Does every polycube tiling imply a regular polycube tiling? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T00:19:42Z http://mathoverflow.net/feeds/question/93365 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93365/does-every-polycube-tiling-imply-a-regular-polycube-tiling Does every polycube tiling imply a regular polycube tiling? Gjergji Zaimi 2012-04-06T22:53:39Z 2012-09-09T23:18:35Z <p>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 <em>regular</em> if all the centers of the hypercubes have integer coordinates (i.e. the hypercubes fit together in a $\mathbb Z^d$ lattice).</p> <p>I haven't been able to make much progress on the following seemingly simple and natural question:</p> <blockquote> <p>If a given d-polycube tiles $\mathbb R^d$, must it also tile this space regularly?</p> </blockquote> <p>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.</p> <p>However this doesn't work in $d\geq 3$ and I'm having trouble coming up with an argument in that case.</p> http://mathoverflow.net/questions/93365/does-every-polycube-tiling-imply-a-regular-polycube-tiling/106672#106672 Answer by domotorp for Does every polycube tiling imply a regular polycube tiling? domotorp 2012-09-08T14:14:08Z 2012-09-09T21:10:36Z <p>I am almost sure that the answer is no. Your question is strongly related to <a href="http://en.wikipedia.org/wiki/Keller%27s_conjecture" rel="nofollow">Keller's conjecture</a> which turned out to be false. There a tiling of $\mathbb Z^{10}$ was given by translates of the unit cube such that the center of each translate is a halfinteger. Moreover, the tiling has a very strong lattice-like structure, 1024 cubes whose center is in $[0,2]^{10}$ are translated by every vector from $2\mathbb Z^{10}$.</p> <p>Of course, the cube won't give a counterexample to your question. But you can divide each cube from this construction to like $3^{99}$ little cubes and then "dig little tunnels" among adjacent cubes to make sure the resulting polyominos only fit in the given way. This leaves many details to work out and I am not 100% sure it can be done but looks like a promising approach.</p>