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Home/ Questions/Q 6682795
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T04:44:50+00:00 2026-05-26T04:44:50+00:00

I’m trying to find out an efficient way of finding the shortest path between

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I’m trying to find out an efficient way of finding the shortest path between 2 nodes in a graph with positive edge costs that goes trough a subset of nodes.

More formally:

Given a graph G = (U, V) where U is the set of all nodes in the graph and V is the set of all edges in the graph, a subset of U called U’ and a cost function say:

    f : UxU -> R+  
    f(x, y) = cost to travel from node x to node y if there exists an edge 
              between node x and node y or 0 otherwise,

I have to find the shortest path between a source node and a target node that goes trough all the nodes in U’.

The order in which I visit the nodes in U’ doesn’t matter and I am allowed to visit a node more than once.

My original idea was to make use of Roy-Floyd algorithm to generate the cost matrix.
Then, for each permutation of the nodes in U’ I would be computing the cost between the source and the target like this: f(source_node, P1) + f(P1, P2) + … + f(Pk, target) saving the configuration for the lowest cost and then reconstructing the path.

The complexity for this approach is O(n3 + k!) O(n3 + k*k!), where n is the number of nodes in the graph and k the number of nodes in the subset U’, which is off limits since I’ll have to deal with graphs with maximum n = 2000 nodes out of which maximum n – 2 nodes will be part of the U’ subset.

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  1. Editorial Team
    Editorial Team
    2026-05-26T04:44:50+00:00Added an answer on May 26, 2026 at 4:44 am

    This is a generalization of travelling salesman. If U’ == U, you get exactly TSP.

    You can use the O(n^2 * 2^n) TSP algorithm, where the exponential factor for full scale TSP (2^n) will reduce to k = |U’|, so you’d get O(n^2 * 2^k).

    This has the DP solution to TSP.

    http://www.lsi.upc.edu/~mjserna/docencia/algofib/P07/dynprog.pdf

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