skia2/experimental/Intersection/QuarticRoot.cpp
skia.committer@gmail.com c1ad022608 Sanitizing source files in Skia_Periodic_House_Keeping
git-svn-id: http://skia.googlecode.com/svn/trunk@5594 2bbb7eff-a529-9590-31e7-b0007b416f81
2012-09-19 02:01:47 +00:00

273 lines
7.8 KiB
C++

// from http://tog.acm.org/resources/GraphicsGems/gems/Roots3And4.c
/*
* 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 <math.h>
#include "CubicUtilities.h"
#include "QuarticRoot.h"
const double PI = 4 * atan(1);
// unlike quadraticRoots in QuadraticUtilities.cpp, this does not discard
// real roots <= 0 or >= 1
static int quadraticRootsX(const double A, const double B, const double C,
double s[2]) {
if (approximately_zero(A)) {
if (approximately_zero(B)) {
s[0] = 0;
return C == 0;
}
s[0] = -C / B;
return 1;
}
/* normal form: x^2 + px + q = 0 */
const double p = B / (2 * A);
const double q = C / A;
const double D = p * p - q;
if (approximately_zero(D)) {
s[0] = -p;
return 1;
} else if (D < 0) {
return 0;
} else {
assert(D > 0);
double sqrt_D = sqrt(D);
s[0] = sqrt_D - p;
s[1] = -sqrt_D - p;
return 2;
}
}
#define USE_GEMS 0
#if USE_GEMS
// unlike cubicRoots in CubicUtilities.cpp, this does not discard
// real roots <= 0 or >= 1
static int cubicRootsX(const double A, const double B, const double C,
const double D, double s[3]) {
int num;
/* normal form: x^3 + Ax^2 + Bx + C = 0 */
const double invA = 1 / A;
const double a = B * invA;
const double b = C * invA;
const double c = D * invA;
/* substitute x = y - a/3 to eliminate quadric term:
x^3 +px + q = 0 */
const double a2 = a * a;
const double Q = (-a2 + b * 3) / 9;
const double R = (2 * a2 * a - 9 * a * b + 27 * c) / 54;
/* use Cardano's formula */
const double Q3 = Q * Q * Q;
const double R2plusQ3 = R * R + Q3;
if (approximately_zero(R2plusQ3)) {
if (approximately_zero(R)) {/* one triple solution */
s[0] = 0;
num = 1;
} else { /* one single and one double solution */
double u = cube_root(-R);
s[0] = 2 * u;
s[1] = -u;
num = 2;
}
}
else if (R2plusQ3 < 0) { /* Casus irreducibilis: three real solutions */
const double theta = acos(-R / sqrt(-Q3)) / 3;
const double _2RootQ = 2 * sqrt(-Q);
s[0] = _2RootQ * cos(theta);
s[1] = -_2RootQ * cos(theta + PI / 3);
s[2] = -_2RootQ * cos(theta - PI / 3);
num = 3;
} else { /* one real solution */
const double sqrt_D = sqrt(R2plusQ3);
const double u = cube_root(sqrt_D - R);
const double v = -cube_root(sqrt_D + R);
s[0] = u + v;
num = 1;
}
/* resubstitute */
const double sub = a / 3;
for (int i = 0; i < num; ++i) {
s[i] -= sub;
}
return num;
}
#else
static int cubicRootsX(double A, double B, double C, double D, double s[3]) {
if (approximately_zero(A)) { // we're just a quadratic
return quadraticRootsX(B, C, D, s);
}
if (approximately_zero(D)) {
int num = quadraticRootsX(A, B, C, s);
for (int i = 0; i < num; ++i) {
if (approximately_zero(s[i])) {
return num;
}
}
s[num++] = 0;
