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- /*
- * 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/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;
- }
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