[docs] Add back in all images and fix broken image links.
Change-Id: I0ffd838880746dbfe534ea615277fc24cd26b7ec No-Try: true Docs-Preview: https://skia.org/docs/dev/design/conical/corollary2.3.3?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/Digests?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/Status?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/Cluster?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/Perf?cl=392736 Docs-Preview: https://skia.org/docs/dev/design/conical/corollary2.3.2?cl=392736 Docs-Preview: https://skia.org/docs/dev/design/conical/lemma4?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/IssueHighlight?cl=392736 Docs-Preview: https://skia.org/docs/dev/design/conical/lemma3.1?cl=392736 Docs-Preview: https://skia.org/docs/dev/tools/tracing?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/BlameView?cl=392736 Docs-Preview: https://skia.org/docs/dev/design/conical/corollary2.2.2?cl=392736 Docs-Preview: https://skia.org/docs/dev/design/conical/?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/download?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/Isolate?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/Search?cl=392736 Docs-Preview: https://skia.org/docs/dev/tools/end?cl=392736 Docs-Preview: https://skia.org/docs/dev/tools/image?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/Grid?cl=392736 Docs-Preview: https://skia.org/docs/dev/design/conical/corollary2.2.1?cl=392736 Docs-Preview: https://skia.org/docs/dev/tools/debugger?cl=392736 Docs-Preview: https://skia.org/docs/dev/design/pdftheory?cl=392736 Docs-Preview: https://skia.org/docs/dev/tools/resources?cl=392736 Docs-Preview: https://skia.org/docs/dev/tools/playcommands?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/DotDiagram?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/Ignores?cl=392736 Docs-Preview: https://skia.org/docs/dev/tools/onlinedebugger?cl=392736 Docs-Preview: https://skia.org/docs/dev/tools/crosshair?cl=392736 Docs-Preview: https://skia.org/docs/dev/tools/frameplayback?cl=392736 Docs-Preview: https://skia.org/docs/dev/design/conical/corollary2.3.1?cl=392736 Docs-Preview: https://skia.org/docs/dev/tools/buttons?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/ClusterConfig?cl=392736 Docs-Preview: https://skia.org/docs/dev/design/conical/lemma3.2?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/skiaperf?cl=392736 Docs-Preview: https://skia.org/docs/dev/contrib/SuggestedReviewers?cl=392736 Docs-Preview: https://skia.org/docs/user/modules/PathKit_effects?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/Regression?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/ByTest?cl=392736 Docs-Preview: https://skia.org/docs/dev/tools/settings?cl=392736 Docs-Preview: https://skia.org/docs/dev/design/conical/lemma1?cl=392736 Docs-Preview: https://skia.org/docs/dev/design/PdfLogicalDocumentStructure?cl=392736 Docs-Preview: https://skia.org/docs/dev/testing/skiagold?cl=392736 Docs-Preview: https://skia.org/docs/dev/tools/expand?cl=392736 Docs-Preview: https://skia.org/docs/dev/tools/tracing_load?cl=392736 Docs-Preview: https://skia.org/docs/dev/tools/layers?cl=392736 Docs-Preview: https://skia.org/docs/dev/tools/gpuop?cl=392736 Reviewed-on: https://skia-review.googlesource.com/c/skia/+/392736 Reviewed-by: Joe Gregorio <jcgregorio@google.com>
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@ -15,11 +15,12 @@ MathJax.Hub.Config({
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(Please refresh the page if you see a lot of dollars instead of math symbols.)
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We present a fast shading algorithm (compared to bruteforcely solving the quadratic equation of
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gradient $t$) for computing the two-point conical gradient (i.e., `createRadialGradient` in
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We present a fast shading algorithm (compared to bruteforcely solving the
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quadratic equation of gradient $t$) for computing the two-point conical gradient
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(i.e., `createRadialGradient` in
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[spec](https://html.spec.whatwg.org/multipage/canvas.html#dom-context-2d-createradialgradient)).
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It reduced the number of multiplications per pixel from ~10 down to 3, and brought a speedup of up to
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26% in our nanobenches.
