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/*
9http://stackoverflow.com/questions/2009160/how-do-i-convert-the-2-control-points-of-a-cubic-curve-to-the-single-control-poi
10*/
11
12/*
13Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2.
14Then for degree elevation, the equations are:
15
16Q0 = P0
17Q1 = 1/3 P0 + 2/3 P1
18Q2 = 2/3 P1 + 1/3 P2
19Q3 = P2
20In 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
23P1 = 3/2 Q1 - 1/2 Q0
24P1 = 3/2 Q2 - 1/2 Q3
25If 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
28P1 = -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
35SkDQuad 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