This question concerns a process that iterates intersection of randomly rotated planar shapes.

Start with a simply connected region $R_0$ in the plane, and let $c_0$ be the centroid of $R_0$. Rotate $R_0$ about $c_0$ a random angle; call the result $R'_0$. Set $R_1 = R_0 \cap R'_0$. And repeat, always rotating about $c_0$, and computing $R_{i+1} = R_i \cap R'_i$. It is not difficult to see that, if $c_0 \in R_0$, then the process converges to a disk:

          Rotation center $c_0$ fixed to centroid of $R_0$: $\rightarrow$ disk. (Scale changes frame-to-frame.)
If $c_0 \not\in R_0$, then eventually the empty set is reached.

My question concerns the process where the rotation center moves each step to $c_i$, the centroid of region $R_i$. Then sometimes, even when $c_0 \not\in R_0$, the process converges to a disk:

          Rotation center $c_i=$ the centroid of $R_i$: $\rightarrow$ disk. (Scale changes frame-to-frame.)
And sometimes, for the same shape, it leads to the empty set:
          Rotation center $c_i=$ the centroid of $R_i$: $\rightarrow \varnothing$ (in the 5th step not shown).
          (Scale changes frame-to-frame.)

For the process that moves the rotation center $c_i$ to the centroid of $R_i$ at each step:

Q. What characteristics do the shapes $R_0$ possess that lead to a disk with high probability? And what characteristics lead to $\varnothing$ with high probability?

For example, I believe that if $R_0$ is convex, then the process always leads to a disk (not generally the same disk as when the center is fixed at $c_0$ throughout). But I am having difficulty seeing any regularity for nonconvex $R_0$.

  • 1
    $\begingroup$ I recommend the following: For a carefully chosen finite set of shapes, for each shape in the set, make a profile curve of the (after rotation and intersection the new) centroid position versus angle of rotation. The profile curve might inform you as to what size the next shape might be. Gerhard "It's A Moduli Space Problem?" Paseman, 2017.03.24. $\endgroup$ – Gerhard Paseman Mar 24 '17 at 16:08

I thought about doing the construction with arbitrary sets , and came up with Cantor's doughnut. Although this does not address the posted question directly, I think it shows more thought is needed in considering the dynamics for arbitrary sets. Also, this example is like a bunch of polygons (up to measure zero and with some curves).

The solid annulus is invariant under Joseph's operation, so let's mix it up a little. For symmetry's sake, map two copies of the Cantor set to [0,2pi], and then remove the ray from the annulus which has angle alpha where alpha is in the mapped Cantor set. This is Cantor's doughnut.

Now perform Joseph's operation on this set. I believe the centroid stays invariant under this operation, and the result gets lighter (in some sense), but for any countable number of iterations, the result is neither empty nor a disk. It makes me wonder if perhaps there are other possibilities for Joseph's dynamic.

Gerhard "Cantor Was A Light Eater" Paseman, 2017.03.24.

  • $\begingroup$ Clever to explore the Cantor doughnut! $\endgroup$ – Joseph O'Rourke Mar 24 '17 at 23:11

Here is a perspective that should help Joseph predict how the dynamic should proceed.

Given a figure F with centroid c, let the rotation trace R be the set of points d such that d is the centroid of the intersection of F with a copy rotated about c. If we keep the perspective of a fixed frame in which F does not move, R becomes hard for me to describe, and it helps me to think of it as a cardioid of some type, even if it is not. We do have the following though: if we rotate a copy of F by angle alpha, and call the location of the centroid of intersection d(alpha) and then rotate the whole diagram by -alpha, we should get the centroid of intersection which occurs by rotating the copy by -alpha, or d(-alpha). I conjecture that there is an axis such that when that is fixed and with respect to that axis one figure is rotated by beta and the other by -beta, the centroid of intersection stays on that axis for all values of beta.

Let us suppose that conjecture holds. (It should, since I think rotation by alpha means the point d(alpha) occupies (s, alpha/2) in a judiciously chosen polar coordinate system.) Now we see if R stays completely inside figure F. Then I make the next conjecture: If R is contained in a disk inside F, then this is preserved across Joseph's dynamic. This makes sense to me since we are taking intersections at each step.

If part of R lies outside the polygon, then I see Joseph's question as determining that fraction that stays outside, and asking for how often the iterated version stays outside. For more than finitely many iterations, I am unsure what to do, but for fixed number m of iterations I see the result as computing an m- dimensional integral, where for m=1 it is the fraction of R inside F according to a suitable measure, for n=2 it is a space R1(alpha) which is an amalgam of the state spaces resulting from each possible alpha, and in general an iterated integral over a space I will not attempt to describe . If one can get this far, then the ambitious can attempt to take limits as m grows large to assign a probability of the empty set occurring for the interesting figures F.

Gerhard "Daily Ambition Level Running Low" Paseman, 2017.03.27.

  • $\begingroup$ If I am right about this axis thing and this other preservation thing, I see a quicker resolution for Joseph. Write a program which takes a figure F, copies it, rotates it and the copy about the centroid in opposite rotational directions, and computes the 'axis' that is R and the angular measure t in which R stays in F. Then the probability of the empty set occurring is like a product of (1-t_i) where for larger iterations t_i may need integration to determine. Gerhard "Deferring Sometimes Makes Things Easier" Paseman, 2017.03.27. $\endgroup$ – Gerhard Paseman Mar 27 '17 at 20:22

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