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It reduced the number of multiplications per pixel from ~10 down to 3, and
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brought a speedup of up to 26% in our nanobenches.
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This document has 3 parts:
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@ -27,16 +28,19 @@ This document has 3 parts:
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2. [Algorithm](#algorithm)
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3. [Appendix](#appendix)
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Part 1 and 2 are self-explanatory. Part 3 shows how to geometrically proves our Theorem 1 in part
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2; it's more complicated but it gives us a nice picture about what's going on.
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Part 1 and 2 are self-explanatory. Part 3 shows how to geometrically proves our
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Theorem 1 in part 2; it's more complicated but it gives us a nice picture about
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what's going on.
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## <span id="problem-statement">Problem Statement and Setup</span>
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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
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point $P = (x, y)$ we want the shader to quickly compute a gradient $t \in \mathbb R$ such that $p$
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is on the linearly interpolated circle with center $C_t = (1-t) \cdot C_0 + t \cdot C_1$ and radius
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$r_t = (1-t) \cdot r_0 + t \cdot r_1 > 0$ (note that radius $r_t$ has to be _positive_). If
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there are multiple (at most 2) solutions of $t$, choose the bigger one.
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Let two circles be $C_0, r_0$ and $C_1, r_1$ where $C$ is the center and $r$ is
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the radius. For any point $P = (x, y)$ we want the shader to quickly compute a
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gradient $t \in \mathbb R$ such that $p$ is on the linearly interpolated circle
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with center $C_t = (1-t) \cdot C_0 + t \cdot C_1$ and radius
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$r_t = (1-t) \cdot r_0 + t \cdot r_1 > 0$ (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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@ -45,90 +49,109 @@ There are two degenerated cases:
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<!-- TODO maybe add some fiddle or images here to illustrate the two degenerated cases -->
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They are easy to handle so we won't cover them here. From now on, we assume $C_0 \neq C_1$ and $r_0
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They are easy to handle so we won't cover them here. From now on, we assume
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$C_0 \neq C_1$ and $r_0
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\neq r_1$.
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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
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corresponding linearly interpolated radius $r_f = (1-f) \cdot r_0 + f \cdot r_1 = 0$.
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Solving the latter equation gets us $f = r_0 / (r_0 - r_1)$.
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As $r_0 \neq r_1$, we can find a focal point
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$C_f = (1-f) \cdot C_0 + f \cdot C_1$ where its corresponding linearly
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interpolated radius $r_f = (1-f) \cdot r_0 + f \cdot r_1 = 0$. Solving the
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latter equation gets us $f = r_0 / (r_0 - r_1)$.
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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
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swap $C_0, r_0$ with $C_1, r_1$, compute swapped gradient $t_s$ as if $r_1 \neq 0$, and finally set
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$t = 1 - t_s$. The only catch here is that with multiple solutions of $t_s$, we shall choose the
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smaller one (so $t$ could be the bigger one).
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As $C_0 \neq C_1$, focal point $C_f$ is different from $C_1$ unless $r_1 = 0$.
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If $r_1 = 0$, we can swap $C_0, r_0$ with $C_1, r_1$, compute swapped gradient
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$t_s$ as if $r_1 \neq 0$, and finally set $t = 1 - t_s$. The only catch here is
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that with multiple solutions of $t_s$, we shall choose the smaller one (so $t$
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could be the bigger one).
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Assuming that we've done swapping if necessary so $C_1 \neq C_f$, we can then do a linear
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transformation to map $C_f, C_1$ to $(0, 0), (1, 0)$. After the transformation:
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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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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 $r_t > 0$ so we're only
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interested positive solution $x_t$. Again, if there are multiple $x_t$ solutions, we may want to
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find the bigger one if $1 - f > 0$, and smaller one if $1 - f < 0$, so the corresponding $t$ is
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always the bigger one (note that $f \neq 1$, 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 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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## <span id="algorithm">Algorithm</span>
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**Theorem 1.** The solution to $x_t$ is
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1. $\frac{x^2 + y^2}{(1 + r_1) x} = \frac{x^2 + y^2}{2 x}$ if $r_1 = 1$
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2. $\left(\sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$ if $r_1 > 1$
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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 $r_1 < 1$.
