src/pathops/SkDQuadLineIntersection.cpp - platform/external/skqp - Gitiles

 /*
  * 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 "SkIntersections.h"
 #include "SkPathOpsLine.h"
 #include "SkPathOpsQuad.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))
          }
         }

 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:
     LineQuadraticIntersections(const SkDQuad& q, const SkDLine& l, SkIntersections* i)
         : quad(q)
         , line(l)
         , intersections(i) {
     }

     int intersectRay(double roots[2]) {
     /*
         solve by rotating line+quad so line is horizontal, then finding the roots
         set up matrix to rotate quad to x-axis
         |cos(a) -sin(a)|
         |sin(a)  cos(a)|
         note that cos(a) = A(djacent) / Hypoteneuse
                   sin(a) = O(pposite) / Hypoteneuse
         since we are computing Ts, we can ignore hypoteneuse, the scale factor:
         |  A     -O    |
         |  O      A    |
         A = line[1].fX - line[0].fX (adjacent side of the right triangle)
         O = line[1].fY - line[0].fY (opposite side of the right triangle)
         for each of the three points (e.g. n = 0 to 2)
         quad[n].fY' = (quad[n].fY - line[0].fY) * A - (quad[n].fX - line[0].fX) * O
     */
         double adj = line[1].fX - line[0].fX;
         double opp = line[1].fY - line[0].fY;
         double r[3];
         for (int n = 0; n < 3; ++n) {
             r[n] = (quad[n].fY - line[0].fY) * adj - (quad[n].fX - line[0].fX) * opp;
         }
         double A = r[2];
         double B = r[1];
         double C = r[0];
         A += C - 2 * B;  // A = a - 2*b + c
         B -= C;  // B = -(b - c)
         return SkDQuad::RootsValidT(A, 2 * B, C, roots);
     }

     int intersect() {
         addEndPoints();
         double rootVals[2];
         int roots = intersectRay(rootVals);
         for (int index = 0; index < roots; ++index) {
             double quadT = rootVals[index];
             double lineT = findLineT(quadT);
             if (PinTs(&quadT, &lineT)) {
                 SkDPoint pt = line.xyAtT(lineT);
                 intersections->insert(quadT, lineT, pt);
             }
         }
         return intersections->used();
     }

     int horizontalIntersect(double axisIntercept, double roots[2]) {
         double D = quad[2].fY;  // f
         double E = quad[1].fY;  // e
         double F = quad[0].fY;  // d
         D += F - 2 * E;         // D = d - 2*e + f
         E -= F;                 // E = -(d - e)
         F -= axisIntercept;
         return SkDQuad::RootsValidT(D, 2 * E, F, roots);
     }

     int horizontalIntersect(double axisIntercept, double left, double right, bool flipped) {
         addHorizontalEndPoints(left, right, axisIntercept);
         double rootVals[2];
         int roots = horizontalIntersect(axisIntercept, rootVals);
         for (int index = 0; index < roots; ++index) {
             double quadT = rootVals[index];
             SkDPoint pt = quad.xyAtT(quadT);
             double lineT = (pt.fX - left) / (right - left);
             if (PinTs(&quadT, &lineT)) {
                 intersections->insert(quadT, lineT, pt);
             }
         }
         if (flipped) {
             intersections->flip();
         }
         return intersections->used();
     }

     int verticalIntersect(double axisIntercept, double roots[2]) {
         double D = quad[2].fX;  // f
         double E = quad[1].fX;  // e
         double F = quad[0].fX;  // d
         D += F - 2 * E;         // D = d - 2*e + f
         E -= F;                 // E = -(d - e)
         F -= axisIntercept;
         return SkDQuad::RootsValidT(D, 2 * E, F, roots);
     }

     int verticalIntersect(double axisIntercept, double top, double bottom, bool flipped) {
         addVerticalEndPoints(top, bottom, axisIntercept);
         double rootVals[2];
         int roots = verticalIntersect(axisIntercept, rootVals);
         for (int index = 0; index < roots; ++index) {
             double quadT = rootVals[index];
             SkDPoint pt = quad.xyAtT(quadT);
             double lineT = (pt.fY - top) / (bottom - top);
             if (PinTs(&quadT, &lineT)) {
                 intersections->insert(quadT, lineT, pt);
             }
         }
         if (flipped) {
             intersections->flip();
         }
         return intersections->used();
     }

 protected:
     // add endpoints first to get zero and one t values exactly
     void addEndPoints() {
         for (int qIndex = 0; qIndex < 3; qIndex += 2) {
             for (int lIndex = 0; lIndex < 2; lIndex++) {
                 if (quad[qIndex] == line[lIndex]) {
                     intersections->insert(qIndex >> 1, lIndex, line[lIndex]);
                 }
             }
         }
     }

