Binets: Fundamental Building Blocks for Phylogenetic Networks
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Abstract
Phylogenetic networks are a generalization of evolutionary trees that are used by biologists to represent the evolution of organisms which have undergone reticulate evolution. Essentially, a phylogenetic network is a directed acyclic graph having a unique root in which the leaves are labelled by a given set of species. Recently, some approaches have been developed to construct phylogenetic networks from collections of networks on 2 and 3leaved networks, which are known as binets and trinets, respectively. Here we study in more depth properties of collections of binets, one of the simplest possible types of networks into which a phylogenetic network can be decomposed. More specifically, we show that if a collection of level1 binets is compatible with some binary network, then it is also compatible with a binary level1 network. Our proofs are based on useful structural results concerning lowest stable ancestors in networks. In addition, we show that, although the binets do not determine the topology of the network, they do determine the number of reticulations in the network, which is one of its most important parameters. We also consider algorithmic questions concerning binets. We show that deciding whether an arbitrary set of binets is compatible with some network is at least as hard as the wellknown graph isomorphism problem. However, if we restrict to level1 binets, it is possible to decide in polynomial time whether there exists a binary network that displays all the binets. We also show that to find a network that displays a maximum number of the binets is NPhard, but that there exists a simple polynomialtime 1/3approximation algorithm for this problem. It is hoped that these results will eventually assist in the development of new methods for constructing phylogenetic networks from collections of smaller networks.
Keywords
Reticulate evolution Phylogenetic network Subnetwork Binet Algorithm1 Introduction
Phylogenetic networks are a generalization of evolutionary trees which biologists use to represent the evolution of species that have undergone reticulate evolution. Such networks are essentially directed acyclic graphs having a unique root in which the leaves are labelled by a set X of species (Huson et al. 2010). In contrast to evolutionary trees, which can only represent speciation events, phylogenetic networks permit the representation of evolutionary events such as gene transfer and hybridization which are known to occur in organisms such as bacteria and plants, respectively. Although theoretical properties of evolutionary trees have been studied since at least the 1970s, phylogenetic networks have been considered from this perspective only more recently, especially the rooted variants which we will focus on in this paper.
One of the most important open questions concerning phylogenetic networks is how to construct them for biological datasets (Bapteste et al. 2013). It is now common practice for biologists to construct evolutionary trees from molecular data, and several computer programs are available for this purpose (Felsenstein 2004). However, the problem of constructing networks from such data is an active area of research, and there are only a limited number of programs available for biologists to perform this task. A survey of some of these methods and the theory underpinning phylogenetic networks may be found in Gusfield (2014), Huson et al. (2010), Morrison (2011).
One approach that has been recently developed for constructing phylogenetic networks involves building them up from smaller networks, using what can be thought of as a divideandconquer approach (Oldman et al. 2016). In particular, for a set X of species, a network is constructed for every subset of X size 3 (called a trinet), and then the trinets are puzzled together to build a network (see Fig. 1 for an example of a trinet). This approach constructs and is based on level1 networks, networks that are slightly more general than evolutionary trees (see Sect. 2 for the definition of such networks).
In light of these considerations some obvious questions immediately arise concerning binets. For example, when is a collection of binets displayed by some phylogenetic network (the compatibility problem), and how much information might we expect to extract concerning a phylogenetic network by just looking at the collection of binets that it displays? In this paper, we shall address these and related algorithmic questions concerning binets. It is hoped that these results will be useful in future for developing improved methods for constructing phylogenetic networks from smaller networks.
We now present a summary of the rest of the paper. After introducing some preliminaries concerning phylogenetic networks in the next section, we derive a key structural result for networks (Corollary 1) which is useful in identifying which of the two possible types of binet is displayed on two leaves within a binary phylogenetic network (that is a network in which all internal vertices have degree 3). Using this theorem, in Sect. 4 we show that the collection of level1 binets displayed by any binary phylogenetic network can always be displayed by some binary level1 network (Theorem 3). This reduces the problem of understanding binets displayed by arbitrary binary networks to level1 networks. To prove this result, we develop a framework which also implies that there is a polynomialtime algorithm in X for deciding whether or not a collection of level1 binets with combined leaf set X can be displayed by some network with leaf set X, and, if it is, gives a level1 network that does this (see Sect. 6). Note that this is related to an algorithm presented in Huber et al. (2015).
In Sect. 5, we turn to the question as to what can be deduced about the features of a phylogenetic network just by considering the collection of binets that it displays. Note that, as might be expected, there are networks—even trinets—that display the same set of binets but that are not equivalent. For example, the two trinets in Fig. 1 both display the same set of binets, but they are not equivalent. Even so, we will show in Theorem 4 that if two level1 networks both display exactly the same collection of binets, then they must have the same number of reticulation vertices (indegree2 vertices). Note that the number of such vertices corresponds to the number of reticulate evolutionary events, such as hybridization, that took place in the evolutionary history of the species labelling the leaves of the network. Consequently, the binets displayed by a network can at least capture a useful coursegrained feature of the network in question.
In Sects. 6 and 7, we consider some algorithmic questions concerning binets. As we have mentioned above, it can be decided in polynomial time in X as to when a collection of binets with combined leaf set X is displayed by some level1 network on X. However, we show that if we consider arbitrary binets (i.e. not necessarily binary or level1) then this decision problem becomes at least as hard as the graph isomorphism problem (see Theorem 5), one of the most famous problems whose complexity is still unknown. In addition, in Sect. 7 we consider a related problem which, for a given collection of binary level1 binets, asks for a network which displays the maximum number of binets in this collection. This is closely related to the maximum rooted triplet consistency problem for evolutionary trees (Byrka et al. 2010). We show that the binet problem is NPcomplete (Theorem 6), by giving a reduction from the feedback arc set problem. However, we also show that the problem is 1/3approximable. In fact, given any collection of binary level1 binets we can always find some network that displays at least 1/3 of the binets (see Theorem 7). We conclude in Sect. 8 with discussion of some possible future research directions, and a brief discussion of a potential application of our results.
