About 37 year ago someone showed me that if you start with a large number of points uniformly distributed in a square and then proceed to remove them in stages via "convex peeling" — the boundaries of the last groups of points remaining are exceedingly circular. ... But I've never seen a proof of this. Any ideas?
I noticed that the furthest point from each point in the set is always part of the convex hull, so the hull is always contained within the intersection of a bunch of circles (where each circle centers at a point and has the furthest point on its circumference). The longest-lived points in the middle in particular draw the smallest circles and tend to confine the hull to a circular shape. Here's an animation of this starting with 2000 points: https://imgur.com/a/Fj8PO And here's the python code I used to generate it: https://pastebin.com/MqPi729n On Tue, Nov 21, 2017 at 10:07 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Suppose you have a finite set of points in the plane.
The process known as "convex peeling" arose as one way of defining the "median" of a finite set of 2-dimensional data, analogous to the usual one for 1-dimensional data:
Let the set of points be X.
Define
X_0 = X
and X_(n+1) = X_n - {p in X_n | p is on the boundary of the convex hull of X_n}
for n >= 0. This sequential "filtration" of X is known as convex peeling.
About 37 year ago someone showed me that if you start with a large number of points uniformly distributed in a square and then proceed to remove them in stages via "convex peeling" — the boundaries of the last groups of points remaining are exceedingly circular.
So that there appears to be a true statement, something like: ----- With probability = 1, the shape of the normalized result of convex peeling [toward the end of the process] approaches [in some sense] a circle, as the number of points approaches oo. -----
But I've never seen a proof of this. Any ideas?
—Dan
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