# Dual Numbers & Automatic Differentiation

In the last post, I talked about imaginary numbers, complex numbers, and how to use them to rotate vectors in 2d.

In this post, I want to share another interesting type of number called a “Dual Number” that uses the symbol ε (epsilon) and has a neat trick of automatically calculating the derivative of a function while you calculate the value of the function at the same time.

Dual numbers are pretty similar to imaginary numbers but there is one important difference. With imaginary numbers, i^2 = -1, but with dual numbers, ε^2 = 0 (and ε is not 0!). That may seem like a small difference, but oddly, that opens up a whole interesting world of mathematical usefulness.

Before we dig into automatic differentiation, I want to go over the mathematical basics for how dual numbers behave.

## Basic Dual Number Math

Adding dual numbers is the same as adding complex numbers; you just add the real and dual parts separately:

(3 + 4ε) + (1 + 2ε) = 4 + 6ε

Subtraction works the same way as well:

(3 + 4ε) – (1 + 2ε) = 2 + 2ε

To multiply dual numbers, you use F.O.I.L. just like you do with complex numbers:

(3 + 4ε) * (1 + 2ε) =
3 + 6ε + 4ε + 8ε^2 =
3 + 10ε + 8ε^2

However, since ε^2 is zero, the last term 8ε^2 disappears:
3 + 10ε

It’s interesting to note that with complex numbers, the i^2 became -1, so the last term changed from imaginary to real, meaning that the imaginary numbers fed back into the real numbers during multiplication. With dual numbers, that isn’t the case, the dual numbers don’t feed back into the real numbers during multiplication.

In both complex and dual numbers the real terms do affect the non real terms during multiplication.

The division operator relates to the conjugate. I have source code for it below, and some of the links at the end of the post go into the details of that and other operations.

## Quick Review: Derivatives (Slope)

If you know the line formula y=mx+b, but you don’t know what a derivative is you are in luck. Remember how “m” is the slope of the line, specifying how steep it is? That is what the derivative is too, it’s just the slope.

Below is a graph of y=2x+1. At every point on that line, the derivative (or slope) is 2. That means that for every step we make on the x axis to the right (positive direction), we make 2 steps up on the y axis (positive direction).

Now, check out this graph of y=x^2-0.2

The derivative (or slope) at every point on this graph is 2x. That means that the slope changes depending on where the x coordinate is!

So, when x=0, the slope is 0. You can see that in the graph where x=0, that it is horizontal, meaning that a step on the x axis becomes no steps on the y axis (only at that point where x is 0, and only if you take an infinitely small step).

When x is 1, the slope is 2, when x is 2, the slope is 4, when x is 3, the slope is 6. Since the numbers increase as we increase x from 0, that tells us that the graph gets steeper as we go to the right, which you can see in the graph.

Alternately, when x is -1, the slope is -2, when x is -2, the slope is -4, and when x is -3, the slope is -6. This shows us that as we decrease x from 0, the graph gets steeper in the opposite direction, which you can see in the graph as well.

## What is Automatic Differentiation?

Let’s say you have a function (possibly a curve) describing the path of a rocket, and you want to make the rocket point down the path that it’s traveling.

One way you might do this is to evaluate your function f(T) to get the current location of your rocket (where T is how long the rocket has been flying), and then calculate the derivative f'(T) to find the slope of the graph at that point so that you can orient the rocket in that direction.

You could calculate the value and slope of the function at time T independently easily enough if you know how to get the derivative of a function (a calculus topic), or use wolframalpha.com.

However, if you have a complex equation, or maybe if the equation is controlled by user input, or game data, it might not be so easy to figure out what the derivative is at run time.

For instance… imagine having a function that rolled random numbers to figure out what mathematical operation it should preform on a number next (if we roll a 0, add 3, if we roll a 1 multiply by 2, if we roll a 2, square the number… etc). It isn’t going to be simple to take the derivative of the same mathematical function.

Here enters automatic differentiation (or AD). AD lets you calculate the derivative WHILE you are calculating the value of the function.

That way, you can do whatever math operations you want on your number, and in the end you will have both the value of f(T) as well as the derivative f'(T).

## Using ε for Automatic Differentiation

You can use dual number operations on numbers to calculate the value of f(x) while also calculating f'(x) at the same time. I’ll show you how with a simple example using addition and multiplication like we went over above.

the first thing we do is convert our 4 into a dual number, using 1 for the dual component, since we are plugging it in for the value of x, which has a derivative of 1.

4+1ε

Next, we want to multiply that by the constant 3, using 0 for the dual component since it is just a constant (and the derivative of a constant is 0)

(4+1ε) * (3 + 0ε) =
12 + 0ε + 3ε + 0ε^2 =
12 + 3e

Lastly, we need to add the constant 2, using 0 again for the dual component since it’s just a constant.
(12 + 3ε) + (2 + 0ε) =
14 + 3ε

In our result, the real number component (14) is the value of f(4) and the dual component (3) is the derivative f'(4), which is correct if you work it out!

Let’s try f(5). First we convert 5 to a dual number, with the dual component being 1.

5 + 1ε

Next we need to multiply it by the constant 3 (which has a dual component of 0)

(5 + 1ε) * (3 + 0e) =
15 + 0ε + 3ε + 0ε^2 =
15 + 3ε

Now, we add the constant 2 (which has a dual component of 0 again since it’s just a constant)
(15 + 3ε) + (2 + 0ε) =
17 + 3ε

So, our answer says that f(5) = 17, and f'(5) = 3, which again you can verify is true!

The example above worked well but it was a linear function. What if we want to do a function like f(x) = 5x^2 + 4x + 1?

Let’s calculate f(2). We are going to first calculate the 5x^2 term, so we need to start by making a dual number for the function parameter x:
(2 + 1ε)

Next, we need to multiply it by itself to make x^2:
(2 + 1ε) * (2 + 1ε) =
4 + 2ε + 2ε + 1ε^2 =
4 + 4ε

(remember that ε^2 is 0, so the last term disappears)

next, we multiply that by the constant 5 to finish making the 5x^2 term:
(4 + 4ε) * (5 + 0ε) =
20 + 0ε + 20ε + 0ε^2 =
20 + 20ε

Now, putting that number aside for a second we need to calculate the “4x” term by multiplying the value we plugged in for x by the constant 4
(2 + 1ε) * (4 + 0ε) =
8 + 0ε + 4ε + 0ε^2 =
8 + 4ε

Next, we need to add the last 2 values together (the 5x^2 term and the 4x term):
(20 + 20ε) + (8 + 4ε) =
28 + 24ε

Lastly, we need to add in the last term, the constant 1
(28 + 24ε) + (1 + 0ε) =
29 + 24e

There is our answer! For the equation y = 5x^2 + 4x + 1, f(2) = 29 and f'(2) = 24. Check it, it’s correct (:

As one last example let’s calculate f(10) and f'(10) with the same function above y = 5x^2 + 4x + 1.

First, to start calculating the 5x^2 term, we need to make 10 into a dual number and multiply it by itself to make x^2:
(10 + 1ε) * (10 + 1ε) =
100 + 10ε + 10ε + 1ε^2 =
100 + 20ε

Next, we multiply by the constant 5 to finish making the 5x^2 term:
(100 + 20ε) * (5 + 0ε) =
500 + 0ε + 100ε + 0ε^2 =
500 + 100ε

Putting that aside, let’s calculate the 4x term by multiplying our x value by the constant 4:
(10 + 1ε) * (4 + 0ε) =
40 + 0ε + 4ε + 0ε^2 =
40 + 4ε

Lastly, let’s add our terms: 5x^2, 4x and the constant 1
(500 + 100ε) + (40 + 4ε) + (1 + 0ε) =
541 + 104ε

The answer tells us that for the equation y = 5x^2 + 4x + 1, f(10) = 541 and f'(10) = 104.

