A hypotetical question. Say I'm given a tree T and a list of pair of nodes (x, y) in T. I'm asked how many of the pairs I can connect simotaneously (connect x with y) using each edge in T at most once.
How would one to that?
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It sounds like you're describing the travelling salesman problem? Where you try and visit as many nodes as possible without using any edge twice? This is NP-hard, and the focus of many-a-research. http://en.wikipedia.org/wiki/Travelling_salesman_problem – TZHX Nov 26 '10 at 20:52
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2@TZHX No, I think by pairs of nodes he means paths. So he wants to maximise the number of given paths that are edge disjoint. – moinudin Nov 26 '10 at 20:55
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@marcog, you're right. That's exactly what i want! – Alexander Nov 26 '10 at 20:58
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Oh damn, just read that the edge disjoint path problem (which it apparently is called) is NP-complete... :( – Alexander Nov 26 '10 at 21:20
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It isn't NP-hard for trees. You can do the following, for example.
Root the tree arbitrarily.
For each vertex v, compute the optimal solution that is confined to v's subtree.
Additionally, for each v, for each path p that includes v's parent edge, compute the optimal solution that is confined to v's subtree except for path p.
(2) and (3) can be computed using the solutions to the smaller problems within v's subtree.
It's probably easier to look at steps 1, 2, and 3, and figure out the recurrence yourself, but just to give you an idea, (2) can be computed by taking the maximum of several solutions: one that sums the solutions for the children of v (i.e., the sum of the solutions produced in step 2 for each child), and one for each path that includes v plus the step 2 solutions for the other children (this will essentially include the sum of one or two solutions produced in step 3 for v's children, plus the sum of the solutions produced in step 2 for the other children).
jonderry
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