When Three Trees Go to War
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 et al. [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 − ε)n. Concretely, we prove tha...
Saved in:
| Main Authors: | , , |
|---|---|
| Format: | Article |
| Language: | English |
| Published: |
Peer Community In
2024-06-01
|
| Series: | Peer Community Journal |
| Online Access: | https://peercommunityjournal.org/articles/10.24072/pcjournal.419/ |
| Tags: |
Add Tag
No Tags, Be the first to tag this record!
|
| Summary: | 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 et al. [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 − ε)n. Concretely, we prove that, for each ε > 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 − ε)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 ε) 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. |
|---|---|
| ISSN: | 2804-3871 |