Re: [math-fun] Statistics for a class of random solids
Fred Lunnon <fred.lunnon@gmail.com> wrote:
Assuming "randomness" defined in a spherically symmetric manner, the resulting polyhedron will presumably resemble a lumpy sphere with occasional mountains. On the plains in between, most polygons will be triangles at 6-valent vertices. Whether mountains dominate plains would depend on the distribution tail --- I'd guess not for any reasonable extension of normal.
To me, "mountain" implies concavity (at the base of the mountain, not at its peak). So by definition a convex hull would have no mountains. That raises the question of how far above the horizontal a feature could be on such a "planet," where horizontal is defined in reference to the local gravity. And where the local gravity is defined as what you would get if the interior has some fixed density and the exterior has zero density. Which raises the question of efficient ways to calculate the surface gravity on a uniform polyhedron.
Computing the "convex hull" of a point set is a typical example of "geometric sorting" problems, to which computer graphics buffs have devoted considerable attention. Googling (ugh!) those terms should locate plenty of references.
Thanks. I see that key words include "Graham scan," "Chan's algorithm," "Quickhull," "gift wrapping," "monotone chain," and "Jarvis march."
I'm going back for the moment to the question about the underlying distribution. When I was posing the problem, I selected a normal distribution for each coordinate independently. Keith expressed a perfectly-understandable doubt about whether the resulting distribution would be anisotropic. I have a vague memory from college that, surprisingly, the resulting distribution in 3-space (and maybe, by generalization, in any Euclidean space) is perfectly spherically symmetric, and that the distance from the origin is a "one-armed normal" distribution. The key is that e^(-x1^2)*e(-x2^2)*e^(-x3^2)*...*e(-x[n]^2) = e (- (x1^2 + x2^2 + x3^2 + ... + x[n]^2)), and by Pythagoras, the exponent is R^2. That means that, to answer another question somebody asked here recently, you can produce points uniformly distributed on a sphere by generating N Gaussian coordinates and normalizing the resulting vector. On Sun, Mar 10, 2019 at 1:58 PM Keith F. Lynch <kfl@keithlynch.net> wrote:
Fred Lunnon <fred.lunnon@gmail.com> wrote:
Assuming "randomness" defined in a spherically symmetric manner, the resulting polyhedron will presumably resemble a lumpy sphere with occasional mountains. On the plains in between, most polygons will be triangles at 6-valent vertices. Whether mountains dominate plains would depend on the distribution tail --- I'd guess not for any reasonable extension of normal.
To me, "mountain" implies concavity (at the base of the mountain, not at its peak). So by definition a convex hull would have no mountains. That raises the question of how far above the horizontal a feature could be on such a "planet," where horizontal is defined in reference to the local gravity. And where the local gravity is defined as what you would get if the interior has some fixed density and the exterior has zero density. Which raises the question of efficient ways to calculate the surface gravity on a uniform polyhedron.
Computing the "convex hull" of a point set is a typical example of "geometric sorting" problems, to which computer graphics buffs have devoted considerable attention. Googling (ugh!) those terms should locate plenty of references.
Thanks. I see that key words include "Graham scan," "Chan's algorithm," "Quickhull," "gift wrapping," "monotone chain," and "Jarvis march."
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participants (2)
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Allan Wechsler -
Keith F. Lynch