1 | /* $Id$ $Revision$ */ |
2 | /* vim:set shiftwidth=4 ts=8: */ |
3 | |
4 | /************************************************************************* |
5 | * Copyright (c) 2011 AT&T Intellectual Property |
6 | * All rights reserved. This program and the accompanying materials |
7 | * are made available under the terms of the Eclipse Public License v1.0 |
8 | * which accompanies this distribution, and is available at |
9 | * http://www.eclipse.org/legal/epl-v10.html |
10 | * |
11 | * Contributors: See CVS logs. Details at http://www.graphviz.org/ |
12 | *************************************************************************/ |
13 | |
14 | |
15 | #include "config.h" |
16 | #include <math.h> |
17 | #include "solvers.h" |
18 | |
19 | #ifdef DMALLOC |
20 | #include "dmalloc.h" |
21 | #endif |
22 | |
23 | #ifndef HAVE_CBRT |
24 | #define cbrt(x) ((x < 0) ? (-1*pow(-x, 1.0/3.0)) : pow (x, 1.0/3.0)) |
25 | #endif |
26 | #ifndef M_PI |
27 | #define M_PI 3.14159265358979323846 |
28 | #endif |
29 | |
30 | #define EPS 1E-7 |
31 | #define AEQ0(x) (((x) < EPS) && ((x) > -EPS)) |
32 | |
33 | int solve3(double *coeff, double *roots) |
34 | { |
35 | double a, b, c, d; |
36 | int rootn, i; |
37 | double p, q, disc, b_over_3a, c_over_a, d_over_a; |
38 | double r, theta, temp, alpha, beta; |
39 | |
40 | a = coeff[3], b = coeff[2], c = coeff[1], d = coeff[0]; |
41 | if (AEQ0(a)) |
42 | return solve2(coeff, roots); |
43 | b_over_3a = b / (3 * a); |
44 | c_over_a = c / a; |
45 | d_over_a = d / a; |
46 | |
47 | p = b_over_3a * b_over_3a; |
48 | q = 2 * b_over_3a * p - b_over_3a * c_over_a + d_over_a; |
49 | p = c_over_a / 3 - p; |
50 | disc = q * q + 4 * p * p * p; |
51 | |
52 | if (disc < 0) { |
53 | r = .5 * sqrt(-disc + q * q); |
54 | theta = atan2(sqrt(-disc), -q); |
55 | temp = 2 * cbrt(r); |
56 | roots[0] = temp * cos(theta / 3); |
57 | roots[1] = temp * cos((theta + M_PI + M_PI) / 3); |
58 | roots[2] = temp * cos((theta - M_PI - M_PI) / 3); |
59 | rootn = 3; |
60 | } else { |
61 | alpha = .5 * (sqrt(disc) - q); |
62 | beta = -q - alpha; |
63 | roots[0] = cbrt(alpha) + cbrt(beta); |
64 | if (disc > 0) |
65 | rootn = 1; |
66 | else |
67 | roots[1] = roots[2] = -.5 * roots[0], rootn = 3; |
68 | } |
69 | |
70 | for (i = 0; i < rootn; i++) |
71 | roots[i] -= b_over_3a; |
72 | |
73 | return rootn; |
74 | } |
75 | |
76 | int solve2(double *coeff, double *roots) |
77 | { |
78 | double a, b, c; |
79 | double disc, b_over_2a, c_over_a; |
80 | |
81 | a = coeff[2], b = coeff[1], c = coeff[0]; |
82 | if (AEQ0(a)) |
83 | return solve1(coeff, roots); |
84 | b_over_2a = b / (2 * a); |
85 | c_over_a = c / a; |
86 | |
87 | disc = b_over_2a * b_over_2a - c_over_a; |
88 | if (disc < 0) |
89 | return 0; |
90 | else if (disc == 0) { |
91 | roots[0] = -b_over_2a; |
92 | return 1; |
93 | } else { |
94 | roots[0] = -b_over_2a + sqrt(disc); |
95 | roots[1] = -2 * b_over_2a - roots[0]; |
96 | return 2; |
97 | } |
98 | } |
99 | |
100 | int solve1(double *coeff, double *roots) |
101 | { |
102 | double a, b; |
103 | |
104 | a = coeff[1], b = coeff[0]; |
105 | if (AEQ0(a)) { |
106 | if (AEQ0(b)) |
107 | return 4; |
108 | else |
109 | return 0; |
110 | } |
111 | roots[0] = -b / a; |
112 | return 1; |
113 | } |
114 | |