/*
* @(#)Spline2D.java
*
* Copyright (c) 2003 Martin Krueger
* Copyright (c) 2005 David Benson
*
*/
import java.awt.geom.Point2D;
import java.util.Arrays;
/**
* Interpolates points given in the 2D plane. The resulting spline
* is a function s: R -> R^2 with parameter t in [0,1].
*
* @author krueger
*/
public class Spline2D {
/**
* Array representing the relative proportion of the total distance
* of each point in the line ( i.e. first point is 0.0, end point is
* 1.0, a point halfway on line is 0.5 ).
*/
private double[] t;
private Spline splineX;
private Spline splineY;
/**
* Total length tracing the points on the spline
*/
private double length;
/**
* Creates a new Spline2D.
* @param points
*/
public Spline2D(Point2D[] points) {
double[] x = new double[points.length];
double[] y = new double[points.length];
for(int i = 0; i< points.length; i++) {
x[i] = points[i].getX();
y[i] = points[i].getY();
}
init(x, y);
}
/**
* Creates a new Spline2D.
* @param x
* @param y
*/
public Spline2D(double[] x, double[] y) {
init(x, y);
}
private void init(double[] x, double[] y) {
if (x.length != y.length) {
throw new IllegalArgumentException("Arrays must have the same length.");
}
if (x.length < 2) {
throw new IllegalArgumentException("Spline edges must have at least two points.");
}
t = new double[x.length];
t[0] = 0.0; // start point is always 0.0
// Calculate the partial proportions of each section between each set
// of points and the total length of sum of all sections
for (int i = 1; i < t.length; i++) {
double lx = x[i] - x[i-1];
double ly = y[i] - y[i-1];
// If either diff is zero there is no point performing the square root
if ( 0.0 == lx ) {
t[i] = Math.abs(ly);
} else if ( 0.0 == ly ) {
t[i] = Math.abs(lx);
} else {
t[i] = Math.sqrt(lx*lx+ly*ly);
}
length += t[i];
t[i] += t[i-1];
}
for(int i = 1; i< (t.length)-1; i++) {
t[i] = t[i] / length;
}
t[(t.length)-1] = 1.0; // end point is always 1.0
splineX = new Spline(t, x);
splineY = new Spline(t, y);
}
/**
* @param t 0 <= t <= 1
*/
public double[] getPoint(double t) {
double[] result = new double[2];
result[0] = splineX.getValue(t);
result[1] = splineY.getValue(t);
return result;
}
/**
* Used to check the correctness of this spline
*/
public boolean checkValues() {
return (splineX.checkValues() && splineY.checkValues());
}
public double getDx(double t) {
return splineX.getDx(t);
}
public double getDy(double t) {
return splineY.getDx(t);
}
public Spline getSplineX() {
return splineX;
}
public Spline getSplineY() {
return splineY;
}
public double getLength() {
return length;
}
}
/* This code is PUBLIC DOMAIN */
/**
* Interpolates given values by B-Splines.
*
* @author krueger
*/
class Spline {
private double[] xx;
private double[] yy;
private double[] a;
private double[] b;
private double[] c;
private double[] d;
/** tracks the last index found since that is mostly commonly the next one used */
private int storageIndex = 0;
/**
* Creates a new Spline.
* @param xx
* @param yy
*/
public Spline(double[] xx, double[] yy) {
setValues(xx, yy);
}
/**
* Set values for this Spline.
* @param xx
* @param yy
*/
public void setValues(double[] xx, double[] yy) {
this.xx = xx;
this.yy = yy;
if (xx.length > 1) {
calculateCoefficients();
}
}
/**
* Returns an interpolated value.
