I've read in Lemma 2 of the paper 1 that if every square region of the plane admits a tiling, then the whole plain admits a tiling, but the proof is omitted. This sounds like a compactness property, but I don't see how easy can be proven.

I added the tag "automata theory" because there is a correspondence between Wang tilings and finite state transducers.


2 Answers 2


Yes, this is a compactness property.

Let $X$ be the region you want to tile, and $T$ the finite set of possible tiles. The space $T^X$ of all assignments of tiles to $X$ is compact by Tychonoff’s theorem. For any finite $X_0\subseteq X$, the set $C_{X_0}\subseteq T^X$ of all correct tilings of $X_0$ is closed (in fact, clopen), and since $C_{X_0\cup X_1}\subseteq C_{X_0}\cap C_{X_1}$, they generate a filter. Thus, assuming every finite subset of $X$ can be tiled, the system $\{C_{X_0}:X_0\subseteq X\text{ finite}\}$ has fip, and by compacteness, its intersection is nonempty. This intersection is the set of all correct tilings of $X$.

Instead of all finite $X_0$, it suffices to use any family of finite subsets which is upwards directed and covers $X$, such as, in your case, the set of all finite square subregions of the plane.

Alternatively, you may set up the argument to use the compacteness of classical propositional logic: let $\{p_{x,t}:x\in X,t\in T\}$ be a set of propositional variables, and let $C$ be the theory consisting of the formulas $$\begin{gather*} \bigvee_{t\in T}p_{x,t},\qquad x\in X,\\ \neg(p_{x,t}\land p_{x,t'}),\qquad x\in X,t\ne t'\in T, \end{gather*}$$ and $$\neg(p_{x,t}\land p_{x',t'})$$ for all $x,x'\in X$ that are neighbours, and $t,t'\in T$ that are incompatible when placed on $x$ and $x'$, respectively. If each finite subset of $X$can be tiled, then $C$ is consistent, and any satisfying assignment gives a tiling of $X$.

  • $\begingroup$ Yes, I thought about an argument using compactness of propositional logic. Yet this argument is in a sense not constructive because implicitly uses the Axiom of Chioce and hides an algorithmic property that should sound as follows: if $t_1: X \to T $ is a tiling and $t_2: Y \to T$ is another one whose with $Y\supseteq X$, then there exists a finite sequence of transformations of $t_2$ yielding a $t_3: Y \to T$ that extends $t_1$. Something similar is detailed in this answer math.stackexchange.com/a/36109/34750 but concerns other domini pieces rather than Wang tiles. $\endgroup$
    – Alberto
    Apr 1, 2019 at 7:23

First, there is an infinite tiling of the plane if and only if there are square tilings $$ T_1 < T_2 < T_3 < \ ... $$ such that $T_i$ is an $i \times i$ square tiling, and $T_i$ is a subtiling of $T_{i+1}$.

Now, for each $T_i$ define $f(T_i)$ to be the size of the largest square tiling that $T_i$ is a subtiling of, where $f(T_i) = \infty$ if for all $n > i$, $T_i$ is a subtiling of some $n\times n$ tiling.

Lemma 1: if for an $i \times i$ tiling $T_i$, $f(T_i) = \infty$, then there is some $(i+1) \times (i+1)$ tiling $T_{i+1}$ with $T_i < T_{i+1}$ and $f(T_{i+1}) = \infty$.

Proof: Suppose not. Then look at the maximum value $n$ of $f(T)$ for $(i+1) \times (i+1)$ tilings containing $T_i$:

$$n = \max_{T > T_i \atop T \mathrm{\ is\ } (i+1) \times (i+1)} f(T). $$

If this is finite, then no tiling containing $T_i$ can be larger than $n \times n$, contradicting $f(T_i) = \infty$. And if it is infinite, one of the $(i +1) \times (i+1)$ tilings $T$ must have $f(T) = \infty$, as otherwise we have $\infty$ is the maximum of a finite number of finite integers (since there are a finite number of $(i+1) \times (i+1)$ tilings).

So at least one of the $(i+1) \times (i+1)$ tilings containing $T$ must have an infinite value for $f$. QED

Now, we can ask whether $f(T_1) = \infty$ for any single $1 \times 1$ tile. If not, then there is a maximum size square that admits a tiling. And if so, then given a tiling $T_1$ with $f(T_1) = \infty$, we can use lemma 1 to find tilings $$ T_1 < T_2 < T_3 < \ ... $$ such that $f(T_i) = \infty$ for all $i$, showing that there is an infinite tiling of the plain.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.