## Sample Code

There are lots of other mathematical operations that you can do with dual numbers. I’ve collected as many as I was able to find and made up some sample code that uses them. The sample code is below, as well as the program output.

#include
#include

#define PI 3.14159265359f

// In production code, this class should probably take a template parameter for
// it's scalar type instead of hard coding to float
class CDualNumber
{
public:
CDualNumber (float real = 0.0f, float dual = 0.0f)
: m_real(real)
, m_dual(dual)
{
}

float Real () const { return m_real; }
float Dual () const { return m_dual; }

private:
float m_real;
float m_dual;
};

//----------------------------------------------------------------------
// Math Operations
//----------------------------------------------------------------------
inline CDualNumber operator + (const CDualNumber &a, const CDualNumber &b)
{
return CDualNumber(a.Real() + b.Real(), a.Dual() + b.Dual());
}

inline CDualNumber operator - (const CDualNumber &a, const CDualNumber &b)
{
return CDualNumber(a.Real() - b.Real(), a.Dual() - b.Dual());
}

inline CDualNumber operator * (const CDualNumber &a, const CDualNumber &b)
{
return CDualNumber(
a.Real() * b.Real(),
a.Real() * b.Dual() + a.Dual() * b.Real()
);
}

inline CDualNumber operator / (const CDualNumber &a, const CDualNumber &b)
{
return CDualNumber(
a.Real() / b.Real(),
(a.Dual() * b.Real() - a.Real() * b.Dual()) / (b.Real() * b.Real())
);
}

inline CDualNumber sqrt (const CDualNumber &a)
{
float sqrtReal = ::sqrt(a.Real());
return CDualNumber(
sqrtReal,
0.5f * a.Dual() / sqrtReal
);
}

inline CDualNumber pow (const CDualNumber &a, float y)
{
return CDualNumber(
::pow(a.Real(), y),
y * a.Dual() * ::pow(a.Real(), y - 1.0f)
);
}

inline CDualNumber sin (const CDualNumber &a)
{
return CDualNumber(
::sin(a.Real()),
a.Dual() * ::cos(a.Real())
);
}

inline CDualNumber cos (const CDualNumber &a)
{
return CDualNumber(
::cos(a.Real()),
-a.Dual() * ::sin(a.Real())
);
}

inline CDualNumber tan (const CDualNumber &a)
{
return CDualNumber(
::tan(a.Real()),
a.Dual() / (::cos(a.Real()) * ::cos(a.Real()))
);
}

inline CDualNumber atan (const CDualNumber &a)
{
return CDualNumber(
::atan(a.Real()),
a.Dual() / (1.0f + a.Real() * a.Real())
);
}

inline CDualNumber SmoothStep (CDualNumber x)
{
// f(x) = 3x^2 - 2x^3
// f'(x) = 6x - 6x^2
return x * x * (CDualNumber(3) - CDualNumber(2) * x);
}

//----------------------------------------------------------------------
// Test Functions
//----------------------------------------------------------------------

void TestSmoothStep (float x)
{
CDualNumber y = SmoothStep(CDualNumber(x, 1.0f));
printf("smoothstep 3x^2-2x^3(%0.4f) = %0.4fn", x, y.Real());
printf("smoothstep 3x^2-2x^3'(%0.4f) = %0.4fnn", x, y.Dual());
}

void TestTrig (float x)
{
CDualNumber y = sin(CDualNumber(x, 1.0f));
printf("sin(%0.4f) = %0.4fn", x, y.Real());
printf("sin'(%0.4f) = %0.4fnn", x, y.Dual());

y = cos(CDualNumber(x, 1.0f));
printf("cos(%0.4f) = %0.4fn", x, y.Real());
printf("cos'(%0.4f) = %0.4fnn", x, y.Dual());

y = tan(CDualNumber(x, 1.0f));
printf("tan(%0.4f) = %0.4fn", x, y.Real());
printf("tan'(%0.4f) = %0.4fnn", x, y.Dual());

y = atan(CDualNumber(x, 1.0f));
printf("atan(%0.4f) = %0.4fn", x, y.Real());
printf("atan'(%0.4f) = %0.4fnn", x, y.Dual());
}

void TestSimple (float x)
{
CDualNumber y = CDualNumber(3.0f) / sqrt(CDualNumber(x, 1.0f));
printf("3/sqrt(%0.4f) = %0.4fn", x, y.Real());
printf("3/sqrt(%0.4f)' = %0.4fnn", x, y.Dual());

y = pow(CDualNumber(x, 1.0f) + CDualNumber(1.0f), 1.337f);
printf("(%0.4f+1)^1.337 = %0.4fn", x, y.Real());
printf("(%0.4f+1)^1.337' = %0.4fnn", x, y.Dual());
}

int main (int argc, char **argv)
{
TestSmoothStep(0.5f);
TestSmoothStep(0.75f);
TestTrig(PI * 0.25f);
TestSimple(3.0f);
return 0;
}

Here is the program output:

## Closing Info

When you are thinking what number ε has to be so that ε^2 is 0 but ε is not 0, you may be tempted to think that it is an imaginary number, just like i (the square root of -1) that doesn’t actually exist. This is actually not how it is… I’ve seen ε described in two ways.

One way I’ve seen it described is that it’s an infinitesimal number. That sort of makes sense to me, but not in a concrete and tangible way.

The way that makes more sense to me is to describe it as a matrix like this:
[0, 1]
[0, 0]

If you multiply that matrix by itself, you will get zero(s) as a result.

In fact, an alternate way to implement the dual numbers is to treat them like a matrix like that.

I also wanted to mention that it’s possible to modify this technique to get the 2nd derivative of a function or the 3rd, or the Nth. It isn’t only limited to the 1st derivative. Check the links at the bottom of this post for more info, but essentially, if you want 1st and 2nd derivative, you need to make it so that ε^3 = 0 instead of ε^2 = 0. There is a way to do that with matrices.

Another neat thing is that you can also extend this into multiple dimensions. This is useful for situations like if you have some terrain described by mathematical functions, when you are walking the grid of terrain to make vertex information, you can get the slope / gradient / surface normal at the same time.

Lastly, I wanted to mention a different kind of number called a hyperbolic number.

The imaginary number i^2 = -1 and we can use it to do 2d rotations.

The dual number ε^2 is 0 (and ε is not 0) and we can use it to do automatic differentiation.

Hyperbolic numbers have j, and j^2 = 1 (and j is not 1). I’m not sure, but I bet they have some interesting usefulness to them too. It would be interesting to research that more sometime. If you know anything about them, please post a comment!

This shadertoy is what got me started looking into dual numbers. It’s a mandelbrot viewer done by iq using dual numbers to estimate a distance from a point to the mandelbrot set (as far as I understand it anyhow, ha!). He uses that estimated distance to color the pixels.

I didn’t get very much into the reasons of why this works (has to do with taylor series terms disappearing if ε^2 is 0), or the rigorous math behind deriving the operators, but here are some great links I found researching this stuff and putting this blog post together.

# Four Ways to Calculate Sine Without Trig

Is it possible to sin without trig? That is a question that has plagued theologians for centuries. As evil as trigonometry is, modern science shows us that yes, it is possible to sin without trig. Here are some ways that I’ve come across.

