On the existence of funneled orientations for classes of rooted phylogenetic networks
Abstract.
Recently, there has been a growing interest in the relationships between unrooted and rooted phylogenetic networks. In this context, a natural question to ask is if an unrooted phylogenetic network can be oriented as a rooted phylogenetic network such that the latter satisfies certain structural properties. In a recent preprint, Bulteau et al. claim that it is computational hard to decide if has a funneled (resp. funneled tree-child) orientation, for when the internal vertices of have degree at most 5. Unfortunately, the proof of their funneled tree-child result appears to be incorrect. In this paper, we present a corrected proof and show that hardness remains for other popular classes of rooted phylogenetic networks such as funneled normal and funneled reticulation-visible. Additionally, our results hold regardless of whether is rooted at an existing vertex or by subdividing an edge with the root.
Key words and phrases:
graph orientation, network classes, phylogenetic network, rooted, unrooted1. Introduction
Phylogenetic networks are commonly used to represent evolutionary relationships between taxa such as species, individuals of a population, or viruses. In general terms, phylogenetic networks are graphs whose vertices represent taxa and edges represent inferred evolutionary relationships. Unrooted phylogenetic networks are undirected graphs that do not contain any explicit information about the direction of evolution such as ancestor-descendant relationships. To include such information, rooted phylogenetic networks are used which are directed acyclic graphs with a single source (the root) and, as in the unrooted case, leaves representing extant taxa, and internal vertices representing hypothetical or extinct taxa. In addition, there also exists a wide range of well-studied classes of phylogenetic networks that are each characterised by certain structural properties such as level- and tree-based networks, which have been defined for rooted and unrooted phylogenetic networks, as well as tree-child and reticulation-visible networks, which have been defined for rooted phylogenetic networks only. For a comprehensive overview of phylogenetic network classes, we refer the interested reader to [9]. Network classes that are used throughout this paper are formally defined in the next section.
Initiated by a paper by Huber et al. [6], there has recently been a growing interest in the relationships between unrooted and rooted phylogenetic networks. More specifically, the authors have investigated several questions in the context of orienting an unrooted phylogenetic network as a rooted phylogenetic network. The process of orienting consists of subdividing an edge of with a new vertex and assigning a direction to each edge such that the resulting directed graph is a rooted phylogenetic network with root . Given , results for the following decision problems have been established in [6]:
-
(i)
Constrained Orientation: If the desired in-degree for each vertex of is given as well as an edge of to be subdivided with , can be oriented as a rooted phylogenetic network with root that satisfies all in-degree constraints?
-
(ii)
-Orientation: Can be oriented such that the resulting digraph belongs to a given class of phylogenetic networks?
More specifically, Huber et al. [6] present a polynomial-time algorithm for Constrained Orientation that computes an orientation of satisfying the given in-degree constraints or outputs that there is none. Furthermore, they have shown that, if is binary (i.e., each internal vertex has degree 3), then it is NP-complete to decide if has a tree-based orientation and, lastly, if is binary and satisfies certain properties, then -Orientation is fixed-parameter tractable with respect to the so-called level of . Notably, the latter result includes the popular class of binary tree-child networks. In the same paper, the authors also state two open questions.
-
Q1
Given an unrooted binary phylogenetic network , can it be decided in polynomial time if has a tree-child orientation?
-
Q2
Given an unrooted phylogenetic network , can it be decided in polynomial time if has a funneled orientation?
The term funneled was coined by Huber et al. [6, p. 26] and refers to the restriction that each reticulation (i.e., a vertex of a rooted phylogenetic network with in-degree at least 2) has out-degree 1. As noted by the authors, it is common in the literature to define reticulations in this way.
Recently Bulteau et al. [1, Cor. 4] have shown that it is NP-complete to decide if has a funneled orientation if each vertex of has degree at most 5 and a vertex of is chosen as root (instead of subdividing an edge of with the root). We remark that Bulteau et al. refer to a funneled orientation as a valid orientation and that their construction to establish NP-completeness allows for vertices of degree 2 in which are explicitly excluded in the definition of such a network in [6] as well as in most other literature on phylogenetic networks because degree-2 vertices do not carry any biological meaning. Relatedly, Garvardt et al. [5] have analysed the parameterised complexity of a variant of the Constraint Orientation problem, where the in-degree of each vertex is not fixed but can take on a value of a list that is associated with . On the positive side, Bulteau et al. [1, Cor. 3] have shown that Q2 can be answered affirmatively if has degree at most 3. In particular, they have established a linear-time algorithm that computes a funneled orientation of if it exists or, otherwise, that correctly concludes that no such orientation exists. This last result is in line with an earlier result by Janssen et al. [8, Lem. 4.13 and 4.14] who characterised those unrooted binary phylogenetic networks that can be oriented as rooted binary phylogenetic networks. Inspecting the proof of their characterisation [8, Lem. 4.13], their result also yields a polynomial-time algorithm for deciding if an unrooted binary phylogenetic network can be oriented as a rooted binary phylogenetic network.
Turning to Q1, Maeda et al. [11] have recently presented several necessary conditions for when an unrooted binary phylogenetic network has a tree-child orientation. It is worth noting that, in the binary case, any orientation as a rooted phylogenetic network is also funneled. Moreover, given an unrooted phylogenetic network whose non-leaf vertices have degree at least 2 and at most 5 and given a designated root vertex of , Bulteau et al. [1, Cor. 5] claim that it is NP-hard to decide if has a funneled tree-child orientation with root . Unfortunately, as we will show later, the proof of this corollary is incorrect. In summary, Q1 remains open.
In this paper, we make a step towards answering Q1 by presenting a corrected proof that establishes NP-completeness for deciding if an unrooted phylogenetic network with maximum degree 5 has a funneled tree-child orientation. In comparison to Bulteau et al. [1], our result does not require a designated root to be part of the input and our construction yields unrooted phylogenetic networks without degree-2 vertices. Although our result builds on Bulteau et al., correcting their proof requires novel gadgets and careful reasoning. We also show that our result extends to several other classes of funneled orientations such as funneled tree-sibling and funneled normal networks.
