Uniform Sampling on Disks, Shells, Balls, and Triangles.

Table of Contents


Uniform sampling of an arbitrary geometry is a common task that you might encounter in your research. This post summarizes the general inverse approach. I learned the approach in the classic PBRT book, which is one of my favorite books even beyond computer graphics.


We start with the basics of sampling a 3D hemisphere shell that is parameterized in the polar coordinate:

  1. We need to find \(p(\theta, \phi)\) so that if we take a sample from this distribution, we are sampling uniformly from the hemisphere.
  2. We then run use the inverse transform sampling method to sample \(\theta\) from \(p(\theta)\) and \(\phi \sim p(\phi \mid \theta)\) with the given \(\theta\).

How do we find \(p(\theta, \phi)\) in the first place? We need to understand that if we are sampling uniformly from a geometry, then

\[p(\theta, \phi)\mathrm{d}\phi \mathrm{d}\theta \propto \mathrm{d}A. \tag{1}\]

This is saying that \(p(\theta, \phi)\mathrm{d}\phi \mathrm{d}\theta\) must be proportional to the area or volume of the infinitesimal region \(\mathrm{d}A\) on the geometry. If this holds, then two infinitesimal regions with the same area will be sampled with equal probability.

Once Equation (1) is crystal clear to you, the rests are just math manipulation. We will derive the result on 2D disks, 3D sphere surfaces, and balls.

1. 3D Hemisphere Shells

To derive the \(\mathrm{d}A\) for a hemisphere, it is important to understand Figure 1:

Infinitesimal Area of a Hemisphere
Figure 1. Infinitesimal area of a hemisphere.

For the infinitesimal area, it is obvious that one side is \(\mathrm{d}\theta\). For the other side, you need to pay attention to its projection on the horizontal plane. The projected arc has radius of \(\sin\theta\) and radian of \(\mathrm{d}\phi\). Assuming \(r=1\), we have

\[\mathrm{d}A = \sin\theta\mathrm{d}\phi\mathrm{d}\theta,\]

which gives

\[p(\theta, \phi) \propto \sin\theta.\]

For \(\phi\), it can be drawn from uniform distribution \(\phi \sim U(0, 2\pi)\) which is trivial. For \(\theta\), we compute its CDF:

\[P(\theta) = \int_0^{\theta} \sin\theta' \mathrm{d}\theta' = 1-\cos\theta. \]

With the inverse sampling approach, we have

\[ \begin{align} \theta &= \cos^{-1}(1-\xi_1) = \cos^{-1}\xi_1' \\ \phi &= 2\pi\xi_2 \end{align} \]

where \(\xi_1',\xi_2\) are randomly sampled from $[0,1]$.

2. 3D Hemisphere Volume

Similar to the hemisphere shell, we have the volume infinitesimal

\[\mathrm{d}V = r^2\sin\theta\mathrm{d}r\mathrm{d}\phi\mathrm{d}\theta,\]

which gives

\[p(r, \theta, \phi) \propto r^2\sin\theta.\]

This gives \(p(r) \propto r^2\) and \(p(\theta\mid r) \propto \sin\theta\). We can compute the CDF of \(p(r)\):

\[P(r) \propto \int_0^{r} r'^2 \mathrm{d}r' \propto r^3 \]

With the inverse sampling approach, we have

\[ \begin{align} r &= \sqrt[3]{\xi_0} \\ \theta &= \cos^{-1}\xi_1' \\ \phi &= 2\pi\xi_2 \end{align}. \]

3. 2D Disks

Sampling 2D disk can be derived similarly to the 3D ball case. The results are

\[ \begin{align} r &= \sqrt[2]{\xi_0} \\ \phi &= 2\pi\xi_2 \end{align}. \]

4. 2D Triangles

To sample on 2D triangles, we use the barycentric coordinate \((u, v)\) such that \(0 \le u+v\le 1\) and \(0 \le u,v\le 1\). Since it is a flat surface, it is obvious that

\[ p(u, v) \propto c. \]

Therefore, we have

\[ \begin{align} p(u) &\propto \int_{0}^{1-u} c \mathrm{d}v \propto 1-u\\ p(v \mid u) &= \frac{p(u,v)}{p(u)} \propto \frac{1}{1-u} \end{align}. \]

The CDFs are

\[ \begin{align} P(u) &= 2u-u^2\\ p(v \mid u) &\propto v = \frac{v}{1-u} \end{align}. \]

Assign the CDFs to uniform distributed random numbers and solve a quadratic equation, we have

\[ \begin{align} u &= 1 - \sqrt{1-\xi_1} \\ v &= \xi_2\sqrt{1-\xi_1} \end{align}. \]

Yichao Zhou
Ph.D. in Computer Science

I am interested in the interdisciplinary field of computer vision and computer graphics.