     void addHorizontalEndPoints(double left, double right, double y) {
         for (int qIndex = 0; qIndex < 3; qIndex += 2) {
             if (quad[qIndex].fY != y) {
                 continue;
             }
             if (quad[qIndex].fX == left) {
                 intersections->insert(qIndex >> 1, 0, quad[qIndex]);
             }
             if (quad[qIndex].fX == right) {
                 intersections->insert(qIndex >> 1, 1, quad[qIndex]);
             }
         }
     }

     void addVerticalEndPoints(double top, double bottom, double x) {
         for (int qIndex = 0; qIndex < 3; qIndex += 2) {
             if (quad[qIndex].fX != x) {
                 continue;
             }
             if (quad[qIndex].fY == top) {
                 intersections->insert(qIndex >> 1, 0, quad[qIndex]);
             }
             if (quad[qIndex].fY == bottom) {
                 intersections->insert(qIndex >> 1, 1, quad[qIndex]);
             }
         }
     }

     double findLineT(double t) {
         SkDPoint xy = quad.xyAtT(t);
         double dx = line[1].fX - line[0].fX;
         double dy = line[1].fY - line[0].fY;
         if (fabs(dx) > fabs(dy)) {
             return (xy.fX - line[0].fX) / dx;
         }
         return (xy.fY - line[0].fY) / dy;
     }

     static bool PinTs(double* quadT, double* lineT) {
         if (!approximately_one_or_less(*lineT)) {
             return false;
         }
         if (!approximately_zero_or_more(*lineT)) {
             return false;
         }
         if (precisely_less_than_zero(*quadT)) {
             *quadT = 0;
         } else if (precisely_greater_than_one(*quadT)) {
             *quadT = 1;
         }
         if (precisely_less_than_zero(*lineT)) {
             *lineT = 0;
         } else if (precisely_greater_than_one(*lineT)) {
             *lineT = 1;
         }
         return true;
     }

 private:
     const SkDQuad& quad;
     const SkDLine& line;
     SkIntersections* intersections;
 };

 // utility for pairs of coincident quads
 static double horizontalIntersect(const SkDQuad& quad, const SkDPoint& pt) {
     LineQuadraticIntersections q(quad, *(static_cast<SkDLine*>(0)),
             static_cast<SkIntersections*>(0));
     double rootVals[2];
     int roots = q.horizontalIntersect(pt.fY, rootVals);
     for (int index = 0; index < roots; ++index) {
         double t = rootVals[index];
         SkDPoint qPt = quad.xyAtT(t);
         if (AlmostEqualUlps(qPt.fX, pt.fX)) {
             return t;
         }
     }
     return -1;
 }

 static double verticalIntersect(const SkDQuad& quad, const SkDPoint& pt) {
     LineQuadraticIntersections q(quad, *(static_cast<SkDLine*>(0)),
             static_cast<SkIntersections*>(0));
     double rootVals[2];
     int roots = q.verticalIntersect(pt.fX, rootVals);
     for (int index = 0; index < roots; ++index) {
         double t = rootVals[index];
         SkDPoint qPt = quad.xyAtT(t);
         if (AlmostEqualUlps(qPt.fY, pt.fY)) {
             return t;
         }
     }
     return -1;
 }

 double SkIntersections::Axial(const SkDQuad& q1, const SkDPoint& p, bool vertical) {
     if (vertical) {
         return verticalIntersect(q1, p);
     }
     return horizontalIntersect(q1, p);
 }

 int SkIntersections::horizontal(const SkDQuad& quad, double left, double right, double y,
                                 bool flipped) {
     LineQuadraticIntersections q(quad, *(static_cast<SkDLine*>(0)), this);
     return q.horizontalIntersect(y, left, right, flipped);
 }

 int SkIntersections::vertical(const SkDQuad& quad, double top, double bottom, double x,
                               bool flipped) {
     LineQuadraticIntersections q(quad, *(static_cast<SkDLine*>(0)), this);
     return q.verticalIntersect(x, top, bottom, flipped);
 }

 int SkIntersections::intersect(const SkDQuad& quad, const SkDLine& line) {
     LineQuadraticIntersections q(quad, line, this);
     return q.intersect();
 }

 int SkIntersections::intersectRay(const SkDQuad& quad, const SkDLine& line) {
     LineQuadraticIntersections q(quad, line, this);
     fUsed = q.intersectRay(fT[0]);
     for (int index = 0; index < fUsed; ++index) {
         fPt[index] = quad.xyAtT(fT[0][index]);
     }
     return fUsed;
 }
	/*
	* 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 "SkIntersections.h"
	#include "SkPathOpsLine.h"
	#include "SkPathOpsQuad.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 + 2b(1 - t)t + c*t^2 - x,
	d(1 - t)^2 + 2e(1 - t)t + ft^2 - gx - 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 - 2e + 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 + 2b(1 - t)t + ct^2 - g'y - h',
	d(1 - t)^2 + 2e(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 - 2b + c) - g'(d - 2*e + f) )
	B = 2(-(a - b ) + g'(d - e ) )
	C = ( (a ) - g'*(d ) - h' )
	*/


