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/*
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, 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);
}
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,
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);
}
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