/* * 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.
"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.
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(5); // allow short partial coincidence plus discrete intersections
}
void checkCoincident() { int last = fIntersections->used() - 1; for (int index = 0; index < last; ) { double quadMidT = ((*fIntersections)[0][index] + (*fIntersections)[0][index + 1]) / 2;
SkDPoint quadMidPt = fQuad.ptAtT(quadMidT); double t = fLine->nearPoint(quadMidPt, nullptr); if (t < 0) {
++index; continue;
} if (fIntersections->isCoincident(index)) {
fIntersections->removeOne(index);
--last;
} elseif (fIntersections->isCoincident(index + 1)) {
fIntersections->removeOne(index + 1);
--last;
} else {
fIntersections->setCoincident(index++);
}
fIntersections->setCoincident(index);
}
}
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();
} 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) && uniqueAnswer(quadT, pt)) {
fIntersections->insert(quadT, lineT, pt);
}
}
checkCoincident(); 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) && uniqueAnswer(quadT, pt)) {
fIntersections->insert(quadT, lineT, pt);
}
} if (flipped) {
fIntersections->flip();
}
checkCoincident(); return fIntersections->used();
}
bool uniqueAnswer(double quadT, const SkDPoint& pt) { for (int inner = 0; inner < fIntersections->used(); ++inner) { if (fIntersections->pt(inner) != pt) { continue;
} double existingQuadT = (*fIntersections)[0][inner]; if (quadT == existingQuadT) { returnfalse;
} // check if midway on quad is also same point. If so, discard this double quadMidT = (existingQuadT + quadT) / 2;
SkDPoint quadMidPt = fQuad.ptAtT(quadMidT); if (quadMidPt.approximatelyEqual(pt)) { returnfalse;
}
} #if ONE_OFF_DEBUG
SkDPoint qPt = fQuad.ptAtT(quadT);
SkDebugf("%s pt=(%1.9g,%1.9g) cPt=(%1.9g,%1.9g)\n", __FUNCTION__, pt.fX, pt.fY,
qPt.fX, qPt.fY); #endif returntrue;
}
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) && uniqueAnswer(quadT, pt)) {
fIntersections->insert(quadT, lineT, pt);
}
} if (flipped) {
fIntersections->flip();
}
checkCoincident(); 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]);
}
}
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