# A list of path tracing shaders

Path tracing is a surprisingly simple technique to render realistic images. This would be my definition if you are unfamiliar with the term. But if you already have experience with various ray tracing techniques, I would probably say that path tracing is a remarkably elegant solution to the rendering equation. You can implement a toy path tracer in a weekend or, if you’ve already done it a few times before, within 25 minutes.

Recently I was documenting myself on path tracing, and some of the techniques that can be used, like next event estimation, bidirectional path tracing, Russian roulette, etc. This is a case where ShaderToy can be an invaluable source of examples and information, and so I was browsing path tracing shaders there. As the number of open tabs was starting to get impractical, I decided to use the “playlist” feature of ShaderToy to bookmark them all.

You can find the list here: Path tracing, on ShaderToy.

The examples of path tracers listed include very naive implementations, hacky ones, rendering features like advanced BRDF, volumetric lighting or spectral rendering, or various noise reduction techniques such as next event estimation, bidirectional path tracing, multiple importance sampling, accumulation over frames with temporal reprojection, screen space blue noise, or convolutional neural network based denoising.

Some of those shaders are meant to be artworks, but even the technical experimentation ones look nice, because the global illumination inherent to path tracing tends to generate images that are pretty.

# Building an artificial window

Several years ago, I mentioned the Italian company CoeLux, which specializes in making artificial windows: light fixtures that look like sunlight in a clear blue sky.

