TITLE: uncountable number of Penrose tiling
QUESTION [8 upvotes]: 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?

REPLY [2 votes]: 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.