Question

I believe that the Hamiltonian cycle problem can be summed up as the following:

Given an undirected graph G = (V, E), a
Hamiltonian circuit is a tour in G passing through
every vertex of G once and only once.

Now, what I would like to do is reduce my problem to this. My problem is:

Given a weighted undirected graph G, integer k, and vertices u, v
both in G, is there a simple path in G from u to v
with total weight of AT LEAST k?

So knowing that the Hamiltonian cycle problem is NP-complete, by reducing this problem to the Hamiltonian, this problem is also proved NP-complete. My issue is the function reducing it to Hamiltonian.

1. The big issue is that the Hamiltonian problem does not deal with edge weights, so I must convert my graph to one that doesn’t have any weights.
2. On top of that, this problem has a designated start and finish (u and v), whereas the Hamiltonian finds a cycle, so any start is the same as the finish.

For (1), I am thinking along the lines of passing a graph with all simple paths of total weight LESS THAN k taken out.
For (2), I am thinking that this is not really an issue, because if there is a Hamiltonian cycle, then the simple path from u to v can be sliced out of it.

So, my real questions are:

1. Is my solution going to give me the right answer?
2. If yes, then how can I take out the edges that will produce simple paths of total weight less than k WITHOUT affecting the possibility that one of those edges may be required for the actual solution? Because if an edge e is taken out because it produces a simple path of weight < k for a subset of E, it can still be used in a simple path with a different combination of edges to produce a path of weight >= k.

Thanks!

Answer 1

More of a hint than an answer:

A unweighted graph can be interpreted as a weighted graph where each edge has weight 1. What would the cost of a Hamiltonian cycle be in a graph like that?