LinearEquation* run_linbcg(LinearEquation *leq,
float tolerance,
int max_iterations) {
int iterations;
double error;
/* Solves A · x = b for x[1..n], given b[1..n], by the iterative biconjugate gradient method. On input x[1..n] should be set to an initial guess of the solution (or all zeros); itol is 1,2,3, or 4, specifying which convergence test is applied (see text); itmax is the maximum number of allowed iterations; and tol is the desired convergence tolerance. On output, x[1..n] is reset to the improved solution, iter is the number of iterations actually taken, and err is the estimated error. The matrix A is referenced only through the user-supplied routines atimes, which computes the product of either A or its transpose on a vector; and asolve, which solves A · x = b or A' · x = b for some preconditioner matrix A (possibly the trivial diagonal part of A). */
linbcg(leq->sa, leq->ija, leq->n, leq->b, leq->x, 1,
tolerance, max_iterations, &iterations, &error);
return leq;
}
// transform gradients to luminance
static void transform_to_luminance(pyramid_t* pp, float* const x, pfstmo_progress_callback progress_cb, const bool bcg, const int itmax, const float tol)
{
pyramid_t* pC = pyramid_allocate(pp->cols, pp->rows);
pyramid_calculate_scale_factor(pp, pC); // calculate (Cx,Cy)
pyramid_scale_gradient(pp, pC); // scale small gradients by (Cx,Cy);
float* b = matrix_alloc(pp->cols * pp->rows);
pyramid_calculate_divergence_sum(pp, b); // calculate the sum of divergences (equal to b)
// calculate luminances from gradients
if (bcg)
linbcg(pp, pC, b, x, itmax, tol, progress_cb);
else
lincg(pp, pC, b, x, itmax, tol, progress_cb);
matrix_free(b);
pyramid_free(pC);
}
int main(void)
{
int i,ii,iter;
double *b,*bcmp,*x,err;
b=dvector(1,NP);
bcmp=dvector(1,NP);
x=dvector(1,NP);
for (i=1;i<=NP;i++) {
x[i]=0.0;
b[i]=1.0;
}
b[1]=3.0;
b[NP] = -1.0;
linbcg(NP,b,x,ITOL,TOL,ITMAX,&iter,&err);
printf("%s %15e\n","Estimated error:",err);
printf("%s %6d\n","Iterations needed:",iter);
printf("\nSolution vector:\n");
for (ii=1;ii<=NP/5;ii++) {
for (i=5*(ii-1)+1;i<=5*ii;i++) printf("%12.6f",x[i]);
printf("\n");
}
for (i=1;i<=(NP % 5);i++)
printf("%12.6f",x[5*(NP/5)+i]);
printf("\n");
dsprsax(sa,ija,x,bcmp,NP);
/* this is a double precision version of sprsax */
printf("\npress RETURN to continue...\n");
(void) getchar();
printf("test of solution vector:\n");
printf("%9s %12s\n","a*x","b");
for (i=1;i<=NP;i++)
printf("%12.6f %12.6f\n",bcmp[i],b[i]);
free_dvector(x,1,NP);
free_dvector(bcmp,1,NP);
free_dvector(b,1,NP);
return 0;
}
