# Two-point Conical Gradient

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We present a fast shading algorithm (compared to bruteforcely solving the quadratic equation of
gradient $t$) for computing the two-point conical gradient (i.e., `createRadialGradient`

in
spec).
It reduced the number of multiplications per pixel from ~10 down to 3, and brought a speedup of up to
26% in our nanobenches.

This document has 3 parts:

Part 1 and 2 are self-explanatory. Part 3 shows how to geometrically proves our Theorem 1 in part 2; it’s more complicated but it gives us a nice picture about what’s going on.

## Problem Statement and Setup

Let two circles be $C_0, r_0$ and $C_1, r_1$ where $C$ is the center and $r$ is the radius. For any
point $P = (x, y)$ we want the shader to quickly compute a gradient $t \in \mathbb R$ such that $p$
is on the linearly interpolated circle with center $C_t = (1-t) \cdot C_0 + t \cdot C_1$ and radius
$r_t = (1-t) \cdot r_0 + t \cdot r_1 > 0$ (note that radius $r_t$ has to be *positive*). If
there are multiple (at most 2) solutions of $t$, choose the bigger one.

There are two degenerated cases:

- $C_0 = C_1$ so the gradient is essentially a simple radial gradient.
- $r_0 = r_1$ so the gradient is a single strip with bandwidth $2 r_0 = 2 r_1$.

They are easy to handle so we won’t cover them here. From now on, we assume $C_0 \neq C_1$ and $r_0 \neq r_1$.

As $r_0 \neq r_1$, we can find a focal point $C_f = (1-f) \cdot C_0 + f \cdot C_1$ where its corresponding linearly interpolated radius $r_f = (1-f) \cdot r_0 + f \cdot r_1 = 0$. Solving the latter equation gets us $f = r_0 / (r_0 - r_1)$.

As $C_0 \neq C_1$, focal point $C_f$ is different from $C_1$ unless $r_1 = 0$. If $r_1 = 0$, we can swap $C_0, r_0$ with $C_1, r_1$, compute swapped gradient $t_s$ as if $r_1 \neq 0$, and finally set $t = 1 - t_s$. The only catch here is that with multiple solutions of $t_s$, we shall choose the smaller one (so $t$ could be the bigger one).

Assuming that we’ve done swapping if necessary so $C_1 \neq C_f$, we can then do a linear transformation to map $C_f, C_1$ to $(0, 0), (1, 0)$. After the transformation:

- All centers $C_t = (x_t, 0)$ must be on the $x$ axis
- The radius $r_t$ is $x_t r_1$.
- Given $x_t$ , we can derive $t = f + (1 - f) x_t$

From now on, we’ll focus on how to quickly computes $x_t$. Note that $r_t > 0$ so we’re only interested positive solution $x_t$. Again, if there are multiple $x_t$ solutions, we may want to find the bigger one if $1 - f > 0$, and smaller one if $1 - f < 0$, so the corresponding $t$ is always the bigger one (note that $f \neq 1$, otherwise we’ll swap $C_0, r_0$ with $C_1, r_1$).

## Algorithm

**Theorem 1.** The solution to $x_t$ is

- $\frac{x^2 + y^2}{(1 + r_1) x} = \frac{x^2 + y^2}{2 x}$ if $r_1 = 1$
- $\left(\sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$ if $r_1 > 1$
- $\left(\pm \sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$ if $r_1 < 1$.

Case 2 always produces a valid $x_t$. Case 1 and 3 requires $x > 0$ to produce valid $x_t > 0$. Case 3 may have no solution at all if $(r_1^2 - 1) y^2 + r_1^2 x^2 < 0$.

*Proof.* Algebriacally, solving the quadratic equation $(x_t - x)^2 + y^2 = (x_t r_1)^2$ and
eliminate negative $x_t$ solutions get us the theorem.

Alternatively, we can also combine Corollary 2., 3., and Lemma 4. in the Appendix to geometrically prove the theorem. $\square$

Theorem 1 by itself is not sufficient for our shader algorithm because:

- we still need to compute $t$ from $x_t$ (remember that $t = f + (1-f) x_t$);
- we still need to handle cases of choosing the bigger/smaller $x_t$;
- we still need to handle the swapped case (we swap $C_0, r_0$ with $C_1, r_1$ if $r_1 = 0$);
- there are way too many multiplications and divisions in Theorem 1 that would slow our shader.

