skia2/experimental/Intersection/QuadraticUtilities.cpp
caryclark@google.com 235f56a92f shape ops work in progress
add quartic solution for intersecting quadratics

git-svn-id: http://skia.googlecode.com/svn/trunk@5541 2bbb7eff-a529-9590-31e7-b0007b416f81
2012-09-14 14:19:30 +00:00

91 lines
2.5 KiB
C++

/*
* Copyright 2012 Google Inc.
*
* Use of this source code is governed by a BSD-style license that can be
* found in the LICENSE file.
*/
#include "QuadraticUtilities.h"
#include <math.h>
/*
Numeric Solutions (5.6) suggests to solve the quadratic by computing
Q = -1/2(B + sgn(B)Sqrt(B^2 - 4 A C))
and using the roots
t1 = Q / A
t2 = C / Q
*/
// note: caller expects multiple results to be sorted smaller first
// note: http://en.wikipedia.org/wiki/Loss_of_significance has an interesting
// analysis of the quadratic equation, suggesting why the following looks at
// the sign of B -- and further suggesting that the greatest loss of precision
// is in b squared less two a c
int quadraticRoots(double A, double B, double C, double t[2]) {
B *= 2;
double square = B * B - 4 * A * C;
if (approximately_negative(square)) {
if (!approximately_positive(square)) {
return 0;
}
square = 0;
}
double squareRt = sqrt(square);
double Q = (B + (B < 0 ? -squareRt : squareRt)) / -2;
int foundRoots = 0;
double ratio = Q / A;
if (approximately_zero_or_more(ratio) && approximately_one_or_less(ratio)) {
if (approximately_less_than_zero(ratio)) {
ratio = 0;
} else if (approximately_greater_than_one(ratio)) {
ratio = 1;
}
t[0] = ratio;
++foundRoots;
}
ratio = C / Q;
if (approximately_zero_or_more(ratio) && approximately_one_or_less(ratio)) {
if (approximately_less_than_zero(ratio)) {
ratio = 0;
} else if (approximately_greater_than_one(ratio)) {
ratio = 1;
}
if (foundRoots == 0 || !approximately_negative(ratio - t[0])) {
t[foundRoots++] = ratio;
} else if (!approximately_negative(t[0] - ratio)) {
t[foundRoots++] = t[0];
t[0] = ratio;
}
}
return foundRoots;
}
void dxdy_at_t(const Quadratic& quad, double t, double& x, double& y) {
double a = t - 1;
double b = 1 - 2 * t;
double c = t;
if (&x) {
x = a * quad[0].x + b * quad[1].x + c * quad[2].x;
}
if (&y) {
y = a * quad[0].y + b * quad[1].y + c * quad[2].y;
}
}
void xy_at_t(const Quadratic& quad, double t, double& x, double& y) {
double one_t = 1 - t;
double a = one_t * one_t;
double b = 2 * one_t * t;
double c = t * t;
if (&x) {
x = a * quad[0].x + b * quad[1].x + c * quad[2].x;
}
if (&y) {
y = a * quad[0].y + b * quad[1].y + c * quad[2].y;
}
}