petsc-3.7.3 2016-08-01

MatGetSubMatrix

Gets a single submatrix on the same number of processors as the original matrix.

Synopsis

#include "petscmat.h" 
PetscErrorCode MatGetSubMatrix(Mat mat,IS isrow,IS iscol,MatReuse cll,Mat *newmat)

Collective on Mat

Input Parameters

	mat	- the original matrix
	isrow	- parallel IS containing the rows this processor should obtain
	iscol	- parallel IS containing all columns you wish to keep. Each process should list the columns that will be in IT's "diagonal part" in the new matrix.
	cll	- either MAT_INITIAL_MATRIX or MAT_REUSE_MATRIX

Output Parameter

newmat -the new submatrix, of the same type as the old

Notes

The submatrix will be able to be multiplied with vectors using the same layout as iscol.

Some matrix types place restrictions on the row and column indices, such as that they be sorted or that they be equal to each other.

The index sets may not have duplicate entries.

The first time this is called you should use a cll of MAT_INITIAL_MATRIX, the MatGetSubMatrix() routine will create the newmat for you. Any additional calls to this routine with a mat of the same nonzero structure and with a call of MAT_REUSE_MATRIX will reuse the matrix generated the first time. You should call MatDestroy() on newmat when you are finished using it.

The communicator of the newly obtained matrix is ALWAYS the same as the communicator of the input matrix.

If iscol is NULL then all columns are obtained (not supported in Fortran).

Example usage

Consider the following 8x8 matrix with 34 non-zero values, that is assembled across 3 processors. Let's assume that proc0 owns 3 rows, proc1 owns 3 rows, proc2 owns 2 rows. This division can be shown

as follows

            1  2  0  |  0  3  0  |  0  4
    Proc0   0  5  6  |  7  0  0  |  8  0
            9  0 10  | 11  0  0  | 12  0
    -------------------------------------
           13  0 14  | 15 16 17  |  0  0
    Proc1   0 18  0  | 19 20 21  |  0  0
            0  0  0  | 22 23  0  | 24  0
    -------------------------------------
    Proc2  25 26 27  |  0  0 28  | 29  0
           30  0  0  | 31 32 33  |  0 34

Suppose isrow = [0 1 | 4 | 6 7] and iscol = [1 2 | 3 4 5 | 6]. The resulting submatrix is

            2  0  |  0  3  0  |  0
    Proc0   5  6  |  7  0  0  |  8
    -------------------------------
    Proc1  18  0  | 19 20 21  |  0
    -------------------------------
    Proc2  26 27  |  0  0 28  | 29
            0  0  | 31 32 33  |  0

Examples

src/ksp/ksp/examples/tutorials/ex49.c.html
src/ksp/ksp/examples/tutorials/ex64.c.html