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caryclark@google.com639df892012-01-10 21:46:10 +00001#include "CubicIntersection.h"
2#include "IntersectionUtilities.h"
3
4/*
5 Given a cubic c, t1, and t2, find a small cubic segment.
6
7 The new cubic is defined as points A, B, C, and D, where
8 s1 = 1 - t1
9 s2 = 1 - t2
10 A = c[0]*s1*s1*s1 + 3*c[1]*s1*s1*t1 + 3*c[2]*s1*t1*t1 + c[3]*t1*t1*t1
11 D = c[0]*s2*s2*s2 + 3*c[1]*s2*s2*t2 + 3*c[2]*s2*t2*t2 + c[3]*t2*t2*t2
12
13 We don't have B or C. So We define two equations to isolate them.
14 First, compute two reference T values 1/3 and 2/3 from t1 to t2:
15
16 c(at (2*t1 + t2)/3) == E
17 c(at (t1 + 2*t2)/3) == F
18
19 Next, compute where those values must be if we know the values of B and C:
20
21 _12 = A*2/3 + B*1/3
22 12_ = A*1/3 + B*2/3
23 _23 = B*2/3 + C*1/3
24 23_ = B*1/3 + C*2/3
25 _34 = C*2/3 + D*1/3
26 34_ = C*1/3 + D*2/3
27 _123 = (A*2/3 + B*1/3)*2/3 + (B*2/3 + C*1/3)*1/3 = A*4/9 + B*4/9 + C*1/9
28 123_ = (A*1/3 + B*2/3)*1/3 + (B*1/3 + C*2/3)*2/3 = A*1/9 + B*4/9 + C*4/9
29 _234 = (B*2/3 + C*1/3)*2/3 + (C*2/3 + D*1/3)*1/3 = B*4/9 + C*4/9 + D*1/9
30 234_ = (B*1/3 + C*2/3)*1/3 + (C*1/3 + D*2/3)*2/3 = B*1/9 + C*4/9 + D*4/9
31 _1234 = (A*4/9 + B*4/9 + C*1/9)*2/3 + (B*4/9 + C*4/9 + D*1/9)*1/3
32 = A*8/27 + B*12/27 + C*6/27 + D*1/27
33 = E
34 1234_ = (A*1/9 + B*4/9 + C*4/9)*1/3 + (B*1/9 + C*4/9 + D*4/9)*2/3
35 = A*1/27 + B*6/27 + C*12/27 + D*8/27
36 = F
37 E*27 = A*8 + B*12 + C*6 + D
38 F*27 = A + B*6 + C*12 + D*8
39
40Group the known values on one side:
41
42 M = E*27 - A*8 - D = B*12 + C* 6
43 N = F*27 - A - D*8 = B* 6 + C*12
44 M*2 - N = B*18
45 N*2 - M = C*18
46 B = (M*2 - N)/18
47 C = (N*2 - M)/18
48 */
49
50static double interp_cubic_coords(const double* src, double t)
51{
52 double ab = interp(src[0], src[2], t);
53 double bc = interp(src[2], src[4], t);
54 double cd = interp(src[4], src[6], t);
55 double abc = interp(ab, bc, t);
56 double bcd = interp(bc, cd, t);
57 double abcd = interp(abc, bcd, t);
58 return abcd;
59}
60
61void sub_divide(const Cubic& src, double t1, double t2, Cubic& dst) {
62 double ax = dst[0].x = interp_cubic_coords(&src[0].x, t1);
63 double ay = dst[0].y = interp_cubic_coords(&src[0].y, t1);
64 double ex = interp_cubic_coords(&src[0].x, (t1*2+t2)/3);
65 double ey = interp_cubic_coords(&src[0].y, (t1*2+t2)/3);
66 double fx = interp_cubic_coords(&src[0].x, (t1+t2*2)/3);
67 double fy = interp_cubic_coords(&src[0].y, (t1+t2*2)/3);
68 double dx = dst[3].x = interp_cubic_coords(&src[0].x, t2);
69 double dy = dst[3].y = interp_cubic_coords(&src[0].y, t2);
70 double mx = ex * 27 - ax * 8 - dx;
71 double my = ey * 27 - ay * 8 - dy;
72 double nx = fx * 27 - ax - dx * 8;
73 double ny = fy * 27 - ay - dy * 8;
74 /* bx = */ dst[1].x = (mx * 2 - nx) / 18;
75 /* by = */ dst[1].y = (my * 2 - ny) / 18;
76 /* cx = */ dst[2].x = (nx * 2 - mx) / 18;
77 /* cy = */ dst[2].y = (ny * 2 - my) / 18;
78}
79
80/* classic one t subdivision */
81static void interp_cubic_coords(const double* src, double* dst, double t)
82{
83 double ab = interp(src[0], src[2], t);
84 double bc = interp(src[2], src[4], t);
85 double cd = interp(src[4], src[6], t);
86 double abc = interp(ab, bc, t);
87 double bcd = interp(bc, cd, t);
88 double abcd = interp(abc, bcd, t);
89
90 dst[0] = src[0];
91 dst[2] = ab;
92 dst[4] = abc;
93 dst[6] = abcd;
94 dst[8] = bcd;
95 dst[10] = cd;
96 dst[12] = src[6];
97}
98
99void chop_at(const Cubic& src, CubicPair& dst, double t)
100{
101 interp_cubic_coords(&src[0].x, &dst.pts[0].x, t);
102 interp_cubic_coords(&src[0].y, &dst.pts[0].y, t);
103}