2 Preliminaries
Throughout this paper, X is a nonempty finite set (which usually represents a set of species or organisms).
2.1 Digraphs
A directed graph, or digraph for short, \(G=(V,E)\) consists of a finite set \(V=V(G)\) of vertices and a set \(E=E(G)\) of arcs, where each arc is an ordered pair (u, v) of vertices in V in which u is said to be a parent of v, denoted by \(u=p(v)\), and v a child of u. All digraphs studied here contain no loops, that is, vertices that are children of themselves. The indegree of vertex u is the number of vertices v in V such that (v, u) is an arc, and the outdegree of u is the number of vertices w with (u, w) being an arc. A root is a vertex with indegree 0. A leaf is a vertex of outdegree 0, and the set of leaves is denoted by L(G). Any vertex in G that is neither a root nor a leaf is referred to as an interior vertex. In addition, an interior vertex is a tree vertex if it has indegree 1, and a reticulation vertex if it has indegree greater than 1.
A directed path or dipath in a digraph is a sequence \(u_0,u_1,\ldots ,u_k\) (\(k\ge 1\)) of vertices such that \((u_{i1},u_i)\) is an arc for \(1\le i \le k\). An acyclic digraph is a digraph that does not contain any directed path starting and ending at the same vertex. If an acyclic digraph G contains a unique root, which is usually designated by \(\rho =\rho (G)\), then it will be referred to as a rooted acyclic digraph.
An acyclic digraph G induces a canonical partial order \(\prec _{G}\) on its vertex set V, that is, \(v\prec _{G} u\) if there exists a directed path from u to v. In this case, we shall say that v is below u. When the digraph G is clear from the context, \(\prec _{G}\) will be written as \(\prec \). In addition, we write \(v \preceq u\) if \(u=v\) or \(u\prec v\). Given a subset U of the vertex set of an acyclic digraph, we say that \(u\in U\) is a lowest vertex in U if there is no \(v\in U\) with \(v\prec u\).
Let \(\underline{G}\) be the undirected graph obtained from digraph G by ignoring the direction of the arcs in G. Then G is connected if \(\underline{G}\) is connected, that is, there exists an undirected path between every pair of distinct vertices in \(\underline{G}\). Note that a rooted acyclic digraph is necessarily connected (since each connected component of an acyclic digraph has at least one root). A cut vertex is a vertex of G whose removal disconnects \(\underline{G}\). Similarly, a cut arc is an arc of G whose removal disconnects \(\underline{G}\). A directed graph is biconnected if it contains no cut vertex, and a biconnected component of G is a maximal biconnected subgraph, which is called trivial if it contains precisely one arc (which is necessarily a cut arc), and nontrivial otherwise.
2.2 Phylogenetic Networks
A phylogenetic network \({N}\) on X is a rooted acyclic digraph whose leaves are bijectively labelled by the elements in X and which does not contain any vertex with indegree one and outdegree one. For simplicity, we will just write \({L(N)=X}\) in case there is no confusion about the labelling. To simplify the argument, throughout this paper we will also assume that all leaves in a phylogenetic network have indegree one. In addition, a phylogenetic network is binary if each tree vertex, as well as the root, has outdegree 2, and each reticulation vertex has indegree 2 and outdegree 1. Finally, we say a binary phylogenetic network is levelk (\({k\ge 0}\)) if each of its biconnected components contains at most k reticulation vertices. To some extent, the concept of the level of a phylogenetic network can be regarded as a measure of its ‘distance’ to being a phylogenetic tree. In particular, a binary phylogenetic network is a phylogenetic tree if and only if it is level0. A phylogenetic network is called simple if it contains precisely one nontrivial biconnected component H and no cut arcs other than the ones leaving H.
Two networks \({{N}_1=\left( V_1,E_1\right) }\) and \({{N}_2=\left( V_2,E_2\right) }\) on X are said to be isomorphic if there exists a bijection \(f: V_1{\rightarrow } V_2\) such that \(f(x)=x\) for all \(x\in X\), and (u, v) is an arc in \({N}_1\) if and only if \(\left( f(u),f(v)\right) \) is an arc in \({N}_2\).
Finally, the cluster of a vertex u, denoted by \({\mathcal {C}}_N(u)={\mathcal {C}}(u)\), is defined as the subset of X consisting of the leaves below u. Here we will use the convention that \({\mathcal {C}}(u)=\{u\}\) if u is a leaf.
2.3 Stable Ancestors and Binets
Given a phylogenetic network \({N}\) on X and a subset \(U\subseteq V(N)\), a stable ancestor of U in N is a vertex v in \(V(N){\setminus } U\) such that every path in N from the root to a vertex in U contains v. Note that for two stable ancestors u and \(u'\) of U, we have either \(u\preceq v\) or \(v\preceq u\). Therefore, there exists a unique lowest vertex in the set of stable ancestors of U, which will be referred to as the lowest stable ancestor of U in N and denoted by \({{\textsc {lsa}}_N(U)={\textsc {lsa}}(U)}\). Note that for a subset Y of X with \(Y\ge 2\), there exist two elements x and y in Y such that \({\textsc {lsa}}(Y)={\textsc {lsa}}\left( \{x,y\}\right) \). For simplicity, we also write \({\textsc {lsa}}\left( \{x,y\}\right) \) as \({\textsc {lsa}}(x,y)\).
The following property of lowest stable ancestors will be useful.
Lemma 1
Suppose that u and v are two vertices in a phylogenetic network such that \(u\prec v\prec {\textsc {lsa}}(u)\), then we have \({\textsc {lsa}}(v)\preceq {\textsc {lsa}}(u)\).
