0b7da433fe
git-svn-id: http://skia.googlecode.com/svn/trunk@6223 2bbb7eff-a529-9590-31e7-b0007b416f81
231 lines
8.0 KiB
C++
231 lines
8.0 KiB
C++
// Another approach is to start with the implicit form of one curve and solve
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// (seek implicit coefficients in QuadraticParameter.cpp
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// by substituting in the parametric form of the other.
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// The downside of this approach is that early rejects are difficult to come by.
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// http://planetmath.org/encyclopedia/GaloisTheoreticDerivationOfTheQuarticFormula.html#step
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#include "CurveIntersection.h"
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#include "Intersections.h"
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#include "QuadraticParameterization.h"
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#include "QuarticRoot.h"
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#include "QuadraticUtilities.h"
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/* given the implicit form 0 = Ax^2 + Bxy + Cy^2 + Dx + Ey + F
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* and given x = at^2 + bt + c (the parameterized form)
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* y = dt^2 + et + f
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* then
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* 0 = A(at^2+bt+c)(at^2+bt+c)+B(at^2+bt+c)(dt^2+et+f)+C(dt^2+et+f)(dt^2+et+f)+D(at^2+bt+c)+E(dt^2+et+f)+F
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*/
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static int findRoots(const QuadImplicitForm& i, const Quadratic& q2, double roots[4]) {
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double a, b, c;
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set_abc(&q2[0].x, a, b, c);
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double d, e, f;
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set_abc(&q2[0].y, d, e, f);
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const double t4 = i.x2() * a * a
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+ i.xy() * a * d
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+ i.y2() * d * d;
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const double t3 = 2 * i.x2() * a * b
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+ i.xy() * (a * e + b * d)
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+ 2 * i.y2() * d * e;
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const double t2 = i.x2() * (b * b + 2 * a * c)
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+ i.xy() * (c * d + b * e + a * f)
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+ i.y2() * (e * e + 2 * d * f)
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+ i.x() * a
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+ i.y() * d;
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const double t1 = 2 * i.x2() * b * c
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+ i.xy() * (c * e + b * f)
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+ 2 * i.y2() * e * f
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+ i.x() * b
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+ i.y() * e;
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const double t0 = i.x2() * c * c
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+ i.xy() * c * f
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+ i.y2() * f * f
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+ i.x() * c
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+ i.y() * f
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+ i.c();
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return quarticRoots(t4, t3, t2, t1, t0, roots);
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}
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static void addValidRoots(const double roots[4], const int count, const int side, Intersections& i) {
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int index;
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for (index = 0; index < count; ++index) {
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if (!approximately_zero_or_more(roots[index]) || !approximately_one_or_less(roots[index])) {
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continue;
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}
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double t = 1 - roots[index];
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if (approximately_less_than_zero(t)) {
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t = 0;
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} else if (approximately_greater_than_one(t)) {
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t = 1;
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}
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i.insertOne(t, side);
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}
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}
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static bool onlyEndPtsInCommon(const Quadratic& q1, const Quadratic& q2, Intersections& i) {
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// the idea here is to see at minimum do a quick reject by rotating all points
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// to either side of the line formed by connecting the endpoints
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// if the opposite curves points are on the line or on the other side, the
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// curves at most intersect at the endpoints
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for (int oddMan = 0; oddMan < 3; ++oddMan) {
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const _Point* endPt[2];
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for (int opp = 1; opp < 3; ++opp) {
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int end = oddMan ^ opp;
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if (end == 3) {
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end = opp;
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}
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endPt[opp - 1] = &q1[end];
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}
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double origX = endPt[0]->x;
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double origY = endPt[0]->y;
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double adj = endPt[1]->x - origX;
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double opp = endPt[1]->y - origY;
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double sign = (q1[oddMan].y - origY) * adj - (q1[oddMan].x - origX) * opp;
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assert(!approximately_zero(sign));
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for (int n = 0; n < 3; ++n) {
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double test = (q2[n].y - origY) * adj - (q2[n].x - origX) * opp;
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if (test * sign > 0) {
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goto tryNextHalfPlane;
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}
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}
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for (int i1 = 0; i1 < 3; i1 += 2) {
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for (int i2 = 0; i2 < 3; i2 += 2) {
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if (q1[i1] == q2[i2]) {
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i.insert(i1 >> 1, i2 >> 1);
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}
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}
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}
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assert(i.fUsed < 3);
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return true;
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tryNextHalfPlane:
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;
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}
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return false;
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}
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bool intersect2(const Quadratic& q1, const Quadratic& q2, Intersections& i) {
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// if the quads share an end point, check to see if they overlap
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if (onlyEndPtsInCommon(q1, q2, i)) {
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return i.intersected();
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}
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QuadImplicitForm i1(q1);
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QuadImplicitForm i2(q2);
