Cubic polynomials and cubic Beziers

So it turns out that for cubic Bezier curves, t values of 0, 1/3, 2/3 and 1 have special meanings. A general cubic polynomial is of the form

y = a0 + a1 * x + a2 * x^2 + a3 * x^3

where ai’s are real constants.

If the variable x is limited to the interval x0 <= x <= x0 + χ (that's the Greek letter Chi), where χ > 0, then it’s equivalent to a special case of cubic Bezier curves. Namely, when the t values are 0, 1/3, 2/3 and 1.

In fact, there’s a mathematical proof of it. Thanks to Professor Samuel Dagan of Tel-Aviv University for writing in and letting me know of his work. Here’s more of his work.

Does the point lie on the Bezier curve?

Someone recently asked me how to tell if a point lies on a Bezier curve.

For the purposes of discussion, it’s a quadratic Bezier curve and all 3 control points are known (or the start and end points and the 1 control point if you prefer). You can read more about the reverse process of finding the control points here, which is the reference point of that person’s question.

The answer is actually straightforward. Substitute everything into the Bezier curve equation and solve for t. Here’s the quadratic Bezier equation:
B(t) = (1-t)^2 * p0 + 2(1-t)t * p1 + t^2 * p2

Let’s say p0 is [1,1] and p1 is [1.5,4.5] and p2 is [2,3]. We’ll keep the points in 2 dimensions to keep the maths working less cumbersome. And let’s say the point you want to check is [1.8,3.4]. We substitute all the points into the equation, and we get this:

[1.8,3.4] = (1-t)^2 * [1,1] + 2(1-t)t * [1.5,4.5] + t^2 * [2,3]

I know, it doesn’t look pretty. But hey, we’re doing this by hand. If you’re writing code to generalise the solution, the code will probably look just a little uglier, but the solution will come out faster. Like probably instantly given the current modern processors.

Because we’re dealing with 2 dimensional points, that equation splits into 2 separate equations (with scalars instead of vectors as coefficients), like so:
1.8 = (1-t)^2 * 1 + 2(1-t)t * 1.5 + t^2 * 2
3.4 = (1-t)^2 * 1 + 2(1-t)t * 4.5 + t^2 * 3

If you have 3 dimensional points, you’d have 3 equations. Note that even then, the degree of your equations remains as 2. The degree of the Bezier curve is independent of the number of dimensions you’re working with.

If you simplify
1.8 = (1-t)^2 * 1 + 2(1-t)t * 1.5 + t^2 * 2

You get t = 0.8. It so happens that in this case, there’s only one solution.

If you simplify
3.4 = (1-t)^2 * 1 + 2(1-t)t * 4.5 + t^2 * 3

You get
5*t^2 – 7*t + 2.4 = 0

and after solving for that, you get t = 0.6 or t = 0.8 (you’re a smart person, you know how to solve a quadratic equation, right?)

Now, the solution t=0.8 appears in the solution sets of both equations. Therefore, the point [1.8,3.4] lies on the Bezier curve. In fact, t=0.8 is the t value.

Multiple solutions

What if you get multiple t values appearing in multiple solution sets of equations?

Consider the case where p0 is [1,1], p1 is [2,3], and p2 is [1,1]. Notice that the start and end points are the same point. Let’s say you want to know if the point [1,1] lies on the curve (yes I know it’s the same point). Substituting all the points, we get:

[1,1] = (1-t)^2 * [1,1] + 2(1-t)t * [2,3] + t^2 * [1,1]

This gives us the 2 equations:
1 = 1 – 2*t + t^2 + 4*t – 4*t^2 + t^2
1 = 1 – 2*t + t^2 + 6*t – 6*t^2 + t^2

They simplify to:
2*t^2 – 2*t = 0
4*t^2 – 4*t = 0

Hey presto! The solution set is t=0 or t=1 for both equations. Therefore, the point [1,1] lies on the curve. In fact, it lies on the curve where t=0 or t=1. And t=0 and t=1 happens to coincide with the start and end points respectively.

