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Penrose tiling with a line drawn through it

Let us consider a Penrose tiling of $\mathbb R^2$. Starting with an arbitrary point on the tiling, draw an arbitrary straight line. Assume that this straight line never overlaps perfectly with a boundary edge of a tile.

Create a sequence by recording what tiles the straight line intersects as it proceeds from its starting point to infinity. As an example, for the above picture the first few terms of the sequence will be $$GBGGBBBGBGGBGBGG...$$ Where $G$ represents the Green skinny rhombus tile, and $B$ represents the fat blue rhombus tile

Is there an algorithm to generate sequences of this type? So far I have been drawing a line and marking the tiles the line intersects by sight, and I am hoping there is an easier way.

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  • $\begingroup$ Hi Darren. This is just a guess, but if you consider this Penrose tiling as a cut and project tiling with a certain total/physical/internal space, then if you lift the tiling to total space, does this line also lift uniquely to a straight line going through total space? So then this sequence is really just the projection/recoding of a higher dimensional cutting sequence through the cubical tiling of $\mathbb{Z}^5$. $\endgroup$
    – Dan Rust
    Commented May 26, 2023 at 15:54
  • $\begingroup$ @DanRust: I think this might work, although we probably want to lift to a hyperplane in $\mathbb R^5$ instead whose shadow is the given line. The objects we're intersecting with are kind of complicated from the 5-dimensional perspective though; I don't see offhand how to describe them in a way that would make this nice. $\endgroup$ Commented May 26, 2023 at 20:26
  • $\begingroup$ @DanRust that is a good idea - I think there is a possibility that the solution could lie in that direction, but as RavenclawPrefect notes the objects are complicated. I'll think about this. The cut-and-project tiling Dan Rust describes is in doi.org/10.1016/1385-7258(81)90017-2 in case that is helpful to anyone $\endgroup$
    – Darren Ong
    Commented May 27, 2023 at 2:11

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