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Let a dodecahedron sit on the plane, with one face's vertices on an origin-centered unit circle. Fix the orientation so that the edge whose indices are $(1,2)$ is horizontal. For any $p \in \mathbb{R}^2$, define the dodecahedral distance $dd(p)$ from $o=(0,0)$ to $p$ to be the fewest number of edge-rolls that will result in a face of the dodecahedron landing on top of $p$. Equivalently, imagine reflecting a regular pentagon over edges, as illustrated below: It takes $4$ rolls/reflections to cover $p=(5,\pi)$:


          Dodeca5pi
          $p=(5,\pi)$, $dd(p)=4$, $s=(3,1,4,2)$.
My main question is:

Q. Given $p$, how can one calculate $dd(p)$?

Greedily choosing, at each step, the roll that is best aligned with the vector $p-o$ does not always succeed.

Could one characterize the sequences of roll indices $s$, where rolling over edge $(i,i+1)$ of the pentagon is index $i\,$? These sequences feel analogous to continued-fraction approximations. What do all the points $p$ of $\mathbb{R}^2$ with $dd(p)=k$ look like, i.e., what is the shape of a $dd$-circle $C_k$?


          Dodeca717
          $p=(18.3,-1.4)$, $dd(p) \le 12$, $s=(2, 4, 1, 3, 5, 2, 4, 1, 3, 5, 2, 5)$.


That the dodecahedral distance is well-defined follows, e.g., from "Thinnest covering of the plane by regular pentagons."

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    $\begingroup$ Looking at the shapes of two, three, and four circles may be informative. Pentagon distance would be a good approximation: roll along an optimal angle to achieve y distance, then go along x to hit point. My guess is Pentagon distance and roll distance will differ by at most one. Gerhard "Try Drawing A Five Sphere" Paseman, 2017.12.15. $\endgroup$ – Gerhard Paseman Dec 15 '17 at 15:17
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    $\begingroup$ The following paper might be helpful: Dmitri V. Alekseevsky, Peter W. Michor, Yurii A. Neretin: Rolling of Coxeter polyhedra along mirrors. In: Geometric Methods in Physics: XXXI Workshop 2012. Pages 67-86. Trends in Mathematics, Birkhauser, 2013. arXiv:0907.3502. $\endgroup$ – Peter Michor Dec 15 '17 at 15:18
  • $\begingroup$ @Joseph O' Rourke Basis.. each time a face rolls through supplement of dihedral $arctan2$ $\endgroup$ – Narasimham Dec 15 '17 at 22:35
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Here are a few trivial lemmas. I won't use anything about the rolling motion, just that the distance is defined by gluing pentagons edge-to-edge:

  • The $dd$-circle of radius $k$, which I'll call $C_k$, is a closed polygonal curve. Let $D_k$ be the closed $dd$-disk of radius $k$; note that $D_k$ may contain holes and $C_k$ is not in general simple! (This happens even at $k=2$.)
  • The orientations of the line segments forming $C_k$ (measured relative to the $+x$ ray) when $k$ is even (odd) take the form $\pi m/5$ with $m$ an odd (even) integer; this is because pentagons are glued to each other in only two orientations (up-pointing or down-pointing) and furthermore, we can only glue down-pointing pentagons to up-pointing ones and vice versa. (Note also that no line segments of $C_k$ lie on $C_{k+1}$).
  • It's not hard to see that $C_k$ is strictly contained within $D^\circ_{k+2}$, the interior of $D_{k+2}$ (the open $dd$-disk). Suppose that $v$ is a simple vertex of $C_k$ so that the interior angle $\alpha$ is defined and $\alpha> 4\pi/5$. Then $v$ is surrounded by the pentagons added to its adjacent segments and therefore lies inside $D^\circ_{k+1}$. Otherwise, if the interior angle $\alpha\leq 4\pi/5$ then $v$ is also a vertex on $C_{k+1}$. However the interior angle at $v$ on $C_{k+1}$ is now $\alpha+6\pi/5>4\pi/5$, so it gets "eaten up" in the next layer.