The whole point (haha!) is that, as long as you have at least one value of t that appears in the solution sets of all the equations, then said point you’re checking lies on the curve.

Higher degree Bezier curves

This is a toughie. If you have a cubic Bezier curve, then you’re solving a degree 3 polynomial (of t). If you have a Bezier curve of degree N, then you’re solving a degree N polynomial.

There are algorithms to solve generic degree polynomials, but they are out of scope here. Assuming the highest degree of Bezier curves you’ll ever work with is 3 (cubic), then this Wikipedia article on cubic functions will help. Remember, cubic Bezier curve equations are still cubic equations.

Higher dimensionality

The number of dimensions you’re working with determines the number of equations you need to solve. If you’re working with 5 dimensional points, then you need to solve for 5 equations.

For example, if you’re working with cubic Bezier curves and using 5 dimensional points, then you need to solve 5 cubic functions. You will have possibly 3 (unique) t values for each equation. Let’s say your solution sets are as follows:
t = -1, 3, 5
t = 0, 1, 3
t = -2, 2, 3
t = 3, 3, 4 (yay repeated values!)
t = 3, 6, 8

The value t=3 appears in all 5 sets of solutions, therefore your point lies on the curve.

Keep it real

In the process of solving your equations, there’s a possibility that you might get imaginary solutions. You know, those involving the square root of -1. Dismiss them.

Your Bezier curve is in the real world. The point you’re checking must therefore also lie in the real world.

Unless you’re working with some abstract imaginary Bezier curves on an advanced maths paper. Then good luck to you! The logic above for solving still applies.

Actual applications

When applying the above, you don’t usually get nice numbers like [1.8,3.4] lying on the curve with t=0.8. You get numbers with lots of numbers behind the decimal point that seems to continue forever. You don’t get exact values.

What if you get a t=0.798 for one equation, and t=0.802 for another equation?

Use your common sense. Set an error margin for what is acceptable.

My suggestion is to NOT use the values of t to check for the margin. Substitute the values of t into the equation, and then check the points if they’re within the error margin.

This means you don’t check the difference between t=0.798 and t=0.802, which is 0.04. Is 0.04 within your error margin? Maybe. But you’re not checking for this.

You substitute t=0.798 and t=0.802 into the equation, and you get 2 points: [1.798,3.40198] and [1.802,3.39798]

Then you say, “Are these points close enough that I consider them to be the same point?” Use whatever you think is appropriate. I think the Euclidean distance norm works fine. Then check if that “close enough” criteria is within your error margin.

If you’re checking for [1.8,3.4], then ask yourself, “Is [1.8,3.4] close enough to [1.798,3.40198]? And is [1.8,3.4] close enough to [1.802,3.39798]?”

Obviously, doing this by hand sucks big time. Good thing you’re a programmer.

Rapidly calculating Bezier points

The standard cubic Bezier curve is given by
B(t) = (1-t)3p0 + 3(1-t)2tp1 + 3(1-t)t2p2 + t3p3
where p0, p1, p2, p3 are the control points and t is in the interval [0, 1].

Very elegant, but not very practical. Too many multiplications and additions to be used in a fast-paced environment such as game development. If it’s a 3D Bezier curve, 3 separate calculations are needed for the x-, y- and z-component for a given t value. What we need is a simplication of the equation.

Expanding the polynomial equation, we have
= (1 – 3t + 3t2 – t3)p0
+ (3t -6t2 + 3t3)p1
+ (3t2 – 3t3)p2
+ t3p3

Rearranging the coefficients of tn, we have
= t3(-p0 + 3p1 – 3p2 + p3)
+ t2(3p0 – 6p1 + 3p2)
+ t(-3p0 + 3p1)
+ p0

The coefficients can now be reduced to constants by precalculating them, and calculation of a Bezier point takes less computation.

In matrix form, the equation looks like
Coefficient matrix form of Bezier curve