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I have a sphere of radius $R_{s}$, and I would like to pick random points in its volume with uniform probability. How can I do so while preventing any sort of clustering around poles or the center of the sphere?

Since I'm unable to answer my own question, here's another solution:

Using the strategy suggested by Wolfram MathWorld for picking points on the surface of a sphere: Let $\theta$ be randomly distributed real numbers over the interval $[0,2\pi]$, let $\phi=\arccos(2v−1)$ where $v$ is a random real number over the interval $[0,1]$, and let $r=R_s (\mathrm{rand}(0,1))^\frac13$. Converting from spherical coordinates, a random point in $(x,y,z)$ inside the sphere would therefore be: $((r\cos(\theta)\sin(\phi)),(r\sin(\theta)\sin(\phi)),(r\cos(\phi)))$.

A quick test with a few thousand points in the unit sphere appears to show no clustering. However, I'd appreciate any feedback if someone sees a problem with this approach.

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Summary: Nate proposed a transformation method, while Kevin proposed a rejection method. –  J. M. Dec 1 '11 at 3:08
I think the solution you took from MathWorld is sound, if a bit slower than the approaches proposed in the answer (evaluating transcendental functions is expensive!). –  J. M. Dec 1 '11 at 3:58
I don't know much but, my approach would be something along the lines of choosing a random pole from the surface to the center and then choosing a point on that line where the probability is greater the closer to the surface, to account for the expansion of the sphere. –  Sebastian Garrido Dec 10 '13 at 20:17

4 Answers 4

up vote 17 down vote accepted

Let's say your sphere is centered at the origin $(0,0,0)$.

For the distance $D$ from the origin of your random pointpoint, note that you want $P(D \le r) = \left(\frac{r}{R_s}\right)^3$. Thus if $U$ is uniformly distributed between 0 and 1, taking $D = R_s U^{1/3}$ will do the trick.

For the direction, a useful fact is that if $X_1, X_2, X_3$ are independent normal random variables with mean 0 and variance 1, then $$\frac{1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} (X_1, X_2, X_3)$$ is uniformly distributed on (the surface of) the unit sphere. You can generate normal random variables from uniform ones in various ways; the Box-Muller algorithm is a nice simple approach.

So then $$\frac{R_s U^{1/3}}{\sqrt{X_1^2 + X_2^2 + X_3^2}} (X_1, X_2, X_3)$$ would produce a uniformly distributed point inside the ball of radius $R_s$.

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This also generalizes to the $n$-sphere (with $D = R_s U^{1/n}$). –  Michael Lugo Dec 1 '11 at 4:58

An alternative method in $3$ dimensions:

Step 1: Take $x, y, $ and $z$ each uniform on $[-r_s, r_s]$.

Step 2: If $x^2+y^2+z^2\leq r_s^2$, stop. If not, throw them away and return to step $1$.

Your success probability each time is given by the volume of the sphere over the volume of the cube, which is about $0.52$. So you'll require slightly more than $2$ samples on average.

If you're in higher dimensions, this is not a very efficient process at all, because in a large number of dimensions a random point from the cube is probably not in the sphere (so you'll have to take many points before you get a success). In that case a modified version of Nate's algorithm would be the way to go.

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Nate and Kevin already answered the two I knew... Recalling this and this, I think that another way to generate a uniform distribution over the sphere surface would be to generate a uniform distribution over the vertical cylinder enclosing the sphere, and then project horizontally.

That is , generate $z \sim U[-R,R]$, $\theta \sim U[0,2\pi]$, and then $x=\sqrt{R^2-z^2} \cos(\theta)$, $y=\sqrt{R^2-z^2} \sin(\theta)$. This (if I'm not mistaken) gives a uniform distribution over the sphere surface. Then, apply Nate's recipe to get a uniform distribution over the sphere volume.

This method is a little simpler (and more efficient) than the accepted answer, though it's not generalizable to other dimensions.

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In case it isn't immediately clear to people, this method works because of Archimedes' theorem on slicing cylinders and spheres: mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html –  Michael Lugo Dec 1 '11 at 5:00
I believe you need a factor of $\sqrt{R^2 - z^2}$ in the choice of $x$ and $y$, in order to be on the sphere's surface. –  Erik P. Feb 21 '13 at 15:57
@ErikP. Of course, you're right, I forgot to do the projection, I'd left the points over the cylinder surface. Fixed, thanks! –  leonbloy Feb 21 '13 at 18:23

I just want to add a small derivation to leonbloy's answer, which uses calculus instead of geometrical intuition. Changing from cartesian $(x,y,z)$ to spherical $(r,\theta,\phi)$ coordinates, we have for the volume element $$dx dy dz =r^2 \sin \theta ~ dr d\theta d\phi$$ The coordinates $(r,\theta,\phi)$ don't work for a uniform distribution because we still have a non-constant factor in front of $dr d\theta d\phi$. Therefore we introduce $$u=-\cos \theta \Rightarrow du= \sin \theta d\theta$$ $$\lambda=r^3/R^3 \Rightarrow d \lambda=\frac{3}{R^3}r^2dr$$ with which we obtain an expression with a constant pre-factor $$dx dy dz= \frac{R^3}{3} d\lambda du d\phi$$ The range of our variables is $\lambda \in [0,1], ~u \in [-1,1), \phi \in [0, 2\pi) $. Choosing those numbers uniformly we get cartesian coordinates

$$ \begin{align} x&=r \sin(\theta) \cos (\phi) =&R \lambda^{1/3} \sqrt{1-u^2}\cos(\phi)\\ y&=r \sin(\theta) \sin (\phi) =&R \lambda^{1/3} \sqrt{1-u^2}\sin(\phi) \\ z&=r \cos (\theta)=&R \lambda^{1/3} u \end{align} $$

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