# Cranberry Lair

## Intro

In our last post, we looked at our SIMD-friendly math library. In this post, we’ll be taking a look at how cranberray builds its BVH. For a primer to BVHs, I recommend going to PBRT’s treatment on the topic here.

Canberray’s BVH is built recursively as a binary tree. It starts with the top level node containing all nodes and recursively splits it into 2 children at each level. Originally, the BVH was split simply by selecting the left and right halves of the children nodes. The next approach was to split according to the median of a particular axis (typically the largest one), and now it makes use of the surface area heuristic.

## SAH

The Surface Area Heuristic or SAH is a well known strategy to build an efficient BVH that tries to minimize the number of intersection tests applied on a tree.

PBRT has an excellent treatment on the topic [2], and I’ve linked one of the original papers referencing the idea [1]. I will try to instead provide an intuitive picture of why and how SAH works.

The idea behind SAH is to discover an efficient partition of a collection that will try to minimize the cost of traversing the tree for any ray.

The SAH function is presented as:

$C_o = C_{trav} + P_A\sum C_{isect}(a_i) + P_B\sum C_{isect}(b_i)$

This cost model is based on probabilities. $C_{trav}$ is the cost of traversing the parent node to determine if any of the two children nodes intersect the ray. In cranberray, this would be the cost of testing the ray against both children AABBs and determining if it intersects either. $C_{isect}$ is the cost of testing the shape itself against the ray. In cranberray, this would be the cost of testing the triangles against the ray. Finally, $P_A$ and $P_B$ are the probabilities that the bounding volumes containing the children nodes intersect the ray.

Now, well make a slight modification and reformulate $P_A$ and $P_B$ as $P(A|C)$ and $P(B|C)$ where C is our parent node. This states that $P(A|C)$ represents the probability of the event A happening given the event C has already happened. In this context, this means that $P(A|C)$ represents the probability that we intersect our volume A given that we’ve already intersected our parent volume C.

With this change, our function is now:

$C_o = C_{trav} + P(A|C)\sum C_{isect}(a_i) + P(B|C)\sum C_{isect}(b_i)$

Now, let’s imagine that we’re a lonesome ray traveling through our BVH. We arrive at a node and we test both of it’s children to see if we intersect it. This takes us 10 units of time ($C_{trav}$). At this point we would intersect A with a probability of $P(A|C)$ and we would intersect B with a probability of $P(B|C)$. In this example, we only intersect with A, and so we test our ray against it’s contained shapes. Let’s say that A has 10 shapes, as a result, we have to pay the cost of testing the intersection with all 10 shapes. This is $\sum_{i=1}^{10} C_{isect}(a_i)$. Finally, let’s say that the cost of $C_{isect}$ is 15 for every child shape.

This gives us the final actual cost of:

$C_o = 10+sum_{i=1}^{10} 15 = 10+150 = 160$

What our original cost function is doing, is calculating the expected cost of doing this operation for any possible ray.

As a result, if we were to build with these values:

$C_{trav}=10$

$C_{isect}=15$

$P(A|C)=1$

$P(B|C)=0$

$A_n = 10$

$B_n = 0$

Then our expected cost for this node would be:

$C_o = 10 + 1*\sum_{i=1}^{10} 15 + 0*\sum_{i=1}^{0} 15 = 10+150 = 160$

Which is exactly the cost of our example above because if our A node has a probability of 1, we would always experience the scenario above.

Of course, if we were to take some more likely value such as:

$P(A|C)=0.5$

$P(B|C)=0.6$

$A_n = 10$

$B_n = 12$

$C_o = 10 + 0.5*\sum_{i=1}^{10} 15 + 0.6*\sum_{i=1}^{12} 15 = 10+0.5*150+0.6*180 = 193$

Which would be the expected value of our node. Notice that our probabilities can add up to more than or less than 1 because it is reasonable that we might be able to intersect with both volumes or none at all.

Now we need to determine how we want to define $P(A|C)$. This probability is the probability of a random ray intersecting with our volume given that we’ve intersected with our parent volume.

$P(A|C)$ is given as the surface area of our volume A divided by the surface area of our parent volume C. As a result, we can define $P(A|C) = \frac{S_A}{S_C}$ as the probability of intersecting A given that we’ve intersected C.

### Derivation In 2D

In 2 dimensions, this makes sense, as we can image a rectangle, embedded in another rectangle. The probability of a random point being placed in the child rectangle given that it is contained in the parent rectangle is equal to the area of the embedded rectangle divided by the area of the parent rectangle.