Proof
Since \(u\prec v\), we know that there exists a dipath P from \(\rho \) to u that contains v. By the definition of lowest stable ancestor, we know that \({\textsc {lsa}}(u)\) and \({\textsc {lsa}}(v)\) are contained in P. Hence, either \({\textsc {lsa}}(v)\preceq {\textsc {lsa}}(u)\) or \({\textsc {lsa}}(u)\prec {\textsc {lsa}}(v)\). If \({\textsc {lsa}}(u)\prec {\textsc {lsa}}(v)\), then we have \(v \prec {\textsc {lsa}}(u)\prec {\textsc {lsa}}(v)\). Then there exists a dipath \(P'\) from \(\rho \) to v that does not contain \({\textsc {lsa}}(u)\) (otherwise \({\textsc {lsa}}(u)\) would be a stable ancestor of v that is below \({\textsc {lsa}}(v)\)). Using that \(u\prec v\prec {\textsc {lsa}}(u)\), it follows that there exists a dipath from \(\rho \) to u that does not contain \({\textsc {lsa}}(u)\), a contradiction. Therefore, \({\textsc {lsa}}(v)\preceq {\textsc {lsa}}(u)\). \(\square \)
For \(Y\subseteq X\), the subnet of \({N}\) on Y, denoted by \({N}_Y\), is defined as the subgraph obtained from \({N}\) by deleting all vertices that are not on any path from \({\textsc {lsa}}(Y)\) to elements in Y and subsequently suppressing all indegree 1 and outdegree 1 vertices and parallel arcs until no such vertices or arcs exist. A network \(N'\) is said to be displayed by network N if \({N'=N_Y}\) for some \(Y\subseteq X\).
Note that, by definition, \({N}_X={N}\) if and only if \({\textsc {lsa}}(X)=\rho ({N})\). In this case, \({N}\) is referred to as a recoverable network. Note that every subnet of \({N}\) is necessarily recoverable. Moreover, a collection of subnets is displayed by some network if and only if it is displayed by some recoverable network. Therefore, we assume all networks in this paper to be recoverable.
3 A Structure Theorem
In this section we present a key result (Corollary 1) concerning the structure of the nontrivial biconnected component of a simple network. Note that a similar result has been obtained for a special collection of (nonbinary) phylogenetic networks in Huber et al. (2016).
Theorem 1
Let N be a binary phylogenetic network on X whose root \(\rho \) is in some nontrivial biconnected component H. Then there exists a lowest vertex in H with \({\textsc {lsa}}(v)=\rho \).
Proof
Let \(\varGamma _0(H)\) be the set of reticulation vertices v in H for which the distance (length of a shortest directed path) between \(\rho \) and \({\textsc {lsa}}(v)\) is minimum over all reticulation vertices in H. Note that \(\varGamma _0(H)\ne \emptyset \).
We first show that \({\textsc {lsa}}(v)=\rho \) for all \(v\in \varGamma _0(H)\). Suppose this were not the case. Then there exists a vertex \(v\in \varGamma _0(H)\) such that \({\textsc {lsa}}(v)\prec \rho \). Note that \({\textsc {lsa}}(v)\) necessarily has outdegree 2 and therefore has indegree 1 since N is binary and \({\textsc {lsa}}(v)\ne \rho \). Denote the parent of \({\textsc {lsa}}(v)\) by \(v^*\). Since H is biconnected, there exists some undirected path from \(\rho \) to \({\textsc {lsa}}(v)\) that does not contain the edge \({e=\left\{ {\textsc {lsa}}(v),v^*\right\} }\). Let \(P_u=v_0,\ldots ,v_t\), where \({v_0=\rho }\) and \({v_t={\textsc {lsa}}(v)}\), be such an undirected path for which \({\textsc {alt}}(P_u)\) is minimum.
We claim that \({\textsc {alt}}(P_u)=1\). To see this, note first that since \({v_0=\rho }\), \({v_t={\textsc {lsa}}(v)}\) and \({v_{t1}\ne v^*}\), we know that \(\left( v_0,v_1\right) \) and \(\left( v_{t},v_{t1}\right) \) are arcs of N. Hence, \({\textsc {alt}}\left( P_u\right) \) is odd and strictly positive. Assume for the sake of contradiction that \({\textsc {alt}}\left( P_u\right) \not =1\), then we have \({\textsc {alt}}\left( P_u\right) \ge 3\). Let \(v_k\) (\({1<k<t}\)) be the second alternating vertex contained in \(P_u\) (when travelling from \(v_0\) to \(v_t\)).
Now fix a directed path \(P_d\) in N from \(\rho \) to \(v_k\).
If the arc \(\left( v^*,{\textsc {lsa}}(v)\right) \) is not contained in \(P_d\), then, we can find an undirected path from \(\rho \) to \({\textsc {lsa}}(v)\) that does not contain e and has fewer alternating vertices than \(P_u\) by following \(P_d\) until we reach a vertex in \(\{v_k,\ldots ,v_t\}\) and then following \(P_u\) to \({\textsc {lsa}}(v)\). This gives a contradiction.
Now assume that the arc \(\left( v^*,{\textsc {lsa}}(v)\right) \) is contained in \(P_d\). Then we can find an undirected path from \(\rho \) to \({\textsc {lsa}}(v)\) that does not contain e and has only one alternating vertex as follows. Follow \(P_u\) up to \(v_k\) and then follow \(P_d\) backward from \(v_k\) to \({\textsc {lsa}}(v)\). Since this path has fewer alternating vertices than \(P_u\), we again obtain a contradiction.
We have thus shown that \({\textsc {alt}}\left( P_u\right) =1\). Denoting this alternating vertex in \(P_u\) by r, then r is necessarily a reticulation by the choice of \(P_u\). Hence, \(P_u\) consists of two directed paths: a directed path from \(\rho \) to r that does not contain \({\textsc {lsa}}(v)\) and a directed path from \({\textsc {lsa}}(v)\) to r. However, this means that \({\textsc {lsa}}(v)\prec {\textsc {lsa}}(r)\), a contradiction to the assumption that \(v\in \varGamma _0(H)\).
Hence, we know that \(\varGamma _0(H)\) is the set of reticulation vertices v of H such that \({\textsc {lsa}}(v)=\rho \) and that \(\varGamma _0(H)\) is not empty.
