dolfin team mailing list archive

["Garth N. Wells" <gnw20@xxxxxxxxx>]

On Mon, Oct 8, 2012 at 1:09 AM, Anders Logg <logg@xxxxxxxxx> wrote:
> 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?
>

The fix/hack was to add special cases to get around the (0, 0)
connectivity case. It's ugly.

I'll see if the (0, 0) case can be made consistent with the (d, d)
case - treating a cell as a  'Point cell' that is connected to a
vertex.

Garth

> --
> 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
>> >



-- 
Garth N. Wells
Department of Engineering, University of Cambridge
http://www.eng.cam.ac.uk/~gnw20