uncountable number of Penrose tiling
I am interested in Penrose tiling, though I am not an expert, so I apologize in advance.
It is known that that the number of different Penrose tilings is uncountable. This may be implied from de Bruijn method, of inducing these tilings from a cross section of a regular (cubic) tiling in 5 dimensions (But maybe I am mixing here between the two areas, since it is not clear for me from de Bruijn method if the size of the tiles themselves vary for each cross-section). However, in Martin Gardner's book "Penrose Tiles to Trapdoor ciphers" (or PDF of first two chpters), the author describes a way of Conway to prove – in another way – that the number of different Penrose tilings is uncountable.
However, this method is completely unclear for me. Here is the description of Gardner for Conway's method:
Conway's proof of the uncountability of Penrose patterns (Penrose had earlier proved it in a different way) can be outlined as follows. On the kite label one side of the axis of symmetry $L$, the other $R$ (for left and right). Do the same on the dart, using $l$ and $r$. Now pick a random point on the tiling. Record the letter that gives its location on the tile. Inflate the pattern one step, note the location of the same point in a second-generation tile and again record the letter. Continuing through higher inflations, you generate an infinite sequence of symbols that is a unique labeling of the original pattern seen, so to speak, from the selected point.
[…] Not all possible sequences of the four symbols can be produced this way, but those that label different patterns can be shown to corespond in number with the number of points on a line.
How does this method induce a bijective function to the points on the line?
geometry tiling
DavidDavid
$\begingroup$ The first 3 pages of this paper outline the proof much more clearly. $\endgroup$ – Dan Rust Sep 5 '14 at 12:06
The set of tilings (which I'll call $T$) corresponds in number to the points on a line, so this essentially claims that $\lvert T\rvert=\lvert\mathbb R\rvert=\mathfrak c$. Instead of actually establishing a bijection, you can also establish two injections.
Upper bound on the cardinality
$\lvert T\rvert\le\lvert\mathbb R\rvert$ is easy: interpret $L,R,l,r$ as digits $0,1,2,3$ and the whole infinite sequence as a decimal fraction. Since the digit $9$ does not occur, you don't have to worry about things like $0.99\!\ldots=1$ so you know that every infinite sequence corresponds exactly to one real number. So you have an injective map from $T$ to $[0,1)\subset\mathbb R$.
Uncountability of sequences
$\lvert T\rvert\ge\lvert\mathbb R\rvert$ is harder. You could interpret rational numbers as fractions with base $4$, so every digit would correspond to one of the letters. You could also do the following preprocessing:
$$f(x):=\begin{cases} L,g(x) &\text{if } x\in[0,1) \\ R,g(1/x) &\text{if } x\in[1,\infty) \\ l,g(-x) &\text{if } x\in(-1,0) \\ r,g(-1/x)&\text{if } x\in(\infty,-1] \end{cases}$$
This will map any $x$ to the range $[0,1]$ which can then be used as input to the second function $g$ that represents that number in base $4$ and thus turns it into a sequence. You could write $1=0.333\!\ldots_4$ or tweak the definition of $f$ a bit (e.g. using $1-g(1/x)$ in the second case) to ensure that you only translate numbers from the range $[0,1)$ so that you can concentrate on the digits after the point.
Uncountability of point labels
So at this point, you have an injective function from $\mathbb R$ to infinite sequences in these four letters. But now you have a technical problem:
Not all possible sequences of the four symbols can be produced this way
You have to work out the exact conditions of which sequences can and which can not be produced in this way. My guess is that you can show that there are only countably many forbidden sequences, so the remaining sequences would still be uncountably many.
Uncountability of tilings
As Daniel reminded me in a comment, what we have worked to so far (at least if you manage to show the countability of the sequences which are not tilings) demonstrates that there are uncountably many sequences describing a pattern seen from a given tile. But there are infinitely many tiles from which you could label the pattern and it would still be the same pattern that should be counted only once.
