[petsc-users] A quick question on 'un-symmetric graph'

[Jed Brown]

On Sat, Jan 5, 2013 at 10:59 PM, Mark F. Adams <mark.adams at columbia.edu>wrote:

> Isolated nodes are removed (so I do not actually do a true MIS) but if the
> graph not symmetric then other other processors think they are talking to a
> BC node and expect it to talk back to them.  Maybe we are not understanding
> each other.


I meant to mark boundary nodes (by looking at rows) and communicate that
status so that other rows don't expect it to "talk back".
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