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2. $\left(\sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$ if
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$r_1 > 1$
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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 valid $x_t > 0$. Case
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3 may have no solution at all if $(r_1^2 - 1) y^2 + r_1^2 x^2 < 0$.
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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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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 $(x_t - x)^2 + y^2 = (x_t r_1)^2$ and
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eliminate negative $x_t$ solutions get us the theorem.
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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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Alternatively, we can also combine Corollary 2., 3., and Lemma 4. in the Appendix to geometrically
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prove the theorem. $\square$
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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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Theorem 1 by itself is not sufficient for our shader algorithm because:
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1. we still need to compute $t$ from $x_t$ (remember that $t = f + (1-f) x_t$);
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2. we still need to handle cases of choosing the bigger/smaller $x_t$;
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3. we still need to handle the swapped case (we swap $C_0, r_0$ with $C_1, r_1$ if $r_1 = 0$);
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4. there are way too many multiplications and divisions in Theorem 1 that would slow our shader.
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3. we still need to handle the swapped case (we swap $C_0, r_0$ with $C_1, r_1$
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if $r_1 = 0$);
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4. there are way too many multiplications and divisions in Theorem 1 that would
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slow our shader.
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Issue 2 and 3 are solved by generating different shader code based on different situations. So they
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are mainly correctness issues rather than performance issues. Issue 1 and 4 are performance
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critical, and they will affect how we handle issue 2 and 3.
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Issue 2 and 3 are solved by generating different shader code based on different
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situations. So they are mainly correctness issues rather than performance
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issues. Issue 1 and 4 are performance critical, and they will affect how we
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handle issue 2 and 3.
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The key to handle 1 and 4 efficiently is to fold as many multiplications and divisions into the
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linear transformation matrix, which the shader has to do anyway (remember our linear transformation
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to map $C_f, C_1$ to $(0, 0), (1, 0)$).
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The key to handle 1 and 4 efficiently is to fold as many multiplications and
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divisions into the linear transformation matrix, which the shader has to do
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anyway (remember our linear transformation to map $C_f, C_1$ to
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$(0, 0), (1, 0)$).
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For example, let $\hat x, \hat y = |1-f|x, |1-f|y$. Computing $\hat x_t$ with respect to $\hat x,
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\hat y$ allow us to have $t = f + (1 - f)x_t = f + \text{sign}(1-f) \cdot \hat x_t$. That saves us
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one multiplication. Applying similar techniques to Theorem 1 gets us:
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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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$t = f + (1 - f)x_t = f + \text{sign}(1-f) \cdot \hat x_t$. That saves us one
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multiplication. Applying similar techniques to Theorem 1 gets us:
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1. If $r_1 = 1$, let $x' = x/2,~ y' = y/2$, then $x_t = (x'^2 + y'^2) / x'$.
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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
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2. If $r_1 > 1$, let
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$x' = r_1 / (r_1^2 - 1) x,~ y' = \frac{\sqrt{r_1^2 - 1}}{r_1^2 - 1} y$, then
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$x_t = \sqrt{x'^2 + y'^2} - x' / r_1$
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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
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3. If $r_1 < 1$, let
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$x' = r_1 / (r_1^2 - 1) x,~ y' = \frac{\sqrt{1 - r_1^2}}{r_1^2 - 1} y$, then
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$x_t = \pm\sqrt{x'^2 - y'^2} - x' / r_1$
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Combining it with the swapping, the equation $t = f + (1-f) x_t$, and the fact that we only want
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positive $x_t > 0$ and bigger $t$, we have our final algorithm:
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Combining it with the swapping, the equation $t = f + (1-f) x_t$, and the fact
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that we only want positive $x_t > 0$ and bigger $t$, we have our final
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algorithm:
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**Algorithm 1.**
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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,
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1. Let $C'_0, r'_0, C'_1, r'_1 = C_0, r_0, C_1, r_1$ if there is no swapping and
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$C'_0,
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r'_0, C'_1, r'_1 = C_1, r_1, C_0, r_0$ if there is swapping.