# 1 – Slope Iteration Method

The above image uses 1024 samples from 0 to 2pi to aproximate sine iteratively using it’s slope. Red is true sin, green is the aproximation, and yellow is where they overlap.

This method comes from Game Programming Gems 8, which you can snag a copy of from amazon below if you are interested. It’s mentioned in chapter 6.1 A Practical DSP Radio Effect (which is a really cool read btw!).
Amazon: Game Programming Gems 8

This method uses calculus but is actually pretty straightforward and intuitive – and very surprising to me that it works so well!

The derivative of sin(x) is cos(x). That means, for any x you plug into sin, the slope of the function at that point is the cosine value of that same x.

In other words, sin(0) is 0, but it has a slope of cos(0) which is 1. Since slope is rise over run (change in y / change in x) that means that at sin(0), if you take an infinitely small step forward on x, you need to take the same sized step on y. That will get you to the next value of sine.

Let’s test that out!
sin(0) = 0 so we start at (0,0) on the graph.

If we try a step size of 0.1, our approximation is:
sin(0.1) = 0.1

The actual value according to google is 0.09983341664. so our error was about 0.0002. That is actually pretty close!

sin(0.25) = 0.25

The real value is 0.24740395925, so our error is about 0.003. We have about 10 times the error that we did at 0.1.

what if we try it with 0.5?
sin(0.5) = 0.5

The actual value is 0.4794255386, which leaves us with an error of about 0.02. Our error is 100 times as much as it was at 0.1. As you can see, the error increases as our step size gets larger.

If we wanted to, we could get the slope (cosine value) at the new x value and take another step. we could continue doing this to get our sine approximation, knowing that the smaller the step that we use, the more accurate our sine approximation would be.

We haven’t actually won anything at this point though, because we are just using cosine to take approximated steps through sine values. We are paying the cost of calculating cosine, so we might as well just calculate sine directly instead.

Well luckily, cosine has a similar relationship with sine; the derivative of cos(x) is -sin(x).

Now, we can use cosine values to step through sine values, and use those same sine values to step through cosine values.

Since we know that cos(0) = 1.0 and sin(0) = 0.0, we can start at an angle of 0 with those values and we can iteratively step through the values.

Here is a sample function in C++

// theta is the angle we want the sine value of.
// higher resolution means smaller step size AKA more accuracy but higher computational cost.
// I used a resolution value of 1024 in the image at the top of this section.
float SineApproximation (float theta, float resolution)
{
// calculate the stepDelta for our angle.
// resolution is the number of samples we calculate from 0 to 2pi radians
const float TwoPi = 6.28318530718f;
const float stepDelta = (TwoPi / resolution);

// initialize our starting values
float angle = 0.0;
float vcos = 1.0;
float vsin = 0.0;

// while we are less than our desired angle
while(angle < theta) {

// calculate our step size on the y axis for our step size on the x axis.
float vcosscaled = vcos * stepDelta;
float vsinscaled = vsin * stepDelta;

// take a step on the x axis
angle += stepDelta;

// take a step on the y axis
vsin += vcosscaled;
vcos -= vsinscaled;
}

// return the value we calculated
return vsin;
}

Note that the higher the angle you use, the more the error rate accumulates. One way to help this would be to make sure that theta was between 0 and 2pi, or you could even just calculate between 0 and pi/2 and mirror / reflect the values for the other quadrants.

This function is quite a bit of work to calculate a single value of sine but it’s real power comes in the fact that it’s iterative. If you ever find yourself in a situation where you need progressive values of sine, and have some fixed angle step size through the sine values, this algorithm just needs to do a couple multiplies and adds to get to the next value of sine.

One great use of this could be in DSP / audio synthesis, for sine wave generation. Another good use could be in efficiently evaluating trigonometry based splines (a future topic I plan to make a post about!).

You can see this in action in this shadertoy or look below at the screenshots:

64 Samples – Red is true sine, green is our approximation, and yellow is where they are the same

128 Samples

256 Samples

1024 Samples

# 2 – Taylor Series Method

Another way to calculate sine is by using an infinite Taylor series. Thanks to my friend Yuval for showing me this method.

You can get the Taylor series for sine by typing “series sin(x)” into wolfram alpha. You can see that here: Wolfram Alpha: series sin(x).

Wolfram alpha says the series is: x-x^3/6+x^5/120-x^7/5040+x^9/362880-x^11/39916800 ….

what this means is that if you plug a value in for x, you will get an approximation of sine for that x value. It’s an infinite series, but you can do as few or as many terms as you want to be able to trade off speed for accuracy.

For instance check out these graphs.

Google: graph y = x-x^3/6+x^5/120-x^7/5040+x^9/362880-x^11/39916800, y = sin(x)
(Note that I had to zoom out a bit to show where it became inaccurate)

When looking at these graphs, you’ll probably notice that very early on, the approximation is pretty good between -Pi/2 and + Pi/2. I leveraged that by only using those values (with modulus) and mirroring them to be able to get a sine value of any angle with more accuracy.

When using just x-x^3/6, there was an obvious problem at the top and bottom of the sine waves:

When i boosted the equation with another term, bringing it to x-x^3/6+x^5/120, my sine approximation was much better:

You can see this method in action in this shadertoy:

# 3 – Smoothstep Method

The third method may be my favorite, due to it’s sheer elegance and simplicity. Thanks to P_Malin on shadertoy.com for sharing this one with me.

There’s a function in graphics called “smoothstep” that is used to take the hard linear edge off of things, and give it a smoother, more organic feel. You can read more about it here: Wikipedia: Smoothstep.

BTW if you haven’t read the last post, I talk about how smooth step is really just a 1d bezier curve with specific control points (One Dimensional Bezier Curves). Also, smoothstep is just this function: y = (3x^2 – 2x^3).

Anyhow, if you have a triangle wave that has values from 0 to 1 on the y axis, and put it through a smoothstep, then scale it to -1 to +1 on the y axis, you get a pretty darn good sine approximation out.

Here is a step by step recipe:

Step 1 – Make a triangle wave that has values on the y axis from 0 to 1

Step 2 – Put that triangle wave through smoothstep (3x^2 – 2x^3)

Step 3 – Scale the result to having values from -1 to +1 on the axis.

That is pretty good isn’t it?

You can see this in action in this shadertoy (thanks to shadertoy’s Dave_Hoskins for some help with improving the code):

After I made that shadertoy, IQ, the creator of shadertoy who is an amazing programmer and an impressive “demoscene” guy, said that he experimented with removing the error from that technique to try to get a better sin/cos aproximation.

You can see that here: Shadertoy: Sincos approximation

Also, I recommend checking out IQ’s website. He has a lot of interesting topics on there: IQ’s Website

# 4 – CORDIC Mathematics

This fourth way is a method that cheaper calculators use to calculate trig functions, and other things as well.

I haven’t taken a whole lot of time to understand the details, but it looks like it’s using imaginary numbers to rotate vectors iteratively, doing a binary search across the search space to find the desired values.

The benefit of this technique is that it can be implemented with VERY little hardware support.

The number of logic gates for the implementation of a CORDIC is roughly comparable to the number required for a multiplier as both require combinations of shifts and additions.
Wikipedia: Coordinate Rotation Digital Computer

# One Dimensional Bezier Curves

I was recently looking at the formula for bezier curves:

A * (1-T)^2 + B * 2 * (1-T) * T + C * T ^2

Cubic Bezier curve:
A*(1-T)^3+3*B*(1-T)^2*T+3*C*(1-T)*T^2+D*T^3

And I wondered… since you can have a Bezier curve in any dimension, what would happen if you made the control points (A,B,C,D) scalar? In other words, what would happen if you made bezier curves 1 dimensional?