The remainder of the paper is organised as follows. Section 2 contains definitions and formal problem statements. In Section 3, we show that, although the proof by Bulteau et al. [1, Cor. 5] is incorrect, the result that it is NP-complete to decide if an unrooted phylogenetic network with degree at most 5 has a funneled tree-child orientation can be recovered. We then show in Section 4 that hardness remains if each non-leaf vertex of has degree exactly 5. This latter result is subsequently used to establish NP-completeness for deciding if has a funneled tree-sibling, funneled reticulation-visible, or funneled normal orientation. We remark that all results hold under two rooting variants: Variant as introduced by Huber et al. [6] subdivides an edge of with the root, and Variant chooses a vertex of to be the root. If is binary, one typically chooses a leaf of to be the root since, otherwise, the resulting orientation is not binary. Indeed, all constructions presented in this paper choose a leaf of as the root when establishing hardness under Variant .
2. Preliminaries
In this section, we introduce notation and terminology that is used throughout the rest of the paper, and formally state the decision problems whose computational complexity we investigate in subsequent sections.
2.1. Phylogenetic networks.
Let be a non-empty finite set. An unrooted phylogenetic network on is a simple undirected graph with no degree-2 vertex and with a bijection between the vertices in that have degree 1 and the set , that is, the leaves of are bijectively labelled with elements from . We will freely refer to leaves using their labels. Two distinct vertices and of are called neighbours if is an edge in .
A rooted phylogenetic network on is a directed acyclic graph with no loop and no parallel arcs that satisfies the following properties:
-
(i)
there is a unique vertex , the root, with in-degree 0 and out-degree at least ,
-
(ii)
a vertex of out-degree 0 has in-degree 1 and the set of vertices with out-degree 0 is , and
-
(iii)
each internal vertex has either in-degree 1 and out-degree at least 2, or in-degree at least 2 and out-degree at least 1.
For technical reasons, we sometimes consider a class of networks that is more general than the class of rooted (resp. unrooted) phylogenetic networks. Specifically, we refer to a network that is a subdivision of a rooted phylogenetic network on (resp. unrooted phylogenetic network on ) as a rooted pseudo network on (resp. unrooted pseudo network on ).
Now, let be a rooted pseudo network on . For an arc in , we say that is a parent of and is a child of . If two distinct vertices and of have a common parent, we say that and are siblings. Moreover, a vertex of is called a tree vertex if it has in-degree and out-degree at least and is called a reticulation if it has in-degree at least and out-degree at least . Lastly, an arc is called a shortcut in if there is a directed path from to in that does not contain .
We next define five classes of rooted pseudo networks. These classes are typically defined in the context of rooted phylogenetic networks. However, their definitions naturally carry over to rooted pseudo networks as follows. Let be a rooted pseudo network. We say that is
-
(i)
tree-child if each non-leaf vertex of has a child that is a tree vertex or a leaf,
-
(ii)
normal if is tree-child and does not contain a shortcut,
-
(iii)
tree-sibling if each reticulation of has a sibling that is a tree vertex or a leaf,
-
(iv)
reticulation-visible if, for each reticulation of , there is a leaf such that each directed path from the root of to contains , and
-
(v)
funneled if each reticulation of has out-degree 1.
Since the above five classes are only defined for rooted pseudo networks, we will omit the adjective ‘rooted’ when referring to them. For more details on these network classes, see [2, 3, 7, 9, 12].
2.2. Orientations.
A connector network with is a graph that can be obtained from an unrooted pseudo network by deleting the label of leaves in . We call a degree-1 vertex of a connector leaf or unlabelled leaf if it has no label, and a non-connector leaf or labelled leaf otherwise. Furthermore, we refer to as the partner network of . Note that is the unique partner network of up to those leaf labels that exist in and not in . Lastly, for , let be a connector network. We say that an unrooted pseudo network contains as a pending subgraph if can be obtained from by identifying its connector leaf with a vertex of some unrooted pseudo network and deleting the label if is a leaf of . For example, the unrooted pseudo network that underlies the network shown in Figure 1(ii) has both and that are shown in (i) of the same figure as a pending subgraph.
For the purpose of the upcoming definitions, let be a connector network with . Furthermore, let be the set of all labelled leaves of , and let be the set of all unlabelled leaves of . We next define a process that assigns a direction to each edge of and, if , also introduces a root vertex .
First, for , an orientation of is obtained by either
Variant . subdividing an edge of with a new root vertex and then assigning a direction to each edge such that has in-degree 0 and each element in has out-degree 0, or
Variant . choosing a vertex of to be the root by setting , deleting the label of if is a labelled leaf, and then assigning a direction to each edge such that has in-degree 0 and each element in if (resp. if ) has out-degree 0.
Note that such an orientation always exists. Now, let be a class of rooted (pseudo) networks and let . For , we say that an unrooted pseudo network has a -orientation or, equivalently, that is -orientable if there exists an orientation of such that the following properties are satisfied.
-
(i)
is obtained from by following Variant ,
-
(ii)
is a network in with root ,
-
(iii)
if , then has leaves, and
-
(iv)
if , then has leaves.
Turning to , we say that a connector network with a unique connector leaf has a -orientation or, equivalently, that is -orientable if its partner network has a -orientation. For , is called -root-forcing if there is no -orientation of that subdivides the (unique) edge incident with with a new root vertex. Similarly, for , is called -root-forcing if there is no -orientation of that chooses to be the root.

To illustrate, see Figure 1 for an example of a connector network with a single connector leaf that is -root-forcing for when is the class of tree-child networks. In Figure 1 as well as in all other figures of this paper, the root is indicated by a small square, connector leaves are indicated by open circles whereas all other vertices are indicated by filled circles. Moreover, we frequently label internal vertices in figures. Their only purpose is to make references. Indeed, they should not be regarded as genuine labels as those used for leaves of pseudo networks and non-connector leaves of connector networks. Furthermore, labels of leaves of pseudo networks and non-connector leaves of connector networks are sometimes omitted from figures to facilitate a clear presentation unless they are of importance (e.g. if they are explicitly mentioned in a proof).