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2. Let $f = r'_0 / (r'_0 - r'_1)$ and $1 - f = r'_1 / (r'_1 - r'_0)$
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3. Let $x' = x/2,~ y' = y/2$ if $r_1 = 1$, and
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$x' = r_1 / (r_1^2 - 1) x,~ y' = \sqrt{|r_1^2 - 1|} / (r_1^2 - 1) y$ if $r_1 \neq 1$
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$x' = r_1 / (r_1^2 - 1) x,~ y' = \sqrt{|r_1^2 - 1|} / (r_1^2 - 1) y$ if
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$r_1 \neq 1$
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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 $r_1 > 1$,
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let $\hat x_t = \sqrt{\hat x^2 + \hat y^2} - \hat x / r_1$
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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 $r_0, r_1$,
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or if $1 - f < 0$
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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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@ -136,74 +159,87 @@ positive $x_t > 0$ and bigger $t$, we have our final algorithm:
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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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In step 7, we try to select either the smaller or bigger $\hat x_t$ based on whether the final $t$
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has a negative or positive relationship with $\hat x_t$. It's negative if we've swapped, or if
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$\text{sign}(1 - f)$ is negative (these two cannot both happen).
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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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It's negative if we've swapped, or if $\text{sign}(1 - f)$ is negative (these
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two cannot both happen).
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Note that all the computations and if decisions not involving $\hat x, \hat y$ can be precomputed
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before the shading stage. The two if decisions $\hat x^2 - \hat y^2 < 0$ and $\hat x^t < 0$ can
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also be omitted by precomputing the shading area that never violates those conditions.
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Note that all the computations and if decisions not involving $\hat x, \hat y$
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can be precomputed before the shading stage. The two if decisions
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$\hat x^2 - \hat y^2 < 0$ and $\hat x^t < 0$ can also be omitted by precomputing
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the shading area that never violates those conditions.
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The number of operations per shading is thus:
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- 1 addition, 2 multiplications, and 1 division if $r_1 = 1$
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- 2 additions, 3 multiplications, and 1 sqrt for $r_1 \neq 1$ (count subtraction as addition;
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dividing $r_1$ is multiplying $1/r_1$)
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- 2 additions, 3 multiplications, and 1 sqrt for $r_1 \neq 1$ (count subtraction
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as addition; dividing $r_1$ is multiplying $1/r_1$)
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- 1 more addition operation if $f \neq 0$
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- 1 more addition operation if swapped.
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In comparison, for $r_1 \neq 1$ case, our current raster pipeline shading algorithm (which shall
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hopefully soon be upgraded to the algorithm described here) mainly uses formula $$t = 0.5 \cdot
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(1/a) \cdot \left(-b \pm \sqrt{b^2 - 4ac}\right)$$ It precomputes $a = 1 - (r_1 - r_0)^2, 1/a, r1 -
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r0$. Number $b = -2 \cdot (x + (r1 - r0) \cdot r0)$ costs 2 multiplications and 1 addition. Number
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$c = x^2 + y^2 - r_0^2$ costs 3 multiplications and 2 additions. And the final $t$ costs 5 more
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multiplications, 1 more sqrt, and 2 more additions. That's a total of 5 additions, 10
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multiplications, and 1 sqrt. (Our algorithm has 2-4 additions, 3 multiplications, and 1 sqrt.) Even
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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 $f$).
|
||||
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 $f$).
|
||||
|
||||
## <span id="appendix">Appendix</span>
|
||||
|
||||
**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 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
|
||||
$C = (x', y')$.
|
||||
_Proof._ Draw a line from $P$ that's parallel to $C_1 P_1$. Let it intersect
|
||||
with $x$-axis on point $C = (x', y')$.
|
||||
|
||||
<img src="conical/lemma1.svg"/>
|
||||
<img src="./lemma1.svg"/>
|
||||
|
||||
Triangle $\triangle C_f C P$ is similar to triangle $\triangle C_f C_1 P_1$.
|
||||
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, $x'
|
||||
= ||C_f C_1|| \cdot (||C_f P|| / ||C_f P_1||) = ||C_f P|| / ||C_f P_1||$. $\square$
|
||||
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 $C_1, r_1$.
