235f56a92f
add quartic solution for intersecting quadratics git-svn-id: http://skia.googlecode.com/svn/trunk@5541 2bbb7eff-a529-9590-31e7-b0007b416f81
319 lines
10 KiB
C++
319 lines
10 KiB
C++
/*
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* Copyright 2012 Google Inc.
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*
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* Use of this source code is governed by a BSD-style license that can be
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* found in the LICENSE file.
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*/
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#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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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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int intersectRay() {
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/*
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solve by rotating line+quad so line is horizontal, then finding the roots
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set up matrix to rotate quad to x-axis
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|cos(a) -sin(a)|
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|sin(a) cos(a)|
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note that cos(a) = A(djacent) / Hypoteneuse
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sin(a) = O(pposite) / Hypoteneuse
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since we are computing Ts, we can ignore hypoteneuse, the scale factor:
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| A -O |
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| O A |
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A = line[1].x - line[0].x (adjacent side of the right triangle)
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O = line[1].y - line[0].y (opposite side of the right triangle)
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for each of the three points (e.g. n = 0 to 2)
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quad[n].y' = (quad[n].y - line[0].y) * A - (quad[n].x - line[0].x) * O
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*/
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double adj = line[1].x - line[0].x;
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double opp = line[1].y - line[0].y;
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double r[3];
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for (int n = 0; n < 3; ++n) {
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r[n] = (quad[n].y - line[0].y) * adj - (quad[n].x - line[0].x) * opp;
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}
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double A = r[2];
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double B = r[1];
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double C = r[0];
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A += C - 2 * B; // A = a - 2*b + c
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B -= C; // B = -(b - c)
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return quadraticRoots(A, B, C, intersections.fT[0]);
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}
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int intersect() {
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int roots = intersectRay();
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for (int index = 0; index < roots; ) {
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double lineT = findLineT(intersections.fT[0][index]);
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if (lineIntersects(lineT, index, roots)) {
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++index;
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}
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}
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return roots;
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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 horizontalIntersect(double axisIntercept, double left, double right) {
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int roots = horizontalIntersect(axisIntercept);
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for (int index = 0; index < roots; ) {
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double x;
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xy_at_t(quad, intersections.fT[0][index], x, *(double*) NULL);
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double lineT = (x - left) / (right - left);
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if (lineIntersects(lineT, index, roots)) {
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++index;
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}
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}
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return roots;
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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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int verticalIntersect(double axisIntercept, double top, double bottom) {
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int roots = verticalIntersect(axisIntercept);
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for (int index = 0; index < roots; ) {
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double y;
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xy_at_t(quad, intersections.fT[0][index], *(double*) NULL, y);
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double lineT = (y - top) / (bottom - top);
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if (lineIntersects(lineT, index, roots)) {
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++index;
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}
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}
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return roots;
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}
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protected:
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double findLineT(double t) {
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double x, y;
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xy_at_t(quad, t, x, y);
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if (approximately_equal(x, line[0].x) && approximately_equal(y, line[0].y)) {
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return 0;
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}
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if (approximately_equal(x, line[1].x) && approximately_equal(y, line[1].y)) {
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return 1;
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}
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double dx = line[1].x - line[0].x;
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double dy = line[1].y - line[0].y;
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if (fabs(dx) > fabs(dy)) {
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return (x - line[0].x) / dx;
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}
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return (y - line[0].y) / dy;
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}
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bool lineIntersects(double lineT, const int x, int& roots) {
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if (!approximately_zero_or_more(lineT) || !approximately_one_or_less(lineT)) {
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if (x < --roots) {
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intersections.fT[0][x] = intersections.fT[0][roots];
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}
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return false;
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}
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if (approximately_less_than_zero(lineT)) {
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lineT = 0;
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} else if (approximately_greater_than_one(lineT)) {
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lineT = 1;
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}
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intersections.fT[1][x] = lineT;
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return true;
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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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};
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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 roots = q.horizontalIntersect(pt.y);
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for (int index = 0; index < roots; ++index) {
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double x;
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double t = intersections.fT[0][index];
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xy_at_t(quad, t, x, *(double*) 0);
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if (approximately_equal(x, pt.x)) {
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return t;
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}
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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 roots = q.verticalIntersect(pt.x);
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for (int index = 0; index < roots; ++index) {
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double y;
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double t = intersections.fT[0][index];
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xy_at_t(quad, t, *(double*) 0, y);
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if (approximately_equal(y, pt.y)) {
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return t;
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}
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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, left, right);
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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, top, bottom);
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if (flipped) {
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// OPTIMIZATION: instead of swapping, pass original line, use [1].y - [0].y
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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 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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int intersectRay(const Quadratic& quad, const _Line& line, Intersections& i) {
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LineQuadraticIntersections q(quad, line, i);
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return q.intersectRay();
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}
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