Well it turns out something pretty interesting happens. If you replace T with x, you end up with f(x) functions, like the below:

y = A * (1-x)^2 + B * 2 * (1-x) * x + C * x ^2

Cubic Bezier curve:
y = A*(1-x)^3+3*B*(1-x)^2*x+3*C*(1-x)*x^2+D*x^3

What makes that so interesting is that most math operations you may want to do on a bezier curve are a lot easier using y=f(x), instead of the parameterized formula Point = F(S,T).

For instance, try to calculate where a line segment intersects with a parameterized bezier curve, and then try it again with a quadratic equation. Or, try calculating the indefinite integral of the parameterized curve so that you can find the area underneath it (like, for Analytic Fog Density). Or, try to even find the given Y location that the curve has for a specific X value. These things are pretty difficult with a parameterized function, but a lot easier with the y=f(x) style function.

This ease of use does come at a price though. Specifically, you can’t move control points freely, you can only move them up and down and cannot move them left and right. If you are ok with that trade off, these 1 dimensional curves can be pretty powerful.

Below is an image of a 1 dimensional cubic bezier curve that has control points A = 0.5 B = 0.25 C = 0.75 D = 0.5. The function of this curve is y = 0.5 * (1-x)^3 + 0.25 * 3x(1-x)^2 + 0.75 * 3x^2(1-x) + 0.5 * x^3.

You can ask google to graph that for you to see that it is in fact the same: Google: graph y = 0.5 * (1-x)^3 + 0.25 * 3x(1-x)^2 + 0.75 * 3x^2(1-x) + 0.5 * x^3

Another benefit to these one dimensional bezier curves is that you could kind of use them as a “curve fitting” method. If you have some data that you wanted to approximate with a quadratic function, you could adjust the control points of a one dimensional quadratic Bezier curve to match your data set. If you need more control points to get a better aproximation of the data, you can just increase the degree of the bezier curve (check this out for more info on how to do that: Bezier Curves Part 2 (and Bezier Surfaces)).

# Smoothstep as a Cubic 1d Bezier Curve

It’s kind of out of the scope of this post to talk about why smoothstep is awesome, but to give you strong evidence that it is, it’s a GLSL built in function. You may have also seen it used in the post I made about distance fields (Distance Field Textures), because one of it’s common uses is to make the edges of things look smoother. Here’s a wikipedia page on it as well if you want more info: Wikipedia: Smoothstep

Anyhow, I had no idea, but apparently the smoothstep equation is the same as if you take a 1d cubic bezier curve and make the first two control points 0.0, and the second two control points 1.0.

The equation for smoothstep is: y = 3*x^2 – 2*x^3

The equation for the bezier curve i mentioned is: y = 0*(1-x)^3+3*0*(1-x)^2*x+3*1*(1-x)*x^2+1*x^3

otherwise known as: y = 3*1*(1-x)*x^2+1*x^3

If you work it out, those are the same equations! Wolfram alpha can verify that for us even: Wolfram Alpha: does 3*x^2 – 2*x^3 = 3*1*(1-x)*x^2+1*x^3.

Kinda neat 😛

# Moving Control Points on the X Axis

There’s a way you could fake moving control points on the X axis (left and right) if you really needed to. What I’m thinking is basically that you could scale X before you plug it into the equation.

For instance, if you moved the last control point to the left so that it was at 0.9 instead of 1.0, the space between the 3rd and 4th control point is now .23 instead of .33 on the x axis. You could simulate that by having some code like this:

if (x > 0.66)
x = 0.66 + (x - 0.66) / 0.33 * 0.23

Basically, we are squishing the X values that are between 0.66 and 1.0 into 0.66 to 0.9. This is the x value we want to use, but we’d still plug the raw, unscaled x value into the function to get the y value for that x.

As another example, let’s say you moved the 3rd control point left from 0.66 to 0.5. In this situation, you would squish the X values that were between 0.33 and 0.66 into 0.33 to 0.5. HOWEVER, you would also need to EXPAND the values that were between 0.66 and 1.0 to be from 0.5 and 1.0. Since you only moved the 3rd control point left, you’d have to squish the section to the left of the control point, and expand the section to the right to make up the difference to keep the 4th control point in the same place. The code for converting X values is a little more complex, but not too bad.

What happens if you move the first control point left or right? Well, basically you have to expand or squish the first section, but you also need to add or subtract an offset for the x as well.

I’ll leave the last 2 conversions as an exercise for whoever might want to give this a shot 😛

Another complication that comes up with the above is, what if you tried to move the 3rd control point to the left, past the 2nd control point? Here are a couple ways I can think of off hand to deal with it, but there are probably other ways too, and the right way to deal with it depends on your needs.

1. Don’t let them do it – If a control point tries to move past another control point, just prevent it from happening
2. Switch the control points – If you move control point 3 to the left past control point 2, secretly have them start controling control point 2 as they drag it left. As far as the user is concerned, it’s doing what they want, but as far as we are concerned, the control points never actually crossed
3. Move both – if you move control point 3 to the left past control point 2, take control point 2 along for the ride, keeping it to the left of control point 3

When allowing this fake x axis movement, it does complicate the math a bit, which might bite you depending on what you are doing with the curve. Intersecting a line segment with it for example is going to be more complex.

An alternative to this would be letting the control points move on the X axis by letting a user control a curve that controls the X axis location of the control points – hopefully this would happen behind the scenes and they would just move points in X & Y, not directly editing the curve that controls X position of control points. This is a step towards making the math simpler again, by modifying the bezier curve function, instead of requiring if statements and loops, but as far as all the possibly functions I can think of, moving one control point on the X axis is probably going to move other control points around. Also, it will probably change the shape of the graph. It might change in a reasonable way, or it might be totally unreasonable and not be a viable alternative. I’m not really sure, but maybe someday I’ll play around with it – or if you do, please post a comment back and let us know how it went for you!

Here are some links to experiment with these guys and see them in action:
HTML5 Interactive 1D Quadratic Bezier Demo
HTML5 Interactive 1D Cubic Bezier Demo
Shadertoy: Interactive 1D Cubic Bezier Demo

# Counting in Binary is a Fractal

I can’t remember for sure but I think I read about this first in A New Kind Of Science by Stephen Wolfram (yes, the same guy who made Wolfram Alpha!). Counting in binary is actually a fractal, check out these 4 sequential images…

# Real Time Demo

To see it in action, check out this WebGL shadertoy:

# Wang Tiling

Wang tiling is a really cool concept… it’s a good way to use 2d tiled graphics in such a way that can look very organic, without discernable patterns.

The basic idea of how they work is that each tile has a type of edge. You start by placing a random tile, and then you start putting down it’s neighboring tiles. When you place a tile, the rule is you can only put down a tile that has compatible edge types (aka the tiles can go together seamlessly). Rinse and repeat and pretty soon you have tile based graphics that don’t look tiled at all.

