/************************************************************************* * * OpenOffice.org - a multi-platform office productivity suite * * $RCSfile: gauss.hxx,v $ * * $Revision: 1.4 $ * * last change: $Author: rt $ $Date: 2005-09-07 20:56:43 $ * * The Contents of this file are made available subject to * the terms of GNU Lesser General Public License Version 2.1. * * * GNU Lesser General Public License Version 2.1 * ============================================= * Copyright 2005 by Sun Microsystems, Inc. * 901 San Antonio Road, Palo Alto, CA 94303, USA * * This library is free software; you can redistribute it and/or * modify it under the terms of the GNU Lesser General Public * License version 2.1, as published by the Free Software Foundation. * * 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 * Lesser General Public License for more details. * * You should have received a copy of the GNU Lesser General Public * License along with this library; if not, write to the Free Software * Foundation, Inc., 59 Temple Place, Suite 330, Boston, * MA 02111-1307 USA * ************************************************************************/ /** This method eliminates elements below main diagonal in the given matrix by gaussian elimination. @param matrix The matrix to operate on. Last column is the result vector (right hand side of the linear equation). After successful termination, the matrix is upper triangular. The matrix is expected to be in row major order. @param rows Number of rows in matrix @param cols Number of columns in matrix @param minPivot If the pivot element gets lesser than minPivot, this method fails, otherwise, elimination succeeds and true is returned. @return true, if elimination succeeded. */ template bool eliminate( Matrix& matrix, int rows, int cols, const BaseType& minPivot ) { BaseType temp; int max, i, j, k; /* *must* be signed, when looping like: j>=0 ! */ /* eliminate below main diagonal */ for(i=0; i fabs(matrix[ max*cols + i ]) ) max = j; /* check pivot value */ if( fabs(matrix[ max*cols + i ]) < minPivot ) return false; /* pivot too small! */ /* interchange rows 'max' and 'i' */ for(k=0; k=i; --k) matrix[ j*cols + k ] -= matrix[ i*cols + k ] * matrix[ j*cols + i ] / matrix[ i*cols + i ]; } /* everything went well */ return true; } /** Retrieve solution vector of linear system by substituting backwards. This operation _relies_ on the previous successful application of eliminate()! @param matrix Matrix in upper diagonal form, as e.g. generated by eliminate() @param rows Number of rows in matrix @param cols Number of columns in matrix @param result Result vector. Given matrix must have space for one column (rows entries). @return true, if back substitution was possible (i.e. no division by zero occured). */ template bool substitute( const Matrix& matrix, int rows, int cols, Vector& result ) { BaseType temp; int j,k; /* *must* be signed, when looping like: j>=0 ! */ /* substitute backwards */ for(j=rows-1; j>=0; --j) { temp = 0.0; for(k=j+1; k bool solve( Matrix& matrix, int rows, int cols, Vector& result, BaseType minPivot ) { if( eliminate(matrix, rows, cols, minPivot) ) return substitute(matrix, rows, cols, result); return false; }