fa0588ff67
in the middle of switching to sortless version git-svn-id: http://skia.googlecode.com/svn/trunk@3768 2bbb7eff-a529-9590-31e7-b0007b416f81
291 lines
8.9 KiB
C++
291 lines
8.9 KiB
C++
#include "CurveIntersection.h"
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#include "Intersections.h"
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#include "LineUtilities.h"
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#include "QuadraticUtilities.h"
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/*
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Find the interection of a line and quadratic by solving for valid t values.
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From http://stackoverflow.com/questions/1853637/how-to-find-the-mathematical-function-defining-a-bezier-curve
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"A Bezier curve is a parametric function. A quadratic Bezier curve (i.e. three
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control points) can be expressed as: F(t) = A(1 - t)^2 + B(1 - t)t + Ct^2 where
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A, B and C are points and t goes from zero to one.
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This will give you two equations:
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x = a(1 - t)^2 + b(1 - t)t + ct^2
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y = d(1 - t)^2 + e(1 - t)t + ft^2
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If you add for instance the line equation (y = kx + m) to that, you'll end up
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with three equations and three unknowns (x, y and t)."
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Similar to above, the quadratic is represented as
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x = a(1-t)^2 + 2b(1-t)t + ct^2
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y = d(1-t)^2 + 2e(1-t)t + ft^2
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and the line as
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y = g*x + h
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Using Mathematica, solve for the values of t where the quadratic intersects the
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line:
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(in) t1 = Resultant[a*(1 - t)^2 + 2*b*(1 - t)*t + c*t^2 - x,
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d*(1 - t)^2 + 2*e*(1 - t)*t + f*t^2 - g*x - h, x]
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(out) -d + h + 2 d t - 2 e t - d t^2 + 2 e t^2 - f t^2 +
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g (a - 2 a t + 2 b t + a t^2 - 2 b t^2 + c t^2)
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(in) Solve[t1 == 0, t]
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(out) {
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{t -> (-2 d + 2 e + 2 a g - 2 b g -
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Sqrt[(2 d - 2 e - 2 a g + 2 b g)^2 -
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4 (-d + 2 e - f + a g - 2 b g + c g) (-d + a g + h)]) /
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(2 (-d + 2 e - f + a g - 2 b g + c g))
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},
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{t -> (-2 d + 2 e + 2 a g - 2 b g +
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Sqrt[(2 d - 2 e - 2 a g + 2 b g)^2 -
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4 (-d + 2 e - f + a g - 2 b g + c g) (-d + a g + h)]) /
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(2 (-d + 2 e - f + a g - 2 b g + c g))
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}
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}
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Numeric Solutions (5.6) suggests to solve the quadratic by computing
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Q = -1/2(B + sgn(B)Sqrt(B^2 - 4 A C))
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and using the roots
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t1 = Q / A
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t2 = C / Q
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Using the results above (when the line tends towards horizontal)
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A = (-(d - 2*e + f) + g*(a - 2*b + c) )
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B = 2*( (d - e ) - g*(a - b ) )
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C = (-(d ) + g*(a ) + h )
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If g goes to infinity, we can rewrite the line in terms of x.
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x = g'*y + h'
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And solve accordingly in Mathematica:
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(in) t2 = Resultant[a*(1 - t)^2 + 2*b*(1 - t)*t + c*t^2 - g'*y - h',
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d*(1 - t)^2 + 2*e*(1 - t)*t + f*t^2 - y, y]
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(out) a - h' - 2 a t + 2 b t + a t^2 - 2 b t^2 + c t^2 -
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g' (d - 2 d t + 2 e t + d t^2 - 2 e t^2 + f t^2)
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(in) Solve[t2 == 0, t]
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(out) {
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{t -> (2 a - 2 b - 2 d g' + 2 e g' -
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Sqrt[(-2 a + 2 b + 2 d g' - 2 e g')^2 -
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4 (a - 2 b + c - d g' + 2 e g' - f g') (a - d g' - h')]) /
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(2 (a - 2 b + c - d g' + 2 e g' - f g'))
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},
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{t -> (2 a - 2 b - 2 d g' + 2 e g' +
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Sqrt[(-2 a + 2 b + 2 d g' - 2 e g')^2 -
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4 (a - 2 b + c - d g' + 2 e g' - f g') (a - d g' - h')])/
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(2 (a - 2 b + c - d g' + 2 e g' - f g'))
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}
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}
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Thus, if the slope of the line tends towards vertical, we use:
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A = ( (a - 2*b + c) - g'*(d - 2*e + f) )
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B = 2*(-(a - b ) + g'*(d - e ) )
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C = ( (a ) - g'*(d ) - h' )
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*/
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class LineQuadraticIntersections : public Intersections {
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public:
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LineQuadraticIntersections(const Quadratic& q, const _Line& l, Intersections& i)
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: quad(q)
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, line(l)
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, intersections(i) {
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}
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bool intersect() {
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double slope;
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double axisIntercept;
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moreHorizontal = implicitLine(line, slope, axisIntercept);
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double A = quad[2].x; // c
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double B = quad[1].x; // b
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double C = quad[0].x; // a
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A += C - 2 * B; // A = a - 2*b + c
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B -= C; // B = -(a - b)
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double D = quad[2].y; // f
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double E = quad[1].y; // e
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double F = quad[0].y; // d
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D += F - 2 * E; // D = d - 2*e + f
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E -= F; // E = -(d - e)
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if (moreHorizontal) {
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A = A * slope - D;
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B = B * slope - E;
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C = C * slope - F + axisIntercept;
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} else {
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A = A - D * slope;
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B = B - E * slope;
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C = C - F * slope - axisIntercept;
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}
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double t[2];
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int roots = quadraticRoots(A, B, C, t);
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for (int x = 0; x < roots; ++x) {
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intersections.add(t[x], findLineT(t[x]));
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}
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return roots > 0;
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}
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int horizontalIntersect(double axisIntercept) {
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double D = quad[2].y; // f
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double E = quad[1].y; // e
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double F = quad[0].y; // d
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D += F - 2 * E; // D = d - 2*e + f
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E -= F; // E = -(d - e)
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F -= axisIntercept;
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return quadraticRoots(D, E, F, intersections.fT[0]);
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}
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int verticalIntersect(double axisIntercept) {
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double D = quad[2].x; // f
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double E = quad[1].x; // e
