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This question is motivated by Escher's series of Metamorphosis woodcuts (see e.g. here), where one tesselating tile is gradually transformed into another. Basically, this is a precise way of asking the following: can we always "metamorphose" between any pair of tesselating shapes? (Now asked in modified form at MO.)


First, some definitions (everything is in $\mathbb{R}^2$ with the usual metric):

  • A shape is a compact connected set $X$ with $X=cl(int(X))$.

  • A tiling is a pair $(\mathscr{S},F)$ where $\mathscr{S}$ is a finite set of shapes and $F$ is a set of functions such that

    • each $f\in F$ is an isometric embedding of some $S\in\mathscr{S}$ into $\mathbb{R}^2$,

    • $\bigcup_{f\in F}ran(f)=\mathbb{R}^2$, and

    • if $f,g\in F$ are distinct with domains $S,T$ respectively and $x\in int(S)$, then $f(x)\not\in ran(g)$ ("shapes only meet at their boundaries").

  • A tile is a shape $X$ such that $(\{X\},F)$ is a tiling for some $F$.

  • $d_H$ is Hausdorff distance. (Note that we could replace $d_H$ with the modified version $d_{H*}(U,V)$ = the infimum over planar isometries $p$ of $d_H(p(U), V)$ without changing the question substantively; all that would change is that the number of tiles needed would shrink, but I'm not looking at that here.)

Now with apologies to Escher, given tiles $A,B$ and $\epsilon>0$ an $\epsilon$-metamorphism from $A$ to $B$ is a tiling $(\mathscr{S},F)$ such that

  • if $f\in F$ and $(x,y)\in ran(f)$ with $x<0$ then $dom(f)=A$,

  • there is some $N$ such that if $g\in F$ and $(x,y)\in ran(g)$ with $x>N$ then $dom(g)=B$ (call the least such $N$ the length of the $\epsilon$-metamorphism), and

  • if $f,g\in F$ and $ran(f)\cap ran(g)\not=0$ then $d_H(dom(f), dom(g))<\epsilon$.

Question: is there always an $\epsilon$-metamorphism between $A$ and $B$ for any tiles $A,B$ and any positive $\epsilon$?

I recall seeing a theorem that the answer is yes, and that moreover we can always find metamorphisms with length "close to" the naive guess $$\max\{diam(A),diam(B)\}\cdot d_H(A,B)\over \epsilon.$$ However, I haven't been able to track it down or prove it myself. I'm also interested in whether the situation changes as we appropriately tweak things to work in $\mathbb{R}^n$ for $n>2$.

Noah Schweber
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  • I'm having difficulty imagining how to do this if $A$ and $B$ are regular hexagons with $A$ having side length $1$ and $B$ having side length $\pi$ or something, but I have no idea how you would prove that this does not work. – James E Hanson Oct 21 '22 at 01:36
  • @JamesHanson: This case is not too bad; you can make the vertical "zigzags" gradually less bendy until you have a series of rectangles, then stretch the rectangles to their new heights and expand the zigzags back out. In general any $A$ and $B$ that tile by translation will be easy to metamorphose into each other. What I worry about is something like the 15th pentagonal tiling; I don't see a way to metamorphose from this into literally any other monohedral tiling (even a scaled vertion of itself). – RavenclawPrefect Oct 29 '22 at 16:32
  • The "Hausdoff Distance" link gives the Metamorphosis I,II,II link instead. I assume that's a copy/paste error. Did you have another site in mind? – CyclotomicField Oct 29 '22 at 18:28
  • Without the length condition, you can always construct a metamorphosis by shrinking the tiles to arbitrarily small sizes so that the Hausdorff distance condition fails to bind. – Jacob Manaker Oct 29 '22 at 18:35
  • @CyclotomicField Fixed, thanks! – Noah Schweber Oct 29 '22 at 18:35
  • @JacobManaker: I thought about this construction, but I don't see how you actually get it to work - how do you avoid the tiles separating arbitrarily far at extreme $y$-values? – RavenclawPrefect Oct 29 '22 at 18:46
  • @JacobManaker I don't quite see the details, but I strongly suspect that that works. I've tweaked the MO version to address this, but I'll leave this version as is. – Noah Schweber Oct 29 '22 at 18:54

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