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Let $T$ be a tiling of the plane. Fix an origin and shoot a ray $r$ from the origin. Mark off points $p_i$ along $r$ separated by unit distance. Compute from $r$ a binary number $0 < b(r) < 1$ that alternates $0$'s and $1$'s for each marked point $p_i$ as the ray enters a new tile of $T$. For example, the square tiling and illustrated ray below lead to $$ .00011011100010011011100 \ldots $$

          BitsRay
          Square tiles side length $=7/3$. Ray slope $= 1/\sqrt{2}$.
To avoid thin tiles, assume every tile includes a disk of diameter $> 1$ so that more than one $p_i$ could land in a tile. One needs a rule when $p_i$ is on the boundary of a tile to make $b(r)$ well-defined, but I think that detail is not relevant to my question.

It is not difficult to find tilings and rays where $b(r)$ is rational, irrational, or transcendental, for example, by selecting the slope appropriately in the above example.

Q1. Is there a tiling $T$ such that every $b(r)$, for all origins and rays $r$, is transcendental?

If the answer to Q1 is No, the following two questions are superfluous:

Q2. What is the class of all such transcendental tilings (if I may coin a term)?

Q3. How does this class relate to the aperiodic tilings?

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    $\begingroup$ As for Q1, I would imagine that a regular rectangular tiling, where the rectangles all have side lengths $e$ and $\pi$ (or something like that) should do the trick. (Proving this works could be a bit of a chore, though.) $\endgroup$
    – Will Brian
    Commented Apr 21, 2017 at 16:03
  • $\begingroup$ @WillBrian: If the rectangles are $\pi \times e$ and the slope of the ray is $e/\pi$, then ... $\endgroup$
    – Joseph O'Rourke
    Commented Apr 21, 2017 at 16:28
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    $\begingroup$ If the slope of the ray is $e / \pi$ and the ray happens to originate at a corner of one of the rectangles, then I think $b(r)$ should be transcendental if and only if $e \cdot \pi$ is. (If the ray originates elsewhere I'm less sure, but I think this should still be true.) It is an open question whether $e \cdot \pi$ is transcendental, so I guess we don't want to work with this tiling (in this case, "a bit of a chore" is a bit of an understatement). What we want instead is rectangles of side length $\alpha$ and $\beta$, where $\alpha$, $\beta$, and $\alpha \cdot \beta$ are all transcendental. $\endgroup$
    – Will Brian
    Commented Apr 21, 2017 at 17:17
  • $\begingroup$ @WillBrian: Oh, you're right; sorry. $\endgroup$
    – Joseph O'Rourke
    Commented Apr 21, 2017 at 18:14
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    $\begingroup$ @JosephO'Rourke I'm not sure this is exactly what you're looking for, but this 2022 paper seems relevant: Substitution tilings with transcendental inflation factor $\endgroup$
    – Max Lonysa Muller
    Commented Apr 30, 2023 at 9:44

1 Answer 1

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Answering Q1, I believe there is a transcendental tiling. Let us begin with this tiling with congruent convex pentagons:

          enter image description here

Notice that one can rearrange any of the triples of pentagons that form the regular hexagon, by rotating any triple we want by 180 degrees. Thus we can have two kinds of triples: pointing up or down. Now, the entire tiling can be viewed as formed by triples, each filling a hexagon, and we can view the tiling by hexagons as the union of concentric "annuli" with disjoint interiors. Make it so that the annuli's thickness (not just diameter) is strictly increasing, if needed, increasing exponentially. Then arrange all triples in the same annulus to be pointing in the same direction, alternating the direction when passing from an annulus to the adjacent one.

HexNested
(Image added by J.O'Rourke.)

Then, although I do not have a complete proof, I believe the tiling will be transcendental. My feeling is based on the example of a binary transcendental number formed by alternating blocks of zeros and ones of strictly increasing length.

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    $\begingroup$ (I incorporated your image directly.) $\endgroup$
    – Joseph O'Rourke
    Commented Apr 21, 2017 at 23:28
  • $\begingroup$ @JosephO'Rourke: If it won't take too much time for you, could you make a larger drawing to show at least a couple of rings (annuli)? $\endgroup$
    – Wlodek Kuperberg
    Commented Apr 21, 2017 at 23:36
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    $\begingroup$ OK: start with "the seed": a single hexagon, say at the origin. The first annulus, of thickness 1, is a single ring of the six hexagons that touch the seed. The second annulus, of thickness 2, consists of two rings: the hexagons OUTSIDE the 1-st annulus that touch it, plus the next ring of hexagons: those that touch the hexagons of the previous ring. In other words, the rings are formed by the hexagons that touch the previous ring, the annuli are the unions of consecutive rings. If we want the annuli to grow exponentially, then the 3-rd annulus should consist of 4 rings, and so on. $\endgroup$
    – Wlodek Kuperberg
    Commented Apr 21, 2017 at 23:53
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    $\begingroup$ Yes, it is. But the asymptotic behavior of the pattern of zeros and ones does not depend on the origin. $\endgroup$
    – Wlodek Kuperberg
    Commented Apr 22, 2017 at 1:42
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    $\begingroup$ Why not tile with triangles? Eg start with a square lattice, and divide every square [x,x+1] \times [y,y+1] into two triangles along the diagonal, using the 45-degree line if max(|x|,|y|) has odd binary length, and using the 135-degree line otherwise. $\endgroup$
    – user44143
    Commented Apr 22, 2017 at 7:57

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