CODEWITHSUNDEEP
lapack
ericj
ericj

Reputation: 2301

When usng rows with LAPACKE_sgetrs, why must ldb=1 (instead of 3, instead of n)?

We want to solve x for Ax=b, 

A=
0  2  3
1  1  -1
0  -1  1

b=
13
0
1

x=
1
2
3

The program below first writes A=P*L*U. It works with columns. It is something like:

 float a[3*3]={
0,1,0,       
2,1,-1,
3,-1,1
};
  float b[8]={
13,
0,
1  

lapack_int n=3,lda=3,ldb=3,nrhs=1,info,piv[3];

info= LAPACKE_sgetrf(LAPACK_COL_MAJOR,n,n,a,lda,piv);

info= LAPACKE_sgetrs(LAPACK_COL_MAJOR,'N',n,nrhs,a,lda,piv,b,ldb);

This works. Now I would like to program with rows:

 float a[3*3]={
   0,2,3,
   1,1,-1,
   0,-1,1
};
  float b[8]={
13,
0,
1  

lapack_int n=3,lda=3,ldb=1,nrhs=1,info,piv[3];

info= LAPACKE_sgetrf(LAPACK_ROW_MAJOR,n,n,a,lda,piv);

info= LAPACKE_sgetrs(LAPACK_ROW_MAJOR,'N',n,nrhs,a,lda,piv,b,ldb);

My question is: why must ldb=1 (instead of 3)?

Upvotes: 1

Views: 99

Answers (1)

francis
francis

Reputation: 9817

This behavior is due to the wrapper LAPACKE.

If LAPACK_COL_MAJOR is used, the wrapper almost directly calls sgetrs() of LAPACK, as shown in the source of LAPACKE. Hence, the leading dimension ldb of the array b must be equal or higher than the number of rows of the matrix a, that is n=3. Therefore, the requirement is LDB >= max(1,N) as in sgetrs().

One the other hand, if LAPACK_ROW_MAJOR is used, b is transposed. Consequently, the leading dimension of the array ldb is now related to the number of right hand sides nrhs=1. The requirement is now LDB >= max(1,NRHS) as tested on line 59 : if( ldb < nrhs ). Then the array b and the matrix are transposed by calling LAPACKE_sge_trans. Finally, sgetrs() is called using lbd=n and the result is transposed back.

Upvotes: 1

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