Question

I saw in this video that computing the clustering coefficient of central node of a star graph using the following algorithm is theta(n^2) and for a clique it is theta(n^3). is that correct?

def clustering_coefficient(G,v):
    neighbors = G[v].keys()
    if len(neighbors) == 1: return 0.0
    links = 0.0
    for w in neighbors:
        for u in neighbors:
            if u in G[w]: links += 0.5
    return 2.0*links/(len(neighbors)*(len(neighbors)-1))

Answer 1

The complexity depends on the density of your graph, and the efficiency of the in predicate.

A naive implementation on a complete graph obviously is O(n^3): two nested loops and one in predicate, each running naively in linear time. If you keep the links in a hashmap (and not in a dense matrix representation!) then the runtime is only O(n^2) – for a single node. But usually, such an algorithm is applied for each node, adding another factor of n to it.

If your graph is not complete (and you use a more efficient in predicate), things get a lot faster. Assuming that every node has sqrt(n) neighbors, the complexity of the algorithm will be O(sqrt(n)^2)*n (for all nodes, that is), which is probably their O(n^2) result.

Assuming that every node has exactly two neighbors. Then the complexity can easily be brought down to O(1) * n. Oh, and if every node has 0 neighbors, it’s even simpler.