Question

Here is an exercise in the Algorithm Design Manual.

Consider the problem of determining whether a given undirected graph G
= (V, E) contains a triangle or cycle of length 3.

(a) Give an O(|V |^3) to find a triangle if one exists.

(b) Improve
your algorithm to run in time O(|V |·|E|). You may assume |V | ≤ |E|.

Observe that these bounds gives you time to convert between the
adjacency matrix and adjacency list representations of G.

Here is my thoughts:

(a) If the graph is given as an adjacency list, I can convert the list to matrix by O(|V|^2). then I do:

for (int i = 0;i < n;i++) 
   for (int j = i+1;j < n;j++) 
   if (matrix[i][j] == 1) 
      for (int k = j+1;k < n;k++) 
         if (matrix[i][k] == 1 && matrix[j][k] == 1)
             return true;

This should give a O(|V|^3) to test the triangle.

(b) My first intuitive is that if the graph is given as an adjacency list, then I will do a bfs. Whenever a cross edge is found, for example, if y-x is a cross edge, then i will check whether parent[y] == parent[x], if true, then a triangle is found.

Could anyone tell me whether my thinking is correct or not?

Also for this (b), I am not sure its complexity. Should it be O(|V| + |E|), right?

How can I do it in O(|V|*|E|)?

Answer 1

A possible O(|V||E|) solution, is the same idea of the brute-force in (a):

for each edge (u, v):
  for each vertex w:
     if (v, w) is an edge and (w, u) is an edge:
          return true
return false

check all edges, and not all vertices pairs – with another vertex that forms a triangle – it is enough information to determine if the edge and vertex form a feasible solution.

Counter example to BFS solution:

       A
     / | \
    /  |  \
   B   C   D
   |   |   |
   |   |   |
   F---G---H
   |       |
   ---------
    (F, H) is also an edge

Note that father[F] != father[G] != father[H], thus the algorithm will return false – but nevertheless, (F, G, H) is a feasible solution!