Index: src/pathops/SkPathOpsQuad.cpp |
=================================================================== |
--- src/pathops/SkPathOpsQuad.cpp (revision 0) |
+++ src/pathops/SkPathOpsQuad.cpp (revision 0) |
@@ -0,0 +1,293 @@ |
+/* |
+ * 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 "SkLineParameters.h" |
+#include "SkPathOpsCubic.h" |
+#include "SkPathOpsQuad.h" |
+#include "SkPathOpsTriangle.h" |
+ |
+// from http://blog.gludion.com/2009/08/distance-to-quadratic-bezier-curve.html |
+// (currently only used by testing) |
+double SkDQuad::nearestT(const SkDPoint& pt) const { |
+ SkDVector pos = fPts[0] - pt; |
+ // search points P of bezier curve with PM.(dP / dt) = 0 |
+ // a calculus leads to a 3d degree equation : |
+ SkDVector A = fPts[1] - fPts[0]; |
+ SkDVector B = fPts[2] - fPts[1]; |
+ B -= A; |
+ double a = B.dot(B); |
+ double b = 3 * A.dot(B); |
+ double c = 2 * A.dot(A) + pos.dot(B); |
+ double d = pos.dot(A); |
+ double ts[3]; |
+ int roots = SkDCubic::RootsValidT(a, b, c, d, ts); |
+ double d0 = pt.distanceSquared(fPts[0]); |
+ double d2 = pt.distanceSquared(fPts[2]); |
+ double distMin = SkTMin(d0, d2); |
+ int bestIndex = -1; |
+ for (int index = 0; index < roots; ++index) { |
+ SkDPoint onQuad = xyAtT(ts[index]); |
+ double dist = pt.distanceSquared(onQuad); |
+ if (distMin > dist) { |
+ distMin = dist; |
+ bestIndex = index; |
+ } |
+ } |
+ if (bestIndex >= 0) { |
+ return ts[bestIndex]; |
+ } |
+ return d0 < d2 ? 0 : 1; |
+} |
+ |
+bool SkDQuad::pointInHull(const SkDPoint& pt) const { |
+ return ((const SkDTriangle&) fPts).contains(pt); |
+} |
+ |
+SkDPoint SkDQuad::top(double startT, double endT) const { |
+ SkDQuad sub = subDivide(startT, endT); |
+ SkDPoint topPt = sub[0]; |
+ if (topPt.fY > sub[2].fY || (topPt.fY == sub[2].fY && topPt.fX > sub[2].fX)) { |
+ topPt = sub[2]; |
+ } |
+ if (!between(sub[0].fY, sub[1].fY, sub[2].fY)) { |
+ double extremeT; |
+ if (FindExtrema(sub[0].fY, sub[1].fY, sub[2].fY, &extremeT)) { |
+ extremeT = startT + (endT - startT) * extremeT; |
+ SkDPoint test = xyAtT(extremeT); |
+ if (topPt.fY > test.fY || (topPt.fY == test.fY && topPt.fX > test.fX)) { |
+ topPt = test; |
+ } |
+ } |
+ } |
+ return topPt; |
+} |
+ |
+int SkDQuad::AddValidTs(double s[], int realRoots, double* t) { |
+ int foundRoots = 0; |
+ for (int index = 0; index < realRoots; ++index) { |
+ double tValue = s[index]; |
+ if (approximately_zero_or_more(tValue) && approximately_one_or_less(tValue)) { |
+ if (approximately_less_than_zero(tValue)) { |
+ tValue = 0; |
+ } else if (approximately_greater_than_one(tValue)) { |
+ tValue = 1; |
+ } |
+ for (int idx2 = 0; idx2 < foundRoots; ++idx2) { |
+ if (approximately_equal(t[idx2], tValue)) { |
+ goto nextRoot; |
+ } |
+ } |
+ t[foundRoots++] = tValue; |
+ } |
+ nextRoot: |
+ {} |
+ } |
+ return foundRoots; |
+} |
+ |
+// note: caller expects multiple results to be sorted smaller first |
+// note: http://en.wikipedia.org/wiki/Loss_of_significance has an interesting |
+// analysis of the quadratic equation, suggesting why the following looks at |