|
||||
**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 $C_1, r_1$. $\square$
|
||||
_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
|
||||
**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'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;
|
||||
|
||||
<img src="conical/corollary2.2.1.svg"/>
|
||||
<img src="conical/corollary2.2.2.svg"/>
|
||||
<img src="./corollary2.2.1.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"
|
||||
case.
|
||||
- when $r_1 < 1$, there may be $0, 1$, or $2$ solutions; this was also
|
||||
previously as the "outside" case.
|
||||
|
||||
<img src="conical/corollary2.3.1.svg" width="30%"/>
|
||||
<img src="conical/corollary2.3.2.svg" width="30%"/>
|
||||
<img src="conical/corollary2.3.3.svg" width="30%"/>
|
||||
<img src="./corollary2.3.1.svg" width="30%"/>
|
||||
<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
|
||||
|
||||
@ -211,47 +247,48 @@ $$
|
||||
x_t = {|| C_f P || \over ||C_f P_1||} = \frac{x^2 + y^2}{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}
|
||||
$$
|
||||
|
||||
_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 $||C_f P_1||$.
|
||||
_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 $x \geq 0$:**
|
||||
|
||||
<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 $C_1
|
||||
P_1$. Triangle $\triangle C_1 H C_f$ is similar to triangle $\triangle P X_P C_f$. Thus
|
||||
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}
|
||||
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
|
||||
$C_f P_1$.
|
||||
Define $X_P$ and $H$ similarly as before except that now $H$ is on ray $P_1 C_f$
|
||||
instead of $C_f P_1$.
|
||||
|
||||
<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
|
||||
$\triangle C_1 H P_1$ is a right triangle, so we have
|
||||
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:
|
||||
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}}
|
||||
@ -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.
|
||||
|
||||
<img src="conical/lemma4.svg"/>
|
||||
<img src="./lemma4.svg"/>
|
||||
|
||||
For case 2, we notice that $||C_f P_1||$ could be
|
||||
|
||||
|
||||
233
site/docs/dev/design/conical/corollary2.2.1.svg
Normal file
233
site/docs/dev/design/conical/corollary2.2.1.svg
Normal file
@ -0,0 +1,233 @@
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|
||||
</g> <!-- drawing style -->
|
||||
</g> <!-- transform -->
|
||||
<g transform="matrix(0.8021505862642581, 0, 0, 0.8021505862642581, 0, 0)">
|
||||
<g stroke-linejoin="round" stroke-linecap="round" fill="none" stroke-opacity="1" stroke="#000000">
|
||||
<path d="M 152 131 C 152 132.65685424949237 150.65685424949237 134 149 134 C 147.34314575050763 134 146 132.65685424949237 146 131 C 146 129.34314575050763 147.34314575050763 128 149 128 C 150.65685424949237 128 152 129.34314575050763 152 131 z"/>
|
||||
</g> <!-- drawing style -->
|
||||
</g> <!-- transform -->
|
||||
<g transform="matrix(0.8021505862642581, 0, 0, 0.8021505862642581, 0, 0)">
|
||||
