Given a rooted, binary phylogenetic network and a rooted, binary phylogenetic tree, can the tree be embedded into the network? This problem, called Tree Containment, arises when validating networks constructed by phylogenetic inference methods. We present the first algorithm for (rooted) Tree Containment using the treewidth t of the input network N as parameter, showing that the problem can be solved in 2O(t2) |N| time and space.@en

We consider the problem of orienting a given, undirected graph into a (directed) acyclic graph such that the in-degree of each vertex ν is in a prescribed list λ(ν). Variants of this problem have been studied for a long time and with various applications, but mostly without the requirement for acyclicity. Without this requirement, the problem is closely related to the classical General Factor problem, which is known to be NP-hard in general, but polynomial-time solvable if no list λ(ν) contains large "gaps" [Cornuejols, J. Comb. Theory B, 1988]. In contrast, we show that deciding if an acyclic orientation exists is NP-hard even in the absence of such "gaps". On the positive side, we design parameterized algorithms for various, natural parameterizations of the acyclic orientation problem. A special case of the orientation problem with degree constraints recently came up in the context of reconstructing evolutionary histories (that is, phylogenetic networks). This phylogenetic setting imposes additional structure onto the problem that can be exploited algorithmically, allowing us to show fixed-parameter tractability when parameterized by either the treewidth of G (a smaller parameter than the frequently employed "level"), by the number of vertices ν for which |λ(ν)| ≥ 2, by the number of vertices ν for which the highest value in λ(ν) is at least 2. While the latter result can be extended to the general degree-constraint acyclic orientation problem, we show that the former cannot unless FPT=W[1].@en

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 formu-lated 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.@en