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#include "CurveIntersection.h"
#include "CubicUtilities.h"
#include "Intersections.h"
#include "LineUtilities.h"
/*
Find the interection of a line and cubic by solving for valid t values.
Analogous to line-quadratic intersection, solve line-cubic intersection by
representing the cubic as:
x = a(1-t)^3 + 2b(1-t)^2t + c(1-t)t^2 + dt^3
y = e(1-t)^3 + 2f(1-t)^2t + g(1-t)t^2 + ht^3
and the line as:
y = i*x + j (if the line is more horizontal)
or:
x = i*y + j (if the line is more vertical)
Then using Mathematica, solve for the values of t where the cubic intersects the
line:
(in) Resultant[
a*(1 - t)^3 + 3*b*(1 - t)^2*t + 3*c*(1 - t)*t^2 + d*t^3 - x,
e*(1 - t)^3 + 3*f*(1 - t)^2*t + 3*g*(1 - t)*t^2 + h*t^3 - i*x - j, x]
(out) -e + j +
3 e t - 3 f t -
3 e t^2 + 6 f t^2 - 3 g t^2 +
e t^3 - 3 f t^3 + 3 g t^3 - h t^3 +
i ( a -
3 a t + 3 b t +
3 a t^2 - 6 b t^2 + 3 c t^2 -
a t^3 + 3 b t^3 - 3 c t^3 + d t^3 )
if i goes to infinity, we can rewrite the line in terms of x. Mathematica:
(in) Resultant[
a*(1 - t)^3 + 3*b*(1 - t)^2*t + 3*c*(1 - t)*t^2 + d*t^3 - i*y - j,
e*(1 - t)^3 + 3*f*(1 - t)^2*t + 3*g*(1 - t)*t^2 + h*t^3 - y, y]
(out) a - j -
3 a t + 3 b t +
3 a t^2 - 6 b t^2 + 3 c t^2 -
a t^3 + 3 b t^3 - 3 c t^3 + d t^3 -
i ( e -
3 e t + 3 f t +
3 e t^2 - 6 f t^2 + 3 g t^2 -
e t^3 + 3 f t^3 - 3 g t^3 + h t^3 )
Solving this with Mathematica produces an expression with hundreds of terms;
instead, use Numeric Solutions recipe to solve the cubic.
The near-horizontal case, in terms of: Ax^3 + Bx^2 + Cx + D == 0
A = (-(-e + 3*f - 3*g + h) + i*(-a + 3*b - 3*c + d) )
B = 3*(-( e - 2*f + g ) + i*( a - 2*b + c ) )
C = 3*(-(-e + f ) + i*(-a + b ) )
D = (-( e ) + i*( a ) + j )
The near-vertical case, in terms of: Ax^3 + Bx^2 + Cx + D == 0
A = ( (-a + 3*b - 3*c + d) - i*(-e + 3*f - 3*g + h) )
B = 3*( ( a - 2*b + c ) - i*( e - 2*f + g ) )
C = 3*( (-a + b ) - i*(-e + f ) )
D = ( ( a ) - i*( e ) - j )
For horizontal lines:
(in) Resultant[
a*(1 - t)^3 + 3*b*(1 - t)^2*t + 3*c*(1 - t)*t^2 + d*t^3 - j,
e*(1 - t)^3 + 3*f*(1 - t)^2*t + 3*g*(1 - t)*t^2 + h*t^3 - y, y]
(out) e - j -
3 e t + 3 f t +
3 e t^2 - 6 f t^2 + 3 g t^2 -
e t^3 + 3 f t^3 - 3 g t^3 + h t^3
So the cubic coefficients are:
*/
class LineCubicIntersections : public Intersections {
public:
LineCubicIntersections(const Cubic& c, const _Line& l, double r[3])
: cubic(c)
, line(l)
, range(r) {
}
int intersect() {
double slope;
double axisIntercept;
moreHorizontal = implicitLine(line, slope, axisIntercept);
double A, B, C, D;
coefficients(&cubic[0].x, A, B, C, D);
double E, F, G, H;
coefficients(&cubic[0].y, E, F, G, H);
if (moreHorizontal) {
A = A * slope - E;
B = B * slope - F;
C = C * slope - G;
D = D * slope - H + axisIntercept;
} else {
A = A - E * slope;
B = B - F * slope;
C = C - G * slope;
D = D - H * slope - axisIntercept;
}
return cubicRoots(A, B, C, D, range);
}
int horizontalIntersect(double axisIntercept) {
double A, B, C, D;
coefficients(&cubic[0].y, A, B, C, D);
D -= axisIntercept;
return cubicRoots(A, B, C, D, range);
}
int verticalIntersect(double axisIntercept) {
double A, B, C, D;