Specifically here is a strategy I like to use for filling a grid with wang tiles:

1. Place any random tile in the upper left corner
2. Put a tile below it that has an edge on it’s top that is compatible with the already placed tile’s bottom edge
3. Continue placing tiles downward until you reach the bottom of the column
4. Now, move back to the top, move over to the next column, and now place a tile such that the left edge is compatible with the right edge of the tile it is next to.
5. Moving down, you now have to find a tile which is compatible with both the tile above it, and the tile to the left. Since there are going to be multiple tiles that fit these constraints, just choose randomly from the ones that do.
6. Rinse and repeat until the grid is filled

There is a lot more info out there (links at bottom of post) so I’ll leave it at that and show you some results I got with some simple tiles.

The tiles I used are very geometric, but if you have more organic looking tiles, the resulting tile grid will look a lot more organic as well.

Also, as the links at the bottom will tell you, if you have wang tiles where each axis has only 2 edge types, even though the number of permutations of tiles in that situation is 16 (XVariation^2 * YVariation^2), you can actually get away with just using 8 tiles (XVariation * YVariation * 2). In my example below I had to use all 16 though because I’m just generating edge types in a pixel shader without deeply analyzing neighboring tiles, and it would be a lot more complex to limit my generation to just the 8 tiles. If you can think of a nice way to generate a wang tile grid using only the 8 tiles though, please let me know!

The 16 wang tiles used:

A resulting grid:

Here’s a more complex set of 16 wang tiles:

And a resulting grid:

Wang Tiling Research Paper

Introduction to Wang Tiles

By the way… something really crazy about wang tiles is that apparently they can be used to do computation and they are turing complete. Seriously? Yes! Check out the link below:

Computing with Tiles

# Temporal supersampling, flipquads and real time raytracing

Follow me on this train of thought 😛

1) There’s this thing called super sampling where you render an image at a larger resolution, so that you can properly downsample it to the right size (the size of your screen for instance) to avoid aliasing problems. The problem here is that you are rendering more pixels, so it is more expensive to render, which is usually a deal breaker for real time applications that are trying to push the envelope of performance – like modern games.

2) There’s a way to get around this with something called Temporal Supersampling where you use the last frame rendered to provide extra information for the current frame, so that in a way, you get supersampled data by spreading it out over 2 frames. (More info on supersampling: Temporal supersampling). You get better results by jittering (offseting) the pixels you render from frame to frame, by a sub-pixel amount. This is the usual monte carlo sampling kind of situation… find some cheap but well behaving pseudorandom number generator you can run in your pixel shader to offset each pixel by, or use a regular pattern of some sort that gives good enough results.

3) That gives you 2 samples if you only compost the last and current frame, but more samples is better of course. You could keep more frames from the past around, but that takes up the precious resource of memory. Apparently, when the hardware does MSAA (multisampling antialiasing), it has different configurations for different numbers of samples and it’s configurable somehow. If you have 2 samples, they may be 2 vertical dots, or 2 horizontal dots. If you have 3 samples, it might look like a “3” on a domino. If you have 5 samples it might look like a “5” on a domino.

4) Sometimes a corner will be sampled so that a sample can be shared across multiple pixels to increase efficiency. There is this really interesting thing called “flipquads” that samples on an edge for that same reason. You can see some info on here: An Extremely Inexpensive Multisampling Scheme. Basically, you only do two samples per pixel, sampling at 2 of the 4 sample locations on the edge of a pixel, so that the pixels that share the edge can use the results. Effectively, you are doing 2 sample per pixel, but getting 4 samples per pixel due to sample sharing.

5) If you combine flipquads with temporal supersampling, it means that you get 4 samples for the cost of 2, amortized over 2 frames. So, you essentially just render the normal amount of pixels (1 sample per pixel), compost frame N against frame N-1, and get the benefit of a 4 tap MSAA. So, it’s really cheap, and yes… it does actually help significantly, despite the fact that so many samples are redundant.

None of the above is anything new… I watched it all in various SIGGRAPH 2014 presentations earlier today from big name modern games – and man am i amazed what people are doing these days!

Now for the new part…

One way for raytracers to get better visual quality is to do multiple rays per sample, doing monte carlo sampling, where each of the rays in the group is perturbed by tiny amounts. Some details here: Advanced Topics in Computer Graphics: Sampling Techniques

In my own personal OpenCL real time raytracer, I don’t have the luxury of doing multiple rays per pixel – and in fact, I have a graphics option that allows you to render only half the screen (top / bottom) each frame alternating, to cut the number of rays down so that it runs faster!

What if a person was able to do temporal supersampling with a realtime raytracer, using flipquads to make it so it could get the information of 4 rays per pixel, while only taking a single ray cast per pixel each frame? Wouldn’t that be something?

There are some technical details to work out but I think there is some real magic here waiting to happen.

The biggest technical problem I foresee is reprojecting the pixels from the last frame to the current frame. This probably would work ok if your rays had a strict projection matrix governing them, but there may be difficulties with reflection and refraction, and honestly, I personally want to distort camera rays for game effects (like being underwater) so wouldn’t want to be stuck with a strict projection matrix. Maybe there’s some clever solution to make it all ok though…

Also – the link to flipquads is actually an explanation of “fliptris” a technique using 1.25 samples per pixel. If that were amortized across 2 frames, that means you would only need to cast 62% of your rays theoretically. That might be a nice performance win, while gaining the benefits of temporal supersampling and ultimately having 3 samples for each pixel!

# Distance Field Textures

A friend recently turned me onto a really cool paper (thanks James!) that Valve wrote that allows you to encode monochromatic (black & white) textures in a way that they can be incredibly low resolution, but when you scale them up, they still look crisp and smooth, not blurry or pixelated.

It is really quite amazing and is perfect for things like fonts or decals.

I recommend reading the paper, but below are some details to help you implement this in your own application, and also some examples of things taken to the extreme.

Here’s a really easy to use program that can turn fonts or SVG files into distance field images: signed-distance-field-font-generator

# Implementation

Ok so, in a signed distance field texture, the alpha value of each pixel is a value of how far that pixel is from the edge of the shape. In a signed distance field, you essentially take the value which is from 0 to 1, and you subtract 0.5 and multiply by 2 so that you change it from 0-1 to -1 to +1. Negative distances mean the pixel is inside the shape, Positive distances mean the pixel is outside the shape.

You only need to do that math if you care about the exact distance though. If you only care about whether the pixel is inside or outside the shape, you can just consider values less than 0.5 to be inside the shape, and values greater than 0.5 to be outside the shape. In other words, you could just do an ALPHA TEST against 0.5 to render these guys.

Here’s an excerpt of some OpenCL code that does this:

float alpha = read_imagef(tex3dIn, g_textureSampler, textureCoords).w;
float3 color = (alpha < 0.5f) ? (float3)(1.0f) : (float3)(0.0f);

I'll refer to that code as the "Alpha Test" code.

Another way to do it would be to use smoothstep to smooth the jaggies out a bit. Here's an excerpt of some OpenCL code that does that:

const float smoothing = 1.0/64.0;
float distance = read_imagef(tex3dIn, g_textureSampler, textureCoords).w;
float alpha = Saturate(smoothstep(0.5 – smoothing, 0.5 + smoothing, distance));
float3 color = (float3)(1.0f – alpha);

In the above, the smoothing constant can be adjusted to change how it smooths out the jaggies.

Note that even though the texture is monochromatic, you could use the color channel in the texture if you wanted to, or multiply the color by some other color to make it a colored image.

Here are the two source images I used. The first one is of the "Comic Sans" font which I doubled vertically since my textures have to be square, and the second one is a mustache SVG vector graphics image I found online. The font image is 512×512 and the mustache is 128×128.