Second, for , an orientation of is obtained by assigning a direction to each edge in . Let be an acyclic orientation of such that each vertex in has out-degree 0, and each vertex that is not in has in-degree at least 1 and out-degree at least 1, then is called
-
(i)
-compatible if each vertex of with in-degree at least 2 has out-degree 1,
-
(ii)
-compatible if each vertex of that is not in has a child that is either in or a vertex with in-degree 1 and out-degree at least 1, and
-
(iii)
strongly -compatible if each vertex of that is not in has a child that is either in or a vertex with in-degree 1 and out-degree at least 1.
2.3. Problem statements
Let be a class of rooted phylogenetic networks, and let . In this paper, we analyse the following decision problem.
Funneled -Orientation
Instance. An unrooted phylogenetic network .
Question. Does there exist a funneled -orientation of ?
If , we emphasise that any vertex of can be chosen as root. In particular, no preselected root is part of the input to Funneled -Orientation. To obtain NP-hardness results for Funneled -Orientation in the next section, we design three gadgets, which are connector networks. Multiple copies of these gadgets are then combined such that the resulting graph is an unrooted pseudo network or an unrooted phylogenetic network . We then orient each gadget separately such that, collectively, these orientations will result in a funneled -orientation of if such an orientation exists. To provide some intuition for some of the more technical definitions given in Section 2.2, if a gadget is a connector network with exactly one connector leaf, then the upcoming hardness constructions enforce that the root of is contained in the subgraph of that is obtained by orienting . On the other hand, if a gadget is a connector network with , then the hardness constructions enforce that the root of is not contained in the subgraph of that is obtained by orienting .
We now turn to the statement of two Boolean satisfiability problems that we later use to establish NP-hardness of Funneled -Orientation. Let be a set of variables. A literal is a variable or its negation , and a clause is a disjunction of a subset of . If a clause contains exactly distinct literals for , then it is called a -clause. We say that a clause is positive if it is a subset of . A Boolean formula in conjunctive normal form (or short, Boolean formula) is a conjunction of clauses, i.e., an expression of the form , where is a clause for all . Now, let be a Boolean formula. We say that is positive if each clause of is positive, i.e., no clause contains an element in . A truth assignment for is a mapping , where represents the truth value True and represents the truth value False. A truth assignment satisfies if at least one literal of each clause evaluates to under . If satisfies and has the additional property that at least one literal of each clause evaluates to , we say that nae-satisfies .
Positive Not-All-Equal -SAT
Instance. A set of variables and a collection of positive clauses over such that each clause consists of either two or three distinct variables.
Question. Is there a truth assignment for that nae-satisfies ?
Positive 1-in-3 SAT
Instance. A set of variables and a collection of positive clauses over such that each clause consists of 3 distinct variables.
Question. Is there a truth assignment for that sets exactly one variable in each clause in to ?
3. Funneled Tree-Child Orientations
In this section, we analyse the aforementioned result by Bulteau et al. [1] and establish NP-completeness of Funneled -Orientation for when an unrooted phylogenetic network with maximum degree 5 is given and is the class of rooted phylogenetic networks that are tree-child.
3.1. Construction by Bulteau et al.
As mentioned in the introduction, an incorrect proof has recently appeared in the literature [1] that claims NP-completeness for the following decision problem which is slightly different from Funneled -Orientation: Given an unrooted pseudo network on with maximum degree 5 and a vertex of , does there exist an orientation of such that is a rooted pseudo network on (resp. if ) with root that is funneled and tree-child?
We next describe the high-level idea of the construction that is presented in [1] and then explain why the tree-child orientation result does not hold. Given an instance of Positive Not-All-Equal -SAT, the authors construct an unrooted pseudo network with leaf set . The construction of is based on the bipartite incidence graph which has an internal variable vertex for each variable, an internal clause vertex for each clause, and an edge connecting with if and only if appears in . Additionally, a leaf is attached to each and a vertex is introduced that is adjacent to each . Lastly, to guarantee that has degree at most 5, additional edges and vertices are added. Importantly, the introduction of these additional edges and vertices does not change the edges that connect a variable vertex with a clause vertex. To keep the upcoming discussion as simple as possible, we omit this last step in the construction from the discussion and instead view as a high-degree vertex. As an example, consider the following yes-instance of Positive Not-All-Equal -SAT, which is also detailed in [1].
(1) |
For this instance, the construction results in the unrooted pseudo network that is shown in Figure 2(a).

Now, let be an unrooted pseudo network obtained from the above construction for some instance of Positive Not-All-Equal -SAT, and let be an orientation of . Bulteau at al. essentially thought to have proved that is a rooted pseudo network on with root that is funneled if and only if is a rooted pseudo network on with root that is funneled and tree-child. Suppose that is a rooted pseudo network on with root that is funneled. We next briefly describe some properties of . Clearly, there are arcs and for each variable vertex with . Since is funneled, the edges that join a variable vertex with a clause vertex are either all directed away from or all directed into . Moreover, there is no clause vertex that has all arcs directed into it or all arcs directed away from it in . Intuitively, can be translated into a truth assignment that nae-satisfies , where arcs directed into a clause vertex correspond to true variables and arcs directed out of a clause vertex correspond to false variables. Hence, each clause contains at least one true variable and at least one false variable. Returning to the Positive Not-All-Equal -SAT instance (1) that is described above, Figure 2(b) shows an orientation that is a rooted pseudo network on four leaves and with root that is funneled. However, this orientation is not tree-child because, for example, has no child that is a tree vertex or a leaf. In general, let be a child of in . Note that for some variable vertex . Since is also a child of , it follows that the in-degree of is at least 2. Hence, is a reticulation and, because the same argument applies to any other child of , it follows that has no child that is a tree vertex or a leaf. The erroneous statement in [1, p. 10] is the following:
“Since each clause contains at least one true variable, each clause vertex has at least one child that is a tree node.”
The correct conclusion would be that each clause vertex has at least one parent that is a tree vertex.
3.2. Funneled tree-child orientations for pseudo networks with maximum degree 5
Let be the class of rooted pseudo networks that are tree-child. In this subsection, we establish hardness for the following decision problem.
Funneled -Orientation (, pseudo)
Instance. An unrooted pseudo network with maximum degree 5.
Question. Does there exist a funneled -orientation of ?
We start by introducing three gadgets that we call the root gadget, connection gadget, and caterpillar gadget. These gadgets are shown on the left-hand side of Figure 3, 4, and 5, respectively. They will play a central role in reducing Positive Not-All-Equal -SAT to Funneled -Orientation (, pseudo), thereby establishing NP-completeness for the latter.
