graph - How to find Minimum Directed Cycle (minimum total weight)?

Question

Here is an excise:

Let G be a weighted directed graph with n vertices and m edges, where all edges have positive weight. A directed cycle is a directe

Answer 1

Is my algorithm correct?

No. Let me give a counter example. Imagine you start DFS from u, there are two paths p1 and p2 from u to v and 1 path p3 from v back to u, p1 is shorter than p2.

Assume you start by taking the p2 path to v, and walk back to u by path p3. One cycle found but apparently it's not minimum. Then you continue exploring u by taking the p1 path, but since v is fully explored, the DFS ends without finding the minimum cycle.

Answer 2

"For each vertex, it might point back to one of its ancestors and thus forms a cycle"

I think it might point back to any of its ancestors which means N

Also, how are u going to mark vertexes when you came out of its dfs, you may come there again from other vertex and its going to be another cycle. So this is not (n+m) dfs anymore.

1. So ur algo is incomplete
2. same here

3. During one dfs, I think the vertex should be either unseen, or check, and for checked u can store the minimum weight for the path to the starting vertex. So if on some other stage u find an edge to that vertex u don't have to search for this path any more. This dfs will find the minimum directed cycle containing first vertex. and it's O(n^2) (O(n+m) if u store the graph as list)

So if to do it from any other vertex its gonna be O(n^3) (O(n*(n+m))

Sorry, for my english and I'm not good at terminology

Answer 3

You can use Floyd-Warshall algorithm here.

The Floyd-Warshall algorithm finds shortest path between all pairs of vertices.

The algorithm is then very simple, go over all pairs (u,v), and find the pair that minimized dist(u,v)+dist(v,u), since this pair indicates on a cycle from u to u with weight dist(u,v)+dist(v,u). If the graph also allows self-loops (an edge (u,u)) , you will also need to check them alone, because those cycles (and only them) were not checked by the algorithm.

pseudo code:

run Floyd Warshall on the graph
min <- infinity
vertex <- None
for each pair of vertices u,v
    if (dist(u,v) + dist(v,u) < min):
           min <- dist(u,v) + dist(v,u)
           pair <- (u,v)
return path(u,v) + path(v,u)

path(u,v) + path(v,u) is actually the path found from u to v and then from v to u, which is a cycle.

The algorithm run time is O(n^3), since floyd-warshall is the bottle neck, since the loop takes O(n^2) time.

I think correctness in here is trivial, but let me know if you disagree with me and I'll try to explain it better.

Answer 4

I did a similar kind of thing but i did not use any visited array for dfs (which was needed for my algorithm to work correctly) and hence i realised that my algorithm was of exponential complexity.

Since, you are finding all cycles it is not possible to find all cycles in less than exponential time since there can be 2^(e-v+1) cycles.