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