Here are pictures of $D_k$ for $k=1,\dots,8$ placed side by side with $D_{k}\setminus D_{k-1}$, (made by "hand" from edited versions of this file from wikipedia):

$D_1$:

$D_2$ and $D_2\setminus D_1$:

$D_3$ and $D_3\setminus D_2$:

$D_4$ and $D_4\setminus D_3$:

$D_5$ and $D_5\setminus D_4$:

$D_6$ and $D_6\setminus D_5$:

$D_7$ and $D_7\setminus D_6$:

$D_8$ and $D_8\setminus D_7$:

From these images, the following patterns are apparent: the "outer frontier" of $C_k$ is determined by a closed chain of $5k$ pentagons joined vertex to vertex. When $k$ is odd, these pentagons are all "down-pointing" and when $k$ is even they all are "up-pointing". From now on assume $k\geq2$. Then the pentagon chain is formed from 10 "segments of pentagons", which I'll call pentasegments, joining 10 corner pentagons where the pentasegments change direction.

There are two types of pentasegments: those where the bases of the constituent pentagons point outwards relative to the interior of $D_k$ (type I):

type I pentasegment

and those where the vertices of the pentagons point outwards (type II):

type II pentasegment

The top pentasegment of $C_k$ is type I if $k$ is odd and type II if $k$ is even and the 10 pentasegments alternate between type I and type II so that there are 5 of each type.

When $k$ is even, there are $k/2-1$ pentagons on each pentasegment lying strictly between the corners (the number of pentagons on each pentasegment including the 2 corners is $k/2+1$).

When $k$ is odd, there are $(k-1)/2$ pentagons on each type I pentasegment between the corners and $(k-3)/2$ pentagons on each type II pentasegment between the corners. There are holes between each of the pentagons lying on the type II pentasegments, thus there are $5\lfloor \frac{k}{2}\rfloor$ holes in total.

I think the above gives a full characterization of $C_k$ and also shouldn't be too hard to prove, though I'm finding it awkward to turn the pictures into words. First, the corners propagate along "zig-zag" paths of pentagons; for example, a formula for the coordinates in $\mathbb{C}$ of the centroid of one of the corners of $C_k$ is:

$$2r_{in}\sum_{j=1}^ke^{i\pi m_j/5},$$

where $r_{in}=\frac{1}{2}\sqrt{1+\frac{2}{\sqrt{5}}}$ is the inradius of the regular pentagon (formula from mathworld), $m_{2l-1}=2$, and $m_{2l}=1$. This simplifies to:

$$\left\lceil\frac{k}{2}\right\rceil e^{2\pi i/5}+\left\lfloor\frac{k}{2}\right\rfloor e^{\pi i/5},$$

and there are 9 other similar formulas for the other corners. Another easy step is seeing that type I pentasegments turn into type II pentasegments (and vice versa) after attaching the next layer of pentagons, and seeing how the number of pentagons along each type of pentasegment changes is also straightforward. Maybe there's a more elegant description which might then lead to a simple formula / algorithm to compute $dd(p)$.

Here's some geometry which may help in writing an algorithm; as you can see, the details are straightforward but rather messy. If I haven't screwed up the law of cosines, the centroids of the corner pentagons of $C_k$ lie on the circle of radius $R_k$ centered at the origin, where:

$$\left(\frac{R_k}{r_{in}}\right)^2=\left\lceil\frac{k}{2}\right\rceil^2+\left\lfloor\frac{k}{2}\right\rfloor^2+2\left\lceil\frac{k}{2}\right\rceil\left\lfloor\frac{k}{2}\right\rfloor\cos\frac{\pi}{5}.$$

The 10-gon $P_k$ formed by these centroids is inscribed in this circle and is determined once we calculate the polar angles of two neighboring centroids, $\alpha_k,\beta_k$. This can be done with the law of sines but I won't write out expressions explicitly here.

Let $S_k$ be the polygonal annulus lying strictly between $P_k$ and $P_{k+1}$. Given a point $p$, it will lie in some $S_k$ or on some $P_k$, and this tells us that $|dd(p)-k|\leq 1$. I don't know what the best algorithm for determining this $k$ is (something using the polar coordinates of $p$?), but I suspect Joseph O'Rourke will know. Once we get $k$, then it's possible to pin down $dd(p)$ exactly by analyzing in more detail where $p$ sits relative to the edges of $C_k$, though again I'm not sure what the most efficient algorithm would be.

