topic Re: Solve band matrix with GBTRS in Intel® oneAPI Math Kernel Library

Solve band matrix with GBTRS

cyfeng — Fri, 10 Oct 2008 02:39:28 GMT

I try to solve a simple case AX=B with subroutine GBTRS.

The matrix B is 6x1: B^T=(3, 1, 11, -3, -5, -1)

The matrix A 6x6 containing 2 subdiagonals and 1 superdiagonal is:

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

Doesthe matrix A in band matrixbe arranged right?

Is the following codecorrect?I can't get the correct answer...:(

REAL::A(6,6),B(6)
INTEGER::IPIV(6), INFO

DATA A/0.,0.,0.,2.,-3.,0., 0.,0.,-1.,1.,0.,0., 0.,0.,-2.,-1.,2.,1., &
& 0.,0.,2.,1.,-1.,-3., 0.,0.,1.,-2.,-3.,0., 0.,0.,0.,1.,0.,0. /
DATA B/3., 1., 11., -3., -5., -1./

CALL GBTRF(A,2,6,IPIV)
CALL GBTRS(A, B, IPIV)

Re: Solve band matrix with GBTRS

Vladimir_Koldakov__I — Fri, 10 Oct 2008 14:19:53 GMT

Quoting - cyfeng

I try to solve a simple case AX=B with subroutine GBTRS.

The matrix B is 6x1: B^T=(3, 1, 11, -3, -5, -1)

The matrix A 6x6 containing 2 subdiagonals and 1 superdiagonal is:

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

Doesthe matrix A in band matrixbe arranged right?

Is the following codecorrect?I can't get the correct answer...:(

REAL::A(6,6),B(6)
INTEGER::IPIV(6), INFO

DATA A/0.,0.,0.,2.,-3.,0., 0.,0.,-1.,1.,0.,0., 0.,0.,-2.,-1.,2.,1., &
& 0.,0.,2.,1.,-1.,-3., 0.,0.,1.,-2.,-3.,0., 0.,0.,0.,1.,0.,0. /
DATA B/3., 1., 11., -3., -5., -1./

CALL GBTRF(A,2,6,IPIV)
CALL GBTRS(A, B, IPIV)

Re: Solve band matrix with GBTRS

Vladimir_Koldakov__I — Fri, 10 Oct 2008 14:28:23 GMT

Hi, cyfeng,

As I see the matrix has 1 upper and one lower subdiagonals.

In this case the correct call isGBTRF(A,1,6,IPIV) as the second argument is KL.

-Vladimir

Re: Solve band matrix with GBTRS

Vladimir_Koldakov__I — Fri, 10 Oct 2008 16:32:06 GMT

Sorry, I was wrong. It is really KL=2. In such a case you should put KL into the second interface:

CALL GBTRS(A, B, IPIV, 2)

Otherwise it is assumed KL=KU=5/3

-Vladimir

Re: Solve band matrix with GBTRS

cyfeng — Fri, 10 Oct 2008 23:20:09 GMT

Quoting - Vladimir Koldakov (Intel)

Sorry, I was wrong. It is really KL=2. In such a case you should put KL into the second interface:

CALL GBTRS(A, B, IPIV, 2)

Otherwise it is assumed KL=KU=5/3

-Vladimir

Hi, Vladimir.

I paid too much attention to the GBTRF to make such a stupid question.

I should read the document more carefully.

Thank you very much :)