Minimum removed nodes required to cut path from A to B algorithm in Python

Question

I am trying to solve a problem related to graph theory but can't seem to remember/find/understand the proper/best approach so I figured I'd ask the experts...

I have a list of paths from two nodes (1 and 10 in example code). I'm trying to find the minimum number of nodes to remove to cut all paths. I'm also only able to remove certain nodes.

I currently have it implemented (below) as a brute force search. This works fine on my test set but is going to be an issue when scaling up to a graphs that have paths in the 100K and available nodes in the 100 (factorial issue). Right now, I'm not caring about the order I remove nodes in, but I will at some point want to take that into account (switch sets to list in code below).

I believe there should be a way to solve this using a max flow/min cut algorithm. Everything I'm reading though is going way over my head in some way. It's been several (SEVERAL) years since doing this type of stuff and I can't seem to remember anything.

So my questions are:

1) Is there a better way to solve this problem other than testing all combinations and taking the smallest set?

2) If so, can you either explain it or, preferably, give pseudo code to help explain? I'm guessing there is probably a library that already does this in some way (I have been looking and using networkX lately but am open to others)

3) If not (or even of so), suggestions for how to multithread/process solution? I want to try to get every bit of performance I can from computer. (I have found a few good threads on this question I just haven't had a chance to implement so figured I'd ask at same time just in chance. I first want to get everything working properly before optimizing.)

4) General suggestions on making code more "Pythonic" (probably will help with performance too). I know there are improvements I can make and am still new to Python.

Thanks for the help.

#!/usr/bin/env python


def bruteForcePaths(paths, availableNodes, setsTested, testCombination, results, loopId):

    #for each node available, we are going to
    # check if we have already tested set with node
    # if true- move to next node
    # if false- remove the paths effected,
    #           if there are paths left,
    #               record combo, continue removing with current combo,
    #           if there are no paths left,
    #               record success, record combo, continue to next node

    #local copy
    currentPaths = list(paths)
    currentAvailableNodes = list(availableNodes)
    currentSetsTested = set(setsTested)
    currentTestCombination= set(testCombination)

    currentLoopId = loopId+1

    print "loop ID: %d" %(currentLoopId)
    print "currentAvailableNodes:"
    for set1 in currentAvailableNodes:
        print "  %s" %(set1)

    for node in currentAvailableNodes:
        #add to the current test set
        print "%d-current node: %s current combo: %s" % (currentLoopId, node, currentTestCombination)
        currentTestCombination.add(node)
        # print "Testing: %s" % currentTestCombination
        # print "Sets tested:"
        # for set1 in currentSetsTested:
        #     print "  %s" % set1

        if currentTestCombination in currentSetsTested:
            #we already tested this combination of nodes so go to next node
            print "Already test: %s" % currentTestCombination
            currentTestCombination.remove(node)
            continue

        #get all the paths that don't have node in it
        currentRemainingPaths = [path for path in currentPaths if not (node in path)]

        #if there are no paths left
        if len(currentRemainingPaths) == 0:
            #save this combination
            print "successful combination: %s" % currentTestCombination
            results.append(frozenset(currentTestCombination))
            #add to remember we tested combo
            currentSetsTested.add(frozenset(currentTestCombination))
            #now remove the node that was add, and go to the next one
            currentTestCombination.remove(node)
        else:
            #this combo didn't work, save it so we don't test it again
            currentSetsTested.add(frozenset(currentTestCombination))
            newAvailableNodes = list(currentAvailableNodes)
            newAvailableNodes.remove(node)
            bruteForcePaths(currentRemainingPaths,
                            newAvailableNodes,
                            currentSetsTested,
                            currentTestCombination,
                            results,
                            currentLoopId)

            currentTestCombination.remove(node)

    print "-------------------"
    #need to pass "up" the tested sets from this loop
    setsTested.update(currentSetsTested)

    return None

if __name__ == '__main__':

    testPaths = [
        [1,2,14,15,16,18,9,10],
        [1,2,24,25,26,28,9,10],
        [1,2,34,35,36,38,9,10],
        [1,3,44,45,46,48,9,10],
        [1,3,54,55,56,58,9,10],
        [1,3,64,65,66,68,9,10],

        [1,2,14,15,16,7,10],
        [1,2,24,7,10],
        [1,3,34,35,7,10],

        [1,3,44,35,6,10],
        ]


    setsTested = set()
    availableNodes = [2, 3, 6, 7, 9]
    results = list()
    currentTestCombination = set()

    bruteForcePaths(testPaths, availableNodes, setsTested, currentTestCombination, results, 0)

    print "results:"
    for result in sorted(results, key=len):
        print result

UPDATE: I reworked the code using itertool for generating the combinations. It make the code cleaner and faster (and should be easier to multiprocess. Now to try to figure out the dominate nodes as suggested and multiprocess function.

def bruteForcePaths3(paths, availableNodes, results):

    #start by taking each combination 2 at a time, then 3, etc
    for i in range(1,len(availableNodes)+1):
        print "combo number: %d" % i

        currentCombos = combinations(availableNodes, i)

        for combo in currentCombos:
            #get a fresh copy of paths for this combiniation
            currentPaths = list(paths)
            currentRemainingPaths = []
            # print combo

            for node in combo:
                #determine better way to remove nodes, for now- if it's in, we remove
                currentRemainingPaths = [path for path in currentPaths if not (node in path)]
                currentPaths = currentRemainingPaths

            #if there are no paths left
            if len(currentRemainingPaths) == 0:
                #save this combination
                print combo
                results.append(frozenset(combo))

    return None

Answer 1

Here is an answer which ignores the list of paths. It just takes a network, a source node, and a target node, and finds the minimum set of nodes within the network, not either source or target, so that removing these nodes disconnects the source from the target.

