package com.vividsolutions.jump.geom; import com.vividsolutions.jump.I18N; /** * Implements some 2D matrix operations * (in particular, solving systems of linear equations). * * @author Martin Davis * */ public class Matrix { private static void swapRows(double[][] m, int i, int j) { if (i == j) return; for (int col = 0; col < m[0].length; col++) { double temp = m[i][col]; m[i][col] = m[j][col]; m[j][col] = temp; } } private static void swapRows(double[] m, int i, int j) { if (i == j) return; double temp = m[i]; m[i] = m[j]; m[j] = temp; } /** * Solves a system of equations using Gaussian Elimination. * In order to avoid overhead the algorithm runs in-place * on A - if A should not be modified the client must supply a copy. * * @param a an nxn matrix in row/column order )modified by this method) * @param b a vector of length n * * @return a vector containing the solution (if any) * @return null if the system has no or no unique solution * * @throws IllegalArgumentException if the matrix is the wrong size */ public static double[] solve( double[][] a, double[] b ) { int n = b.length; if ( a.length != n || a[0].length != n ) throw new IllegalArgumentException(I18N.get("jump.geom.Matrix.Matrix-A-is-incorrectly-sized")); // Use Gaussian Elimination with partial pivoting. // Iterate over each row for (int i = 0; i < n; i++ ) { // Find the largest pivot in the rows below the current one. int maxElementRow = i; for (int j = i + 1; j < n; j++ ) if ( Math.abs( a[j][i] ) > Math.abs( a[maxElementRow][i] ) ) maxElementRow = j; if ( a[maxElementRow][i] == 0.0 ) return null; // Exchange current row and maxElementRow in A and b. swapRows(a, i, maxElementRow ); swapRows(b, i, maxElementRow ); // Eliminate using row i for (int j = i + 1; j < n; j++ ) { double rowFactor = a[j][i] / a[i][i]; for (int k = n - 1; k >= i; k-- ) a[j][k] -= a[i][k] * rowFactor; b[j] -= b[i] * rowFactor; } } /** * A is now (virtually) in upper-triangular form. * The solution vector is determined by back-substitution. */ double[] solution = new double[n]; for (int j = n - 1; j >= 0; j-- ) { double t = 0.0; for (int k = j + 1; k < n; k++ ) t += a[j][k] * solution[k]; solution[j] = ( b[j] - t ) / a[j][j]; } return solution; } }