# Distance Field Textures in Action

Here’s a shot of the texture usages rendered from a distance:

# Font in Action

Here’s a shot of the text close up with the alpha test code:

Here’s the same shot, using the smooth step code. Keep in mind that the “8” you are looking at is about 32×32 pixels 😛

Here’s the text taken from 512×512 down to 256×256, rendered with the alpha test code. You can already see degradation unfortunately but the look at the pictures above and remember that the full font texture is essentially 512×256 (I doubled it because my textures have to be square) and looks great up close:

Here’s the 256×256 font texture again, this time rendered with smooth step. A little bit better, but still pretty bad (but not bad for the resolution of the source font texture!):

# Decal in Action

Here’s the mustache decal, which has a source image size of 128×128, rendered with the alpha test code:

Here’s the mustache rendered with the smooth step code:

Now it starts to get interesting. Here it is at 64×64 with alpha test code:

And now 64×64 with smooth step:

Here’s 32×32 with alpha test:

Here’s 32×32 with smooth step:

Here’s 16×16 with alpha test:

And lastly, here’s 16×16 with smooth step. Not freaking bad for a 16×16 texture right??!!!

Apparently another great use for these is to encode a shadow map as a distance field texture. This does a great job of keeping your shadow line smooth, effectively letting you use a much lower resolution texture to store the shadow maps.

This is a no brainer for static shadows, but dynamic shadows this may not be as useful, as it seems like you’d need to generate the full sized texture to make the distance field texture, so would require some extra memory and processing when generated at runtime. There may be some clever tricks to avoiding that though, not sure.

# Analytic Fog Density

There are a number of ways to implement the effect of fog with modern real time rendered graphics. This blog post will explain how to render fog that has varying density, based on a function of X,Y,Z location in space, like in the picture above.

# Faked Fog

One way is to “fake it” and do something like set the color of a pixel on an object to be based on it’s height. For instance you might say that pixels with a y axis value above 15 are unfogged, pixels with y axis values between 15 and 10 progressively get more fogged as they get closer to 10, and pixels with y axis values less than 10 are completely fogged. That can make some fog that looks like this:

A strange side effect of doing that though, is if you go down “into” the fog, and look out of the fog, things that should be fogged won’t. For instance, looking up at a mountain from inside the fog, the mountain won’t be fogged at all, even though it should be because you are inside of the fog.

A better way to do it, if you intend for the camera to be able to go into the fog, is to calculate a fogging amount for a pixel based on how far away it is from the view point, and how dense the fog is between the view point and the destination point.

If you are doing ray based rendering, like ray tracing or ray marching, you might find yourself trying to find how much fog is between points that don’t involve the view point – like if you are calculating the reflection portion of a ray. In this case, you are just finding out how much fog there is between the point where the reflection happened and the closest intersection. You can consider the point of reflection as the “view point” for the purpose of fogging.

Sometimes, the entire scene might not be in fog. In this case, you have to find where the fog begins and ends, instead of the total distance between the view point and the destination point.

In any case, the first thing you need to do when fogging is figure out the point where the fog begins, and the point where the fog ends. Then, you can figure out how much fog there is based on how the fog density works.

# Constant Density Fog

The simplest sort of fog is fog that has the same density all throughout it.

What you do in this case is just multiply the fog density by the distance spent in the fog to come up with a final fog value.

As an example, your fog density might be “0.04” and if you are fogging a pixel 10 units away, you multiply density by distance. FogAmount = 0.04 * 10.0 = 0.4.

Doing this, you know the pixel should be 40% fogged, so you interpolate the pixel’s color 40% towards the fog color. You should make sure to clamp the fog amount to be between 0 and 1 to avoid strange visual anomolies.

The image below shows a constant fog density of 0.04.

Here’s an image of the same constant density fog as viewed from inside the fog:

A problem with constant fog density though, is that if you view it from edge on, you’ll get a very noticeable hard edge where the fog begins, like you can see in the image below:

# Linear Density Fog

With linear fog density, the fog gets denser linearly, the farther you go into the fog.

With a fog plane, you can get the density of the fog for a specified point by doing a standard “distance from plane to point” calculation and multiplying that by how much the fog density grows per unit of distance. If your plane is defined by A*x+B*y+C*y+D = 0, and your point is defined as X,Y,Z, you just do a dot product between the plane and the point, giving the point a W component of one.

In other words…

FogDensity(Point, Plane) = (Plane.NormX * Point.X + Plane.NormY * Point.Y + Plane.NormZ * Point.Z + Plane.D * 1.0) * FogGrowthFactor

Here’s a picture of linear fog with a fog growth factor of 0.01:

The same fog viewed from the inside:

And lastly, the fog viewed edge on to show that the “hard line” problem of linear fog is gone (dramatic difference isn’t it?!):

# Analytic Fog Density – Integrals

Taking a couple steps further, you might want to use equations to define fog density with some function FogDensity = f(x,y,z,).

How could you possibly figure out how much fog there is between two given points when the density between them varies based on some random function?

One way would be to take multiple samples along the line segment between the view point and the destination point, and either calculate the fog amount in each section, or maybe average the densities you calculate and multiply the result by the total distance. You might have to take a lot of samples to make this look correct, causing low frame rate, or accepting low visual quality as a compromise.

If you look at the graphs for the previous fog types, you might notice that we are trying to find the area under the graphs between points A and B. For constant density fog, the shape is a rectangle, so we just multiply width (time in fog) by height (the constant fog density) to get the fog amount. For linear density fog, the shape is a trapezoid, so we use the trapezoid area formula which is height (in this case, the distance in the fog) times the sum of the base lengths (the fog densities at points A and B) divided by 2.

How can we get the area under the graph between A and B for an arbitrary formula though? Well, a way exists luckily, using integrals (thanks to my buddy “Danny The Physicist” for educating me on the basics of integrals!).

There’s a way to transform a formula to get an “indefinite integral”, which itself is also a formula. I won’t go into the details of how to do that, but you can easily get the indefinite integral of a function by typing it into Wolfram Alpha.

Once you have the indefinite integral (let’s call it G(x)) of the fog density formula (let’s call it F(x)), if you calculate G(B) – G(A), that will give you the area under the graph in F(X) between A and B. Yes, seriously, that gives us the area under the graph between our points, thus giving us the amount of fog that exists between the two points for an arbitrary fog density function!

Note that when you plug a value into the indefinite integral and get a number out, that number is called the definite integral.

# Analytic Fog Density – Implementation Details

Now that the theory is worked out let’s talk about implementation details.

First off, coming from an additive audio synthesis type of angle, I figured I might have some good luck adding together sine waves of various frequencies and amplitudes, so I started with this:

sin(x*F) * A

F is a frequency multiplier that controls how long the sine wave is. A is an amplitude multiplier that controls how dense the fog gets max.

Next, I knew that I needed a fog density function that never goes below zero, because that would mean if you looked through a patch of negative fog density, it would make the other fog you were looking through be less dense. That is just weird, and doesn’t exist in reality (but maybe there is some interesting visual effect hiding in there somewhere??), so the formula evolved to this, making sure the function never went below zero:

(1 + sin(x*F)) * A

Plugging that equation into wolfram alpha, it says the indefinite integral is:

(x – (cos(x*F)) / F) * A

You can check that out here:
Wolfram Alpha: (1 + sin(x*F)) * A.

