Plan 9 from Bell Labs’s /usr/web/sources/contrib/steve/root/sys/src/cmd/graphviz/pathplan/solvers.c

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Distributed under the MIT License.
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/*
    This software may only be used by you under license from AT&T Corp.
    ("AT&T").  A copy of AT&T's Source Code Agreement is available at
    AT&T's Internet website having the URL:
    <http://www.research.att.com/sw/tools/graphviz/license/source.html>
    If you received this software without first entering into a license
    with AT&T, you have an infringing copy of this software and cannot use
    it without violating AT&T's intellectual property rights.
*/
#pragma prototyped

#ifdef HAVE_CONFIG_H
#include "config.h"
#endif
#include <math.h>
#include "solvers.h"

#ifdef DMALLOC
#include "dmalloc.h"
#endif

#ifndef HAVE_CBRT
#define cbrt(x) ((x < 0) ? (-1*pow(-x, 1.0/3.0)) : pow (x, 1.0/3.0))
#endif
#ifndef M_PI
#define M_PI 3.14159265358979323846
#endif

#define EPS 1E-7
#define AEQ0(x) (((x) < EPS) && ((x) > -EPS))

int solve3 (double *coeff, double *roots) {
    double a, b, c, d;
    int rootn, i;
    double p, q, disc, b_over_3a, c_over_a, d_over_a;
    double r, theta, temp, alpha, beta;

    a = coeff[3], b = coeff[2], c = coeff[1], d = coeff[0];
    if (AEQ0 (a))
        return solve2 (coeff, roots);
    b_over_3a = b / (3 * a);
    c_over_a = c / a;
    d_over_a = d / a;

    p = b_over_3a * b_over_3a;
    q = 2 * b_over_3a * p - b_over_3a * c_over_a + d_over_a;
    p = c_over_a / 3 - p;
    disc = q * q + 4 * p * p * p;

    if (disc < 0) {
        r = .5 * sqrt (-disc + q * q);
        theta = atan2 (sqrt (-disc), -q);
        temp = 2 * cbrt (r);
        roots[0] = temp * cos (theta / 3);
        roots[1] = temp * cos ((theta + M_PI + M_PI) / 3);
        roots[2] = temp * cos ((theta - M_PI - M_PI) / 3);
        rootn = 3;
    } else {
        alpha = .5 * (sqrt (disc) - q);
        beta = -q - alpha;
        roots[0] = cbrt (alpha) + cbrt (beta);
        if (disc > 0)
            rootn = 1;
        else
            roots[1] = roots[2] = -.5 * roots[0], rootn = 3;
    }

    for (i = 0; i < rootn; i++)
        roots[i] -= b_over_3a;

    return rootn;
}

int solve2 (double *coeff, double *roots) {
    double a, b, c;
    double disc, b_over_2a, c_over_a;

    a = coeff[2], b = coeff[1], c = coeff[0];
    if (AEQ0(a))
        return solve1 (coeff, roots);
    b_over_2a = b / (2 * a);
    c_over_a = c / a;

    disc = b_over_2a * b_over_2a - c_over_a;
    if (disc < 0)
        return 0;
    else if (disc == 0) {
        roots[0] = -b_over_2a;
        return 1;
    } else {
        roots[0] = -b_over_2a + sqrt (disc);
        roots[1] = -2 * b_over_2a - roots[0];
        return 2;
    }
}

int solve1 (double *coeff, double *roots) {
    double a, b;

    a = coeff[1], b = coeff[0];
    if (AEQ0 (a)) {
        if (AEQ0 (b))
            return 4;
        else
            return 0;
    }
    roots[0] = -b / a;
    return 1;
}

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