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@ -15,11 +15,12 @@ MathJax.Hub.Config({
(Please refresh the page if you see a lot of dollars instead of math symbols.) (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 We present a fast shading algorithm (compared to bruteforcely solving the
gradient $t$) for computing the two-point conical gradient (i.e., `createRadialGradient` in quadratic equation of gradient $t$) for computing the two-point conical gradient
(i.e., `createRadialGradient` in
[spec](https://html.spec.whatwg.org/multipage/canvas.html#dom-context-2d-createradialgradient)). [spec](https://html.spec.whatwg.org/multipage/canvas.html#dom-context-2d-createradialgradient)).
It reduced the number of multiplications per pixel from ~10 down to 3, and brought a speedup of up to It reduced the number of multiplications per pixel from ~10 down to 3, and
26% in our nanobenches. brought a speedup of up to 26% in our nanobenches.
This document has 3 parts: This document has 3 parts:
@ -27,16 +28,19 @@ This document has 3 parts:
2. [Algorithm](#algorithm) 2. [Algorithm](#algorithm)
3. [Appendix](#appendix) 3. [Appendix](#appendix)
Part 1 and 2 are self-explanatory. Part 3 shows how to geometrically proves our Theorem 1 in part Part 1 and 2 are self-explanatory. Part 3 shows how to geometrically proves our
2; it's more complicated but it gives us a nice picture about what's going on. Theorem 1 in part 2; it's more complicated but it gives us a nice picture about
what's going on.
## <span id="problem-statement">Problem Statement and Setup</span> ## <span id="problem-statement">Problem Statement and Setup</span>
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 Let two circles be $C_0, r_0$ and $C_1, r_1$ where $C$ is the center and $r$ is
point $P = (x, y)$ we want the shader to quickly compute a gradient $t \in \mathbb R$ such that $p$ the radius. For any point $P = (x, y)$ we want the shader to quickly compute a
is on the linearly interpolated circle with center $C_t = (1-t) \cdot C_0 + t \cdot C_1$ and radius gradient $t \in \mathbb R$ such that $p$ is on the linearly interpolated circle
$r_t = (1-t) \cdot r_0 + t \cdot r_1 > 0$ (note that radius $r_t$ has to be _positive_). If with center $C_t = (1-t) \cdot C_0 + t \cdot C_1$ and radius
there are multiple (at most 2) solutions of $t$, choose the bigger one. $r_t = (1-t) \cdot r_0 + t \cdot r_1 > 0$ (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: There are two degenerated cases:
@ -45,90 +49,109 @@ There are two degenerated cases:
<!-- TODO maybe add some fiddle or images here to illustrate the two degenerated cases --> <!-- TODO maybe add some fiddle or images here to illustrate the 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 They are easy to handle so we won't cover them here. From now on, we assume
$C_0 \neq C_1$ and $r_0
\neq r_1$. \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 As $r_0 \neq r_1$, we can find a focal point
corresponding linearly interpolated radius $r_f = (1-f) \cdot r_0 + f \cdot r_1 = 0$. $C_f = (1-f) \cdot C_0 + f \cdot C_1$ where its corresponding linearly
Solving the latter equation gets us $f = r_0 / (r_0 - r_1)$. 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 As $C_0 \neq C_1$, focal point $C_f$ is different from $C_1$ unless $r_1 = 0$.
swap $C_0, r_0$ with $C_1, r_1$, compute swapped gradient $t_s$ as if $r_1 \neq 0$, and finally set If $r_1 = 0$, we can swap $C_0, r_0$ with $C_1, r_1$, compute swapped gradient
$t = 1 - t_s$. The only catch here is that with multiple solutions of $t_s$, we shall choose the $t_s$ as if $r_1 \neq 0$, and finally set $t = 1 - t_s$. The only catch here is
smaller one (so $t$ could be the bigger one). 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 Assuming that we've done swapping if necessary so $C_1 \neq C_f$, we can then do
transformation to map $C_f, C_1$ to $(0, 0), (1, 0)$. After the transformation: a linear transformation to map $C_f, C_1$ to $(0, 0), (1, 0)$. After the
transformation:
1. All centers $C_t = (x_t, 0)$ must be on the $x$ axis 1. All centers $C_t = (x_t, 0)$ must be on the $x$ axis
2. The radius $r_t$ is $x_t r_1$. 2. The radius $r_t$ is $x_t r_1$.
3. Given $x_t$ , we can derive $t = f + (1 - f) x_t$ 3. 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 From now on, we'll focus on how to quickly computes $x_t$. Note that $r_t > 0$
interested positive solution $x_t$. Again, if there are multiple $x_t$ solutions, we may want to so we're only interested positive solution $x_t$. Again, if there are multiple
find the bigger one if $1 - f > 0$, and smaller one if $1 - f < 0$, so the corresponding $t$ is $x_t$ solutions, we may want to find the bigger one if $1 - f > 0$, and smaller
always the bigger one (note that $f \neq 1$, otherwise we'll swap $C_0, r_0$ with $C_1, r_1$). 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 $C_1, r_1$).
