问题
I'm stuck at the following problem: Given a weighted digraph G, I'd like to construct the minimal subgraph of G that contains all negative (simple) cycles of G.
I do know how to find a negative cycle using Bellman-Ford, and I know that the number of simple cycles in a directed graph is exponential.
One naive way to approach the problem would be to simply iterate all simple cycles and pick those that are negative, but I have the feeling that there might be a polynomial-time algorithm. Most articles that I found through Google were about finding a (rather than all) negative cycle.
I'm hoping to find some experts here on stackoverflow that may give some hints towards a polynomial-time solution, or hints towards proving that it can't be solved in polynomial time.
Many thanks in advance!
Cheers, Robert
回答1:
For anyone interested in or stuck at a similar problem: it's NP-complete. Thanks to wich for pointing me to the thread in cstheory.
To see why it's NP-complete, first of all observe that the problem may be stated as follows: given a weighted directed graph G with N verices and an edge E on G, find out whether E lies on a (simple) negative cycle. If it does, E should be in the subgraph H. If it does not, it should not be in H.
Now, let edge E be E = (u, v) with weight w. We'd like to know whether there's a path from v to u with total weight W such that W + w < 0. If we could do this in polynomial time, we could also solve the Hamiltonian Cycle problem in polynomial time:
Assign to edge E a weight of N - 1.00001. Assign to all other edges in the graph a weight of -1. Now the graph's only negative cycle on which E lies, is the cycle that contains all vertices (that cycle has weight -0.00001) and is thus a Hamiltonian Cycle.
Many thanks for thinking along!
来源：https://stackoverflow.com/questions/12364815/finding-the-minimal-subgraph-that-contains-all-negative-cycles