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    $\begingroup$ Beautiful! That there are holes in the $dd$-disk even for $k=2$ is an interesting surprise. $\endgroup$ – Joseph O'Rourke Dec 15 '17 at 23:48
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    $\begingroup$ The periodic packing shown in figure 7 of this article by Craig Kaplan plus.maths.org/content/trouble-five (see also jiggerwit.wordpress.com/2016/09/16/pentagon-packings by Tom Hales and Wöden Kusner, who attribute it to Dürer) seems relevant to determining the shape of $C_k$ but I haven't got a clear idea yet. $\endgroup$ – j.c. Dec 15 '17 at 23:55
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    $\begingroup$ If it's not much trouble, could you add more of these pictures? They might reveal more about those holes. $\endgroup$ – domotorp Dec 16 '17 at 4:03
  • $\begingroup$ I wonder if your $C_k$ analysis could be turned into an algorithm for computing $dd(p)$, perhaps by binary search? $\endgroup$ – Joseph O'Rourke Dec 17 '17 at 0:51
  • $\begingroup$ Thank you for this amazingly detailed analysis! I will think about the algorithmic issues. $\endgroup$ – Joseph O'Rourke Dec 17 '17 at 20:41
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Here are a few pictures and speculations to add to the the great ones from @j.c.

Here are the pentagons which arise from starting with the central green one and doing the reflections which do not overlap any previous ones.

enter image description here

These are just the outer frontiers of the $C_i.$ (Specifically, this shows the frontiers for $i \le 7$ and half the frontier for $i=8$.) It seems clear (though I do not claim a proof) that for a target point inside one of the pentagons one does best to roll out to it using just these pentagons. For a target point in one of the rhombi one should roll out to a neighboring pentagon and at the last step reflect over a previously unused (for reflections) edge. One can use one of the bordering pentagons closest to the center for the closer half of the rhombus and a small part of the further half.


Here is a central portion of the same thing with lines connecting the centers of pentagons which share an edge. The resulting hexagons tile the plane with a $5$-fold rotational symmetry.

enter image description here


Finally, here is a picture of just these hexagons going out further. For each of the $5$ orientations, the hexagons of that orientation fall into $2$ connected components. One orientation is highlighted in blue.

enter image description here

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  • $\begingroup$ I like your conjecture that only in the last step might one need to roll into a rhomb. My explorations (e.g., my 2nd example) support this hypothesis. I find your pentagon-rhomb tiling (your 1st image) quite beautiful. Might it be "new" in some sense? $\endgroup$ – Joseph O'Rourke Dec 18 '17 at 13:22
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    $\begingroup$ Albrecht Dürer had it first ecademy.agnesscott.edu/~lriddle/ifs/pentagon/durer.htm. it takes a bigger patch (for me) to see what is going on. $\endgroup$ – Aaron Meyerowitz Dec 18 '17 at 13:54
  • $\begingroup$ Dürer's "narrow lozenges" = rhombi. Of course that's a translation from 16th-century German. But "rhombus" is in Euclid. It would be interesting to see the original Dürer language... $\endgroup$ – Joseph O'Rourke Dec 18 '17 at 14:35
  • $\begingroup$ Re: the first image, the attribution to Dürer is also in the article by Craig Kaplan that I linked to in a comment to my own answer, see Fig. 8. of plus.maths.org/content/trouble-five . The original medieval German text can be found here de.wikisource.org/wiki/… (see pages [66] and [67], the text beginning with "Fvrbas wil jch ein fuenf ...") $\endgroup$ – j.c. Dec 18 '17 at 15:52
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    $\begingroup$ The relevant passage from Aaron Meyerowitz's link above seems to be the paragraph beginning: "Zum fuenftenn setz..." and I believe the words translated as "five narrow lozenges" by Strauss are "fuennf schmal rauten". As far as I can tell (though I am not a native speaker), Raute also means rhombus in modern German as well. $\endgroup$ – j.c. Dec 18 '17 at 15:58

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