* @param x
* @return the interpolated value
*/
public double getValue(double x) {
if (xx.length == 0) {
return Double.NaN;
}
if (xx.length == 1) {
if (xx[0] == x) {
return yy[0];
} else {
return Double.NaN;
}
}
int index = Arrays.binarySearch(xx, x);
if (index > 0) {
return yy[index];
}
index = - (index + 1) - 1;
//TODO linear interpolation or extrapolation
if (index < 0) {
return yy[0];
}
return a[index]
+ b[index] * (x - xx[index])
+ c[index] * Math.pow(x - xx[index], 2)
+ d[index] * Math.pow(x - xx[index], 3);
}
/**
* Returns an interpolated value. To be used when a long sequence of values
* are required in order, but ensure checkValues() is called beforehand to
* ensure the boundary checks from getValue() are made
* @param x
* @return the interpolated value
*/
public double getFastValue(double x) {
// Fast check to see if previous index is still valid
if (storageIndex > -1 && storageIndex < xx.length-1 && x > xx[storageIndex] && x < xx[storageIndex + 1]) {
} else {
int index = Arrays.binarySearch(xx, x);
if (index > 0) {
return yy[index];
}
index = - (index + 1) - 1;
storageIndex = index;
}
//TODO linear interpolation or extrapolation
if (storageIndex < 0) {
return yy[0];
}
double value = x - xx[storageIndex];
return a[storageIndex]
+ b[storageIndex] * value
+ c[storageIndex] * (value * value)
+ d[storageIndex] * (value * value * value);
}
/**
* Used to check the correctness of this spline
*/
public boolean checkValues() {
if (xx.length < 2) {
return false;
} else {
return true;
}
}
/**
* Returns the first derivation at x.
* @param x
* @return the first derivation at x
*/
public double getDx(double x) {
if (xx.length == 0 || xx.length == 1) {
return 0;
}
int index = Arrays.binarySearch(xx, x);
if (index < 0) {
index = - (index + 1) - 1;
}
return b[index]
+ 2 * c[index] * (x - xx[index])
+ 3 * d[index] * Math.pow(x - xx[index], 2);
}
/**
* Calculates the Spline coefficients.
*/
private void calculateCoefficients() {
int N = yy.length;
a = new double[N];
b = new double[N];
c = new double[N];
d = new double[N];
if (N == 2) {
a[0] = yy[0];
b[0] = yy[1] - yy[0];
return;
}
double[] h = new double[N - 1];
for (int i = 0; i < N - 1; i++) {
a[i] = yy[i];
h[i] = xx[i + 1] - xx[i];
// h[i] is used for division later, avoid a NaN
if (h[i] == 0.0) {
h[i] = 0.01;
}
}
a[N - 1] = yy[N - 1];
double[][] A = new double[N - 2][N - 2];
double[] y = new double[N - 2];
for (int i = 0; i < N - 2; i++) {
y[i] =
3
* ((yy[i + 2] - yy[i + 1]) / h[i
+ 1]
- (yy[i + 1] - yy[i]) / h[i]);
A[i][i] = 2 * (h[i] + h[i + 1]);
if (i > 0) {
A[i][i - 1] = h[i];
}
if (i < N - 3) {
A[i][i + 1] = h[i + 1];
}
}
solve(A, y);
for (int i = 0; i < N - 2; i++) {
c[i + 1] = y[i];
b[i] = (a[i + 1] - a[i]) / h[i] - (2 * c[i] + c[i + 1]) / 3 * h[i];
d[i] = (c[i + 1] - c[i]) / (3 * h[i]);
}
b[N - 2] =
(a[N - 1] - a[N - 2]) / h[N
- 2]
- (2 * c[N - 2] + c[N - 1]) / 3 * h[N
- 2];
d[N - 2] = (c[N - 1] - c[N - 2]) / (3 * h[N - 2]);
}
/**
* Solves Ax=b and stores the solution in b.
*/
public void solve(double[][] A, double[] b) {
int n = b.length;
for (int i = 1; i < n; i++) {
A[i][i - 1] = A[i][i - 1] / A[i - 1][i - 1];
A[i][i] = A[i][i] - A[i - 1][i] * A[i][i - 1];
b[i] = b[i] - A[i][i - 1] * b[i - 1];
}
b[n - 1] = b[n - 1] / A[n - 1][n - 1];
for (int i = b.length - 2; i >= 0; i--) {
b[i] = (b[i] - A[i][i + 1] * b[i + 1]) / A[i][i];
}
}
}
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