Intersection work in progress
Review URL: https://codereview.appspot.com/5576043
git-svn-id: http://skia.googlecode.com/svn/trunk@3087 2bbb7eff-a529-9590-31e7-b0007b416f81
diff --git a/experimental/Intersection/LineQuadraticIntersection.cpp b/experimental/Intersection/LineQuadraticIntersection.cpp
new file mode 100644
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+++ b/experimental/Intersection/LineQuadraticIntersection.cpp
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+#include "CubicIntersection.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;
+}
+
+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;
+
+};
+
+bool intersectStart(const Quadratic& quad, const _Line& line, Intersections& i) {
+ LineQuadraticIntersections q(quad, line, i);
+ return q.intersect();
+}