<g fill-opacity="1" fill-rule="nonzero" stroke="none" fill="#404040">
|
||||
<path d="M 154.171875 116.25 L 155.359375 116.25 L 155.359375 119.84375 L 159.65625 119.84375 L 159.65625 116.25 L 160.84375 116.25 L 160.84375 125 L 159.65625 125 L 159.65625 120.828125 L 155.359375 120.828125 L 155.359375 125 L 154.171875 125 L 154.171875 116.25 z"/>
|
||||
</g> <!-- drawing style -->
|
||||
</g> <!-- transform -->
|
||||
<g transform="matrix(0.8021505862642581, 0, 0, 0.8021505862642581, 0, 0)">
|
||||
<g fill-opacity="1" fill-rule="nonzero" stroke="none" fill="#404040">
|
||||
<path d="M 235 198 C 235 199.65685424949237 233.65685424949237 201 232 201 C 230.34314575050763 201 229 199.65685424949237 229 198 C 229 196.34314575050763 230.34314575050763 195 232 195 C 233.65685424949237 195 235 196.34314575050763 235 198 z"/>
|
||||
</g> <!-- drawing style -->
|
||||
</g> <!-- transform -->
|
||||
<g transform="matrix(0.8021505862642581, 0, 0, 0.8021505862642581, 0, 0)">
|
||||
<g stroke-linejoin="round" stroke-linecap="round" fill="none" stroke-opacity="1" stroke="#000000">
|
||||
<path d="M 235 198 C 235 199.65685424949237 233.65685424949237 201 232 201 C 230.34314575050763 201 229 199.65685424949237 229 198 C 229 196.34314575050763 230.34314575050763 195 232 195 C 233.65685424949237 195 235 196.34314575050763 235 198 z"/>
|
||||
</g> <!-- drawing style -->
|
||||
</g> <!-- transform -->
|
||||
<g transform="matrix(0.8021505862642581, 0, 0, 0.8021505862642581, 0, 0)">
|
||||
<g fill-opacity="1" fill-rule="nonzero" stroke="none" fill="#404040">
|
||||
<path d="M 236.75 183.25 L 238.03125 183.25 L 240.203125 186.5 L 242.390625 183.25 L 243.65625 183.25 L 240.84375 187.453125 L 243.84375 192 L 242.578125 192 L 240.109375 188.28125 L 237.640625 192 L 236.359375 192 L 239.484375 187.328125 L 236.75 183.25 z"/>
|
||||
</g> <!-- drawing style -->
|
||||
</g> <!-- transform -->
|
||||
<g transform="matrix(0.8021505862642581, 0, 0, 0.8021505862642581, 0, 0)">
|
||||
<g fill-opacity="1" fill-rule="nonzero" stroke="none" fill="#404040">
|
||||
<path d="M 246.189453125 190.515625 L 246.189453125 193.265625 L 247.423828125 193.265625 Q 248.111328125 193.265625 248.486328125 192.90625 Q 248.876953125 192.546875 248.876953125 191.890625 Q 248.876953125 191.234375 248.486328125 190.875 Q 248.111328125 190.515625 247.423828125 190.515625 L 246.189453125 190.515625 z M 245.205078125 189.703125 L 247.423828125 189.703125 Q 248.658203125 189.703125 249.283203125 190.265625 Q 249.908203125 190.8125 249.908203125 191.890625 Q 249.908203125 192.96875 249.283203125 193.515625 Q 248.658203125 194.0625 247.423828125 194.0625 L 246.189453125 194.0625 L 246.189453125 197 L 245.205078125 197 L 245.205078125 189.703125 z"/>
|
||||
</g> <!-- drawing style -->
|
||||
</g> <!-- transform -->
|
||||
</g><!-- layer0 -->
|
||||
</g> <!-- default stroke -->
|
||||
</svg> <!-- bounding box -->
|
||||
|
After Width: | Height: | Size: 38 KiB |
@ -49,7 +49,7 @@ serializing the PDF file.
