1 /* 2 * Copyright 2015 Google Inc. 3 * 4 * Use of this source code is governed by a BSD-style license that can be 5 * found in the LICENSE file. 6 */ 7 8 /* 9 http://stackoverflow.com/questions/2009160/how-do-i-convert-the-2-control-points-of-a-cubic-curve-to-the-single-control-poi 10 */ 11 12 /* 13 Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2. 14 Then for degree elevation, the equations are: 15 16 Q0 = P0 17 Q1 = 1/3 P0 + 2/3 P1 18 Q2 = 2/3 P1 + 1/3 P2 19 Q3 = P2 20 In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways to compute P1 from 21 the equations above: 22 23 P1 = 3/2 Q1 - 1/2 Q0 24 P1 = 3/2 Q2 - 1/2 Q3 25 If this is a degree-elevated cubic, then both equations will give the same answer for P1. Since 26 it's likely not, your best bet is to average them. So, 27 28 P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3 29 */ 30 31 #include "src/pathops/SkPathOpsCubic.h" 32 #include "src/pathops/SkPathOpsQuad.h" 33 34 // used for testing only toQuad() const35SkDQuad SkDCubic::toQuad() const { 36 SkDQuad quad; 37 quad[0] = fPts[0]; 38 const SkDPoint fromC1 = {(3 * fPts[1].fX - fPts[0].fX) / 2, (3 * fPts[1].fY - fPts[0].fY) / 2}; 39 const SkDPoint fromC2 = {(3 * fPts[2].fX - fPts[3].fX) / 2, (3 * fPts[2].fY - fPts[3].fY) / 2}; 40 quad[1].fX = (fromC1.fX + fromC2.fX) / 2; 41 quad[1].fY = (fromC1.fY + fromC2.fY) / 2; 42 quad[2] = fPts[3]; 43 return quad; 44 } 45