View source for Classification of (6,12,18,4,0) sets

A set with statistics. (6,12,18,4,0) cannot have two points of type d with the same c-statistic. If it does say the coordinate of both points which is not equal to 2 is one then slice on that coordinate there result will be that both side slices will have their center point and can have at most three points of type c. So the remaining 12 must be in the center slice. But there they will block all points of type d but the only remaining spaces for these points are the outside slices and there are only two such spaces there and we need two so there is a contradiction.

Now without loss of generality taking in account the above we can take the points of type d to be (1,2,2,2), (2,1,2,2),(2,2,1,2) and (2,2,2,1) . We slice on the first coordinate. Then the cube with first coordinate equal to one contains its center point and can only have three points of type c. The center cube contains the remaining points of type d and hence can contain 9 of the remaining points of type d. The rest must be in the remaining slice. It must contain all its points of type c. But we could have cut along any coordinate thus the set contains all points of type c with one coordinate equal to 3. This accounts for all 18 of its points.

Now for the points of type a. There can only be two points of type a in the slice with first coordinate equal to three as any two points that don’t block a line block the remaining points. So there must be four in the other side slice. Now if the point (1,1,3,3) is in the set then it will block all points except (1,1,1,1), (1,1,3,1), (1,1,1,3) and (1,3,3,3) and (3,1,1,1), (3,1,3,1) and (3,1,1,3) and (3,3,3,3). To have four points in the slice with first coordinate equal to one it must contain one of (1,1,3,1) and (1,1,1,3)
But either of these together with (1,1,3,3) will block all but one of the points in the slice with first coordinate equal to three but we need to so we have a contradiction. And by a similar argument we can deal with all points of the which are permutations of (1,1,3,3).

If the point (1,3,3,3) is in the set it will block all points which are permutations of (1,1,1,3) and (1,1,1,1) and to get four points in the slice with first coordinate equal to one we will need points which are permutations of (1,1,3,3) but that leads to contradictions as noted above.

So the only way to have four points in the side slice with first coordinate equal to one is to have the points (1,1,1,1), (1,1,1,3), (1,1,3,1) and (1,1,1,3).

These points block all but he points (3,1,1,1) and (3,3,3,3) these can be included and we need them to have six points so we add them and we have our six points of type a.

Now we need to get the points of type b. We can include all points which are permutations of (1,3,3,2) and get 12 points so we need to show that any deviation from this leads to less points. Let us look at the 8 points of type a which have the first coordinate equal to 2 we will show that the only way to have three points in that set is if they all are permutations of (1,3,3,2). Assume we have a point with two ones say (2,1,1,3) we will have the points (2,2,1,3) and (2,1,2,3) in the set from the above result on points of type b. They will block (2,3,1,3) and (2,1,3,3). The points (2,1,2,2) and (2,2,1,2) will be in the set from our discussion of points of type d. They will block (2,1,3,1) and (2,3,1,1). So the only points left are (2,3,3,1), (2,3,3,3)
And (2,1,1,1) but (2,1,1,1) is blocked by (1,1,1,1) and (3,1,1,1).
So we must have both of (2,3,3,1) and (2,3,3,3) but (2,3,3,2) is in our set as noted in our discussion of points of type c. So we can have only one of the two. So the only way we can have three points in the set is if we have the three permutations of (1,3,3,2) in the set.
Similar arguments apply to the other sets with one fixed coordinate equal to two and the rest not equal to two. Thus the only set of points of type b that works is the permutations of the points (1,3,3,2) and we are done.