|
||||
|
||||
## <span id="PDF_Objects_and_Document_Structure">PDF Objects and Document Structure</span>
|
||||
|
||||
|
||||
|
||||
|
||||
**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
|
||||
|
||||
BIN
site/docs/dev/testing/BlameView.png
Normal file
BIN
site/docs/dev/testing/BlameView.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 188 KiB |
BIN
site/docs/dev/testing/ByTest.png
Normal file
BIN
site/docs/dev/testing/ByTest.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 199 KiB |
BIN
site/docs/dev/testing/Cluster.png
Normal file
BIN
site/docs/dev/testing/Cluster.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 58 KiB |
BIN
site/docs/dev/testing/ClusterConfig.png
Normal file
BIN
site/docs/dev/testing/ClusterConfig.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 213 KiB |
BIN
site/docs/dev/testing/Digests.png
Normal file
BIN
site/docs/dev/testing/Digests.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 408 KiB |
BIN
site/docs/dev/testing/DotDiagram.png
Normal file
BIN
site/docs/dev/testing/DotDiagram.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 312 KiB |
BIN
site/docs/dev/testing/Grid.png
Normal file
BIN
site/docs/dev/testing/Grid.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 321 KiB |
BIN
site/docs/dev/testing/Ignores.png
Normal file
BIN
site/docs/dev/testing/Ignores.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 254 KiB |
BIN
site/docs/dev/testing/Isolate.png
Normal file
BIN
site/docs/dev/testing/Isolate.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 30 KiB |
BIN
site/docs/dev/testing/IssueHighlight.png
Normal file
BIN
site/docs/dev/testing/IssueHighlight.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 189 KiB |
BIN
site/docs/dev/testing/Perf.png
Normal file
BIN
site/docs/dev/testing/Perf.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 169 KiB |
BIN
site/docs/dev/testing/Regression.png
Normal file
BIN
site/docs/dev/testing/Regression.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 64 KiB |
BIN
site/docs/dev/testing/Search.png
Normal file
BIN
site/docs/dev/testing/Search.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 206 KiB |
BIN
site/docs/dev/testing/Status.png
Normal file
BIN
site/docs/dev/testing/Status.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 43 KiB |
@ -19,13 +19,13 @@ Add the checkout location to your $PATH.
|
||||
To download the isolated files for a test first visit
|
||||
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:
|
||||
|
||||
|
||||
<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:
|
||||
|
||||
|
||||
@ -48,14 +48,14 @@ Solution today:
|
||||
- Blame is not sorted in any particular order
|
||||
- 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">
|
||||
|
||||
- Select digests for triage
|
||||
- Digests will be listed in order with largest difference first
|
||||
- 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">
|
||||
|
||||
- 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
|
||||
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>
|
||||
@ -88,7 +88,7 @@ To find your results:
|
||||
- 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.
|
||||
|
||||
<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">
|
||||
|
||||
To rebaseline images:
|
||||
@ -96,7 +96,7 @@ To rebaseline images:
|
||||
- Access the Ignores view and create a new, short-interval (hours) ignore for
|
||||
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">
|
||||
|
||||
- Click on the Ignore to bring up a search view filtered by the affected
|
||||
@ -141,7 +141,7 @@ Solution:
|
||||
|
||||
- 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">
|
||||
|
||||
- Click the magnifier to filter by configuration
|
||||
@ -150,12 +150,12 @@ Solution:
|
||||
- Click on configurations under “parameters” to highlight data points and
|
||||
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">
|
||||
|
||||
- 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">
|
||||
|
||||
- Access the Dot diagram to see history of commits for the trace
|
||||
@ -163,7 +163,7 @@ Solution:
|
||||
- Each line represents a configuration
|
||||
- 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>
|
||||
@ -181,5 +181,5 @@ Solution:
|
||||
- Access the Search view
|
||||
- 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">
|
||||
|
||||
@ -9,7 +9,7 @@ linkTitle: "Skia Perf"
|
||||
[Skia Perf](https://perf.skia.org) is a web application for analyzing and
|
||||
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
|
||||
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.
|
||||
|
||||
<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:
|
||||
|
||||
<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
|
||||
|
||||
BIN
site/docs/dev/tools/buttons.png
Normal file
BIN
site/docs/dev/tools/buttons.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 1.1 KiB |
BIN
site/docs/dev/tools/crosshair.png
Normal file
BIN
site/docs/dev/tools/crosshair.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 28 KiB |
@ -20,7 +20,7 @@ Features:
|
||||
- Android offscreen layer visualization
|
||||
- Shared resource viewer
|
||||
|
||||
<img src="/dev/tools/onlinedebugger.png" style="display: inline-block;" />
|
||||
<img src="../onlinedebugger.png" style="display: inline-block;" />
|
||||
|
||||
## User Guide
|
||||
|
||||
@ -33,8 +33,8 @@ capture one from an android device using the
|
||||
### Command Playback and Filters
|
||||
|
||||
Try playing back the commands within the current frame using the lower play
|
||||
button <img src="/dev/tools/playcommands.png" style="display: inline-block;" />,
|
||||
(the one not in a circle) You should see the image built up one draw at a time.
