/* http://stackoverflow.com/questions/2009160/how-do-i-convert-the-2-control-points-of-a-cubic-curve-to-the-single-control-poi */ /* Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2. Then for degree elevation, the equations are: Q0 = P0 Q1 = 1/3 P0 + 2/3 P1 Q2 = 2/3 P1 + 1/3 P2 Q3 = P2 In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways to compute P1 from the equations above: P1 = 3/2 Q1 - 1/2 Q0 P1 = 3/2 Q2 - 1/2 Q3 If this is a degree-elevated cubic, then both equations will give the same answer for P1. Since it's likely not, your best bet is to average them. So, P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3 SkDCubic defined by: P1/2 - anchor points, C1/C2 control points |x| is the euclidean norm of x mid-point approx of cubic: a quad that shares the same anchors with the cubic and has the control point at C = (3·C2 - P2 + 3·C1 - P1)/4 Algorithm pick an absolute precision (prec) Compute the Tdiv as the root of (cubic) equation sqrt(3)/18 · |P2 - 3·C2 + 3·C1 - P1|/2 · Tdiv ^ 3 = prec if Tdiv < 0.5 divide the cubic at Tdiv. First segment [0..Tdiv] can be approximated with by a quadratic, with a defect less than prec, by the mid-point approximation. Repeat from step 2 with the second resulted segment (corresponding to 1-Tdiv) 0.5<=Tdiv<1 - simply divide the cubic in two. The two halves can be approximated by the mid-point approximation Tdiv>=1 - the entire cubic can be approximated by the mid-point approximation confirmed by (maybe stolen from) http://www.caffeineowl.com/graphics/2d/vectorial/cubic2quad01.html // maybe in turn derived from http://www.cccg.ca/proceedings/2004/36.pdf // also stored at http://www.cis.usouthal.edu/~hain/general/Publications/Bezier/bezier%20cccg04%20paper.pdf */ #include "SkPathOpsCubic.h" #include "SkPathOpsLine.h" #include "SkPathOpsQuad.h" #include "SkReduceOrder.h" #include "SkTArray.h" #include "SkTSort.h" #define USE_CUBIC_END_POINTS 1 static double calc_t_div(const SkDCubic& cubic, double precision, double start) { const double adjust = sqrt(3.) / 36; SkDCubic sub; const SkDCubic* cPtr; if (start == 0) { cPtr = &cubic; } else { // OPTIMIZE: special-case half-split ? sub = cubic.subDivide(start, 1); cPtr = ⊂ } const SkDCubic& c = *cPtr; double dx = c[3].fX - 3 * (c[2].fX - c[1].fX) - c[0].fX; double dy = c[3].fY - 3 * (c[2].fY - c[1].fY) - c[0].fY; double dist = sqrt(dx * dx + dy * dy); double tDiv3 = precision / (adjust * dist); double t = SkDCubeRoot(tDiv3); if (start > 0) { t = start + (1 - start) * t; } return t; } SkDQuad SkDCubic::toQuad() const { SkDQuad quad; quad[0] = fPts[0]; const SkDPoint fromC1 = {(3 * fPts[1].fX - fPts[0].fX) / 2, (3 * fPts[1].fY - fPts[0].fY) / 2}; const SkDPoint fromC2 = {(3 * fPts[2].fX - fPts[3].fX) / 2, (3 * fPts[2].fY - fPts[3].fY) / 2}; quad[1].fX = (fromC1.fX + fromC2.fX) / 2; quad[1].fY = (fromC1.fY + fromC2.fY) / 2; quad[2] = fPts[3]; return quad; } static bool add_simple_ts(const SkDCubic& cubic, double precision, SkTArray<double, true>* ts) { double tDiv = calc_t_div(cubic, precision, 0); if (tDiv >= 1) { return true; } if (tDiv >= 0.5) { ts->push_back(0.5); return true; } return false; } static void addTs(const SkDCubic& cubic, double precision, double start, double end, SkTArray<double, true>* ts) { double tDiv = calc_t_div(cubic, precision, 0); double parts = ceil(1.0 / tDiv); for (double index = 0; index < parts; ++index) { double newT = start + (index / parts) * (end - start); if (newT > 0 && newT < 1) { ts->push_back(newT); } } } // flavor that returns T values only, deferring computing the quads until they are needed // FIXME: when called from recursive intersect 2, this could take the original cubic // and do a more precise job when calling chop at and sub divide by computing the fractional ts. // it would still take the prechopped cubic for reduce order and find cubic inflections void SkDCubic::toQuadraticTs(double precision, SkTArray<double, true>* ts) const { SkReduceOrder reducer; int order = reducer.reduce(*this, SkReduceOrder::kAllow_Quadratics); if (order < 3) { return; } double inflectT[5]; int inflections = findInflections(inflectT); SkASSERT(inflections <= 2); if (!endsAreExtremaInXOrY()) { inflections += findMaxCurvature(&inflectT[inflections]); SkASSERT(inflections <= 5); } SkTQSort<double>(inflectT, &inflectT[inflections - 1]); // OPTIMIZATION: is this filtering common enough that it needs to be pulled out into its // own subroutine? while (inflections && approximately_less_than_zero(inflectT[0])) { memmove(inflectT, &inflectT[1], sizeof(inflectT[0]) * --inflections); } int start = 0; do { int next = start + 1; if (next >= inflections) { break; } if (!approximately_equal(inflectT[start], inflectT[next])) { ++start; continue; } memmove(&inflectT[start], &inflectT[next], sizeof(inflectT[0]) * (--inflections - start)); } while (true); while (inflections && approximately_greater_than_one(inflectT[inflections - 1])) { --inflections; } SkDCubicPair pair; if (inflections == 1) { pair = chopAt(inflectT[0]); int orderP1 = reducer.reduce(pair.first(), SkReduceOrder::kNo_Quadratics); if (orderP1 < 2) { --inflections; } else { int orderP2 = reducer.reduce(pair.second(), SkReduceOrder::kNo_Quadratics); if (orderP2 < 2) { --inflections; } } } if (inflections == 0 && add_simple_ts(*this, precision, ts)) { return; } if (inflections == 1) { pair = chopAt(inflectT[0]); addTs(pair.first(), precision, 0, inflectT[0], ts); addTs(pair.second(), precision, inflectT[0], 1, ts); return; } if (inflections > 1) { SkDCubic part = subDivide(0, inflectT[0]); addTs(part, precision, 0, inflectT[0], ts); int last = inflections - 1; for (int idx = 0; idx < last; ++idx) { part = subDivide(inflectT[idx], inflectT[idx + 1]); addTs(part, precision, inflectT[idx], inflectT[idx + 1], ts); } part = subDivide(inflectT[last], 1); addTs(part, precision, inflectT[last], 1, ts); return; } addTs(*this, precision, 0, 1, ts); }