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I know several specific variations of the domino tiling problem has been determined to be decidable or undecidable, such as the seed domino problem. I have a variation which I have not been able to find, but as I am not familiar with this subject, I thought someone here might know better.

Given a Wang tile $T=(T_N,T_E,T_S,T_W)$ with colors $\mathcal{C}$, each entry corresponds to the color in the north, east, south or west sides. Given two Wang tiles $T^{(1)},T^{(2)}$ that agree on opposite sides, their fusion would be gluing them on the side on which they agree. For example, if $T^{(1)}_N=T^{(2)}_S$, their fusion along South-North side is a new Wang tile

$$T^{(1)} \oplus_{SN} T^{(2)}=\Big( T^{(2)}_N, (T^{(2)}_E,T^{(1)}_E), T^{(1)}_S, (T^{(2)}_W,T^{(1)}_W) \Big),$$

over a different set of colors obviously. We can similarly define $T^{(1)}\oplus_{WE} T^{(2)}$ when they satisfy $T^{(1)}_E=T^{(2)}_W$. If we take a Wang tile set $\mathcal{T}$ with colors $\mathcal{C}$, we can obtain new Wang tiles sets

$$ \mathcal{T_1} = \big\{ T \oplus_{SN} T' : T,T' \in \mathcal{T}, T_N=T'_S \big\} $$ and

$$ \mathcal{T_2} = \big\{ T \oplus_{WE} T' : T,T' \in \mathcal{T}_1, T_E=T'_W \big\}. $$

$\mathcal{T}_2$ will again be a collection of Wang tiles over the new color set $\mathcal{C}\times \mathcal{C}$. I can consider two related questions:

  1. Does $\mathcal{T}_2$ tile the plane periodically?
  2. Does $\mathcal{T}_1$ tile the plane periodically?

Is it known whether question (1) being un\decidable imply question (2) is un\decidable? I am assuming that $\mathcal{T}_1$ tiles the plane.

This seems like a natural question to me, but I really have a bad intuition on the subject, so it may be obviously resolvable. I would be glad to find appropriate references, or even solutions.

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2 Answers 2

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Each tiling with tiles from $\mathcal{T}$ gives rise to exactly two tilings with tiles from $\mathcal{T}_1$ and exactly two tilings with tiles from $\mathcal{T}_2$, given by taking the two choices of fusing neighbouring pairs (either $2n\longleftrightarrow 2n+1$ or $2n-1 \longleftrightarrow 2n$). This process also preserves periodicity.

Up to translation, this correspondence is also reversible and again preserves periodicity. So $\mathcal{T}$ admits a periodic tiling if and only if $\mathcal{T}_1$ admits a periodic tiling if and only if $\mathcal{T}_2$ admits a periodic tiling.

The same argument works for non-periodic tilings of course.

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  • $\begingroup$ Thanks for your answer. Do you know what happens when one wishes to consider subsets of $\mathcal{T}_2$ instead? I am interested in whether given a Wang tile set $\mathcal{T}$, which admits a periodic tiling, one can break the 'periodicity' one level higher so that it does not have a periodic tiling? $\endgroup$ Commented Jan 16, 2023 at 15:24
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    $\begingroup$ @Keen-ameteur I'm not sure. Certainly, removing tiles from your tile set can remove admissible tilings (including all periodic ones), and may also make it so that it is undecidable if the set of tiles can tile at all. Actually deciding when that happens is probably example dependant and it's unlikely to be a decidable problem in general. $\endgroup$
    – Dan Rust
    Commented Jan 16, 2023 at 16:23
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Dynamically, the set of tilings $X = X_{\mathcal{T}} \subset \mathcal{T}^{\mathbb{Z}^2}$ by a set of Wang tiles $\mathcal{T}$ is a compact dynamical system under the action of $\mathbb{Z}^2$, with the product topology on $\mathcal{T}^{\mathbb{Z}^2}$.

If $H \leq \mathbb{Z}^2$ is a finite-index subgroup, then $H$ is isomorphic to $\mathbb{Z}^2$, and it is well-known that $(X, H)$ is itself topologically conjugate (in bijection through a shift-commuting homeomorphic) with a set of tilings. Your $\mathcal{T}_1$ realizes this construction for the subgroup $H = \mathbb{Z} \times 2\mathbb{Z}$, i.e. we have $(X_{\mathcal{T}_1}, \mathbb{Z}^2) \cong (X_{\mathcal{T}}, H)$ where $H$ is identified in the obvious way with $\mathbb{Z}^2$. In other words, as far as dynamical properties are concerned, we can just think of $X_{\mathcal{T}_1}$ as the same space $X$ with a different action.

Now, a totally periodic point is a point which is fixed by a finite-index subgroup $K$ of the acting group. If $x \in X$ is fixed by $K$, then it is also fixed by the finite index subgroup $K \cap H$, thus total periodicity is equivalent whether or not we consider the $H$-action or the $\mathbb{Z}^2$-action.

(This is all completely general, and works for any acting group.)

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  • $\begingroup$ Why do I have an upvote after about 10 seconds? Is it a bot or someone was just waiting for this answer? $\endgroup$
    – Ville Salo
    Commented Jan 16, 2023 at 16:30
  • $\begingroup$ I was reading Dan's comment when your answer popped up. $\endgroup$ Commented Jan 16, 2023 at 16:33
  • $\begingroup$ Fair enough! Was just very fast $\endgroup$
    – Ville Salo
    Commented Jan 16, 2023 at 16:33

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