It’s also kind of fun to ask google to graph these functions so you can see what they do to help understand how they work. Here are the graphs for A = 0.01 and F = 0.6:
Fog Density: graph (1 + sin(x*0.6)) * 0.01
Indefinite Integral: graph (x – (cos(x*0.6)) / 0.6) * 0.01

So, if you have point A and B where the fogging begins and ends, you might think you can do this to get the right answer:
FogAmount = G(B.x) – G(A.x)

Nope! There’s a catch. That would work if A and B had no difference on the y or z axis, but since they probably do, you need to jump through some hoops. In essence, you need to stretch your answer across the entire length of the line segment between A and B.

To do that, firstly you need to get that fog amount down to unit length. You do that by modifying the formula like so:
FogAmount = (G(B.x) – G(A.x)) / (B.x – A.x)

This also has a secondary benefit of making it so that your fog amount is always positive (so long as your fog density formula F(X) can’t ever go negative!), which saves an abs() call. Making it always positive ensures that this works when viewing fog both from the left and the right.

Now that we have the fog amount down to unit length, we need to scale it to be the length of the line segment, which makes the formula into this:
FogAmount = (G(B.x) – G(A.x)) * Length(B-A)/(B.x – A.x)

That formula will now give you the correct fog amount.

But, one axis of fog wasn’t enough to look very good, so I wanted to make sure and do one sine wave on each axis. I used 0.01 amplitude for each axis, but for the X axis i used a frequency of 0.6, for the Y axis i used a frequency of 1.2 and for the Z axis i used a frequency of 0.9.

Also, I wanted to give a little bit of baseline fog, so I added some constant density fog in as well, with a constant density of 0.1.

As a bonus, I also gave each axis a “movement factor” that made the sine waves move over time. X axis had a factor of 2.0, Y axis had a factor of 1.4 and Z axis had a factor of 2.2.

Putting all of this together, here is the final fog equation (GLSL pixel shader code) for finding the fog amount between any two points at a specific point in time:

//=======================================================================================
float DefiniteIntegral (in float x, in float amplitude, in float frequency, in float motionFactor)
{
// Fog density on an axis:
// (1 + sin(x*F)) * A
//
// indefinite integral:
// (x - cos(F * x)/F) * A
//
// ... plus a constant (but when subtracting, the constant disappears)
//
x += iGlobalTime * motionFactor;
return (x - cos(frequency * x)/ frequency) * amplitude;
}

//=======================================================================================
float AreaUnderCurveUnitLength (in float a, in float b, in float amplitude, in float frequency, in float motionFactor)
{
// we calculate the definite integral at a and b and get the area under the curve
// but we are only doing it on one axis, so the "width" of our area bounding shape is
// not correct.  So, we divide it by the length from a to b so that the area is as
// if the length is 1 (normalized... also this has the effect of making sure it's positive
// so it works from left OR right viewing).  The caller can then multiply the shape
// by the actual length of the ray in the fog to "stretch" it across the ray like it
// really is.
return (DefiniteIntegral(a, amplitude, frequency, motionFactor) - DefiniteIntegral(b, amplitude, frequency, motionFactor)) / (a - b);
}

//=======================================================================================
float FogAmount (in vec3 src, in vec3 dest)
{
float len = length(dest - src);

// calculate base fog amount (constant density over distance)
float amount = len * 0.1;

// calculate definite integrals across axes to get moving fog adjustments
adjust += AreaUnderCurveUnitLength(dest.x, src.x, 0.01, 0.6, 2.0);
adjust += AreaUnderCurveUnitLength(dest.y, src.y, 0.01, 1.2, 1.4);
adjust += AreaUnderCurveUnitLength(dest.z, src.z, 0.01, 0.9, 2.2);

// make sure and not go over 1 for fog amount!
}

I ended up only using one sine wave per axis, but I think with more sine waves, or perhaps different functions entirely, you could get some more convincing looking fog.

At some point in the future, I’d like to play around with exponential fog density (instead of linear) where the exponential power is a parameter.

I also think that maybe squaring the sine waves could make them have sharper density changes perhaps…

One thing that bugs me in the above screenshots is the obvious “hard line” in both constant and linear fog where it seems fog crosses a threshold and gets a lot denser. I’m not really sure how to fix that yet. In traditional rasterized graphics you could put the fog amount on a curve, to give it a smoother transition, but in ray based rendering, that could make things a bit odd – like you could end up with an exponential curve butting up against the start of a different exponential curve (due to reflection or refraction or similar). The fog density would end up looking like log graph paper which would probably not look so great – although honestly I haven’t tried it to see yet!

Here’s a good read on fog defined by a plane, that also gets into how to make branchless calculations for the fog amounts.
Unified Distance Formulas for Halfspace Fog

Interactive ShaderToy.com demo with GLSL pixel shader source code that you can also edit in real time with WebGL:

# Bezier Curves Part 2 (and Bezier Surfaces)

This is a follow up post to Bezier Curves. My plan was to write a post about b-splines and nurbs next, but after looking into them deeper, I found out they aren’t going to work for my needs so I’m scratching that.

Here’s some basic info on b-splines and nurbs though before diving deeper into Bezier curves and surfaces.

## B-Splines (Basis Splines)

Bezier curves are nice, but the more control points you add, the more complex the math gets because the degree of the curve function increases with each control point added. You can put multiple Bezier curves end to end to be able to have more intricate curves, but another option is to use B-Splines.

B-Splines are basically Bezier curves which let you specify more control points without raising the degree of the Bezier curve. They do this by having control points only affect part of the total curve.

This way, you could make a quadratic b-spline which had 10 control points. Only a few control points control any given point on the curve, so the curve stays quadratic (and so does the math), but you get a lot more control points. A “Knot Vector” is what controls which parts of the curve the control points control.

A Bezier curve is actually a special case of B-Spline where all control points affect the entire curve.

## Nurbs (Non Uniform Rational B-Spline)

Sometimes when working with curves, you want some control points to be stronger that others. You can accomplish this in Bezier curves and B-splines by doubling up or trippling up control points in the same location to make that control point twice, or three times as strong respectively.

What if you want a control point to be 1.3 times stronger though? That gets a lot more complicated.

Nurbs solve that problem by letting you specify a weight per control point.

Just like Bezier curves are a special case of B-Splines, B-Splines are a special case of nurbs. A B-Spline could be thought of as nurbs that has the same weighting for all control points.

## Back to Bezier!

My end goal is to find a curve / surface type that is flexible enough to be used to make a variety of shapes by artists, but is efficient at doing line segment tests against on the GPU. To this end, B-Splines and Nurbs add algorithmic and mathematical complexity over Bezier curves, and seem to be out of the running unless I can’t find anything more promising.

My best bet right now looks like a Bezier Triangle. Specifically, a quadratic Bezier triangle, where each side of the triangle is a quadratic Bezier curve that has 3 control points. When I get those details fully worked out, I’ll report back, but for now, here’s some interesting info I found about generalizing bezier curves both in order (linear, quadratic, cubic, quartic, etc) as well as in the number of dimensions (line, curve, triangle, tetrahedron, etc).

## Bezier Generalized

I found the generalized equation on the wikipedia page for Bezier triangles and am super glad i found it, it is very cool!

I want to show you some specifics to explain the generalization by example.

(A * S + B * T) ^ 2

Expanding that gives you:
A^2 * S^2 + A * B * 2 * S * T + B^2 * T^2

In the above, S and T are Barycentric Coordinates in a 1 dimensional Simplex. Since we know that barycentric coordinates always add up to 1, we can replace S with (1-T) to get the below:

A^2 * (1-T)^2 + A * B * 2 * (1-T) * T + B^2 * T^2

Now, ignoring T and the constants, and only looking at A and B, we have 3 forms: A^2, AB and B^2. Those are our 3 control points! Let’s replace them with A,B and C to get the below:

A * (1-T)^2 + B * 2 * (1-T) * T + C * T ^2

And there we go, there’s the quadratic Bezier curve formula seen in the previous post.