## <span id="algorithm">Algorithm</span> ## <span id="algorithm">Algorithm</span>
**Theorem 1.** The solution to $x_t$ is **Theorem 1.** The solution to $x_t$ is
1. $\frac{x^2 + y^2}{(1 + r_1) x} = \frac{x^2 + y^2}{2 x}$ if $r_1 = 1$ 1. $\frac{x^2 + y^2}{(1 + r_1) x} = \frac{x^2 + y^2}{2 x}$ if $r_1 = 1$
2. $\left(\sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$ if $r_1 > 1$ 2. $\left(\sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$ if
3. $\left(\pm \sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$ if $r_1 < 1$. $r_1 > 1$
3. $\left(\pm \sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$ 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 Case 2 always produces a valid $x_t$. Case 1 and 3 requires $x > 0$ to produce
3 may have no solution at all if $(r_1^2 - 1) y^2 + r_1^2 x^2 < 0$. 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 _Proof._ Algebriacally, solving the quadratic equation
eliminate negative $x_t$ solutions get us the theorem. $(x_t - x)^2 + y^2 = (x_t r_1)^2$ 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 Alternatively, we can also combine Corollary 2., 3., and Lemma 4. in the
prove the theorem. $\square$ Appendix to geometrically prove the theorem. $\square$
Theorem 1 by itself is not sufficient for our shader algorithm because: Theorem 1 by itself is not sufficient for our shader algorithm because:
1. we still need to compute $t$ from $x_t$ (remember that $t = f + (1-f) x_t$); 1. we still need to compute $t$ from $x_t$ (remember that $t = f + (1-f) x_t$);
2. we still need to handle cases of choosing the bigger/smaller $x_t$; 2. we still need to handle cases of choosing the bigger/smaller $x_t$;
3. we still need to handle the swapped case (we swap $C_0, r_0$ with $C_1, r_1$ if $r_1 = 0$); 3. we still need to handle the swapped case (we swap $C_0, r_0$ with $C_1, r_1$
4. there are way too many multiplications and divisions in Theorem 1 that would slow our shader. if $r_1 = 0$);
4. 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 Issue 2 and 3 are solved by generating different shader code based on different
are mainly correctness issues rather than performance issues. Issue 1 and 4 are performance situations. So they are mainly correctness issues rather than performance
critical, and they will affect how we handle issue 2 and 3. 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 The key to handle 1 and 4 efficiently is to fold as many multiplications and
linear transformation matrix, which the shader has to do anyway (remember our linear transformation divisions into the linear transformation matrix, which the shader has to do
to map $C_f, C_1$ to $(0, 0), (1, 0)$). 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, For example, let $\hat x, \hat y = |1-f|x, |1-f|y$. Computing $\hat x_t$ with
\hat y$ allow us to have $t = f + (1 - f)x_t = f + \text{sign}(1-f) \cdot \hat x_t$. That saves us respect to $\hat x,
one multiplication. Applying similar techniques to Theorem 1 gets us: \hat y$ 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:
1. If $r_1 = 1$, let $x' = x/2,~ y' = y/2$, then $x_t = (x'^2 + y'^2) / x'$. 1. If $r_1 = 1$, let $x' = x/2,~ y' = y/2$, then $x_t = (x'^2 + y'^2) / x'$.
2. 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 2. 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
$x_t = \sqrt{x'^2 + y'^2} - x' / r_1$ $x_t = \sqrt{x'^2 + y'^2} - x' / r_1$
3. 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 3. 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
$x_t = \pm\sqrt{x'^2 - y'^2} - x' / r_1$ $x_t = \pm\sqrt{x'^2 - y'^2} - x' / r_1$
Combining it with the swapping, the equation $t = f + (1-f) x_t$, and the fact that we only want Combining it with the swapping, the equation $t = f + (1-f) x_t$, and the fact
positive $x_t > 0$ and bigger $t$, we have our final algorithm: that we only want positive $x_t > 0$ and bigger $t$, we have our final
algorithm:
**Algorithm 1.** **Algorithm 1.**
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, 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. r'_0, C'_1, r'_1 = C_1, r_1, C_0, r_0$ if there is swapping.
