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//
// Didasko Tutorial Package
// Copyright (2005) Sandia Corporation
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// Basic definition of communicator.
// This code should be run with one process
#include "Didasko_ConfigDefs.h"
#if defined(HAVE_DIDASKO_EPETRA)
#include <iostream>
#include "Epetra_ConfigDefs.h"
#ifdef HAVE_MPI
#include "mpi.h"
#include "Epetra_MpiComm.h"
#else
#include "Epetra_SerialComm.h"
#endif
#include "Epetra_SerialDenseVector.h"
#include "Epetra_SerialDenseMatrix.h"
#include "Epetra_SerialDenseSolver.h"
int main(int argc, char *argv[])
{
#ifdef HAVE_MPI
MPI_Init(&argc, &argv);
Epetra_MpiComm Comm(MPI_COMM_WORLD);
#else
#endif
// Total number of elements in vectors, can be any positive number
int NumRows = 5;
x.Size( NumRows );
b.Size( NumRows );
// set the elements of the vector
for( int i=0 ; i<NumRows ; ++i ) b[i] = 1.0, x[i]=0.0;
A.Shape( NumRows, NumRows );
A2.Shape( NumRows, NumRows ); // A2 is a copy of A
// Hilbert matrix (ill-conditioned)
for( int i=0 ; i<NumRows ; ++i )
for( int j=0 ; j<NumRows ; ++j )
A(i,j) = 1.0/(i+j+2);
cout<< A;
// set up the solver
Problem.SetMatrix( A );
Problem.SetVectors( x, b );
A2 = A;
// we make a copy of A because Problem.Solve() will
// overwrite A with its LU decomposition. Try with
// cout << A after the following invocation
b.Multiply('N','N',1.0, A2, x, 0.0);
cout << "A * x = \n" << b;
double rcond;
cout << "The (estimated) condition number of A is " << 1/rcond << endl;
Problem.SetMatrix( A2 );
Problem.Invert();
cout << "The inverse of A is\n";
cout << A2;
#ifdef HAVE_MPI
MPI_Finalize();
#endif
} /* main */
#else
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char *argv[])
{
puts("Please configure Didasko with:\n"
"--enable-epetra");
return 0;
}
#endif