The next three lemmas establish properties of the root, connection, and caterpillar gadget. Throughout the proofs, we use the vertex labels as shown in Figures 3, 4, and 5, when referring to vertices of these gadgets.

Lemma 3.1.
Let be the root gadget, and let be the class of rooted pseudo networks that are tree-child. Then
-
(i)
has a -orientation whose root subdivides , and
-
(ii)
is -root-forcing.
Proof.
The orientation that is shown on the right-hand side of Figure 3 is a -orientation of the partner network of which only differs from by viewing as a non-connector leaf. Hence, has a -orientation whose root subdivides . This establishes (i). Now, assume towards a contradiction that does not satisfy (ii). Then there exists a -orientation of whose root subdivides . Then is an arc in . Furthermore, as and are non-connector leaves, and are arcs in . We next distinguish two cases.
First, suppose that has in-degree 1 and out-degree 2 in . Then, is the parent of , and and are the children of . Observe that either or is an arc in . By symmetry, we may assume that is an arc in . This implies that has in-degree 2 and out-degree 1. Moreover, since is a -orientation of , it follows that each of and has in-degree 1 and out-degree 2. In turn and are arcs in . Hence, has in-degree 2 and, thus, both children and of have in-degree 2; thereby contradicting that is a -orientation of . Second, suppose that has in-degree 2 and out-degree 1 in . By symmetry, we may assume that is the second parent of and is the child of . Again, as is a -orientation of , has in-degree 1 and out-degree 2. This implies that is an arc in and, so, and are the arcs of a directed cycle; thereby contradicting that is acyclic. Combining both cases establishes the lemma. ∎

The next corollary is a generalisation of Lemma 3.1(ii).
Corollary 3.2.
Let be the class of rooted pseudo networks that are tree-child. Let be an unrooted pseudo network that contains a pending subgraph isomorphic to the root gadget (up to the labelling of the two non-connector leaves of that gadget). If has a funneled -orientation, then the root subdivides an edge of the root gadget that is not .
Proof.
Observe that the connector leaf is either an internal vertex or a leaf of . Assume that there exists a funneled -orientation of whose root subdivides or an edge that is not an edge of the root gadget. Since there exists a directed path from to leaf of the root gadget, it follows that is an arc of . Now the proof can be established as the proof of Lemma 3.1(ii). ∎
Lemma 3.3.
Let be the connection gadget. Then satisfies the following properties:
-
(i)
There exists an orientation of that is -compatible and strongly -compatible in which each of and has in-degree 1 and out-degree 2, and and are arcs.
-
(ii)
There exists an orientation of that is -compatible and strongly -compatible in which each of and has in-degree 1 and out-degree 2, and and are arcs.
-
(iii)
Each -compatible orientation of has either arcs and or arcs and .
Proof.
Since has two connector leaves, recall that, by definition, an orientation of is obtained by assigning a direction to each edge of . The orientations of that are shown in Figure 4(b) and (c) establish (i) and (ii), respectively. Next we show that also satisfies (iii). Let be a -compatible orientation of . Due to (i), such an orientation exists. Furthermore, by definition of -compatibility, and are arcs in .
First, suppose that and are arcs in . By symmetry, we may assume that is an arc in . In turn, this implies the arc exists because, otherwise, has in-degree 0. Using the same argument again, and are also arcs in . Now and are the arcs of a directed cycle in ; thereby contradicting that is acyclic. Second, suppose that and are arcs in . By symmetry, we may again assume that is an arc in . Then, is an arc in because, otherwise has out-degree 0. Since is -compatible and has in-degree 2 and out-degree 1, has in-degree 1 and out-degree 2. Hence is an arc in . Lastly, is an arc in because, otherwise, has no child that is in or has in-degree 1 and out-degree at least 1. Now and are again the arcs of a directed cycle in ; another contradiction. By combining both cases, we deduce that, each -compatible orientation of has either arcs and , or arcs and . This establishes (iii) and, therefore, the lemma. ∎
While the root and connection gadgets are new to this paper, the caterpillar gadget is based on what Bulteau et al. [1] refer to as Rule 1. This rule is a graph operation that reduces the degree of an unrooted phylogenetic pseudo network while preserving the existence of certain orientations of . Essentially, Rule 1 is based on the observation that a vertex with two adjacent leaves and an arc that is directed into forces directions on all other edges incident with in every funneled orientation. More precisely, we have the following:
Observation 3.4.
Let be an -compatible orientation of a connector network with . Furthermore, let be a vertex of that is incident with arcs such that one arc is directed into and two arcs are directed out of . Then has in-degree 1 and out-degree .
Before establishing properties of the caterpillar gadget in the next lemma, we note that the non-connector leaf of this gadget is not necessary to show that the lemma holds. Its only purpose is to turn into a degree-5 vertex.

Lemma 3.5.
For , let be the caterpillar gadget with degree-5 vertices . Then has a unique -compatible orientation with arc . Furthermore,
-
(i)
is strongly -compatible and
-
(ii)
for each , is an arc in .
Proof.
Suppose that is an arc in an -compatible orientation of . We first show that is unique. For each , let and be the two non-connector leaves adjacent to (these leaf labels are omitted in Figure 5). Since , , and are arcs in and is -compatible, it follows from Observation 3.4 that and are arcs in . Now, since , , and are arcs in , it again follows by Observation 3.4 that and are also arcs in . Continuing in this way and noting that has, in addition to and , a third non-connector leaf , it is easily checked that there exists exactly one -compatible orientation for with arc . Moreover, this orientation, which is illustrated on the right-hand side of Figure 5, also satisfies (ii) and, as no vertex in has in-degree at least 2, satisfies (i). ∎
The proof of the following theorem uses a construction that is based on ideas presented in Bulteau et al. [1] and summarised in Section 3.1. However, we use a different root gadget and introduce a connection gadget that enables the tree-child property for orientations corresponding to nae-satisfying truth assignments.
Theorem 3.6.
Let be the class of rooted pseudo networks that are tree-child. Then Funneled -Orientation (, pseudo) is NP-complete.
Proof.
