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I am applying the classic deletion contraction algorithm to a Graph G of "n" vertices and "m" edges.

Z(G) = Z(G-e) + Z(G/e)

In Wikipedia, http://en.wikipedia.org/wiki/Chromatic_polynomial#Deletion.E2.80.93contraction

They say that complexity is: O(1.6180^(n+m)). Mi main question is why they included the number of vertices in the complexity ?? when is clear that the recursion only depends on the number of edges.

The closest reference to deletion-contraction is fibonacci sequence, which its computing complexity is demonstrated in Herbert S. Wilf's Algorithms and Complexity book http://www.math.upenn.edu/~wilf/AlgComp3.html pages 18-19.

All help is welcome.

Brian Tompsett - 汤莱恩
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labotsirc
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1 Answers1

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Look at page 46 of the pdf version. Deletion and contraction each reduce the number of edges by 1, so a recurrence in edges only shows that Z(G) is O(2m), which is worse than O(Fib(n + m)) for all but the sparsest graphs. The improvement in considering vertices as well as edges is that, when a self-loop is formed, we know immediately that the chromatic polynomial is zero.

scowl
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  • Scowl, forgot to say thanks for the indication to the other page, the min( O(2^m), O(fib^(n+m)) ) should be what i am looking for. Because i am also applying optimizations to self loops and single bridges, among other ones too (n-serial and n-parallel edges). The only thing unclear is if its a constant that justifies the diference i pointed out on the triangle example for the 2^m cost, or an error on the asumptions i made. – labotsirc Jan 11 '12 at 03:01