|
||||
button <img src="../playcommands.png" style="display: inline-block;" />, (the
|
||||
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
|
||||
when run. Try filtering these out by pasting
|
||||
@ -52,18 +52,18 @@ using `,` (comma) and `.` (period).
|
||||
> at the beginning.
|
||||
|
||||
Any command can be expanded using the
|
||||
<img src="/dev/tools/expand.png" style="display: inline-block;" /> icon to see
|
||||
all of the parameters that were recorded with that command.
|
||||
<img src="../expand.png" style="display: inline-block;" /> icon to see all of
|
||||
the parameters that were recorded with that command.
|
||||
|
||||
Commands can be disabled or enabled with the checkbox that becomes available
|
||||
after expanding the command's detail view.
|
||||
|
||||
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
|
||||
|
||||
<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
|
||||
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
|
||||
|
||||
<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
|
||||
Dec 2019, this only shows images.
|
||||
@ -97,7 +97,7 @@ ids in the process that recorded the SKP.
|
||||
|
||||
### 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
|
||||
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
|
||||
|
||||
<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
|
||||
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
|
||||
|
||||
<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
|
||||
<img src="/dev/tools/gpuop.png" style="display: inline-block;" />. This is the
|
||||
GPU operation id. When multiple commands share a GPU op id, this indicates that
|
||||
they were batched together when sent to the GPU. In the WASM debugger, this goes
|
||||
<img src="../gpuop.png" style="display: inline-block;" />. This is the GPU
|
||||
operation id. When multiple commands share a GPU op id, this indicates that they
|
||||
were batched together when sent to the GPU. In the WASM debugger, this goes
|
||||
though WebGL.
|
||||
|
||||
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.
|
||||
|
||||
<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:
|
||||
|
||||
|
||||
BIN
site/docs/dev/tools/debugger.png
Normal file
BIN
site/docs/dev/tools/debugger.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 176 KiB |
BIN
site/docs/dev/tools/end.png
Normal file
BIN
site/docs/dev/tools/end.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 228 B |
BIN
site/docs/dev/tools/expand.png
Normal file
BIN
site/docs/dev/tools/expand.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 225 B |
BIN
site/docs/dev/tools/frameplayback.png
Normal file
BIN
site/docs/dev/tools/frameplayback.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 1.9 KiB |
BIN
site/docs/dev/tools/gpuop.png
Normal file
BIN
site/docs/dev/tools/gpuop.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 678 B |
BIN
site/docs/dev/tools/image.png
Normal file
BIN
site/docs/dev/tools/image.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 346 B |
BIN
site/docs/dev/tools/layers.png
Normal file
BIN
site/docs/dev/tools/layers.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 24 KiB |
BIN
site/docs/dev/tools/onlinedebugger.png
Normal file
BIN
site/docs/dev/tools/onlinedebugger.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 445 KiB |
BIN
site/docs/dev/tools/playcommands.png
Normal file
BIN
site/docs/dev/tools/playcommands.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 243 B |
BIN
site/docs/dev/tools/resources.png
Normal file
BIN
site/docs/dev/tools/resources.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 2.7 KiB |
BIN
site/docs/dev/tools/settings.png
Normal file
BIN
site/docs/dev/tools/settings.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 7.8 KiB |
@ -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
|
||||
[chrome://tracing](chrome://tracing), click Load:
|
||||
|
||||
|
||||
|
||||
|
||||
... 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
|
||||
controls.
|
||||
|
||||
|
||||
|
||||
|
||||
Android ATrace
|
||||
--------------
|
||||
|
||||
BIN
site/docs/dev/tools/tracing.png
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BIN
site/docs/dev/tools/tracing.png
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After Width: | Height: | Size: 56 KiB |
BIN
site/docs/dev/tools/tracing_load.png
Normal file
BIN
site/docs/dev/tools/tracing_load.png
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|
After Width: | Height: | Size: 6.0 KiB |
BIN
site/docs/user/modules/PathKit_effects.png
Normal file
BIN
site/docs/user/modules/PathKit_effects.png
Normal file
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|
After Width: | Height: | Size: 111 KiB |
Loading…
Reference in New Issue
Block a user