \geq \binom{n+1}{q} 2^{n-q} }[/math]

where q is the nearest integer to [math]\displaystyle{ n/3 }[/math], formed by taking all strings with q 2s, together with all strings with q-1 2s and an odd number of 1s. This for instance gives the lower bound [math]\displaystyle{ c'_5 \geq 120 }[/math], which compares with the upper bound [math]\displaystyle{ c'_5 \leq 4 c'_4 = 124 }[/math].

Using DHJ(3), we have the upper bound

[math]\displaystyle{ c'_n = o(3^n) }[/math],

but no effective decay rate is known. It would be good to have a combinatorial proof of this fact (which is weaker than DHJ(3), but implies Roth's theorem).

n=3

We have [math]\displaystyle{ c'_3 = 16 }[/math]. The lower bound can be seen for instance by taking all the strings with one 2, and half the strings with no 2 (e.g. the strings with an odd number of 1s). The upper bound can be deduced from the corresponding lower and upper bounds for [math]\displaystyle{ c_3 = 18 }[/math]; the 17-point and 18-point line-free sets each contain a geometric line.

If a Moser set in [math]\displaystyle{ [3]^3 }[/math] contains 222, then it can have at most 14 points, since the remaining 26 points in the cube split into 13 antipodal pairs, and at most one of each pair can lie in the set. By exhausting over the [math]\displaystyle{ 2^{13} = 8192 }[/math] possibilities, it can be shown that it is impossible for a 14-point set to exist; any Moser set containing 222 must in fact omit at least one antipodal pair completely and thus have only 13 points.

n=4

A computer search has obtained all extremisers to [math]\displaystyle{ c'_4=43 }[/math]. The 42-point solutions can be found here.

Given a subset of [math]\displaystyle{ [3]^4 }[/math], let a be the number of points with no 2s, b be the number of points with 1 2, and so forth. The quintuple (a,b,c,d,e) thus lies between (0,0,0,0,0) and (16,32,24,8,1).

The 43-point solutions have distributions (a,b,c,d,e) as follows:

• (5,20,18,0,0) [16 solutions]
• (4,16,23,0,0) [768 solutions]
• (3,16,24,0,0) [512 solutions]
• (4,15,24,0,0) [256 solutions]

The 42-point solutions are distributed as follows:

• (6,24,12,0,0) [8 solutions]
• (5,20,17,0,0) [576 solutions]
• (5,19,18,0,0) [384 solutions]
• (6,16,18,2,0) [192 solutions]
• (4,20,18,0,0) [272 solutions]
• (5,17,20,0,0) [192 solutions]
• (5,16,21,0,0) [3584 solutions]
• (4,17,21,0,0) [768 solutions]
• (4,16,22,0,0) [26880 solutions]
• (5,15,22,0,0) [1536 solutions]
• (4,15,23,0,0) [22272 solutions]
• (3,16,23,0,0) [15744 solutions]
• (4,14,24,0,0) [4224 solutions]
• (3,15,24,0,0) [8704 solutions]
• (2,16,24,0,0) [896 solutions]

Note how c is usually quite large, and d quite low.

If a Moser set in [math]\displaystyle{ [3]^4 }[/math] contains 2222, then by the n=3 theory, any middle slice (i.e. 2***, *2**, **2*, or ***2) is missing at least one antipodal pair. But each antipodal pair belongs to at most three middle slices, thus two of the 40 antipodal pairs must be completely missing. As a consequence, any Moser set containing 2222 can have at most 39 points.

n=5

I think that c_5′ must be 128 or less. There are only 24 points possible with two 2’s so if there are three in a row the first one must have at least 18 of these points the second cube must also have at least 18 so there must be at least 12 in both final the third must have at least 18 so there must be at least 6 in all three which results in a line. So the maximum value for c_5′ is 128 or less.

