Moser's cube problem

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

[math]c'_0 = 1; c'_1 = 2; c'_2 = 6; c'_3 = 16; c'_4 = 43; c'_5 = 124.[/math]

Beyond this point, we only have some upper and lower bounds, in particular [math]353 \leq c'_6 \leq 361[/math]; see this spreadsheet for the latest bounds.

The best known asymptotic lower bound for [math]c'_n[/math] is

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

formed by fixing the number of 2s to a single value near n/3. Compare this to the DHJ(2) or Sperner limit of [math]2^n/\sqrt{n}[/math]. Is it possible to do any better? Note that we have a significantly better bound for the DHJ(3) [math]c_n[/math]:

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

A more precise lower bound is

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

where q is the nearest integer to [math]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]c'_5 \geq 120[/math], which compares with the upper bound [math]c'_5 \leq 3 c'_4 = 129[/math].

Using DHJ(3), we have the upper bound

[math]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).

Notation

Given a Moser set A in [math][3]^n[/math], we let a be the number of points in A with no 2s, b be the number of points in A with one 2, c the number of points with two 2s, etc. We call (a,b,c,...) the statistics of A. Given a slice S of [math][3]^n[/math], we let a(S), b(S), etc. denote the statistics of that slice (dropping the fixed coordinates). Thus for instance if A = {11, 12, 22, 32}, then (a,b,c) = (1,2,1), and (a(1*),b(1*)) = (1,1).

We call a statistic (a,b,c,...) attainable if it is attained by a Moser set. We say that an attainable statistic is Pareto-optimal if it cannot be pointwise dominated by any other attainable statistic (a',b',c',...) (thus [math]a' \geq a[/math], [math]b' \geq b[/math], etc.) We say that it is extremal if it is not a convex combination of any other attainable statistic (this is a stronger property than Pareto-optimal). For the purposes of maximising linear scores of attainable statistics, it suffices to check extremal statistics.

We let [math](\alpha_0,\alpha_1,\ldots)[/math] be the normalized version of [math](a,b,\ldots)[/math], in which one divides the number of points of a certain type in the set by the total number of points in the set. Thus for instance [math]\alpha_0 = a/2^n, \alpha_1 = b/(n 2^{n-1}), \alpha_3 = c/(\binom{n}{2} 2^{n-2})[/math], etc., and the [math]\alpha_i[/math] range between 0 and 1. Averaging arguments show that any linear inequality obeyed by the [math]\alpha_i[/math] at one dimension is automatically inherited by higher dimensions, as are shifted versions of this inequality (in which [math]\alpha_i[/math] is replaced by [math]\alpha_{i+1}[/math].

The idea in 'c-statistics' is to identify 1s and 3s but leave the 2s intact. Let’s use x to denote letters that are either 1 or 3, then the 81 points in [math][3]^4[/math] get split up into 16 groups: 2222 (1 point), 222x, 22×2, 2×22, x222 (two points each), 22xx, 2×2x, x22x, 2xx2, x2×2, xx22 (four points each), 2xxx, x2xx, xx2x, xxx2 (eight points each), xxxx (sixteen points). Let c(w) denote the number of points inside a group w, e.g. c(xx22) is the number of points of the form xx22 inside the set, and is thus an integer from 0 to 4.

n=0

We trivially have [math]c'_0=1.[/math]

n=1

We trivially have [math]c'_1=2.[/math] The Pareto-optimal values of the statistics (a,b) are (1,1) and (2,0); these are also the extremals. We thus have the inequality [math]a+b \leq 2[/math], or in normalized notation

[math]2\alpha_0 + \alpha_1 \leq 2[/math].

n=2

We have [math]c'_2 = 6[/math]; the upper bound follows since [math]c'_2 \leq 3 c'_1[/math], and the lower bound follows by deleting one of the two diagonals from [math][3]^2[/math] (these are the only extremisers).

The extremiser has statistics (a,b,c) = (2,4,0), which are of course Pareto-optimal. If c=1 then we must have a, b at most 2 (look at the lines through 22). This is attainable (e.g. {11, 12, 22, 23, 31}, and so (2,2,1) is another Pareto-optimal statistic. If a=4, then b and c must be 0, so we get another Pareto-optimal statistic (4,0,0) (attainable by {11, 13, 31, 33} of course). If a=3, then c=0 and b is at most 2, giving another Pareto-optimal statistic (3,2,0); but this is a convex combination of (4,0,0) and (2,4,0) and is thus not extremal. Thus the complete set of extremal statistics are

(4,0,0), (2,4,0), (2,2,1).

The sharp linear inequalities obeyed by a,b,c (other than the trivial ones [math]a,b,c \geq 0[/math]) are then

  • [math]2a+b+2c \leq 8[/math]
  • [math]b+2c \leq 4[/math]
  • [math]a+2c \leq 4[/math]
  • [math]c \leq 1[/math].

In normalized notation, we have

  • [math]4\alpha_0 + 2\alpha_1 + \alpha_2 \leq 4[/math]
  • [math]2\alpha_1 + \alpha_2 \leq 2[/math]
  • [math]2\alpha_0 + \alpha_2 \leq 2[/math]
  • [math]\alpha_2 \leq 1[/math].

The Pareto optimizers for c-statistics are (c(22),c(2x),c(x2),c(xx)) = (1112),(0222),(0004), which are covered by these linear inequalities:

  • [math]c(22)+c(2x)+c(xx) \le 4[/math],
  • [math]c(22)+c(x2)+c(xx) \le 4[/math].

n=3

We have [math]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 upper and lower bounds for [math]c_3 = 18[/math]; the 17-point and 18-point line-free sets each contain a geometric line.

If a Moser set in [math][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]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. (A human proof of this fact can be found here.)

The Pareto-optimal statistics are

(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(3,6,5,0),(4,4,5,0),(3,7,4,0),(4,6,4,0), (3,9,3,0),(4,7,3,0),(5,4,3,0),(4,9,2,0),(5,6,2,0),(6,3,2,0),(3,10,1,0),(5,7,1,0), (6,4,1,0),(4,12,0,0),(5,9,0,0),(6,6,0,0),(7,3,0,0),(8,0,0,0).

These were found from a search of the [math]2^{27}[/math] subsets of the cube. A spreadsheet containing these statistics can be found here.

The extremal statistics are

(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(4,4,5,0),(4,6,4,0),(4,12,0,0),(8,0,0,0)

The sharp linear bounds are :

  • [math]2a+b+2c+4d \leq 22[/math]
  • [math]3a+2b+3c+6d \leq 36[/math]
  • [math]7a+2b+4c+8d \leq 56[/math]
  • [math]6a+2b+3c+6d \leq 48[/math]
  • [math]b+c+3d \leq 12[/math]
  • [math]a+2c+4d \leq 14[/math]
  • [math]5a+4c+8d \leq 40[/math]
  • [math]a+4d \leq 8[/math]
  • [math]b+6d \leq 12[/math]
  • [math]c+3d \leq 6[/math]
  • [math]d \leq 1[/math]

In normalized notation,

  • [math]8\alpha_0+ 6\alpha_1 + 6\alpha_2 + 2\alpha_3 \leq 11[/math]
  • [math]4\alpha_0+4\alpha_1+3\alpha_2+\alpha_3 \leq 6[/math]
  • [math]7\alpha_0+3\alpha_1+3\alpha_2+\alpha_3 \leq 7[/math]
  • [math]8\alpha_0+4\alpha_1+3\alpha_2+\alpha_3 \leq 8[/math]
  • [math]4\alpha_1+2\alpha_2+\alpha_3 \leq 4[/math]
  • [math]4\alpha_0+6\alpha_2+2\alpha_3 \leq 7[/math]
  • [math]5\alpha_0+3\alpha_2+\alpha_3 \leq 5[/math]
  • [math]2\alpha_0+\alpha_3 \leq 2[/math]
  • [math]2\alpha_1+\alpha_3 \leq 2[/math]
  • [math]2\alpha_2+\alpha_3 \leq 2[/math]
  • [math]\alpha_3 \leq 1[/math]

The c-statistics can also be found from an exhaustive search of Moser sets. The resulting Pareto sets and linear inequalities can be found on Sheet 8 of this spreadsheet

The following are Pareto optimal (3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(3,6,5,0),(4,4,5,0),(3,7,4,0),(4,6,4,0), (3,9,3,0),(4,7,3,0),(5,4,3,0),(4,9,2,0),(5,6,2,0),(6,3,2,0),(3,10,1,0),(5,7,1,0), (6,4,1,0),(4,12,0,0),(5,9,0,0),(6,6,0,0),(7,3,0,0),(8,0,0,0).

We first compute the Pareto-minimal counterexamples (i.e. the (a,b,c,d) which are larger than these but end up in the list after removing one from any of a,b,c,d) and eliminate each of them. From the n=2 inequalities (and using the planes xy1, xy2, and xxx, averaged over symmetries) we have a number of linear constraints

[math]a/4+c/6 \leq 2; a/4 + d \leq 2; b/6+c/6 \leq 2; b/6+d \leq 2; c/3+d \leq 2; a/2+b/6+c/6 \leq 4; b/3+c/3+d \leq 4[/math]

which eliminate many of these; these inequalities, by the way, together with the trivial inequality d \leq 1, are given as Y3 in the maple computations on the Wiki. It seems that the Pareto-minimal counterexamples that are consistent with these inequalities are (4,5,3,1), (3,11,1,0), (4,10,1,0), (5,8,1,0), (6,5,1,0), (5,7,2,0), (6,4,2,0), (4,8,3,0), (5,5,3,0), (6,0,3,0), (3,8,4,0), (4,7,4,0), (5,0,4,0), (3,7,5,0), (4,5,5,0), (3,6,6,0), so one has to eliminate 16 possibilities.

