# 4D Moser sets with d at least 4 have at most 40 points

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

$4\alpha_0+4\alpha_1+3\alpha_2+\alpha_3 \leq 6 (1)$
$7\alpha_0+3\alpha_1+3\alpha_2+\alpha_3 \leq 7 (2)$

which when averaged on middle slices gives the 4D inequalities

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

Averaging (1) on side slices similarly gives

$a/4 + b/8 + c/8 + d/8 \leq 6. (5)$

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

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

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) $\square$

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. $\square$

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. $\square$

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. $\square$

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. $\square$