1 files changed, 45 insertions, 0 deletions
diff --git a/gfx/skia/skia/src/pathops/SkDCubicToQuads.cpp b/gfx/skia/skia/src/pathops/SkDCubicToQuads.cpp
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index 0000000000..c7c7944580
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+++ b/gfx/skia/skia/src/pathops/SkDCubicToQuads.cpp
@@ -0,0 +1,45 @@
+/*
+ * Copyright 2015 Google Inc.
+ *
+ * Use of this source code is governed by a BSD-style license that can be
+ * found in the LICENSE file.
+ */
+
+/*
+http://stackoverflow.com/questions/2009160/how-do-i-convert-the-2-control-points-of-a-cubic-curve-to-the-single-control-poi
+*/
+
+/*
+Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2.
+Then for degree elevation, the equations are:
+
+Q0 = P0
+Q1 = 1/3 P0 + 2/3 P1
+Q2 = 2/3 P1 + 1/3 P2
+Q3 = P2
+In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways to compute P1 from
+ the equations above:
+
+P1 = 3/2 Q1 - 1/2 Q0
+P1 = 3/2 Q2 - 1/2 Q3
+If this is a degree-elevated cubic, then both equations will give the same answer for P1. Since
+ it's likely not, your best bet is to average them. So,
+
+P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3
+*/
+
+#include "src/pathops/SkPathOpsCubic.h"
+#include "src/pathops/SkPathOpsPoint.h"
+#include "src/pathops/SkPathOpsQuad.h"
+
+// used for testing only
+SkDQuad SkDCubic::toQuad() const {
+    SkDQuad quad;
+    quad[0] = fPts[0];
+    const SkDPoint fromC1 = {(3 * fPts[1].fX - fPts[0].fX) / 2, (3 * fPts[1].fY - fPts[0].fY) / 2};
+    const SkDPoint fromC2 = {(3 * fPts[2].fX - fPts[3].fX) / 2, (3 * fPts[2].fY - fPts[3].fY) / 2};
+    quad[1].fX = (fromC1.fX + fromC2.fX) / 2;
+    quad[1].fY = (fromC1.fY + fromC2.fY) / 2;
+    quad[2] = fPts[3];
+    return quad;
+}