org.jgrasstools.gears.utils.math.interpolation.splines.NatCubicClosed Maven / Gradle / Ivy

/*
 * JGrass - Free Open Source Java GIS http://www.jgrass.org 
 * (C) HydroloGIS - www.hydrologis.com 
 * 
 * This library is free software; you can redistribute it and/or modify it under
 * the terms of the GNU Library General Public License as published by the Free
 * Software Foundation; either version 2 of the License, or (at your option) any
 * later version.
 * 
 * This library 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 Library General Public License for more
 * details.
 * 
 * You should have received a copy of the GNU Library General Public License
 * along with this library; if not, write to the Free Foundation, Inc., 59
 * Temple Place, Suite 330, Boston, MA 02111-1307 USA
 */
package org.jgrasstools.gears.utils.math.interpolation.splines;

/**
 * This is adapted from: http://www.cse.unsw.edu.au/~lambert/splines/natcubicclosed.html
 * 
 * @author Tim Lambert (http://www.cse.unsw.edu.au/~lambert/)
 * @author Andrea Antonello (www.hydrologis.com)
 */
public class NatCubicClosed extends NatCubic {

    /* calculates the closed natural cubic spline that interpolates
       x[0], x[1], ... x[n]
       The first segment is returned as
       C[0].a + C[0].b*u + C[0].c*u^2 + C[0].d*u^3 0<=u <1
       the other segments are in C[1], C[2], ...  C[n] */

    public Cubic[] calcNaturalCubic( int n, double[] x ) {
        double[] w = new double[n + 1];
        double[] v = new double[n + 1];
        double[] y = new double[n + 1];
        double[] D = new double[n + 1];
        double z, F, G, H;
        int k;
        /* We solve the equation
           [4 1      1] [D[0]]   [3(x[1] - x[n])  ]
           |1 4 1     | |D[1]|   |3(x[2] - x[0])  |
           |  1 4 1   | | .  | = |      .         |
           |    ..... | | .  |   |      .         |
           |     1 4 1| | .  |   |3(x[n] - x[n-2])|
           [1      1 4] [D[n]]   [3(x[0] - x[n-1])]
           
           by decomposing the matrix into upper triangular and lower matrices
           and then back sustitution.  See Spath "Spline Algorithms for Curves
           and Surfaces" pp 19--21. The D[i] are the derivatives at the knots.
           */
        w[1] = v[1] = z = 1.0f / 4.0f;
        y[0] = z * 3 * (x[1] - x[n]);
        H = 4;
        F = 3 * (x[0] - x[n - 1]);
        G = 1;
        for( k = 1; k < n; k++ ) {
            v[k + 1] = z = 1 / (4 - v[k]);
            w[k + 1] = -z * w[k];
            y[k] = z * (3 * (x[k + 1] - x[k - 1]) - y[k - 1]);
            H = H - G * w[k];
            F = F - G * y[k - 1];
            G = -v[k] * G;
        }
        H = H - (G + 1) * (v[n] + w[n]);
        y[n] = F - (G + 1) * y[n - 1];

        D[n] = y[n] / H;
        D[n - 1] = y[n - 1] - (v[n] + w[n]) * D[n]; /* This equation is WRONG! in my copy of Spath */
        for( k = n - 2; k >= 0; k-- ) {
            D[k] = y[k] - v[k + 1] * D[k + 1] - w[k + 1] * D[n];
        }

        /* now compute the coefficients of the cubics */
        Cubic[] C = new Cubic[n + 1];
        for( k = 0; k < n; k++ ) {
            C[k] = new Cubic((double) x[k], D[k], 3 * (x[k + 1] - x[k]) - 2 * D[k] - D[k + 1], 2 * (x[k] - x[k + 1]) + D[k]
                    + D[k + 1]);
        }
        C[n] = new Cubic((double) x[n], D[n], 3 * (x[0] - x[n]) - 2 * D[n] - D[0], 2 * (x[n] - x[0]) + D[n] + D[0]);
        return C;
    }

}