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I was recently asked if I could find an algorithm to compute the minimum cost spanning tree of a given graph, where the total cost of the spanning tree is given by the product of the edge costs rather than by their sum.

There are several algorihms to compute the regular minium spanning tree, but I am unsure of how to tweak them for the case mentioned above. Any ideas?

Thank you.

AlexTex
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Since log(product of edge costs) = sum (log(edge costs)), just log-transform the edge-weights, and find the minimum cost spanning tree for these weights.

Aniko
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    oooh good answer! ill just mention the constraints if it isnt obvious - the costs MUST all be positive, and SHOULD all be > 1. – jon_darkstar Nov 18 '10 at 19:56
  • @jon_darkstar You are right, they do have to be all positive, but I believe values 0-1 are OK. The resulting tree will not be the minimum cost subgraph (as adding extra edges could potentially decrease the cost), but it will still be the minimum cost _tree_. – Aniko Nov 18 '10 at 20:00
  • yes you're right, 'tree' constraint already bars such silliness. – jon_darkstar Nov 18 '10 at 20:02
  • @Aniko: Very interesting approach. – AlexTex Nov 18 '10 at 20:04
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    I believe you don't even need to make the log transformation. The standard MST algorithms do not use anything other than comparisons among the edge weights, and log is a monotonic function, so it does not change the ordering. – jonderry Nov 18 '10 at 20:25
  • @jonderry I also think you are correct - you can use the original weights. However to prove this one does need the log transformation :) – Aniko Nov 18 '10 at 21:42
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    @Aniko, (0-1] is an acceptable range for logs, but AlexTex must beware of zero-weighted edges, since log(0) is undefined. – Dijkstra Nov 19 '10 at 11:02
  • @Dijkstra Good point. Of couse, if there is a 0-weight edge, then the problem is trivial, because any spanning tree containing it will be product-minimal with product 0. – Aniko Nov 19 '10 at 14:06
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Since a logarithm is a monotone transformation, the minimum spanning tree will be exactly the same when you take the log of all weights and when you leave all weights the way they are. So: there is NO difference in finding the MST according to the sum of all weights and according to the product of all weights.

For the article of the proof of the fact that a minimum spanning tree of a graph is invariant towards monotone transformation of the weights in the graoh, type The Saga of Minimum Spanning Tree in google. And the first link will be the one you need. Page 167, monotone isomorphism.

Yulia
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My best idea - Find ALL minimal (meaning to unnecessary edges) spanning trees by brute force, pick the one with smallest product.

Most (or all) more efficient solutions no longer apply - mainly bc optimal solutions NO LONGER necessarily contain optimal sub problems. (What are the restrictions? Very important - edges of cost less than 1 are actually negative cost, edges of length 1 are free. they BETTER be all positive!)

I'm not sure if this question is really meaningful. For one, you must either give self-loop costs (or assume 1) because we can't take product with zero. Splitting a path works differently, traveling the same path twice costs c^2? Furthermore, I feel like this should be a different quality of a path with a different name than 'cost'

jon_darkstar
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  • For the sake of the discussion, we could assume that all costs are positive numbers greater than or equal to 1. Thanks for the idea. – AlexTex Nov 18 '10 at 19:59