Find which tetrahedron in a Triangulation contains a Given Facet
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I have a Delaunay tetrahedralization, defined by a Coordinate Vector "X" (of size n x 3) and tetrahedra IDs "TRI"(of size m x 4). I want to know which tetrahedron in the triangulation contains a given triangular facet, based on IDs of the 3 vertices. MWE:
TRI = [1 2 3 4;5 6 7 8; 9 10 11 12; 13 14 15 16];
Facet = [5 7 8];
and I am obviously looking for
ans = 2
but since the number of columns of TRI and Facet are not the same, I can't use the ismember function. Any other ideas? Thanks.
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John D'Errico
2017년 3월 1일
편집: John D'Errico
2017년 3월 1일
Um, can you say easy, peasy?
1. Sort the facet indices in your target facet.
2. Sort the facet indices in your triangulation.
3. Form a list of all facets. Each tetrahedron has 4 facets, so there will be 4 facets for each simplex in the complex.
FACET = sort(FACET);
TRI = sort(TRI,2);
facetlist = [TRI(:,[1 2 3]);TRI(:,[1 2 4]);TRI(:,[1 3 4]);TRI(:,[2 3 4])];
Now a simple application of ismember will find the facet of interest. Note that the facet may appear once or twice. This depends upon whether the simplex shares that facet with another simplex in the tessellation.
How do you identify which simplex the facets you have just identified came from? One simple solution would be ind2sub, but mod will also suffice.
For example:
tri = delaunayn(rand(7,3))
tri =
3 1 6 5
2 6 1 5
7 3 4 5
7 4 2 5
7 1 3 5
7 2 1 5
tri = sort(tri,2)
tri =
1 3 5 6
1 2 5 6
3 4 5 7
2 4 5 7
1 3 5 7
1 2 5 7
facet = [1 5 7];
facetlist = [tri(:,[1 2 3]);tri(:,[1 2 4]);tri(:,[1 3 4]);tri(:,[2 3 4])];
[~,ind] = ismember(facetlist,facet,'rows');
ind = find(ind)
ind =
17
18
Now, which simplex did each of those facets arise from?
ntri = size(tri,1);
mod(ind-1,ntri) + 1
ans =
5
6
That particular facet was shared between simplex 5 and simplex 6 in the list. If you look back at the original triangulation, it did appear in each of those places. As I said, if a facet is on the boundary of the tessellation, then it will appear only once.
Now you should see why the sorts were necessary, as otherwise the ismember search may well fail.
As I said, easy. A few simple lines of code.
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John D'Errico
2017년 3월 1일
You get used to this type of computations when you work with tessellations enough. They get natural before long.
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