[petsc-users] Storage space for symmetric (SBAIJ) matrix

[Jed Brown]

On Fri, Sep 17, 2010 at 13:15, Daniel Langr <daniel.langr at gmail.com> wrote:
> Hi all,
>
> I do not understand much how PETSc works with symmetric matrices. I tried
> some tests with a symmetric matrix, which have nonzeroes only
> in the main diagonal and in the last column/row, e.g., the following
> pattern:
>
> * 0 0 *
> 0 * 0 *
> 0 0 * *
> * * * *
>
> for a 4 x 4 matrix. Since PETSc requires to set values only for the upper
> triangular part of a matrix, I set two values for every row except of the
> last one with only one value. My code looks like:
>
> MatCreate(PETSC_COMM_WORLD, &A);
> MatSetSizes(A, PETSC_DECIDE, PETSC_DECIDE, n, n);
> MatSetType(A, MATMPISBAIJ);
> MatMPISBAIJSetPreallocation(A, 1, 1, PETSC_NULL, 1, PETSC_NULL);
> MatGetOwnershipRange(A, &first, &last);
> last--;
>
> for (i = first; i <= last; i++) {
>  MatSetValue(A, i, i, a, INSERT_VALUES);
>  if (i != (n - 1))
>    MatSetValue(A, i, n - 1, a, INSERT_VALUES);
> }
>
> MatAssemblyBegin(A, MAT_FINAL_ASSEMBLY);
> MatAssemblyEnd(A, MAT_FINAL_ASSEMBLY);
>
> (The value of the variable "a" changes for every nonzero entry. I omitted
> this code in the above printout together with an error checking for better
> readability.)
>
> Resulting matrix works quite fine but when I check the matrix info (e.g.
> using -mat_view_info_detailed option) for matrix size n=10000 and 2 MPI
> processes I get:
>
> // First and last row for every MPI process
> Rank: 0, Fist row: 0, Last row: 4999
> Rank: 1, Fist row: 5000, Last row: 9999
>
> Matrix Object:
>  type=mpisbaij, rows=10000, cols=10000
>  total: nonzeros=19999, allocated nonzeros=69990
>    [0] Local rows 10000 nz 10000 nz alloced 10000 bs 1 mem 254496
>    [0] on-diagonal part: nz 5000
>    [0] off-diagonal part: nz 5000
>    [1] Local rows 10000 nz 9999 nz alloced 59990 bs 1 mem 264494
>    [1] on-diagonal part: nz 9999
>    [1] off-diagonal part: nz 0

How were you preallocating this matrix?

Jed