Let be an unrooted pseudo network, and let be an orientation of following Variant . Since it can be checked in polynomial time if is a -orientation of , it follows that Funneled -Orientation (, pseudo) is in NP. We next establish NP-hardness using a polynomial-time reduction from the variant of Positive Not-All-Equal (2,3)-SAT, where each variable appears exactly three times (see Dehghan et al. [4] for a proof that the problem remains NP-complete under this restriction).
Let be an instance of Positive Not-All-Equal (2,3)-SAT such that each variable appears in exactly three clauses. Let (resp. ) be the number of clauses in that contain exactly two (resp. three) distinct variables. Let be the caterpillar gadget with connector leaves , let be the root gadget with connector leaf , and let be copies of the connection gadget each with connector leaves and . We next construct an unrooted pseudo network in the following way.

-
(1)
Identify the connector leaf of with the connector leaf of . We will continue to refer to the resulting degree-2 vertex as .
-
(2)
Attach a new leaf to each connector leaf with via a new edge.
-
(3)
Let be new vertices. Furthermore, let , and let . For each and such that is a variable of , add one of the connection gadgets by identifying the connector leaf with and identifying the connector leaf with .
Let be the resulting unrooted pseudo network. We may assume for the remainder of the proof that all leaves of have a distinct leaf label and that the leaf set of is . Note that . By construction and since each variable appears exactly three times in , the maximum degree of a vertex in is 5. Figure 6 shows an example of this construction for the Boolean formula given in (1).
3.6.1.
is a yes-instance if and only if has a funneled -orientation.
Proof.
First, suppose that is a yes-instance. Let be a truth assignment that nae-satisfies each clause in . We construct a funneled -orientation of . Instead of referring to edges of , we refer to edges of the root, connection, and caterpillar gadgets as shown in Figures 3–5 when assigning directions because each edge in , except for the edges incident with with , corresponds to a unique edge in one of the gadgets that make up . We start by subdividing the edge of with a new root vertex and direct the edges of the resulting root gadget as shown on the right-hand side of Figure 3. In particular, we have the arc in . Turning to the edges of , we direct the edges of this gadget as shown on the right-hand side of Figure 5. In particular, we have the arc in . Now, for each , let , and be the three neighbours of in that are vertices of three distinct connection gadgets. We direct the edge into and, depending on , we direct the three edges , , and in one of the following two ways:
-
(i)
if , and
-
(ii)
if .
At this point, each connection gadget in has exactly one edge that is already assigned a direction and this edge is (where is the vertex as shown in Figure 4). If is directed into in , then direct the remaining edges of the connection gadget as shown in Figure 4(b) and, otherwise, as shown in Figure 4(c). This completes the assignment of directions to the edges of and, therefore, the construction of . Note that the set of vertices with out-degree 0 is and each such vertex has in-degree 1, and the unique vertex with in-degree 0 is . We next show that is indeed a funneled -orientation of . In preparation for the upcoming arguments, let
Intuitively, contains each vertex in that is a connector leaf in one of the gadgets that are used in the construction of . Furthermore, observe that each vertex in with has either in-degree 1 and out-degree 4, or in-degree 4 and out-degree 1.
We start by establishing that is acyclic. Since the orientations that are shown in Figures 3–5 are acyclic, it is sufficient to show that no vertex in lies on a directed cycle. Let be the vertex of that is a neighbour of in . Since there is no directed path from to in , it follows that does not lie on a directed cycle. Assume towards a contradiction that lies on a directed cycle for some . Then, by the observation at the end of the last paragraph, has in-degree 1 and out-degree 4 and there is a directed path from to . Hence, there is an arc for some ; a contradiction. Finally, if lies on a directed cycle for some , then some lies on a directed cycle for some ; again a contradiction. We conclude that is acyclic and, thus, is a rooted pseudo network on with root .
We next argue that is funneled, i.e., each vertex in with in-degree at least 2 has out-degree 1. It follows from the orientations shown in Figures 3–5 that each vertex of that is not in satisfies this property. Turning to the vertices in , has in-degree 1 and out-degree 1. Moreover, for each , has in-degree 1 if and out-degree 1 if . Finally, recall that each clause in has either two or three distinct variables. Since nae-satisfies , it follows that has either in-degree 1 and out-degree is 2, or in-degree 2 and out-degree 1 for each . Hence, is funneled.
We complete this part of the proof by showing that is tree-child. Since has in-degree 1 and out-degree 1 in , it follows from Lemmas 3.1(i), 3.3(i)–(ii), and 3.5(i) that each vertex of that is not a leaf and not in has a child that is a leaf or a tree vertex. Furthermore, has a unique child that is a tree vertex, and each with is adjacent to a leaf . Now consider a vertex in with . Each child of is a vertex of a connection gadget. The edges of such a connection gadget are directed as shown in Figure 4(c). Importantly, is a tree vertex. We conclude that is a pseudo network on that is funneled and tree-child.
Second, let be a funneled -orientation of with root . By Corollary 3.2, subdivides an edge of that is not . Moreover, since is funneled and there is a directed path from to each leaf in , and are arcs in . By Lemma 3.5(ii), it now follows that is an arc in for each . Hence, each has a parent and a child in . We now define a truth assignment as follows. For each , we set if is a tree vertex and if is a reticulation in .
Towards a contradiction, assume that does not nae-satisfy each clause in . We distinguish two cases. First, suppose that there exists a clause or for some whose variables are all set to under . We continue by assuming that and note that the same argument applies for when . As and are tree vertices in , each arc that joins a vertex in with a vertex of a connection gadget is directed towards the latter. Now consider a directed path in from a vertex in to , and let be the last arc on . Since is a -orientation, the direction of the arcs restricted to a single connection gadget in is -compatible. Hence, Lemma 3.3(iii) implies that is directed into . It now follows that has in-degree 3 and out-degree 0; a contradiction. Second, suppose that there exists a clause whose variables are all set to under . Then, by an analogous argument, has in-degree 0 and out-degree-3; also a contradiction. Thus, nae-satisfies each clause in , that is, is a yes-instance. ∎
Noting that can be constructed in polynomial time and has a size that is polynomial in and , we conclude the proof of the theorem. ∎
3.3. Hardness remains for phylogenetic networks
Theorem 3.6 shows that the erroneous result by Bulteau et al. [1, Cor. 5] can be corrected by introducing additional gadgets. However, the construction results in unrooted and rooted pseudo networks. In this short subsection, we outline two modifications to the construction that avoid degree-2 vertices, thereby showing that the following decision problem is NP-complete for when is the class of tree-child networks.
