src/pathops/SkDQuadLineIntersection.cpp - platform/external/skia - 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:
     enum PinTPoint {
         kPointUninitialized,
         kPointInitialized
     };

     LineQuadraticIntersections(const SkDQuad& q, const SkDLine& l, SkIntersections* i)
         : fQuad(q)
         , fLine(l)
         , fIntersections(i)
         , fAllowNear(true) {
         i->setMax(2);
     }

     void allowNear(bool allow) {
         fAllowNear = allow;
     }

     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 = fLine[1].fX - fLine[0].fX;
         double opp = fLine[1].fY - fLine[0].fY;
         double r[3];
         for (int n = 0; n < 3; ++n) {
             r[n] = (fQuad[n].fY - fLine[0].fY) * adj - (fQuad[n].fX - fLine[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() {
         addExactEndPoints();
         if (fAllowNear) {
             addNearEndPoints();
         }
         if (fIntersections->used() == 2) {
             // FIXME : need sharable code that turns spans into coincident if middle point is on
         } else {
             double rootVals[2];
             int roots = intersectRay(rootVals);
             for (int index = 0; index < roots; ++index) {
                 double quadT = rootVals[index];
                 double lineT = findLineT(quadT);
                 SkDPoint pt;
                 if (pinTs(&quadT, &lineT, &pt, kPointUninitialized)) {
                     fIntersections->insert(quadT, lineT, pt);
                 }
             }
         }
         return fIntersections->used();
     }

     int horizontalIntersect(double axisIntercept, double roots[2]) {
         double D = fQuad[2].fY;  // f
         double E = fQuad[1].fY;  // e
         double F = fQuad[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) {
         addExactHorizontalEndPoints(left, right, axisIntercept);
         if (fAllowNear) {
             addNearHorizontalEndPoints(left, right, axisIntercept);
         }
         double rootVals[2];
         int roots = horizontalIntersect(axisIntercept, rootVals);
         for (int index = 0; index < roots; ++index) {
             double quadT = rootVals[index];
             SkDPoint pt = fQuad.ptAtT(quadT);
             double lineT = (pt.fX - left) / (right - left);
             if (pinTs(&quadT, &lineT, &pt, kPointInitialized)) {
                 fIntersections->insert(quadT, lineT, pt);
             }
         }
         if (flipped) {
             fIntersections->flip();
         }
         return fIntersections->used();
     }

     int verticalIntersect(double axisIntercept, double roots[2]) {
         double D = fQuad[2].fX;  // f
         double E = fQuad[1].fX;  // e
         double F = fQuad[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) {
         addExactVerticalEndPoints(top, bottom, axisIntercept);
         if (fAllowNear) {
             addNearVerticalEndPoints(top, bottom, axisIntercept);
         }
         double rootVals[2];
         int roots = verticalIntersect(axisIntercept, rootVals);
         for (int index = 0; index < roots; ++index) {
             double quadT = rootVals[index];
             SkDPoint pt = fQuad.ptAtT(quadT);
             double lineT = (pt.fY - top) / (bottom - top);
             if (pinTs(&quadT, &lineT, &pt, kPointInitialized)) {
                 fIntersections->insert(quadT, lineT, pt);
             }
         }
         if (flipped) {
             fIntersections->flip();
         }
         return fIntersections->used();
     }

 protected:
     // add endpoints first to get zero and one t values exactly
     void addExactEndPoints() {
         for (int qIndex = 0; qIndex < 3; qIndex += 2) {
             double lineT = fLine.exactPoint(fQuad[qIndex]);
             if (lineT < 0) {
                 continue;
             }
             double quadT = (double) (qIndex >> 1);
             fIntersections->insert(quadT, lineT, fQuad[qIndex]);
         }
     }