2. Let $f = r'_0 / (r'_0 - r'_1)$ and $1 - f = r'_1 / (r'_1 - r'_0)$ 2. Let $f = r'_0 / (r'_0 - r'_1)$ and $1 - f = r'_1 / (r'_1 - r'_0)$
3. Let $x' = x/2,~ y' = y/2$ if $r_1 = 1$, and 3. 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 $r_1 \neq 1$ $x' = r_1 / (r_1^2 - 1) x,~ y' = \sqrt{|r_1^2 - 1|} / (r_1^2 - 1) y$ if
$r_1 \neq 1$
4. Let $\hat x = |1 - f|x', \hat y = |1 - f|y'$ 4. Let $\hat x = |1 - f|x', \hat y = |1 - f|y'$
5. If $r_1 = 1$, let $\hat x_t = (\hat x^2 + \hat y^2) / \hat x$ 5. If $r_1 = 1$, let $\hat x_t = (\hat x^2 + \hat y^2) / \hat x$
6. If $r_1 > 1$, 6. If $r_1 > 1$, let $\hat x_t = \sqrt{\hat x^2 + \hat y^2} - \hat x / r_1$
let $\hat x_t = \sqrt{\hat x^2 + \hat y^2} - \hat x / r_1$
7. If $r_1 < 1$ 7. If $r_1 < 1$
8. return invalid if $\hat x^2 - \hat y^2 < 0$ 8. return invalid if $\hat x^2 - \hat y^2 < 0$
9. let $\hat x_t = -\sqrt{\hat x^2 - \hat y^2} - \hat x / r_1$ if we've swapped $r_0, r_1$, 9. let $\hat x_t = -\sqrt{\hat x^2 - \hat y^2} - \hat x / r_1$ if we've swapped
or if $1 - f < 0$ $r_0, r_1$, or if $1 - f < 0$
10. let $\hat x_t = \sqrt{\hat x^2 - \hat y^2} - \hat x / r_1$ otherwise 10. let $\hat x_t = \sqrt{\hat x^2 - \hat y^2} - \hat x / r_1$ otherwise
@ -136,74 +159,87 @@ positive $x_t > 0$ and bigger $t$, we have our final algorithm:
12. Let $t = f + \text{sign}(1 - f) \hat x_t$ 12. Let $t = f + \text{sign}(1 - f) \hat x_t$
13. If swapped, let $t = 1 - t$ 13. 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 $t$ In step 7, we try to select either the smaller or bigger $\hat x_t$ based on
has a negative or positive relationship with $\hat x_t$. It's negative if we've swapped, or if whether the final $t$ has a negative or positive relationship with $\hat x_t$.
$\text{sign}(1 - f)$ is negative (these two cannot both happen). 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 Note that all the computations and if decisions not involving $\hat x, \hat y$
before the shading stage. The two if decisions $\hat x^2 - \hat y^2 < 0$ and $\hat x^t < 0$ can can be precomputed before the shading stage. The two if decisions
also be omitted by precomputing the shading area that never violates those conditions. $\hat x^2 - \hat y^2 < 0$ 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: The number of operations per shading is thus:
- 1 addition, 2 multiplications, and 1 division if $r_1 = 1$ - 1 addition, 2 multiplications, and 1 division if $r_1 = 1$
- 2 additions, 3 multiplications, and 1 sqrt for $r_1 \neq 1$ (count subtraction as addition; - 2 additions, 3 multiplications, and 1 sqrt for $r_1 \neq 1$ (count subtraction
dividing $r_1$ is multiplying $1/r_1$) as addition; dividing $r_1$ is multiplying $1/r_1$)
- 1 more addition operation if $f \neq 0$ - 1 more addition operation if $f \neq 0$
- 1 more addition operation if swapped. - 1 more addition operation if swapped.
In comparison, for $r_1 \neq 1$ case, our current raster pipeline shading algorithm (which shall In comparison, for $r_1 \neq 1$ case, our current raster pipeline shading
hopefully soon be upgraded to the algorithm described here) mainly uses formula $$t = 0.5 \cdot algorithm (which shall hopefully soon be upgraded to the algorithm described
(1/a) \cdot \left(-b \pm \sqrt{b^2 - 4ac}\right)$$ It precomputes $a = 1 - (r_1 - r_0)^2, 1/a, r1 - here) mainly uses formula
r0$. Number $b = -2 \cdot (x + (r1 - r0) \cdot r0)$ costs 2 multiplications and 1 addition. Number $$t = 0.5 \cdot
$c = x^2 + y^2 - r_0^2$ costs 3 multiplications and 2 additions. And the final $t$ costs 5 more (1/a) \cdot \left(-b \pm \sqrt{b^2 - 4ac}\right)$$ It precomputes
multiplications, 1 more sqrt, and 2 more additions. That's a total of 5 additions, 10 $a = 1 - (r_1 - r_0)^2, 1/a, r1 -
multiplications, and 1 sqrt. (Our algorithm has 2-4 additions, 3 multiplications, and 1 sqrt.) Even r0$. Number
if it saves the $0.5 \cdot (1/a), 4a, r_0^2$ and $(r_1 - r_0) r_0$ multiplications, there are still $b = -2 \cdot (x + (r1 - r0) \cdot r0)$ costs 2 multiplications and 1 addition.
