# Does every polyomino tile R^n for some n?

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 any polyomino $P$ 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.

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It's a cute question, Tim, but I would guess the answer is no, perhaps already for 1-dim polyominos. Try ooooooooo oo ooo oooooooo and see how that works out –  Igor Pak Dec 20 '10 at 1:07
The interval in $\mathbb R$ of length 25 minus three intervals of length 1, starting at 9, 12, and 16. –  Allen Knutson Dec 20 '10 at 4:39
@Tim & others: Sorry for the confusion. I am not claiming that this tile is a counterexample (though I did check 2- and 3-dim). All I am saying that since the problem is very much unclear in rather simple special cases, hoping that the answer is always yes is perhaps too optimistic. The connectivity is not mentioned in the question and doesn't seem terribly relevant; I agree with Tim on this. Anyhow, there is no need to trust my intuition - see for yourself if you can tile the space with this tile. –  Igor Pak Dec 20 '10 at 18:36
@all How can a 5x5 square with the middle square removed, tile any R^n? Which n ? –  fastforward Feb 11 '11 at 12:37
@fastforward: I don't know, but if your worry is that the hole can't be filled, it's enough to lift to $\mathbb{R}^4$. Suppose the first tile is in a horizontal plane, centered at the origin. In "our" 3D slice of $\mathbb{R}^4$, the second tile will be partially visible as a vertical stack of five cubes going through the hole. Shift over to the "next" 3D slice of $\mathbb{R}^4$ and the first tile will disappear but you'll see another vertical stack of five cubes. Shift over again and you'll see four cubes (the middle cube will be missing). The next two shifts will have five cubes again. –  Timothy Chow Feb 11 '11 at 14:59

Here I prove that it does not matter whether we consider only connected dominoes or not, as there was a lot of discussion about it.

Suppose that the original polyomino, P, is d dimensional. We will construct a 2d dimensional connected polyomino, Q, that can be tiled with P. Clearly, this proves the statement, as if it is impossible to tile any space with P, it is also impossible to do so with Q.

Denote a large enough d dimensional brick that contains P by R. Take the 2d dimensional polyomino P x R, so here every original cube of P is replaced by a 2d brick, 1 x R. Note that P x R is contained in an R x R brick. Fill in the missing parts of this R x R brick by 1 x P polyominos. Notice that this means that R x P will be also filled up completely. This polyomino, Q, will be connected, as we can freely move anywhere in the first d coordinates in R x P and in the last d coordinates in P x R.

Note: The complement of the set obtained this way is R\P x R\P. If we repeat this, then it can be achieved that our polyomino is arbitrarily dense, i.e. it fills out at least 99% of a brick.

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Looks nice, but did you mean $R \times P \cup P \times R$ at the end? I wonder if it would be useful to push that idea further and get things like $R \times P \times P \cup P \times R \times P \cup P \times P \times R$ (maybe possible or not, maybe useful or not, I haven't thought yet) –  Dave Pritchard Jan 19 '11 at 18:18
Oh, I forgot to put their complement, now I fixed it, thx. I was trying for a while but could not finish the proof from here, but it is a very interesting question. I think we can get all sorts of things that you write, but we will always have a little gap of the form (R\P)^k this way. –  domotorp Jan 20 '11 at 8:56

@Tim: Thanks for posting the question.

@All: I am allowing all translations and rotations. I am also allowing reflections, but that's irrelevant to the question, because you can always get a reflection by rotating in one dimension higher. I do not intend to impose that a polyomino be connected, it's just a finite collection of unit cubes with vertices at $Z^n\subset R^n$.

For example, "o o" tiles in 1 dimension, "oo o" tiles in 1 dimension (allowing reflections) or 2 (not allowing reflections), "oo oo" tiles in 2 dimensions (entertaining exercise) and "oo ooo" tiles in 2 dimensions (easier). I don't know the answer for "ooo ooo".

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That answer was supposed to be a comment, but I can't figure out how to do that. Sorry. –  Adam Chalcraft Dec 20 '10 at 22:17
Adam: Since this is your first post, you don't have enough reputation to comment yet. What you've done is the right thing under the circumstances. –  Timothy Chow Dec 20 '10 at 22:40
It is known that any set of 4 cells tiles the plane. Not what I would have guessed –  Aaron Meyerowitz Dec 21 '10 at 0:25
Tiling "oo oo" in two dimensions is entertaining, as you promised. For those who want to play around with it, go to omnium-gatherum.appspot.com/pages/tile.html. –  I. J. Kennedy Dec 21 '10 at 5:58
To I. J. Kennedy: tiling the plane with ooo..o is even more entertaining. –  Zsbán Ambrus Dec 27 '10 at 14:07

@Erich: It's certainly not true that every polyomino tiles some hypercuboid. Here's one way to see this. Consider the $n_1\times\ldots\times n_d$ hypercuboid. Index the cells by $(x_1,\ldots,x_d)$ where $0\leq x_i\leq n_i-1$. Give cell $(x_1,\ldots,x_d)$ the value $t^{x_1+\ldots+x_d}$, where $t$ is an indeterminate. Summed over the whole hypercuboid, this is $$\prod_{i=1}^d(1+t+\ldots+t^{n_i-1}),$$ so its complex roots are all on the unit circle. Now take a symmetric 1-d polyomino, and put it in the hypercuboid so that its value has minimal degree in $t$. Its value is some polynomial $p(t)$. Now wherever you put the polyomino, its value is $t^k p(t)$ for some $k\geq 0$. If you tile anything with the polyomino, the total value is $q(t)p(t)$ for some polynomial $q(t)$. So if you can tile the hypercuboid, all the roots of $q(t)p(t)$ must lie on the unit circle. In particular, all the roots of $p(t)$ must lie on the unit circle. Choose a symmetric 1-d polyomino for which this is false. For example, 1101011 (1 is a square, 0 is a hole).

I am endebted to Imre Leader for the idea behind this proof.

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The best-known results for tiling rectangles in 2-D can be found here:

http://www2.stetson.edu/~efriedma/mathmagic/0299.html

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Yes, that's a problem I overlooked. It is a bit hard to visualize tilings in $R^3$, not to say in higher dimensions. I'll see if I can fix it or do something else. –  fedja Dec 21 '10 at 0:26