How can I find a negative weighted cycle of 3 edges in a graph?

Question

I have a directed graph with about 10,000 nodes. All edges are weighted. I want to find a negative cycle containing only 3 edges. Is there any algorithm quicker than O(n^3)?

a sample code: (g is my graph)

if (DETAILS) std::printf  ("Calculating cycle of length 3.\n");
for (int i=0;i<NObjects;i++)
{
    for (int j=i+1;j<NObjects;j++)
    {
        for (int k=j+1;k<NObjects;k++)
        {
            if ((d= g[i][j]+g[j][k]+g[k][i])<0)
            {
                results[count][0] = i;
                results[count][1] = j;
                results[count][2] = k;
                results[count][3] = d;
                count++;
                if (count>=MAX_OUTPUT_SIZE3)
                    goto finish3;
            }

            if ((d= g[i][k]+g[k][j]+g[j][i])<0)
            {
                results[count][0] = j;
                results[count][1] = i;
                results[count][2] = k;
                results[count][3] = d;
                count++;
                if (count>=MAX_OUTPUT_SIZE3)
                    goto finish3;
            }
        }

    }
}
finish3:

Also, you ask whether there is any algorithm faster than an O(n^3) algorithm, this means you have one already: care to put a link to that algorithm? — fge, Dec 29 '12 at 13:25
How do you represent your graph? Given 10K nodes, I assume it's an adjacency list, not a matrix, right? — Sergey Kalinichenko, Dec 29 '12 at 13:26
@remo: If the number of edges is in `Ω(n²)`, I don't think you can do much better then `n³` — Niklas B., Dec 29 '12 at 13:32
Yes, I wonder if there is a faster one. Please note I want at most one such cycles (if any) — remo, Dec 29 '12 at 13:32
let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/21863/discussion-between-remo-and-niklas-b) — remo, Dec 29 '12 at 13:34
You cannot do it faster than `O(N^3)` because your graph is fully connected (each vertex is connected to all other vertices). — Sergey Kalinichenko, Dec 29 '12 at 13:35
I've read the chat discussion, and it's still not clear whether there actually are `Ω(n²)` edges or not! Also in the chat the OP implies that the edges are directed, but his/her presented code only checks 1 direction for each edge (a directed triangle can be "oriented" 2 different ways). — j_random_hacker, Dec 29 '12 at 16:33

score 4 · Answer 1 · answered Dec 29 '12 at 16:55

I cannot think of any algorithm with definite comlexity lower than O(n³), but also the constant factor is important in practice. The following algorithm allows pruning to speed up finding a cycle of length 3 with negative sum of weights - or checking that there is no such cycle.

sort the (directed) edges according to their weight
take the edge with the lowest weight as starting edge.
try all edges connected to the end vertex of the starting edge with a weight not lower than the starting edge (1st pruning) and check the sum of weights when you close the cycle. If you find a cycle with negative sum you are done.
continue with the edge with the next lowest weight as the starting edge. If its weight is negative goto 3 - otherwise you are done (2nd pruning)

The idea is that at least one of the edges of a cylce with negative sum must have a negative weight. And that we can start a cycle at the edge with lowest weight in the cycle.

If you know that the number edges with negative weights is O(n) then this algorithm will be O(n² ld n) since the algorithm will then be dominated by step 1 (= sorting the edges according to their weight).

Thank you very much. This is a very good starting point for me. {How can I check the chat room again?} — remo, Dec 30 '12 at 11:06
@remo If you click on 'show more comments' below your question you will see the relevant link. If you are just looking for chat in general there is a little link on top of the page. — Dennis Jaheruddin, Dec 30 '12 at 15:43

How can I find a negative weighted cycle of 3 edges in a graph?

1 Answers1