Now fix a vertex v in \(\varGamma _0(H)\) that is lowest over all vertices of \(\varGamma _0(H)\), that is, there does not exist a vertex u in \(\varGamma _0(H)\) such that \(v\prec u\). It remains to show that v is lowest over all vertices of H. Assume that this is not the case. Then the child c of v is also in H. If c were a reticulation then, by Lemma 1, \({\textsc {lsa}}(v)\preceq {\textsc {lsa}}(c)\). However, this would imply that \({{\textsc {lsa}}(c)=\rho }\), contradicting the choice of v. Hence, c is a tree vertex.
Since H is biconnected, there exists some undirected path from \(\rho \) to c that does not contain v. Let \(P_u=w_0,\ldots ,w_t\) be such a path such that \({\textsc {alt}}\left( P_u\right) \) is minimum. Note that we have \({w_0=\rho }\) and \({w_t=c}\).
Since c is a tree vertex and \(P_u\) does not contain its parent v, \(\left( w_{t},w_{t1}\right) \) is an arc of N. Together with \(\left( w_0,w_1\right) \) being an arc in N, we know that \({\textsc {alt}}\left( P_u\right) \) is odd and strictly positive. We now show, using a similar proof as above, that \({\textsc {alt}}\left( P_u\right) =1\). If this were not the case, then we would have \({\textsc {alt}}\left( P_u\right) \ge 3\). Let \(w_k\) (\({1<k<t}\)) be the second alternating vertex contained in \(P_u\). We know that \(\left( w_k,w_{k1}\right) \) and \(\left( w_k,w_{k+1}\right) \) are two arcs contained in N. Now fix a directed path \(P_d\) in N from \(\rho \) to \(w_k\).
If the vertex v is not contained in \(P_d\), then we can find an undirected path from \(\rho \) to c that does not contain v and has fewer alternating vertices than \(P_u\) by following \(P_d\) from \(\rho \) it reaches a vertex from \(\{w_k,\ldots ,w_t\}\) and then following \(P_u\) up to c. If v is contained in \(P_d\), then we follow \(P_u\) from \(\rho \) to \(w_k\) and then follow \(P_d\) from \(w_k\) to c and obtain an undirected path from \(\rho \) to c that does not contain v and has one alternating vertices, which is less than the number of alternating vertices in \(P_u\). In either case, we obtain a contradiction.
We have thus shown that \({\textsc {alt}}\left( P_u\right) =1\). Denoting this alternating vertex in \(P_u\) by r, then r is necessarily a reticulation by the choice of \(P_u\). Hence, \(P_u\) consists of two directed paths: a directed path from \(\rho \) to r that does not contain v and a directed path from c to r. However, this means that \(v \prec {\textsc {lsa}}(r)\), and hence \({\textsc {lsa}}(v)\preceq {\textsc {lsa}}(r)\) in view of Lemma 1. This implies that \(r\in \varGamma _0(H)\), a contradiction to the assumption that v is lowest among \(\varGamma _0(H)\). \(\square \)
The following is a direct consequence of the above theorem.
Corollary 1
Suppose that N is a simple binary phylogenetic network. Let H be the unique nontrivial biconnected component of N. Then there exists a lowest vertex v of H such that there exist two arcdisjoint directed paths from the root of N to v.
4 Displaying Binets by Binary Networks
A collection of binary level1 binets is compatible if there exists some binary network that displays all binets from the collection. In this section, we study the compatibility of binets. Our main result in this section (Theorem 3) shows that when studying the compatibility of binets, we can restrict to binary level1 networks.
We will restrict ourselves throughout this section to thin collections of binets, i.e. collections containing at most one binet on x and y for all distinct \(x,y\in X\). Clearly, any collection of binets that is not thin is not compatible.

Type I if both A and B are sink sets (i.e. there is no arc from any element in A to any element in B or vice versa);

Type II if either A or B (but not both) is a sink set; and

Type III if for all \(x\in A,y\in B\) (x, y) is an arc in G if and only if (y, x) is an arc in G.
For a collection \({\mathcal {B}}\) of binary level1 binets on X, we introduce the digraph \(D({\mathcal {B}})\) with vertex set X and (x, y) being an arc in \(D\left( {\mathcal {B}}\right) \) if \(T(x,y)\in {\mathcal {B}}\) or \(R(x;y)\in {\mathcal {B}}\). See Fig. 4 for an example.
The following two lemmas show important properties of typed splits that will be used to establish Theorems 2 and 3.
Lemma 2
Suppose that \({\mathcal {B}}\) and \({\mathcal {B}}'\) are two thin collections of binary level1 binets on X with \({\mathcal {B}}\subseteq {\mathcal {B}}'\). Then each typed split of \(D\left( {\mathcal {B}}'\right) \) is a typed split of \(D\left( {\mathcal {B}}\right) \).
Proof
Suppose that AB is a typed split of \(D\left( {\mathcal {B}}'\right) \). If AB is of Type I in \(D\left( {\mathcal {B}}'\right) \), then it is of Type I in \(D\left( {\mathcal {B}}\right) \) since \(D\left( {\mathcal {B}}\right) \) is a subgraph of \(D\left( {\mathcal {B}}'\right) \). Similarly, if AB is of Type II in \(D\left( {\mathcal {B}}'\right) \), then it is of Type I or II in \(D\left( {\mathcal {B}}\right) \). If AB is of Type III in \(D\left( {\mathcal {B}}'\right) \) then (since \({\mathcal {B}}'\) is thin) any binet on x and y with \(x\in A\) and \(y\in B\) is T(x, y). Therefore, AB is of Type I or III in \(D\left( {\mathcal {B}}\right) \). \(\square \)
Lemma 3
Suppose that \({\mathcal {B}}\) is a thin collection of binary level1 binets on X. If \({\mathcal {B}}\) is displayed by a binary network, then \(D\left( {\mathcal {B}}\right) \) has a typed split.
Proof
Suppose that \({\mathcal {B}}\) is displayed by a binary network. Then \({\mathcal {B}}\) is displayed by a binary recoverable network N. Let \({\mathcal {B}}'\) be the set of binary level1 binets contained in \({\mathcal {B}}(N)\). Then we have \({\mathcal {B}}\subseteq {\mathcal {B}}' \subseteq {\mathcal {B}}(N)\). By Lemma 2, it suffices to show that \(D\left( {\mathcal {B}}'\right) \) has a typed split.
