Abstract

BackgroundThe supertree problem, i.e., the task of finding a common refinement of a set of rooted trees is an important topic in mathematical phylogenetics. The special case of a common leaf set L is known to be solvable in linear time. Existing approaches refine one input tree using information of the others and then test whether the results are isomorphic.ResultsAn O(k|L|) algorithm, LinCR, for constructing the common refinement T of k input trees with a common leaf set L is proposed that explicitly computes the parent function of T in a bottom-up approach.ConclusionLinCR is simpler to implement than other asymptotically optimal algorithms for the problem and outperforms the alternatives in empirical comparisons.AvailabilityAn implementation of LinCR in Python is freely available at https://github.com/david-schaller/tralda.

Highlights

  • Given a collection of rooted phylogenetic trees T1, T2, ...Tk, the supertree problem in phylogenetics consists in determining whether there is a common tree T that “displays” all input treesTi, 1 ≤ i ≤ k, and if so, a supertree T is to be constructed [1, 2]

  • Availability: An implementation of LinCR in Python is freely available at https://github.com/david-schaller/tralda

  • |L(Ti)|, and ki=ki1=L1(|TLi()T . iW)|2r,ittihnigs problem is solved by the algorithm of Aho et al [3], which is commonly called BUILD in the the phylogenetic literature [4], in O(Nn) time for binary trees and O(Rn) time in general

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Summary

Introduction

Given a collection of rooted phylogenetic trees T1 , T2 , ...Tk , the supertree problem in phylogenetics consists in determining whether there is a common tree T that “displays” all input treesTi , 1 ≤ i ≤ k , and if so, a supertree T is to be constructed [1, 2]. I.e., O(|L|) time, algorithms for the common refinement of two input trees T1 and T2 with a common leaf set have become available. The existence of a common refinement is verified by checking that the parent function defines a tree T and, if so, that T displays each of the input trees Tj .

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