Finding cycle of 3 nodes ( or triangles) in a graph

Question

I am working with complex networks. I want to find group of nodes which forms a cycle of 3 nodes (or triangles) in a given graph. As my graph contains about million edges, u

Answer 1

I am working on the same problem of counting number of triangles on undirected graph and wisty's solution works really well in my case. I have modified it a bit so only undirected triangles are counted.

    #### function for counting undirected cycles
    def generate_triangles(nodes):
        visited_ids = set() # mark visited node
        for node_a_id in nodes:
            temp_visited = set() # to get undirected triangles
            for node_b_id in nodes[node_a_id]:
                if node_b_id == node_a_id:
                    raise ValueError # to prevent self-loops, if your graph allows self-loops then you don't need this condition
                if node_b_id in visited_ids:
                    continue
                for node_c_id in nodes[node_b_id]:
                    if node_c_id in visited_ids:
                        continue    
                    if node_c_id in temp_visited:
                        continue
                    if node_a_id in nodes[node_c_id]:
                        yield(node_a_id, node_b_id, node_c_id)
                    else:
                        continue
                temp_visited.add(node_b_id)
            visited_ids.add(node_a_id)

Of course, you need to use a dictionary for example

    #### Test cycles ####

    nodes = {}

    nodes[0] = [1, 2, 3]
    nodes[1] = [0, 2]
    nodes[2] = [0, 1, 3]
    nodes[3] = [1]

    cycles = list(generate_triangles(nodes))
    print cycles

Using the code of Wisty, the triangles found will be [(0, 1, 2), (0, 2, 1), (0, 3, 1), (1, 2, 3)]

which counted the triangle (0, 1, 2) and (0, 2, 1) as two different triangles. With the code I modified, these are counted as only one triangle.

I used this with a relatively small dictionary of under 100 keys and each key has on average 50 values.