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Boost Library, how to get determinant from lu_factorize()?

I am trying to calculate a determinant using the boost c++ libraries. I found the code for the function InvertMatrix() which I have copied below. Every time I calculate this inverse, I want the determinant as well. I have a good idea how to calculate, by multiplying down the diagonal of the U matrix from the LU decomposition. There is one problem, I am able to calculate the determinant properly, except for the sign. Depending on the pivoting I get the sign incorrect half of the time. Does anyone have a suggestion on how to get the sign right every time? Thanks in advance.

template<class T>
bool InvertMatrix(const ublas::matrix<T>& input, ublas::matrix<T>& inverse)
{
 using namespace boost::numeric::ublas;
 typedef permutation_matrix<std::size_t> pmatrix;
 // create a working copy of the input
 matrix<T> A(input);
 // create a permutation matrix for the LU-factorization
 pmatrix pm(A.size1());

 // perform LU-factorization
 int res = lu_factorize(A,pm);
 if( res != 0 ) return false;

Here is where I inserted my best shot at calculating the determinant.

 T determinant = 1;

 for(int i = 0; i < A.size1(); i++)
 {
  determinant *= A(i,i);
 }

End my portion of the code.

 // create identity matrix of "inverse"
 inverse.assign(ublas::identity_matrix<T>(A.size1()));

 // backsubstitute to get the inverse
 lu_substitute(A, pm, inverse);

 return true;
}
like image 820
phoganuci Avatar asked Oct 14 '22 13:10

phoganuci


1 Answers

The permutation matrix pm contains the information you'll need to determine the sign change: you'll want to multiply your determinant by the determinant of the permutation matrix.

Perusing the source file lu.hpp we find a function called swap_rows which tells how to apply a permutation matrix to a matrix. It's easily modified to yield the determinant of the permutation matrix (the sign of the permutation), given that each actual swap contributes a factor of -1:

template <typename size_type, typename A>
int determinant(const permutation_matrix<size_type,A>& pm)
{
    int pm_sign=1;
    size_type size=pm.size();
    for (size_type i = 0; i < size; ++i)
        if (i != pm(i))
            pm_sign* = -1; // swap_rows would swap a pair of rows here, so we change sign
    return pm_sign;
}

Another alternative would be to use the lu_factorize and lu_substitute methods which don't do any pivoting (consult the source, but basically drop the pm in the calls to lu_factorize and lu_substitute). That change would make your determinant calculation work as-is. Be careful, however: removing pivoting will make the algorithm less numerically stable.

like image 166
Managu Avatar answered Oct 20 '22 18:10

Managu