	class LineQuadraticIntersections {
	public:
	LineQuadraticIntersections(const SkDQuad& q, const SkDLine& l, SkIntersections* i)
	: quad(q)
	, line(l)
	, intersections(i) {
	}

	int intersectRay(double roots[2]) {
	/*
	solve by rotating line+quad so line is horizontal, then finding the roots
	set up matrix to rotate quad to x-axis
	\|cos(a) -sin(a)\|
	\|sin(a) cos(a)\|
	note that cos(a) = A(djacent) / Hypoteneuse
	sin(a) = O(pposite) / Hypoteneuse
	since we are computing Ts, we can ignore hypoteneuse, the scale factor:
	\| A -O \|
	\| O A \|
	A = line[1].fX - line[0].fX (adjacent side of the right triangle)
	O = line[1].fY - line[0].fY (opposite side of the right triangle)
	for each of the three points (e.g. n = 0 to 2)
	quad[n].fY' = (quad[n].fY - line[0].fY) * A - (quad[n].fX - line[0].fX) * O
	*/
	double adj = line[1].fX - line[0].fX;
	double opp = line[1].fY - line[0].fY;
	double r[3];
	for (int n = 0; n < 3; ++n) {
	r[n] = (quad[n].fY - line[0].fY) * adj - (quad[n].fX - line[0].fX) * opp;
	}
	double A = r[2];
	double B = r[1];
	double C = r[0];
	A += C - 2 * B; // A = a - 2*b + c
	B -= C; // B = -(b - c)
	return SkDQuad::RootsValidT(A, 2 * B, C, roots);
	}

	int intersect() {
	addEndPoints();
	double rootVals[2];
	int roots = intersectRay(rootVals);
	for (int index = 0; index < roots; ++index) {
	double quadT = rootVals[index];
	double lineT = findLineT(quadT);
	if (PinTs(&quadT, &lineT)) {
	SkDPoint pt = line.xyAtT(lineT);
	intersections->insert(quadT, lineT, pt);
	}
	}
	return intersections->used();
	}

	int horizontalIntersect(double axisIntercept, double roots[2]) {
	double D = quad[2].fY; // f
	double E = quad[1].fY; // e
	double F = quad[0].fY; // d
	D += F - 2 * E; // D = d - 2*e + f
	E -= F; // E = -(d - e)
	F -= axisIntercept;
	return SkDQuad::RootsValidT(D, 2 * E, F, roots);
	}

	int horizontalIntersect(double axisIntercept, double left, double right, bool flipped) {
	addHorizontalEndPoints(left, right, axisIntercept);
	double rootVals[2];
	int roots = horizontalIntersect(axisIntercept, rootVals);
	for (int index = 0; index < roots; ++index) {
	double quadT = rootVals[index];
	SkDPoint pt = quad.xyAtT(quadT);
	double lineT = (pt.fX - left) / (right - left);
	if (PinTs(&quadT, &lineT)) {
	intersections->insert(quadT, lineT, pt);
	}
	}
	if (flipped) {
	intersections->flip();
	}
	return intersections->used();
	}

	int verticalIntersect(double axisIntercept, double roots[2]) {
	double D = quad[2].fX; // f
	double E = quad[1].fX; // e
	double F = quad[0].fX; // d
	D += F - 2 * E; // D = d - 2*e + f
	E -= F; // E = -(d - e)
	F -= axisIntercept;
	return SkDQuad::RootsValidT(D, 2 * E, F, roots);
	}

	int verticalIntersect(double axisIntercept, double top, double bottom, bool flipped) {
	addVerticalEndPoints(top, bottom, axisIntercept);
	double rootVals[2];
	int roots = verticalIntersect(axisIntercept, rootVals);
	for (int index = 0; index < roots; ++index) {
	double quadT = rootVals[index];
	SkDPoint pt = quad.xyAtT(quadT);
	double lineT = (pt.fY - top) / (bottom - top);
	if (PinTs(&quadT, &lineT)) {
	intersections->insert(quadT, lineT, pt);
	}
	}
	if (flipped) {
	intersections->flip();
	}
	return intersections->used();
	}