Two sequences denote the same pattern if they agree after some point $k$. So you can generate alternate labels for the same pattern by subsequently modifying elements of the sequence, starting at the first. You have four options for the first sequence item, and for each of them four for the second, and so on. You could explore all possible sequence modifications in a breadth first search, therefore there are only countably many (i.e. $\lvert\mathbb N\rvert=\aleph_0$) of these.
So why does that not break the uncountability of $T$? I'm not sure how you feel, but (at least at the moment) a counter-argument works best for me. Suppose there were only countably many tilings in $T$, represented by any suitable sequence for each. Then you could use the common $\mathbb N^2$ iteration: Take the original label of the first tiling, then the original label of the second tiling followed by the first modification of the first tiling, then the original label of the thirs tiling followed by the first modification of the second tiling followed by the second modification of the first tiling, and so on. In the end you'd enumerate all point labels, so there could only be $\lvert\mathbb N^2\rvert=\lvert\mathbb N\rvert=\aleph_0$ of these. This is a contradiction to the uncountability of label sequences we showed before.
Now that I think about it, this only shows $\lvert T\rvert>\aleph_0$ but one would need the continuum hypothesis to deduce $\lvert T\rvert=\mathfrak c$ from this. So if anyone has a better explanation (which still works on an intuitive level) of why $\mathfrak c/\aleph_0=\mathfrak c$ feel free to edit this post, comment, or provide your own answer.
MvGMvG
$\begingroup$ There is a slight caveat in this which Gardner seems to ignore. If you choose a different point in your tiling, it will correspond to a different sequence of symbols. We should really be quotienting out by the relation that two sequences are equivalent if and only if they are eventually equal for all terms past some $k$th term. One can show that these equivalence classes are independent of the choice of starting point, and only rely on the tiling. Because these equivalence classes are countable, this does not affect the final counting argument. $\endgroup$ – Dan Rust Sep 5 '14 at 12:50
$\begingroup$ Thanks. The article of Soljanin suggests that the set of Penrose tilings has the same cardinality of the set of all sequences $\{x_n \}$ composed of $0$ and $1$ s.t. if $x_n=1$ then $x_{n+1}=0$. So the problem reduces to prove that this set has cardinality $\aleph$. $\endgroup$ – David Sep 5 '14 at 12:52
$\begingroup$ @Daniel: Thanks for reminding me! I had thought about that at some point, but forgot about it while reading the paper you suggested. $\endgroup$ – MvG Sep 5 '14 at 12:52
$\begingroup$ @MvG: Thanks for extended answer. However, the set of the sequences which are not tilings (using Soljanin notation) is the set of the sequences of 0 and 1 which contain the sequence 11. I am pretty sure that this set is uncountable also, right? (just take the sequences that start with 11, and you may have any sequence you'd like). $\endgroup$ – David Sep 5 '14 at 13:24
$\begingroup$ (...cont.) For instance if $s=00100010010100001\ldots$ then $d(s)=(2,3,2,1,4,\ldots)$. If $s$ begins with a $1$ then we set $n_0=0$. It's clear that $d(s)$ completely determines $s$, and every possible sequence of positive natural numbers is associated to an admissible $s$. The set of all sequences of positive naturals is uncountable, and so then must the sequence of admissible sequences. $\endgroup$ – Dan Rust Sep 5 '14 at 13:58
Here is a nice general statement:
Claim: If $T$ is an FLC, aperiodic, repetitive tiling, then there are uncountably many tilings (taken up to translation) which are locally isomorphic to $T$.
Any given Penrose tiling $T$ is FLC, aperiodic and repetitive, and any other tiling $T'$ is a Penrose tiling if and only if $T$ and $T'$ are locally isomorphic.
Some explanation of these terms is in order.
By a tiling I mean a covering of some Euclidean space $\mathbb{R}^d$ by tiles - each of which are bounded subsets of $\mathbb{R}^d$ and are the closures of their interiors - which do not intersect on anything more than their boundaries.