In fact, I can now rule out 128-point sets as follows. If one had a 128-point set, then the set must slice as 43+43+42 (or a permutation thereof) in every direction. But all 43-point slices have d=0 and all 42-point slices have d <= 2. Thus, there are at most two points amongst the sixteen points x222y, x22y2, x2y22, xy222 with x,y = 1,3 that lie in the set. Averaging this over all orientations we see that at most one eighth of all points with exactly 3 2s lie in the set. Thus, by the pigeonhole principle and a double counting argument, there exists a middle slice of the set with this property, i.e. a slice with c <= 24/8 = 3, but we know from the 43-point and 42-point distribution data that this is impossible.

A lower bound [math]\displaystyle{ c'_5 \geq 124 }[/math] can be established by taking

• all points with two 1s;
• all points with one 1 and an even number of 0s; and
• Four points with no 1s and two or three 0s. Any two of these four points differ in three places, except for one pair of points that differ in one place.

This suggests further solutions for c'_N

q 1s, all points from A(N-q,1) q-1 1s, points from A(N-q+1,2) q-2 1s, points from A(N-q+2,3) etc.

where A(m,d) is a subset of [0,2]^m for which any two points differ from each other in at least d places.

Mathworld’s entry on error-correcting codes suggests it might be NP-complete to find the size of A(m,d) in general.

|A(m,1)| = 2^m because it includes all points in [0,2]^m |A(m,2)| = 2^{m-1} because it can include all points in [0,2]^m with an odd number of 0s

Lemma 1: Any Moser set containing an element with at least four 2s has at most 124 points.

Proof If the Moser set contains 22222, then each of the 121 antipodal pairs can have at most one point in the set, leading to only 122 points. So we may assume that 22222 is not in the set.

By the n=4 theory, the 1**** slice now has at most 39 points. In order to make at least 125=39+43+43 points, the 2**** and 3**** slice must now have exactly 43 points. By the n=4 theory, this implies that there are at least 18 points of the form 222xy, 22xy2, 2xy22, or 2x22y, where the x, y denote 1 or 3. This gives 36 "xy" wildcards in all in four coordinate slots; by the pigeonhole principle one of the slots sees at least 9 of the wildcards. By symmetry, we may assume that the second coordinate slot sees at least 9 of these wildcards. In other words, there are at least 9 points of the form 2x22y, 2x2y2, and 2xy22. The x=1 and x=3 cases can each absorb at most six of these, thus we see that there are at least three points of the form 2122y, 212y2, 21y22, and three points of the form 2322y, 232y2, 23y22.

The *1*** slice now has d >= 3 and thus has at most 41 points; similarly for the *3*** slice. Meanwhile, the *2*** slice contains 12222 and thus has d >= 1, leading to at most 42 points. This leads to at most 41+41+42=124 points in all.[math]\displaystyle{ \Box }[/math]

Variants

A lower bound for Moser’s cube k=4 (values 0,1,2,3) is: q entries are 1 or 2; or q-1 entries are 1 or 2 and an odd number of entries are 0. This gives a lower bound of

[math]\displaystyle{ \binom{n}{n/2} 2^n + \binom{n}{n/2-1} 2^{n-1} }[/math]

which is comparable to [math]\displaystyle{ 4^n/\sqrt{n} }[/math] by Stirling's formula.

For k=5 (values 0,1,2,3,4) If A, B, C, D, and E denote the numbers of 0-s, 1-s, 2-s, 3-s, and 4-s then the first three points of a geometric line form a 3-term arithmetic progression in A+E+2(B+D)+3C. So, for k=5 we have a similar lower bound for the Moser’s problem as for DHJ k=3, i.e. [math]\displaystyle{ 5^{n - O(\sqrt{\log n})} }[/math].

The k=6 version of Moser implies DHJ(3). Indeed, any k=3 combinatorial line-free set can be "doubled up" into a k=6 geometric line-free set of the same density by pulling back the set from the map [math]\displaystyle{ \phi: [6]^n \to [3]^n }[/math] that maps 1, 2, 3, 4, 5, 6 to 1, 2, 3, 3, 2, 1 respectively; note that this map sends k=6 geometric lines to k=3 combinatorial lines.