Many resources are needed to download a project. Please understand that we have to compensate our server costs. Thank you in advance. Project price only 1 $
You can buy this project and download/modify it how often you want.
/*
* Copyright (c) 2007, 2018, Oracle and/or its affiliates. All rights reserved.
* DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
*
* This code is free software; you can redistribute it and/or modify it
* under the terms of the GNU General Public License version 2 only, as
* published by the Free Software Foundation. Oracle designates this
* particular file as subject to the "Classpath" exception as provided
* by Oracle in the LICENSE file that accompanied this code.
*
* This code is distributed in the hope that it will be useful, but WITHOUT
* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
* FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
* version 2 for more details (a copy is included in the LICENSE file that
* accompanied this code).
*
* You should have received a copy of the GNU General Public License version
* 2 along with this work; if not, write to the Free Software Foundation,
* Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
*
* Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA
* or visit www.oracle.com if you need additional information or have any
* questions.
*/
package com.sun.marlin;
final class DCurve {
double ax, ay, bx, by, cx, cy, dx, dy;
double dax, day, dbx, dby;
DCurve() {
}
void set(final double[] points, final int type) {
// if instead of switch (perf + most probable cases first)
if (type == 8) {
set(points[0], points[1],
points[2], points[3],
points[4], points[5],
points[6], points[7]);
} else if (type == 4) {
set(points[0], points[1],
points[2], points[3]);
} else {
set(points[0], points[1],
points[2], points[3],
points[4], points[5]);
}
}
void set(final double x1, final double y1,
final double x2, final double y2,
final double x3, final double y3,
final double x4, final double y4)
{
final double dx32 = 3.0d * (x3 - x2);
final double dy32 = 3.0d * (y3 - y2);
final double dx21 = 3.0d * (x2 - x1);
final double dy21 = 3.0d * (y2 - y1);
ax = (x4 - x1) - dx32; // A = P3 - P0 - 3 (P2 - P1) = (P3 - P0) + 3 (P1 - P2)
ay = (y4 - y1) - dy32;
bx = (dx32 - dx21); // B = 3 (P2 - P1) - 3(P1 - P0) = 3 (P2 + P0) - 6 P1
by = (dy32 - dy21);
cx = dx21; // C = 3 (P1 - P0)
cy = dy21;
dx = x1; // D = P0
dy = y1;
dax = 3.0d * ax;
day = 3.0d * ay;
dbx = 2.0d * bx;
dby = 2.0d * by;
}
void set(final double x1, final double y1,
final double x2, final double y2,
final double x3, final double y3)
{
final double dx21 = (x2 - x1);
final double dy21 = (y2 - y1);
ax = 0.0d; // A = 0
ay = 0.0d;
bx = (x3 - x2) - dx21; // B = P3 - P0 - 2 P2
by = (y3 - y2) - dy21;
cx = 2.0d * dx21; // C = 2 (P2 - P1)
cy = 2.0d * dy21;
dx = x1; // D = P1
dy = y1;
dax = 0.0d;
day = 0.0d;
dbx = 2.0d * bx;
dby = 2.0d * by;
}
void set(final double x1, final double y1,
final double x2, final double y2)
{
final double dx21 = (x2 - x1);
final double dy21 = (y2 - y1);
ax = 0.0d; // A = 0
ay = 0.0d;
bx = 0.0d; // B = 0
by = 0.0d;
cx = dx21; // C = (P2 - P1)
cy = dy21;
dx = x1; // D = P1
dy = y1;
dax = 0.0d;
day = 0.0d;
dbx = 0.0d;
dby = 0.0d;
}
int dxRoots(final double[] roots, final int off) {
return DHelpers.quadraticRoots(dax, dbx, cx, roots, off);
}
int dyRoots(final double[] roots, final int off) {
return DHelpers.quadraticRoots(day, dby, cy, roots, off);
}
int infPoints(final double[] pts, final int off) {
// inflection point at t if -f'(t)x*f''(t)y + f'(t)y*f''(t)x == 0
// Fortunately, this turns out to be quadratic, so there are at
// most 2 inflection points.
final double a = dax * dby - dbx * day;
final double b = 2.0d * (cy * dax - day * cx);
final double c = cy * dbx - cx * dby;
return DHelpers.quadraticRoots(a, b, c, pts, off);
}
int xPoints(final double[] ts, final int off, final double x)
{
return DHelpers.cubicRootsInAB(ax, bx, cx, dx - x, ts, off, 0.0d, 1.0d);
}
int yPoints(final double[] ts, final int off, final double y)
{
return DHelpers.cubicRootsInAB(ay, by, cy, dy - y, ts, off, 0.0d, 1.0d);
}
// finds points where the first and second derivative are
// perpendicular. This happens when g(t) = f'(t)*f''(t) == 0 (where
// * is a dot product). Unfortunately, we have to solve a cubic.