return num;
}
double a, b, c;
{
double invA = 1 / A;
a = B * invA;
b = C * invA;
c = D * invA;
}
double a2 = a * a;
double Q = (a2 - b * 3) / 9;
double R = (2 * a2 * a - 9 * a * b + 27 * c) / 54;
double Q3 = Q * Q * Q;
double R2MinusQ3 = R * R - Q3;
double adiv3 = a / 3;
double r;
double* roots = s;
if (R2MinusQ3 > -FLT_EPSILON / 10 && R2MinusQ3 < FLT_EPSILON / 10 ) {
if (approximately_zero(R)) {/* one triple solution */
*roots++ = -adiv3;
} else { /* one single and one double solution */
double u = cube_root(-R);
*roots++ = 2 * u - adiv3;
*roots++ = -u - adiv3;
}
}
else if (R2MinusQ3 < 0) // we have 3 real roots
{
double theta = acos(R / sqrt(Q3));
double neg2RootQ = -2 * sqrt(Q);
r = neg2RootQ * cos(theta / 3) - adiv3;
*roots++ = r;
r = neg2RootQ * cos((theta + 2 * PI) / 3) - adiv3;
*roots++ = r;
r = neg2RootQ * cos((theta - 2 * PI) / 3) - adiv3;
*roots++ = r;
}
else // we have 1 real root
{
double A = fabs(R) + sqrt(R2MinusQ3);
A = cube_root(A);
if (R > 0) {
A = -A;
}
if (A != 0) {
A += Q / A;
}
r = A - adiv3;
*roots++ = r;
}
return (int)(roots - s);
}
#endif
int quarticRoots(const double A, const double B, const double C, const double D,
const double E, double s[4]) {
if (approximately_zero(A)) {
if (approximately_zero(B)) {
return quadraticRootsX(C, D, E, s);
}
return cubicRootsX(B, C, D, E, s);
}
int num;
int i;
if (approximately_zero(E)) {
num = cubicRootsX(A, B, C, D, s);
for (i = 0; i < num; ++i) {
if (approximately_zero(s[i])) {
return num;
}
}
s[num++] = 0;
return num;
}
double u, v;
/* normal form: x^4 + Ax^3 + Bx^2 + Cx + D = 0 */
const double invA = 1 / A;
const double a = B * invA;
const double b = C * invA;
const double c = D * invA;
const double d = E * invA;
/* substitute x = y - a/4 to eliminate cubic term:
x^4 + px^2 + qx + r = 0 */
const double a2 = a * a;
const double p = -3 * a2 / 8 + b;
const double q = a2 * a / 8 - a * b / 2 + c;
const double r = -3 * a2 * a2 / 256 + a2 * b / 16 - a * c / 4 + d;
if (approximately_zero(r)) {
/* no absolute term: y(y^3 + py + q) = 0 */
num = cubicRootsX(1, 0, p, q, s);
s[num++] = 0;
} else {
/* solve the resolvent cubic ... */
(void) cubicRootsX(1, -p / 2, -r, r * p / 2 - q * q / 8, s);
/* ... and take the one real solution ... */
const double z = s[0];
/* ... to build two quadric equations */
u = z * z - r;
v = 2 * z - p;
if (approximately_zero(u)) {
u = 0;
} else if (u > 0) {
u = sqrt(u);
} else {
return 0;
}
if (approximately_zero(v)) {
v = 0;
} else if (v > 0) {
v = sqrt(v);
} else {
return 0;
}
num = quadraticRootsX(1, q < 0 ? -v : v, z - u, s);
num += quadraticRootsX(1, q < 0 ? v : -v, z + u, s + num);
}
// eliminate duplicates
for (i = 0; i < num - 1; ++i) {
for (int j = i + 1; j < num; ) {
if (approximately_equal(s[i], s[j])) {
if (j < --num) {
s[j] = s[num];
}
} else {
++j;
}
}
}
/* resubstitute */
const double sub = a / 4;
for (i = 0; i < num; ++i) {
s[i] -= sub;
}
return num;
}