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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 "SkPathOpsCubic.h"
32 #include "SkPathOpsQuad.h"
33 
34 // used for testing only
toQuad() const35 SkDQuad 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