# Can we enforce a P3 Penrose tiling using unmarked rhombi with a small set of local matching rules?

Suppose I've got some rhombi that I want my friend to construct a P3 Penrose tiling out of:

However, the edge markings on my tiles have worn off, so I need to give my friend instructions about how to place unmarked tiles edge-to-edge such that they are constrained to only produce valid Penrose tilings from the set.

One local condition that holds of a Penrose tiling is that no rhombus shares an edge with a congruent rhombus in the same orientation - we could try giving my friend this rule to follow. However, that's not enough, because they could still screw up and make something like this:

Is there a small number of additional "simple" rules we can impose that will constrain the possible tilings nicely? Of course something like "all tile arrangements within radius 10 of a vertex must be one of the following forms" should suffice, but I'm curious if there's a compact characterization of Penrose tilings that enforces the necessary matching rules without relying on correspondences between distant tiles or lots of casework. Apologies for the subjectivity of this question, but I hope it's clear what I'm gesturing at.

A recent paper by Lutfalla and Fernique provided one possible answer to this question. The restrictions imposed by the possible vertex figures do not suffice, as demonstrated in this counterexample:

But if we require the tiles touching every edge emanating from a given vertex to come from a valid Penrose tiling, i.e. one of the following 15 configurations:

then it is forced to be a valid Penrose tiling.

This is not that compactly describable, though, so I'm not accepting this self-answer yet in hopes that a better classification of unmarked Penrose tilings is posted.

Edit 2023-05-14: Theorem 6.1 of M. Seneschal's Quasicrystals and Geometry states that any tiling whose vertex figures ("stars") are one of the seven allowable configurations and such that no two stars that share a rhomb are related by a half-turn about the center of that rhomb is a Penrose tiling. (Note that this second condition fails to obtain in the above counterexample - there are two kinds of vertices with the aforementioned 180 degree symmetry.)

After writing some code to explore this more, I've found a characterization that I like.

In any tiling of the plane by rhombi, we can cross from an edge to the parallel edge on the opposite side of a face, and repeat this procedure to get a "ladder" of parallel edges, which also gives us a ribbon of rhombs joined at opposite edges. An example of this phenomenon:

In order to enforce that our tiling be a Penrose tiling, it suffices to enforce the following two rules:

• For a given type of rhomb (thick or thin), the directions they point in as you go along a ladder should alternate. So if a vertical ribbon has a thin rhomb pointing left, the next thin rhomb on that ladder should point right.

• No vertex should have all four angles ($$\frac\pi5, \frac{2\pi}5, \frac{3\pi}5, \frac{4\pi}5$$) meet.

This amounts to forbidding the following ten configurations (in fact the first rule forbids infinitely many, but the below suffice to force the tiling to obey the matching rules):