     void addNearEndPoints() {
         for (int qIndex = 0; qIndex < 3; qIndex += 2) {
             double quadT = (double) (qIndex >> 1);
             if (fIntersections->hasT(quadT)) {
                 continue;
             }
             double lineT = fLine.nearPoint(fQuad[qIndex]);
             if (lineT < 0) {
                 continue;
             }
             fIntersections->insert(quadT, lineT, fQuad[qIndex]);
         }
         // FIXME: see if line end is nearly on quad
     }

     void addExactHorizontalEndPoints(double left, double right, double y) {
         for (int qIndex = 0; qIndex < 3; qIndex += 2) {
             double lineT = SkDLine::ExactPointH(fQuad[qIndex], left, right, y);
             if (lineT < 0) {
                 continue;
             }
             double quadT = (double) (qIndex >> 1);
             fIntersections->insert(quadT, lineT, fQuad[qIndex]);
         }
     }

     void addNearHorizontalEndPoints(double left, double right, double y) {
         for (int qIndex = 0; qIndex < 3; qIndex += 2) {
             double quadT = (double) (qIndex >> 1);
             if (fIntersections->hasT(quadT)) {
                 continue;
             }
             double lineT = SkDLine::NearPointH(fQuad[qIndex], left, right, y);
             if (lineT < 0) {
                 continue;
             }
             fIntersections->insert(quadT, lineT, fQuad[qIndex]);
         }
         // FIXME: see if line end is nearly on quad
     }

     void addExactVerticalEndPoints(double top, double bottom, double x) {
         for (int qIndex = 0; qIndex < 3; qIndex += 2) {
             double lineT = SkDLine::ExactPointV(fQuad[qIndex], top, bottom, x);
             if (lineT < 0) {
                 continue;
             }
             double quadT = (double) (qIndex >> 1);
             fIntersections->insert(quadT, lineT, fQuad[qIndex]);
         }
     }

     void addNearVerticalEndPoints(double top, double bottom, double x) {
         for (int qIndex = 0; qIndex < 3; qIndex += 2) {
             double quadT = (double) (qIndex >> 1);
             if (fIntersections->hasT(quadT)) {
                 continue;
             }
             double lineT = SkDLine::NearPointV(fQuad[qIndex], top, bottom, x);
             if (lineT < 0) {
                 continue;
             }
             fIntersections->insert(quadT, lineT, fQuad[qIndex]);
         }
         // FIXME: see if line end is nearly on quad
     }

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

     bool pinTs(double* quadT, double* lineT, SkDPoint* pt, PinTPoint ptSet) {
         if (!approximately_one_or_less(*lineT)) {
             return false;
         }
         if (!approximately_zero_or_more(*lineT)) {
             return false;
         }
         double qT = *quadT = SkPinT(*quadT);
         double lT = *lineT = SkPinT(*lineT);
         if (lT == 0 || lT == 1 || (ptSet == kPointUninitialized && qT != 0 && qT != 1)) {
             *pt = fLine.ptAtT(lT);
         } else if (ptSet == kPointUninitialized) {
             *pt = fQuad.ptAtT(qT);
         }
         SkPoint gridPt = pt->asSkPoint();
         if (gridPt == fLine[0].asSkPoint()) {
             *lineT = 0;
         } else if (gridPt == fLine[1].asSkPoint()) {
             *lineT = 1;
         }
         if (gridPt == fQuad[0].asSkPoint()) {
             *quadT = 0;
         } else if (gridPt == fQuad[2].asSkPoint()) {
             *quadT = 1;
         }
         return true;
     }

 private:
     const SkDQuad& fQuad;
     const SkDLine& fLine;
     SkIntersections* fIntersections;
     bool fAllowNear;
 };

 // 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.ptAtT(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.ptAtT(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) {
     SkDLine line = {{{ left, y }, { right, y }}};
     LineQuadraticIntersections q(quad, line, this);
     return q.horizontalIntersect(y, left, right, flipped);
 }

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

 int SkIntersections::intersect(const SkDQuad& quad, const SkDLine& line) {
     LineQuadraticIntersections q(quad, line, this);
     q.allowNear(fAllowNear);
     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.ptAtT(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:
	enum PinTPoint {
	kPointUninitialized,
	kPointInitialized
	};

