On 4/25/23 09:44, Lucas Myers wrote:
Re: FEPointEvaluation and RemotePointEvaluation, I think I actually need
more than that because I need the quadrature points on each of the cells
that I'm querying, so I can just do the standard procedure with FEValues
and the corresponding Quadrature object.
In that case, the way I would implement this is that every process makes
a list of circles: points at which it wants to evaluate the convolution,
and the radius of the convolution.
In a second step, you then determine which of these circles require
communication. This isn't actually entirely trivial, but a good
criterion is probably whether a circle has overlap with an artificial
cell. So:
std::set<Circle> all_circles = ...;
std::set<Circle> circles_requiring_communication;
for (cell=...)
if (cell->is_artificial())
for (c : all_circles)
if (c not already in circles_requiring_communication)
if (c overlaps with cell)
circles_requiring_communication.insert (c);
In a third step, you can either send all of these circles to all other
processes (via Utilities::MPI::all_gather()), or you can be a bit
smarter and first get a bounding box for each of the other processes'
locally owned cells (see the function in GridTools for that); in that
latter case, you'd only send a circle to a process if it overlaps with
that other process's bounding box, using one of the ConsensusAlgorithms
functions.
At this point, each process has a list of circles that overlap with its
own subdomain. Now you work on that by looping over all of its locally
owned cells, on each cell query all circles, and do what you need to do.
This results in a list of contributions that need to be sent back to the
original process requesting that information. This could probably be
done nicely using the request-answer system of the ConsensusAlgorithms
functions.
You'd need to be careful with cases where a process gets sent a circle
that has overlap with its bounding box, but not with any of its cells.
In that case, the reply is simply a zero contribution to the convolution
integral.
I hope this makes sense!
Best
W.
--
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Wolfgang Bangerth email: bange...@colostate.edu
www: http://www.math.colostate.edu/~bangerth/
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