[petsc-users] Understanding MatCreate bsize parameter

[Florian Lindner]

Am Donnerstag, 26. März 2015, 07:34:27 schrieb Jed Brown:
> Florian Lindner <mailinglists at xgm.de> writes:
> 
> > Hello,
> >
> > I'm using petsc with petsc4py.
> >
> > A matrix is created like that
> >
> >     MPIrank = MPI.COMM_WORLD.Get_rank()
> >     MPIsize = MPI.COMM_WORLD.Get_size()
> >     print("MPI Rank = ", MPIrank)
> >     print("MPI Size = ", MPIsize)
> >     parts = partitions()
> >     
> >     print("Dimension= ", nSupport + dimension, "bsize = ", len(parts[MPIrank]))
> >
> >     MPI.COMM_WORLD.Barrier() # Just to keep the output together
> >     A = PETSc.Mat(); A.createDense( (nSupport + dimension, nSupport + dimension), bsize = len(parts[MPIrank]) ) # <-- crash here
> 
> bsize is collective (must be the same on all processes).  It is used for
> vector-valued problems (like elasticity -- bs=3 in 3 dimensions).

It seems I'm still misunderstanding the bsize parameter.

If I distribute a 10x10 matrix on three ranks I need to have a non-homogenous distribution, and thats what petsc does itself:

A.createDense( (n, n) )

print("Rank = ", rank, "Range    = ", A.owner_range, "Size = ", A.owner_range[1] - A.owner_range[0])
print("Rank = ", rank, "ColRange = ", A.getOwnershipRangeColumn(), "Size = ", A.getOwnershipRangeColumn()[1] - A.getOwnershipRangeColumn()[0])

gives:

Rank =  2 Range    =  (7, 10) Size =  3
Rank =  2 ColRange =  (7, 10) Size =  3
Rank =  0 Range    =  (0, 4)  Size =  4
Rank =  0 ColRange =  (0, 4)  Size =  4
Rank =  1 Range    =  (4, 7)  Size =  3
Rank =  1 ColRange =  (4, 7)  Size =  3


How can I manually set a distribution of rows like above? My approach was to call create with bsize = [3,3,4][rank] but that obviously is not the way...

Thanks,
Florian