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/* simq.c
*
* Solution of simultaneous linear equations AX = B
* by Gaussian elimination with partial pivoting
*
*
*
* SYNOPSIS:
*
* double A[n*n], B[n], X[n];
* int n, flag;
* int IPS[];
* int simq();
*
* ercode = simq( A, B, X, n, flag, IPS );
*
*
*
* DESCRIPTION:
*
* B, X, IPS are vectors of length n.
* A is an n x n matrix (i.e., a vector of length n*n),
* stored row-wise: that is, A(i,j) = A[ij],
* where ij = i*n + j, which is the transpose of the normal
* column-wise storage.
*
* The contents of matrix A are destroyed.
*
* Set flag=0 to solve.
* Set flag=-1 to do a new back substitution for different B vector
* using the same A matrix previously reduced when flag=0.
*
* The routine returns nonzero on error; messages are printed.
*
*
* ACCURACY:
*
* Depends on the conditioning (range of eigenvalues) of matrix A.
*
*
* REFERENCE:
*
* Computer Solution of Linear Algebraic Systems,
* by George E. Forsythe and Cleve B. Moler; Prentice-Hall, 1967.
*
*/
/* simq 2 */
#include <stdio.h>
#define fabs(x) ((x) < 0 ? -(x) : (x))
int simq( A, B, X, n, flag, IPS )
double A[], B[], X[];
int n, flag;
int IPS[];
{
int i, j, ij, ip, ipj, ipk, ipn;
int idxpiv, iback;
int k, kp, kp1, kpk, kpn;
int nip, nkp, nm1;
double em, q, rownrm, big, size, pivot, sum;
nm1 = n-1;
if( flag < 0 )
goto solve;
/* Initialize IPS and X */
ij=0;
for( i=0; i<n; i++ )
{
IPS[i] = i;
rownrm = 0.0;
for( j=0; j<n; j++ )
{
q = fabs( A[ij] );
if( rownrm < q )
rownrm = q;
++ij;
}
if( rownrm == 0.0 )
{
printf("SIMQ ROWNRM=0");
return(1);
}
X[i] = 1.0/rownrm;
}
/* simq 3 */
/* Gaussian elimination with partial pivoting */
for( k=0; k<nm1; k++ )
{
big= 0.0;
idxpiv = 0;
for( i=k; i<n; i++ )
{
ip = IPS[i];
ipk = n*ip + k;
size = fabs( A[ipk] ) * X[ip];
if( size > big )
{
big = size;
idxpiv = i;
}
}
if( big == 0.0 )
{
printf( "SIMQ BIG=0" );
return(2);
}
if( idxpiv != k )
{
j = IPS[k];
IPS[k] = IPS[idxpiv];
IPS[idxpiv] = j;
}
kp = IPS[k];
kpk = n*kp + k;
pivot = A[kpk];
kp1 = k+1;
for( i=kp1; i<n; i++ )
{
ip = IPS[i];
ipk = n*ip + k;
em = -A[ipk]/pivot;
A[ipk] = -em;
nip = n*ip;
nkp = n*kp;
for( j=kp1; j<n; j++ )
{
ipj = nip + j;
A[ipj] = A[ipj] + em * A[nkp + j];
}
}
}
kpn = n * IPS[n-1] + n - 1; /* last element of IPS[n] th row */
if( A[kpn] == 0.0 )
{
printf( "SIMQ A[kpn]=0");
return(3);
}
/* simq 4 */
/* back substitution */
solve:
ip = IPS[0];
X[0] = B[ip];
for( i=1; i<n; i++ )
{
ip = IPS[i];
ipj = n * ip;
sum = 0.0;
for( j=0; j<i; j++ )
{
sum += A[ipj] * X[j];
++ipj;
}
X[i] = B[ip] - sum;
}
ipn = n * IPS[n-1] + n - 1;
X[n-1] = X[n-1]/A[ipn];
for( iback=1; iback<n; iback++ )
{
/* i goes (n-1),...,1 */
i = nm1 - iback;
ip = IPS[i];
nip = n*ip;
sum = 0.0;
for( j=i+1; j<n; j++ )
sum += A[nip+j] * X[j];
X[i] = (X[i] - sum)/A[nip+i];
}
return(0);
}
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