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#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))
}
}
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
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 Intersections {
public:
LineQuadraticIntersections(const Quadratic& q, const _Line& l, Intersections& i)
: quad(q)
, line(l)
, intersections(i) {
}
bool intersect() {
double slope;
double axisIntercept;
moreHorizontal = implicitLine(line, slope, axisIntercept);
double A = quad[2].x; // c
double B = quad[1].x; // b
double C = quad[0].x; // a
A += C - 2 * B; // A = a - 2*b + c
B -= C; // B = -(a - b)
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)
if (moreHorizontal) {
A = A * slope - D;
B = B * slope - E;
C = C * slope - F + axisIntercept;
} else {
A = A - D * slope;
B = B - E * slope;
C = C - F * slope - axisIntercept;
}
double t[2];
int roots = quadraticRoots(A, B, C, t);
for (int x = 0; x < roots; ++x) {
intersections.add(t[x], findLineT(t[x]));
}
return roots > 0;
}
int horizontalIntersect(double axisIntercept) {
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 quadraticRoots(D, E, F, intersections.fT[0]);
}
protected:
double findLineT(double t) {
const double* qPtr;
const double* lPtr;
if (moreHorizontal) {
qPtr = &quad[0].x;
lPtr = &line[0].x;
} else {
qPtr = &quad[0].y;
lPtr = &line[0].y;
}
double s = 1 - t;
double quadVal = qPtr[0] * s * s + 2 * qPtr[2] * s * t + qPtr[4] * t * t;
return (quadVal - lPtr[0]) / (lPtr[2] - lPtr[0]);
}
private:
const Quadratic& quad;
const _Line& line;
Intersections& intersections;
bool moreHorizontal;
};
int horizontalIntersect(const Quadratic& quad, double left, double right,
double y, double tRange[2]) {
Intersections i;
LineQuadraticIntersections q(quad, *((_Line*) 0), i);
int result = q.horizontalIntersect(y);
int tCount = 0;
for (int index = 0; index < result; ++index) {
double x, y;
xy_at_t(quad, i.fT[0][index], x, y);
if (x < left || x > right) {
continue;
}
tRange[tCount++] = i.fT[0][index];
}
return tCount;
}
bool intersect(const Quadratic& quad, const _Line& line, Intersections& i) {
LineQuadraticIntersections q(quad, line, i);
return q.intersect();
}
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