This is a question posed by Adam Chalcraft. I am posting it here because I think it deserves wider circulation, and because maybe someone already knows the answer.

A *polyomino* is usually defined to be a finite set of unit squares, glued together edge-to-edge. Here I generalize it to mean a finite set of unit hypercubes, glued together facet-to-facet.

Given a polyomino $P$ in $\mathbb{R}^m$, I can *lift* it to a polyomino in a higher-dimensional Euclidean space $\mathbb{R}^{m+n}$ by crossing it with a unit $n$-cube: the lifted polyomino is just $P\times [0,1]^n$.

Obviously, not all polyominos tile space.

Is it true that given

anypolyomino $P$ in $\mathbb{R}^m$, there exists some $n$ such that the lifted polyomino $P\times [0,1]^n$ tiles $\mathbb{R}^{m+n}$?

Many people's first instinct is that multiply-connected polyominos (those with "holes" in them) can't possibly tile, but you can get inside holes if you lift to a high enough dimension.

ooooooooo oo ooo ooooooooand see how that works out $\endgroup$ – Igor Pak Dec 20 '10 at 1:0711more comments