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I'm interested in tilings of the plane by squares, with labels on the edges. It's well known that (1) the question "can one tile the plane with the following finite set of tiles?" is undecidable, and (2) there are finite sets of tiles that tile the plane, but only aperiodically. Also, (1) implies (2).

Is the question "can one tile the plane periodically with the following finite set of tiles?" decidable?

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  • $\begingroup$ By periodic, do you mean periodic in both directions? $\endgroup$ Commented Feb 11, 2013 at 15:02
  • $\begingroup$ The tilability question is undecidable even if we restrict the tiles to be rectangles. $\endgroup$ Commented Feb 11, 2013 at 16:16
  • $\begingroup$ My question was whether it would count as periodic if the pattern repeats only in the horizontal direction, but the columns are not periodic vertically. I assume not. If periodic in both directions, then you can assume period is the same via least-common-multiple. $\endgroup$ Commented Feb 11, 2013 at 16:20
  • $\begingroup$ @Joel: both questions are valid. I'm interested in understanding the current status, so extra restrictions are also welcome: periodic in one or two directions; or require deterministic tiles (for every corner, horizontal and vertical label, there exists at most 1 tile with these labels in that corner). I'll vote you up if nobody comes up with a complete argument within a few days :) $\endgroup$
    – grok
    Commented Feb 13, 2013 at 3:21
  • $\begingroup$ Joel's two questions are equivalent: if there is a horizontally periodic point, you can find a totally periodic point by the pigeonhole principle by finding two equal rows and repeating the area in between. $\endgroup$
    – Ville Salo
    Commented Aug 9 at 3:10

2 Answers 2

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Deciding whether a set of tiles admits a periodic tiling or no tiling at all is undecidable as well.

This has been shown in Y.S. Gurevich, I.I. Koryakov, Remarks on Berger's paper on the domino problem, J Sib Math J 13, 319–321 (1972). The results can also be found in the book The Classical Decision Problem by Egon Börger, Erich Grädel, Yuri Gurevich, where Theorem 3.1.7 states

The set of domino systems that admit, respectively, no tiling and a periodic tiling of $\mathbb Z\times \mathbb Z$ or $\mathbb N\times \mathbb N$ are recursively inseparable.

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The answer appears to be no.

Consider first the case of the anchor-tile periodic tiling problem, where we insist that a particular anchor tile is used. Let's modify the usual Wang tile argument, due to Berger, for the oringal tiling problem. That argument shows that for any Turing machine program $p$, we can create a set of Wang tile types (square tiles with labels on the edges, to be matched up in a tiling) such that the operation of program $p$ corresponds directly to the pattern of tiles appearing in any tiling. Basically, the anchor row displays a complete picture of the Turing machine in the start configuration, and each subsequent row shows the configuraton after one additional step of computation using program $p$. The main idea of the scheme is that the tiling can continue as long as the program keeps running, and so there is a tiling if and only if the program does not halt. This is why the original anchor tiling problem is undecidable. (And the anchor feature was removed by a separate argument.)

But essentially the same idea can be used to solve your problem. We can modify the tiles involving the halt state so that a halting computation will support a periodic tiling. Basically, what is needed are tiles that will lead from a tile on which the program has halted to transform the next row to the initial configuration again. It would be a detailed thing to check, but after one has spent time using these Wang tiles to simulate Turing machine computations, this seems to be easily handled in the same way.

The end situation will be that if $p$ halts, then there will be a periodic tiling using the desired anchor tile, and if $p$ does not halt, then there will be no periodic tiling using that anchor tile. So the question will be undecidable.

Finally, it seems to me that one can remove the anchor tile requirement in the same way that is done with the original Wang tile argument (but this is a complication).

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