Doc typos and formatting
Change-Id: I727681b43c4603471d7e3973a114a75006f70b79
No-Try: true
Reviewed-on: https://skia-review.googlesource.com/c/skia/+/540744
Reviewed-by: Robert Phillips <robertphillips@google.com>
Commit-Queue: James Godfrey-Kittle <jamesgk@google.com>
diff --git a/site/docs/dev/design/conical/_index.md b/site/docs/dev/design/conical/_index.md
index c5c6a98..d7ab05f 100644
--- a/site/docs/dev/design/conical/_index.md
+++ b/site/docs/dev/design/conical/_index.md
@@ -42,12 +42,12 @@
_positive_). If there are multiple (at most 2) solutions of $t$, choose the
bigger one.
-There are two degenerated cases:
+There are two degenerate cases:
1. $C_0 = C_1$ so the gradient is essentially a simple radial gradient.
2. $r_0 = r_1$ so the gradient is a single strip with bandwidth $2 r_0 = 2 r_1$.
-<!-- 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 degenerate 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
@@ -68,15 +68,15 @@
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 $x_t r_1$.
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 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 $C_1, r_1$).
+From now on, we'll focus on how to quickly compute $x_t$. Note that $r_t > 0$ so
+we're only interested in positive solutions for $x_t$. 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
@@ -88,13 +88,13 @@
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
+Case 2 always produces a valid $x_t$. Case 1 and 3 require $x > 0$ 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.
+_Proof._ Algebraically, solving the quadratic equation
+$(x_t - x)^2 + y^2 = (x_t r_1)^2$ and eliminating negative $x_t$ solutions gets
+us the theorem.
Alternatively, we can also combine Corollary 2., 3., and Lemma 4. in the
Appendix to geometrically prove the theorem. $\square$
@@ -120,7 +120,7 @@
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
+\hat y$ allows 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:
@@ -148,16 +148,14 @@
4. Let $\hat x = |1 - f|x', \hat y = |1 - f|y'$
5. If $r_1 = 1$, let $\hat x_t = (\hat x^2 + \hat y^2) / \hat x$
6. If $r_1 > 1$, let $\hat x_t = \sqrt{\hat x^2 + \hat y^2} - \hat x / r_1$
-7. If $r_1 < 1$
-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
- $r_0, r_1$, or if $1 - f < 0$
-
-10. let $\hat x_t = \sqrt{\hat x^2 - \hat y^2} - \hat x / r_1$ otherwise
-
-11. $t$ is invalid if $\hat x_t < 0$ (this check is unnecessary if $r_1 > 1$)
-12. Let $t = f + \text{sign}(1 - f) \hat x_t$
-13. If swapped, let $t = 1 - t$
+7. If $r_1 < 1$:
+ 1. return invalid if $\hat x^2 - \hat y^2 < 0$
+ 2. 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 $1 - f < 0$
+ 3. let $\hat x_t = \sqrt{\hat x^2 - \hat y^2} - \hat x / r_1$ otherwise
+8. $t$ is invalid if $\hat x_t < 0$ (this check is unnecessary if $r_1 > 1$)
+9. Let $t = f + \text{sign}(1 - f) \hat x_t$
+10. 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 $\hat x_t$.
@@ -192,18 +190,18 @@
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
+still 6 multiplications. Moreover, it sends in 4 uniforms 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, $x_t = || C_f P || / ||C_f P_1||$
+intersection point $P_1$ between that ray and circle $C_1 = (1, 0), r_1$, there
+exists an $x_t$ that equals the length of segment $C_f P$ over the 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 $C = (x', y')$.
+with the $x$-axis on point $C = (x', y')$.
<img src="./lemma1.svg"/>
@@ -243,16 +241,13 @@
<img src="./corollary2.3.2.svg" width="30%"/>
<img src="./corollary2.3.3.svg" width="30%"/>
-**Lemma 3.** When solution exists, one such solution is
-
+**Lemma 3.** When solutions 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 $||C_f P|| = \sqrt(x^2 + y^2)$.
+_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 $x \geq 0$:**
@@ -260,7 +255,7 @@
<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
+perpendicular to $C_f
P_1$. Triangle $\triangle C_1 H C_f$ is similar to triangle
$\triangle P X_P C_f$. Thus
$$||C_f H|| = ||C_f C_1|| \cdot (||C_f X_P|| / ||C_f P||) = x / \sqrt{x^2 + y^2}$$
@@ -295,21 +290,15 @@
$-x$ 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
@@ -323,14 +312,26 @@
_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}
+\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
@@ -363,6 +364,3 @@
As $r_1 \neq 1$, we can apply the similar transformation in Corollary 3. to get
the two formula in the lemma. $\square$
-
-$$
-$$
