Chromium Code Reviews
chromiumcodereview-hr@appspot.gserviceaccount.com (chromiumcodereview-hr) | Please choose your nickname with Settings | Help | Chromium Project | Gerrit Changes | Sign out
(2)

Unified Diff: src/pathops/SkDCubicToQuads.cpp

Issue 12880016: Add intersections for path ops (Closed) Base URL: http://skia.googlecode.com/svn/trunk/
Patch Set: Created 7 years, 9 months ago
Use n/p to move between diff chunks; N/P to move between comments. Draft comments are only viewable by you.
Jump to:
View side-by-side diff with in-line comments
Download patch
« no previous file with comments | « src/pathops/SkDCubicLineIntersection.cpp ('k') | src/pathops/SkDLineIntersection.cpp » ('j') | no next file with comments »
Expand Comments ('e') | Collapse Comments ('c') | Show Comments Hide Comments ('s')
Index: src/pathops/SkDCubicToQuads.cpp
===================================================================
--- src/pathops/SkDCubicToQuads.cpp (revision 0)
+++ src/pathops/SkDCubicToQuads.cpp (revision 0)
@@ -0,0 +1,190 @@
+/*
+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 "SkTDArray.h"
+#include "TSearch.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 = &sub;
+ }
+ 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, SkTDArray<double>* ts) {
+ double tDiv = calc_t_div(cubic, precision, 0);
+ if (tDiv >= 1) {
+ return true;
+ }
+ if (tDiv >= 0.5) {
+ *ts->append() = 0.5;
+ return true;
+ }
+ return false;
+}
+
+static void addTs(const SkDCubic& cubic, double precision, double start, double end,
+ SkTDArray<double>* 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->append() = 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, SkTDArray<double>* ts) const {
+ SkReduceOrder reducer;
+ int order = reducer.reduce(*this, SkReduceOrder::kAllow_Quadratics, SkReduceOrder::kFill_Style);
+ if (order < 3) {
+ return;
+ }
+ double inflectT[5];
+ int inflections = findInflections(inflectT);
+ SkASSERT(inflections <= 2);
+ if (!endsAreExtremaInXOrY()) {
+ inflections += findMaxCurvature(&inflectT[inflections]);
+ SkASSERT(inflections <= 5);
+ }
+ QSort<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])) {
+ memcpy(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;
+ }
+ memcpy(&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,
+ SkReduceOrder::kFill_Style);
+ if (orderP1 < 2) {
+ --inflections;
+ } else {
+ int orderP2 = reducer.reduce(pair.second(), SkReduceOrder::kNo_Quadratics,
+ SkReduceOrder::kFill_Style);
+ 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);
+}
« no previous file with comments | « src/pathops/SkDCubicLineIntersection.cpp ('k') | src/pathops/SkDLineIntersection.cpp » ('j') | no next file with comments »

Powered by Google App Engine
This is Rietveld 408576698