## Saturday, October 20, 2018

### Flavors of Sampling in Ray Tracing

Most "ray tracers" these days are what we used to call "path tracers" or "Monte Carlo ray tracers".   Because they all have about the same core architecture, this short hand is taking over.

These ray tracers have variants but a big one is whether the ray tracing is:
• Batch: run your ray tracer in the background until done (e.g., movie renderers)
• Realtime: run what you can in 30ms or some other frame time (e.g. game renderers go RTX!)
• Progressive: just keep updating the image to make it less noisy as you trace more rays (e.g., a lighting program used by at artist in a studio or engineering firm)
For batch rendering let's pretend for now we are only getting noise from antialiasing the pixels so we have a fundamentally 2D program.   This typically has something like the following for each pixel (i,j):

pixel_color = (0,0,0)
for s = 0 to N-1
u =  random_from_0_to_1()
v =  random_from_0_to_1()
pixel_color += sample_ color(i+u,j+v)
pixel_color /= N

That "sample_color()" does whatever it does to sample a font or whatever.    The first thing that bites you is the diminishing return associated with Monte Carlo methods:   error = constant / sqrt(num_samples).     So to halve the error we need 4X the samples.

Don Mitchell has a nice paper that explains that if you take "stratified" samples, you get better error rates for most functions.    In the case of 2D with edges the error = constant / num_samples.   That is LOADS better.   In graphics a common what to stratify samples is called "jittering" where you usually take a perfect square number of samples in each pixel: N = n*n.   This yields the code:

pixel_color = (0,0,0)
for s = 0 to n-1
for t = 0 to n-1
u =  (s + random_from_0_to_1()) / n
v =  (t + random_from_0_to_1()) / n
pixel_color += sample_ color(i+u,j+v)

Visually the difference in the pattern of the samples is more obvious.   There is a very nice blog post on the details of this sampling here, and it includes this nice figure from Suffern's book:

It turns out that you can replace the random samples not only with jittered sample, but with a regular grid and you will converge to the right answer.   But better still you can use quasi-random samples which makes this a quasi-Monte Carlo method (QMC).   The code is largely the same!   Just replace the (u,v) part above.    The theory that justifies it is a bit different, but the key thing is the points need to be "uniform" and not tend to clump anywhere.   These QMC methods have been investigated in detail by my friend and coworker Alex Keller.   The simplest QMC method is, like the jittering, best accomplished if you know N (but it doesn't need to be restricted to a perfect square).   A famous and good one is the Hammersley Point set.   Here's a picture from Wolfram:
It's regular and uniform, but not too regular.

In 2D these sampling patterns, jittering and QMC are way way better than pure random.   However, there is no free lunch.   In a general path tracer, it is more than a 2D problem.   The program might look like this:

For every pixel
for example sample
pick u,v for screen
pick a time t
pick an a,b on lens
if ray hits something
pick a u',v' for light sampling
pick a u",v" for BRDF sampling
recurse

If you track your random number generation and you take 3 diffuse bounces with shadow rays that will be nine random numbers.    You could think of a ray path through the scene as something that gets generated by a function:

ray_path get_ray_path(u, u', u", u'", u"", u""', ....)

So you sample a nine dimensional hypercube randomly and map those nine-dimensional points to ray paths.   Really this happens in some procedural recursive process, but abstractly it's a mapping.   This means we run into the  CURSE OF DIMENSIONALITY.   Once the integral is high dimensional, if the integrand is complex, STRATIFIED SAMPLING DOESN'T HELP.    However, you will notice (almost?) all serious production ray tracers do add stratified sampling.   Why?

The reason is that for many pixels, the integrand is mostly constant except for two of the dimensions.   For example, consider a pixel that is:

• In focus (so lens sample doesn't matter)
• Not moving (so time sample doesn't matter)
• Doesn't have an edge (so pixel sample doesn't matter)
• Is fully in (or out of) shadow (so light sample doesn't matter)
What does matter is the diffuse bounce.   So it acts like a 2D problem.   So we need the BRDF samples, let's call them u7 and u8, to be well stratified.

