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Pré-Publication, Document De Travail Année : 2023

When Three Trees Go to War

Résumé

How many reticulations are needed for a phylogenetic network to display a given set of k phylogenetic trees on n leaves? For k = 2, Baroni, Semple, and Steel [Ann. Comb. 8, 391-408 (2005)] showed that the answer is n − 2. Here, we show that, for k ≥ 3 the answer is at least (3 /2 − epsilon)n. Concretely, we prove that, for each epsilon > 0, there is some n ∈ N such that three n-leaf caterpillar trees can be constructed in such a way that any network displaying these caterpillars contains at least (3 /2 − epsilon)n reticulations. The case of three trees is interesting since it is the easiest case that cannot be equivalently formulated in terms of agreement forests. Instead, we base the result on a surprising lower bound for multilabelled trees (MUL-trees) displaying the caterpillars. Indeed, we show that one cannot do (more than an epsilon) better than the trivial MUL-tree resulting from a simple concatenation of the given caterpillars. The results are relevant for the development of methods for the Hybridization Number problem on more than two trees. This fundamental problem asks to construct a binary phylogenetic network with a minimum number of reticulations displaying a given set of phylogenetic trees.
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Licence : CC BY - Paternité

Dates et versions

hal-04013152 , version 1 (07-03-2023)
hal-04013152 , version 2 (17-08-2023)
hal-04013152 , version 3 (19-09-2023)

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Paternité

Identifiants

  • HAL Id : hal-04013152 , version 3

Recommandation PCI

Recommandé par :

Simone Linz. Bounding the reticulation number for three phylogenetic trees.
Peer Community in Mathematical and Computational Biology. 2023-10-12
https://doi.org/10.24072/pci.mcb.100187

Citer

Leo van Iersel, Mark Jones, Mathias Weller. When Three Trees Go to War. 2023. ⟨hal-04013152v3⟩

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