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Consider the following iterative process. Start with a planar region $R=R_0$ of $\mathbb{R}^2$. I am thinking of $R$ as connected, but it may become disconnected. In the example below, $R$ starts as a vertical rectangle. Let $c_i$ be the centroid of $R_i$. In each step, do the following.

  • Chose a randomly oriented, directed line $L_i$ through $c_i$.
  • Partition $R_i$ into two "halves," one on each side of $L_i$: $R_i^+$ right of $L_i$ and $R_i^-$ left of $R_i$.
  • Reflect the positive (right) half $R_i^+$ about the normal $N_i$ to $L_i$ through $c_i$. Call that $R_i^{'+}$.
  • The untwisted-half $\cup$ the twisted-half constitutes $R_{i+1}$, i.e., $R_{i+1} =R_i^- \cup R_i^{'+}$. Update $c_{i+1}$ and repeat.


          FlipRect
          $R_i$ is the vertical rectangle. $R_{i+1}$ is the blue $\cup$ yellow nonconvex shape: $R_i^- \cup R_i^{'+}$.


My question is:

Q. Does every start shape $R_0$ tend to some common limit shape? Or rather are the results fundamentally dependent on the (arbitrary) start shape?

Note that the area of $R$ is fixed, as the unions are always disjoint except along the boundary line $L_i$. It is clear that a disk is a fixed point of this process.

I know this question could likely be answered (at least conjecturally) by an implementation, but I have not yet invested in that. Can anyone see what would be the results, without the benefit of an implementation? Or might someone see a path to a not-too-difficult implementation?

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    $\begingroup$ I am confused by your sentences right after the question. I do not see volume preservation or a disk as a fixed point. Please rework the process definition and/or the two claims to consistency. After that, I am willing to take a guess on the answer. Gerhard "Will Wait Here For It" Paseman, 2018.07.15. $\endgroup$ – Gerhard Paseman Jul 15 '18 at 21:56
  • $\begingroup$ @GerhardPaseman: Thanks for attending (as always!). Disk as fixed point: Centroid is the center of the disk. $L$ cuts off semidisk. Reflecting that results in the identical semidisk. So $R_{i+1} = R_i$. $\endgroup$ – Joseph O'Rourke Jul 15 '18 at 21:59
  • $\begingroup$ @GerhardPaseman: Re same area. Let $R_i^+$ be the "half" to the right of $L_i$. Because the reflection is w.r.t. $N_i$ orthogonal to $L_i$, the reflection of $R_i^+$ remains entirely to the right of $L_i$. So we are just twisting $R_i^+$ about $N_i$, and it will not intersect $R_i^-$ except in a set of measure zero along $L_i$. $\endgroup$ – Joseph O'Rourke Jul 15 '18 at 22:02
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    $\begingroup$ Going in the opposite direction to a circle, looking at how this transforms a straight line might give some of the other side of the picture. A Cantor set? If so then it should lie in a circle, symmetric about every line through the circle's center, if it converges to a fixed shape. $\endgroup$ – David Hughes Jul 16 '18 at 18:48
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    $\begingroup$ Here is a simple implementation: three equal point masses. I think you can show that the sum of the three side bisectors reduces with each move, giving convergence of the three points toward each other. After that, you can try four point masses. Gerhard "Two Is Way Too Easy" Paseman, 2018.07.19. $\endgroup$ – Gerhard Paseman Jul 20 '18 at 5:27
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(1) Not convergent case for some sequence : Consider a line $l\ :\ y=f(x)=L(x-1) + \frac{1}{2}$ where $L>0$ is large.

If $A=(0,0),\ (2,0),\ (2,1),\ (0,1)$, then assume that convex hull of four points is $Y_1$. Here $l$ divides $Y_1$ into $X$ and $X_1$ where $A\in X$.

Here if we have one process at a center $c_1$, wrt a line $l$, about $X_1$, then we have $Y_2=X\bigcup X_2$. Note that if $c_i\in l$ is a center of $Y_i=X\bigcup X_i$, then $d(c_1,A)>d(c_2,A)$.

If we have one process on $Y_2$, then we have $Y_3 $ s.t. $X_3$ is a translation of $X_1$ along a line $l$. Hence $Y_4=Y_2$.

(2) Convergent case : $$X=\{(x,y)|x^2+(y-1)^2\leq 1,\ x\leq 0\},\ X_1=\{ (x,y)| 0\leq x\leq L,\ 2-\epsilon\leq y\leq 2\}$$ Assume that $Y_1= X\bigcup X_1$ has a center $c_1\in l$ where $l\ :\ x=0$.

When centers $c_i$ of $Y_i$ is in $l$, then $Y_n$ converges.

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This strikes me as similar to one of your earlier questions, and it might be equivalent. (Unfortunately, I don't remember which one. I don't think it was the Cantor doughnut one though.). A key observation is that you shift the center of mass in a particular way. Pick a line through the original centroid, and compute the centroid of each half piece. The new centroid shifts in the direction of the line at a distance related to cosine of the angle of the line joining the centroid halves with the chosen line, and moves the two halves closer together in a centroid sense. I think a certain convergence is going on here, but it would behoove you to look at how the centroids of the halves vary with the bisecting line. If after a few iterations you do not find the variance growing, there is a good chance for convergence of something.

Gerhard "Not Looking For Bad Chances" Paseman, 2018.07.15.

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  • $\begingroup$ Here's an idea. If you can prove that Max d(x,centroid) over all x in a region gets reduced regardless of which line you pick, then you can get some kind of convergence. I don't think the limiting shape will be a disk, but it will be contained in a disk whose size you probably can compute. I recommend considering computing centroids and extreme of d(x,centroid). Gerhard "Now We Can Get Somewhere" Paseman, 2018.07.15. $\endgroup$ – Gerhard Paseman Jul 16 '18 at 2:20
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The rectangle is an interesting shape to start intuition, but to build up the intuition, it is better to look at regions derived from the rectangle by this process.

Suppose we had a long skinny rectangle, with two nearby vertices A and B. After several iterations of this process, it is likely that A and B will still be close because none of the centroid lines chosen during the iteration will have separated A from B. So some portion of the rectangle end will be recognizable.

I was initially misled by the rectangle because I thought that for every iteration, the centroid of the halves had to be equidistant from the centroid of the untwisted figure. This is not so. Smaller halves can be farther removed. Twisting the figure does bring the centroid of the halves closer together, but the size of the containing disk may increase, especially if the disk has its center at the (before and after) centroid.

As in the Cantor doughnut question, I suggest looking at a centroid locus. In this case, as one rotates the bisecting line, consider how the centroid halves vary in distance from the centroid. It is tempting to say that the locus before a twist is larger than the locus after a twist, but I am unsure and do not have a suggestion of a proof at this time. However, this locus seems to me to be a key item to investigate. If the tempting bit is true, then one can say that the mass will tend to concentrate toward a center. Not that it will form a disk, but that it will approximate a regular polygon or more likely a snowflake. An interesting (if true) idea is that the mass of the region in an annulus around the centroid does not vary, or varies little and predictably, after a twist. If so, then I can imagine an approximation of the dynamic as cutting the region into washer pieces and then rotating the pieces around in some fashion to minimize some quantity. I don't think this interesting idea actually is true for this dynamic, but something like it should be true.

Gerhard "Time For Scissors And Glue" Paseman, 2018.07.15.

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