+// the sign of B -- and further suggesting that the greatest loss of precision |
+// is in b squared less two a c |
+int SkDQuad::RootsValidT(double A, double B, double C, double t[2]) { |
+ double s[2]; |
+ int realRoots = RootsReal(A, B, C, s); |
+ int foundRoots = AddValidTs(s, realRoots, t); |
+ return foundRoots; |
+} |
+ |
+/* |
+Numeric Solutions (5.6) suggests to solve the quadratic by computing |
+ Q = -1/2(B + sgn(B)Sqrt(B^2 - 4 A C)) |
+and using the roots |
+ t1 = Q / A |
+ t2 = C / Q |
+*/ |
+// this does not discard real roots <= 0 or >= 1 |
+int SkDQuad::RootsReal(const double A, const double B, const double C, double s[2]) { |
+ const double p = B / (2 * A); |
+ const double q = C / A; |
+ if (approximately_zero(A) && (approximately_zero_inverse(p) || approximately_zero_inverse(q))) { |
+ if (approximately_zero(B)) { |
+ s[0] = 0; |
+ return C == 0; |
+ } |
+ s[0] = -C / B; |
+ return 1; |
+ } |
+ /* normal form: x^2 + px + q = 0 */ |
+ const double p2 = p * p; |
+ if (!AlmostEqualUlps(p2, q) && p2 < q) { |
+ return 0; |
+ } |
+ double sqrt_D = 0; |
+ if (p2 > q) { |
+ sqrt_D = sqrt(p2 - q); |
+ } |
+ s[0] = sqrt_D - p; |
+ s[1] = -sqrt_D - p; |
+ return 1 + !AlmostEqualUlps(s[0], s[1]); |
+} |
+ |
+bool SkDQuad::isLinear(int startIndex, int endIndex) const { |
+ SkLineParameters lineParameters; |
+ lineParameters.quadEndPoints(*this, startIndex, endIndex); |
+ // FIXME: maybe it's possible to avoid this and compare non-normalized |
+ lineParameters.normalize(); |
+ double distance = lineParameters.controlPtDistance(*this); |
+ return approximately_zero(distance); |
+} |
+ |
+SkDCubic SkDQuad::toCubic() const { |
+ SkDCubic cubic; |
+ cubic[0] = fPts[0]; |
+ cubic[2] = fPts[1]; |
+ cubic[3] = fPts[2]; |
+ cubic[1].fX = (cubic[0].fX + cubic[2].fX * 2) / 3; |
+ cubic[1].fY = (cubic[0].fY + cubic[2].fY * 2) / 3; |
+ cubic[2].fX = (cubic[3].fX + cubic[2].fX * 2) / 3; |
+ cubic[2].fY = (cubic[3].fY + cubic[2].fY * 2) / 3; |
+ return cubic; |
+} |
+ |
+SkDVector SkDQuad::dxdyAtT(double t) const { |
+ double a = t - 1; |
+ double b = 1 - 2 * t; |
+ double c = t; |
+ SkDVector result = { a * fPts[0].fX + b * fPts[1].fX + c * fPts[2].fX, |
+ a * fPts[0].fY + b * fPts[1].fY + c * fPts[2].fY }; |
+ return result; |
+} |
+ |
+SkDPoint SkDQuad::xyAtT(double t) const { |
+ double one_t = 1 - t; |
+ double a = one_t * one_t; |
+ double b = 2 * one_t * t; |
+ double c = t * t; |
+ SkDPoint result = { a * fPts[0].fX + b * fPts[1].fX + c * fPts[2].fX, |
+ a * fPts[0].fY + b * fPts[1].fY + c * fPts[2].fY }; |
+ return result; |
+} |
+ |
+/* |
+Given a quadratic q, t1, and t2, find a small quadratic segment. |
+ |
+The new quadratic is defined by A, B, and C, where |
+ A = c[0]*(1 - t1)*(1 - t1) + 2*c[1]*t1*(1 - t1) + c[2]*t1*t1 |
+ C = c[3]*(1 - t1)*(1 - t1) + 2*c[2]*t1*(1 - t1) + c[1]*t1*t1 |
+ |
+To find B, compute the point halfway between t1 and t2: |
+ |
+q(at (t1 + t2)/2) == D |
+ |
+Next, compute where D must be if we know the value of B: |
