skia2/experimental/Intersection/CubicToQuadratics.cpp
caryclark@google.com d68bc309bb shape ops work in progress
cubic to quadratic experiment

git-svn-id: http://skia.googlecode.com/svn/trunk@7046 2bbb7eff-a529-9590-31e7-b0007b416f81
2013-01-07 13:17:18 +00:00

116 lines
3.8 KiB
C++

/*
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
Cubic 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
*/
#include "CubicUtilities.h"
#include "CurveIntersection.h"
static double calcTDiv(const Cubic& cubic, double precision, double start) {
const double adjust = sqrt(3) / 36;
Cubic sub;
const Cubic* cPtr;
if (start == 0) {
cPtr = &cubic;
} else {
// OPTIMIZE: special-case half-split ?
sub_divide(cubic, start, 1, sub);
cPtr = &sub;
}
const Cubic& c = *cPtr;
double dx = c[3].x - 3 * (c[2].x - c[1].x) - c[0].x;
double dy = c[3].y - 3 * (c[2].y - c[1].y) - c[0].y;
double dist = sqrt(dx * dx + dy * dy);
double tDiv3 = precision / (adjust * dist);
return cube_root(tDiv3);
}
static void demote(const Cubic& cubic, Quadratic& quad) {
quad[0] = cubic[0];
quad[1].x = (cubic[3].x - 3 * (cubic[2].x - cubic[1].x) - cubic[0].x) / 4;
quad[1].y = (cubic[3].y - 3 * (cubic[2].y - cubic[1].y) - cubic[0].y) / 4;
quad[2] = cubic[3];
}
int cubic_to_quadratics(const Cubic& cubic, double precision, SkTDArray<Quadratic>& quadratics) {
quadratics.setCount(1); // FIXME: every place I have setCount(), I also want setReserve()
Cubic reduced;
int order = reduceOrder(cubic, reduced, kReduceOrder_QuadraticsAllowed);
if (order < 3) {
memcpy(quadratics[0], reduced, order * sizeof(_Point));
return order;
}
double tDiv = calcTDiv(cubic, precision, 0);
if (approximately_greater_than_one(tDiv)) {
demote(cubic, quadratics[0]);
return 3;
}
if (tDiv >= 0.5) {
CubicPair pair;
chop_at(cubic, pair, 0.5);
quadratics.setCount(2);
demote(pair.first(), quadratics[0]);
demote(pair.second(), quadratics[1]);
return 3;
}
double start = 0;
int index = -1;
// if we iterate but make little progress, check to see if the curve is flat
// or if the control points are tangent to each other
do {
Cubic part;
sub_divide(cubic, start, tDiv, part);
quadratics.append();
demote(part, quadratics[++index]);
if (tDiv == 1) {
break;
}
start = tDiv;
tDiv = calcTDiv(cubic, precision, start);
if (tDiv > 1) {
tDiv = 1;
}
} while (true);
return 3;
}