Consider the root \(\rho \) of N, which is equal to \({\textsc {lsa}}(X)\) since N is recoverable. Denote the two children of \(\rho \) by \(u_1\) and \(u_2\). We consider two cases.
The first case is that at least one arc incident with \(\rho \) is a cut arc. Then the other arc incident with \(\rho \) is also a cut arc. Then let \(A={\mathcal {C}}\left( u_1\right) \) and \(B={\mathcal {C}}\left( u_2\right) \). Note that AB is a split because neither A nor B is empty. In addition, for all \(x\in A, y\in B\) we have \(N_{\{x,y\}}=T(x,y)\) and hence AB is a Type III split with respect to \(D\left( {\mathcal {B}}'\right) \).
In the second case, both arcs incident with \(\rho \) are not cut arcs. Hence, the root \(\rho \) is contained in a nontrivial biconnected component H containing \(u_1\) and \(u_2\). By Corollary 1, there exists a lowest vertex v in H with two arcdisjoint paths \(P_1,P_2\) from \(\rho \) to v. Since v is a lowest vertex in H, we know that v is a reticulation vertex and the arc leaving v is a cut arc. Let \(B={\mathcal {C}}(v)\) and \(A=X{\setminus } B\). Then B is clearly nonempty. In addition, A is nonempty, as otherwise \({\textsc {lsa}}(X)\preceq v\), a contradiction to the fact that \({{\textsc {lsa}}(X)=\rho }\) (as N is recoverable). Therefore, AB is a split.
Consider \({x\in A}\) and \({y\in B}\) and the subnetwork \(N_{\{x,y\}}\). There is at least one directed path from \(\rho \) to x, and each such path contains at least one arc of \(P_1\) or \(P_2\). Hence, in the process of obtaining \(N_{\{x,y\}}\) from N, the paths \(P_1,P_2\) do not become parallel arcs. Therefore, \(N_{\{x,y\}}\) contains two arcdisjoint paths from \(\rho \) to v and we can conclude that \({N_{\{x,y\}} \not =T(x,y)}\).
Therefore, if \(N_{\{x,y\}}\in {\mathcal {B}}^*\), that is, \(N_{\{x,y\}}\) is level1, then \(N_{\{x,y\}}=R(x;y)\). This implies that there is no arc (y, x). Therefore, AB is a Type I or Type II split of \(D\left( {\mathcal {B}}^*\right) \). \(\square \)
Theorem 2
Suppose that \({\mathcal {B}}\) is a thin collection of binary level1 binets on X. If there exists a typed split AB of \(D\left( {\mathcal {B}}\right) \) such that \({\mathcal {B}}_A\) and \({\mathcal {B}}_B\) are both displayed by some binary level1 network, then \({\mathcal {B}}\) is displayed by a binary level1 network. Moreover, if \({\mathcal {B}}\) is displayed by a binary level1 network, then there exists at least one typed split of \(D\left( {\mathcal {B}}\right) \) and, for each typed split AB of \(D\left( {\mathcal {B}}\right) \), \({\mathcal {B}}_A\) and \({\mathcal {B}}_B\) are both displayed by some binary level1 network.
Proof
First suppose that there exists a typed split AB of \(D\left( {\mathcal {B}}\right) \) such that \({\mathcal {B}}_A\) and \({\mathcal {B}}_B\) are displayed by binary level1 networks \(N_A\) and \(N_B\), respectively.
Now suppose that \({\mathcal {B}}\) is displayed by a binary level1 network N. By Lemma 3, there exists a typed split AB of \(D\left( {\mathcal {B}}\right) \). Then \({\mathcal {B}}_A\subseteq {\mathcal {B}}\left( N_A\right) \) and \({\mathcal {B}}_B\subseteq {\mathcal {B}}\left( N_B\right) \). \(\square \)
We now prove the main result of this section.
Theorem 3
Suppose that \({\mathcal {B}}\) is a thin collection of binary level1 binets on X. Then \({\mathcal {B}}\) is displayed by a binary level1 network if and only if it is displayed by a binary network.
Proof
Suppose that \({\mathcal {B}}\) is displayed by a binary network. We claim that \({\mathcal {B}}\) is also displayed by a binary level1 network. We shall establish this claim by induction on X.
If \(X=2\), then \({\mathcal {B}}\) contains at most one binet, which has leaf set X. Therefore we know that \({\mathcal {B}}\) is displayed by a binary level1 network.
Now assume that \(X> 2\), and the claim holds for all sets \(X'\) with \(2\le X'<X\). Let N be a binary network on X with \({\mathcal {B}}\subseteq {\mathcal {B}}(N)\). By Lemma 3, there exists a typed split AB of \(D\left( {\mathcal {B}}\right) \). Note that \({\mathcal {B}}_A \subseteq {\mathcal {B}}\left( N_A\right) \) and \({\mathcal {B}}_B \subseteq {\mathcal {B}}\left( N_B\right) \). Therefore, by induction, each of \({\mathcal {B}}_A\) and \({\mathcal {B}}_B\) is displayed by a binary level1 network. By Theorem 2, it follows that \({\mathcal {B}}\) is displayed by a binary level1 network. \(\square \)
5 Binets Determine the Number of Reticulations of a Binary Level1 Network
In this section we show that, although the collection of binets displayed by a level1 network does not necessarily determine the network (see Fig. 1), it does in fact determine the number of reticulations in the network. We begin by showing that it suffices to consider level1 networks in which all cycles (in the underlying undirected graph) have length 3.
First, we introduce some further notation. A semicycle C of an acyclicdirected graph is the union of two nonidentical, internally vertexdisjoint, directed paths from s to t, with \({s=s(C)}\) and \({t=t(C)}\) two distinct vertices that are referred to as the source and terminal of C, respectively. The length of a semicycle is the number of distinct vertices that it contains.
We now show that we may restrict to networks in which all semicycles have length 3.
Lemma 4
If N is a binary level1 network, then there exists a binary level1 network \(N'\) in which every semicycle has length 3, such that \({\mathcal {B}}\left( N'\right) ={\mathcal {B}}(N)\) and N and \(N'\) have the same number of reticulation vertices.