6 multiplications. Moreover, it sends in 4 unitofmrs to the shader while our algorithm only needs 2 Number $c = x^2 + y^2 - r_0^2$ costs 3 multiplications and 2 additions. And the
uniforms ($1/r_1$ and $f$). 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 $f$).
## <span id="appendix">Appendix</span> ## <span id="appendix">Appendix</span>
**Lemma 1.** Draw a ray from $C_f = (0, 0)$ to $P = (x, y)$. For every **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 $x_t$ intersection points $P_1$ between that ray and circle $C_1 = (1, 0), r_1$, there
that equals to the length of segment $C_f P$ over length of segment $C_f P_1$. That is, exists an $x_t$ that equals to the length of segment $C_f P$ over length of
$x_t = || C_f P || / ||C_f P_1||$ 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 _Proof._ Draw a line from $P$ that's parallel to $C_1 P_1$. Let it intersect
$C = (x', y')$. with $x$-axis on point $C = (x', y')$.
<img src="conical/lemma1.svg"/> <img src="./lemma1.svg"/>
Triangle $\triangle C_f C P$ is similar to triangle $\triangle C_f C_1 P_1$. 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 Therefore $||P C|| = ||P_1 C_1|| \cdot (||C_f C|| / ||C_f C_1||) = r_1 x'$. Thus
to $x_t$. Because triangle $\triangle C_f C P$ and triangle $\triangle C_f C_1 P_1$ are similar, $x' $x'$ is a solution to $x_t$. Because triangle $\triangle C_f C P$ and triangle
= ||C_f C_1|| \cdot (||C_f P|| / ||C_f P_1||) = ||C_f P|| / ||C_f P_1||$. $\square$ $\triangle C_f C_1 P_1$ 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 **Lemma 2.** For every solution $x_t$, if we extend/shrink segment $C_f P$ to
$1 / x_t$ (i.e., find $P_1$ on ray $C_f P$ such that $||C_f P_1|| / ||C_f P|| = 1 / x_t$), then $C_f P_1$ with ratio $1 / x_t$ (i.e., find $P_1$ on ray $C_f P$ such that
$P_1$ must be on circle $C_1, r_1$. $||C_f P_1|| / ||C_f P|| = 1 / x_t$), 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 _Proof._ Let $C_t = (x_t, 0)$. Triangle $\triangle C_f C_t P$ is similar to
$||C_1 P_1|| = r_1$ and $P_1$ is on circle $C_1, r_1$. $\square$ $C_f C_1 P_1$. 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 **Corollary 1.** By lemma 1. and 2., we conclude that the number of solutions
number of intersections between ray $C_f P$ and circle $C_1, r_1$. Therefore $x_t$ 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 - when $r_1 > 1$, there's always one unique intersection/solution; we call this
was previously known as the "inside" case; "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 - when $r_1 = 1$, there's either one or zero intersection/solution (excluding
on the circle); we call this "focal-on-circle"; this was previously known as the "edge" case; $C_f$ which is always on the circle); we call this "focal-on-circle"; this was
previously known as the "edge" case;
<img src="conical/corollary2.2.1.svg"/> <img src="./corollary2.2.1.svg"/>
<img src="conical/corollary2.2.2.svg"/> <img src="./corollary2.2.2.svg"/>
- when $r_1 < 1$, there may be $0, 1$, or $2$ solutions; this was also previously as the "outside" - when $r_1 < 1$, there may be $0, 1$, or $2$ solutions; this was also
case. previously as the "outside" case.
<img src="conical/corollary2.3.1.svg" width="30%"/> <img src="./corollary2.3.1.svg" width="30%"/>
<img src="conical/corollary2.3.2.svg" width="30%"/> <img src="./corollary2.3.2.svg" width="30%"/>
<img src="conical/corollary2.3.3.svg" width="30%"/> <img src="./corollary2.3.3.svg" width="30%"/>
**Lemma 3.** When solution exists, one such solution is **Lemma 3.** When solution exists, one such solution is
@ -211,47 +247,48 @@ $$
x_t = {|| C_f P || \over ||C_f P_1||} = \frac{x^2 + y^2}{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}} x_t = {|| C_f P || \over ||C_f P_1||} = \frac{x^2 + y^2}{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}
$$ $$
_Proof._ As $C_f = (0, 0), P = (x, y)$, we have $||C_f P|| = \sqrt(x^2 + y^2)$. So we'll mainly _Proof._ As $C_f = (0, 0), P = (x, y)$, we have $||C_f P|| = \sqrt(x^2 + y^2)$.
focus on how to compute $||C_f P_1||$. So we'll mainly focus on how to compute $||C_f P_1||$.