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double F = quad[0].x; // d
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D += F - 2 * E; // D = d - 2*e + f
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E -= F; // E = -(d - e)
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F -= axisIntercept;
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return quadraticRoots(D, E, F, intersections.fT[0]);
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}
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protected:
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double findLineT(double t) {
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const double* qPtr;
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const double* lPtr;
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if (moreHorizontal) {
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qPtr = &quad[0].x;
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lPtr = &line[0].x;
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} else {
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qPtr = &quad[0].y;
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lPtr = &line[0].y;
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}
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double s = 1 - t;
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double quadVal = qPtr[0] * s * s + 2 * qPtr[2] * s * t + qPtr[4] * t * t;
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return (quadVal - lPtr[0]) / (lPtr[2] - lPtr[0]);
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}
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private:
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const Quadratic& quad;
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const _Line& line;
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Intersections& intersections;
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bool moreHorizontal;
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};
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// utility for pairs of coincident quads
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static double horizontalIntersect(const Quadratic& quad, const _Point& pt) {
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Intersections intersections;
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LineQuadraticIntersections q(quad, *((_Line*) 0), intersections);
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int result = q.horizontalIntersect(pt.y);
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if (result == 0) {
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return -1;
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}
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assert(result == 1);
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double x, y;
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xy_at_t(quad, intersections.fT[0][0], x, y);
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if (approximately_equal(x, pt.x)) {
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return intersections.fT[0][0];
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}
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return -1;
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}
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static double verticalIntersect(const Quadratic& quad, const _Point& pt) {
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Intersections intersections;
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LineQuadraticIntersections q(quad, *((_Line*) 0), intersections);
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int result = q.horizontalIntersect(pt.x);
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if (result == 0) {
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return -1;
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}
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assert(result == 1);
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double x, y;
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xy_at_t(quad, intersections.fT[0][0], x, y);
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if (approximately_equal(y, pt.y)) {
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return intersections.fT[0][0];
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}
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return -1;
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}
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double axialIntersect(const Quadratic& q1, const _Point& p, bool vertical) {
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if (vertical) {
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return verticalIntersect(q1, p);
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}
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return horizontalIntersect(q1, p);
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}
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int horizontalIntersect(const Quadratic& quad, double left, double right,
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double y, double tRange[2]) {
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Intersections i;
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LineQuadraticIntersections q(quad, *((_Line*) 0), i);
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int result = q.horizontalIntersect(y);
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int tCount = 0;
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for (int index = 0; index < result; ++index) {
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double x, y;
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xy_at_t(quad, i.fT[0][index], x, y);
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if (x < left || x > right) {
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continue;
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}
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tRange[tCount++] = i.fT[0][index];
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}
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return tCount;
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}
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int horizontalIntersect(const Quadratic& quad, double left, double right, double y,
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bool flipped, Intersections& intersections) {
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LineQuadraticIntersections q(quad, *((_Line*) 0), intersections);
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int result = q.horizontalIntersect(y);
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for (int index = 0; index < result; ) {
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double x, y;
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xy_at_t(quad, intersections.fT[0][index], x, y);
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if (x < left || x > right) {
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if (--result > index) {
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intersections.fT[0][index] = intersections.fT[0][result];
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}
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continue;
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}
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intersections.fT[0][index] = (x - left) / (right - left);
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++index;
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}
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if (flipped) {
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// OPTIMIZATION: instead of swapping, pass original line, use [1].x - [0].x
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for (int index = 0; index < result; ++index) {
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intersections.fT[1][index] = 1 - intersections.fT[1][index];
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}
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}
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return result;
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}
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int verticalIntersect(const Quadratic& quad, double top, double bottom, double x,
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bool flipped, Intersections& intersections) {
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LineQuadraticIntersections q(quad, *((_Line*) 0), intersections);
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int result = q.verticalIntersect(x);
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for (int index = 0; index < result; ) {
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double x, y;
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xy_at_t(quad, intersections.fT[0][index], x, y);
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if (y < top || y > bottom) {
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if (--result > index) {
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intersections.fT[0][index] = intersections.fT[0][result];
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}
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continue;
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}
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intersections.fT[0][index] = (y - top) / (bottom - top);
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++index;
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}
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if (flipped) {
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// OPTIMIZATION: instead of swapping, pass original line, use [1].x - [0].x
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for (int index = 0; index < result; ++index) {
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intersections.fT[1][index] = 1 - intersections.fT[1][index];
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}
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}
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return result;
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}
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bool intersect(const Quadratic& quad, const _Line& line, Intersections& i) {
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LineQuadraticIntersections q(quad, line, i);
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return q.intersect();
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}
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