caryclark@google.com | 07393ca | 2013-04-08 11:47:37 +0000 | [diff] [blame] | 1 | /* |
| 2 | http://stackoverflow.com/questions/2009160/how-do-i-convert-the-2-control-points-of-a-cubic-curve-to-the-single-control-poi |
| 3 | */ |
| 4 | |
| 5 | /* |
| 6 | Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2. |
| 7 | Then for degree elevation, the equations are: |
| 8 | |
| 9 | Q0 = P0 |
| 10 | Q1 = 1/3 P0 + 2/3 P1 |
| 11 | Q2 = 2/3 P1 + 1/3 P2 |
| 12 | Q3 = P2 |
| 13 | In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways to compute P1 from |
| 14 | the equations above: |
| 15 | |
| 16 | P1 = 3/2 Q1 - 1/2 Q0 |
| 17 | P1 = 3/2 Q2 - 1/2 Q3 |
| 18 | If this is a degree-elevated cubic, then both equations will give the same answer for P1. Since |
| 19 | it's likely not, your best bet is to average them. So, |
| 20 | |
| 21 | P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3 |
caryclark@google.com | 07393ca | 2013-04-08 11:47:37 +0000 | [diff] [blame] | 22 | */ |
| 23 | |
| 24 | #include "SkPathOpsCubic.h" |
caryclark@google.com | 07393ca | 2013-04-08 11:47:37 +0000 | [diff] [blame] | 25 | #include "SkPathOpsQuad.h" |
caryclark@google.com | 07393ca | 2013-04-08 11:47:37 +0000 | [diff] [blame] | 26 | |
| 27 | SkDQuad SkDCubic::toQuad() const { |
| 28 | SkDQuad quad; |
| 29 | quad[0] = fPts[0]; |
| 30 | const SkDPoint fromC1 = {(3 * fPts[1].fX - fPts[0].fX) / 2, (3 * fPts[1].fY - fPts[0].fY) / 2}; |
| 31 | const SkDPoint fromC2 = {(3 * fPts[2].fX - fPts[3].fX) / 2, (3 * fPts[2].fY - fPts[3].fY) / 2}; |
| 32 | quad[1].fX = (fromC1.fX + fromC2.fX) / 2; |
| 33 | quad[1].fY = (fromC1.fY + fromC2.fY) / 2; |
| 34 | quad[2] = fPts[3]; |
| 35 | return quad; |
| 36 | } |