Shape Transformation Via Medial Axis Transform

Mathematics Asked by fweth on December 22, 2020

Let $P_1,dots,P_n$ be simple polygons which don’t intersect each other and $Ssubseteqmathbf{R}^2$ the set of points lying in the interior of an odd number of the $P_i$, so $S$ can be thought as the interior of a finite number of simple polygons, each with a finite number of holes, also described by simple polygons. I want to perform the following transformation to $S$:

  • Apply medial axis transform, i.e. get the set $textsf{MAT}(S)$ of triples $(x,y,r)$ such that the open ball $B_r(x,y)$ is contained in $S$ but not strictly contained in another open ball $B_{r’}(x’,y’)subseteq S$.
  • Given a bijective linear map $f:mathbf{R}^2rightarrowmathbf{R}^2$, construct the shape $$S_f:=f^{-1}(cup_{(x,y,r)intextsf{MAT}(S)}B_r(f(x,y)).$$ In other words, for each $(x,y,r)intextsf{MAT}(S)$, draw the squeezed ball $f^{-1}(B_1(0,0))$ centered at $(x,y)$ and scaled by $r$ on the plane.

The lines in the medial axis transform don’t have to be straight, but I have the hunch that $S_f$ can again be described via polygons like $S$. If that’s correct, is there an easy way to directly compute the line segments of $S_f$, without computing the medial axis transform first? What if we look at balls in another $p$-norm?

[EDIT] No, $S_f$ can’t be described via polygons in general. But maybe for certain $f$?

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