	protected:
	// add endpoints first to get zero and one t values exactly
	void addEndPoints() {
	for (int qIndex = 0; qIndex < 3; qIndex += 2) {
	for (int lIndex = 0; lIndex < 2; lIndex++) {
	if (quad[qIndex] == line[lIndex]) {
	intersections->insert(qIndex >> 1, lIndex, line[lIndex]);
	}
	}
	}
	}

	void addHorizontalEndPoints(double left, double right, double y) {
	for (int qIndex = 0; qIndex < 3; qIndex += 2) {
	if (quad[qIndex].fY != y) {
	continue;
	}
	if (quad[qIndex].fX == left) {
	intersections->insert(qIndex >> 1, 0, quad[qIndex]);
	}
	if (quad[qIndex].fX == right) {
	intersections->insert(qIndex >> 1, 1, quad[qIndex]);
	}
	}
	}

	void addVerticalEndPoints(double top, double bottom, double x) {
	for (int qIndex = 0; qIndex < 3; qIndex += 2) {
	if (quad[qIndex].fX != x) {
	continue;
	}
	if (quad[qIndex].fY == top) {
	intersections->insert(qIndex >> 1, 0, quad[qIndex]);
	}
	if (quad[qIndex].fY == bottom) {
	intersections->insert(qIndex >> 1, 1, quad[qIndex]);
	}
	}
	}

	double findLineT(double t) {
	SkDPoint xy = quad.xyAtT(t);
	double dx = line[1].fX - line[0].fX;
	double dy = line[1].fY - line[0].fY;
	if (fabs(dx) > fabs(dy)) {
	return (xy.fX - line[0].fX) / dx;
	}
	return (xy.fY - line[0].fY) / dy;
	}

	static bool PinTs(double* quadT, double* lineT) {
	if (!approximately_one_or_less(*lineT)) {
	return false;
	}
	if (!approximately_zero_or_more(*lineT)) {
	return false;
	}
	if (precisely_less_than_zero(*quadT)) {
	*quadT = 0;
	} else if (precisely_greater_than_one(*quadT)) {
	*quadT = 1;
	}
	if (precisely_less_than_zero(*lineT)) {
	*lineT = 0;
	} else if (precisely_greater_than_one(*lineT)) {
	*lineT = 1;
	}
	return true;
	}

	private:
	const SkDQuad& quad;
	const SkDLine& line;
	SkIntersections* intersections;
	};

	// utility for pairs of coincident quads
	static double horizontalIntersect(const SkDQuad& quad, const SkDPoint& pt) {
	LineQuadraticIntersections q(quad, (static_cast<SkDLine>(0)),
	static_cast<SkIntersections*>(0));
	double rootVals[2];
	int roots = q.horizontalIntersect(pt.fY, rootVals);
	for (int index = 0; index < roots; ++index) {
	double t = rootVals[index];
	SkDPoint qPt = quad.xyAtT(t);
	if (AlmostEqualUlps(qPt.fX, pt.fX)) {
	return t;
	}
	}
	return -1;
	}

	static double verticalIntersect(const SkDQuad& quad, const SkDPoint& pt) {
	LineQuadraticIntersections q(quad, (static_cast<SkDLine>(0)),
	static_cast<SkIntersections*>(0));
	double rootVals[2];
	int roots = q.verticalIntersect(pt.fX, rootVals);
	for (int index = 0; index < roots; ++index) {
	double t = rootVals[index];
	SkDPoint qPt = quad.xyAtT(t);
	if (AlmostEqualUlps(qPt.fY, pt.fY)) {
	return t;
	}
	}
	return -1;
	}

	double SkIntersections::Axial(const SkDQuad& q1, const SkDPoint& p, bool vertical) {
	if (vertical) {
	return verticalIntersect(q1, p);
	}
	return horizontalIntersect(q1, p);
	}

	int SkIntersections::horizontal(const SkDQuad& quad, double left, double right, double y,
	bool flipped) {
	LineQuadraticIntersections q(quad, (static_cast<SkDLine>(0)), this);
	return q.horizontalIntersect(y, left, right, flipped);
	}

	int SkIntersections::vertical(const SkDQuad& quad, double top, double bottom, double x,
	bool flipped) {
	LineQuadraticIntersections q(quad, (static_cast<SkDLine>(0)), this);
	return q.verticalIntersect(x, top, bottom, flipped);
	}

	int SkIntersections::intersect(const SkDQuad& quad, const SkDLine& line) {
	LineQuadraticIntersections q(quad, line, this);
	return q.intersect();
	}

	int SkIntersections::intersectRay(const SkDQuad& quad, const SkDLine& line) {
	LineQuadraticIntersections q(quad, line, this);
	fUsed = q.intersectRay(fT[0]);
	for (int index = 0; index < fUsed; ++index) {
	fPt[index] = quad.xyAtT(fT[0][index]);
	}
	return fUsed;
	}