A short note on exponential-time algorithms for hybridization number
Abstract
In this short note we prove that, given two (not necessarily binary) rooted phylogenetic trees T1, T2 on the same set of taxa X, where |X|=n, the hybridization number of T1 and T2 can be computed in time O*(2n) i.e. O(2n poly(n)). The result also means that a Maximum Acyclic Agreement Forest (MAAF) can be computed within the same time bound.
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