A short note on exponential-time algorithms for hybridization number

TL;DR

This paper proves that the hybridization number for two rooted phylogenetic trees can be computed in exponential time, specifically O^{*}(2^n), enabling the calculation of a Maximum Acyclic Agreement Forest within the same bound.

Contribution

It establishes an exponential-time algorithm for computing the hybridization number and MAAF for two rooted phylogenetic trees, improving understanding of computational complexity in this area.

Findings

Hybridization number can be computed in O^{*}(2^n) time.

Maximum Acyclic Agreement Forest can be computed within the same time bound.

Provides complexity bounds for phylogenetic tree comparison.

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.