The distribution (4,5,3,1) cannot occur. Assume that we have set with these statistics. Note that no two points with two 2’s can have the same c-statistic. Without loss of generality we can assume that these points in the Moser set are (1,2,2), (2,1,2) and (2,2,1). Next we note that the points (1,1,1) and (1,1,3) cannot be in the set. If (1,1,1) it blocks (3,3,3), (1,3,3) and (3,1,3) and also (3,3,1). This leaves (1,1,3) and one of (1,3,1) and (1,1,3) and prevents four points with four twos being in the set giving a contradiction.

If (1,1,3) is in the set it blocks all points with no 2’s whose third coordinate is 1 except (1,1,1). However (1,1,1) is forbidden by the previous paragraph. The only way for there to be 4 points of type a is if all point of type a with a final coordinate equal to 3 are in the set. However this blocks all points of type b whose final coordinate is 3. The points (1,2,2) and (2,2,2) limit the the number of points of type b whose final coordinates are one and two to two each. This prevents 5 such points and we have another contradiction.

Now in order to have 4 points of type a we must have two among (1,1,1),(1,1,3),(1,3,3) and (3,3,3) but by the above we cannot have (1,1,1) or (1,1,3) so we must have (1,3,3) and (3,3,3). Now we must have one of (1,3,1) and (1,1,3) without loss of generality say (1,3,1) this will block all other points of type a except (1,3,3) recalling that (1,1,1) and (3,3,3) cannot be in the set so to have four points of type a we must have (1,3,3). (1,3,1) and (1,3,3) block (1,3,2). (3,3,1) and (3,3,3) block (3,3,2). (1,3,3) and (3,3,3) block (2,3,3). (1,3,1) and (3,3,1) block (2,3,1) Furthermore because of the formation of potential lines we can only have one of the following pairs: (1,1,2) and (1,3,2), (1,2,1) and (1,3,2), and (1,2,1) and (1,2,3) because of this and because we need 8 points of type b we must have both (3,2,1) and(3,2,3) to bring the total to 5 but this blocks both (1,2,1) and (1,2,3) and there is no way to get 5 points of type a. So this gives a contradiction and we are done.

The distribution (3,11,1,0) cannot occur as every point of type c lies on two lines whose other points are of type b with no points of type be lying on both lines. 11 points of type b implies the removal of one from the complete set of such points. Thus for every point of type c at most one line will be affected and the other will block the point of type c thus making it impossible.

The distribution (4,10,1,0) cannot occur because the two missing points of type b must lie on one face, the face that contains the element c, The face opposite it contains all its points of type b. Hence it at most two points of type a which must be on a diagonal. These points combined with the points of type b not on either of the two faces which must all be in the set since the missing points are on the original face block two points on a diagonal in the original face so the remaining possibilities for points on the original face must be on the other diagonal but c will limit the number of these to one. Thus the maximum of points of type a are one on the original face and two on the opposite face which gives a maximum of three but we must have four so this distribution cannot be realized.

The distribution (5,8,1,0) cannot be realized. The point of type c must be on a face. That face can contain at most two points of type a and two of type b to avoid lines with the point of type c which is the center point of the face. If the opposite face has all four points of type a then these will block all point of type b in that face the center point of the original face will block two more allowing a maximum of 6 which is less than 8 so the opposite face must have three points and the original face must have two.

Then there are only two points of type b on the original face the opposite face must have three of type a these must block two points of type b so there must be only two points of type b on the opposite face. To get eight points of type b the remaining four points in the center slice between the face and its opposite must have all four of its points of type b. These points are at the corners of the center slice and together with the three points of type a in the opposite face they block three of the slots for corner points for the original face, Thus the original face can only have one point of type a and together with the three of the opposite face we get a maximum of four points of type a and we are done.

The distribution (6,5,1,0) cannot be realized. The point of type c is the center point of a face which must have at most two points of type a so the opposite face must have four points of type a to give six. These four points block all points of type b in the opposite face. The presence of a center point in the original face blocks two points of type b so the original face has at most two points of type b. The remaining slots for points of type b are in the center at the corner positions of the center slice three of the must be filled and with the four points of the opposite slice they block three of the slots for points of type a in the original slice that leaves one open slot allowing a maximum of one point of type a and a total of at most 5 points of type a which is less than 6 so we get a contradiction and we are done.

A configuration of type (5,7,2,0) cannot be realized. First we see that two points of type c cannot have the same c-statistic. If they did we could slice the cube so there are two opposite faces with the center point filled. these forces would each have at most two points of type a for a maximum of 4 which is less than 5 which gives a contradiction. So we can assume without loss of generality the points of type c are (1,2,2) and (2,1,2).

Slice along the first coordinate then the square with first coordinate with value one has its center point and hence has at most 2 points each of type a and type b. If the slice with first coordinate 3 has four points of type a then it can have no points of type b and the maximum number of points of type b are four in the square with first coordinate 2 and two in the square with first coordinate 1 for a total of 6 which is less than 7 which gives a contradiction. So the square with first coordinate three must have at most three points of type a, and the square with first coordinate one must have at most two of type a. So the only way 5 such points can be realized is if the the square with first coordinate 1 has two points of type a and the square with first coordinate three has three such points.

The three points of type a in the square with first coordinate three block two points of type b so only two points of type b can be in that square. Since the center spot is filled in the square with first coordinate one that square can have at most two points of type b. Thus for there to be 7 points of type b there must be three in the square with first coordinate 2. Because of the presence of the point with coordinates (2,1,2) only one of the points (2,1,3) and (2,1,1) can be in the set which means that both (2,3,3) and (2,3,1) must be in the set. The presence of these points means that only two of the points (1,3,1), (3,3,1), (1,3,3), and (3,3,3) can be in the set. To get five points of type a three of (1,1,1),(1,1,3),(3,1,1) and (3,1,3) must be in the set. But these points can be divided into two pairs (1,1,1) and (3,1,3), and (1,1,3) and (3,1,3) which form lines with (2,1,2) thus there can be at most two of these points and at most 4 points of type a but we need 5 such points so we are done.

The configuration of type (6,4,2,0) cannot be realized. Assume we have such a configuration. The two points of type c in the configuration cannot have the same c-statistic. If they did we could slice along a coordinate and get the two side squares to have their center points thus they could each have only two points of type a, limiting the points of type a to 4 which is less than 6 giving a contradiction. Without loss of generality we can assume the two points of type c are (1,2,2) and (2,1,2).

Slice on the first coordinate then the square of points with first coordinate equal to one has its center points and so can only have two points of type a. The square with first coordinate three must have all of its points of type a to make six. The square with first coordinate three must have no points of type b as its four coordinates of type a blocks them. The square with first coordinate one can have at most two of these points because it has its center point. That means that the square with first coordinate two must have two such points.

The square with third coordinate equal to three contains all of its points of type a from above. In particular it contains (3,1,1) and (3,1,3). These together with point (2,1,2) block points (1,1,1) and (1,1,3) int the square with first coordinate equal to one. Since that square contains two points of type a it must contain the remaining two, (1,3,1) and (1,3,3). These two together with (3,3,1) and (3,3,3) block (2,3,1) and (2,3,3) in the square with two as its first coordinate. But the only other possibilities for a point of type b in this square are (2,1,1) and (2,1,3) but (2,1,2) prevents both from being in the set. Thus this square can only have one point of type b, the square with one as its first coordinate has its center point and can only have one and as noted above the square with first coordinate equal to three has no such points. So the maximum number of points of type b in the set is 3 which is less than 4 and we are done.

A Moser configuration with n=3 cannot have statistics (4,8,3,0). First we show that no two points of type c can have the same c-statistics. If they do then we can slice on a a coordinate and get the two side slices that both have their center slice then each can only have two points of type b which means to have 8 such points the center slice must have all four of its points of type b and hence can have no points of type c, but since the two side slices only each have one there are less than three points of type c and we are done. Without loss of generality we than can take the three center points to be (1,2,2), (2,1,2) and (2,2,1). We slice along the first coordinate.

The square with its first coordinate equal to 3 cannot have four points. If it did it would have no points of type b and the other two slices would have to have all of their points of this type but the square with first coordinate equal to one has its center point equal to one has its center point so it can only have two and we get a contradiction. The square with its first coordinate equal to three cannot have three points if it did it would have at most two points of type b or a line would be formed. The square with its first coordinate equal to one would have at most two. So for there to be 8 such points there must four such points in the square with first coordinate equal to 2 but that would block all points of type c in the square and it has (2,1,2) and (2,2,1) by the above. So the square with its first coordinate equal to three must have at most two points of type a and since the square with first coordinate equal to one can have at most two due to the presence of its center point it must have two and the other side slice must have two.

Now the square with its first coordinate equal to three cannot contain the point (3,1,1) because along with the points (2,1,2) and (2,2,1) it would block (1,1,3) and (1,3,1) but this would limit the possible points of type a to those on a diagonal which due to the presence of the center square can only one point which would give a contradiction to the conclusion of the above paragraph. The square with its first coordinate equal to one cannot contain the point (1,1,1) because along with the points (2,1,2) and (2,2,1) it would block (3,1,3) and (3,3,1) this would force the two points of the square with first coordinate equal to three to have its points on the other diagonal but this then it would have to have both points on this diagonal and hence contain the point (3,1,1) which leads to a contradiction as noted previously.

The square with first coordinate equal to one must have two points and hence it must have a point on each diagonal as the center point prevents the presence of two. Since (1,1,1)is forbidden by the above it must have (1,3,3). If the square with first coordinate equal to three contained (3,3,3) then with (1,3,3) it would block the point (2,3,3) the points (2,1,2) and (2,2,1) would block an additional point of type b in the center slice and the center slice would have at most two points of type b, due to the presence of its center point the square with first coordinate equal to one would have at most two such points. So to get a total of 8 the square with first coordinate equal to three would have to have all four of its points of type b but together with the point (3,3,3) these would block (3,1,3) and (3,3,1).

