在环(环)内生成均匀随机点 [英] Generate a uniformly random point within an annulus (ring)

问题描述

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找到的其他解决方案进行比较:

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Create random number within an annulus

I would like to obtain a uniformly obtained random point within an annulus, that is, the area that lies inside a circle of radius R1, but outside a circle of radius R2, where R1 > R2 and both circles are centered at the same point. I would like to avoid using rejection sampling.

If possible, I would like the solution to be similar to this one —used for calculating random points within a circle— which I find it extremely elegant and intuitive. That is, I would also like to avoid using the square root.

解决方案

EDIT: Please note that this solution might not be uniform. See comments by Mark Dickinson below.

Ok, I think I figured it out. Note that this solution is heavily inspired in this answer, and that r1 = R1/R1 and r2 = R2/R1.

Pseudo-code:

t = 2*pi*random()
u = random()+random()
r = if u>1 then 2-u else u
r = if r<r2 then r2+r*((R1-R2)/R2) else r
[r*cos(t), r*sin(t)]

Here it is in Mathematica.

f[] := Block[{u, t, r}, u = Random[] + Random[];
r1 = 1; r2 = 0.3;
t = Random[] 2 Pi;
r = If[u > 1, 2 - u, u];
r = If[r < r2, r2 + r*((R1 - R2)/R2), r];
{r Cos[t], r Sin[t]}]

ListPlot[Table[f[], {10000}], AspectRatio -> Automatic]

What it does is to remap all the numbers falling inside the inner circle into the annulus, spreading them evenly. If somebody finds a problem regarding the uniformity of this solution please comment.

Compare with this other solution found here:

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