How to find the nodes of a triangle from an adjacency matrix in Octave - octave

Question

I know how to find the number of triangles in an adjacency matrix.
tri = trace(A^3) / 6
But i require to find the nodes so that i can finally find the value of the edges from adjacency matrix since it's a sign graph. Is there already existing function which does that?

Answer 1

Taking the power of the adjacency matrix loses information about the intermediate nodes. Instead of a 2-dimensional matrix, we need 3 dimensions.
Given a graph:
and its adjacency matrix:
A =
0 0 0 0 1 1 0 1 0 0
0 0 0 1 0 1 0 0 0 0
0 0 0 1 0 0 0 1 0 1
0 1 1 0 1 0 1 0 0 0
1 0 0 1 0 0 1 0 0 0
1 1 0 0 0 0 0 1 1 0
0 0 0 1 1 0 0 0 1 0
1 0 1 0 0 1 0 0 0 0
0 0 0 0 0 1 1 0 0 0
0 0 1 0 0 0 0 0 0 0
Compute the 3d matrix T such that T(i,j,k) == 1 iff there is a path in the graph i=>j=>k=>i.
T = and(A, permute(A, [3 1 2]))
This is the equivalent of squaring the adjacency matrix, but keeping the path information. and is used here instead of multiplication in case A is a weighted adjacency matrix. If you sum along the 2nd dimension, you'll get A^2:
>> isequal(squeeze(sum(T,2)), A^2)
ans = 1
Now that we've got the paths of length 2, we just need to filter so we keep only the paths that return to their starting points.
T = and(T, permute(A.', [1 3 2])); % Transpose A in case graph is directed
Now, if T(i,j,k) == 1, then there is a triangle starting at node i, through nodes j and k and returning to node i. If you want to find all such paths:
[M,N,P] = ind2sub(size(T), find(T));
P = [M,N,P];
P will be a list of all triangular paths:
P =
8 6 1
6 8 1
7 5 4
5 7 4
7 4 5
4 7 5
8 1 6
1 8 6
5 4 7
4 5 7
6 1 8
1 6 8
In this case we get 12 paths. All paths in an undirected graph have 6 duplicates: one starting at each triangle point, times 2 directions. This gives the same results as trace:
>> trace(A^3)
ans = 12
If you want to remove the duplicates, the simplest way for triangles is to simply sort the vertex ordering and then take the unique rows of the list. This works for triangles only because all permutations of the nodes in the cycle are present. For longer cycles, this will not work.
P = unique(sort(P, 2), 'rows');
P =
1 6 8
4 5 7

Answer 2

Here is a solution using matrix multiplication:
C = (A * A.') & A;
[x, y] = find(tril(C));
n = numel(x);
D = sparse([x; y], [1:n 1:n].', 1, size(A,1), n);
[X, ~, V] = find(C * D);
tri = [x y X(V == 2)]
tri = unique(sort(tri, 2), 'rows');
First we need to know what are triangle nodes. Two nodes are triangle nodes if they have a common neighbor and both of them are neighbor of each other.
We take the definition to compute an adjacency matrix C that only contains triangle nodes and all other node are removed.
The expression A * A.' selects nodes that have common neighbors and the & A operator says that those nodes that have common neighbors should by neighbor of each other.
Now we can use [x, y] = find(tril(C)); to extract the first and the second points of each triangle as x and y respectively.
For the third node we need to find a node that has x and y as its neighbors. As before we can use the multiplication of boolean matrix trick to speed up the computation.
Finally the result tri has duplicates that should be remove using unique and sort.