/*
* 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 "CurveIntersection.h"
#include "Intersections.h"
#include "LineUtilities.h"
#include "QuadraticUtilities.h"
/*
Find the interection of a line and quadratic by solving for valid t values.
From http://stackoverflow.com/questions/1853637/how-to-find-the-mathematical-function-defining-a-bezier-curve
"A Bezier curve is a parametric function. A quadratic Bezier curve (i.e. three
control points) can be expressed as: F(t) = A(1 - t)^2 + B(1 - t)t + Ct^2 where
A, B and C are points and t goes from zero to one.
This will give you two equations:
x = a(1 - t)^2 + b(1 - t)t + ct^2
y = d(1 - t)^2 + e(1 - t)t + ft^2
If you add for instance the line equation (y = kx + m) to that, you'll end up
with three equations and three unknowns (x, y and t)."
Similar to above, the quadratic is represented as
x = a(1-t)^2 + 2b(1-t)t + ct^2
y = d(1-t)^2 + 2e(1-t)t + ft^2
and the line as
y = g*x + h
Using Mathematica, solve for the values of t where the quadratic intersects the
line:
(in) t1 = Resultant[a*(1 - t)^2 + 2*b*(1 - t)*t + c*t^2 - x,
d*(1 - t)^2 + 2*e*(1 - t)*t + f*t^2 - g*x - h, x]
(out) -d + h + 2 d t - 2 e t - d t^2 + 2 e t^2 - f t^2 +
g (a - 2 a t + 2 b t + a t^2 - 2 b t^2 + c t^2)
(in) Solve[t1 == 0, t]
(out) {
{t -> (-2 d + 2 e + 2 a g - 2 b g -
Sqrt[(2 d - 2 e - 2 a g + 2 b g)^2 -
4 (-d + 2 e - f + a g - 2 b g + c g) (-d + a g + h)]) /
(2 (-d + 2 e - f + a g - 2 b g + c g))
},
{t -> (-2 d + 2 e + 2 a g - 2 b g +
Sqrt[(2 d - 2 e - 2 a g + 2 b g)^2 -
4 (-d + 2 e - f + a g - 2 b g + c g) (-d + a g + h)]) /
(2 (-d + 2 e - f + a g - 2 b g + c g))
}
}
Using the results above (when the line tends towards horizontal)
A = (-(d - 2*e + f) + g*(a - 2*b + c) )
B = 2*( (d - e ) - g*(a - b ) )
C = (-(d ) + g*(a ) + h )
If g goes to infinity, we can rewrite the line in terms of x.
x = g'*y + h'
And solve accordingly in Mathematica:
(in) t2 = Resultant[a*(1 - t)^2 + 2*b*(1 - t)*t + c*t^2 - g'*y - h',
d*(1 - t)^2 + 2*e*(1 - t)*t + f*t^2 - y, y]
(out) a - h' - 2 a t + 2 b t + a t^2 - 2 b t^2 + c t^2 -
g' (d - 2 d t + 2 e t + d t^2 - 2 e t^2 + f t^2)
(in) Solve[t2 == 0, t]
(out) {
{t -> (2 a - 2 b - 2 d g' + 2 e g' -
Sqrt[(-2 a + 2 b + 2 d g' - 2 e g')^2 -
4 (a - 2 b + c - d g' + 2 e g' - f g') (a - d g' - h')]) /
(2 (a - 2 b + c - d g' + 2 e g' - f g'))
},
{t -> (2 a - 2 b - 2 d g' + 2 e g' +
Sqrt[(-2 a + 2 b + 2 d g' - 2 e g')^2 -
4 (a - 2 b + c - d g' + 2 e g' - f g') (a - d g' - h')])/
(2 (a - 2 b + c - d g' + 2 e g' - f g'))
}
}
Thus, if the slope of the line tends towards vertical, we use:
A = ( (a - 2*b + c) - g'*(d - 2*e + f) )
B = 2*(-(a - b ) + g'*(d - e ) )
C = ( (a ) - g'*(d ) - h' )
*/
class LineQuadraticIntersections {
public:
LineQuadraticIntersections(const Quadratic& q, const _Line& l, Intersections& i)
: quad(q)
, line(l)
, intersections(i) {
}
int intersectRay(double roots[2]) {
/*
solve by rotating line+quad so line is horizontal, then finding the roots
set up matrix to rotate quad to x-axis
|cos(a) -sin(a)|
|sin(a) cos(a)|
note that cos(a) = A(djacent) / Hypoteneuse
sin(a) = O(pposite) / Hypoteneuse
since we are computing Ts, we can ignore hypoteneuse, the scale factor:
| A -O |
| O A |
A = line[1].x - line[0].x (adjacent side of the right triangle)