Issue 2 and 3 are solved by generating different shader code based on different situations. So they are mainly correctness issues rather than performance issues. Issue 1 and 4 are performance critical, and they will affect how we handle issue 2 and 3.

The key to handle 1 and 4 efficiently is to fold as many multiplications and divisions into the linear transformation matrix, which the shader has to do anyway (remember our linear transformation to map $C_f, C_1$ to $(0, 0), (1, 0)$).

For example, let $\hat x, \hat y = |1-f|x, |1-f|y$. Computing $\hat x_t$ with respect to $\hat x, \hat y$ allow us to have $t = f + (1 - f)x_t = f + \text{sign}(1-f) \cdot \hat x_t$. That saves us one multiplication. Applying similar techniques to Theorem 1 gets us:

- If $r_1 = 1$, let $x’ = x/2,~ y’ = y/2$, then $x_t = (x’^2 + y’^2) / x’$.
- If $r_1 > 1$, let $x’ = r_1 / (r_1^2 - 1) x,~ y’ = \frac{\sqrt{r_1^2 - 1}}{r_1^2 - 1} y$, then $x_t = \sqrt{x’^2 + y’^2} - x’ / r_1$
- If $r_1 < 1$, let $x’ = r_1 / (r_1^2 - 1) x,~ y’ = \frac{\sqrt{1 - r_1^2}}{r_1^2 - 1} y$, then $x_t = \pm\sqrt{x’^2 - y’^2} - x’ / r_1$

Combining it with the swapping, the equation $t = f + (1-f) x_t$, and the fact that we only want positive $x_t > 0$ and bigger $t$, we have our final algorithm:

**Algorithm 1.**

Let $C’_0, r’_0, C’_1, r’_1 = C_0, r_0, C_1, r_1$ if there is no swapping and $C’_0, r’_0, C’_1, r’_1 = C_1, r_1, C_0, r_0$ if there is swapping.

Let $f = r’_0 / (r’_0 - r’_1)$ and $1 - f = r’_1 / (r’_1 - r’_0)$

Let $x’ = x/2,~ y’ = y/2$ if $r_1 = 1$, and $x’ = r_1 / (r_1^2 - 1) x,~ y’ = \sqrt{|r_1^2 - 1|} / (r_1^2 - 1) y$ if $r_1 \neq 1$

Let $\hat x = |1 - f|x’, \hat y = |1 - f|y’$

If $r_1 = 1$, let $\hat x_t = (\hat x^2 + \hat y^2) / \hat x$

If $r_1 > 1$, let $\hat x_t = \sqrt{\hat x^2 + \hat y^2} - \hat x / r_1$

If $r_1 < 1$

return invalid if $\hat x^2 - \hat y^2 < 0$

let $\hat x_t = -\sqrt{\hat x^2 - \hat y^2} - \hat x / r_1$ if we’ve swapped $r_0, r_1$, or if $1 - f < 0$

let $\hat x_t = \sqrt{\hat x^2 - \hat y^2} - \hat x / r_1$ otherwise

$t$ is invalid if $\hat x_t < 0$ (this check is unnecessary if $r_1 > 1$)

Let $t = f + \text{sign}(1 - f) \hat x_t$

If swapped, let $t = 1 - t$

In step 7, we try to select either the smaller or bigger $\hat x_t$ based on whether the final $t$ has a negative or positive relationship with $\hat x_t$. It’s negative if we’ve swapped, or if $\text{sign}(1 - f)$ is negative (these two cannot both happen).

Note that all the computations and if decisions not involving $\hat x, \hat y$ can be precomputed before the shading stage. The two if decisions $\hat x^2 - \hat y^2 < 0$ and $\hat x^t < 0$ can also be omitted by precomputing the shading area that never violates those conditions.

The number of operations per shading is thus:

- 1 addition, 2 multiplications, and 1 division if $r_1 = 1$
- 2 additions, 3 multiplications, and 1 sqrt for $r_1 \neq 1$ (count substraction as addition; dividing $r_1$ is multiplying $1/r_1$)
- 1 more addition operation if $f \neq 0$
- 1 more addition operation if swapped.