Funneled -Orientation ()
Instance. An unrooted phylogenetic network such that each non-leaf vertex has degree 3 or 5.
Question. Does there exist a funneled -orientation of ?
Let be an unrooted pseudo network on that is constructed from an instance of Positive Not-All-Equal (2,3)-SAT as described in the proof of Theorem 3.6. By construction, each non-leaf vertex of has degree 2, 3, or 5. Moreover, the degree-2 vertices in are and for each such that is a clause with exactly two distinct variables. Consider a 2-clause with and . Observe that can be viewed as an XOR constraint on and , that is, exactly one of the two variables is set to in a truth assignment that nae-satisfies . In , this is simulated by the clause vertex being adjacent to two vertices that are part of two distinct connection gadgets, say and . Without loss of generality, we assume that the vertex of has been identified with in the construction of . We refer to the operation of deleting all vertices of except for and , and then identifying with as the 2-clause suppression of . See Figure 7 for an example of a -clause suppression.

Lemma 3.7.
Let be the class of rooted pseudo networks that are tree-child. Let be an unrooted pseudo network that is constructed from an instance of Positive Not-All-Equal (2,3)-SAT as described in the proof of Theorem 3.6. Let be the unrooted phylogenetic network obtained from by suppressing and applying the 2-clause suppression to each clause vertex that has degree 2 in . Then has a funneled -orientation if and only if has a funneled -orientation.
Proof.
We first note that if an unrooted phylogenetic network has a -orientation, then this orientation is necessarily a tree-child network (without any vertex of in-degree 1 and out-degree 1). Furthermore, by construction, observe that, except for , each edge in corresponds to a unique edge in . In particular, for each -clause , the edge in , where is a vertex of a connection gadget in corresponds to the edge . These two edges are indicated by thick lines in Figure 7.
Now, first, suppose that is a funneled -orientation of . By Corollary 3.2, the root of subdivides an edge of the root gadget such that . Let be the orientation obtained from by subdividing in with a root vertex, directing the edge into , and then directing each remaining edge as it is directed in . Let be a -clause. Then one of and , say , is a tree vertex and is a reticulation in and . It follows that each with is a tree vertex (resp. reticulation) in if and only if is a tree vertex (resp. reticulation) in , and the same applies to . Noting that the unique child of in is a tree vertex, it now follows that, as is a funneled -orientation of , is such an orientation of .
Second, suppose that is a funneled -orientation of . Let be the edge in that contains the root in . By Corollary 3.2, is an edge of the root gadget such that . Hence, is also an edge of . We now obtain an orientation for by subdividing with , directing the two edges that are incident with away from and directing the remaining edges as follows. For each edge such that and is in and , has the same direction in as in . Furthermore, for each leaf that is in but not in , the unique edge incident with is directed towards . Since there exists a directed path from to each leaf in , it follows that contains the arc . For the same reason, and, consequently, are arcs in . Now, the only edges in that have not been assigned a direction yet are edges of the connection gadgets whose non-leaf vertices lie on paths between and that do not contain for each 2-clause . By construction, is a tree vertex in if and only if is a reticulation in . Assume that is a tree vertex in . Then is an arc in and , where is the neighbour of in a connection gadget . Furthermore, by Lemma 3.3(iii), is an arc in , where is the neighbour of in . In , there exists a connection gadget whose non-leaf vertices lie on paths from to that do not contain . Since is a degree-2 vertex in and is an arc of (recall that in ), we direct the edges of as shown in Figure 4(c) which yields a strongly -compatible orientation for . Then has in-degree 1 and out-degree 1 in , and has a child in that is a tree vertex. Replacing Figure 4(c) with Figure 4(b), the case for when is a reticulation is similar and omitted. As is a funneled -orientation of , it now follows that we obtain a funneled -orientation of by repeating the above steps for each -clause. ∎
Theorem 3.8.
Let be the class of tree-child networks. Then Funneled -Orientation () is NP-complete if each non-leaf vertex has degree 3 or 5.
4. Hardness for Other Classes of Funneled Phylogenetic Networks
In this section, we establish NP-completeness for the following decision problem.
Funneled -Orientation ()
Instance. An unrooted phylogenetic network such that each non-leaf vertex has degree 5.
Question. Does there exist a funneled -orientation of ?
In particular, we consider the four classes of tree-child, tree-sibling, normal, and reticulation-visible networks and both rooting variants and .


We start by establishing properties of the degree- root gadget that is shown on the left-hand side of Figure 8.
Lemma 4.1.
Let be the degree-5 root gadget. Let be the class of tree-child networks, and let the class of funneled phylogenetic networks. Then
-
(i)
has a funneled -orientation such that subdivides ,
-
(ii)
is -root-forcing,
-
(iii)
has a funneled -orientation such that , and
-
(iv)
is -root-forcing.
Proof.
The orientation that is shown in Figures 8 (right-hand side) and 9 is a funneled -orientation and a funneled -orientation, respectively, of the partner network of which only differs from by viewing as a non-connector leaf. This establishes (i) and (iii). Now, assume towards a contradiction that does not satisfy (ii). Then there exists a -orientation of such that subdivides . Then is an arc in . Furthermore, each vertex in has in-degree 1 and out-degree 4 since each such vertex has at least two adjacent leaves and is a -orientation. We next distinguish two cases.
First, suppose that has in-degree 1 and out-degree 4 in . Then and are arcs in . Since either or is an arc in , or has in-degree at least and out-degree at least ; thereby contradicting that is a -orientation of . Second, suppose that has in-degree 4 and out-degree 1. By symmetry, we may assume that is an arc in . Since and have in-degree 1 and out-degree 4, it follows that , and are arcs in . Now, as has in-degree 1 and out-degree 4, we have that either and , or and are arcs in . In the former case, we get the directed cycle . Hence, we may assume that and are arcs in . As has in-degree 4 and out-degree 1, is an arc in . This implies that contains the directed cycle ; thereby again contradicting that is a -orientation of . Combining both cases establishes (ii).