	LineQuadraticIntersections(const SkDQuad& q, const SkDLine& l, SkIntersections* i)
	: fQuad(q)
	, fLine(l)
	, fIntersections(i)
	, fAllowNear(true) {
	i->setMax(2);
	}

	void allowNear(bool allow) {
	fAllowNear = allow;
	}

	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 = fLine[1].fX - fLine[0].fX;
	double opp = fLine[1].fY - fLine[0].fY;
	double r[3];
	for (int n = 0; n < 3; ++n) {
	r[n] = (fQuad[n].fY - fLine[0].fY) * adj - (fQuad[n].fX - fLine[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() {
	addExactEndPoints();
	if (fAllowNear) {
	addNearEndPoints();
	}
	if (fIntersections->used() == 2) {
	// FIXME : need sharable code that turns spans into coincident if middle point is on
	} else {
	double rootVals[2];
	int roots = intersectRay(rootVals);
	for (int index = 0; index < roots; ++index) {
	double quadT = rootVals[index];
	double lineT = findLineT(quadT);
	SkDPoint pt;
	if (pinTs(&quadT, &lineT, &pt, kPointUninitialized)) {
	fIntersections->insert(quadT, lineT, pt);
	}
	}
	}
	return fIntersections->used();
	}

	int horizontalIntersect(double axisIntercept, double roots[2]) {
	double D = fQuad[2].fY; // f
	double E = fQuad[1].fY; // e
	double F = fQuad[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) {
	addExactHorizontalEndPoints(left, right, axisIntercept);
	if (fAllowNear) {
	addNearHorizontalEndPoints(left, right, axisIntercept);
	}
	double rootVals[2];
	int roots = horizontalIntersect(axisIntercept, rootVals);
	for (int index = 0; index < roots; ++index) {
	double quadT = rootVals[index];
	SkDPoint pt = fQuad.ptAtT(quadT);
	double lineT = (pt.fX - left) / (right - left);
	if (pinTs(&quadT, &lineT, &pt, kPointInitialized)) {
	fIntersections->insert(quadT, lineT, pt);
	}
	}
	if (flipped) {
	fIntersections->flip();
	}
	return fIntersections->used();
	}

	int verticalIntersect(double axisIntercept, double roots[2]) {
	double D = fQuad[2].fX; // f
	double E = fQuad[1].fX; // e
	double F = fQuad[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) {
	addExactVerticalEndPoints(top, bottom, axisIntercept);
	if (fAllowNear) {
	addNearVerticalEndPoints(top, bottom, axisIntercept);
	}
	double rootVals[2];
	int roots = verticalIntersect(axisIntercept, rootVals);
	for (int index = 0; index < roots; ++index) {
	double quadT = rootVals[index];
	SkDPoint pt = fQuad.ptAtT(quadT);
	double lineT = (pt.fY - top) / (bottom - top);
	if (pinTs(&quadT, &lineT, &pt, kPointInitialized)) {
	fIntersections->insert(quadT, lineT, pt);
	}
	}
	if (flipped) {
	fIntersections->flip();
	}
	return fIntersections->used();
	}

	protected:
	// add endpoints first to get zero and one t values exactly
	void addExactEndPoints() {
	for (int qIndex = 0; qIndex < 3; qIndex += 2) {
	double lineT = fLine.exactPoint(fQuad[qIndex]);
	if (lineT < 0) {
	continue;
	}
	double quadT = (double) (qIndex >> 1);
	fIntersections->insert(quadT, lineT, fQuad[qIndex]);
	}
	}

	void addNearEndPoints() {
	for (int qIndex = 0; qIndex < 3; qIndex += 2) {
	double quadT = (double) (qIndex >> 1);
	if (fIntersections->hasT(quadT)) {
	continue;
	}
	double lineT = fLine.nearPoint(fQuad[qIndex]);
	if (lineT < 0) {
	continue;
	}
	fIntersections->insert(quadT, lineT, fQuad[qIndex]);
	}
	// FIXME: see if line end is nearly on quad
	}

