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POReduction algorithm from the Hamiltonian cycle
 

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    <p>I believe that the Hamiltonian cycle problem can be summed up as the following:</p> <blockquote> <p>Given an undirected graph <code>G = (V, E)</code>, a Hamiltonian circuit is a tour in <code>G</code> passing through every vertex of <code>G</code> once and only once.</p> </blockquote> <p>Now, what I would like to do is reduce my problem to this. My problem is:</p> <blockquote> <p>Given a weighted undirected graph <code>G</code>, integer <code>k</code>, and vertices <code>u, v</code> both in <code>G</code>, is there a simple path in <code>G</code> from <code>u</code> to <code>v</code> with total weight of AT LEAST <code>k</code>?</p> </blockquote> <p>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.</p> <ol> <li>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.</li> <li>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.</li> </ol> <p>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.</p> <p>So, my real questions are:</p> <ol> <li>Is my solution going to give me the right answer?</li> <li>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 &lt; 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.</li> </ol> <p>Thanks!</p>
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