Proof
Consider a semicycle of N with source s and terminal t and length at least 4. Let \(\left( u_1,v_1\right) \), \(\ldots \), \(\left( u_k,v_k\right) \), (t, w) be the arcs leaving the semicycle. Then \(k\ge 2\). Let \(N^*\) be a network obtained from a binary tree on \(\{v_1,\ldots ,v_k\}\) by replacing \(v_i\) by the subgraph of N rooted at \(v_i\), for \(i=1,\ldots ,k\). Let \(N_w\) be the subgraph of N rooted at w. Then we construct \(N'\) from N by replacing the subgraph of N rooted at s by the network \(R\left( N^*;N_w\right) \). It is straightforward to see that \(N'\) is a binary level1 network with the required properties. \(\square \)
We now establish the main result of this section.
Theorem 4
If \(N_1\) and \(N_2\) are rooted binary level1 phylogenetic networks on X with \({\mathcal {B}}\left( N_{1}\right) ={\mathcal {B}}\left( N_{2}\right) \) then \(N_1\) and \(N_2\) have the same number of reticulation vertices.
Proof
The proof is by induction on the number of leaves X. The induction basis for \(X=2\) is clear. Now suppose that \(N_1\) and \(N_2\) are two nonisomorphic rooted binary level1 phylogenetic networks on \(X\ge 3\) with \({\mathcal {B}}\left( N_{1}\right) ={\mathcal {B}}\left( N_{2}\right) \) but with different numbers of reticulation vertices. We add an outdegree1 root to each of \(N_1\) and \(N_2\) with an arc to the original root. By Lemma 4, we may assume that all semicycles in \(N_1\) and \(N_2\) have length 3.
Choose an arbitrary leaf \(x\in X\) and let \(X'=X\setminus \{x\}\). Let \(N_1'\) and \(N_2'\) be the networks obtained from \(N_1_{X'}\) and \(N_2_{X'}\), respectively, by adding an outdegree1 root with an arc to the original root. Then \(N_1'\) and \(N_2'\) have the same number of reticulation vertices by induction.
Since all semicycles in \(N_1\) and \(N_2\) are assumed to have length 3, there are three cases for the location of x in each of the networks \(N_1,N_2\), illustrated in Fig. 6.
If the parent of x is in a semicycle in \(N_i\), let \(v_i\) be the source of this semicycle, and let \(v_i\) be the parent of x otherwise. Let \(B_i := {\mathcal {C}}(v_i){\setminus } \{x\}\) and \(A_i := X'{\setminus } B_i\) (recall that \({\mathcal {C}}(v_i)\) denotes the cluster of \(v_i\)).
The first case is that the parent of x is not in a semicycle in \(N_1\) but is the terminal of a semicycle in \(N_2\). First suppose that \(B_1\cap B_2 \ne \emptyset \). Then choose an arbitrary vertex \(y\in B_1\cap B_2\). Then \(N_1_{\{x,y\}}=T(x,y)\) while \({N_2_{\{x,y\}}=R(y;x)}\), a contradiction. Hence, we may assume that \({B_1\cap B_2 =\emptyset }\). Then \({B_1=A_2}\) and \({B_2=A_1}\). Clearly, \({B_1,B_2\ne \emptyset }\). Take \(y\in B_1=A_2\) and \(z\in B_2=A_1\). Then \(N_1_{\{x,y\}}=T(x,y)\) and hence \(N_2_{\{x,y\}}=T(x,y)\), from which we can deduce that \(N_2_{\{z,y\}}=T(z,y)\). In addition, \(N_2_{\{z,x\}}=R(z;x)\) and hence \(N_1_{\{z,x\}}=R(z;x)\), from which we can deduce that \(N_1_{\{z,y\}}=R(z;y)\). This leads to a contradiction since \(N_2_{\{z,y\}}=T(z,y)\).
The second case is that the parent of x is not in a semicycle in \(N_1\) but is the nonterminal nonsource vertex of a semicycle in \(N_2\). First suppose that \({B_1\cap B_2 \ne \emptyset }\). Then choose an arbitrary vertex \({y\in B_1\cap B_2}\). Then \(N_1_{\{x,y\}}=T(x,y)\) while \(N_2_{\{x,y\}}=R(x;y)\), a contradiction. Hence, we may assume that \(B_1\cap B_2 =\emptyset \). Then, as in the previous case, \(B_1=A_2\ne \emptyset \) and \(A_1=B_2\ne \emptyset \). Take \(y\in B_1=A_2\) and \(z\in B_2=A_1\). Then, similar to the previous case, \(N_1_{\{x,y\}}=T(x,y)\) and hence \(N_2_{\{x,y\}}=T(x,y)\), from which we can deduce that \({N_2_{\{z,y\}}=T(z,y)}\). In addition, \(N_2_{\{z,x\}}=R(x;z)\) and hence \({N_1_{\{z,x\}}=R(x;z)}\), from which we can deduce that \({N_1_{\{z,y\}}=R(y;z)}\). This again leads to a contradiction since \({N_2_{\{z,y\}}=T(z,y)}\). \(\square \)
6 Complexity of Binet Compatibility
A direct consequence of Theorem 2 is that there exists a simple polynomialtime algorithm to decide whether there exists a binary level1 network displaying a given collection \({\mathcal {B}}\) of binary level1 binets (see Huber et al. 2015 for a related algorithm). In particular, a sink set of \(D\left( {\mathcal {B}}\right) \) can be found in polynomial time by computing the strongly connected components of \(D\left( {\mathcal {B}}\right) \) (Tarjan 1972) and checking for each of them whether it is a sink set. This can be used to find a typed split, if it exists. If such a split does not exist, then \({\mathcal {B}}\) is not compatible. Otherwise, we can try to construct networks for \({\mathcal {B}}_A\) and \({\mathcal {B}}_B\) recursively and combine them as described in the proof of Theorem 2. This algorithm is similar to the Aho algorithm for deciding whether a set of rooted trees can be displayed by some rooted tree (Aho et al. 1981).