**When $x \geq 0$:** **When $x \geq 0$:**
<img src="conical/lemma3.1.svg"/> <img src="./lemma3.1.svg"/>
Let $X_P = (x, 0)$ and $H$ be a point on $C_f P_1$ such that $C_1 H$ is perpendicular to $C_1 Let $X_P = (x, 0)$ and $H$ be a point on $C_f P_1$ such that $C_1 H$ is
P_1$. Triangle $\triangle C_1 H C_f$ is similar to triangle $\triangle P X_P C_f$. Thus perpendicular to $C_1
P_1$. Triangle $\triangle C_1 H C_f$ is similar to triangle
$\triangle P X_P C_f$. Thus
$$||C_f H|| = ||C_f C_1|| \cdot (||C_f X_P|| / ||C_f P||) = x / \sqrt{x^2 + y^2}$$ $$||C_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}$$ $$||C_1 H|| = ||C_f C_1|| \cdot (||P X_P|| / ||C_f P||) = y / \sqrt{x^2 + y^2}$$
Triangle $\triangle C_1 H P_1$ is a right triangle with hypotenuse $r_1$. Hence Triangle $\triangle C_1 H P_1$ is a right triangle with hypotenuse $r_1$. Hence
$$ ||H P_1|| = \sqrt{r_1^2 - ||C_1 H||^2} = \sqrt{r_1^2 - y^2 / (x^2 + y^2)} $$ $$ ||H P_1|| = \sqrt{r_1^2 - ||C_1 H||^2} = \sqrt{r_1^2 - y^2 / (x^2 + y^2)} $$
We have We have \begin{align} ||C_f P_1|| &= ||C_f H|| + ||H P_1|| \\\\\\ &= x /
\begin{align} \sqrt{x^2 + y^2} + \sqrt{r_1^2 - y^2 / (x^2 + y^2)} \\\\\\ &= \frac{x +
||C_f P_1|| &= ||C_f H|| + ||H P_1|| \\\\\\ \sqrt{r_1^2 (x^2 + y^2) - y^2}}{\sqrt{x^2 + y^2}} \\\\\\ &= \frac{x +
&= x / \sqrt{x^2 + y^2} + \sqrt{r_1^2 - y^2 / (x^2 + y^2)} \\\\\\ \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}{\sqrt{x^2 + y^2}} \end{align}
&= \frac{x + \sqrt{r_1^2 (x^2 + y^2) - y^2}}{\sqrt{x^2 + y^2}} \\\\\\
&= \frac{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}{\sqrt{x^2 + y^2}}
\end{align}
**When $x < 0$:** **When $x < 0$:**
Define $X_P$ and $H$ similarly as before except that now $H$ is on ray $P_1 C_f$ instead of Define $X_P$ and $H$ similarly as before except that now $H$ is on ray $P_1 C_f$
$C_f P_1$. instead of $C_f P_1$.
<img src="conical/lemma3.2.svg"/> <img src="./lemma3.2.svg"/>
As before, triangle $\triangle C_1 H C_f$ is similar to triangle $\triangle P X_P C_f$, and triangle As before, triangle $\triangle C_1 H C_f$ is similar to triangle
$\triangle C_1 H P_1$ is a right triangle, so we have $\triangle P X_P C_f$, and triangle $\triangle C_1 H P_1$ is a right triangle,
so we have
$$||C_f H|| = ||C_f C_1|| \cdot (||C_f X_P|| / ||C_f P||) = -x / \sqrt{x^2 + y^2}$$ $$||C_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}$$ $$||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)} $$ $$ ||H P_1|| = \sqrt{r_1^2 - ||C_1 H||^2} = \sqrt{r_1^2 - y^2 / (x^2 + y^2)} $$
Note that the only difference is changing $x$ to $-x$ because $x$ is negative. Note that the only difference is changing $x$ to $-x$ because $x$ is negative.
