This is the same problem here but with one more condition that the sum of the distance cannot be a negative integer.

The full description of the problem is: Is it possible to find a simple path (no repeated vertices) between two vertices in a graph such that the sum of the weights of its constituent edges (the weight of an edge can be negative) is less than K, in which K is a positive number.

I have seen other reductions from the Hamilton Path problem, like

Those questions are different from the one I listed here. How to reduce the Hamilton path problem to this problem? Or is it possible to achieve this?

  • $\begingroup$ I'm confused. Please state the problem you want solved in a self-contained manner, in the question. Do you want to require that the sum of the weights be non-negative, or not? Such a condition doesn't appear in the second paragraph. $\endgroup$
    – D.W.
    Commented Apr 28 at 4:05
  • $\begingroup$ @D.W. Hi, yes I want to require that the sum of the weights be non-negative. In the 2nd paragraph, it is stated as the sum of the weights .... K is a positive number. $\endgroup$
    – Lebecca
    Commented Apr 30 at 16:18

1 Answer 1


There is an easy reduction from s-t-Hamilton path, to prove your problem is NP-hard:

Given a Graph (V, E) and two vertices s and t (the s-t-Hamilton path instance), construct an instance for your problem in the following way: Add a vertex s' and an edge (s', s), set the cost of this edge to n (where n = |V(G)|) and the cost of all other edges to -1 and K to 1.

Now there exists a simple s'-t path in your new instance with cost 1 iff there is a Hamilton s-t-path in the original instance.

As this is clearly a special case of a problem known to be in NP, your problem is therefore NP-complete.


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