[orxan.shibli]

Hi all.

I want to know is there an efficient way of determining cell neighbours on unstructured grid other than the way in which nodes of (say) hexahedra compared one by one. I mean,

hex_1 : node_1, node_2, node_3, node_4
hex_2 : node_5, node_6, node_7, node_8

if hex_1 and hex_2 have 3 nodes in common then they are neighbours.

So is there a better way?

[sbaffini]

Well, i think it depends on which kind of information you have. In a cell-based solver i think it's much more efficient to deal with faces. In fluent .msh files you have the following kind of information:

nodes:
x1 y1 z1
x2 y2 z2
.
.
.
xn yn zn

faces:
nf1 n_1 n_2 ... n_nf1 c0_f1 c1_f1
nf2 n_1 n_2 ... n_nf2 c0_f2 c1_f2
.
.
.
nfm n_1 n_2 ... n_nfm c0_fm c1_fm

where there are n total nodes, m total faces and for the j-th face the total number of nodes is nfj

In this kind of file you have, for each face, the two neighboring cells (c0_fj c1_fj). Hence, to obtain the information you're looking for, you just need to invert such relationship (once for all). This can be done by allocating an array like:

f1_1 f2_1 . . . fk_1
f1_2 f2_2 . . . fk_2
.
.
.
f1_s f2_s . . . fk_s

where each row is relative to a cell of the grid. Of course not all the cells will have the same number of faces, so some memory could be wasted.

the array is built by a single loop over the faces and, once you have it, you can retrieve the neighbor information trough the 2 previous arrays.

If you have a node based solver, this is probably not going to work. In that case you can find some information on the book:

R. Lohner: Applied CFD Techniques. An Introduction based on Finite Element Methods. Wiley, 2nd Ed.

P.S. I can't think of hexas made up by 4 nodes and sharing each other only 3 nodes. I would change these numbers in 8 and 4 respectively... or maybe hexas in tetra.

[orxan.shibli]

Method is cell-centered. Meshing program is GMSH. The file is msh file. Something like this:

-total number of nodes -
-total number of elements -

node_1 x y z
.
.
.

line_1 node_# node_# (these lines belong to boundary only!)
.
.
.

triangle_1 node_# node_# node_#
.
.
.

tetra_1 node_# node_# node_# node_#
.
.
.

Thats all.

[sbaffini]

I really suggest you to read the book mentioned above, which has algorthmic examples to treat this specific problem. Nonetheless, in this case, you are right: the only way is to check for common nodes as this is the only information you have.

[orxan.shibli]

Thank you very much for answering my questions.