Since (3,1,1) is not possible due to the above the square with first coordinate equal to three would have at most one point of type a which contradicts the fact it must have two proved above. So the point (3,3,3) cannot occur in this slice. Since we have already proved it cannot contain (3,1,1) and it must have two points of type a it must contain (3,1,3) and (3,3,1).

Now the square with first coordinate equal to one contains (1,1,1) by the above it cannot contain (1,3,3) and since it must have two points of type a it must include one of (1,3,1) or (1,1,3). Since by the above the other side slice must contain both of the points (3,3,1) and (3,1,3) one of the points (2,3,1) or (2,1,3) will be blocked. The points (2,1,2) and (2,2,1) would block an additional point of type b in the center slice and it will have at most two points of type b. The square with first coordinate equal to one will have at most two points of type b. So the square with first coordinate equal to three must contain all of its points of type b for there to be 8 such points. With (2,1,2) and (2,2,1) they would block the points (1,1,2) and (1,2,1). Since the square with first coordinate equal to one must have first coordinate equal to one must have two points of type b it must contain (1,3,2) and (1,2,3) but with (1,3,3) they would block both (1,3,1) and (1,1,3) but the square with first coordinate equal to one must have one of these points as shown previously so are done.

A Moser set with statistics (5,5,3,0) cannot be realized. First we note that two of the points of type c cannot have the same c-statistic. If they did then we could slice on the coordinate where the two points do not equal 2. Then both side squares would have their center points and the number of points of type a on both would be limited to 2 which is less than 5. So without loss of generality we can assume the three points of type c are (1,2,2), (2,1,2) and (2,2,1). We cut along the first coordinate. If the the square with first coordinate equal to three has four points of type a then these points will block all the points of type b in the square and since the other side square has its center point it can only have two points of type b there must be three points of type b in the center square of the slice. But since that square contains (2,1,2) and (2,2,1) it cannot contain (2,1,1) as then it would block two points of type a and we need three of these in the center slice. So the points in the center slice of type a must be the other three besides (2,1,1) and in combination with the four points of type a in the square with first coordinate equal to three they block all points in the square with first coordinate equal to one except (1,1,1). But one together with the points (2,1,2) and (2,2,1) it blocks two points of type a in the square with first coordinate equal to three but that square must have all of its points of type a and we get a contradiction and we are done.

Then since the square with first coordinate equal to one has its center point and only contains at most two points of type a the other side slice must have at least three points of this type. But since we have ruled out four of these points above the only distribution left is 2 points of type a in the square with first coordinate equal to 1 and three of this type in the other side slice. If the square with first coordinate equal to one has contains the point (1,1,1) the with (2,1,2) and (2,2,1) it will block two points of type a in the other side slice which gives a contradiction.

Since the square with first coordinate equal to 1 must have two points of type a it must have at least one on each diagonal so it must have the point (1,3,3). Now if the square with first coordinate equal to three contains (3,1,1) then with the points (2,1,2) and (2,2,1) it will block two points of type a in a diagonal on the other side cube but that will limit the number of points of type a in that cube to one which gives a contradiction so it cannot contain the point (3,1,1) and so it must contain all other points of type a. These include (3,3,3) which together with (1,3,3) which we have also shown to be in the Moser set blocks (2,3,3). Now if the center slice contains (1,2,2) it will block all points except (2,3,3) which is already blocked so the only way the center slice can have two points with one 2's is if has (2,1,3) and (2,3,1) but these together with (3,1,3) and (3,3,1) block all the points of type a in a diagonal in the square with coordinate one. This forces two points in the other diagonal in order to have two points of type a but this forms a line with the center point of that square. So there is only one point of type b in the center slice.

The square with coordinate 1 can only have at most two points of type b since it has its center point. The other side slice can only have two as we know what the points of type a in that square are and the block two points of type b. The center square can only have one point of type b. Given the above the only way 5 points of type b can be reached is for each of the side slices to have two such points and the center slice to have one. The square with first coordinate three must have the points (3,1,2) and (3,2,1) as all the others are blocked by points of type a. These with the points (2,1,2) and (2,2,1) block the points (1,1,2) and (1,2,1). For the square with first coordinate one to have two points of type b it must contain (1,3,2) and (1,2,3) which together with the point (1,3,3) block the points (1,3,1) and (1,1,3) which together with the fact already shown that (1,1,1) cannot be in the set limits the number of points of type a in the square with first coordinate equal to one to one which eliminates the last case and we are done.

A Moser set with statistics (6,0,3,0) cannot be realized. First we note that two points of type c cannot have the same c-statistic. If they did we could slice along the coordinate not equal to 2 in both points and get two side slices with their center points both would have two or less points of type a but we must have six such points contradiction. So without loss of generality we can take these points to be (1,2,2), (2,1,2) and (2,2,1). We slice along the first coordinate. If the square with first coordinate equal to three has four points of type a then it must contain (3,1,1) this with (2,1,2) and (2,2,1) blocks (1,1,3) and (1,3,1) but this forces the points of type a in the square with first coordinate equal to one to lie and a diagonal and due to the presence of the center point limits there number to one and the total number of such points to 5 but we must have six so we have a contradiction, so the square of side three must have three or less such points but the other side square is limited to two such points due to the presence of its center point. This limits the number of points of type a to 5 but we need six so we have disposed of all cases and this set cannot be realized.

A Moser set with statistics (3,8,4,0) cannot be realized. Since there are 4 points of type c two must have the same c-statistic. Without loss of generality say these points are (1,2,2) and (3,2,2). Slice along the first coordinate. The two side squares can only have two points of type b each since they both have their center points. To have 8 points of type b the center slice must have all four of its corner points but these block all points of type c in the center slice and since there are a maximum of two such points in the side slices we have less than four and we are done.

A Moser set with statistics (4,7,4,0) cannot be realized. Since there are 4 types of type c two must have the same c-statistics. Without loss of generality say these points are (1,2,2) and (3,2,2). Slice along the first coordinate. There are at most two points of type b in both side slices for a maximum total of four. Thus there must be three points of this type in the center slice which means that it must have three of its corner points in the set. The only possible spots for points of type c in the central slice are the two spots adjacent to the missing corner point of the center slice. These must be filled as we have 4 points of type c and only two in the side slices. Without loss of generality say the missing corner point is (2,1,1) and the two adjacent points of type c are (2,2,1) and (2,1,2). Then of the pairs (3,1,3) and (1,1,3), (3,1,3) and (1,1,3), (1,3,1) and (3,3,1) the set can have only one point in each pair because the three corners in the center slice would form a line with any complete pair. Thus the set must have one of the two points (1,1,1) or (3,1,1). If it has (1,1,1) with (2,2,1) and (2,1,2) it would block all the points of type a in a diagonal in the square with first coordinate equal to 3. But that would force points of type a in that square to a diagonal and limit there number to one limiting the total number of such points in the set to three giving a contradiction. If it has (3,1,1) with (2,2,1) and (2,1,2) it would block all the points of type a in a diagonal in the square with first coordinate equal to 1. But that would force points of type a in that square to a diagonal and limit there number to one limiting the total number of such points in the set to three giving a contradiction. So we have exhausted all cases and we are done.

A Moser set with statistics (5,0,4,0) cannot be realized. Since there are 4 points of type c two must have the same c-statistic. Without loss of generality we can say these points are (1,2,2) and (3,2,2). Slice along the first coordinate then the two side slices have their center points and hence have at most 2 points of type a each but we need five such points so have a contradiction and we are done.






n=4

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

Proof that [math]c'_4 \leq 43[/math]: When e=1 (i.e. the 4D set contains 2222) then we have at most 41 points (in fact at most 39) by counting antipodal points, so assume e=0.

Define the score of a 3D slice to be a/4+b/3+c/2+d. Observe from double counting that the size of a 4D set is the sum of the scores of its eight side slices.

But by looking at the extremals we see that the largest score is 44/8, attained at only one point, namely when (a,b,c,d) = (2,6,6,0). So the only way one can have a 44-point set is if all side slices are (2,6,6,0), or equivalently if the whole set has statistics (a,b,c,d,e) = (4,16,24,0,0). But then we have all the points with two 2s, which means that the four "a" points cannot be separated by Hamming distance 2. We conclude that we must have an antipodal pair among the "a" points with an odd number of 1s, and an antipodal pair among the "a" points with an even number of 1s. By the symmetries of the cube, we may take the a-set to then be 1111, 3333, 1113, 3331. But then the "b" set must exclude both 1112 and 3332, and so can have at most three points in the eight-point set xyz2 (with x,y,z=1,3) rather than four (to get four points one must alternate in a checkerboard pattern). Adding this to the at most four points of the form xy2z, x2yz, 2xyz we see that b is at most 15, a contradiction. [math]\Box[/math]

Given a subset of [math][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.

One of the (6,24,12,0,0) solutions is [math]\Gamma_{220}+\Gamma_{202}+\Gamma_{022}+\Gamma_{112}+\Gamma_{211}[/math] (i.e. the set of points containing exactly two 1s, and/or exactly two 3s). The other seven are reflections of this set.

There are 2,765,200 41-point solutions, listed here. The statistics for such points can be found here. Noteworthy features of the statistics:

Statistics for the 41-point, 42-point, and 43-point solutions can be found here.

If a Moser set in [math][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. (A more refined analysis can be found at found here.)