If I wanted to find the minimum set of edges, I could find out how just by searching for Max-Flow min-cut. Note that the Wikipedia article at http://en.wikipedia.org/wiki/Max-flow_min-cut_theorem#Generalized_max-flow_min-cut_theorem states that there is a generalized max-flow min-cut theorem which considers vertex capacity as well as edge capacity, which is at least encouraging. Note also that edge capacities are given as Cuv, where Cuv is the maximum capacity from u to v. In the diagram they seem to be drawn as u/v. So the edge capacity in the forward direction can be different from the edge capacity in the backward direction.

To disguise a minimum vertex cut problem as a minimum edge cut problem I propose to make use of this asymmetry. First of all give all the existing edges a huge capacity - for example 100 times the number of nodes in the graph. Now replace every vertex X with two vertices Xi and Xo, which I will call the incoming and outgoing vertices. For every edge between X and Y create an edge between Xo and Yi with the existing capacity going forwards but 0 capacity going backwards - these are one-way edges. Now create an edge between Xi and Xo for each X with capacity 1 going forwards and capacity 0 going backwards.

Now run max-flow min-cut on the resulting graph. Because all the original links have huge capacity, the min cut must all be made up of the capacity 1 links (actually the min cut is defined as a division of the set of nodes into two: what you really want is the set of pairs of nodes Xi, Xo with Xi in one half and Xo in the other half, but you can easily get one from the other). If you break these links you disconnect the graph into two parts, as with standard max-flow min-cut, so deleting these nodes will disconnect the source from the target. Because you have the minimum cut, this is the smallest such set of nodes.

If you can find code for max-flow min-cut, such as those pointed to by http://www.cs.sunysb.edu/~algorith/files/network-flow.shtml I would expect that it will give you the min-cut. If not, for instance if you do it by solving a linear programming problem because you happen to have a linear programming solver handy, notice for example from http://www.cse.yorku.ca/~aaw/Wang/MaxFlowMinCutAlg.html that one half of the min cut is the set of nodes reachable from the source when the graph has been modifies to subtract out the edge capacities actually used by the solution - so given just the edge capacities used at max flow you can find it pretty easily.

Answer 2

If the paths were not provided as part of the problem I would agree that there should be some way to do this via http://en.wikipedia.org/wiki/Max-flow_min-cut_theorem, given a sufficiently ingenious network construction. However, because you haven't given any indication as to what is a reasonable path and what is not I am left to worry that a sufficiently malicious opponent might be able to find strange collections of paths which don't arise from any possible network.

In the worst case, this might make your problem as difficult as http://en.wikipedia.org/wiki/Set_cover_problem, in the sense that somebody, given a problem in Set Cover, might be able to find a set of paths and nodes that produces a path-cut problem whose solution can be turned into a solution of the original Set Cover problem.

If so - and I haven't even attempted to prove it - your problem is NP-Complete, but since you have only 100 nodes it is possible that some of the many papers you can find on Set Cover will point at an approach that will work in practice, or can provide a good enough approximation for you. Apart from the Wikipedia article, http://www.cs.sunysb.edu/~algorith/files/set-cover.shtml points you at two implementations, and a quick search finds the following summary at the start of a paper in http://www.ise.ufl.edu/glan/files/2011/12/EJORpaper.pdf:

The SCP is an NP-hard problem in the strong sense (Garey and Johnson, 1979) and many algorithms have been developed for solving the SCP. The exact algorithms (Fisher and Kedia, 1990; Beasley and JØrnsten, 1992; Balas and Carrera, 1996) are mostly based on branch-and-bound and branch-and-cut. Caprara et al. (2000) compared different exact algorithms for the SCP. They show that the best exact algorithm for the SCP is CPLEX. Since exact methods require substantial computational effort to solve large-scale SCP instances, heuristic algorithms are often used to find a good or near-optimal solution in a reasonable time. Greedy algorithms may be the most natural heuristic approach for quickly solving large combinatorial problems. As for the SCP, the simplest such approach is the greedy algorithm of Chvatal (1979). Although simple, fast and easy to code, greedy algorithms could rarely generate solutions of good quality....

Answer 3

Edit: If you want to destroy in fact all paths, and not those from a given list, then max-flow techniques as explained by mcdowella is much better than this approach.

As mentioned by mcdowella, the problem is NP-hard in general. However, the way your example looks, an exact approach might be feasible.

First, you can delete all vertices from the paths that are not available for deletion. Then, reduce the instance by eliminating dominated vertices. For example, every path that contains 15 also contains 2, so it never makes sense to delete 15. In the example if all vertices were available, 2, 3, 9, and 35 dominate all other vertices, so you'd have the problem down to 4 vertices.

Then take a vertex from the shortest path and branch recursively into two cases: delete it (remove all paths containing it) or don't delete it (delete it from all paths). (If the path has length one, omit the second case.) You can then check for dominance again.

This is exponential in the worst case, but might be sufficient for your examples.