	void addExactHorizontalEndPoints(double left, double right, double y) {
	for (int qIndex = 0; qIndex < 3; qIndex += 2) {
	double lineT = SkDLine::ExactPointH(fQuad[qIndex], left, right, y);
	if (lineT < 0) {
	continue;
	}
	double quadT = (double) (qIndex >> 1);
	fIntersections->insert(quadT, lineT, fQuad[qIndex]);
	}
	}

	void addNearHorizontalEndPoints(double left, double right, double y) {
	for (int qIndex = 0; qIndex < 3; qIndex += 2) {
	double quadT = (double) (qIndex >> 1);
	if (fIntersections->hasT(quadT)) {
	continue;
	}
	double lineT = SkDLine::NearPointH(fQuad[qIndex], left, right, y);
	if (lineT < 0) {
	continue;
	}
	fIntersections->insert(quadT, lineT, fQuad[qIndex]);
	}
	// FIXME: see if line end is nearly on quad
	}

	void addExactVerticalEndPoints(double top, double bottom, double x) {
	for (int qIndex = 0; qIndex < 3; qIndex += 2) {
	double lineT = SkDLine::ExactPointV(fQuad[qIndex], top, bottom, x);
	if (lineT < 0) {
	continue;
	}
	double quadT = (double) (qIndex >> 1);
	fIntersections->insert(quadT, lineT, fQuad[qIndex]);
	}
	}

	void addNearVerticalEndPoints(double top, double bottom, double x) {
	for (int qIndex = 0; qIndex < 3; qIndex += 2) {
	double quadT = (double) (qIndex >> 1);
	if (fIntersections->hasT(quadT)) {
	continue;
	}
	double lineT = SkDLine::NearPointV(fQuad[qIndex], top, bottom, x);
	if (lineT < 0) {
	continue;
	}
	fIntersections->insert(quadT, lineT, fQuad[qIndex]);
	}
	// FIXME: see if line end is nearly on quad
	}

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

	bool pinTs(double* quadT, double* lineT, SkDPoint* pt, PinTPoint ptSet) {
	if (!approximately_one_or_less(*lineT)) {
	return false;
	}
	if (!approximately_zero_or_more(*lineT)) {
	return false;
	}
	double qT = quadT = SkPinT(quadT);
	double lT = lineT = SkPinT(lineT);
	if (lT == 0 \|\| lT == 1 \|\| (ptSet == kPointUninitialized && qT != 0 && qT != 1)) {
	*pt = fLine.ptAtT(lT);
	} else if (ptSet == kPointUninitialized) {
	*pt = fQuad.ptAtT(qT);
	}
	SkPoint gridPt = pt->asSkPoint();
	if (gridPt == fLine[0].asSkPoint()) {
	*lineT = 0;
	} else if (gridPt == fLine[1].asSkPoint()) {
	*lineT = 1;
	}
	if (gridPt == fQuad[0].asSkPoint()) {
	*quadT = 0;
	} else if (gridPt == fQuad[2].asSkPoint()) {
	*quadT = 1;
	}
	return true;
	}

	private:
	const SkDQuad& fQuad;
	const SkDLine& fLine;
	SkIntersections* fIntersections;
	bool fAllowNear;
	};

	// 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.ptAtT(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.ptAtT(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) {
	SkDLine line = {{{ left, y }, { right, y }}};
	LineQuadraticIntersections q(quad, line, this);
	return q.horizontalIntersect(y, left, right, flipped);
	}

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

	int SkIntersections::intersect(const SkDQuad& quad, const SkDLine& line) {
	LineQuadraticIntersections q(quad, line, this);
	q.allowNear(fAllowNear);
	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.ptAtT(fT[0][index]);
	}
	return fUsed;
	}