We have the following inequalities connecting a,b,c,d,e:

  • [math]4a+b \leq 64[/math]: There are 32 lines connecting two "a" points with a "b" point; each "a" point belongs to four of these lines, and each "b" point belongs to one. But each such line can have at most two points in the set, and the claim follows.
  • This can be refined to [math]4a+b+\frac{2}{3} c \leq 64[/math]: There are 24 planes connecting four "a" points, four "b" points, and one "c" point; each "a" point belongs to six of these, each "b" point belongs to three, and each "c" point belongs to one. For each of these planes, we have [math]2a + b + 2c \leq 8[/math] from the n=2 theory, and the claim follows.
  • [math]6a+2c \leq 96[/math]: There are 96 lines connecting two "a" points with a "c" point; each "a" point belongs to six of these lines, and each "c" point belongs to two. But each such line can have at most two points in the set, and the claim follows.
  • [math]3b+2c \leq 96[/math]: There are 48 lines connecting two "b" points to a "c" point; each "b" point belongs to three of these points, and each "c" point belongs to two. But each such line can have at most two points in the set, and the claim follows.

The inequalities for n=3 imply inequalities for n=4. Indeed, there are eight side slices of [math][3]^4[/math]; each "a" point belongs to four of these, each "b" point belongs to three, each "c" point belongs to two, and each "d" point belongs to one. Thus, any inequality of the form

[math] \alpha a + \beta b + \gamma c + \delta d \leq M[/math]

in three dimensions implies the inequality

[math] 4 \alpha a + 3 \beta b + 2 \gamma c + \delta d \leq 8M[/math]

in four dimensions. Thus we have

  • [math]8a+3b+4c+4d \leq 176[/math]
  • [math]2a+b+c+d \leq 48[/math]
  • [math]14a+3b+4c+4d \leq 244[/math]
  • [math]4a+b+c+d \leq 64[/math]
  • [math]3b+2c+3d \leq 96[/math]
  • [math]a+c+d \leq 28[/math]
  • [math]5a+2c+2d \leq 80[/math]
  • [math]a+d \leq 16[/math]
  • [math]b+2d \leq 32[/math]
  • [math]2c+3d \leq 48[/math]

Cubes also sit diagonally in the 4-dimensional cube. They may have coordinates xxyz, where xx runs over (11,22,33) or (13,22,31), while y and z run over (1,2,3). These cubes have a different distribution of 2s than the ordinary slices: (a,b,c,d,e) = (8,8,6,4,1) instead of (8,12,6,1,0) for the side slices and (0,8,12,6,1) for the middle slices. So a different set of inequalities arise. Apply the same procedure as described above for n=3: (Run through the [math]2^{27}[/math] subsets of the cube; identify those without combinatorial lines; calculate their statistics in the new xxyz arrangement; retain the Pareto-optimal statistics; retain the extremal statistics; find what inequalities they satisfy.) The inequalities that arise are

  • [math]2a+b+2c+2d+4e \le 24[/math] and
  • [math]4a+b+2c+2d+4e \le 32[/math]

within the 3D cube, which when averaged becomes

  • [math]4a+b+2c+4d+16e \le 96[/math] and
  • [math]8a+b+2c+4d+16e \le 128[/math]

for the 4D cube. A spreadsheet containing these statistics can be found here on sheet 2. Other sheets of this spreadsheet contain results for a 3D cube sitting diagonally in a 5D, 6D or 7D cube. Notice the similarity of the equations that arise for the xxxyz, xxyyz and xxxxyz diagonals. Also notice the (a,b,c,...) statistics for the xxxxyyz diagonals have the same Pareto sets and linear inequalities as the cube's c-statistics.

The c-statistics for the 3D cube are bounded by a set of 20 inequalities. By considering the fourteen ways a 3D cube sits in the 4D cube, the result is a set of 239 inequalities for the 4D cube's c-statistics. However, all 239 inequalities are satisfied by a 44-point solution given by c(xxxx) = c(2xxx) = c(22xx) = 4, and permutations.

If a Moser set for n=4 has 4 or more points with three 2’s it has less than 41 points.

We first note that, as shown above, if a Moser set includes 2222, it has at most 39 points. So any Moser set with 41 or more points has no point with four 2's. So e=0.

If a Moser set for n=4 has six or more points with three 2’s it must have less than 41 points. We have 5 points with exactly three 2’s and one coordinate not equal to 2. That gives 5 values not equal to 2 and four coordinates one coordinate must have the coordinate value not equal to 2 from 2 of the five points. One value must be three the other one. We slice at this point and get two cubes with the center point filled which by the n=3 section of the wiki on Moser sets must have 13 points or less. Since there are six or more points with three 2’s the center slice must have the remaining four or more. Now if we have 41 or more points it must have a center slice equal to 15 points or more. However by the Pareto-optimal statistics in the section n=3 of the wiki we see that a cube with c greater than three can have value at most 14. So there at most 13+14+13=40 points and we are done.


On the 3D section of the Wiki, we have the inequalities

[math]4\alpha_0+4\alpha_1+3\alpha_2+\alpha_3 \leq 6 (1)[/math]
[math]7\alpha_0+3\alpha_1+3\alpha_2+\alpha_3 \leq 7 (2)[/math]

which when averaged on middle slices gives the 4D inequalities

[math]b/8 + c/6 + 3d/8 + e \leq 6 (3) [/math]
[math]7b/32 + c/8 + 3d/8 + e \leq 7. (4) [/math]

Averaging (1) on side slices similarly gives

[math]a/4 + b/8 + c/8 + d/8 \leq 6. (5) [/math]

Computing 9/4 *(3) + (4) + 4*(5) gives

[math]a+b+c+d+e \leq 89/2 - 23 d / 32 - 9 e / 4[/math]

and so if a+b+c+d+e ≥ 41 then we must have d ≤ 4.

Lemma 0 I used maple to find the integer solutions to d=4, a+b+c+d+e ≥ 41 that were consistent with all known inequalities. They are: (6,16,15,4,0,0), (7,16,14,4,0,0), (7,17,13,4,0,0), and (7,18,12,4,0,0) [math]\square[/math]

Lemma 1 If a Moser set has 4 points with 3 2’s and its size is 41 or more then it must not have two pairs of points, each pair with three twos and the same c statistic.

Proof Assume the Moser set has two such pairs. Without loss of generality, they are 1222,3222, and 2122,2322. Cut at the first coordinate. Both side slices include their central point, and so must have 13 or fewer points. So the center slice must have 15 or more points.

Recall the two points of the form say 2×22 which are in the center cube. We can slice at the second coordinate, and the side squares have the center spot occupied. Consider the points in the center cube with one coordinate equal to 2 in the cube (and two coordinates in the 4D configuration). There must be at most 8: two each from the side squares since the center spot is occupied and possibly all 4 from the center square.

From the Pareto optimal statistics in section n=3 of this Moser wiki, and the fact the center cube has 15 points and has c=2, the statistics of the center slice must be (4,9,2,0). But from the above we have at most 8 points with one coordinate equal to 2, so we are done. [math]\square[/math]

Lemma 2 If a Moser set has 4 points with 3 2’s and its size is 41 or more then it must not have any set of two points with three twos with the same c statistic.

Proof We start by assuming we have such a pair. By Lemma 1, there is only one such pair. Suppose they are 1222 and 3222. Without loss of generality, the other two points with three 2s are 2122 and 2212.

Slice on the first coordinate. We get two slide slices with 13 or less points, so the center slice must have 15 points or more. It cannot have sixteen points as it has at least one point with three twos in the configuration, which becomes a point with two 2’s in the central slice; that prevents the only realization of sixteen points from occurring since it has no points with two 2’s. Further, from the Pareto optimal statistics for Moser sets for n=3 at the Moser wiki, we see its distribution must be (4,9,2,0).

Slice the center cube along the second coordinate. One outside slice, with the point 2122, must have that point at its center, and hence has a maximum of 5 points and at most two points with two twos. The center slice 22xy has no center point and one point with three twos (2212). Because of this the center slice can have at most 4 points and it can contain at most three points with two twos. So the remaining slice 23xy must have all four points with two twos.

The two side slices 1xyz and 3xyz, since they have thirteen points including the central point, must have statistics (3,6,3,1) or (4,6,2,1) each. So each of the diagonals connecting points with one two in the slice overall must have at least one point and that means that if we cut each of the side cubes along the same coordinate we cut the center cube we get there are a total four points with one two in the two side slices and if they are divided 2, 2. We can pick two pairs one point in a pair in each slice such there is a correspondence between the pairs of the points in that the coordinates j are equal. If they are divided 1,3 again there is a correspondence between the one point and one of the three such that the coordinates except for j are equal. Now in the case of division 2-2 or 1-3 in the above we get a contradiction as follows be we have the above correspondences plus another one in the cube j=1 since it has its center spot occupied combined they give there are one or two points with the same center coordinates except for i in the cube j equals three but the center square of j=3 contains all its points with two two’s which block all lines between points with identical coordinates except for j and i equal to three and we get a line so we must have a four zero split, and since that split must have four points in each side cube and the center slice contains of j=3 contains all of it line so must have at most four of these point and the cube j=1 contains its center point and contains at most four of these points, these points must lie on either the squares i=1,j=3 and i=3,j=1 or i=1,j=1 and i=3,j=3 but then they will form a diagonal cube with i=2,j=2 which contains one point with three twos and this forms a line with one of the pairs of these four points on the diagonal cube and we are done. [math]\square[/math]

Lemma 3 If a 4D Moser set has more than 40 points and 4 points with three 2’s it must have the following statistics: (6,16,15,4,0) or (7,16,14,4,0).

