Picking random points in the volume of sphere with uniform probability

Question

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?

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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.

Answer 1

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 if you choose $U$ uniformly distributed between 0 and 1, and $X_1, X_2, X_3$ iid standard normal and independent of $U$, 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 $R_s$.

Answer 2

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.

Answer 3

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.

Answer 4

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 $dr d\theta d\phi$ (see "EDIT" at the bottom, if you do not see why they don't work). 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

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

EDIT: I want to add an argument why we want a constant prefactor in front of $d\lambda du d\phi$.

Consider the one dimensional case (uniform distribution of points on the line $[0,L]$). For $0<x<L$ the probability to find a point in $(x,x+dx)$ is $P(x)dx$. Since we assume a uniform probability, $P(x)$ has to be $P(x)=1/L$, and hence the probability $P(x)dx=dx/L$ is directly proportional to the volume element $dV=dx$.

Now consider we have a variable $y$, for which we do not know the probability density $Q(y)$ but we know that the volume element is $dV=dx=c dy$ with some constant $c$. Furthermore, we know that $Q(y)dy$ has to be $P(x)dx$ (by definition of probability density). Hence $Q(y)=P(x)dx/dy=c$.

In summary we have shown:

"Variable $y$ is uniformly distributed" $\Leftrightarrow$ "The volume element is $dV=c dy$ for some constant $c$. (For the correct normalization of the probability density the value of $c$ is not arbitrary)

Answer 5

Much simpler way would be to pair surfaces at different distances from the center.

The surface of a sphere of radius R has area 4 π R^2

So just pair up various surfaces that add up to the same area.

For example:

Surface of sphere with r = 0 paired with surface of sphere with r = R to have a total area of 4π R^2

Surface of sphere with r = R/2 paired with surface of sphere with r = sqrt(3)R/2 to have a total area of 4π R^2

To get the random point, do the following steps:

1) For choosing theta (it can be uniformly chosen over its entire range, since all values of theta are equally likely):

Choose a random "theta" in range [0, 2π]

2) For choosing phi (Use same strategy as for areas, but this time it will apply by pairing circumference of circle for a phi value 2π R cos(phi) with its complement to produce total length of 2π R):

Choose a random "a" in range [0, π/2]

Choose a random "b" in range [0, 1]

Choose a random "c" to be either 1 or -1 (up or down)

if b <= cos(a) then
    phi = c a
else
    phi = c inv_cos(1 - cos(a))

3) For choosing r (Pair up every possible radius value with its complement, in such a way that the sum of the surface areas of the spheres will be equal to 4π R^2):

Choose a random "d" in range [0, 1]

Choose a random "e" in range [0, 1]

One r value is "d R" the complementary r value is "sqrt(1 - d^2) R".

Total area = 4π (d R)^2 + 4π (sqrt(1 - d^2) R)^2

Total area = 4π (d^2 R^2 + (1 - d^2) R^2)

Total area = 4π (d^2 R^2 + R^2 - d^2 R^2)

Total area = 4π R^2 (as we wanted)

if e <= d^2 then
    r = d R
else
    r = sqrt(1 - d^2) R

Your random point uniformly chosen inside the sphere in radial coordinates is:

(r, theta, phi)