Lastly, assume towards a contradiction that does not satisfy (iv). Then there exists a -orientation of such that is chosen to be the root of . Then is an arc in . The proof of (iv) can now be established in the same way as the proof of (ii). ∎
To establish the next theorem, the construction from Positive Not-All-Equal (2,3)-SAT as established in the last section cannot be used. Instead, we reduce from Positive 1-in-3 SAT, replace the binary root gadget as shown in Figure 3 with a degree-5 root gadget, omit the connection gadget as shown in Figure 4 altogether, and use clause vertices that each have two adjacent leaves and three adjacent variable vertices. Roughly speaking, the construction results in a phylogenetic network such that any funneled orientation of the constructed unrooted phylogenetic network enforces all clause vertices to be tree vertices by Observation 3.4.
Theorem 4.2.
Let be the class of rooted phylogenetic networks. Then Funneled -Orientation is NP-complete.
Proof.
Let be an unrooted pseudo network, and let be an orientation of following Variant . Since it can be checked in polynomial time if is a -orientation of , it follows that Funneled -Orientation is in NP. We next establish NP-hardness using a polynomial-time reduction from the variant of Positive 1-in-3 SAT, where each variable appears exactly three times (see, e.g., Kratochvíl [10, Cor. 1] for a proof that the problem remains NP-complete under this restriction).
Let be an instance of Positive 1-in-3 SAT such that each variable appears in exactly three clauses. Let be the caterpillar gadget with connector leaves , and let be the degree-5 root gadget with connector leaf . We next construct an unrooted phylogenetic network in the following way.
-
(1)
Identify the connector leaf of with the connector leaf of , and suppress the resulting degree-2 vertex.
-
(2)
Attach a new leaf to each connector leaf with via a new edge.
-
(3)
Let be new vertices, let , and let . For each and such that is a variable of , add the edge . Furthermore, attach two new leaves to each .
We may assume for the remainder of the proof that all leaves of have a distinct label and that the leaf set of is . By construction, each non-leaf vertex in has degree 5. Figure 10 shows and example of this construction for the following Boolean formula
(2) |

To complete the proof, we follow a similar approach as in the proof of Theorem 3.6.
4.2.1.
is a yes-instance if and only if has a funneled -orientation.
Proof.
First, let be a truth assignment that sets exactly one variable of each clause in to . We next construct a funneled -orientation of . As in the proof of Theorem 3.6, instead of referring to edges of , we sometimes refer to edges of and when assigning a direction. We start by subdividing the edge of with a new root vertex and direct the edges of the resulting root gadget as shown on the right-hand side of Figure 8. Noting that the edge in corresponds to the edge in , this assignment of directions to edges implies that the arc is in . We next direct the edges of (except for the edge which also corresponds to the edge in and has already been directed) as shown on the right-hand side of Figure 5. Now, for each , let , , and be the clause vertices that are neighbours of in . Depending on we direct the three edges , , and as follows.
-
(i)
, , and if , and
-
(ii)
, , and if .
At this point, the only edges of that have not yet been assigned a direction are incident with a leaf and are directed towards that leaf in . By construction of , the only vertex with in-degree 0 is and the set of vertices with out-degree 0 is . Since sets exactly one variable of each clause to , each clause vertex is a tree vertex in with in-degree 1 and out-degree 4. Furthermore, each variable vertex is either a tree vertex with in-degree 1 and out-degree 4, or a reticulation with in-degree 4 and out-degree 1 in . It now follows from Lemmas 3.5 and 4.1(i) that is funneled. To see that is also acyclic, we can show that no vertex in lies on a directed cycle of by using the same ‘acyclicity argument’ as that in the proof of 3.6.1. Hence, is a funneled -orientation of .
Second, let be a funneled -orientation of . Using the degree-5 root gadget instead of the root gadget that is shown in Figure 3, and Lemma 4.1(ii) instead of Lemma 3.1(ii) in the proof of Corollary 3.2, it is straightforward to show that subdivides an edge of that is an edge of and not . Hence, is an arc in . It then follows by Lemma 3.5(ii) that, for each , is an arc in . We now define a truth assignment as follows. For each , we set if is a tree vertex and if is a reticulation in . Assume towards a contradiction that there exists a clause with and such that does not set exactly one variable of to . We distinguish three cases:
-
(i)
If , then and are tree vertices and, thus, , and are arcs in . Since the other two vertices adjacent to are leaves, it follows that has in-degree 3 and out-degree 2; a contradiction to being a funneled orientation.
-
(ii)
If and , then are tree vertices and is a reticulation and, thus, , and are arcs in . But then, due to the two adjacent leaves of , it follows that has in-degree 2 and out-degree 3; again a contradiction.
-
(iii)
If , then and are reticulations and, thus, , and are arcs in . But then, due to the two adjacent leaves of , it follows that has in-degree 0 and out-degree 5; a final contradiction.
Hence, sets exactly one variable of each clause in to . ∎
We conclude the proof of the theorem by noting that can be constructed in polynomial time and has a size that is polynomial in and . ∎
We now turn to Funneled -Orientation , where is again the class of rooted phylogenetic networks and the root is placed following Variant . More precisely, given an unrooted phylogenetic network , this variant chooses a vertex of to be the root and assigns a direction to each edge. Consider the degree-5 root gadget as shown on the left-hand side of Figure 8. Suppose that contains as a pending subgraph. First, by Lemma 4.1(iv), any funneled -orientation of chooses a vertex of that is not as the root. Second, Lemma 4.1(iii) shows that has a funneled -orientation. The next corollary, which is a slight strengthening of a result by Bulteau et al. [1, Cor. 4], shows NP-hardness for graphs with maximum degree 5, now follows from an argument that is analogous to that in the proof of Theorem 4.2.
Corollary 4.3.
Let be the class of rooted phylogenetic networks. Then Funneled -Orientation is NP-complete.
We remark that the root gadget as shown in Figure 3 is not -root-forcing for each and can therefore not be used to establish Theorem 4.2 or Corollary 4.3 for networks whose internal vertices have degree 3 or 5.
The following lemma is the key ingredient in showing that Funneled -Orientation is NP-complete for the class of tree-child networks as well as for any class of rooted phylogenetic networks that contains the class of tree-child networks.