Now, instead we’ll imagine uniformally random ray instead of a random line. To select a random ray, we can select a random direction and then select a random ray along the plane defined by that direction.

With this in mind, we want to calculate the probability that our ray will intersect the children volume given that it has intersected the parent volume.

To do this, we want to calculate the average projected area of our shape.

Imagine that we’ve selected our random direction for our ray, we can see that our ray will have intersected our shape if it’s projection onto the plane intersects our random ray.

Now if we add a second volume embedded in the first volume, we can see that the probability of intersecting the first volume would be given by the projected area of the child volume divided by the projected area of the parent volume.

This example only works for a single direction. However, since we’re looking for a probability given any ray direction, we need to calculate the average of this projection across the sphere of all directions.

In 2 dimensions, the projected area of a circle is always it’s diameter. We will use this fact to test our work.

To calculate the average projected area of an arbitrary convex shape[4][5], we first want to split that shape into a series of small discrete patches.

This will be our integration domain.

Once we’ve split our shape into discrete patches, we want to calculate the average projected area of a single patch.

Now we’ll integrate across the circle of directions containing the patch and divide it by the area of our integration. This will effectively give us the average projected area for a single patch of our shape. This works because we will have calculated the projection for every single direction.

$dA_p = \frac{1}{2\pi}\int_{\Omega} |cos\theta| dA d\theta$

Integrating with an absolute value is tricky. Instead due to the symmetry of the problem, we can look at only a quarter of our circle and multiply by 4 and remove our absolute term.

$dA_p = \frac{2}{\pi} dA \int_{0}^{\frac{\pi}{2}} cos\theta d\theta$

Since $\int_{0}^{\frac{\pi}{2}} cos\theta d\theta = 1$ our expression reduces to

$dA_p = \frac{2}{\pi} dA$

Finally, we want to divide our value by 2, this is because we’ve now counted our projected area twice.

$dA_p = \frac{1}{\pi} dA$

Using this formulation, we can calculate the average projected area of our shape by adding all of our average projected areas together.

$A_p = \int_A dA_p$

$A_p = \int_A \frac{1}{\pi} dA$

$A_p = \frac{1}{\pi} A$

And there you have it, the average projected area of a 2 dimensional convex object is $\frac{1}{\pi}$ the area of said object.

Notice that if we apply this to the surface area of a circle with the formula $2\pi r$ we get $A_p = \frac{2\pi r}{\pi} = 2r$ which is our diameter which matches up with our expectation.

### Derivation In 3D

We can take the same steps in 3D.

Calculate the average projection over our sphere of directions:

$dA_p = \frac{1}{4\pi} \int_0^{2\pi} \int_0^{\pi} |cos\theta| sin\theta dA d\theta d\phi$

Integrating across the positive hemisphere and multiplying by 2

$dA_p = \frac{1}{2\pi} dA \int_0^{2\pi} \int_0^{\frac{\pi}{2}} cos\theta sin\theta d\theta d\phi$

Since $\int_0^{\frac{\pi}{2}} cos\theta sin\theta d\theta = \frac{1}{2}$

$dA_p = \frac{1}{2\pi} dA \int_0^{2\pi} \frac{1}{2} d\phi$

$dA_p = \frac{1}{4\pi} dA \int_0^{2\pi} d\phi$

$dA_p = \frac{1}{2} dA$

Finally, dividing by 2 for our double projection

$dA_p = \frac{1}{4} dA$

And plugging into our surface area calculation

$A_p = \int_A dA_p$

$A_p = \frac{1}{4} \int_A dA$

$A_p = \frac{1}{4} A$

### Putting It Together

Finally we can see that our average projected area in 3D is $\frac{1}{4}$ it’s surface area.

To calculate our probability, we simply want to divide our parent’s projected area (C) divided by our child’s projected area (A)

$P(A|C) = \frac{A_p}{C_p}$

$P(A|C) = \frac{0.25*A}{0.25*C}$

$P(A|C) = \frac{A}{C}$

Where A and C is the surface area of our volumes. And voila, that’s how we got that original formula.

## In Depth View

Now that we have all the pieces, we can take a look at the construction of our BVH.

Cranberray builds it’s BVH from the top down. Starting from a containing bounding volume and splitting it in 2 recursively.

