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roots.c
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1992-04-28
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/*
* Roots3And4.c
*
* Utility functions to find cubic and quartic roots,
* coefficients are passed like this:
*
* c[0] + c[1]*x + c[2]*x^2 + c[3]*x^3 + c[4]*x^4 = 0
*
* The functions return the number of non-complex roots and
* put the values into the s array.
*
* Author: Jochen Schwarze (schwarze@isa.de)
*
* Jan 26, 1990 Version for Graphics Gems
* Oct 11, 1990 Fixed sign problem for negative q's in SolveQuartic
* (reported by Mark Podlipec),
* Old-style function definitions,
* IsZero() as a macro
* Nov 23, 1990 Some systems do not declare acos() and cbrt() in
* <math.h>, though the functions exist in the library.
* If large coefficients are used, EQN_EPS should be
* reduced considerably (e.g. to 1E-30), results will be
* correct but multiple roots might be reported more
* than once.
*/
#include "common.h"
#ifndef __WATCOMC__
extern double sqrt(), cbrt(), cos(), acos();
#endif
/* epsilon surrounding for near zero values */
/*
* In case M_PI isn't defined in math.h
*/
#ifndef M_PI
#define M_PI PI
#endif
#define EQN_EPS 1e-9
#define IsZero(x) ((x) > -EQN_EPS && (x) < EQN_EPS)
#ifndef CBRT
#define cbrt(x) ((x) > 0.0 ? pow((double)(x), 1.0/3.0) : \
((x) < 0.0 ? -pow((double)-(x), 1.0/3.0) : 0.0))
#endif
int SolveQuadric(c, s)
double c[ 3 ];
double s[ 2 ];
{
double p, q, D;
/* normal form: x^2 + px + q = 0 */
p = c[ 1 ] / (2 * c[ 2 ]);
q = c[ 0 ] / c[ 2 ];
D = p * p - q;
if (IsZero(D))
{
s[ 0 ] = - p;
return 1;
}
else if (D > 0)
{
double sqrt_D = sqrt(D);
s[ 0 ] = sqrt_D - p;
s[ 1 ] = - sqrt_D - p;
return 2;
}
else /* if (D < 0) */
return 0;
}
int SolveCubic(c, s)
double c[ 4 ];
double s[ 3 ];
{
int i, num;
double sub;
double A, B, C;
double sq_A, p, q;
double cb_p, D;
/* normal form: x^3 + Ax^2 + Bx + C = 0 */
A = c[ 2 ] / c[ 3 ];
B = c[ 1 ] / c[ 3 ];
C = c[ 0 ] / c[ 3 ];
/* substitute x = y - A/3 to eliminate quadric term:
x^3 +px + q = 0 */
sq_A = A * A;
p = 1.0/3 * (- 1.0/3 * sq_A + B);
q = 1.0/2 * (2.0/27 * A * sq_A - 1.0/3 * A * B + C);
/* use Cardano's formula */
cb_p = p * p * p;
D = q * q + cb_p;
if (IsZero(D))
{
if (IsZero(q)) /* one triple solution */
{
s[ 0 ] = 0;
num = 1;
}
else /* one single and one double solution */
{
double u = cbrt(-q);
s[ 0 ] = 2 * u;
s[ 1 ] = - u;
num = 2;
}
}
else if (D < 0) /* Casus irreducibilis: three real solutions */
{
double phi = 1.0/3 * acos(-q / sqrt(-cb_p));
double t = 2 * sqrt(-p);
s[ 0 ] = t * cos(phi);
s[ 1 ] = - t * cos(phi + M_PI / 3);
s[ 2 ] = - t * cos(phi - M_PI / 3);
num = 3;
}
else /* one real solution */
{
double sqrt_D = sqrt(D);
double u = cbrt(sqrt_D - q);
double v = - cbrt(sqrt_D + q);
s[ 0 ] = u + v;
num = 1;
}
/* resubstitute */
sub = 1.0/3 * A;
for (i = 0; i < num; ++i)
s[ i ] -= sub;
return num;
}
int SolveQuartic(c, s)
double c[ 5 ];
double s[ 4 ];
{
double coeffs[ 4 ];
double z, u, v, sub;
double A, B, C, D;
double sq_A, p, q, r;
int i, num;
/* normal form: x^4 + Ax^3 + Bx^2 + Cx + D = 0 */
A = c[ 3 ] / c[ 4 ];
B = c[ 2 ] / c[ 4 ];
C = c[ 1 ] / c[ 4 ];
D = c[ 0 ] / c[ 4 ];
/* substitute x = y - A/4 to eliminate cubic term:
x^4 + px^2 + qx + r = 0 */
sq_A = A * A;
p = - 3.0/8 * sq_A + B;
q = 1.0/8 * sq_A * A - 1.0/2 * A * B + C;
r = - 3.0/256*sq_A*sq_A + 1.0/16*sq_A*B - 1.0/4*A*C + D;
if (IsZero(r))
{
/* no absolute term: y(y^3 + py + q) = 0 */
coeffs[ 0 ] = q;
coeffs[ 1 ] = p;
coeffs[ 2 ] = 0;
coeffs[ 3 ] = 1;
num = SolveCubic(coeffs, s);
s[ num++ ] = 0;
}
else
{
/* solve the resolvent cubic ... */
coeffs[ 0 ] = 1.0/2 * r * p - 1.0/8 * q * q;
coeffs[ 1 ] = - r;
coeffs[ 2 ] = - 1.0/2 * p;
coeffs[ 3 ] = 1;
(void) SolveCubic(coeffs, s);
/* ... and take the one real solution ... */
z = s[ 0 ];
/* ... to build two quadric equations */
u = z * z - r;
v = 2 * z - p;
if (IsZero(u))
u = 0;
else if (u > 0)
u = sqrt(u);
else
return 0;
if (IsZero(v))
v = 0;
else if (v > 0)
v = sqrt(v);
else
return 0;
coeffs[ 0 ] = z - u;
coeffs[ 1 ] = q < 0 ? -v : v;
coeffs[ 2 ] = 1;
num = SolveQuadric(coeffs, s);
coeffs[ 0 ]= z + u;
coeffs[ 1 ] = q < 0 ? v : -v;
coeffs[ 2 ] = 1;
num += SolveQuadric(coeffs, s + num);
}
/* resubstitute */
sub = 1.0/4 * A;
for (i = 0; i < num; ++i)
s[ i ] -= sub;
return num;
}