Abstract
In this short note we prove that, given two (not necessarily binary) rooted phylogenetic trees T_1, T_2 on the same set of taxa X, where |X|=n, the hybridization number of T_1 and T_2 can be computed in time O^{*}(2^n) i.e. O(2^{n} 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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