Also note that now $||C_f P_1|| = -||C_f H|| + ||H P_1||$ and we have $-||C_f H||$ instead of Also note that now $||C_f P_1|| = -||C_f H|| + ||H P_1||$ and we have
$||C_f H||$. That negation cancels out the negation of $-x$ so we get the same equation $-||C_f H||$ instead of $||C_f H||$. That negation cancels out the negation of
of $||C_f P_1||$ for both $x \geq 0$ and $x < 0$ cases: $-x$ so we get the same equation of $||C_f P_1||$ for both $x \geq 0$ and
$x < 0$ cases:
$$ $$
||C_f P_1|| = \frac{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}{\sqrt{x^2 + y^2}} ||C_f P_1|| = \frac{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}{\sqrt{x^2 + y^2}}
@ -312,7 +349,7 @@ $x_t > 0 \Leftrightarrow x > 0$ if the solution exists.)
*Proof.* Case 1 follows naturally from Lemma 3. and Corollary 1. *Proof.* Case 1 follows naturally from Lemma 3. and Corollary 1.
<img src="conical/lemma4.svg"/> <img src="./lemma4.svg"/>
For case 2, we notice that $||C_f P_1||$ could be For case 2, we notice that $||C_f P_1||$ could be

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@ -49,7 +49,7 @@ serializing the PDF file.
## <span id="PDF_Objects_and_Document_Structure">PDF Objects and Document Structure</span> ## <span id="PDF_Objects_and_Document_Structure">PDF Objects and Document Structure</span>
![PDF Logical Document Structure](/dev/design/PdfLogicalDocumentStructure.png) ![PDF Logical Document Structure](../PdfLogicalDocumentStructure.png)
**Background**: The PDF file format has a header, a set of objects and then a **Background**: The PDF file format has a header, a set of objects and then a
footer that contains a table of contents for all of the objects in the document footer that contains a table of contents for all of the objects in the document

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4
site/docs/dev/testing/download.md
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@ -19,13 +19,13 @@ Add the checkout location to your $PATH.
To download the isolated files for a test first visit To download the isolated files for a test first visit
the build status page and find the "isolated output" link: the build status page and find the "isolated output" link:
<img src="Status.png" style="margin-left:30px" width=576 height=271 > <img src="../Status.png" style="margin-left:30px" width=576 height=271 >
Follow that link to find the hash of the isolated outputs: Follow that link to find the hash of the isolated outputs:
<img src="Isolate.png" style="margin-left:30px" width=451 height=301 > <img src="../Isolate.png" style="margin-left:30px" width=451 height=301 >
Then run `isolateserver.py` with --isolated set to that hash: Then run `isolateserver.py` with --isolated set to that hash:

20
site/docs/dev/testing/skiagold.md
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@ -48,14 +48,14 @@ Solution today:
- Blame is not sorted in any particular order - Blame is not sorted in any particular order
- Digests are clustered by runs and the most minimal set of blame - Digests are clustered by runs and the most minimal set of blame
<img src=BlameView.png style="margin-left:30px" align="left" width="800"/> <img src=../BlameView.png style="margin-left:30px" align="left" width="800"/>
<br clear="left"> <br clear="left">
- Select digests for triage - Select digests for triage
- Digests will be listed in order with largest difference first - Digests will be listed in order with largest difference first
- Click to open the digest view with detailed information - Click to open the digest view with detailed information
<img src=Digests.png style="margin-left:40px" align="left" width="780"/> <img src=../Digests.png style="margin-left:40px" align="left" width="780"/>
<br clear="left"> <br clear="left">
- Open bugs for identified owner(s) - Open bugs for identified owner(s)
@ -65,7 +65,7 @@ Solution today:
- The URL reference to the digest in Issue Tracker will link the bug to the - The URL reference to the digest in Issue Tracker will link the bug to the
digest in Gold digest in Gold
<img src="IssueHighlight.png" style="margin-left:60px" align="left" width="720" border=1/> <img src="../IssueHighlight.png" style="margin-left:60px" align="left" width="720" border=1/>
<br clear="left"> <br clear="left">
<br> <br>
@ -88,7 +88,7 @@ To find your results:
- Note: It is not yet implemented in the UI but possible to filter the view by - Note: It is not yet implemented in the UI but possible to filter the view by
CL. Delete hashes in the URL to only include the hash for your CL. CL. Delete hashes in the URL to only include the hash for your CL.
