2

I have a graph G = (V, E) where each edge is bidirectional with positive weight. I want to find the shortest path from vertex s to vertex t using at least k edges but less than 20. No vertex may be repeated in a path.

I am aware of this problem: Dijkstra's algorithm to compute shortest paths using k edges?

That problem wants to find the shortest paths using at most k edges.

My current idea is to create the product graph via V' = V x {0, 1, 2, ..., 20} and then ignore any shortest paths from (s, 0) to (t, n) where n < k. However, because the product construction essentially duplicates nodes, I am not aware of a way to ignore duplicate vertices in the path-finding algorithm.

P.S. If k were very large, say, k = |V|, wouldn't this be the Hamiltonian path problem and therefore NP-complete?

cs questions
  • 21
  • 3
  • 3
    For the PS: yes, it does include Hamiltonian Path Problem as a subproblem. – John Dvorak Apr 09 '19 at 17:10
  • @John Dvorak the linked problem gave a worst case complexity of (⋅(||+||)lg|) and it includes the Hamiltonian Path Problem if k = |V| and there are no paths for k less than this. Am I mistaken? – cs questions Apr 09 '19 at 17:46
  • 1
    @csquestions, if you are not allowed cycles, then this is NP-complete. See this question, specifically the first and last bullet point. Also see last paragraph of this answer to that question. – ryan Apr 09 '19 at 17:52
  • 1
    if you allow cycles, then you can either replicate the graph $k$ times, or apply some dynamic programming approach. Since you don't, it's NP-complete – lox Apr 09 '19 at 17:57

1 Answers1

2

Unfortunately, your problem is NP Complete, since you require that no vertex will be visited twice.

Consider we had a polynomial time solution for some $k$. For clarification; given an undirected graph $G=(V,E)$ and two vertices $s,t$, we can find a shortest path from $s$ to $t$ of length $\geq k$ in polynomial time. Then, by extension we can solve in polynomial time for $k=n-1$ and thus find a path that visits eeach vertex in $V$ exactly once. In other words, we can "solve" Hamiltonian Path.

If you know in advance that $k \leq 20$ then you can exhaustive search all paths of length at most 20, and therefore solve for small $k$.

An exhaustive search, at worst, is to search every vertex order starting from $s$ that ends in $t$ (corresponding to existing edges), so in total you will search: $$(n-1)(n-2)(n-3)...(n-20) = O(n^{20})$$

lox
  • 1,669
  • 1
  • 10
  • 16