A tiling is finitely locally complex, or FLC, (with respect to translations) if, for any given radius $R$, there are only finitely many $R$-patches. By an $R$-patch I just mean some set of tiles of $T$ given as precisely the set of tiles intersecting some ball of radius $R$. You should be able to easily convince yourself that a sufficient condition for FLC is that the tiles of $T$ are polygonal, meet face-to-face, and that there are only finitely many patches consisting of two tiles meeting along a common face (up to translation).
A tiling is aperiodic, here, if there does not exist some non-zero vector $x$ such that $T=T+x$.
A tiling is repetitive (with respect to translations) if, for any given $R$, there exists some $R'$ such that every $R$-patch, up to translation, can be found in every $R'$-ball. Of course, assuming that the tiles can't be arbitrarily small, this easily implies that the tiling is FLC. In this case repetitivity asks that, in addition to only having finitely many $R$-patches, these $R$-patches can also be found relatively densely within the tiling (this is a nice notion of "order" of the tiling: being repetitive and aperiodic is special!).
We say that $T$ and $T'$ are locally isomorphic if every $R$-patch of $T$ is an $R$-patch of $T'$, and vice versa. In other words, if you lived in either $T$ or $T'$, then you would not be able to rule one of them out using only local information to any given finite radius. Note that if a tiling is admitted by some local matching rules, so that tiles meet in some agreed way, then of course the same is true of any locally isomorphic tiling (so if your definition of a Penrose tiling is anything which can be constructed from Penrose tiles, then it is clear that any tiling locally isomorphic to a Penrose tiling is also a Penrose tiling).
Proof of claim: Let $T$ be given, satisfying the conditions of the claim. First, let's pick a control point in the interior of each translation-class of tile. We shall only consider tilings for which a control point corresponding to the central tile lies on the origin.
Of course, any locally isomorphic tiling corresponds to some sequence of $1$-patch centred at the origin, followed by a $2$-patch centred at the origin, followed by ..., where the $n$-patch centred at the origin is an extension of the $(n-1)$-patch. Conversely, given such a sequence, it isn't too hard to see that it must be locally isomorphic to $T$ (by repetitivity, every finite sub-patch of $T$ must be found in the tiling defined by such a sequence). Let's count the number of such sequences.
Suppose that for each $n$-patch $P$ there exist at least two distinct ways of extending it to some $(n+c_P)$-patch, for some $c_P>0$. Since there are only finitely many $n$-patches by FLC, there exist at least two distinct ways of extending any given $n$-patch to an $(n+c_n)$-patch, where $c_n>0$ is just the maximum of the $c_P$ ranging over $n$-patches $P$. We see that the number of sequences is at least as large as the set of countable sequences over a two-element set, and $|2^{\mathbb{N}}|$ is uncountable. Note that there are only countably many control points in any given tiling, so the number of tilings taken up to translation must still be uncountable.
So we see that we just need to show that any given $n$-patch of $P$ has at least two distinct extensions. Suppose not. Then there exists some $P$ which can be only extended in some unique way. By repetitivity, there exist at least two distinct control points $x$ and $y$ in $T$ whose $R$-patches are translates of $P$. But since $P$ can only be extended uniquely, we must have that $T-x$ and $T-y$ are equal, so $T = T + (y-x)$, which contradicts aperiodicity.
Let me just note a couple of things following that proof. Firstly, of course, aperiodicity is necessary to have uncountably many locally isomorphic tilings. What we seem to see from the proof is that, in fact, we only need the slightly stronger condition (let's call it (C)) that every translation-class of finite patch has at least two distinct extensions. Actually, this doesn't quite follow from this proof, because a consistent sequence of patches of the tiling need not be locally isomorphic to the original (but it will be "locally derivable" from it: all of its finite patches are finite patches of $T$). So replacing repetitive with (C) and locally isomorphic with locally derivable still allows for the same simple proof.
