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>
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James Godfrey-Kittle
SkCQ
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@ -42,12 +42,12 @@ $r_t = (1-t) \cdot r_0 + t \cdot r_1 > 0$ (note that radius $r_t$ has to be
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_positive_). If there are multiple (at most 2) solutions of $t$, choose the
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bigger one.
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There are two degenerated cases:
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There are two degenerate cases:
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1. $C_0 = C_1$ so the gradient is essentially a simple radial gradient.
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2. $r_0 = r_1$ so the gradient is a single strip with bandwidth $2 r_0 = 2 r_1$.
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<!-- TODO maybe add some fiddle or images here to illustrate the two degenerated cases -->
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<!-- TODO maybe add some fiddle or images here to illustrate the two degenerate cases -->
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They are easy to handle so we won't cover them here. From now on, we assume
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$C_0 \neq C_1$ and $r_0
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@ -68,15 +68,15 @@ Assuming that we've done swapping if necessary so $C_1 \neq C_f$, we can then do
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a linear transformation to map $C_f, C_1$ to $(0, 0), (1, 0)$. After the
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transformation:
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1. All centers $C_t = (x_t, 0)$ must be on the $x$ axis
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1. All centers $C_t = (x_t, 0)$ must be on the $x$-axis
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2. The radius $r_t$ is $x_t r_1$.
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3. Given $x_t$ , we can derive $t = f + (1 - f) x_t$
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From now on, we'll focus on how to quickly computes $x_t$. Note that $r_t > 0$
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so we're only interested positive solution $x_t$. Again, if there are multiple
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$x_t$ solutions, we may want to find the bigger one if $1 - f > 0$, and smaller
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one if $1 - f < 0$, so the corresponding $t$ is always the bigger one (note that
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$f \neq 1$, otherwise we'll swap $C_0, r_0$ with $C_1, r_1$).
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From now on, we'll focus on how to quickly compute $x_t$. Note that $r_t > 0$ so
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we're only interested in positive solutions for $x_t$. Again, if there are
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multiple $x_t$ solutions, we may want to find the bigger one if $1 - f > 0$, and
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smaller one if $1 - f < 0$, so the corresponding $t$ is always the bigger one
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(note that $f \neq 1$, otherwise we'll swap $C_0, r_0$ with $C_1, r_1$).
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## Algorithm
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@ -88,13 +88,13 @@ $f \neq 1$, otherwise we'll swap $C_0, r_0$ with $C_1, r_1$).
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3. $\left(\pm \sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$ if
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$r_1 < 1$.
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Case 2 always produces a valid $x_t$. Case 1 and 3 requires $x > 0$ to produce
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Case 2 always produces a valid $x_t$. Case 1 and 3 require $x > 0$ to produce
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valid $x_t > 0$. Case 3 may have no solution at all if
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$(r_1^2 - 1) y^2 + r_1^2 x^2 < 0$.
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_Proof._ Algebriacally, solving the quadratic equation
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$(x_t - x)^2 + y^2 = (x_t r_1)^2$ and eliminate negative $x_t$ solutions get us
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the theorem.
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_Proof._ Algebraically, solving the quadratic equation
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$(x_t - x)^2 + y^2 = (x_t r_1)^2$ and eliminating negative $x_t$ solutions gets
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us the theorem.
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Alternatively, we can also combine Corollary 2., 3., and Lemma 4. in the
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Appendix to geometrically prove the theorem. $\square$
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@ -120,7 +120,7 @@ $(0, 0), (1, 0)$).
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For example, let $\hat x, \hat y = |1-f|x, |1-f|y$. Computing $\hat x_t$ with
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respect to $\hat x,
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\hat y$ allow us to have
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\hat y$ allows us to have
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$t = f + (1 - f)x_t = f + \text{sign}(1-f) \cdot \hat x_t$. That saves us one
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multiplication. Applying similar techniques to Theorem 1 gets us:
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@ -148,16 +148,14 @@ algorithm:
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4. Let $\hat x = |1 - f|x', \hat y = |1 - f|y'$
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5. If $r_1 = 1$, let $\hat x_t = (\hat x^2 + \hat y^2) / \hat x$
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6. If $r_1 > 1$, let $\hat x_t = \sqrt{\hat x^2 + \hat y^2} - \hat x / r_1$
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7. If $r_1 < 1$
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8. return invalid if $\hat x^2 - \hat y^2 < 0$
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9. let $\hat x_t = -\sqrt{\hat x^2 - \hat y^2} - \hat x / r_1$ if we've swapped
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$r_0, r_1$, or if $1 - f < 0$
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10. let $\hat x_t = \sqrt{\hat x^2 - \hat y^2} - \hat x / r_1$ otherwise
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11. $t$ is invalid if $\hat x_t < 0$ (this check is unnecessary if $r_1 > 1$)
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12. Let $t = f + \text{sign}(1 - f) \hat x_t$
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13. If swapped, let $t = 1 - t$
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7. If $r_1 < 1$:
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1. return invalid if $\hat x^2 - \hat y^2 < 0$
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2. let $\hat x_t = -\sqrt{\hat x^2 - \hat y^2} - \hat x / r_1$ if we've
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swapped $r_0, r_1$, or if $1 - f < 0$
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3. let $\hat x_t = \sqrt{\hat x^2 - \hat y^2} - \hat x / r_1$ otherwise
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8. $t$ is invalid if $\hat x_t < 0$ (this check is unnecessary if $r_1 > 1$)
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9. Let $t = f + \text{sign}(1 - f) \hat x_t$
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10. If swapped, let $t = 1 - t$
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In step 7, we try to select either the smaller or bigger $\hat x_t$ based on
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whether the final $t$ has a negative or positive relationship with $\hat x_t$.