Cranberray keeps a queue of bounding volumes to process as a ring buffer. This makes management of front popping and back pushing very simple but makes resizing the queue trickier. As a result, Cranberray simply allocates a large buffer. We could likely allocate the maximum possible number of elements in the ring buffer instead (something along the lines of $2n-1$ where n is the next largest power of 2)

Cranberray then selects the axis with the widest breadth of centroids. The code for this looks like so:

cv2 axisSpan[3];
for (uint32_t axis = 0; axis < 3; axis++)
{
for (uint32_t i = 0; i < count; i++)
{
axisSpan[axis].x = fminf(axisSpan[axis].x, caabb_centroid(start[i].bound, axis));
axisSpan[axis].y = fmaxf(axisSpan[axis].y, caabb_centroid(start[i].bound, axis));
}
}

uint32_t axis;
if (axisSpan[0].y - axisSpan[0].x > axisSpan[1].y - axisSpan[1].x && axisSpan[0].y - axisSpan[0].x > axisSpan[2].y - axisSpan[2].x)
{
axis = 0;
}
else if (axisSpan[1].y - axisSpan[1].x > axisSpan[2].y - axisSpan[2].x)
{
axis = 1;
}
else
{
axis = 2;
}


Once we’ve selected our axis, we split our axis in 12 distinct buckets as. (See PBRT [2] for more info on this approach)

We then calculate the cost of each seperation by adding up all the volumes on the left of the seperation and all the buckets on the right of the seperation.

We then store the cost of each seperation and select the seperation with the minimal cost as our split.

We then continue in this recursion until we’ve run out of items in our queue. (When we’ve partitioned all of our child volumes into leaf nodes).

Finally, our BVH is restructured somewhat for an improvement in memory usage.

Our BVH is stored in 2 arrays, a “jump index” array and a bounds array. This allows us to load the bounds array without having to load the jump indices into memory until we absolutely need them.

We read from our bounds memory much more frequently than our jump memory, as a result, splitting them allows us to make more effective use of our caches.

Our BVH structure looks like this:


typedef struct
{
union
{
struct
{
uint32_t left;
uint32_t right;
} jumpIndices;

struct
{
uint32_t index;
} leaf;
};
} bvh_jump_t;

typedef struct
{
caabb* bounds;
bvh_jump_t* jumps;
uint32_t count;
uint32_t leafCount;
} bvh_t;



The final special format of our BVH is that all the leaf nodes in our tree are stored at the end of our array. This allows us to test if a node is a leaf node by simply comparing the index in our array against the size of our array minus the number of leaves contained in the tree. This allows us to use data that we’ve already loaded into memory instead of requiring us to load extra data to use in our branch introducing a data dependency.

You can find the source for BVH construction here: https://github.com/AlexSabourinDev/cranberries/blob/5fe9c25e1df23d558b7ef8b5475717d9e67a19fc/cranberray.c#L963

We’ll be taking a look at the BVH traversal next. Until then, happy coding!

## Future Work

The primary addition to this BVH construction algorithm would likely be to look into parallelizing it’s construction. [6]

## Introduction

I’ve recently been diving into the world of importance sampling and I decided to share my derivation for importance sampling the cosine lobe.

When we’re shading a point in path tracing, we typically shoot a ray from our surface in a uniformity random direction contained on a hemisphere centered about our normal. This has the downside of introducing quite a bit of variance into our renders.

Imagine that we have a very bright light that only occupies a very small projected area from our shaded point.

We would be very likely to miss this light with most of our samples and our render could turn out much darker than we would expect.

This is where importance sampling comes in.

If you imagine that your illumination is a polar function

If we were to sample a random variable with a distribution that matches this function, we would be much more likely to hit the important points of our function. (Hence importance sampling)

I won’t dive deeply into this topic, as there are a variety of excellent resources detailing this topic. [1]

The essence of it however, is that you want to find a Probability Density Function (PDF) that matches the shape of your illumination function. Once you’ve define this PDF, you can sample it using the Cumulative Density Function (CDF).

## Derivation

Since our cosine lobe illumination will look like this:

We will use this as the basis to derive our distribution since we’re most likely to get the most light from directions arriving parallel to our normal.

Thankfully, our cosine lobe has an analytical formula that we can use as our PDF.