O = line[1].y - line[0].y (opposite side of the right triangle)
for each of the three points (e.g. n = 0 to 2)
quad[n].y' = (quad[n].y - line[0].y) * A - (quad[n].x - line[0].x) * O
*/
double adj = line[1].x - line[0].x;
double opp = line[1].y - line[0].y;
double r[3];
for (int n = 0; n < 3; ++n) {
r[n] = (quad[n].y - line[0].y) * adj - (quad[n].x - line[0].x) * opp;
}
double A = r[2];
double B = r[1];
double C = r[0];
A += C - 2 * B; // A = a - 2*b + c
B -= C; // B = -(b - c)
return quadraticRootsValidT(A, 2 * B, C, roots);
}
int intersect() {
addEndPoints();
double rootVals[2];
int roots = intersectRay(rootVals);
for (int index = 0; index < roots; ++index) {
double quadT = rootVals[index];
double lineT = findLineT(quadT);
if (pinTs(quadT, lineT)) {
_Point pt;
xy_at_t(line, lineT, pt.x, pt.y);
intersections.insert(quadT, lineT, pt);
}
}
return intersections.fUsed;
}
int horizontalIntersect(double axisIntercept, double roots[2]) {
double D = quad[2].y; // f
double E = quad[1].y; // e
double F = quad[0].y; // d
D += F - 2 * E; // D = d - 2*e + f
E -= F; // E = -(d - e)
F -= axisIntercept;
return quadraticRootsValidT(D, 2 * E, F, roots);
}
int horizontalIntersect(double axisIntercept, double left, double right, bool flipped) {
addHorizontalEndPoints(left, right, axisIntercept);
double rootVals[2];
int roots = horizontalIntersect(axisIntercept, rootVals);
for (int index = 0; index < roots; ++index) {
_Point pt;
double quadT = rootVals[index];
xy_at_t(quad, quadT, pt.x, pt.y);
double lineT = (pt.x - left) / (right - left);
if (pinTs(quadT, lineT)) {
intersections.insert(quadT, lineT, pt);
}
}
if (flipped) {
flip();
}
return intersections.fUsed;
}
int verticalIntersect(double axisIntercept, double roots[2]) {
double D = quad[2].x; // f
double E = quad[1].x; // e
double F = quad[0].x; // d
D += F - 2 * E; // D = d - 2*e + f
E -= F; // E = -(d - e)
F -= axisIntercept;
return quadraticRootsValidT(D, 2 * E, F, roots);
}
int verticalIntersect(double axisIntercept, double top, double bottom, bool flipped) {
addVerticalEndPoints(top, bottom, axisIntercept);
double rootVals[2];
int roots = verticalIntersect(axisIntercept, rootVals);
for (int index = 0; index < roots; ++index) {
_Point pt;
double quadT = rootVals[index];
xy_at_t(quad, quadT, pt.x, pt.y);
double lineT = (pt.y - top) / (bottom - top);
if (pinTs(quadT, lineT)) {
intersections.insert(quadT, lineT, pt);
}
}
if (flipped) {
flip();
}
return intersections.fUsed;
}
protected:
// add endpoints first to get zero and one t values exactly
void addEndPoints()
{
for (int qIndex = 0; qIndex < 3; qIndex += 2) {
for (int lIndex = 0; lIndex < 2; lIndex++) {
if (quad[qIndex] == line[lIndex]) {
intersections.insert(qIndex >> 1, lIndex, line[lIndex]);
}
}
}
}
void addHorizontalEndPoints(double left, double right, double y)
{
for (int qIndex = 0; qIndex < 3; qIndex += 2) {
if (quad[qIndex].y != y) {
continue;
}
if (quad[qIndex].x == left) {
intersections.insert(qIndex >> 1, 0, quad[qIndex]);
}
if (quad[qIndex].x == right) {
intersections.insert(qIndex >> 1, 1, quad[qIndex]);
}
}
}
void addVerticalEndPoints(double top, double bottom, double x)
{
for (int qIndex = 0; qIndex < 3; qIndex += 2) {
if (quad[qIndex].x != x) {
continue;
}
if (quad[qIndex].y == top) {
intersections.insert(qIndex >> 1, 0, quad[qIndex]);
}
