/* -*- Mode: C++; tab-width: 4; indent-tabs-mode: nil; c-basic-offset: 4 -*- */ /* * This file is part of the LibreOffice project. * * This Source Code Form is subject to the terms of the Mozilla Public * License, v. 2.0. If a copy of the MPL was not distributed with this * file, You can obtain one at http://mozilla.org/MPL/2.0/. * * This file incorporates work covered by the following license notice: * * Licensed to the Apache Software Foundation (ASF) under one or more * contributor license agreements. See the NOTICE file distributed * with this work for additional information regarding copyright * ownership. The ASF licenses this file to you under the Apache * License, Version 2.0 (the "License"); you may not use this file * except in compliance with the License. You may obtain a copy of * the License at http://www.apache.org/licenses/LICENSE-2.0 . */ /** 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. */ #pragma once template bool eliminate( Matrix& matrix, int rows, int cols, const BaseType& minPivot ) { BaseType temp; /* i, j, k *must* be signed, when looping like: j>=0 ! */ /* eliminate below main diagonal */ for(int 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(int 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 occurred). */ template bool substitute( const Matrix& matrix, int rows, int cols, Vector& result ) { BaseType temp; /* j, k *must* be signed, when looping like: j>=0 ! */ /* substitute backwards */ for(int j=rows-1; j>=0; --j) { temp = 0.0; for(int 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; } /* vim:set shiftwidth=4 softtabstop=4 expandtab: */