dolfin team mailing list archive

[Anders Logg <logg@xxxxxxxxx>]

On Sun, Oct 07, 2012 at 08:16:33PM +0100, Garth N. Wells wrote:

> Is this consistent?

I'm not sure, it's just a choice we needed to make. It makes sense
that being connected means sharing a common vertex, but this obviously
fails for the case (0, 0) so then I thought it made sense for two
vertices to be neighbors if they share a cell. For simplices it's the
same as sharing an edge, so when quads are now being added we might
need to rethink this.

> I have a feeling that it's not because I'm having to write a bunch
> of special cases to make BoundaryMesh and the mesh partitioning work
> in 1D because facets and vertices are of the same topological
> dimension.
>
> Would it work if
>
>    CellIterator c(cell)
>
> yielded *c == cell? (and likewise for other topological dimensions)

You mean a cell is only neighbor to itself? I think that would break
quite a few algorithms currently in DOLFIN that rely on looking at
neighboring cells.

Did you fix the 1D boundary mesh bug? What was the fix?

--
Anders


> Garth
>
> >> In the above case, each cell will be connected to itself and the other
> >> cell, so the loop should print only this:
> >>
> >>   In cell loop
> >>   In cell loop
> >>
> >> Why do you get 6? I tried this (in Python) and get only 2.
> >>
> >
> > Sorry, try
> >
> >      UnitSquare mesh(2, 2);
> >
> > I used mesh(1, 1) in testing, and indeed in only prints twice.
> >
> >>> In what sense are entities of the same dimension 'connected'? The
> >>> present behaviour is causing a problem when computing a BoundaryMesh
> >>> in 1D because it picks up the end vertices, but then iterates along
> >>> the lines of
> >>>
> >>>         for (VertexIterator v(vertex); !v.end(); ++v)
> >>>         {
> >>>
> >>>         }
> >>>
> >>> which then yields the vertex that is one place in from the boundary.
> >>
> >> The algorithm for BoundaryMesh is to iterate over all facets of the
> >> mesh, in this case all vertices, and then check for each one if it is
> >> connected to exactly one entity of dimension D (cells) which in this
> >> case is edges (intervals). So something clearly goes wrong in
> >> TopologyComputation in 1D.
> >>
> >> I'll look at it later.
> >>
> >
> > The problem is what I describe above with the iterators over the (d,
> > d) connectivity with d=0. The BoundaryMesh code finds a facet on the
> > boiundary (vertex in the case of 1D), and then iterates over the
> > vertices of the facet:
> >
> >      for (VertexIterator v(facet); !v.end(); ++v)
> >      {
> >
> >      }
> >
> > When facet is not a vertex, this computes all the vertices that lie on
> > the boundary. When facet is a vertex, the above returns vertices that
> > share the cell with facet, but which do not lie on the boundary.
> >
> >
> > Garth
> >