| * 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: |

| In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways to compute P1 from |

| 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" |

| SkDQuad SkDCubic::toQuad() const { |

| 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; |