Binet compatibility (BC)

Input: a set \({\mathcal {B}}\) of binary level1 binets.

Question: is \({\mathcal {B}}\) compatible, i.e. does there exist a binary network N with \({\mathcal {B}}\subseteq {\mathcal {B}}(N)\)?
Theorem 5
Deciding whether there exists a phylogenetic network displaying a given thin set \({\mathcal {B}}\) of binets is GIhard.
Proof
We claim that \(G_1\) and \(G_2\) are isomorphic if and only if there exists a network displaying \(B_1,B_2\) and \(B_3\).
First assume that \(G_1\) and \(G_2\) are isomorphic. Then we can construct a network displaying \(B_1,B_2\) and \(B_3\) as follows. Take \(B_1\) and subdivide the arc \((u_1,y)\) by a new vertex \(u_1'\) and add leaf z with an arc \(\left( u_1',z\right) \). The obtained network clearly displays \(B_1\) and \(B_3\) and it also displays \(B_2\) since \(G_1\) and \(G_2\) are isomorphic.
Now assume that there exists some network N displaying \(B_1,B_2\) and \(B_3\). Then \(N_{\{x,y\}}=B_1\). Hence, N contains a cycle (in the underlying undirected graph) containing a reticulation v, such that x and the image of \(G_1\) are below the arc leaving v, while y is below some other arc leaving the cycle. Since \(N_{\{y,z\}}=T(y,z)\), leaf z is not below v in N. Therefore, deleting v, x and the parent of x from the subgraph of N rooted at v gives \(G_1\).
Similarly, N contains a cycle containing a reticulation \(v'\), such that x and the image of \(G_2\) are below the arc leaving \(v'\), while z is below some other arc leaving the cycle. Since \(N_{\{y,z\}}=T(y,z)\), leaf y is not below \(v'\) in N. Therefore, deleting \(v'\), x and the parent of x from the subgraph of N rooted at \(v'\) gives \(G_2\).
Moreover, \(v=v'\) since \(N_{\{y,z\}}=T(y,z)\). Hence, \(G_1\) and \(G_2\) are isomorphic. \(\square \)
7 Maximum Binet Compatibility

Maximum Binet compatibility (MBC)

Input: a set \({\mathcal {B}}\) of binary level1 binets and an integer k.

Question: does there exist a compatible subset \({\mathcal {B}}'\) of \({\mathcal {B}}\) with \({\mathcal {B}}' \ge k\)?
Lemma 5
If the binet R(x; y) is displayed by a binary level1 network N, then \({\textsc {lsa}}(x,y)\) is the source of a semicycle C in N. In addition, y is below t(C) and x is not below t(C).
Proof
Let \(u={\textsc {lsa}}(x,y)\). Note that u is not a reticulation vertex, as otherwise the child of u would be a stable ancestor of x and y that is below u. Hence, u has two children, denoted by \(u_1\) and \(u_2\).
Observe that neither \((u,u_1)\) nor \((u,u_2)\) is a cut arc, since otherwise we would have \(N_{\{x,y\}}=T(x,y)\), while by the assumption of the lemma \(N_{\{x,y\}}=R(x;y)\). Hence, u is the source of a semicycle C. Let \(v:=t(C)\) be the terminal of C. If neither x nor y is below v, then \(N_{\{x,y\}}=T(x,y)\), a contradiction. If both x and y are below v, then v is a stable ancestor of x and y, a contradiction to \({\textsc {lsa}}(x,y)=u\). Therefore, precisely one of x and y is below v. If x is below v and y is not, then \(N_{\{x,y\}}=R(y;x)\), a contradiction. Therefore, y is below v and x is not. \(\square \)
In view of the last lemma, for each binet \(R(x;y)=N_{\{x,y\}}\), there exists a unique semicycle \(C_N(x;y)\) containing \({\textsc {lsa}}(x,y)\).
Lemma 6
Proof
Let \(C_1=C_N(x;y)\) and \(C_2=C_N(y;z)\). By Lemma 5, \(y\prec t\left( C_1\right) \) but y is not below \(t\left( C_2\right) \), from which we know that \(C_1\not =C_2\). Since \(s\left( C_1\right) \) and \(s\left( C_2\right) \) are stable ancestors of y in view of Lemma 5, we have either \(s\left( C_1\right) \prec s\left( C_2\right) \) or \(s\left( C_2\right) \prec s\left( C_1\right) \) but not both.
Note that if \(s\left( C_1\right) \prec s\left( C_2\right) \), then \(s\left( C_1\right) \prec t\left( C_2\right) \) and hence \(y\prec s\left( C_1\right) \prec t\left( C_2\right) \), a contradiction. Thus \(s\left( C_2\right) \prec s\left( C_1\right) \), from which it follows that \(s\left( C_2\right) \prec t\left( C_1\right) \). \(\square \)
Given a digraph G, let \({\mathcal {R}}(G)\) be the collection of binets \(\{R(x;y)\,\,(x,y) \in E(G)\}\) induced by G. Note that \({\mathcal {R}}(G)\) is a binet set on V(G), i.e. the leaves of the binets in \({\mathcal {R}}(G)\) correspond to the vertices of G.
Proposition 1
Let G be a digraph. Then G is acyclic if and only if \({\mathcal {R}}(G)\) is compatible.
Proof
A set of binets \({\mathcal {B}}\) on X is said to be dense if for each pair of distinct elements x and y in X, there exists precisely one binet on \(\{x,y\}\) in \({\mathcal {B}}\). Hence, a dense set of binets is always thin.
Theorem 6
The problem MBC is NPcomplete, even if the given set of binets is dense.
Proof
We reduce from the NPhard problem feedback arc set in tournaments (FAST) (Alon 2006; Charbit et al. 2007), which is defined as follows. Given a tournament, i.e. a digraph \(G=(V,E)\) with either \((a,b)\in E\) or \((b,a)\in E\) (but not both) for each pair of distinct elements a and b in V, and given a positive integer \(k'\), does there exist a subset \(F\subseteq E\) of at most \(k'\) arcs whose removal makes G acyclic. If such an arc set exists, then we call it a feedback arc set of G.