In comparison, for $r_1 \neq 1$ case, our current raster pipeline shading algorithm (which shall hopefully soon be upgraded to the algorithm described here) mainly uses formula $$t = 0.5 \cdot (1/a) \cdot \left(-b \pm \sqrt{b^2 - 4ac}\right)$$ It precomputes $a = 1 - (r_1 - r_0)^2, 1/a, r1 - r0$. Number $b = -2 \cdot (x + (r1 - r0) \cdot r0)$ costs 2 multiplications and 1 addition. Number $c = x^2 + y^2 - r_0^2$ costs 3 multiplications and 2 additions. And the final $t$ costs 5 more multiplications, 1 more sqrt, and 2 more additions. That’s a total of 5 additions, 10 multiplications, and 1 sqrt. (Our algorithm has 2-4 additions, 3 multiplications, and 1 sqrt.) Even if it saves the $0.5 \cdot (1/a), 4a, r_0^2$ and $(r_1 - r_0) r_0$ multiplications, there are still 6 multiplications. Moreover, it sends in 4 unitofmrs to the shader while our algorithm only needs 2 uniforms ($1/r_1$ and $f$).

## Appendix

**Lemma 1.** Draw a ray from $C_f = (0, 0)$ to $P = (x, y)$. For every
intersection points $P_1$ between that ray and circle $C_1 = (1, 0), r_1$, there exists an $x_t$
that equals to the length of segment $C_f P$ over length of segment $C_f P_1$. That is,
$x_t = || C_f P || / ||C_f P_1||$

*Proof.* Draw a line from $P$ that’s parallel to $C_1 P_1$. Let it intersect with $x$-axis on point
$C = (x’, y’)$.

Triangle $\triangle C_f C P$ is similar to triangle $\triangle C_f C_1 P_1$. Therefore $||P C|| = ||P_1 C_1|| \cdot (||C_f C|| / ||C_f C_1||) = r_1 x’$. Thus $x’$ is a solution to $x_t$. Because triangle $\triangle C_f C P$ and triangle $\triangle C_f C_1 P_1$ are similar, $x’ = ||C_f C_1|| \cdot (||C_f P|| / ||C_f P_1||) = ||C_f P|| / ||C_f P_1||$. $\square$

**Lemma 2.** For every solution $x_t$, if we extend/shrink segment $C_f P$ to $C_f P_1$ with ratio
$1 / x_t$ (i.e., find $P_1$ on ray $C_f P$ such that $||C_f P_1|| / ||C_f P|| = 1 / x_t$), then
$P_1$ must be on circle $C_1, r_1$.

*Proof.* Let $C_t = (x_t, 0)$. Triangle $\triangle C_f C_t P$ is similar to $C_f C_1 P_1$. Therefore
$||C_1 P_1|| = r_1$ and $P_1$ is on circle $C_1, r_1$. $\square$

**Corollary 1.** By lemma 1. and 2., we conclude that the number of solutions $x_t$ is equal to the
number of intersections between ray $C_f P$ and circle $C_1, r_1$. Therefore

- when $r_1 > 1$, there’s always one unique intersection/solution; we call this “well-behaved”; this was previously known as the “inside” case;
- when $r_1 = 1$, there’s either one or zero intersection/solution (excluding $C_f$ which is always on the circle); we call this “focal-on-circle”; this was previously known as the “edge” case;

- when $r_1 < 1$, there may be $0, 1$, or $2$ solutions; this was also previously as the “outside” case.

**Lemma 3.** When solution exists, one such solution is
$$
x_t = {|| C_f P || \over ||C_f P_1||} = \frac{x^2 + y^2}{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}
$$

*Proof.* As $C_f = (0, 0), P = (x, y)$, we have $||C_f P|| = \sqrt(x^2 + y^2)$. So we’ll mainly
focus on how to compute $||C_f P_1||$.

**When $x \geq 0$:**

Let $X_P = (x, 0)$ and $H$ be a point on $C_f P_1$ such that $C_1 H$ is perpendicular to $C_1 P_1$. Triangle $\triangle C_1 H C_f$ is similar to triangle $\triangle P X_P C_f$. Thus $$||C_f H|| = ||C_f C_1|| \cdot (||C_f X_P|| / ||C_f P||) = x / \sqrt{x^2 + y^2}$$ $$||C_1 H|| = ||C_f C_1|| \cdot (||P X_P|| / ||C_f P||) = y / \sqrt{x^2 + y^2}$$

Triangle $\triangle C_1 H P_1$ is a right triangle with hypotenuse $r_1$. Hence $$ ||H P_1|| = \sqrt{r_1^2 - ||C_1 H||^2} = \sqrt{r_1^2 - y^2 / (x^2 + y^2)} $$

