Dan Asimov <dasimov@earthlink.net> wrote:
One method I did not see listed in that linked MathWorld article is to choose the points from a standard normal distribution in the plane, i.e., with density given by
p(x,y) = (2?)^(-1) exp(-(x^2 + y^2)/2).
This is of course circularly symmetric, but it's not at all clear to me that four points picked from this distribution will have the same chance of their convex hull being a quadrilateral as four points picked from the uniform distribution on a disk.
It seems to be different. I've only started, but the first 3 billion give me 0.64905, as contrasted with 0.704480 for a circle and 0.694444 for a square. I'll let it run for a couple days. To make sure I'm generating a standard normal distribution in the plane correctly, please tell me what the means of the r values (distances from the origin) should be. I'm measuring the arithmetic, quadratic ("RMS"), and cubic means. In researching this, I found https://www.jstor.org/stable/pdf/30037457.pdf which says Show that the expected number of vertices of the convex hull of n points sampled from the standard normal distribution in the plane is asymptotically 2 sqrt(2 pi ln n). That doesn't answer your question, but it does answer one of mine (albeit for a different distribution). I guessed, in a post here on Sunday, that it might be proportional to either the square root of n or to the natural log of n. Apparently it's a combination of both.
Affine transformations preserve ratios between areas.
Right. And also preserve straight lines. What I didn't realize was that all triangles are affine transformations of each other. I noticed this for rectangles when I realized that changing the aspect ratio of a rectangle in my program would mean multiplying by a number and then immediately dividing by the same number. So the answer has to be the same for all rectangles. Also on Sunday, I posted: For the circular five-point case I calculated 0.3561808623 and posted 0.356181. That page doesn't give the solution. Assuming the solution is in the same form as the circular four-point case, 1 - (i / (j pi^2)), where i and j are integers, (35 and 12 in the 4-point case), if I try 1 - (305 / (48 pi^2)), I get 0.356188312... which differs from my answer by 7 in the sixth digit. Does anyone know if that's the correct formula? That was with about 9 billion sets. With 59 billion (and counting) I get 0.35618796, so it's looking like it really is 1 - (305 / (48 pi^2)), as it differs from that formula by only 3 in the *7th* digit. That's a hell of a way to find a formula.