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The set of possible kit-and-dart Penrose tilings is uncountably infinite. It would be very helpful to have some natural probability distribution $\mu$ over this set; such a distribution would allow us to make statements like "Viewed as an infinite planar graph, the expected degree of a Penrose tiling is...." and be a little more precise (see, e.g., papers like this).

To address some symmetry issues, let us restrict to the set of Penrose tilings with a tile corner at the origin and an edge between tiles along the positive x-axis ("an edge laying due east"). We will refer to this (also uncountably infinite) set as $\mathcal{P}$.

We would really like $\mu$ to cover all of $\mathcal{P}$ in some sense. So, does there exist a probability distribution $\mu$ over $\mathcal{P}$ that satisfies the following?

Consider $F$, an arbitrarily large, finite subset of tiles in the plane selected from any Penrose tiling in $\mathcal{P}$. Let $P_F\subseteq \mathcal{P}$ be the set of Penrose tilings that contains that fragment in the same location. Then $\mu(P_F)>0$.

If there were an efficient algorithm for sampling from $\mu$, that would be even more helpful, but perhaps I'm hoping for too much.

Edit: Per Steven Stadnicki's and Mateusz Kwaśnicki's comment's, I've clarified that the fragment $F$ occurs in the same location.

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  • $\begingroup$ I am not knowledgeable in that topic, but I do remember Alain Connes's Noncommutative Geometry book mentioning Penrose tilings as one of the examples. Maybe this helps? $\endgroup$
    – Malkoun
    May 20, 2021 at 15:27
  • $\begingroup$ My guess is that the (compact - with the usual topology where "being close enough" means being equal on a large centred ball) space of such tiling has a unique ergodic measure with respect to translations, but at this moment I have nothing to back this up. "Inflation-deflation" approach would then give a way to sample from this distribution. $\endgroup$ May 20, 2021 at 15:55
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    $\begingroup$ Any arbitrary large finite subset of tiles selected from any Penrose tiling appears in every Penrose tiling infinitely often, so your measure is essentially trivial. This is in fact the point of the tilings' appearance in Connes' book, IIRC — traditional measures (in both senses) are insufficient because there is no finitary way to distinguish any two of the uncountably many tilings. $\endgroup$ May 20, 2021 at 16:02
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    $\begingroup$ The tiling space of the penrose tilings is uniquely ergodic with respect to the translation action. As is any reasonable notion of the 'canonical transversal' which is essentially the one you describe as the set of tilings with a vertex at the origin (the action on this set is a little more difficult to describe though and really needs groupoids). This is all in 'Aperiodic Order, Vol. 1' by Baake and Grimm. $\endgroup$
    – Dan Rust
    May 21, 2021 at 0:03
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    $\begingroup$ By Birkhoff's ergodic theorem then, the unique measure is the one which assigns to a set of tilings with a patch $P$ at the origin, the frequency of that patch $P$ in a penrose tiling (which is independant of the choice of tiling) $\endgroup$
    – Dan Rust
    May 21, 2021 at 0:06

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At stated in the comments, all Penrose tilings contain any finite patch infinitely often, so your criterion doesn't narrow things down much. But judging from the sort of intuitive notion you're describing, it sounds like the pentagrid method is what you're looking for? It produces every Penrose tiling using a 5-tuple of real numbers in $[0,1]$ (up to a measure-0 set of invalid choices), so it is very easy to generate samples from.

The idea is to place down infinite "ladders" of parallel lines of constant spacing, one each at angles of $0^\circ, 72^\circ,\ldots,288^\circ$:

                                        enter image description here

We then associate to every intersection of two lines a skinny rhomb (if the intersecting lines differ by $36^\circ$) or a fat rhomb (if they differ by $72^\circ$). This describes the adjacency graph of the tiles, and with a little more work you can place them at the right coordinates. (From there, of course, it's easy to transform into a kite-and-dart tiling.) By normalizing and rotating, you can ensure your tiling is drawn from $\mathcal{P}$.

You may find this MSE post useful to get more information about pentagrids and constructing tilings from them - it contains many informative links about the process.

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  • $\begingroup$ Thank you! This sounds perfect. $\endgroup$ May 21, 2021 at 0:04
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The tiling space of the penrose tilings is uniquely ergodic with respect to the translation action. As is any reasonable notion of the 'canonical transversal' (also sometimes called the discrete hull) which is essentially the one you describe as the set of tilings with a vertex at the origin (the action on this set is a little more difficult to describe though and really needs groupoids). This is all in Aperiodic Order, Vol. 1 by Baake and Grimm.

By Birkhoff's ergodic theorem then, the unique measure is the one which assigns to the set of tilings with a patch $P$ at the origin, the frequency of that patch $P$ in a Penrose tiling (which is independent of the choice of tiling). These frequencies can all be calculated as entries of the corresponding right Perron-Frobenius eigenvector of the associated substitution matrix (or the matrix of the collared substitution with collaring radius suitably large to contain the patch). It's easier to do this with the Robinson triangles because they give an exact Stone inflation, but the same method does work for the kite-and-dart substitution.

Another method for calculating the frequencies is to put everything in terms of the cut-and-project method, in which case the relative volume of the 'acceptance domain' of a particular patch $P$ (compared to the volume of the entire window) is the frequency of the patch $P$.

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