On 2019-11-25 7:54 p.m., Matthew Knepley wrote:
On Mon, Nov 25, 2019 at 6:25 PM Swarnava Ghosh <[email protected]
<mailto:[email protected]>> wrote:
Dear PETSc users and developers,
I am working with dmplex to distribute a 3D unstructured mesh made
of tetrahedrons in a cuboidal domain. I had a few queries:
1) Is there any way of ensuring load balancing based on the number
of vertices per MPI process.
You can now call DMPlexRebalanceSharedPoints() to try and get better
balance of vertices.
Hi Matt,
I just want to follow up if this new function can help to solve the
"Strange Partition in PETSc 3.11" problem I mentioned before. Would you
please let me know when shall I call this function? Right before
DMPlexDistribute?
call DMPlexCreateFromCellList
call DMPlexGetPartitioner
call PetscPartitionerSetFromOptions
call DMPlexDistribute
Thanks,
Danyang
2) As the global domain is cuboidal, is the resulting domain
decomposition also cuboidal on every MPI process? If not, is there
a way to ensure this? For example in DMDA, the default domain
decomposition for a cuboidal domain is cuboidal.
It sounds like you do not want something that is actually
unstructured. Rather, it seems like you want to
take a DMDA type thing and split it into tets. You can get a cuboidal
decomposition of a hex mesh easily.
Call DMPlexCreateBoxMesh() with one cell for every process,
distribute, and then uniformly refine. This
will not quite work for tets since the mesh partitioner will tend to
violate that constraint. You could:
a) Prescribe the distribution yourself using the Shell partitioner type
or
b) Write a refiner that turns hexes into tets
We already have a refiner that turns tets into hexes, but we never
wrote the other direction because it was not clear
that it was useful.
Thanks,
Matt
Sincerely,
SG
--
What most experimenters take for granted before they begin their
experiments is infinitely more interesting than any results to which
their experiments lead.
-- Norbert Wiener
https://www.cse.buffalo.edu/~knepley/
<http://www.cse.buffalo.edu/~knepley/>