<img src=BlameView.png style="margin-left:30px" align="left" width="800"/> <img src=../BlameView.png style="margin-left:30px" align="left" width="800"/>
<br clear="left"> <br clear="left">
To rebaseline images: To rebaseline images:
@ -96,7 +96,7 @@ To rebaseline images:
- Access the Ignores view and create a new, short-interval (hours) ignore for - Access the Ignores view and create a new, short-interval (hours) ignore for
the most affected configuration(s) the most affected configuration(s)
<img src=Ignores.png style="margin-left:30px" align="left" width="800"/> <img src=../Ignores.png style="margin-left:30px" align="left" width="800"/>
<br clear="left"> <br clear="left">
- Click on the Ignore to bring up a search view filtered by the affected - Click on the Ignore to bring up a search view filtered by the affected
@ -141,7 +141,7 @@ Solution:
- Access the By Test view - Access the By Test view
<img src=ByTest.png style="margin-left:30px" align="left" width="800"/> <img src=../ByTest.png style="margin-left:30px" align="left" width="800"/>
<br clear="left"> <br clear="left">
- Click the magnifier to filter by configuration - Click the magnifier to filter by configuration
@ -150,12 +150,12 @@ Solution:
- Click on configurations under “parameters” to highlight data points and - Click on configurations under “parameters” to highlight data points and
compare compare
<img src=ClusterConfig.png style="margin-left:30px" align="left" width="800"/> <img src=../ClusterConfig.png style="margin-left:30px" align="left" width="800"/>
<br clear="left"> <br clear="left">
- Access the Grid view to see NxN diffs - Access the Grid view to see NxN diffs
<img src=Grid.png style="margin-left:30px" align="left" width="800"/> <img src=../Grid.png style="margin-left:30px" align="left" width="800"/>
<br clear="left"> <br clear="left">
- Access the Dot diagram to see history of commits for the trace - Access the Dot diagram to see history of commits for the trace
@ -163,7 +163,7 @@ Solution:
- Each line represents a configuration - Each line represents a configuration
- Dot colors distinguish between digests - Dot colors distinguish between digests
<img src=DotDiagram.png style="margin-left:30px" align="left" width="800"/> <img src=../DotDiagram.png style="margin-left:30px" align="left" width="800"/>
<br clear="left"> <br clear="left">
<br> <br>
@ -181,5 +181,5 @@ Solution:
- Access the Search view - Access the Search view
- Select any parameters desired to search across tests - Select any parameters desired to search across tests
<img src=Search.png style="margin-left:30px" align="left" width="800"/> <img src=../Search.png style="margin-left:30px" align="left" width="800"/>
<br clear="left"> <br clear="left">

6
site/docs/dev/testing/skiaperf.md
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@ -9,7 +9,7 @@ linkTitle: "Skia Perf"
[Skia Perf](https://perf.skia.org) is a web application for analyzing and [Skia Perf](https://perf.skia.org) is a web application for analyzing and
viewing performance metrics produced by Skia's testing infrastructure. viewing performance metrics produced by Skia's testing infrastructure.
<img src=Perf.png style="margin-left:30px" align="left" width="800"/> <br clear="left"> <img src=../Perf.png style="margin-left:30px" align="left" width="800"/> <br clear="left">
Skia tests across a large number of platforms and configurations, and each Skia tests across a large number of platforms and configurations, and each
commit to Skia generates more than 400,000 individual values that are sent to commit to Skia generates more than 400,000 individual values that are sent to
@ -18,11 +18,11 @@ memory and coverage data.
Perf offers clustering, which is a tool to pick out trends and patterns in large sets of traces. Perf offers clustering, which is a tool to pick out trends and patterns in large sets of traces.
<img src=Cluster.png style="margin-left:30px" align="left" width="400"/> <br clear="left"> <img src=../Cluster.png style="margin-left:30px" align="left" width="400"/> <br clear="left">
And can generate alerts when those trends spot a regression: And can generate alerts when those trends spot a regression:
<img src=Regression.png style="margin-left:30px" align="left" width="800"/> <br clear="left"> <img src=../Regression.png style="margin-left:30px" align="left" width="800"/> <br clear="left">
## Calculations ## Calculations

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30
site/docs/dev/tools/debugger.md
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@ -20,7 +20,7 @@ Features:
- Android offscreen layer visualization - Android offscreen layer visualization
- Shared resource viewer - Shared resource viewer
<img src="/dev/tools/onlinedebugger.png" style="display: inline-block;" /> <img src="../onlinedebugger.png" style="display: inline-block;" />
## User Guide ## User Guide
@ -33,8 +33,8 @@ capture one from an android device using the
### Command Playback and Filters ### Command Playback and Filters
Try playing back the commands within the current frame using the lower play Try playing back the commands within the current frame using the lower play
button <img src="/dev/tools/playcommands.png" style="display: inline-block;" />, button <img src="../playcommands.png" style="display: inline-block;" />, (the
(the one not in a circle) You should see the image built up one draw at a time. one not in a circle) You should see the image built up one draw at a time.
Many commands manipulate the matrix or clip but don't make any visible change Many commands manipulate the matrix or clip but don't make any visible change
when run. Try filtering these out by pasting when run. Try filtering these out by pasting
@ -52,18 +52,18 @@ using `,` (comma) and `.` (period).
> at the beginning. > at the beginning.