Cubic Curve:
(A * S + B * T) ^ 3

To make a cubic curve, you just change the power from 2 to 3, that’s all! If you expand that equation, you get:
A^3*S^3+3*A^2*B*S^2*T+3*A*B^2*S*T^2+B^3*T^3

We can swap S with (1-T) to get:

A^3*(1-T)^3+3*A^2*B*(1-T)^2*T+3*A*B^2*(1-T)*T^2+B^3*T^3

Looking at A/B terms we see that there is more this time: A^3, A^2B, AB^2 and B^3. Those are our 4 control points that we can replace with A,B,C,D to get:
A*(1-T)^3+3*B*(1-T)^2*T+3*C*(1-T)*T^2+D*T^3

There is the cubic Bezier curve equation from the previous chapter.

Linear Curve:
(A * S + B * T) ^ 1

To expand that, we just throw away the exponent. After we replace S with (1-T) we get:
A * (1-T) + B * T

That is the formula for linear interpolation between 2 points – which you could think of as the 2 control points of the curve.

One more example before we can generalize.

(A * S + B * T + C * U) ^ 2

If you expand that you get this:
A^2*S^2+2*A*B*S*T+2*A*C*S*U+B^2*T^2+2*B*C*T*U+C^2*U^2

Looking at combinations of A,B & C you have: A^2, AB, AC, B^2, BC, C^2. Once again, these are your control points, and their names tell you where they lie on the triangle. A Bezier triangle is a triangle where the 3 sides of the triangle are bezier curves. A quadratic bezier triangle has quadratic bezier curves for it’s edges which mean that each side has 3 control points. Those 3 control points are made up of the 3 corners of the triangle, and then 3 more control points, each one being between end points. A^2, B^2 and C^2 represent the 3 corners of the triangle. AB is the third control point for the bezier curve on the edge AB. BC and AC follow that pattern as well! Super easy to remember.

In a cubic Bezier triangle, you get a lot more control points, but a new class of control point too: ABC. This control point is in the middle of the triangle like the name would imply.

Anyways, in the expanded quadratic bezier triangle equation above, when you replace the control points with A,B,C for the triangle corner control points (the squares) and D,E,F for the inbetween control points, you get the bezier triangle equation below:

A*S^2+2*D*S*T+2*E*S*U+B*T^2+2*F*T*U+C*U^2

Note that we are dealing with a simplex in 3d now, so once again, instead of needing ALL Barycentric coordinates (S,T,U) we could pick one and replace it. For instance, we could replace U with (1-S-T) to have one less variable floating around.

## All Done for Now

You can use this pattern to expand either in “surface dimension”, or in the dimension of adding more control points (and increasing the order of the equation). I love it because it’s super simple to remember that simple equation, and then just re-calculate the equation you need for whatever your specific usage case is.

If this stuff is confusing, check out the wiki page for Bezier Triangles, it has a great graphic that really shows you what I’m trying to explain:
Bezier Triangle

Next up I either want to make an HTML5 interactive app for messing around with Bezier triangles, or if I can figure out how to intersect a line segment with a quadratic Bezier triangle, i’ll probably just have some real cool looking screenshots to post along w/ the equation I ended up using (;

Special thanks to wolfram alpha for crunching some of these equations. Check it out, it’s really cool!
Wolfram Alpha – Cubic Bezier Curve Expansion

For more bezier fun check out my next Bezier post: One Dimensional Bezier Curves.

# Bezier Curves

Bezier curves are pretty cool. They were invented in the 1950s by Pierre Bezier while he was working at the car company Renault. He created them as a succinct way of describing curves mathematically that could be shared easily with other people, or programmed into machines to make curves that matched the ones created by human designers.

I’m only going to go over bezier curves at the very high level, and give some links to html5 demos I’ve made to let you play around with them and understand how they work, so you too can implement them easily in your own software.

If you want more detailed information, I strongly recommend this book: Focus on Curves and Surfaces

Quadratic bezier curves have 3 control points. The first control point is where the curve begins, the second control point is a true control point to influence the curve, and the third control point is where the curve ends. Click the image below to be taken to my quadratic bezier curve demo.

A quadratic bezier curve has the following parameters:

• t – the “time” parameter, this parameter goes from 0 to 1 to get the points of the curve.
• A – the first control point, which is also where the curve begins.
• B – the second control point.
• C – the third control point, which is also where the curve ends.

To calculate a point on the curve given those parameters, you just sum up the result of these 3 functions:

1. A * (1-t)^2
2. B * 2t(1-t)
3. C * t^2

In otherwords, the equation looks like this:

CurvePoint = A*(1-t)^2 + B*2t(1-t) + C*t^2

To make an entire curve, you would start with t=0 to get the starting point, t=1 to get the end point, and a bunch of values in between to get the points on the curve itself.

## Cubic Bezier Curves

Cubic bezier curves have 4 control points. The first control point is where the curve begins, the second and third control points are true control point to influence the curve, and the fourth control point is where the curve ends. Click the image below to be taken to my cubic bezier curve demo.

A cubic bezier curve has the following parameters:

• t – the “time” parameter, this parameter goes from 0 to 1 to get the points of the curve.
• A – the first control point, which is also where the curve begins.
• B – the second control point.
• C – the second control point.
• D – the fourth control point, which is also where the curve ends.

To calculate a point on the curve given those parameters, you just sum up the result of these 4 functions:

1. A * (1-t)^3
2. B * 3t(1-t)^2
3. C * 3t^2(1-t)
4. D * t^3

In otherwords, the equation looks like this:

CurvePoint = A*(1-t)^3 + B*3t(1-t)^2 + C*3t^2(1-t) + D*t^3

## Math

You might think the math behind these curves has to be pretty complex and non intuitive but that is not the case at all – seriously! The curves are based entirely on linear interpolation.

Here are 2 ways you may have seen linear interpolation before.

1. value = min + percent * (max – min)
2. value = percent * max + (1 – percent) * min

We are going to use the 2nd form and replace “percent” with “t” but they have the same meaning.

Ok so considering quadratic bezier curves, we have 3 control points: A, B and C.

The formula for linearly interpolating between point A and B is this:
point = t * B + (1-t) * A

The formula for linearly interpolating between point B and C is this:
point = t * C + (1-t) * B

Now, here’s where the magic comes in. What’s the formula for interpolating between the AB formula and the BC formulas above? Well, let’s use the AB formula as min, and the BC formula as max. If you plug the formulas into the linear interpolation formula you get this:

point = t * (t * C + (1-t) * B) + (1-t) * (t * B + (1-t) * A)

if you expand that and simplify it you will end up with this equation:
point = A*(1-t)^2 + B*2t(1-t) + C*t^2

which as you may remember is the formula for a quadratic bezier curve. There you have it… a quadratic bezier curve is just a linear interpolation between 2 other linear interpolations.

Cubic bezier curves work in a similar way, there is just a 4th point to deal with.

## Next Up

The demos above are in 2d, but you could easily move to 3d (or higher dimensions!) and use the same equations. Also, there are higher order bezier curves (more control points), but as you add control points, the computational complexity increases, so people usually stick to quadratic or cubic bezier curves, and just string them together. When you put curves end to end like that, they call it a spline.

Next up, be on the look out for posts and demos for b-splines and nurbs!