Proof We have from Lemma 0(post 1113) that only the following statistics are possible: (6,16,15,4,0), (7,16,14,4,0), (7,17,13,4,0), and (7,18,12,4,0). and from Lemma 2 (If a Moser set has 4 points with 3 2’s and its size is 41 or more then it must not have any set of two points with three twos with the same c statistic. )

We look at the number of points with one two. Each one must appear in exactly one of the four possible middle cubes from the four possible coordinate slices. If the number of these points is greater than 16 then there must be one such middle cube with five of these points. By Lemma 2 we know that there must be three points with three twos in the cube as well as otherwise there would be two with the same c statistic. We check the pareto optimizers from the wiki and find there is only one satisfying (5,p,3,q) and that is (5,4,3,0). So the center has 12 points and distribution (5,4,3,0). The side cube with the remaining point with three twos has thirteen points, and distribution (3,6,3,1) or (4,6,2,1). The remaining side cube must have 41-12-13=16 points and hence distribution (4,12,0,0)

But this gives a total of 5 + 12+ 6 points with one two which is too many for any of the possible sets of statistics. So the only distribution that is possible is the one with 16 points with one two namely (6,16,15,4,0)or (7,16,14,4,0) and we are done. [math]\square[/math]

Lemma 4 A 4D Moser set with 4 points with three threes has 40 points or less.

ProofBy the reasoning of Lemma 3 each slice of a Moser set must have 4 points or less with one 2 in the center cube. Since from the above we must have 16 of these points and each point only appears in one coordinate slice the division must be exactly 4 4 4 4.

No extremal slice can have 16 points; if so it has distribution (4,12,0,0) by the Pareto optimizers and hence since by the above the distribution of points with one two must include exactly 4 in the center slice we have a total of 16 and hence none in the other extremal set which contains its center point then by looking at the Pareto optimizers we see that we have to remove 6 points from a thirteen point or four points from a 11 point set in order to realize this and this gives at most 8 points then we have 8 + 16 points for the two side slices and we need 17 points in the center slice which cannot be realized.

So we have the side slice without the center point must have 15 points or less. Next we show that the center slice must have 14 points or less. By the above it must have 4 points with one two but this gives a distribution of (4 x y z) note the points with one two in this slice count as points with no 2 in the internal statistics of the cube. The center slice must have y = to three because we have already established that there are no two points with the same c-statistic so we must have statistics (4 x 3 0) which looking at the Pareto optimizers must have 14 points or less.

Now we recall that now two points with three twos can have the same c-statistic then we can assume without loss of generality that all such points have the coordinate not equal to 2 equal to one. Then we look at the points (1 1 3 2) (1 3 1 2) (3 1 1 2) Only one of these can be in the Moser set because otherwise a line will be formed with the points with three 2’s under the above assumption. Now that means that under one of the coordinate cuts must have a diagonal connecting two points with one two. empty and that means that it can have at most 5 points with one two since each of the other diagonals have at most one point. That implies that the there must be at most 12 points in this cube since the other points have at most 14 and 15 points we must have the distribution 12 14 15 or we will have less than 41 points. Then the statistics of the cube must be (3 5 3 1) or ( 4 5 2 1) and the other side cube must have statistics ( 3 9 3 1) (4 9 2 1) or (4 11 0 0) or (3 12 0 0) but if we have the statistics (6 15 16 4 0) we can only have the choice (3 5 3 1) and the choices (3 9 3 1) and (3 12 0 0) or we would have more than 6 points with no 2’s and we can discard the possiblity (3 12 0 0) because it would give more than 16 points with one 2. So we must have (3 5 3 1) and (3 9 3 1) and this gives center slice (2 9 3 0) but this has less than four points with two twos and will force another slice to have more than 4 and so we are done.

If we have the statistics: (7,16,14,4,0) then recall we must have the distribution 12 14 15 or we will have less than 41 points. Also recall that the statistics of the side cube with its center point must be (3 5 3 1) or ( 4 5 2 1) then the statistics of the other side cube must be ( 3 9 3 0) (4 9 2 0) or (4 11 0 0) or (3 12 0 0) Since the statistics are (7,16,14,4,0) we must have one of the following four pairs of side cubes:

(3 5 3 1) and (4 9 2 0) (3 5 3 1) and (4 11 0 0) ( 4 5 2 1) and (3 9 3 0) ( 4 5 2 1) and (3 12 0 0)

If we have (3 5 3 1) and (4 9 2 0) forces the middle slice to have statistics (0 2 9 3) which has less than 4 points with one two hence forces another slice to have more than four which gives a contradiction

If we have (3 5 3 1) and (4 11 0 0) forces the middle slice to have statistics (0 0 11 3) which has less than 4 points with one two hence forces another slice to have more which gives a contradiction

If we have ( 4 5 2 1) and (3 9 3 0) forces the middle slice to have statistics (0 2 9 3) which has less than 4 points with one two hence forces another slice to have more which gives a contradiction

If we have ( 4 5 2 1) and (3 12 0 0) has 17 points with one two which contradicts the statistics of the entire set. So none of the cases work and we are done. [math]\square[/math]

If a Moser set for n=4 has 1 or more points with three 2’s it has less than 43 points.

Proof: Assume that such a set exists. We have already proved that it contains no point of type e. So we can assume that. This set does not contain such a point.

This set since it has 43 points must have at least 18 points of type c. See http://michaelnielsen.org/polymath1/index.php?title=Maple_calculations. Since there is a point of type d we can slice the set so that one side cube has its center slot filled and so must have 13 points or less. The other cubes must sum to 30. The other side cube must not have 16 points since then by the list of Pareto optimizers its distribution would be (4, 12, 0, 0) and it would contribute no points of type c to the Moser set. Then the other side cube could contribute at most 3 points of type c. So for there to be 18 the central cube must contribute 15 which is not possible.

This means the center cube must have at least 15 points for there to be a total of 43. Now we will show that the configuration has at most one point of type d.

Assume that there are more than one. Then no two can have the same c-statistic or we can slice and get two side cubes with center points filled which give each a total of at most 13 points which would force the center slice to have 17 points which is impossible.

So we can get a slice that has one point of type d in a side cube and the other in the center cube. That means the side cube must have its center slot filled and it must have 13 points, the center cube contains one point of type c(type d in the overall Moser set) and thus must have 15 points or less as the only Pareto optimal set with 16 points contains none of type c. The remaining side cube must have 15 points or more.

The only distribution which works is 13 points in one side cube and 15 in the other two. Now since there is no Moser three set with one point of type c and a total of fifteen points. There must be at least two points of type c in the center cube in giving at least 3 for the overall set. From this and the Pareto optimizers we see that the distribution of the side slice with 13 points must be (3, 6,3,1) or (4,6,2,1) and of the center and side cubes (3,9,3,0) or (4,9,2,1) This means we have at least 17 points of type b. Every one of these points appears in the center cube in exactly one slice. There are 4 slices so there is one slice with 5 of these points in the center cube.

Now since we have at least three points of type d and we can’t have two of them in the side cubes we must have at least one in the center cube. This means that that cube must have at most 13 points. No if either of the side cubes has a point of type d its total will be at most 13 and the third cube must be 17 which gives a contradiction so all three center points must be in the center cube.

This lowers the total of the center cube to at most 12. This in fact forces the center cube to have distribution (5,4,3,0) or that distribution with one point removed.. This also forces the sides cubes to have total 31. This means that one will have 16 points and thus have statistics (4,12,0,0). The other will have 15 points and hence have 9 point s of type b giving a total of 25 of type b if the center cube has 12 points.

If it has 11 points then it will have the distribution (5, 3, 3 as that is the only subset of the Pareto optimizers satisfying the conditions. Then both side cubes will have 16 points which means that neither will contain points of type c which means that there must be 18 points of type c in the center cube which gives a contradiction.

So the center cube has 12 points and the total number of points of type b is 25 so since each point of type b occurs in one slice and there are four possible slices there must be one slice with 6 points of type b since we get a contradiction as above if there are less than 11 points. So there must be 12 points the only distribution that works is (6,6,0,0).

But then the middle slice contains none of the points of type d and one must be on a slide slice lowering the total of that slice to 13 or less forcing the other side slice to be above 18. So we have reached a contradiction in assuming that we have more than one point of type d the remaining case is that there is exactly one point of type d.

If there is exactly one slice so it is in one of the side cubes. That cube will have at most 13 points. The other side cube can have at most 15 points because if it has 16 it will have no points of type c and the other side cube will have at most 3 of type c and since the total size is 18 we must have 15 of type c in the center cube which gives a contradiction.

So the center cube must have at least 15 points. Furthermore since it has no points of type c since we have our single point of type d in one of the side cubes it must be a subset of (4,12,0,0) and must have at least 11 points of type b. To avoid forming a monochromatic line the points of type b of the side cubes must sum to 13 or less.

If the side cube without the center cube has 15 or more points it must have at least 9 points of type b forcing at most 4 of type be in the other side cube which forces its size to 11 or less which forces both of the other cubes to have more than 16 which contradicts the fact that no side cube can have more than 15 points proved above. So the side cube without the center point must have at most 14 points. The other side cube can have at most 13 points so the center cube must have 16 points and since it cannot have 17 we have the distribution (13,16,14).

This forces the center cube to have all its points of type b there are 12 of these. From the Pareto optimizers the side cube with 13 points with the center spot filled must have at exactly 6 points of type b. We will get a line unless the remaining side cube has more than six points of type b. The only distributions with at least 14 points that satisfies this is (3,6,5,0) and (2,6,6,0). From the Pareto optimal statistics we can limit the possibilities for the other cubes of slice and get the following:

So one center cube has statistics (4,12,0,0) the side cube with thirteen points and the center filled has distribution (3,6,3,1) or (4,6,2,1) and final cube has statistics (3,6,5,0) and (2,6,6,0). So the statistics of the Moser set must be (6,16, 21,1,0), (7,16, 20,1,0), or (8,16, 19,1,0). So the exact number of points of type b must be 16.

