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sonata_greenIt seems to me that the key feature of the hexagonal tiling is that there are no cells that share a vertex but not an edge. I think the proper generalization to 3D would then be a honeycomb whose vertex figure is a tetrahedron. There are multiple such, but I'm currently looking at the bitruncated cubic honeycomb, which has the virtue of being cell-transitive.
I'm having trouble wrapping my head around it. Ultimately the question is what "planes" exist[1], but for now I just want to try to understand the nature of the base pattern.
The relationship with the cubic honeycomb seems to mean that a cell at (0,0,0) has cube-face neighbors at the permutations of (±2,0,0) and cube-vertex neighbors at (±1,±1,±1). Which locations are therefore full/empty? For (x,0,0) we have full cube-cells at even x, but it's not simple parity because the vertex operation is odd (and (1,0,0) and (1,1,0) are both empty).
Are (0,1,z) all empty? ...It looks like the cube-edge neighbors of a full cell are necessarily empty; likewise face connections "skip over" an empty cell, so yes.
The question is what invariant (±1,±1,±1) maintains. It's not parity, though it seems like it should have something to do with parity; it's not the mod-3 sum of the coordinates... oh! The rule is that all three coordinates must have the same parity. This explains why (0,1,z) are all empty. This further implies that 2/8=1/4 of the cells are full, which accords with the observation that each edge has exactly one full cube.
[1] I don't quite have a rigorous definition, but it seems to me that a candidate diagonal should separate the space into two congruent half-honeycombs, with the post-smoothing diagonal plane passing through every "surface" cell. However, this is clearly not a sufficient condition, since in 2D this would accept diagonals of every rational slope. Maybe also require the surface to be cell-transitive?
However, the bitruncated cubic honeycomb specifically seems deeply connected to the cubic-octahedral something-or-other, so probably we can just assume that all the diagonals we want have 45-degree slopes. This gives two fundamental diagonal types, the two-coordinate (wide-stair) diagonal and the three-coordinate (qbert) diagonal. ...and the zero-coordinate (orthogonal) plane.
(Why no one-coordinate? Because "diagonal" is a relationship between two coordinates. Except I'm having trouble making that rigorous... something to come back to later, maybe.)
I'm having trouble wrapping my head around it. Ultimately the question is what "planes" exist[1], but for now I just want to try to understand the nature of the base pattern.
The relationship with the cubic honeycomb seems to mean that a cell at (0,0,0) has cube-face neighbors at the permutations of (±2,0,0) and cube-vertex neighbors at (±1,±1,±1). Which locations are therefore full/empty? For (x,0,0) we have full cube-cells at even x, but it's not simple parity because the vertex operation is odd (and (1,0,0) and (1,1,0) are both empty).
Are (0,1,z) all empty? ...It looks like the cube-edge neighbors of a full cell are necessarily empty; likewise face connections "skip over" an empty cell, so yes.
The question is what invariant (±1,±1,±1) maintains. It's not parity, though it seems like it should have something to do with parity; it's not the mod-3 sum of the coordinates... oh! The rule is that all three coordinates must have the same parity. This explains why (0,1,z) are all empty. This further implies that 2/8=1/4 of the cells are full, which accords with the observation that each edge has exactly one full cube.
[1] I don't quite have a rigorous definition, but it seems to me that a candidate diagonal should separate the space into two congruent half-honeycombs, with the post-smoothing diagonal plane passing through every "surface" cell. However, this is clearly not a sufficient condition, since in 2D this would accept diagonals of every rational slope. Maybe also require the surface to be cell-transitive?
However, the bitruncated cubic honeycomb specifically seems deeply connected to the cubic-octahedral something-or-other, so probably we can just assume that all the diagonals we want have 45-degree slopes. This gives two fundamental diagonal types, the two-coordinate (wide-stair) diagonal and the three-coordinate (qbert) diagonal. ...and the zero-coordinate (orthogonal) plane.
(Why no one-coordinate? Because "diagonal" is a relationship between two coordinates. Except I'm having trouble making that rigorous... something to come back to later, maybe.)
