# 4D Moser sets with d at least 3 has at most 41 points

Jump to: navigation, search

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.