We have \begin{align} ||C_f P_1|| &= ||C_f H|| + ||H P_1|| \\\ &= x / \sqrt{x^2 + y^2} + \sqrt{r_1^2 - y^2 / (x^2 + y^2)} \\\ &= \frac{x + \sqrt{r_1^2 (x^2 + y^2) - y^2}}{\sqrt{x^2 + y^2}} \\\ &= \frac{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}{\sqrt{x^2 + y^2}} \end{align}

**When $x < 0$:**

Define $X_P$ and $H$ similarly as before except that now $H$ is on ray $P_1 C_f$ instead of $C_f P_1$.

As before, triangle $\triangle C_1 H C_f$ is similar to triangle $\triangle P X_P C_f$, and triangle $\triangle C_1 H P_1$ is a right triangle, so we have $$||C_f H|| = ||C_f C_1|| \cdot (||C_f X_P|| / ||C_f P||) = -x / \sqrt{x^2 + y^2}$$ $$||C_1 H|| = ||C_f C_1|| \cdot (||P X_P|| / ||C_f P||) = y / \sqrt{x^2 + y^2}$$ $$ ||H P_1|| = \sqrt{r_1^2 - ||C_1 H||^2} = \sqrt{r_1^2 - y^2 / (x^2 + y^2)} $$

Note that the only difference is changing $x$ to $-x$ because $x$ is negative.

Also note that now $||C_f P_1|| = -||C_f H|| + ||H P_1||$ and we have $-||C_f H||$ instead of $||C_f H||$. That negation cancels out the negation of $-x$ so we get the same equation of $||C_f P_1||$ for both $x \geq 0$ and $x < 0$ cases:

$$ ||C_f P_1|| = \frac{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}{\sqrt{x^2 + y^2}} $$

Finally $$ x_t = \frac{||C_f P||}{||C_f P_1||} = \frac{\sqrt{x^2 + y^2}}{||C_f P_1||} = \frac{x^2 + y^2}{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}} $$ $\square$

**Corollary 2.** If $r_1 = 1$, then the solution $x_t = \frac{x^2 + y^2}{(1 + r_1) x}$, and
it’s valide (i.e., $x_t > 0$) iff $x > 0$.

*Proof.* Simply plug $r_1 = 1$ into the formula of Lemma 3. $\square$

**Corollary 3.** If $r_1 > 1$, then the unique solution is
$x_t = \left(\sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$.

*Proof.* From Lemma 3., we have

\begin{align} x_t &= \frac{x^2 + y^2}{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}} \\\ &= { (x^2 + y^2) \left ( -x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2} \right ) \over \left (x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2} \right ) \left (-x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2} \right ) } \\\ &= { (x^2 + y^2) \left ( -x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2} \right ) \over -x^2 + (r_1^2 - 1) y^2 + r_1^2 x^2 } \\\ &= { (x^2 + y^2) \left ( -x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2} \right ) \over (r_1^2 - 1) (x^2 + y^2) } \\\ &= \left(\sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1) \end{align}

The transformation above (multiplying $-x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}$ to enumerator and denomenator) is always valid because $r_1 > 1$ and it’s the unique solution due to Corollary 1. $\square$

**Lemma 4.** If $r_1 < 1$, then

- there’s no solution to $x_t$ if $(r_1^2 - 1) y^2 + r_1^2 x^2 < 0$
- otherwise, the solutions are $x_t = \left(\sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$, or $x_t = \left(-\sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$.

(Note that solution $x_t$ still has to be nonnegative to be valid; also note that $x_t > 0 \Leftrightarrow x > 0$ if the solution exists.)

*Proof.* Case 1 follows naturally from Lemma 3. and Corollary 1.

For case 2, we notice that $||C_f P_1||$ could be

- either $||C_f H|| + ||H P_1||$ or $||C_f H|| - ||H P_1||$ if $x \geq 0$,
- either $-||C_f H|| + ||H P_1||$ or $-||C_f H|| - ||H P_1||$ if $x < 0$.

By analysis similar to Lemma 3., the solution to $x_t$ does not depend on the sign of $x$ and they are either $\frac{x^2 + y^2}{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}$ or $\frac{x^2 + y^2}{x - \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}$.

As $r_1 \neq 1$, we can apply the similar transformation in Corollary 3. to get the two formula in the lemma. $\square$