(Please refresh the page if you see a lot of dollars instead of math symbols.)
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.
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:
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:
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 $C_1, r_1$).
Theorem 1. The solution to $x_t$ is
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:
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:
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:
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 $f$).
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
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
(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
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. $\square$