[petsc-dev] proposed minor PetscPartitioner changes

[Matthew Knepley]

On Sat, Nov 11, 2017 at 1:33 PM, Jed Brown <jed at jedbrown.org> wrote:

> Matthew Knepley <knepley at gmail.com> writes:
>
> > On Sat, Nov 11, 2017 at 1:12 PM, Jed Brown <jed at jedbrown.org> wrote:
> >
> >> Matthew Knepley <knepley at gmail.com> writes:
> >>
> >> >> Matrix and graph are equivalent concepts.
> >> >
> >> >
> >> > This is clearly wrong. A matrix is the coordinate representation of a
> >> > linear operator, and thus has a specific
> >> > behavior under coordinate transformations. A graph is just
> connectivity,
> >> > and really just a relation. I cannot
> >> > count the number of times Barry has ranted about this on petsc-maint
> >> > (usually about Vecs and arrays). The
> >> > mathematical object is not its data structure.
> >>
> >> A graph Laplacian certainly does transform under coordinate
> >> transformation and indeed, we use that property to design effective
> >> coarsening strategies.  That one basis strikes you as intrinsically
> >> "more canonical" does not mean it isn't a linear operator.
> >>
> >
> > That is one operator. This is argument by anecdote. An arbitrary graph
> > is not a linear operator, but an arbitrary matrix definitely is (the
> > coordinate representation of one).
>
> Dude, we solve linear systems and eigenproblems for arbitrary graphs.
> It isn't an anecdote.
>
> Barry (rightly) objects to a 2D array representing a function on a grid
> being considered a Matrix.  We don't "apply" it as a linear operator.
> There is no "vector" on which it operates.
>
> But we absolutely do with a graph.  Our vectors are functions at the
> vertices of the graph.  Applying the graph Laplacian tells us about
> local compatibility of the field over the vertices.  It is entirely
> analogous to fields over a grid.  You don't need a concept of "grid
> refinement" to have matrices.
>

I don't think makes sense. You are saying, because a linear operator (the
graph Laplacian) can be defined using the graph, then the graph is identical
with this operator. I do not agree. Whereas the matrix means nothing else
but the linear operator which it represents.

   Matt

-- 
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.caam.rice.edu/~mk51/>
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