/* * Copyright 2012 Google Inc. * * Use of this source code is governed by a BSD-style license that can be * found in the LICENSE file. */ #include "include/core/SkPath.h" #include "include/core/SkPoint.h" #include "include/core/SkScalar.h" #include "src/pathops/SkIntersections.h" #include "src/pathops/SkPathOpsCurve.h" #include "src/pathops/SkPathOpsDebug.h" #include "src/pathops/SkPathOpsLine.h" #include "src/pathops/SkPathOpsPoint.h" #include "src/pathops/SkPathOpsQuad.h" #include "src/pathops/SkPathOpsTypes.h" #include <cmath> /* Find the intersection 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)) } } 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 { … }; int SkIntersections::horizontal(const SkDQuad& quad, double left, double right, double y, bool flipped) { … } int SkIntersections::vertical(const SkDQuad& quad, double top, double bottom, double x, bool flipped) { … } int SkIntersections::intersect(const SkDQuad& quad, const SkDLine& line) { … } int SkIntersections::intersectRay(const SkDQuad& quad, const SkDLine& line) { … } int SkIntersections::HorizontalIntercept(const SkDQuad& quad, SkScalar y, double* roots) { … } int SkIntersections::VerticalIntercept(const SkDQuad& quad, SkScalar x, double* roots) { … } // SkDQuad accessors to Intersection utilities int SkDQuad::horizontalIntersect(double yIntercept, double roots[2]) const { … } int SkDQuad::verticalIntersect(double xIntercept, double roots[2]) const { … }
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