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@ -192,18 +190,18 @@ final $t$ costs 5 more multiplications, 1 more sqrt, and 2 more additions.
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That's a total of 5 additions, 10 multiplications, and 1 sqrt. (Our algorithm
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has 2-4 additions, 3 multiplications, and 1 sqrt.) Even if it saves the
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$0.5 \cdot (1/a), 4a, r_0^2$ and $(r_1 - r_0) r_0$ multiplications, there are
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still 6 multiplications. Moreover, it sends in 4 unitofmrs to the shader while
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still 6 multiplications. Moreover, it sends in 4 uniforms to the shader while
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our algorithm only needs 2 uniforms ($1/r_1$ and $f$).
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## Appendix
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**Lemma 1.** Draw a ray from $C_f = (0, 0)$ to $P = (x, y)$. For every
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intersection points $P_1$ between that ray and circle $C_1 = (1, 0), r_1$, there
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exists an $x_t$ that equals to the length of segment $C_f P$ over length of
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segment $C_f P_1$. That is, $x_t = || C_f P || / ||C_f P_1||$
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intersection point $P_1$ between that ray and circle $C_1 = (1, 0), r_1$, there
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exists an $x_t$ that equals the length of segment $C_f P$ over the length of
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segment $C_f P_1$. That is, $x_t = || C_f P || / ||C_f P_1||$.
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_Proof._ Draw a line from $P$ that's parallel to $C_1 P_1$. Let it intersect
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with $x$-axis on point $C = (x', y')$.
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with the $x$-axis on point $C = (x', y')$.
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<img src="./lemma1.svg"/>
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@ -243,16 +241,13 @@ $C_1, r_1$. Therefore
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<img src="./corollary2.3.2.svg" width="30%"/>
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<img src="./corollary2.3.3.svg" width="30%"/>
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**Lemma 3.** When solution exists, one such solution is
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**Lemma 3.** When solutions exists, one such solution is
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$$
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x_t = {|| C_f P || \over ||C_f P_1||} = \frac{x^2 + y^2}{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}
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$$
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_Proof._ As $C_f = (0, 0), P = (x, y)$, we have $||C_f P|| = \sqrt(x^2 + y^2)$.
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_Proof._ As $C_f = (0, 0), P = (x, y)$, we have $||C_f P|| = \sqrt{x^2 + y^2}$.
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So we'll mainly focus on how to compute $||C_f P_1||$.
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**When $x \geq 0$:**
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@ -260,7 +255,7 @@ So we'll mainly focus on how to compute $||C_f P_1||$.
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<img src="./lemma3.1.svg"/>
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Let $X_P = (x, 0)$ and $H$ be a point on $C_f P_1$ such that $C_1 H$ is
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perpendicular to $C_1
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perpendicular to $C_f
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P_1$. Triangle $\triangle C_1 H C_f$ is similar to triangle
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$\triangle P X_P C_f$. Thus
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$$||C_f H|| = ||C_f C_1|| \cdot (||C_f X_P|| / ||C_f P||) = x / \sqrt{x^2 + y^2}$$
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@ -295,21 +290,15 @@ $-||C_f H||$ instead of $||C_f H||$. That negation cancels out the negation of
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$-x$ so we get the same equation of $||C_f P_1||$ for both $x \geq 0$ and
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$x < 0$ cases:
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$$
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||C_f P_1|| = \frac{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}{\sqrt{x^2 + y^2}}
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$$
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Finally
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$$
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x_t = \frac{||C_f P||}{||C_f P_1||} = \frac{\sqrt{x^2 + y^2}}{||C_f P_1||}
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= \frac{x^2 + y^2}{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}
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$$ $\square$
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**Corollary 2.** If $r_1 = 1$, then the solution
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@ -323,14 +312,26 @@ $x_t = \left(\sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$.
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_Proof._ From Lemma 3., we have
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\begin{align} x_t &= \frac{x^2 + y^2}{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}
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\\\\\\ &= { (x^2 + y^2) \left ( -x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2} \right )
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\over \left (x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2} \right ) \left (-x +
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\sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2} \right ) } \\\\\\ &= { (x^2 + y^2) \left (
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-x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2} \right ) \over -x^2 + (r_1^2 - 1) y^2 +
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r_1^2 x^2 } \\\\\\ &= { (x^2 + y^2) \left ( -x + \sqrt{(r_1^2 - 1) y^2 + r_1^2
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x^2} \right ) \over (r_1^2 - 1) (x^2 + y^2) } \\\\\\ &= \left(\sqrt{(r_1^2 - 1)
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y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1) \end{align}
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\begin{align}
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x_t &= \frac{x^2 + y^2}{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}} \\\\\\
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&= {
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(x^2 + y^2) \left ( -x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2} \right )
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\over
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\left (x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2} \right )
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\left (-x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2} \right )
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} \\\\\\
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&= {
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(x^2 + y^2) \left ( -x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2} \right )
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\over
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-x^2 + (r_1^2 - 1) y^2 + r_1^2 x^2
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} \\\\\\
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&= {
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(x^2 + y^2) \left ( -x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2} \right )
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\over
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(r_1^2 - 1) (x^2 + y^2)
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} \\\\\\
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&= \left(\sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)
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\end{align}
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The transformation above (multiplying $-x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}$
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to enumerator and denomenator) is always valid because $r_1 > 1$ and it's the
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@ -363,6 +364,3 @@ $\frac{x^2 + y^2}{x - \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}$.
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As $r_1 \neq 1$, we can apply the similar transformation in Corollary 3. to get
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the two formula in the lemma. $\square$
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$$
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$$
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