| /* |
| * 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; |
| } |