LAPACK + C, weird behaviour - c

Question

I am trying to solve a simple linear equations system using LAPACK. I use dbsvg method which is optimised for banded matrices. I've obsereved a realy strange behaviour. When I fill the AT matrix this way:
for(i=0; i<DIM;i++) AB[0][i] = -1;
for(i=0; i<DIM;i++) AB[1][i] = 2;
for(i=0; i<DIM;i++) AB[2][i] = -1;
for(i=0; i<3; i++)
for(j=0;j<DIM;j++) {
AT[i*DIM+j]=AB[i][j];
}
And call:
dgbsv_(&N, &KL, &KU, &NRHS, AT, &LDAB, myIpiv, x, &LDB, &INFO);
It works perfectly. However, when I do it this way:
for(i=0; i<DIM;i++) AT[i] = -1;
for(i=0; i<DIM;i++) AT[DIM+i] = 2;
for(i=0; i<DIM;i++) AT[2*DIM+i] = -1;
It results with a vector filled with NaN. Here are the declarations:
double AB[3][DIM], AT[3*DIM];
double x[DIM];
int myIpiv[DIM];
int N=DIM, KL=1, KU=1, NRHS=1, LDAB=DIM, LDB=DIM, INFO;
Any ideas?

Answer 1

You're not laying out the entries in the band storage properly; it was working before by a happy accident. The LAPACK docs say:
On entry, the matrix A in band storage, in rows KL+1 to
2*KL+KU+1; rows 1 to KL of the array need not be set.
The j-th column of A is stored in the j-th column of the
array AB as follows:
AB(KL+KU+1+i-j,j) = A(i,j) for max(1,j-KU)<=i<=min(N,j+KL)
On exit, details of the factorization: U is stored as an
upper triangular band matrix with KL+KU superdiagonals in
rows 1 to KL+KU+1, and the multipliers used during the
factorization are stored in rows KL+KU+2 to 2*KL+KU+1.
See below for further details.
So if you want a tridiagonal matrix with 2 on the diagonal and -1 above and below, the layout should be:
* * * * * * * ... * * * *
* -1 -1 -1 -1 -1 -1 ... -1 -1 -1 -1
2 2 2 2 2 2 2 ... 2 2 2 2
-1 -1 -1 -1 -1 -1 -1 ... -1 -1 -1 *
LDAB should be 4 in this case. Bear in mind that LAPACK uses a column-major layout, so the actual array should be look like this in memory:
{ *, *, 2.0, -1.0, *, -1.0, 2.0, -1.0, *, -1.0, 2.0, -1.0, ... }
dgbsv was giving different results for the two identical arrays because it was reading off the ends of the arrays that you had laid out.

Answer 2

Is this the exact code you used or just an example? I ran this code here (just cut and pasted from your posts, with a change of AT to AT2 in the second loop:
const int DIM=10;
double AB[DIM][DIM], AT[3*DIM], AT2[3*DIM];
int i,j;
for(i=0; i<DIM;i++) AB[0][i] = -1;
for(i=0; i<DIM;i++) AB[1][i] = 2;
for(i=0; i<DIM;i++) AB[2][i] = -1;
for(i=0; i<3; i++)
for(j=0;j<DIM;j++) {
AT[i*DIM+j]=AB[i][j];
}
printf("AT:");
for (i=0;i<3*DIM;++i) printf("%lf ",AT[i]);
printf("\n\n");
for(i=0; i<DIM;i++) AT2[i] = -1;
for(i=0; i<DIM;i++) AT2[DIM+i] = 2;
for(i=0; i<DIM;i++) AT2[2*DIM+i] = -1;
printf("AT2:");
for (i=0;i<3*DIM;++i) printf("%lf ",AT2[i]);
printf("\n\n");
printf("Diff:");
for (i=0;i<3*DIM;++i) printf("%lf ",AT[i]-AT2[i]);
printf("\n\n");
and got this output
AT:-1.000000 -1.000000 -1.000000 -1.000000 -1.000000 -1.000000 -1.000000 -1.0000
00 -1.000000 -1.000000 2.000000 2.000000 2.000000 2.000000 2.000000 2.000000 2.0
00000 2.000000 2.000000 2.000000 -1.000000 -1.000000 -1.000000 -1.000000 -1.0000
00 -1.000000 -1.000000 -1.000000 -1.000000 -1.000000
AT2:-1.000000 -1.000000 -1.000000 -1.000000 -1.000000 -1.000000 -1.000000 -1.000
000 -1.000000 -1.000000 2.000000 2.000000 2.000000 2.000000 2.000000 2.000000 2.
000000 2.000000 2.000000 2.000000 -1.000000 -1.000000 -1.000000 -1.000000 -1.000
000 -1.000000 -1.000000 -1.000000 -1.000000 -1.000000
Diff:0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.0
00000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.
000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0
.000000 0.000000 0.000000 0.000000
Apparently AT and AT2 are the same. Which I would expect.