Now slice the set so that the point of type d is in the center slice. Then the center cube contains a point of type c and has at most 10 points of type b. It must have at most 14 points that means that the other two cubes must total at least 29. If one is 16 then it contains no points of type c and since the center cube contributes at most 10 the other side cube must have 8 such points. This gives a contradiction which means one side cube must be of size 15 and the other of size 14. The one of size 15 contains at most three of type c. The center contains at most 10 so the remaining cube must have 5 or more.

Again from the Pareto statistics we can limit the statistics of each of the cubes the side slice of size 15 must be (3,9,3,0) (4,9,2,0) otherwise it will have no points of type c and we will get a contradiction as above. The side slice must have at least 5 points of type c and at least 14 points so it must have statistics (2,6,6,0) or (3,6,5,0). The center slice must have 1 point of type c in its statistics and at least 9 points of type b in its statistics which are statistics of type in the whole Moser set. Thus it statistics are either (4,9,1,0) which is a Pareto optimal set with one point removed or (3,10,1,0). Then the distributions are either (5, 19,18,1), (5,18,19,1), (6,19,17,1,0),(6,18,18,1,0), (7,19,16,1,0) or (7,18,17,1,0). Since we must have 18 points of type c the only possible distributions are (5, 19,18,1), (5,18,19,1) or (5,18,19,1). But we have shown the exact number of points of type b must be 16 which is a contradiction and we are done. So every Moser set with d=4 cannot contain a point with three two’s.

If a Moser set for n=4 has 3 or more points with three 2’s it has less than 42 points.

If a Moser set has 42 points it will have less than three points with three twos. Assume that there is a set with more than two such points. If it has 4 or more it will have two with the same c-statistic. We can slice along that coordinate and get two side cubes with the center cube occupied which implies they have 13 points or less. The center cube will have one such point and hence have 15 points or less. So the total number of points is 41 or less which is a contradiction. So we can assume we have exactly three such points. We can show no two of these points can have the same c-statistic. If they do again we can cut along the coordinate and the two side cubes will have their center slots occupied and will have 13 or less points and the center cube will have one such point and will have less than 16 points so the total will be less than 42. So we can assume that no two have the same c-statistic and without loss of generality we can assume that the points are (1,2,2,2), (2,1,2,2) and (2,2,1,2).

Look at the points (1,1,2,2), (2,1,1,2) and (1,2,1,2). No two of them can be in the Moser set because they would form a line with (1,2,2,2), (2,1,2,2) and (2,2,1,2). So one pair must be missing. That pair will have a one say in coordinate a in common and hence. Will be in one side cube of a side slice resulting from the cut along coordinate a that contains its center point. This will mean that the side slice has 5 points of type b and hence contains 12 or less points.

If it contains 9 points then the other two slices must have at least 33 and that will result in one having 17 which leads to a contradiction. If it has 10 points the other two must have at least 32 which means that both must have 16 points which means that the center cube and the other side cube have no points with three 2’s which mean there are less than three such points which leads to a contradiction.

If it has 11 points then the remaining two cubes must total 31. Then one must total 16 and the other 15. If the center cube totals 16 then there can be no points with three 2’s inside it and since only one such point can be in either of the side cubes we have at most 2 which gives a contradiction. If one of the side cubes totals 16 it must have 12 points with one two. The center cube if it totals 15 must contribute three such points and the side cube with 12 points must have statistics (4,4,3,1) or one of the following with two points removed: (3,6,3,1) or (4,6,2,1). In any of these cases it will have at least 4 points with one two. This gives at least 19 points with one 2 in the set.

So we can assume the side cube with center spot occupied has exactly 12 points and it contain at least 5 points with one two..If the other side slice contains its center point the two side slices will total less than 26 and the total in the set will be less than 42 and we will have a contradiction. So the center cube must have two points with three 2’s in the set and hence it will have at most 15 points and it will have at least three points with one two. If it has 16 points it will contain no points with three 2’s and since the side slices can contain at most one we will have a contradiction. So it must contain exactly 15 points which implies the other side slice must have at least 15 and hence 9 points with one two. So we have at lest 5+9+3 points with one two in the set. In the case in the above paragraph we have at least 19 so we can assume at least 17.

Now since each point with one two occurs in exactly one of the cuts along a coordinate in the center slice there must be one slice with 5 points in the center slice with exactly two 2’s and from the Pareto optimizers we see the center slice can have at most 14 points. If both side slices contain their center points then the total number of points would be less than 13+13+14 which is less than 42. So only one can have its center point and the center cube must have two such points and again by the Pareto optimizers it must have at most 13 points. Then if one of the side cube had its center spot occupied it would have at most 13 points and since the center cube has at most 13 points the other side cube must have 16 points. Then it would have no points of type c. The side cube could contribute at most 3 such points and the center cube at most 6 but we must have 12 such points see http://michaelnielsen.org/polymath1/index.php?title=Maple_calculations. So we have a contradiction so all three points with three 2’s must be in the center cube that means from the Pareto optimizers that the center must be a subset of (5,4,3,0) and have at most 12 points. That means that the total of the side cubes must be 30. If one is 16 it has no points with two 2’s in the set. The center has at most 4 such points that means since there are at least 12 the other cube must have at least 8 but by the Pareto statistics a 14 point set can have at most 6 such points. If both are fifteen they can contribute at most 6 such points the center cube can contribute at most 4 and we have a contradiction.


Proof that [math]c'_5[/math] = 124

Let (A,B,C,D,E,F) be the statistics of a five-dimensional Moser set, thus (A,B,C,D,E,F) varies between (0,0,0,0,0,0) and (32,80,80,40,10,1).

There are several Moser sets with the statistics (4,40,80,0,0,0), which thus have 124 points. Indeed, one can take

  • all points with two 2s;
  • all points with one 2 and an even number of 1s; and
  • 11111, 11333, 33311, 33331. Any two of these four points differ in three places, except for one pair of points that differ in one place.

For the rest of this section, we assume that the Moser set [math]{\mathcal A}[/math] is a 125-point Moser set. We will prove that [math]{\mathcal A}[/math] cannot exist.

Lemma 1: F=0.

Proof If F is non-zero, then 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. [math]\Box[/math]

Lemma 2: Every middle slice of [math]{\mathcal A}[/math] has at most 41 points.

Proof Without loss of generality we may consider the 2**** slice. There are two cases, depending on the value of c(2****).

Suppose first that c is at least 17; thus there are at least 17 points of the form 222xy, 22x2y, 22xy2, 2x2y2, 2xy22, or 2x22y, where the x, y denote 1 or 3. This gives 34 "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, thus there are at least 9 points of the form 2x22y, 2x2y2, 2xy22. The x=1, x=3 cases can absorb at most six of these, thus each of these cases must absorb at least three points, with at least one absorbing at least five. Let's say that it's the x=3 case that absorbs 5; thus [math]d(*1***) \geq 3[/math] and [math]d(*3***) \geq 5[/math]. From the n=4 theory this means that the *1*** slice has at most 41 points, and the *3*** slice has at most 40. Meanwhile, the middle slice has at most 43, leading to 41+41+42=124 points in all.

Now suppose c is less than 17; then by the n=4 theory the middle slice is one of the eight (6,24,12,0,0) sets. Without loss of generality we may take it to be [math]\Gamma_{220}+\Gamma_{202}+\Gamma_{022}+\Gamma_{112}+\Gamma_{211}[/math]; in particular, the middle slice contains the points 21122 21212 21221 23322 23232 23223. In particular, the *1*** and *3*** slices have a "d" value of at least three, and so have at most 41 points. If the *2*** slice has at most 42 points, then we are at 41+42+41=124 points as needed, but if we have 43 or more, then we are back in the first case (as [math]c(*2***) \geq 17[/math]) after permuting the indices. [math]\Box[/math]

Since 125=41+41+43, we thus have

Corollary 1: Every side slice of [math]{\mathcal A}[/math] has at least 41 points. If one side slice does have 41 points, then the other has 43.

Combining this with the n=4 statistics, we conclude

Corollary 2: Every side slice of [math]{\mathcal A}[/math] has e=0, and [math]d \leq 3[/math]. Given two opposite side slices, e.g. 1**** and 3****, we have [math]d(1****)+d(3****) \leq 4[/math].
Corollary 3: E=0.
Corollary 4: Any middle slice has a "c" value of at most 8.

Proof Let's work with the 2**** slice. By Corollary 2, there are at most four contributions to c(2****) of the form 21*** or 23***, and similarly for the other three positions. Double counting then gives the claim. [math]\Box[/math]

Lemma 3: [math]2B+C \leq 160[/math].

Proof: There are 160 lines connecting one "C" point to two "B" points (e.g. 11112, 11122, 11132); each "C" point lies in two of these, and each "B" point lies on four. A Moser set can have at most two points out of each of these lines. Double counting then gives the claim. [math]\Box[/math]

Lemma 4 Given m A-points with [math]m \geq 5[/math], there exists at least m-4 pairs (a,b) of such A-points with Hamming distance exactly two (i.e. b differs from a in exactly two places).

Proof It suffices to check this for m=5, since the case of larger m then follows by locating a pair, removing a point that contributes to that pair, and using the induction hypothesis. Given 5 A-points, we may assume by the pigeonhole principle and symmetry that at least three of them have an odd number of 1s. Suppose 11111 is one of the points, and that no pair has Hamming distance 2. All points with two 3s are excluded, so the only points allowed with an odd number of 1s are those with four 3s. But all those points differ from each other in two positions, so at most one of them is allowed. [math]\Box[/math]

Corollary 5 [math]C \leq 79[/math].