coefficients(&cubic[0].x, A, B, C, D);
D -= axisIntercept;
return cubicRoots(A, B, C, D, range);
}
double findLineT(double t) {
const double* cPtr;
const double* lPtr;
if (moreHorizontal) {
cPtr = &cubic[0].x;
lPtr = &line[0].x;
} else {
cPtr = &cubic[0].y;
lPtr = &line[0].y;
}
// FIXME: should fold the following in with TestUtilities.cpp xy_at_t()
double s = 1 - t;
double cubicVal = cPtr[0] * s * s * s + 3 * cPtr[2] * s * s * t
+ 3 * cPtr[4] * s * t * t + cPtr[6] * t * t * t;
return (cubicVal - lPtr[0]) / (lPtr[2] - lPtr[0]);
}
private:
const Cubic& cubic;
const _Line& line;
double* range;
bool moreHorizontal;
};
int horizontalIntersect(const Cubic& cubic, double y, double tRange[3]) {
LineCubicIntersections c(cubic, *((_Line*) 0), tRange);
return c.horizontalIntersect(y);
}
int horizontalIntersect(const Cubic& cubic, double left, double right, double y,
double tRange[3]) {
LineCubicIntersections c(cubic, *((_Line*) 0), tRange);
int result = c.horizontalIntersect(y);
for (int index = 0; index < result; ) {
double x, y;
xy_at_t(cubic, tRange[index], x, y);
if (x < left || x > right) {
if (--result > index) {
tRange[index] = tRange[result];
}
continue;
}
++index;
}
return result;
}
int horizontalIntersect(const Cubic& cubic, double left, double right, double y,
bool flipped, Intersections& intersections) {
LineCubicIntersections c(cubic, *((_Line*) 0), intersections.fT[0]);
int result = c.horizontalIntersect(y);
for (int index = 0; index < result; ) {
double x, y;
xy_at_t(cubic, intersections.fT[0][index], x, y);
if (x < left || x > right) {
if (--result > index) {
intersections.fT[0][index] = intersections.fT[0][result];
}
continue;
}
intersections.fT[0][index] = (x - left) / (right - left);
++index;
}
if (flipped) {
// OPTIMIZATION: instead of swapping, pass original line, use [1].x - [0].x
for (int index = 0; index < result; ++index) {
intersections.fT[1][index] = 1 - intersections.fT[1][index];
}
}
return result;
}
int verticalIntersect(const Cubic& cubic, double top, double bottom, double x,
bool flipped, Intersections& intersections) {
LineCubicIntersections c(cubic, *((_Line*) 0), intersections.fT[0]);
int result = c.verticalIntersect(x);
for (int index = 0; index < result; ) {
double x, y;
xy_at_t(cubic, intersections.fT[0][index], x, y);
if (y < top || y > bottom) {
if (--result > index) {
intersections.fT[0][index] = intersections.fT[0][result];
}
continue;
}
intersections.fT[0][index] = (y - top) / (bottom - top);
++index;
}
if (flipped) {
// OPTIMIZATION: instead of swapping, pass original line, use [1].x - [0].x
for (int index = 0; index < result; ++index) {
intersections.fT[1][index] = 1 - intersections.fT[1][index];
}
}
return result;
}
int intersect(const Cubic& cubic, const _Line& line, double cRange[3], double lRange[3]) {
LineCubicIntersections c(cubic, line, cRange);
int roots;
if (approximately_equal(line[0].y, line[1].y)) {
roots = c.horizontalIntersect(line[0].y);
} else {
roots = c.intersect();
}
for (int index = 0; index < roots; ++index) {
lRange[index] = c.findLineT(cRange[index]);
}
return roots;
}
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