Lemma 4.4.
Let be the class of rooted phylogenetic networks, let be the class of tree-child networks. Furthermore, let be an unrooted phylogenetic network that is constructed from an instance of Positive 1-in-3 SAT as described in the proof of Theorem 4.2. Then, for each , has a funneled -orientation if and only if has a funneled -orientation.
Proof.
We establish the proof for . The proof for is similar and omitted. Clearly, any funneled -orientation of is also a funneled -orientation of . So suppose that is a funneled -orientation of . The only vertex of that is not adjacent to a leaf is the vertex of the degree-5 root gadget. Since is an arc of as argued in the last paragraph of the proof of 4.2.1, it follows that is a child of . Noting that has two adjacent leaves, it now follows that has in-degree 1 and out-degree 4 because is funneled. Thus has a child that is a tree vertex and is a funneled -orientation of . ∎
Since the last lemma does not only hold for being the class of tree-child networks but also for any class of rooted phylogenetic networks that contains the class of tree child networks, the next corollary is an immediate consequence of Theorem 4.2, Corollary 4.3, and Lemma 4.4.
Corollary 4.5.
Let be the class of tree-child, tree-sibling, or reticulation-visible networks. Then Funneled -Orientation is NP-complete for each
Lastly, we establish NP-completeness of Funneled -Orientation for the class of normal networks. Let be an edge of an unrooted phylogenetic network on . We refer to the operation of subdividing with a new vertex and adding the three edges , , and such that , , and are leaves that are not contained in as attaching three leaves to . Now, let be an unrooted phylogenetic network that is constructed from an instance of Positive 1-in-3 SAT as described in the proof of Theorem 4.2. Obtain an unrooted phylogenetic network from by attaching three leaves to each edge that is incident with a vertex in and not incident with a leaf as well as to the edges , , and , where , , and are the vertices as shown on the left-hand side of Figure 8. We refer to as the network pair associated with .
Lemma 4.6.
Let be the class of rooted phylogenetic networks, and let be the class of normal networks. Furthermore, let be the network pair associated with an instance of Positive 1-in-3 SAT. Then, for each , has a funneled -orientation if and only if has a funneled -orientation.
Proof.
Let be the subset of edges of that precisely contains each edge that is subdivided in obtaining from by attaching three leaves. Clearly, each edge that is in and not in is also an edge of . Moreover, for each edge in , there exist edges , and in , where , and are leaves. We refer to those five edges as five edges associated with in . Let be a -orientation of . If , then it follows from the last paragraph of the proof of 4.2.1 that is an arc of . Similarly, if , then follows from the paragraph prior to Corollary 4.3 that is again an arc in . We freely use the existence of the arc throughout the remainder of the proof.
First, suppose that has a funneled -orientation . Let be the edge of that is subdivided in obtaining from . Apply the following three steps to obtain an orientation of .
-
(1)
If , then subdivide of with a vertex and direct the two edges incident with so that has in-degree 0. On the other hand, if is an edge in , recall that is a path of length two in such that is adjacent to three leaves , and . Then subdivide with a vertex and let and be arcs in .
-
(2)
Direct each edge of that is also an edge in such that it has the same direction as in .
-
(3)
For each edge in , direct the five edges associated with in in the following way. If is an arc in , then , and are arcs in and, otherwise, , and are arcs in .
Since is a funneled -orientation of , it follows that is a funneled -orientation of . Moreover, as is an arc of , there is a directed path from to of length two in whose middle vertex is a tree vertex. Since is funneled and, each vertex except for is adjacent to a leaf, it follows that is a funneled -orientation for that is also tree-child. To see that is in fact a funneled -orientation of , it remains to argue that has no shortcut. Let be a reticulation of , and let and be two distinct parents of . Since it is straightforward to check that and are two vertices of each with three adjacent leaves. Assume that there is a directed path from to in . Then there exists an arc such that is not a leaf and . This implies that has in-degree 0 and out-degree 4; a contradiction. It now follows that is not a shortcut and, by a symmetric argument, is not a shortcut either. Thus, is a funneled -orientation of .
Second, suppose that has a funneled -orientation . The root of is a vertex of and . Obtain an orientation of by following Steps (2) and (3) so that Step (3) is applied to each edge in . Using the same argument as in the previous paragraph, it follows that is a funneled -orientation of .
Third, suppose that has a funneled -orientation . Let be the edge of that is subdivided in obtaining from . Obtain an orientation of by reversing Steps (1)–(3) from above as follows.
-
(1’)
If is an edge of , subdivide in with a vertex and direct the two edges incident with such that has in-degree 0. On the other hand, if subdivides one of the five edges associated with an edge , then subdivide in with a vertex and direct the two edges incident with such that has in-degree 0.
-
(2’)
Direct each edge of that is also an edge in such that it has the same direction as in .
-
(3’)
For each in , consider the five edges associated with in . If , and are arcs in , then is an arc in and, if , and are arcs in , then is an arc in .
Since is funneled, one of the two cases described in Step (3’) applies. Furthermore, since is a funneled -orientation of , it follows by construction that is a funneled -orientation of .
Lastly, suppose that has a funneled -orientation . Let be the root of . We first show that is a vertex of . Assume towards a contradiction that is not a vertex of . Let be the unique edge of , where is as shown on the left hand-side of Figure 8 and is the unique internal vertex introduced by attaching three leaves to . By arguments similar to the ones used to establish Lemma 4.1(ii), it follows that is an arc in . Hence is a vertex that is introduced in attaching three leaves to or . By symmetry, we may assume that is one of the four vertices and , where is the unique internal vertex introduced by attaching three leaves to . Regardless of which vertex is chosen as , and are arcs in because is funneled. Furthermore, as and have in-degree 1 and out-degree 4, it follows that and, in turn, are arcs in . Thus, is a shortcut and therefore is not a -orientation of ; a contradiction. Hence, is a vertex of , and we obtain an orientation of by following Steps (2’) and (3’) such that Step (3’) is applied to each edge in . Then it follows again by construction that is a funneled -orientation of . ∎
Corollary 4.7.
Let be the class of normal networks. Then Funneled -Orientation is NP-complete for each .
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