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if (i1.implicit_match(i2)) {
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// FIXME: compute T values
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// compute the intersections of the ends to find the coincident span
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bool useVertical = fabs(q1[0].x - q1[2].x) < fabs(q1[0].y - q1[2].y);
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double t;
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if ((t = axialIntersect(q1, q2[0], useVertical)) >= 0) {
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i.addCoincident(t, 0);
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}
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if ((t = axialIntersect(q1, q2[2], useVertical)) >= 0) {
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i.addCoincident(t, 1);
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}
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useVertical = fabs(q2[0].x - q2[2].x) < fabs(q2[0].y - q2[2].y);
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if ((t = axialIntersect(q2, q1[0], useVertical)) >= 0) {
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i.addCoincident(0, t);
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}
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if ((t = axialIntersect(q2, q1[2], useVertical)) >= 0) {
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i.addCoincident(1, t);
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}
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assert(i.fCoincidentUsed <= 2);
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return i.fCoincidentUsed > 0;
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}
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double roots1[4], roots2[4];
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int rootCount = findRoots(i2, q1, roots1);
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// OPTIMIZATION: could short circuit here if all roots are < 0 or > 1
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#ifndef NDEBUG
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int rootCount2 =
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#endif
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findRoots(i1, q2, roots2);
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assert(rootCount == rootCount2);
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addValidRoots(roots1, rootCount, 0, i);
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addValidRoots(roots2, rootCount, 1, i);
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if (i.insertBalanced() && i.fUsed <= 1) {
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if (i.fUsed == 1) {
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_Point xy1, xy2;
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xy_at_t(q1, i.fT[0][0], xy1.x, xy1.y);
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xy_at_t(q2, i.fT[1][0], xy2.x, xy2.y);
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if (!xy1.approximatelyEqual(xy2)) {
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--i.fUsed;
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--i.fUsed2;
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}
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}
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return i.intersected();
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}
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_Point pts[4];
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bool matches[4];
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int flipCheck[4];
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int closest[4];
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double dist[4];
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int index, ndex2;
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int flipIndex = 0;
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for (ndex2 = 0; ndex2 < i.fUsed2; ++ndex2) {
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xy_at_t(q2, i.fT[1][ndex2], pts[ndex2].x, pts[ndex2].y);
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matches[ndex2] = false;
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}
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for (index = 0; index < i.fUsed; ++index) {
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_Point xy;
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xy_at_t(q1, i.fT[0][index], xy.x, xy.y);
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dist[index] = DBL_MAX;
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closest[index] = -1;
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for (ndex2 = 0; ndex2 < i.fUsed2; ++ndex2) {
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if (!pts[ndex2].approximatelyEqual(xy)) {
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continue;
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}
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double dx = pts[ndex2].x - xy.x;
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double dy = pts[ndex2].y - xy.y;
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double distance = dx * dx + dy * dy;
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if (dist[index] <= distance) {
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continue;
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}
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for (int outer = 0; outer < index; ++outer) {
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if (closest[outer] != ndex2) {
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continue;
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}
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if (dist[outer] < distance) {
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goto next;
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}
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closest[outer] = -1;
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}
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dist[index] = distance;
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closest[index] = ndex2;
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next:
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;
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}
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}
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for (index = 0; index < i.fUsed; ) {
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for (ndex2 = 0; ndex2 < i.fUsed2; ++ndex2) {
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if (closest[index] == ndex2) {
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assert(flipIndex < 4);
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flipCheck[flipIndex++] = ndex2;
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matches[ndex2] = true;
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goto next2;
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}
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}
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if (--i.fUsed > index) {
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memmove(&i.fT[0][index], &i.fT[0][index + 1], (i.fUsed - index) * sizeof(i.fT[0][0]));
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memmove(&closest[index], &closest[index + 1], (i.fUsed - index) * sizeof(closest[0]));
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continue;
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}
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next2:
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++index;
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}
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for (ndex2 = 0; ndex2 < i.fUsed2; ) {
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if (!matches[ndex2]) {
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if (--i.fUsed2 > ndex2) {
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memmove(&i.fT[1][ndex2], &i.fT[1][ndex2 + 1], (i.fUsed2 - ndex2) * sizeof(i.fT[1][0]));
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memmove(&matches[ndex2], &matches[ndex2 + 1], (i.fUsed2 - ndex2) * sizeof(matches[0]));
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continue;
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}
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}
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++ndex2;
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}
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i.fFlip = i.fUsed >= 2 && flipCheck[0] > flipCheck[1];
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assert(i.insertBalanced());
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return i.intersected();
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}
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