The reduction is as follows. For each instance \((G,k')\) of FAST, consider the corresponding instance \(\left( {\mathcal {R}}(G),k\right) \) of MBC with \(k={\mathcal {R}}(G)k'\). Since the set \({\mathcal {R}}(G)\) of binets induced by G can be constructed in polynomial time, it suffices to show that G contains a feedback arc set with size at most \(k'\) if and only if there exists a compatible subset of \({\mathcal {R}}(G)\) of size at least k.
First assume that there exists a feedback arc set \(E'\) of G with size at most \(k'\). That is, \(E'\le k'\), and the digraph \(G^*\) obtained from G by deleting the arcs in \(E'\) is acyclic. Consider the set of binets \({{\mathcal {B}}'=\left\{ R(x;y)\,:\,(x,y)\in E{\setminus } E'\right\} }\). This set contains at least k binets. In addition, since \({\mathcal {B}}'={\mathcal {R}}(G^*)\), it follows by Proposition 1 that \({\mathcal {B}}'\) is compatible.
Now assume that there exists a compatible binet set \({\mathcal {B}}'\subseteq {\mathcal {R}}(G)\) with \({\mathcal {B}}'\ge k\). Consider the set \({E'=\left\{ (x,y)\,:\, R(x,y)\in {\mathcal {R}}(G){\setminus } {\mathcal {B}}'\right\} }\) of arcs of G. Then by Proposition 1, it follows that \(E'\) is a feedback arc set. Moreover, \(E'\le k'\), which completes the proof. \(\square \)
We complete the section by showing that there exists a polynomialtime 1 / 3approximation algorithm for the MBC problem, which follows directly from the next theorem and its proof.
Theorem 7
Suppose that \({\mathcal {B}}\) is a set of binary level1 binets on X. Then there exists a binary level1 network N such that \({\mathcal {B}}(N)\cap {\mathcal {B}}\ge {\mathcal {B}}/3\).
Proof
If at least a third of the binets in \({\mathcal {B}}\) are tree type, then take N to be any binary tree on X and we are done. Hence we may assume that at least two thirds of the binets are reticulate type.
Impose an arbitrary ordering on the elements in X, that is, write \(X=\{x_1,\ldots ,x_n\}\). Let \({\mathcal {B}}_1={\mathcal {B}}\cap \left\{ R\left( x_i;x_j\right) \,:\,1\le i<j\le n\right\} \) and \({\mathcal {B}}_2={\mathcal {B}}\cap \left\{ R\left( x_j;x_i\right) \,:\,1\le i<j\le n\right\} \). Without loss of generality, we may assume that \({\mathcal {B}}_1\ge {\mathcal {B}}_2\) (as the other case can be established in a similar way). Since at least two thirds of the binets are reticulate type, and each of those is contained in either \({\mathcal {B}}_1\) or \({\mathcal {B}}_2\) (but not both), we know that \({\mathcal {B}}_1\ge {\mathcal {B}}/3\). Now consider the network \(N_*\) in Fig. 9, then clearly we have \({\mathcal {B}}_1\subseteq {\mathcal {B}}\left( N_*\right) \). Thus we have \({\mathcal {B}}\left( N_*\right) \cap {\mathcal {B}}\ge {\mathcal {B}}_1 \ge {\mathcal {B}}/3\), from which the theorem follows. \(\square \)
8 Discussion
In this paper we have developed some combinatorial results concerning collections of level1 binets. Several interesting questions arise from these results. For example, we have shown that the collection of level1 binets displayed by a binary phylogenetic network can be displayed by some level1 network, but is there some canonical level1 network that could be used to display such a collection? In addition, can we count the number of binary level1 networks that display a dense compatible collection of binets? We have also seen that the collection of binets displayed by a binary level1 network determine its reticulation number. Therefore it is natural to ask which properties of a phylogenetic network in general are determined by its binets?
We have also studied some algorithmic questions concerning binets. Concerning the maximum binet compatibilty problem, note that the constant 1 / 3 is sharp in Theorem 7. For example, consider the binet collection \(\left\{ R(x;y),T(x,y),R(y;x)\right\} \). However, can a better bound be achieved by restricting to thin collections of binets, and can improved approximation algorithms also be found?
In another direction, it would be interesting to know whether similar results to those proven in this paper might hold for higherlevel networks. For example, what can be said about properties of collections of level2 binets, and does Theorem 4 hold also for higherlevel networks? Also, we could try to generalize some of our results to knets, i.e. networks on k leaves, \(k \ge 2\). For example, does Theorem 3 hold for trinets? In general, it would be interesting to know what additional information the collection of knets displayed by a network might contain for \(k\ge 3\). Note that it has been shown that trinets do not completely determine rooted networks in general Huber et al. (2015). However, do they determine properties of networks such as the number of reticulations?
Similarly, it would be interesting to extend some of our algorithmic results to higherlevel networks and knets. For example, it is known that the compatibility problem is NPcomplete for collections of level1 trinets (Huber et al. 2015). However, to date the maximum trinet compatibility problem has not been studied.
Eventually, it is hoped that new results in these directions could be useful for developing novel methods to construct phylogenetic networks from higherlevel networks and knets. For example, using our results it may be possible to develop approaches to build a consensus network for a collection of phylogenetic trees or networks. Note that consensus networks have already proven themselves useful in the unrooted setting, where they are used to summarize key features displayed by a collection of trees or networks (see e.g. Holland et al. 2004). A consensus method based on binets could work by breaking each of the given networks down into a collection of binets, and then developing methods to pool together the information contained in the resulting binets so as to construct some consensus network, or at least some constraints that any such network should satisfy. Note that similar approaches have been developed to build consensus trees for a collection of phylogenetic trees by breaking each of the trees down into a collection of triplets [see e.g. Bryant (2003, Sect. 2)]. Probably it would be of some interest to first consider how to construct a level1 consensus network for a collection of level1 networks by breaking each of them down into level1 binets. This is already likely to be quite challenging in view of our result concerning NPcompleteness of MBC.
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