Any command can be expanded using the Any command can be expanded using the
<img src="/dev/tools/expand.png" style="display: inline-block;" /> icon to see <img src="../expand.png" style="display: inline-block;" /> icon to see all of
all of the parameters that were recorded with that command. the parameters that were recorded with that command.
Commands can be disabled or enabled with the checkbox that becomes available Commands can be disabled or enabled with the checkbox that becomes available
after expanding the command's detail view. after expanding the command's detail view.
Jog the command playhead to the end of the list with the Jog the command playhead to the end of the list with the
<img src="/dev/tools/end.png" style="display: inline-block;" /> button. <img src="../end.png" style="display: inline-block;" /> button.
### Frame playback ### Frame playback
<img src="/dev/tools/frameplayback.png" style="display: inline-block;" /> <img src="../frameplayback.png" style="display: inline-block;" />
The sample file contains multiple frames. Use the encircled play button to play The sample file contains multiple frames. Use the encircled play button to play
back the frames. The current frame is indictated by the slider position, and the back the frames. The current frame is indictated by the slider position, and the
@ -77,7 +77,7 @@ the end of its list. If the command playhead is somewhere in the middle, say
### Resources Tab ### Resources Tab
<img src="/dev/tools/resources.png" style="display: inline-block;" /> <img src="../resources.png" style="display: inline-block;" />
Any resources that were referenced by commands in the file appear here. As of Any resources that were referenced by commands in the file appear here. As of
Dec 2019, this only shows images. Dec 2019, this only shows images.
@ -97,7 +97,7 @@ ids in the process that recorded the SKP.
### Android Layers ### Android Layers
<img src="/dev/tools/layers.png" style="display: inline-block;" /> <img src="../layers.png" style="display: inline-block;" />
When MSKPs are recorded in Android, Extra information about offscreen hardware When MSKPs are recorded in Android, Extra information about offscreen hardware
layers is recorded. The sample google calendar mskp linked above contains this layers is recorded. The sample google calendar mskp linked above contains this
@ -121,7 +121,7 @@ by clicking the `Exit` button on the layer box.
### Crosshair and Breakpoints ### Crosshair and Breakpoints
<img src="/dev/tools/crosshair.png" style="display: inline-block;" /> <img src="../crosshair.png" style="display: inline-block;" />
Clicking any point in the main view will toggle a red crosshair for selecting Clicking any point in the main view will toggle a red crosshair for selecting
pixels. the selected pixel's color is shown in several formats on the right pixels. the selected pixel's color is shown in several formats on the right
@ -135,12 +135,12 @@ command that draws something you see in the viewer.
### GPU Op Bounds and Other settings ### GPU Op Bounds and Other settings
<img src="/dev/tools/settings.png" style="display: inline-block;" /> <img src="../settings.png" style="display: inline-block;" />
Each of the filtered commands from above has a colored number to its right Each of the filtered commands from above has a colored number to its right
<img src="/dev/tools/gpuop.png" style="display: inline-block;" />. This is the <img src="../gpuop.png" style="display: inline-block;" />. This is the GPU
GPU operation id. When multiple commands share a GPU op id, this indicates that operation id. When multiple commands share a GPU op id, this indicates that they
they were batched together when sent to the GPU. In the WASM debugger, this goes were batched together when sent to the GPU. In the WASM debugger, this goes
though WebGL. though WebGL.
There is a "Display GPU Op Bounds" toggle in the upper right of the interface. There is a "Display GPU Op Bounds" toggle in the upper right of the interface.
@ -162,7 +162,7 @@ the pixel was drawn to more than once.
### Image fit and download buttons. ### Image fit and download buttons.
<img src="/dev/tools/settings.png" style="display: inline-block;" /> <img src="../settings.png" style="display: inline-block;" />
These buttons resize the main view. they are, from left to right: These buttons resize the main view. they are, from left to right:

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site/docs/dev/tools/tracing.md
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@ -35,13 +35,13 @@ clutter and slowdown in the interface), it's best to run a small number of tests
tracing. Once you have generated a file in this way, go to tracing. Once you have generated a file in this way, go to
[chrome://tracing](chrome://tracing), click Load: [chrome://tracing](chrome://tracing), click Load:
![Load Button](tracing_load.png) ![Load Button](../tracing_load.png)
... then select the JSON file. The data will be loaded and can be navigated/inspected using the ... then select the JSON file. The data will be loaded and can be navigated/inspected using the
tracing tools. Tip: press '?' for a help screen explaining the available keyboard and mouse tracing tools. Tip: press '?' for a help screen explaining the available keyboard and mouse
controls. controls.
![Tracing interface](tracing.png) ![Tracing interface](../tracing.png)
Android ATrace Android ATrace
-------------- --------------

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