As an example of an aperiodic tiling not satisfying (C), take a periodic square tiling, but then colour one unique square differently to all of the others (or add a "notch" to it, if you want to force everything geometrically). Then this tiling is FLC and aperiodic but doesn't satisfy (C). And, of course, there are only countably many tilings which are locally isomorphic (or even derivable) to it.
[Edit: I guess that I've only shown that there are at least $|2^{\mathbb{N}}| = |\mathbb{R}|$ translation-classes of tilings in the above proof. But you should easily be able to see that there can only be at most $|\mathbb{N}^{\mathbb{N}}| = |\mathbb{R}|$. Indeed, a locally isomorphic tiling with control point at the origin is described by a finite sequence of choices of central $n$-patch, with $n \in \mathbb{N}$.]
Finally, let me note that the Penrose tilings satisfy the conditions of the claim. This follows easily from their description as either a substitution (or hierarchical) tiling, or as a cut-and-project tiling.
Jamie WaltonJamie Walton
Ok, the first thing we need here, which is actually a common, but sometimes hard to prove, property for a substitution tiling, is that the substitution (or expansion rule) is recognizable. What does recognizable mean? It means that the substitution has an inverse - this can be made precise if we place a metric on the space of all tilings (See my answer to this question) - but for our purposes we'll just say that a recognizable tiling is a tiling where there is a way of taking a tiling, deleting some edges and deflating so that we again end up with a tiling by the same prototile set, and if we compose this process with the substitution then it acts as the identity on the tiling.
What is it that recognizability affords us? Well it means that if we have a tiling $\mathcal{T}$ and we choose some tile $T$ in $\mathcal{T}$, there is a well defined, what is known as a, $1$-supertile $T_1$ containing $T=T_0$, and because we can continue the process, also a $2$-supertile $T_2$, $3$-supertile $T_3$, etc. The $1$-supertile is just the new expanded tile that $T$ is a subset of when we delete the edges during the inverse substitution process.
That the Penrose tilings are recognisable is a non-trivial problem that requires proof. There may be an elementary way which is special to the Penrose tiling, but there is also a theorem of Mossé, and generalised by Solomyak$^{[1]}$, which says that any aperiodic, locally-finite, primitive substitution tiling of $\mathbb{R}^n$ is recognizable (the Penrose tilings satisfy these properties).
Now that we know every tile $T$ in a Penrose tiling $\mathcal{T}$ has a well defined sequence of $n$-supertiles $T_n$, we can associate to every pair $(\mathcal{T},T)$ that sequence. We'll call it $s(\mathcal{T},T)$. The first thing we need to show is that for all $T,T'\in\mathcal{T}$, we have that there exists a $k\geq 0$ such that $$\sigma^k(s(\mathcal{T},T))=\sigma^k(s(\mathcal{T},T'))$$ where here $\sigma(a,b,c,d,\ldots)=(b,c,d,\ldots)$ is the shift function on sequences. This is easy to see though because if $T$ and $T'$ are tiles in a fixed Penrose tiling $\mathcal{T}$, then they are both contained in some ball $B$ in the plane. I claim (and you should prove) that $B$ is fully contained in some $k$-supertile $\hat{T}$. It follows from recognizability that $T_{k}=\hat{T}=T'_{k}$ and so $T_l=T'_l$ for all $l\geq k$, hence $\sigma^k(s(\mathcal{T},T))=\sigma^k(s(\mathcal{T},T'))$.
The next thing we need to figure out is when a sequence $s$ is actually associated to a Penrose tiling. As discussed in Soljanin's paper$^{[2]}$, If we assign one tile the symbol $0$, and the other the symbol $1$, then it can be shown that the only admissible sequences (those associated to a Penrose tiling) are those which do not have the substring $11$ appearing in the sequence. Let's call the set of admissible sequences $A=\{s\mid 11 \mbox{ does not appear in }s\}$.