QMC methods here typically automatically ensure that various projections of the high dimensional samples are themselves well stratified.    Monte Carlo methods, on the other hand, typically try to get this property for some projections, namely the 2 pixel dimensions, the 2 light dimensions, the 2 lens dimensions, etc.   They typically do this as follows for the 4D case:

1. Create N "good" light samples on [0,1)^2
2. Create N "good" pixel samples on [0,1)^2
3. Create a permutation of the integers 0, ..., N-1
4. Create a 4D pattern using the permutation where the ith sample is light1[i], light2[i],pixel[permute[i]], pixel2[permute[i]].
So a pain :)   There are bunches of papers on doing Monte Carlo, QMC, or "blue noise" uniform random point sets.   But we are yet quite done.

First, we don't know how many dimensions we have!   The program recurses and dies by hitting a dark surface, exiting the scene, or Russian Roulette.    Most programs degenerate to pure random after a few bounces to make it so we can sort of know the dimensionality.

Second, we might want a progressive preview on the renderer where it gets better as you wait.   Here's a nice example.

So you don't know what N is in advance!   You want to be able to add samples potentially forever.  This is easy if the samples are purely random, but not so obvious if doing QMC or stratified.   The QMC default answer are to use Halton Points.    These are designed to be progressive!     Alex Keller at NVIDIA and his collaborators have found even better ways to do this with QMC.   Per Christensen and  Andrew Kensler and Charlie Kilpatrick at Pixar have a new Monte Carlo sampling method that is making waves in the movie industry.   I have not ever implemented these and would love to hear your experiences if you do (or have!)

## Monday, June 4, 2018

### Sticking a thin lens in a ray tracer

I need to stick a " ideal thin lens" in my ray tracer.   Rather than a real lens with some material, coating, and shape, it's an idealized version like we get in Physics1.

The thin lens has some basic properties that have some redundancy but they are the ones I remember and deploy when needed:
1. a ray through the lens center is not bent
2. all rays through a point that then pass through the lens converge at some other point
3. a ray through the focal point (a point on the optical axis at distance from lens f, the focal length of the lens) will be parallel to the optical axis after being bent by the lens
4. all rays from a point a distance A from the lens will converge at a distance B on the other side of the lens and obey the thin lens law: 1/A + 1/B = 1/f
Here's a sketch of those rules:

So if I have a (purple) ray with origin a that hits the lens at point m, how does it bend?   It bends towards point b no matter what the ray direction is.   So the new ray is:

p(t) = m + t (b-m)

So what is point b?

It is in the direction of point c, but extended by some distance.  We know the center of the lens c, so we can use the ratios of the segments to extend to that:

b = a + (c-a)  (B+A) / A?

So what is (B+A) /A?

We know 1/A + 1/B = 1/f, so B = 1/(1/f-1/A).   So the point b is:

b = a + (c-a) (1/(1/f-1/A) + A)/A =
b = a + (c-a) (1/(A/f - 1) + 1) =
b = a + (c-a) (A/(A - f))

OK lets try a couple of trivial cases.   What if  A = 2f?

b = a + (c-a) (2f/(2f-f)) = b = a + 2(c-a)

That looks right (symmetric case-- A = B there)

So final answer, given a ray with origin a that hits a lens with center c and focal length f at point m, the refracted ray is:

p(t) = m + t( a + (c-a) (A/(A - f)) - m)

There is a catch.   What if B < 0?   This happens when A < f.  Address that case when it comes up :)

## Thursday, May 31, 2018

### Generating uniform Random rays that hit an axis aligned box

For some tests you want the set of "all" rays that hit a box.   If you want to stratify this is somewhat involved (and I don't know that I have done it nor seen it).   Chris Wyman and I did a jcgt paper on doing this stratified in a square in 2D.   But often stratification isn't needed.   When in doubt I never do it-- the software is always easier to deal with un-stratified, and as soon as dimension gets high most people dont bother because of the Curse of Dimensionality.

We would like to do this with as little math as possible.   First lets consider any side of the box.   (this would apply to any convex polyhedron if we wanted something more general).    If the box in embedded in all possible uniform rays, any ray that hits the box will enter at exactly one point on the surface of the box, and all points are equally likely.   So our first job is to pick a uniform point on the surface.   We can use the technique Greg Turk used to seed points for texture generation on a triangular mesh:

Probability of each side with box of side lengths X, Y, Z is side area of total area.   The side areas are:

XY, YZ, ZX (2 of each)

We can do cumulative area and stuff it in an array of length 6:

c_area[0] = XY
c_area[1] = area[0] + XY
c_area[2] = area[1] + YZ
c_area[3] = area[2] + YZ
c_area[4] = area[3] + ZX
c_area[5] = area[4] + ZX

Now normalize it so it is a cumulative fraction:

for (int i = 0; i < 6; i++)
c_area[i] /= c_area[5]

No take a uniform random real r() in [0,1)

int candidate = 0;
float ra = r();
while (c_area[candidate] < ra) candidate++;

Now candidate is the index to the side.

