[petsc-users] Understanding MatCreate bsize parameter

[Florian Lindner]

Am Freitag, 27. März 2015, 07:26:11 schrieb Matthew Knepley:
> On Fri, Mar 27, 2015 at 4:28 AM, Florian Lindner <mailinglists at xgm.de>
> wrote:
> 
> > 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:
> >
> 
> blockSize really means the uniform block size of the matrix, thus is HAS to
> divide the global size. If it does not,
> you do not have a uniform block size, you have a bunch of different sized
> blocks.

But how can I set a parallel layout when the size of the matrix is not divisable by the number of ranks? When I omit bsize Petsc does that for me, by using block sizes of 4, 3 and 3 on the three different ranks. How can I set such a parallel layout manually?

Thanks,
Florian


> > 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
> >
> 
> 
> 
>