I am designing animations for a string of addressable LEDs. Deterministic patterns generally have a nice regularity to them, but their visual interest diminishes after a while. Random patterns can have a bit more visual interest, since they are different every time. If we just assign a random value to each pixel, the result is very busy, loud, disjointed. It would be better to have something changing in a random way, but with a smooth, regular structure so that the eye can actually take it all in.

One nice way to get a smooth, organic-feeling function is to use a polynomial spline. We can specify values to achieve at certain times (in animation these are called "key frames") and use a polynomial to interpolate between them. By choosing the spline carefully, we can get a motion which is not only continuous, but smooth in both time and space. It will have no jumps, and no times or places where the color suddenly starts changing in a different way. In this post we will walk through a procedure for generating those splines using linear algebra.

# The setting

We have a one-dimensional string of LEDs. We can specify the location of each LED by a number on an interval. To make the numbers work out nice, let's put them on the interval \([0, 3]\). For symmetry, let's describe time in the same way, as an interval \([0, 3]\). For reasons which will become clear, we will focus on the time subinterval \([1, 2]\). We will put the key frames at \(t=0\), \(t=1\), \(t=2\), and \(t=3\). We will specify each key frame by the values at \(s=0\), \(s=1\), \(s=2\), and \(s=3\). That is, we begin with a function defined on \(\{0, 1, 2, 3\}\times\{0, 1, 2, 3\}\), and we wish to extend it to a function on \([1, 2]\times[0, 3]\).

Why use this specific number of key values? I want to make a cubic polynomial, which we can write as \(f(t, s) = \sum_{i, j=0}^3 A_{ij}t^i s^j\). This has sixteen degrees of freedom. By specifying the function at sixteen points, we can make a system which is neither over-constrained nor under-constrained.

Why go for a cubic function? I want to get a curve which is smooth in time and space. On a given interval, I want it to hit the key frames, but I also want it to be smooth at both ends. That makes four constraints, so a third degree polynomial is just right.

Why look at times between 1 and 2? I want to think of time as starting at 1, and proceeding to 2. When the time hits 2, we can extend \(f\) by defining \(f(t, s) = g(t-1, s)\) for \(t\in\{1\}\cup[2,3]\). In this way, \(g\) has the same form as \(f\), with values of \(g(t)\) specified for \(t\in \{0, 1, 2\}\). By dropping the values of \(f\) at \(t=0\) and making new values for \(g\) at \(t=3\), the program repeats with \(g\) taking the role of \(f\).

# The clever bit

When we restrict \(f\) to \(t=2\), we get a polynomial in \(s\). Since it interpolates four values at \(s=0, 1, 2, 3\), it is completely specified. Since \(g\) restricted to \(t=1\) is also a cubic interpolant, we can see that \(f\) is continuous not only at \(\{2\}\times \{0, 1, 2, 3\}\) but on all of \(\{2\}\times [0, 3]\). If we are clever about how we specify \(f\), we can do even better. Notice that \(f_t\) is also a cubic polynomial in \(s\) when restricted to \(t=2\). If we can specify \(f_t(2, s)=g_t(1, s)\) then we can make \(f'\) continuous as well!

How to do this? Since \(f\) is never directly evaluated at \(t=3\), we will use the funciton values there to instead specify \(f_t\) at \(t=2\). That is, we will enforce \(f_t(2, s) = (1-c) (f(3, s)-f(1, s))/2\), where \(c\) is a "tension" term allowing us to tweak the behavior of the interpolant. By assigning the same rule for \(g\), that is, \(g_t(1, s) = (1-c)(f(3, s)-f(1, s))/2\), we get a function which is continuous and smooth.

# Computing it

We began by writing \(f(t, s)=\sum_{i, j=0}^3 A_{ij}t^i s^j\). We can write this as a matrix product,

to keep things tidy, let's define \(P_x = [1, x, x^2, x^3]^T\). Then we can write this as \(f(t, s)=P_t^T A P_s\). The goal then is to specify the matrix \(A\), in terms of the data, the key-frame values of \(f\) on \(\{0, 1, 2, 3\}^2\). Write \(F=[f(i, j)]_{i, j=0}^3\), so we are looking for how to write \(A\) in terms of \(F\).

## Continuity

We use the values at \(t=1, 2\) to enforce the continuity of the spline. This is straightforward: we want \(P_1^TAP_s = F_{1s}\) for \(s=0, 1, 2, 3\). Writing \(P_S = [P_0P_1P_2P_3]\) we can write this more compactly as \(P_1^TAP_S = F_1\). Similarly, we want \(P_2^TAP_S = F_2\). Together, these give

Not enough to solve for \(A\) yet, but we have not used smoothness yet.

## Smoothness

Differentiation yields

Write \(D_x=\frac{d}{dx} P_x\), so we can write this as \(f_t = D_t^T A P_s\). We want \(f(1, s) = \frac{1-c}{2}(f(2, s) - f(0, s))\). As a matrix equation, this becomes

Repeating for the \(t=2\) case yields

Writing it this way shows how to combine with the continuity equation,

## Solving for \(A\)

Now all the matrices are square, so we are in business. We have enough to solve for \(A\). Writing \(U=[P_1,P_2,D_1,D_2]^T\) we find

We can also compute

Multiplying on both sides yields

## Applying the formula

Now that we have a formula for \(f(t, s)=P_t^TAP_s\), we will want to apply it to our light string, which is discrete. Choose some discrete times \(T = [1, 1+\Delta t, 1+2\Delta t, \dots, 2]\) and string positions \(S=[0,\Delta s, 2\Delta s, ..., 3]\). Form them into polynomials \(P_T\) and \(P_S\) as before, then multiply them on the outside in advance; the coefficient matrices we derived above will not change, only F will change. This can save us a lot of multiplication, since much of it can be done in advance.

# Conclusion

I won't belabor this any further here. You can see how this algorithm is implemented in the repository I have shared. I think this shows how linear algebra allows us to express some pretty complicated formulas quite compactly, which lets us think about more-complicated things than would otherwise be possible.