The price of their products is apparently in the range of several tens of thousands of dollars (I’ve heard prices like $20k to 50k), which makes it out of reach for most individuals. Not many details about their invention are available either (from the promotion material: LED powered, several hundred watts of electrical power, a solid diffuse material, and a thickness around 1 meter), and I was left wondering what was the secret sauce to their intriguing technology. The YouTube channel DIY Perks has been working on day light projects for a while now, improving at each iteration. Yesterday they published a video explaining how to build a light that seems to give very similar results as CoeLux’s product, from some basic materials that are fairly simple to find. Since their solution takes roughly the same volume, it’s tempting to think it uses the same technique It’s extremely satisfying to finally see how this works and, despite the practical aspects, quite tempting to try if only to see how it looks in real life. # Intersection of a ray and a plane I previously showed the derivation of how to determine the intersection of a plane and a cone. At the time I had to solve that equation, so after doing so I decided to publish it for anyone to use. Given the positive feedback, it seems this was useful, so I might as well continue with a few more. Here is probably the most basic intersection: a ray and a plane. Solving it is straightforward, which I hope can be seen below. Like last time, I am using vector notation. 1. We define a ray with its origin$O$and its direction as a unit vector$\hat{D}$. Any point$X$on the ray at a signed distance$t$from the origin of the ray verifies:$\vec{X} = \vec{O} + t\vec{D}$. When$t$is positive$X$is in the direction of the ray, and when$t$is negative$X$is in the opposite direction. 2. We define a plane with a point$S$on that plane and the normal unit vector$\hat{N}$, perpendicular to the plane. The distance between any point$X$and the plane is$d = \lvert (\vec{X} – \vec{S}) \cdot \vec{N} \rvert$. If this equality is not obvious for you, you can think of it as the distance between$X$and$S$along the$\vec{N}$direction. When$d=0$, it means$X$is on the plane itself. 3. We define$P$the intersection or the ray and the plane, and which we are interested in finding. Since$P$is both on the ray and on the plane, we can write: $$\left\{ \begin{array}{l} \vec{P}=\vec{O} + t\vec{D} \\ \lvert (\vec{P} – \vec{S}) \cdot \vec{N} \rvert = 0 \end{array} \right.$$ Because the distance$d$from the plane is$0$, the absolute value is irrelevant here. We can just write: $$\left\{ \begin{array}{l} \vec{P}=\vec{O} + t\vec{D} \\ (\vec{P} – \vec{S}) \cdot \vec{N} = 0 \end{array} \right.$$ All we have to do is replace$P$with$\vec{O} + t\vec{D}$in the second equation, and reorder the terms to get$t$on one side. $$(\vec{O} + t\vec{D} – \vec{S}) \cdot \vec{N} = 0$$ $$\vec{O} \cdot \vec{N} + t\vec{D} \cdot \vec{N} – \vec{S} \cdot \vec{N} = 0$$ $$t\vec{D} \cdot \vec{N} = \vec{S} \cdot \vec{N} – \vec{O} \cdot \vec{N}$$ $$t = \frac{(\vec{S} – \vec{O}) \cdot \vec{N}}{ \vec{D} \cdot \vec{N} }$$ A question to ask ourselves is: what about the division by$0$? Looking at the diagram, we can see that$\vec{D} \cdot \vec{N} = 0$means the ray is parallel to the plane, and there is no solution unless$O$is already on the plane. Otherwise, the ray intersects the plane for the value of$t$written above. That’s it, we’re done. Note: There are several, equivalent, ways of representing a plane. If your plane is not defined by a point$S$and a normal vector$\hat{N}$, but rather with a distance to the origin$s$and a normal vector$\hat{N}$, you can notice that$s = \vec{S} \cdot \vec{N}$and simplify the result above, which becomes: $$t = \frac{s – \vec{O} \cdot \vec{N}}{ \vec{D} \cdot \vec{N} }$$ ## Signed distance to a plane For the sake of simplicity, in the above we defined the distance to the plane as an absolute value. It is possible however to define it as a signed value:$d = (\vec{X} – \vec{S}) \cdot \vec{N}$. In this case$d>0$means$X$is somewhere on the side of the plane pointed by$\vec{N}$, while$d<0$means$X$is on the opposite side of the plane. Distances that can be negative are called signed distances, and they are a foundation of Signed Distance Fields (SDF). # Long hiatus Last week I was lucky enough to attend SIGGRAPH 2018, in Vancouver. My colleagues and I were presenting on a booth the work we had done, a VR story with a distinctive comic book look. I was also invited to participate to a panel session on demoscene, where I shared some lessons learned while making the 64k intro H – Immersion. The event brought a certain sense of conclusion to this work, aside from filling me with inspiration and motivation to try new things. It has been a long time since I last posted anything here. For the last two years the majority of my spare time went into making that 64k intro. In fact the last post, “Intersection of a ray and a cone”, was related to it. I was implementing volumetric lighting for the underwater scenes, and wanted to resolve cones of light with ray tracing, before marching inside those cones. LLB and I have talked about the creation process in two making-of articles: “A dive into the making of Immersion”, and “Texturing in a 64kB intro”. During that time, a lot of new things have happened in the computer graphics community. It has been difficult to keep track of everything. The last topic I started experimenting with is point cloud and mesh capture from photos; I might expend on it here in the future. I also want to experiment with DIY motion capture. Anyway, it’s time to resume posting here. # Intersection of a ray and a cone Some time ago I needed to solve analytically the intersection of a ray and a cone. I was surprised to see that there are not that many resources available; there are some, but not nearly as many as on the intersection of a ray and a sphere for example. Add to it that they all use their own notation and that I lack math exercise, after a bit of browsing I decided I needed to write a proof by myself to get a good grasp of the result. So here goes, the solution to the intersection of a ray and a cone, in vector notation. 