gauss_jordan.h

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// -*- c++ -*-

//    Copyright (C) 2009 Ed Rosten (er258@cam.ac.uk)
//
// This file is part of the TooN Library.  This library is free
// software; you can redistribute it and/or modify it under the
// terms of the GNU General Public License as published by the
// Free Software Foundation; either version 2, 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 General Public License for more details.

// You should have received a copy of the GNU General Public License along
// with this library; see the file COPYING.  If not, write to the Free
// Software Foundation, 59 Temple Place - Suite 330, Boston, MA 02111-1307,
// USA.

// As a special exception, you may use this file as part of a free software
// library without restriction.  Specifically, if other files instantiate
// templates or use macros or inline functions from this file, or you compile
// this file and link it with other files to produce an executable, this
// file does not by itself cause the resulting executable to be covered by
// the GNU General Public License.  This exception does not however
// invalidate any other reasons why the executable file might be covered by
// the GNU General Public License.


#ifndef TOON_INC_GAUSS_JORDAN_H
#define TOON_INC_GAUSS_JORDAN_H

#include <utility>
#include <cmath>
#include <TooN/TooN.h>

namespace TooN
{
template<int R, int C, class Precision, class Base> void gauss_jordan(Matrix<R, C, Precision, Base>& m)
{
        using std::swap;

        //Loop over columns to reduce.
        for(int col=0; col < m.num_rows(); col++)
        {
                //Reduce the current column to a single element


                //Search down the current column in the lower triangle for the largest
                //absolute element (pivot).  Then swap the pivot row, so that the pivot
                //element is on the diagonal. The benchmarks show that it is actually
                //faster to swap whole rows than it is to access the rows via indirection 
                //and swap the indirection element. This holds for both pointer indirection
                //and using a permutation vector over rows.
                {
                  using std::abs;
                        int pivotpos = col;
                        double pivotval = abs(m[pivotpos][col]);
                        for(int p=col+1; p <m.num_rows(); p++)
                          if(abs(m[p][col]) > pivotval)
                                {
                                        pivotpos = p;
                                        pivotval = abs(m[pivotpos][col]);
                                }
                        
                        if(col != pivotpos)
                                for(int c=0; c < m.num_cols(); c++)
                                        swap(m[col][c], m[pivotpos][c]);
                }

                //Reduce the current column in every row to zero, excluding elements on
                //the leading diagonal.
                for(int row = 0; row < m.num_rows(); row++)
                {
                        if(row != col)
                        {
                                double multiple = m[row][col] / m[col][col];
                
                                //We could eliminate some of the computations in the augmented
                                //matrix, if the augmented half is the identity. In general, it
                                //is not. 

                                //Subtract the pivot row from all other rows, to make 
                                //column col zero.
                                m[row][col] = 0;
                                for(int c=col+1; c < m.num_cols(); c++)
                                        m[row][c] = m[row][c] - m[col][c] * multiple;
                        }
                }
        }
        
        //Final pass to make diagonal elements one. Performing this in a final
        //pass allows us to avoid any significant computations on the left-hand
        //square matrix, since it is diagonal, and ends up as the identity.
        for(int row=0;row < m.num_rows(); row++)
        {
                double mul = 1/m[row][row];

                m[row][row] = 1;

                for(int col=m.num_rows(); col < m.num_cols(); col++)
                        m[row][col] *= mul;
        }
}

}
#endif

---

libtoon
Author(s): Florian Weisshardt
autogenerated on Fri Jan 11 10:09:37 2013