$PDF(\omega) = C*cos(\theta)$ (1)

Our PDF must integrate to 1, we integrate the PDF across our hemisphere

$\int_{\Omega}PDF(\omega)d\omega$

$\int_0^{2\pi}\int_0^{\frac{\pi}{2}}PDF(\omega)sin\theta d\theta d\phi$

Plug in (1)

$\int_0^{2\pi}\int_0^{\frac{\pi}{2}}C*cos\theta sin\theta d\theta d\phi$

$C*\int_0^{2\pi}\int_0^{\frac{\pi}{2}}cos\theta sin\theta d\theta d\phi$

$\int cos\theta sin\theta d\theta = -\frac{1}{4}cos2\theta$

$\int_0^{\frac{\pi}{2}}cos\theta sin\theta d\theta$

$-\frac{1}{4}cos\pi+ \frac{1}{4}cos0$

$\frac{1}{4}+\frac{1}{4}$

$\int_0^{\frac{\pi}{2}}cos\theta sin\theta d\theta=\frac{1}{2}$ (2)

Plug in (2)

$C*\int_0^{2\pi}\frac{1}{2} d\phi$

$C*\frac{1}{2}*2\pi$

$C*\int_0^{2\pi}\int_0^{\frac{\pi}{2}}cos\theta sin\theta d\theta d\phi=C*\pi$ (3)

Since our PDF has to integrate to 1,

$\int_0^{2\pi}\int_0^{\frac{\pi}{2}}PDF(\omega)sin\theta d\theta d\phi = 1$

Plug in (3),

$C*\pi=1$

$C=\frac{1}{pi}$ (4)

Finally, plug in (4) into our PDF,

$PDF(\omega) = \frac{cos(\theta)}{\pi}$ (5)

Now that we have our PDF, we can calculate our PDF in terms of $\theta$ and $\phi$.

$PDF(\theta,\phi)d\theta d\phi = PDF(\omega)d\omega$

$PDF(\theta,\phi)d\theta d\phi = PDF(\omega)sin\theta d\theta d\phi$

$PDF(\theta,\phi)=PDF(\omega)sin\theta$

$PDF(\theta,\phi)=\frac{cos\theta sin\theta}{\pi}$ (6)

Now we integrate with respect to $\phi$ to get $PDF(\theta)$

$\int_0^{2\pi}\frac{cos\theta sin\theta}{\pi}d\phi = 2cos\theta sin\theta$

$PDF(\theta)=2cos\theta sin\theta$

And then to get $PDF(\phi)$,

$\frac{PDF(\theta,\phi)}{PDF(\theta)}=PDF(\phi)$

$\frac{cos\theta sin\theta}{2cos\theta sin\theta \pi}=\frac{1}{2\pi}$

$PDF(\phi)=\frac{1}{2\phi}$

Now we want to calculate the CDF of each function,

$CDF(\theta)=\int_0^\theta PDF(\theta) d\theta$

$CDF(\theta)=\int_0^\theta 2cos(\theta)sin(\theta) d\theta$

$CDF(\theta)=\int_0^\theta sin(2\theta) d\theta$

$CDF(\theta)=\frac{1}{2}-\frac{cos(2\theta)}{2}$

$CDF(\phi)=\int_0^\phi PDF(\phi) d\phi$

$CDF(\phi)=\int_0^\phi\frac{1}{2\pi} d\phi$

$CDF(\phi)=\frac{\phi}{2\pi}$

Now we want to invert our CDF to sample it using our random variable $y$,

$y=CDF(\theta)$

$y=\frac{1}{2}-\frac{cos(2\theta)}{2}$

$\frac{1}{2}-y=\frac{cos(2\theta)}{2}$

$1-2y=cos(2\theta)$

$\frac{cos^{-1}(1-2y)}{2}=\theta$ (7)

For $\phi$,

$y=CDF(\phi)$

$y=\frac{\phi}{2\pi}$

$y*2\pi = \phi$ (8)

Now we have our CDFs and our PDFs, we can finally calculate our direction.

In pseudo code you can simply do:

$\theta=\frac{cos^{-1}(1-2rand01())}{2}$

$\phi=rand01()*2\pi$

With these directions, you can now sample your scene:

$\frac{SampleScene(SphericalTo3D(\theta, \phi))}{PDF(\omega)}$

Plug in (5)

$\frac{SampleScene(SphericalTo3D(\theta, \phi))\pi}{cos\theta}$

## Conclusion

That’s it! These formulas will sample the hemisphere where it is receiving more light defined by the cosine lobe. The results are pretty awesome.

## Introduction

Recently, I’ve been wrestling with attempting to define the mathematics behind the illumination of a triangular area light. Although I have not succeeded yet, I figured I would chronicle my recent attempts. Here goes!