Proof: Suppose for contradiction that C=80, then D=0 and [math]B \leq 40[/math]. From Lemma 4 we also see that A cannot be 5 or more, leading to the contradiction. [math]\Box[/math]

Define the score of a Moser set in [math][3]^4[/math] to be the quantity [math]a + 5b/4 + 5c/3 + 5d/2 + 5e[/math]. Double-counting (and Lemma 2) gives

Lemma 5 The total score of all the ten side-slices of [math]{\mathcal A}[/math] is equal to [math]5|{\mathcal A}| = 5 \times 125[/math]. In particular, there exists a pair of opposite side-slices whose scores add up to at least 125.

By Lemma 5 and symmetry, we may assume that the 1**** and 3**** slices have score adding up to at least 125. By Lemma 2, the 2**** slice has at most 41 points, which imply that the 1**** and 3**** have 41, 42, or 43 points. From the n=4 statistics we know that all 41-point, 42-point, 43-point slices have score less than 62, with the following exceptions:

  1. (2,16,24,0,0) [42 points, score: 62]
  2. (4,16,23,0,0) [43 points, score: 62 1/3]
  3. (4,15,24,0,0) [43 points, score: 62 3/4]
  4. (3,16,24,0,0) [43 points, score: 63]

Thus the 1**** and 3**** slices must come from the above list. Furthermore, if one of the slices is of type 1, then the other must be of type 4, and if one slice is of type 2, then the other must be of type 3 or 4.

Lemma 6 There exists one cut in which the side slices have total score strictly greater than 125 (i.e. they thus involve only Type 2, Type 3, and Type 4 slices, with at least one side slice not equal to Type 2, and the cut here is of the form 43+39+43).

Proof If not, then all cuts have side slices exactly equal to 125, which by the above table implies that one is Type 1 and one is Type 4, in particular all side slices have c=24. But this forces C=80, contradicting Corollary 5. [math]\Box[/math]

Adding up the corners we conclude

Corollary 6 [math]6 \leq A \leq 8[/math].
Lemma 7 [math]D=0[/math].

Proof Firstly, from Lemma 6 we have a cut in which the two side slices are omitting at most one C-point between them (and have no D-point), which forces the middle slice to have at most one D-point; thus D is at most 1.

Now suppose instead that D=1 (e.g. if 11222 was in the set); then there would be two choices of coordinates in which one of the side slices would have d=1 (e.g. 1**** and *1****). But the n=4 statistics show that such slices have a score of at most 59 7/12, so that the total score from those two coordinates is at most 63 + 59 7/12 = 122 7/12. On the other hand, the other three slices have a net score of at most 63+63 = 126. This averages out to at most 124.633... < 125, a contradiction. [math]\Box[/math]

Corollary 7 Every middle slice has at most 40 points.

Proof From Lemma 7 we see that the middle slice must have c=0, but from the n=4 statistics this is not possible for any slice of size 41 or higher (alternatively, one can use the inequalities [math]4a+b \leq 64[/math] and [math]b \leq 32[/math]). [math]\Box[/math]

Each middle slice now has 39 or 40 points, with c=d=e=0, and so from the inequalities [math]4a+b \leq 64[/math], [math]b \leq 32[/math] must have statistics (8,32,0,0,0),(7,32,0,0,0), or (8,31,0,0,0). In particular the "a" index of the middle slices is at most 8. Summing over all middle slices we conclude

Lemma 9 B is at most 40.
Lemma 10 C is equal to 78 or 79.

Proof By Corollary 5, it suffices to show that [math]C \geq 78[/math].

As 125=43+39+43, we see that every center slice must have at least 39 points. By Lemma 2 and Lemma 7 the center slice has c=d=e=0, thus the center slice has a+b >= 39. On the other hand, from the n=4 theory we have 4a+b <= 64, which forces b >= 31.

By double counting, we see that 2C is equal to the sum of the b's of all the five center slices. Thus C >= 5*31/2 = 77.5 and the claim follows. [math]\Box[/math]

From Lemmas 9 and 10 we see that [math]B+C \leq 119[/math], and thus [math]A \geq 6[/math]. Also, if [math]A \geq 7[/math], then by Lemma 4 we have at least three pairs of A points with Hamming distance 2. At most two of these pairs eliminate the same C point, so we would have C=78 in that case.

Putting all the above facts together, we see that (A,B,C,D,E,F) must be one of the following triples:

  • (6,40,79,0,0,0)
  • (7,40,78,0,0,0)
  • (8,39,78,0,0,0)

All three cases can be eliminated, giving [math]c'_5=124[/math].

Elimination of (6,40,79,0,0,0)

Look at the 6 A points. From Lemma 4 we have at least two pairs (a,b), (c,d) of A-points that have Hamming separation 2.

Now look at the midpoints of these two pairs; these midpoints are C-points cannot lie in the set. But we have exactly one C-point missing from the set, thus the midpoints must be the same. By symmetry, we may thus assume that the two pairs are (11111,11133) and (11113,11131). Thus 11111,11133, 11113, 11131 are in the set, and so every C-point other than 11122 is in the set. On the other hand, the B-points 11121, 11123, 11112, 11132 lie outside the set.

At most one of 11312, 11332 lie in the set (since 11322 lies in the set). Suppose that 11312 lies outside the set, then we have a pair (xy1z2, xy3z2) with x,y,z = 1,3 that is totally omitted from the set, namely (11112,11312). On the other hand, every other pair of this form can have at most one point in the set, thus there are at most seven points in the set of the form (xyzw2) with x,y,z,w = 1,3. Similarly there are at most 8 points of the form xyz2w, or of xy2zw, x2yzw, 2xyzw, leading to at most 39 B-points in all, contradiction.

Elimination of (7,40,78,0,0,0)

By Lemma 4, we have at least three pairs of A-points of distance two apart that lie in the set. The midpoints of these pairs are C-points that do not lie in the set; but there are only two such C-points, thus two pairs must have the same midpoint, so we may assume as before that 111xy lies in the set for x,y=1,3, which implies that 1112* and 111*2 lie outside the set.

Now consider the 160 lines between 2 B points and one C point (cf. Lemma 3). The sum of all the points in each such line (counting multiplicity) is 4B+2C = 316. On the other hand, every one of the 160 lines can have at most two points, and two of these lines (namely 1112*, 111*2) have no points. Thus all the other lines must have exactly two points.

We know that the C-point 11122 is missing from the set; there is one other missing C-point. Since 1112x, 111x2 lie outside the set, we conclude from the previous paragraph that 1132x, 113x2 and 1312x, 131x2 lie in the set. Taking midpoints we conclude that 11322 and 13122 lie outside the set. But this is now three C-points missing (together with 11122), a contradiction.

Elimination of (8,39,78,0,0,0)

By Lemma 4 we have at least four pairs of A-points of distance two apart that lie in the set. The midpoints of these pairs are C-points that do not lie in the set; but there are only two such C-points, thus two pairs (a,b), (c,d) must have the same midpoint p, and the other two pairs (a',b'), (c',d') must also have the same midpoint p'. (Note that every C-point is the midpoint of at most two such pairs.)

Now consider the 160 lines between 2 B points and one C point (cf. Lemma 3). The sum of all the points in each such line (counting multiplicity) is 4B+2C = 312. Every one of the 160 lines can have at most two points, and four of these (those in the plane of (a,b,c,d) or of (a',b',c',d') have no points. Thus all other lines must have exactly two points.

Without loss of generality we have (a,b)=(11111,11133), (c,d) = (11113,11131), thus p = 11122. By permuting the first three indices, we may assume that p' is not of the form x2y2z, x2yz2, xy22z, xy2z2. Then 1112x lies outside the set and 1122x lies in the set, so by the above paragraph 1132x lies in the set; similarly for 113x2, 1312x, 131x2. This implies that 13122, 11322 lie outside the set, but this (together with 11122) shows that at least three C-points are missing, a contradiction.

General n

General solution for [math]c'_N[/math]. For any q, the union of the following sets is a Moser set. The size of this Moser set is maximized when q is near N/3, in which case it is [math]O(3^n/\sqrt{n})[/math]. Most of the points are in the layers with q 2s and q-1 2s.

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

where A(m,d) is a subset of [math][1,3]^m[/math] 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 maximum size of A(m,d) in general. However, the size of A(m,d) can be bounded by sphere-packing arguments. For example, points in A(m,3) are surrounded by non-intersecting spheres of Hamming radius 1, and points in A(m,5) are surrounded by non-intersecting spheres of Hamming radius 2.

  • [math]|A(m,1)| = 2^m[/math] because it includes all points in [math][1,3]^m[/math]
  • [math]|A(m,2)| = 2^{m-1}[/math] because it can include all points in [math][1,3]^m[/math] with an odd number of ones.
  • [math]|A(m,3)| \le 2^m/(m+1)[/math] because the size of a Hamming sphere is m+1.

The integer programming routine from Maple 12 was used to obtain upper bounds for [math]c'_6[/math] and [math]c'_7[/math]. A large number of linear inequalities, such as those described above in sections (n=3) and (n=4), were combined. The details are in Maple calculations. The results were that [math]c'_6 \le 361[/math] and [math]c'_7 \le 1071[/math].

A genetic algorithm has provided the following examples:

Brute-force calculations for [math][3]{}^4[/math]

An attempt was made to find extremal statistics for the four-dimensional case by brute force. Since a simple routine would check 2^81 possibilities, which is infeasible, shortcuts would be needed.

The outline of a suggested routine is as follows: There are just over 3.8 million line-free subsets of [3]^3. Since the cube has 3!2^3 = 48 symmetries, they fall into 83158 equivalence classes. To consider all possibilities, it is enough to do the following:

  • Let the 1*** slice be a representative from one of the 83158 classes.
  • Let the 3*** slice be one of the 3.8 million subsets.
  • Identify which points in 2*** would not complete a vertical or sloping line from the 1*** slice to the 3*** slice.
  • Go to a lookup table, and find the Pareto-optimal statistics of all line-free subsets of 2*** with the restriction from the previous step. This lookup table would need 2^27 rows, or 2^27/48 if one made use of the cube's symmetries.
  • Deduce a set of possible statistics for the combined **** cube. Update the list of [3]^4 Pareto statistics
  • Loop through the 83158 1*** slices and 3.8 million 3*** slices.