1. We define a ray with its origin$O$and its direction as a unit vector$\hat{D}$. Any point$X$on the ray at a signed distance$t$from the origin of the ray verifies:$\vec{X} = \vec{O} + t\vec{D}$. When$t$is positive$X$is in the direction of the ray, and when$t$is negative$X$is in the opposite direction. 2. We define a cone with its tip$C$, its axis as a unit vector$\hat{V}$in the direction of increasing radius, and$\theta$the half angle between the axis and the surface. Any point$X$on the cone verifies:$(\vec{X} – \vec{C}) \cdot \vec{V} = \lVert \vec{X} – \vec{C} \rVert \cos\theta$3. Finally we define$P$the intersection or the ray and the cone, and which we are interested in finding.$P$verifies both equations, so we can write: $$\left\{ \begin{array}{l} \vec{P}=\vec{O} + t\vec{D} \\ \frac{ \vec{P} – \vec{C} }{\lVert \vec{P} – \vec{C} \rVert} \vec{V} = \cos\theta \end{array} \right.$$ We can multiply the second equation by itself to work with it, then reorder things a bit. $$\left\{ \begin{array}{l} \vec{P}=\vec{O} + t\vec{D} \\ \frac{ ((\vec{P} – \vec{C}) \cdot \vec{V})^2 }{ (\vec{P} – \vec{C}) \cdot (\vec{P} – \vec{C}) } = \cos^2\theta \end{array} \right.$$ $$\left\{ \begin{array}{l} \vec{P}=\vec{O} + t\vec{D} \\ ((\vec{P} – \vec{C}) \cdot \vec{V})^2 – (\vec{P} – \vec{C}) \cdot (\vec{P} – \vec{C}) \cos^2\theta = 0 \end{array} \right.$$ Remember the mouthful earlier about$\hat{V}$being in the direction of increasing radius? By elevating$\cos\theta$to square, we’re making negative values of$\cos$positive: values of$\theta$beyond 90° become indistinguishable from values below 90°. This has the side effect of turning it into the equation of not one, but two cones sharing the same axis, tip and angle, but in opposite directions. We’ll fix that later. We replace$\vec{P}$with$\vec{O} + t\vec{D}$and work the equation until we get a good old quadratic function that we can solve. $$\require{cancel} ((\vec{O} + t\vec{D} – \vec{C})\cdot\vec{V})^2 – (\vec{O} + t\vec{D} – \vec{C}) \cdot (\vec{O} + t\vec{D} – \vec{C}) \cos^2\theta = 0$$ $$((t\vec{D} + \vec{CO})\cdot\vec{V})^2 – (t\vec{D} + \vec{CO}) \cdot (t\vec{D} + \vec{CO}) \cos^2\theta = 0$$ $$(t\vec{D}\cdot\vec{V} + \vec{CO}\cdot\vec{V})^2 – (t^2\cancel{\vec{D}\cdot\vec{D}} + 2t\vec{D}\cdot\vec{CO} + \vec{CO}\cdot\vec{CO}) \cos^2\theta = 0$$ $$(t^2(\vec{D}\cdot\vec{V})^2 + 2t(\vec{D}\cdot\vec{V})(\vec{CO}\cdot\vec{V}) + (\vec{CO}\cdot\vec{V})^2) – (t^2 + 2t\vec{D}\cdot\vec{CO} + \vec{CO}\cdot\vec{CO}) \cos^2\theta = 0$$ $$t^2(\vec{D}\cdot\vec{V})^2 + 2t(\vec{D}\cdot\vec{V})(\vec{CO}\cdot\vec{V}) + (\vec{CO}\cdot\vec{V})^2 – t^2\cos^2\theta – 2t\vec{D}\cdot\vec{CO}\cos^2\theta – \vec{CO}\cdot\vec{CO}\cos^2\theta = 0$$ Reorder a bit: $$t^2((\vec{D}\cdot\vec{V})^2 – \cos^2\theta) + 2t((\vec{D}\cdot\vec{V})(\vec{CO}\cdot\vec{V}) – \vec{D}\cdot\vec{CO}\cos^2\theta) + (\vec{CO}\cdot\vec{V})^2 – \vec{CO}\cdot\vec{CO}\cos^2\theta = 0$$ There we go, we have our$at^2 + bt + c = 0$equation, with: $$\left\{ \begin{array}{l} a = (\vec{D}\cdot\vec{V})^2 – \cos^2\theta \\ b = 2\Big((\vec{D}\cdot\vec{V})(\vec{CO}\cdot\vec{V}) – \vec{D}\cdot\vec{CO}\cos^2\theta\Big) \\ c = (\vec{CO}\cdot\vec{V})^2 – \vec{CO}\cdot\vec{CO}\cos^2\theta \end{array} \right.$$ From there, you know the drill: calculate the determinant$\Delta = b^2 – 4ac$then depending on its value: • If$\Delta < 0$, the ray is not intersecting the cone. • If$\Delta = 0$, the ray is intersecting the cone once at$t = \frac{-b}{2a}$. • If$\Delta > 0$, the ray is intersecting the cone twice, at$t_1 = \frac{-b – \sqrt{\Delta}}{2a}$and$t_2 = \frac{-b + \sqrt{\Delta}}{2a}$. But wait! We don’t have one cone but two, so we have to reject solutions that intersect with the shadow cone.$P$must still verify$\frac{ \vec{P} – \vec{C} }{\lVert \vec{P} – \vec{C} \rVert} \vec{V} = \cos\theta$, or simply, if$\theta < 90°$:$(\vec{P} – \vec{C})\cdot\vec{V} > 0$. Note that there is also the corner case of the ray tangent to the cone and having an infinity of solutions to consider. I’ve completely swept it under the rug since it doesn’t matter in the context I was, but if it does to you, you’ve been warned about it. Also remember to check the sign of$t$to know whether$P$is in the direction of the ray. You may need to determine which of$t_1$or$t_2$you want to use, which depends on your use case. For example is your ray origin inside or outside of the cone? Now for a little sanity test, let’s consider the corner case$C=O$, where the ray origin is the tip of the cone (thanks Rubix for the suggestion!). We have$b=0$and$c=0$thus$\Delta=0$and$t=\frac{-b}{2a}=0$which is the expected result. I also tried the cases$\theta=0$and$\theta=\pi/2$, but expanding$\Delta\$ proved too tedious to proceed to the end. So this is left as an exercise, as they say. :)

Finally, to demonstrate that the result is indeed correct, here is a glorious ray traced cone scene on ShaderToy:

I hope this can prove useful to others too.

Oh, and Happy New Year by the way!

# Users don’t read error messages

My CHI professor used to repeat that users didn’t read error messages, and that you should avoid them when possible. Think about it: how often do you close error messages without even reading whatever is written? And when you do read them, how often do you find them both clear and relevant?

I was recently shown this picture (source unknown unfortunately; adapted from these ones) of a slide that captures perfectly the problem with most error dialogs.

Every error message.