private int perpendiculardfddf(final double[] pts, final int off) {
assert pts.length >= off + 4;
// these are the coefficients of some multiple of g(t) (not g(t),
// because the roots of a polynomial are not changed after multiplication
// by a constant, and this way we save a few multiplications).
final double a = 2.0d * (dax * dax + day * day);
final double b = 3.0d * (dax * dbx + day * dby);
final double c = 2.0d * (dax * cx + day * cy) + dbx * dbx + dby * dby;
final double d = dbx * cx + dby * cy;
return DHelpers.cubicRootsInAB(a, b, c, d, pts, off, 0.0d, 1.0d);
}
// Tries to find the roots of the function ROC(t)-w in [0, 1). It uses
// a variant of the false position algorithm to find the roots. False
// position requires that 2 initial values x0,x1 be given, and that the
// function must have opposite signs at those values. To find such
// values, we need the local extrema of the ROC function, for which we
// need the roots of its derivative; however, it's harder to find the
// roots of the derivative in this case than it is to find the roots
// of the original function. So, we find all points where this curve's
// first and second derivative are perpendicular, and we pretend these
// are our local extrema. There are at most 3 of these, so we will check
// at most 4 sub-intervals of (0,1). ROC has asymptotes at inflection
// points, so roc-w can have at least 6 roots. This shouldn't be a
// problem for what we're trying to do (draw a nice looking curve).
int rootsOfROCMinusW(final double[] roots, final int off, final double w2, final double err) {
// no OOB exception, because by now off<=6, and roots.length >= 10
assert off <= 6 && roots.length >= 10;
int ret = off;
final int end = off + perpendiculardfddf(roots, off);
roots[end] = 1.0d; // always check interval end points
double t0 = 0.0d, ft0 = ROCsq(t0) - w2;
for (int i = off; i <= end; i++) {
double t1 = roots[i], ft1 = ROCsq(t1) - w2;
if (ft0 == 0.0d) {
roots[ret++] = t0;
} else if (ft1 * ft0 < 0.0d) { // have opposite signs
// (ROC(t)^2 == w^2) == (ROC(t) == w) is true because
// ROC(t) >= 0 for all t.
roots[ret++] = falsePositionROCsqMinusX(t0, t1, w2, err);
}
t0 = t1;
ft0 = ft1;
}
return ret - off;
}
private static double eliminateInf(final double x) {
return (x == Double.POSITIVE_INFINITY ? Double.MAX_VALUE :
(x == Double.NEGATIVE_INFINITY ? Double.MIN_VALUE : x));
}
// A slight modification of the false position algorithm on wikipedia.
// This only works for the ROCsq-x functions. It might be nice to have
// the function as an argument, but that would be awkward in java6.
// TODO: It is something to consider for java8 (or whenever lambda
// expressions make it into the language), depending on how closures
// and turn out. Same goes for the newton's method
// algorithm in DHelpers.java
private double falsePositionROCsqMinusX(final double t0, final double t1,
final double w2, final double err)
{
final int iterLimit = 100;
int side = 0;
double t = t1, ft = eliminateInf(ROCsq(t) - w2);
double s = t0, fs = eliminateInf(ROCsq(s) - w2);
double r = s, fr;
for (int i = 0; i < iterLimit && Math.abs(t - s) > err * Math.abs(t + s); i++) {
r = (fs * t - ft * s) / (fs - ft);
fr = ROCsq(r) - w2;
if (sameSign(fr, ft)) {
ft = fr; t = r;
if (side < 0) {
fs /= (1 << (-side));
side--;
} else {
side = -1;
}
} else if (fr * fs > 0.0d) {
fs = fr; s = r;
if (side > 0) {
ft /= (1 << side);
side++;
} else {
side = 1;
}
} else {
break;
}
}
return r;
}
private static boolean sameSign(final double x, final double y) {
// another way is to test if x*y > 0. This is bad for small x, y.
return (x < 0.0d && y < 0.0d) || (x > 0.0d && y > 0.0d);
}
// returns the radius of curvature squared at t of this curve
// see http://en.wikipedia.org/wiki/Radius_of_curvature_(applications)
private double ROCsq(final double t) {
final double dx = t * (t * dax + dbx) + cx;
final double dy = t * (t * day + dby) + cy;
final double ddx = 2.0d * dax * t + dbx;
final double ddy = 2.0d * day * t + dby;
final double dx2dy2 = dx * dx + dy * dy;
final double ddx2ddy2 = ddx * ddx + ddy * ddy;
final double ddxdxddydy = ddx * dx + ddy * dy;
return dx2dy2 * ((dx2dy2 * dx2dy2) / (dx2dy2 * ddx2ddy2 - ddxdxddydy * ddxdxddydy));
}
}