+ |
+_12 = A/2 + B/2 |
+12_ = B/2 + C/2 |
+123 = A/4 + B/2 + C/4 |
+ = D |
+ |
+Group the known values on one side: |
+ |
+B = D*2 - A/2 - C/2 |
+*/ |
+ |
+static double interp_quad_coords(const double* src, double t) { |
+ double ab = SkDInterp(src[0], src[2], t); |
+ double bc = SkDInterp(src[2], src[4], t); |
+ double abc = SkDInterp(ab, bc, t); |
+ return abc; |
+} |
+ |
+bool SkDQuad::monotonicInY() const { |
+ return between(fPts[0].fY, fPts[1].fY, fPts[2].fY); |
+} |
+ |
+SkDQuad SkDQuad::subDivide(double t1, double t2) const { |
+ SkDQuad dst; |
+ double ax = dst[0].fX = interp_quad_coords(&fPts[0].fX, t1); |
+ double ay = dst[0].fY = interp_quad_coords(&fPts[0].fY, t1); |
+ double dx = interp_quad_coords(&fPts[0].fX, (t1 + t2) / 2); |
+ double dy = interp_quad_coords(&fPts[0].fY, (t1 + t2) / 2); |
+ double cx = dst[2].fX = interp_quad_coords(&fPts[0].fX, t2); |
+ double cy = dst[2].fY = interp_quad_coords(&fPts[0].fY, t2); |
+ /* bx = */ dst[1].fX = 2*dx - (ax + cx)/2; |
+ /* by = */ dst[1].fY = 2*dy - (ay + cy)/2; |
+ return dst; |
+} |
+ |
+SkDPoint SkDQuad::subDivide(const SkDPoint& a, const SkDPoint& c, double t1, double t2) const { |
+ SkDPoint b; |
+ double dx = interp_quad_coords(&fPts[0].fX, (t1 + t2) / 2); |
+ double dy = interp_quad_coords(&fPts[0].fY, (t1 + t2) / 2); |
+ b.fX = 2 * dx - (a.fX + c.fX) / 2; |
+ b.fY = 2 * dy - (a.fY + c.fY) / 2; |
+ return b; |
+} |
+ |
+/* classic one t subdivision */ |
+static void interp_quad_coords(const double* src, double* dst, double t) { |
+ double ab = SkDInterp(src[0], src[2], t); |
+ double bc = SkDInterp(src[2], src[4], t); |
+ dst[0] = src[0]; |
+ dst[2] = ab; |
+ dst[4] = SkDInterp(ab, bc, t); |
+ dst[6] = bc; |
+ dst[8] = src[4]; |
+} |
+ |
+SkDQuadPair SkDQuad::chopAt(double t) const |
+{ |
+ SkDQuadPair dst; |
+ interp_quad_coords(&fPts[0].fX, &dst.pts[0].fX, t); |
+ interp_quad_coords(&fPts[0].fY, &dst.pts[0].fY, t); |
+ return dst; |
+} |
+ |
+static int valid_unit_divide(double numer, double denom, double* ratio) |
+{ |
+ if (numer < 0) { |
+ numer = -numer; |
+ denom = -denom; |
+ } |
+ if (denom == 0 || numer == 0 || numer >= denom) { |
+ return 0; |
+ } |
+ double r = numer / denom; |
+ if (r == 0) { // catch underflow if numer <<<< denom |
+ return 0; |
+ } |
+ *ratio = r; |
+ return 1; |
+} |
+ |
+/** Quad'(t) = At + B, where |
+ A = 2(a - 2b + c) |
+ B = 2(b - a) |
+ Solve for t, only if it fits between 0 < t < 1 |
+*/ |
+int SkDQuad::FindExtrema(double a, double b, double c, double tValue[1]) { |
+ /* At + B == 0 |
+ t = -B / A |
+ */ |
+ return valid_unit_divide(a - b, a - b - b + c, tValue); |
+} |
+ |
+/* Parameterization form, given A*t*t + 2*B*t*(1-t) + C*(1-t)*(1-t) |
+ * |
+ * a = A - 2*B + C |
+ * b = 2*B - 2*C |
+ * c = C |
+ */ |
+void SkDQuad::SetABC(const double* quad, double* a, double* b, double* c) { |
+ *a = quad[0]; // a = A |
+ *b = 2 * quad[2]; // b = 2*B |
+ *c = quad[4]; // c = C |
+ *b -= *c; // b = 2*B - C |
+ *a -= *b; // a = A - 2*B + C |
+ *b -= *c; // b = 2*B - 2*C |
+} |