A large part of this calculation is to build the lookup table. The script given below, written in Matlab, is intended to carry it out. An input, DVec, is a list of 27-bit numbers. Each number represents one case of the forbidden 2*** points. The output is ParetoCell, which lists the Pareto statistics for each DVec value. A more compact form of the output is ParetoInd.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% DATE : 6 June, 2009
% AUTHOR : Michael Peake
% PROJECT : Polymath1
% Lookup table 
% There is assumed a vector of 27-bit DISALLOWED numbers in DVEC
% Its output is PARETOCELL which lists the Pareto statistics 
% for each DISALLOWED number
if ~exist('eee');
  
  % List of Disallowed 27-bit numbers
  if ~exist('DVec')
     DVec = 0;
  end;
  
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  
  % Find the 230 line-free subsets of [3]^2
  [a,b,c,d,e,f,g,h,k] = ndgrid(0:1);
  abcd = [a(:) b(:) c(:) d(:) e(:) f(:) g(:) h(:) k(:)];
  All512 = abcd;
  % Delete those with complete lines in them
  for n=512:-1:1,
     v=abcd(n,:);
     if any(all(v([1 2 3;4 5 6;7 8 9;1 4 7;2 5 8;3 6 9;1 5 9;3 5 7]),2)),
        abcd(n,:)=[];
     end;
  end;
  
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  % Some numbers needed repeatedly, so precalculate them
  TwoEight = 2.^[0:8];
  Eight = abcd*TwoEight';
  TwoFourEight = 256*Eight;
  % Student Matlab won't allow 230x230 array, but will allow cell arrays
  Eights = cell(1,230);
  % 27-bit numbers built from first layer and third layer
  for n=1:230,Eights{n}=Eight(n)+65536*Eight;end;
  
  % find symmetry group of each one
  %for n=1:230,
  %v=reshape(abcd(n,:),3,3);
  %w=v';
  %x=flipud(v);
  %y=fliplr(v);
  %z=flipud(fliplr(v));
  %a=flipud(w);
  %b=fliplr(w);
  %c=flipud(fliplr(w));
  %e=[v(:) w(:) x(:) y(:) z(:) a(:) b(:) c(:)];
  %same(n) = sum(all(e==(v(:)*ones(1,8))));
  %end;
  
  % calculate the statistics for each line-free square
  for n=1:230,
     v=reshape(abcd(n,:),3,3);
     Stats2(n,:) = [sum(v([1 3 7 9])) sum(v([2 4 6 8])) v(5)];   
  end;
  
  % Convert the stats to a unique index
  Stats2Ind = Stats2*[1;9;9*13];
  % The centre-slice has a different statistic
  Stats2Shift = Stats2*[9;9*13;9*13*7];

  % count the restricted patterns
  for n=1:512,
     WithinCell{n} = find(all(abcd <= All512(n*ones(1,230),:),2));
     Within(n)     = length(WithinCell{n});
     % Sort them so that Stats2Shift is increasing;
     % will make it easier to ignore repetitions later
     [a,b] = sort(Stats2Shift(WithinCell{n}));
     WithinCell{n} = WithinCell{n}(b);
  end;
  Total = 0;Count = zeros(1,48);
  
  % Compare every Statistics for the cube with every other Statistics, 
  % to see if one dominates the other - to keep track of Pareto maxima
  Beaters = cell(9,13,7,2);
  for a=1:9,for b=1:13,for c=1:7,for d=1:2,
              Beaters{a,b,c,d} = zeros(9,13,7,2);
              for e=1:9,
                 for f=1:13,
                    for g=1:7,
                       for h=1:2,
                          if all([a b c d]>=[e f g h]),
                             Beaters{a,b,c,d}(e,f,g,h)=1;
                          elseif all([a b c d]<=[e f g h])
                             Beaters{a,b,c,d}(e,f,g,h)=-1;
                          end;
                       end;end;end;end;
              Beaters{a,b,c,d}(a,b,c,d)=-1;
           end;end;end;end;
  
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  
  % For each Layer1 and Layer3, precalculate which cells 
  % are forbidden in Layer2
  for a=1:230,  
     % Convert square of 01s to 9-bit number
     Layer1 = abcd(a,:);
     Num1 = TwoEight*Layer1';
     % Some bits are needed for sloping lines
     X123 = bitand(Num1,7)*8;
     X789 = bitand(Num1,8*56)/8;
     X147 = bitand(Num1,73)*2;
     X369 = bitand(Num1,4*73)/2;
     
     for b=1:230,
        
        Layer3 = abcd(b,:);
        Num2 = TwoEight*Layer3';
        
        % Check for vertical lines
        Out = bitand(Num1,Num2);         
        
        % Set up for sloping lines
        Y123 = bitand(Num2,7);Y123=Y123*8;
        Y789 = bitand(Num2,448);Y789=Y789/8;
        Y147 = bitand(Num2,73);Y147=Y147*2;
        Y369 = bitand(Num2,292);Y369=Y369/2;
        
        % Check for sloping lines
        Out = bitor(Out,bitand(X123,Y789));
        Out = bitor(Out,bitand(X789,Y123));
        Out = bitor(Out,bitand(X147,Y369));
        Out = bitor(Out,bitand(X369,Y147));
        
        % Check for major diagonals, and bit 5
        v = 16*any(Layer1([1 3 7 9]) & Layer3([9 7 3 1]));
        Out = bitor(Out,v);
        
        NotOut = bitcmp(Out,9);
        % Index
        In=NotOut+1;
        
        % 9-bit number shows which bits are allowed
        Allowed{a}(b) = In;
        
     end;end;%for a,for b
  
  
  %%%%%%%%%%%%%%%%%%%%%%%%%
  eee = 'done';
end;% if ~exist('eee');


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Set up enough memory to keep track of Pareto sets
% These are the pareto-maxima of the second slice,
% will be subject to the DISALLOWED restriction 

% Count of Disallowed 27-bit numbers
nd = length(DVec);

% List of Pareto-max Layer2s
ParetoInd = cell(1,nd);

% Provide enough space
[ParetoInd{:}] = deal(ones(1,230));

% Initialize
[ParetoInd{:}] = deal(ones(1));
NPareto = ones(1,nd);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% DATE : 6 June, 2009
% AUTHOR : Michael Peake
% PROJECT : Polymath1
% Lookup table 
% There is assumed a vector of 27-bit DISALLOWED numbers in DVEC
% Its output is PARETOCELL which lists the Pareto statistics 
% for each DISALLOWED number
for a=1:230,
  
  InFirst = Allowed{a};
  NN0Vec = Eights{a};
  
  for b=1:230,   
     
     % Statistics of Layer1 and Layer3 put together
     Stats0Ind = Stats2Ind(a)+Stats2Ind(b)+1;
     
     % List of possible Layer2s
     Rvec0 = WithinCell{InFirst(b)};
     
     % Stats of all three Layers put together
     Stats3Ind0 = Stats0Ind+Stats2Shift(Rvec0);
     
     % 27-bit number of Layer1 and Layer3
     NN0 = NN0Vec(b);
     
     % Pick which Disallowed are dead already
     DOk = find(~bitand( NN0, DVec));
     
     % 27-bit number including various possible Layer2s
     NN = NN0+TwoFourEight(Rvec0);
     
     % Only the sequel changes for different DISALLOWEDS
     for di = DOk
        
        % Pick a 27-bit number of DISALLOWED bits
        Disallowed = DVec(di);
        
        % Which Layer2s give a cube entirely within allowed bits
        NOk = find(~bitand(NN,Disallowed));
        
        % Don't bother if none apply
        if length(NOk)
           
           Stats3Ind = Stats3Ind0(NOk);
           
           % Remove repeats (values are already sorted)
           for c=[1; 1+find(diff(Stats3Ind))]',
              
              NewStat = Stats3Ind(c);
              % To compare this new Statistic with current Pareto set, look up the 
              % pre-calculated comparisons
              Mat = Beaters{NewStat};
              % Fetch the list of Pareto sets, to prevent repeated indexing
              PInd = ParetoInd{di};
              Comparison = Mat(PInd);
              
              if (all(Comparison>=0)),
                 % Then this is a new Pareto set
                 
                 % remove the dominated Pareto sets
                 Old = find(Comparison);
                 PInd(Old) = [];
                 % include the new Pareto set
                 PInd(end+1) = NewStat;
                 NPareto(di) = length(PInd);
                 % Put the updated Pareto list back in the cell array
                 ParetoInd{di} = PInd;
              end;%if Comparison
           end;%for c[1; 1+find(diff(Stats3Ind))]
        end;%if length(NOk)
     end;%for di
     %disp(num2str([floor(toc) a b Total Count(find(Count))]));
  end;%for b
  %disp(num2str([a NPareto]));
end;% for a

% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Convert from Pareto indexes to lists of Pareto statistics

for di = 1:nd,
  [a,b,c,d] = ind2sub([9 13 7 2],ParetoInd{di}(1:NPareto(di)));
  ParetoCell{di} = [a;b;c;d]'-1;
end;

Larger sides (k>3)

The following set gives a lower bound for Moser’s cube [math][4]^n[/math] (values 1,2,3,4): Pick all points where q entries are 2 or 3; and also pick those where q-1 entries are 2 or 3 and an odd number of entries are 1. This is maximized when q is near n/2, giving a lower bound of

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

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

For k=5 (values 1,2,3,4,5) If A, B, C, D, and E denote the numbers of 1-s, 2-s, 3-s, 4-s and 5-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]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]\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.