Minimum cost strongly connected digraph

问题

I have a digraph which is strongly connected (i.e. there is a path from i to j and j to i for each pair of nodes (i, j) in the graph G). I wish to find a strongly connected graph out of this graph such that the sum of all edges is the least.

To put it differently, I need to get rid of edges in such a way that after removing them, the graph will still be strongly connected and of least cost for the sum of edges.

I think it's an NP hard problem. I'm looking for an optimal solution, not approximation, for a small set of data like 20 nodes.

Edit

A more general description: Given a grap G(V,E) find a graph G'(V,E') such that if there exists a path from v1 to v2 in G than there also exists a path between v1 and v2 in G' and sum of each ei in E' is the least possible. so its similar to finding a minimum equivalent graph, only here we want to minimize the sum of edge weights rather than sum of edges.

Edit:

My approach so far: I thought of solving it using TSP with multiple visits, but it is not correct. My goal here is to cover each city but using a minimum cost path. So, it's more like the cover set problem, I guess, but I'm not exactly sure. I'm required to cover each and every city using paths whose total cost is minimum, so visiting already visited paths multiple times does not add to the cost.

回答1:

Your problem is known as minimum spanning strong sub(di)graph (MSSS) or, more generally, minimum cost spanning sub(di)graph and is NP-hard indeed. See also another book: page 501 and page 480.

I'd start with removing all edges that don't satisfy the triangle inequality - you can remove edge a -> c if going a -> b -> c is cheaper. This reminds me of TSP, but don't know if that leads anywhere.

My previous answer was to use the Chu-Liu/Edmonds algorithm which solves Arborescence problem; as Kazoom and ShreevatsaR pointed out, this doesn't help.

回答2:

I would try this in a dynamic programming kind of way.

0- put the graph into a list

1- make a list of new subgraphs of each graph in the previous list, where you remove one different edge for each of the new subgraphs

2- remove duplicates from the new list

3- remove all graphs from the new list that are not strongly connected

4- compare the best graph from the new list with the current best, if better, set new current best

5- if the new list is empty, the current best is the solution, otherwise, recurse/loop/goto 1

In Lisp, it could perhaps look like this:

(defun best-subgraph (digraphs &optional (current-best (best digraphs)))
  (let* ((new-list (remove-if-not #'strongly-connected
                                  (remove-duplicates (list-subgraphs-1 digraphs)
                                                     :test #'digraph-equal)))
         (this-best (best (cons current-best new-list))))
    (if (null new-list)
        this-best
        (best-subgraph new-list this-best))))

The definitions of strongly-connected, list-subgraphs-1, digraph-equal, best, and better are left as an exercise for the reader.

回答3:

This problem is equivalent to the problem described here: http://www.facebook.com/careers/puzzles.php?puzzle_id=1

回答4:

Few ideas that helped me to solve the famous facebull puzzle:

Preprocessing step:

Pruning: remove all edges a-b if there are cheaper or having the same cost path, for example: a-c-b.
Graph decomposition: you can solve subproblems if the graph has articulation points
Merge vertexes into one virtual vertex if there are only one outgoing edge.

Calculation step:

Get approximate solution using the directed TSP with repeated visits. Use Floyd Warshall and then solve Assignment problem O(N^3) using hungarian method. If we got once cycle - it's directed TSP solution, if not - use branch and bound TSP. After that we have upper bound value - the cycle of the minimum cost.
Exact solution - branch and bound approach. We remove the vertexes from the shortest cycle and try build strongly connected graph with less cost, than upper bound.

That's all folks. If you want to test your solutions - try it here: http://codercharts.com/puzzle/caribbean-salesman

回答5:

Sounds like you want to use the Dijkstra algorithm

回答6:

Seems like what you basically want is an optimal solution for traveling-salesman where it is permitted for nodes to be visited more than once.

Edit:

Hmm. Could you solve this by essentially iterating over each node i and then doing a minimum spanning tree of all the edges pointing to that node i, unioned with another minimum spanning tree of all the edges pointing away from that node?

回答7:

A 2-approximation to the minimal strongly connected subgraph is obtained by taking a union of a minimal in-branching and minimal out-branching, both rooted at the same (but arbitrary) vertex.

An out-branching, also known as arborescence, is a directed tree rooted at a single vertex spanning all vertexes. An in-branching is the same with reverse edges. These can be found by Edmonds' algorithm in time O(VE), and there are speedups to O(E log(V)) (see the wiki page). There is even an open source implementation.

The original reference for the 2-approximation result is the paper by JaJa and Frederickson, but the paper is not freely accessible.

There is even a 3/2 approximation by Adrian Vetta (PDF), but more complicated than the above.

来源：https://stackoverflow.com/questions/1536067/minimum-cost-strongly-connected-digraph

标签

algorithm

graph

traveling-salesman

np-hard