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**Chapter 6: Generating Random Directions**
In this and the next two chapters let’s harden our understanding and tools and figure out which
Cornell Box is right. Let’s first figure out how to generate random directions, but to simplify
things let’s assume the z-axis is the surface normal and $\theta$ is the angle from the normal.
We’ll get them oriented to the surface normal vector in the next chapter. We will only deal with
distributions that are rotationally symmetric about $z$. So $p(direction) = f(\theta)$. If you
have had advanced calculus, you may recall that on the sphere in spherical coordinates
$dA = \sin(\theta) \cdot d\theta \cdot d\phi$. If you haven’t, you’ll have to take my word for the
next step, but you’ll get it when you take advanced calc.
Given a directional pdf, $p(direction) = f(\theta)$ on the sphere, the 1D pdfs on $\theta$ and
$\phi$ are:
$$ a(\phi) = \frac{1}{2\pi} $$
(uniform)
$$ b(\theta) = 2*\pi*f(\theta)\sin(\theta) $$
For uniform random numbers $r_1$ and $r_2$, the material presented in Chapter 2 leads to:
$$ r_1 = \int_{0}^{\phi} \frac{1}{2\pi} = \frac{\phi}{2\pi} $$
Solving for $\phi$ we get:
$$ \phi = 2 \pi \cdot r_1 $$
For theta we have:
$$ r_2 = \int_{0}^{\theta} 2 \pi f(t) \sin(t) $$
Here, $t$ is a dummy variable. Let’s try some different functions for $f()$. Let’s first try a
uniform density on the sphere. The area of the unit sphere is $4\pi$ , so a uniform $p(direction) =
\frac{1}{4\pi}$ on the unit sphere.
$$ r_2 = \frac{-\cos(\theta)}{2} - \frac{-\cos(0)}{2} = \frac{1 - \cos(\theta)}{2} $$
Solving for $\cos(\theta)$ gives:
$$ \cos(\theta) = 1 - 2 r_2 $$
We don’t solve for theta because we probably only need to know $\cos(\theta)$ anyway and don’t want
needless $\arccos()$ calls running around.
To generate a unit vector direction toward $(\theta,\phi)$ we convert to Cartesian coordinates:
$$ x = \cos(\phi) \cdot \sin(\theta) $$
$$ y = \sin(\phi) \cdot \sin(\theta) $$
$$ z = \cos(\theta) $$
And using the identity that $\cos^2 + \sin^2 = 1$, we get the following in terms of random
$(r_1,r_2)$:
$$ x = \cos(2\pi \cdot r_1)\sqrt{1 - (1-2 r_2)^2} $$
$$ y = \sin(2\pi \cdot r_1)\sqrt{1 - (1-2 r_2)^2} $$
$$ z = 1 - 2 r_2 $$
Simplifying a little, $(1 - 2 r_2)^2 = 1 - 4r_2 + 4r_2^2$, so:
$$ x = \cos(2 \pi r_1) \cdot 2 \sqrt{r_2(1 - r_2)} $$
$$ y = \sin(2 \pi r_1) \cdot 2 \sqrt{r_2(1 - r_2)} $$
$$ z = 1 - 2 r_2 $$
We can output some of these:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ C++
int main() {
for (int i = 0; i < 200; i++) {
float r1 = drand48();
float r2 = drand48();
float x = cos(2*M_PI*r1)*2*sqrt(r2*(1-r2));
float y = sin(2*M_PI*r1)*2*sqrt(r2*(1-r2));
float z = 1 - r2;
std::cout << x << " " << y << " " << z << "\n";
}
}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
And plot them for free on plot.ly (a great site with 3D scatterplot support):
On the plot.ly website you can rotate that around and see that it appears uniform.
Now let’s derive uniform on the hemisphere . The density being uniform on the hemisphere means
$p(direction) = 1 / (2 \pi)$ and just changing the constant in the theta equations yields:
$$ \cos(\theta) = 1 - r_2 $$
It is comforting that $\cos(\theta)$ will vary from 1 to 0, and thus theta will vary from 0 to
$\pi/2$. Rather than plot it, let’s do a 2D integral with a known solution. Let’s integrate cosine
cubed over the hemisphere (just picking something arbitrary with a known solution). First let’s do
it by hand:
$$ \int \cos^3 = 2 \pi \int_{0}^{\pi/2} \cos^3 \sin = \frac{\pi}{2} $$
Now for integration with importance sampling. $p(direction) = 1/(2 \pi)$, so we average $f/p$
which is $\cos^3 / (1/(2 \pi))$, and we can test this:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ C++
int main() {
int N = 1000000;
float sum = 0.0;
for (int i = 0; i < N; i++) {
float r1 = drand48();
float r2 = drand48();
float x = cos(2*M_PI*r1)*2*sqrt(r2*(1-r2));
float y = sin(2*M_PI*r1)*2*sqrt(r2*(1-r2));
float z = 1 - r2;
sum += z*z*z / (1.0/(2.0*M_PI));
}
std::cout << "PI/2 = " << M_PI/2 << "\n";
std::cout << "Estimate = " << sum/N << "\n";
}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now let’s generate directions with $p(directions) = \cos(\theta) / \pi$.
$$ r_2 = \int_{0}^{\theta} 2 \pi \frac{\cos(t)}{\pi} \sin(t) = 1 - \cos^2(\theta) $$
So,
$$ \cos(\theta) = \sqrt{1 - r_2} $$
We can save a little algebra on specific cases by noting
$$ z = \cos(\theta) = \sqrt{1 - r_2} $$
$$ x = \cos(\phi) \sin(\theta) = \cos(2 \pi r_1) \sqrt{1 - z^2} = \cos(2 \pi r_1) \sqrt{r_2} $$
$$ y = \sin(\phi) \sin(\theta) = \sin(2 \pi r_1) \sqrt{1 - z^2} = \sin(2 \pi r_1) \sqrt{r_2} $$
Let’s also start generating them as random vectors:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ C++
inline vec3 random_cosine_direction() {
float r1 = drand48();
float r2 = drand48();
float z = sqrt(1-r2);
float phi = 2*M_PI*r1;
float x = cos(phi)*sqrt(r2);
float y = sin(phi)*sqrt(r2);
return vec3(x, y, z);
}
int main() {
int N = 1000000;
float sum = 0.0;
for (int i = 0; i < N; i++) {
vec3 v = random_cosine_direction();
sum += v.z()*v.z()*v.z() / (v.z()/(M_PI));
}
std::cout << "PI/2 = " << M_PI/2 << "\n";
std::cout << "Estimate = " << sum/N << "\n";
}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
We can generate other densities later as we need them. In the next chapter we’ll get them aligned to
the surface normal vector.