| 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 | |