if (quad[qIndex].y == bottom) {
intersections.insert(qIndex >> 1, 1, quad[qIndex]);
}
}
}
double findLineT(double t) {
double x, y;
xy_at_t(quad, t, x, y);
double dx = line[1].x - line[0].x;
double dy = line[1].y - line[0].y;
if (fabs(dx) > fabs(dy)) {
return (x - line[0].x) / dx;
}
return (y - line[0].y) / dy;
}
void flip() {
// OPTIMIZATION: instead of swapping, pass original line, use [1].y - [0].y
int roots = intersections.fUsed;
for (int index = 0; index < roots; ++index) {
intersections.fT[1][index] = 1 - intersections.fT[1][index];
}
}
static bool pinTs(double& quadT, double& lineT) {
if (!approximately_one_or_less(lineT)) {
return false;
}
if (!approximately_zero_or_more(lineT)) {
return false;
}
if (precisely_less_than_zero(quadT)) {
quadT = 0;
} else if (precisely_greater_than_one(quadT)) {
quadT = 1;
}
if (precisely_less_than_zero(lineT)) {
lineT = 0;
} else if (precisely_greater_than_one(lineT)) {
lineT = 1;
}
return true;
}
private:
const Quadratic& quad;
const _Line& line;
Intersections& intersections;
};
// utility for pairs of coincident quads
static double horizontalIntersect(const Quadratic& quad, const _Point& pt) {
LineQuadraticIntersections q(quad, *((_Line*) 0), *((Intersections*) 0));
double rootVals[2];
int roots = q.horizontalIntersect(pt.y, rootVals);
for (int index = 0; index < roots; ++index) {
double x;
double t = rootVals[index];
xy_at_t(quad, t, x, *(double*) 0);
if (AlmostEqualUlps(x, pt.x)) {
return t;
}
}
return -1;
}
static double verticalIntersect(const Quadratic& quad, const _Point& pt) {
LineQuadraticIntersections q(quad, *((_Line*) 0), *((Intersections*) 0));
double rootVals[2];
int roots = q.verticalIntersect(pt.x, rootVals);
for (int index = 0; index < roots; ++index) {
double y;
double t = rootVals[index];
xy_at_t(quad, t, *(double*) 0, y);
if (AlmostEqualUlps(y, pt.y)) {
return t;
}
}
return -1;
}
double axialIntersect(const Quadratic& q1, const _Point& p, bool vertical) {
if (vertical) {
return verticalIntersect(q1, p);
}
return horizontalIntersect(q1, p);
}
int horizontalIntersect(const Quadratic& quad, double left, double right,
double y, double tRange[2]) {
LineQuadraticIntersections q(quad, *((_Line*) 0), *((Intersections*) 0));
double rootVals[2];
int result = q.horizontalIntersect(y, rootVals);
int tCount = 0;
for (int index = 0; index < result; ++index) {
double x, y;
xy_at_t(quad, rootVals[index], x, y);
if (x < left || x > right) {
continue;
}
tRange[tCount++] = rootVals[index];
}
return tCount;
}
int horizontalIntersect(const Quadratic& quad, double left, double right, double y,
bool flipped, Intersections& intersections) {
LineQuadraticIntersections q(quad, *((_Line*) 0), intersections);
return q.horizontalIntersect(y, left, right, flipped);
}
int verticalIntersect(const Quadratic& quad, double top, double bottom, double x,
bool flipped, Intersections& intersections) {
LineQuadraticIntersections q(quad, *((_Line*) 0), intersections);
return q.verticalIntersect(x, top, bottom, flipped);
}
int intersect(const Quadratic& quad, const _Line& line, Intersections& i) {
LineQuadraticIntersections q(quad, line, i);
return q.intersect();
}
int intersectRay(const Quadratic& quad, const _Line& line, Intersections& i) {
LineQuadraticIntersections q(quad, line, i);
return q.intersectRay(i.fT[0]);
}