A non-trivial question to ask is, given an admissible sequence, is its associated Penrose tiling unique? The answer is yes, and the proof is quite obvious. An admissible sequence tells you exactly how to lay down tiles to build your tiling. It's essentially a set of instructions for building a Penrose tiling. The tile $T_0$ is what you place at the origin, $T_1$ is the $1$-supertile you cut up and extend away from $T_0$, and so on. The caveat here is that it's possible these supertiles will only extend off in one direction and not cover the entire plane. I won't go into details as to why this isn't a problem, but just note that we can always fix this problem using a technique known as collaring (See Anderson and Putnam's paper$^{[3]}$ for more details).
So, we have a well defined bijection $A\to\{(\mathcal{T},T) \mid \mathcal{T} \mbox{ is a penrose tiling}, T\in \mathcal{T}\}$. From our discussion about choosing different tiles as the 'seed tile' $T_0$ for a specific tiling, we then also get a bijection $A/{\sim}\to\{\mathcal{T} \mid \mathcal{T}\mbox{ is a Penrose tiling}\}$ where two sequences $s,s'$ are related by $\sim$ here if there exists a $k$ such that $\sigma^k(s)=\sigma^k (s')$, so $A/{\sim}$ is the set of classes of eventually equal admissible sequences.
Counting Argument
Claim: $A$ is uncountable.
Proof: Let $s$ be in $A$ (and also just to make life easier assume that there is no $k$ such that $\sigma^k(s)\neq 000\ldots$ - in the end this only accounts for a single equivalence class so we can ignore it). Then $s$ has an associated sequence $d(s)$ of positive natural numbers $n_0,n_1,n_2,\ldots$ corresponding to the number of consecutive $0$s that occur between the $1$s which appear in $s$. For instance if $$s=00100010010100001\ldots$$ then $$d(s)=(2,3,2,1,4,\ldots).$$ If $s$ begins with a $1$ then we set $n_0=0$. It's clear that $d(s)$ completely determines $s$, and every possible sequence of positive natural numbers is associated to an admissible $s$. The set of all sequences of positive naturals is uncountable, and so then must the sequence of admissible sequences $A$.
Claim: $A/{\sim}$ is uncountable.
Proof: Let $s$ represent an equivalence class in $A/{\sim}$. Let $F$ be the set of all finite strings of $0$s and $1$s. There is an injective function $[s]_{\sim}\to F$ given by assigning to $s'\in[s]_{\sim}$, the shortest initial substring that can be replaced in $s'$ to recover the sequence $s$. As $F$ is countable, it follows that $[s]_{\sim}$ is countable for all $s\in A$.
Now, suppose that $A/{\sim}$ is not uncountable, then it is countable and so $A$ can be written as a countable disjoint union of countable sets. This would imply that $A$ is countable which contradicts the previous claim. It follows that $A/{\sim}$ must be uncountable.
Conclusion: As a corollary we conclude that $\{\mathcal{T} \mid \mathcal{T}\mbox{ is a Penrose tiling}\}$ is uncountable.
$[1]$ B. Solomyak. Nonperiodicity implies unique composition for self-similar translationally finite tilings. Discrete Comput. Geom. 20 (1998), 265–279.
$[2]$ E. Soljanin. Writing Sequences on the Plane. IEEE Trans. Inf. Thy. 48 (2002), 1344-1354
$[3]$ J. E. Anderson and I. F. Putnam. Topological invariants for substitution tilings and their associated $C^*$-algebras. Ergodic Theory Dynam. Systems, 18(3) (1998), 509-537.
Dan RustDan Rust
$\begingroup$ I actually have another question: since the $d(s)$ seq' for the SUN and the STAR patters are resp. 2,3,3,3... and 0,3,3,3,... can one induce from these patterns the fact that these are the only tilings with 5-rotational symmetry (just by looking on the $d(s)$'s? $\endgroup$ – David Sep 10 '14 at 7:47
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