Let's say the side is in the xy plane and x goes from 0 to X and y goes from 0 to Y.   Now pick a uniform random point on that side:

vec3 ray_entry(X*r(), Y*r(), 0);

Now we have a ray origin.   What is the ray direction?  It is not uniform in all directions.   These are the rays that hit the side.   So the density is proportional to the cosine to the normal-- so they are Lambertian!   This is not obvious.   I will punt on justifying that for now.

So for the xy plane one above, the normal is +z and the ray direction is a uniform random point on a disk projected onto the hemisphere:

float theta = 2*M_PI*r();
float z = sqrt(1-x*x-y*y);
ray_direction = vec3(x,y,z);

Now we need that for each of the six sides.

We could probably find symmetries to have 3 cases, or maybe even a loop, but I personally would probably not bother because me trying to be clever usually ends poorly...

## Wednesday, March 14, 2018

### Egyptian estimates of PI

I saw a neat tweet on how the estimate the Egyptians used for PI.

This is all my speculation, and maybe a math history buff can enlighten me, but the D^2 dependence they should discover pretty naturally.   Including the constant before squaring is, I would argue, just as natural as having it outside the parentheses, so let's go with that for now.   So was there a nearby better fraction?   How well did the Egyptians do?   A brute force program should tell us.

We will use the ancient programming language C:

#include
#include
int main() {
double min_error = 10;
for (int denom = 1; denom < 10000; denom++) {
for (int num = 1; num < denom; num++) {
double approx = 2*double(num)/double(denom);
approx = approx*approx;
double error2 = M_PI-approx;
error2 = error2*error2;
if (error2 < min_error) {
min_error = error2;
printf("%d/%d %f\n", num, denom, 4*float(num*num)/float(denom*denom));
}
}
}
}

This produces output:

1/2 1.000000
2/3 1.777778
3/4 2.250000
4/5 2.560000
5/6 2.777778
6/7 2.938776
7/8 3.062500
8/9 3.160494
23/26 3.130177
31/35 3.137959
39/44 3.142562
109/123 3.141252
148/167 3.141597
4401/4966 3.141588
4549/5133 3.141589
4697/5300 3.141589
4845/5467 3.141589
4993/5634 3.141589
5141/5801 3.141590
5289/5968 3.141590
5437/6135 3.141590
5585/6302 3.141590
5733/6469 3.141590
5881/6636 3.141591
6029/6803 3.141591
6177/6970 3.141591
6325/7137 3.141591
6473/7304 3.141591
6621/7471 3.141591
6769/7638 3.141591
6917/7805 3.141592
7065/7972 3.141592
7213/8139 3.141592
7361/8306 3.141592
7509/8473 3.141592
7657/8640 3.141592
7805/8807 3.141592
7953/8974 3.141593
8101/9141 3.141593
8249/9308 3.141593
8397/9475 3.141593
8545/9642 3.141593

So 7/8 was already pretty good, and you need to get to 23/26 before you do any better!   I'd say the Egyptians did extremely well.

What if they had put the constants outside the parens?   How well could they have done?   We can change two of the lines above to:

double approx = 4*double(num)/double(denom);//approx = approx*approx;

and the printf to:

printf("%d/%d %f\n", num, denom, 4*float(num)/float(denom));

And we get:

1/2 2.000000
2/3 2.666667
3/4 3.000000
4/5 3.200000
7/9 3.111111
11/14 3.142857
95/121 3.140496
106/135 3.140741
117/149 3.140940
128/163 3.141104
139/177 3.141243
150/191 3.141361
161/205 3.141464
172/219 3.141552
183/233 3.141631
355/452 3.141593

So 7/9 is not bad!  And 11/14 even better.   So no clear winner here on whether the rational constant should be inside the parens or not.