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Let's define d-polycubes to be a union of unit hypercubes from the $\mathbb Z^d$ tiling of d-dimensional Euclidean space which has connected interior. Given a tiling of $\mathbb R^d$ by identical copies of a d-polycube, we call this tiling regular if all the centers of the hypercubes have integer coordinates (i.e. the hypercubes fit together in a $\mathbb Z^d$ lattice).Here is an example of an irregular tiling:

I haven't been able to make much progress on the following seemingly simple and natural question:

If a given d-polycube tiles $\mathbb R^d$, must it also tile this space regularly?

I believe I have the $d=2$ case by a simple argument. As long as one has a connected tiled region with a concave boundary one can show that the adjacent 2-polycube (a.k.a. polyomino) at that corner must be placed to fit next to the squares of the already placed polyominoes so that edges meet edges and vertices meet vertices. We continue this until we tile the entire plane or we reach a convex region. Since the allowed angles are $\pi/2, \pi, 3\pi/2$ then we must have a rectangle. Now we can tile the space with translations of this rectangle.

However this doesn't work in $d\geq 3$ and I'm having trouble coming up with an argument in that case.

2 added 169 characters in body

Let's define d-polycubes to be a union of unit hypercubes from the $\mathbb Z^d$ tiling of d-dimensional Euclidean space which has connected interior. Given a tiling of $\mathbb R^d$ by identical copies of a d-polycube, we call this tiling regular if all the centers of the hypercubes have integer coordinates (i.e. the hypercubes fit together in a $\mathbb Z^d$ lattice). Here is an example of an irregular tiling:

I haven't been able to make much progress on the following seemingly simple and natural question:

If a given d-polycube tiles $\mathbb R^d$, must it also tile this space regularly?

I believe I have the $d=2$ case by a simple argument. As long as one has a connected tiled region with a concave boundary one can show that the adjacent 2-polycube (a.k.a. polyomino) at that corner must be placed to fit next to the squares of the already placed polyominoes so that edges meet edges and vertices meet vertices. We continue this until we tile the entire plane or we reach a convex region. Since the allowed angles are $\pi/2, \pi, 3\pi/2$ then we must have a rectangle. Now we can tile the space with translations of this rectangle.

However this doesn't work in $d\geq 3$ and I'm having trouble coming up with an argument in that case.

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# Does every polycube tiling imply a regular polycube tiling?

Let's define d-polycubes to be a union of unit hypercubes from the $\mathbb Z^d$ tiling of d-dimensional Euclidean space which has connected interior. Given a tiling of $\mathbb R^d$ by identical copies of a d-polycube, we call this tiling regular if all the centers of the hypercubes have integer coordinates (i.e. the hypercubes fit together in a $\mathbb Z^d$ lattice). I haven't been able to make much progress on the following seemingly simple and natural question:

If a given d-polycube tiles $\mathbb R^d$, must it also tile this space regularly?

I believe I have the $d=2$ case by a simple argument. As long as one has a connected tiled region with a concave boundary one can show that the adjacent 2-polycube (a.k.a. polyomino) at that corner must be placed to fit next to the squares of the already placed polyominoes so that edges meet edges and vertices meet vertices. We continue this until we tile the entire plane or we reach a convex region. Since the allowed angles are $\pi/2, \pi, 3\pi/2$ then we must have a rectangle. Now we can tile the space with translations of this rectangle.

However this doesn't work in $d\geq 3$ and I'm having trouble coming up with an argument in that case.