Not all phylogenetic networks are leaf-reconstructible

Unrooted phylogenetic networks are graphs used to represent reticulate evolutionary relationships. Accurately reconstructing such networks is of great relevance for evolutionary biology. It has recently been conjectured that all unrooted phylogenetic networks for at least five taxa can be uniquely reconstructed from their subnetworks obtained by deleting a single taxon. Here, we show that this conjecture is false, by presenting a counter-example for each possible number of taxa that is at least 4. Moreover, we show that the conjecture is still false when restricted to binary networks. This means that, even if we are able to reconstruct the unrooted evolutionary history of each proper subset of some taxon set, this still does not give us enough information to reconstruct their full unrooted evolutionary history.


Introduction
The reconstruction conjecture, introduced in 1941 by Kelly and Ulam (see Bondy and Hemminger 1977), conjectures that each graph with at least three vertices is uniquely reconstructable from its multiset of vertex-deleted subgraphs. Despite more than seven decades of research, the conjecture is still open.
Recently, a variant of this conjecture was introduced that is relevant for the field of phylogenetics, the study of evolutionary relationships. Such relationships among a set X of entities (e.g. biological species or languages) are traditionally described by a tree with no degree-2 vertices and its leaves bijectively labelled by the elements of X ; this is called a phylogenetic tree on X . More recently, evolutionary histories are more and more often described by phylogenetic networks (Bapteste et al. 2013), which are basically (directed or undirected) graphs with their leaves bijectively labelled by the elements of X . These networks are able to describe more complex evolutionary relationships than trees.
To find out whether it may be possible to accurately reconstruct phylogenetic networks, an important question to answer is which substructures uniquely define a phylogenetic network. For example, although there is much research directed at reconstructing rooted phylogenetic networks from embedded trees [see e.g. Van Iersel et al. (2016) and Whidden et al. (2013)], these trees do not uniquely define a network [see e.g. Pardi and Scornavacca (2015)]. Hence, no method based on embedded trees can be guaranteed to reconstruct the right network, even when it gets error-free and complete trees as input. Moreover, it has recently been shown that rooted phylogenetic networks also cannot be reconstructed uniquely from their subnetworks obtained by deleting one or more leaves and transforming the result into a valid rooted phylogenetic network (Huber et al. 2014). A similar reconstruction question for pedigrees has also been answered negatively (Thatte 2008).
Here, we focus on unrooted phylogenetic networks, which are undirected graphs with leaves labelled by the elements of some taxon set X . Although real evolutionary histories are rooted, it is not always possible to identify the root location and the directions of all arcs. Therefore, just like unrooted phylogenetic trees are studied in addition to rooted phylogenetic trees, unrooted phylogenetic networks are studied increasingly. Van Iersel and Moulton (2018) studied reconstructing such networks from their X -deck, which consists of the graphs obtained by deleting a single taxon from the network (see Fig. 1 for an example). Several promising results were obtained, including a proof that all phylogenetic trees and all decomposable networks (i.e. networks that can be decomposed into two nontrivial subnetworks by deleting a single edge) are reconstructable from their X -deck, assuming |X | ≥ 5. Moreover, the same was shown for networks that can be turned into a tree by deleting at most four edges, and for all networks with sufficiently many leaves. The only known networks not reconstructible from their |X |-decks were ones for which |X | ≤ 4. It was conjectured that all unrooted phylogenetic networks on X , with |X | ≥ 5, can be uniquely reconstructed from their X -deck.
Here, we show that this conjecture is false. To do so, we present, for each finite set X containing at least four elements, two unrooted phylogenetic networks on X that are not isomorphic but have the same X -deck. Moreover, we also give binary networks with x 4 x 2 x 3 x 4 x 1 x 1 x 2 Example of a phylogenetic network N and its X -deck these properties, hence showing that the conjecture restricted to binary networks is still false. These results can be seen as the unrooted counterpart to the results from Huber et al. (2014). However, we also note that there are important differences between the rooted and unrooted case, which make it impossible to directly transform the rooted counter-examples to the unrooted case, see Sect. 2.1.
Our result may have consequences for developing "supernetwork" methods, which attempt to reconstruct phylogenetic networks from subnetworks. Supertree methods work well for phylogenetic trees, which can be explained from the fact that a phylogenetic tree is uniquely determined by its induced set of four-leaved trees (or three-leaved trees in the case of rooted trees). Since phylogenetic networks are not uniquely determined by their subnetworks, developing supernetwork methods will be significantly more challenging than in the tree-case, even for unrooted networks.
The structure of the paper is as follows. We start off by giving formal definitions related to phylogenetic networks and binary sequences, which are central to the construction of our counter-examples, in Sect. 2. In Sect. 2.1, we explain why unrooting the counter-example for the rooted case from Huber et al. (2014) does not give a counter-example for the conjecture considered here. Then, in Sect. 3, we present our counter-examples for the unrooted, non-binary case. Finally, in Sect. 4 we show how these can be transformed into counter-examples for the unrooted, binary case.

Preliminaries
A phylogenetic tree on X is an undirected simple tree, with no degree-2 vertices, such that each leaf is bijectively labelled by an element from X . A biconnected component of a graph is a maximal 2-edge-connected subgraph and it is called a blob if it contains at least two edges. Let X be a finite set with |X | ≥ 2, and let N be an undirected simple graph in which the leaves (degree-1 vertices) are bijectively labelled by the elements of X . We say N is an unrooted phylogenetic network on X if contracting each blob into a single vertex gives phylogenetic tree (or equivalently, each cut-edge induces a unique partition of the leaves). In addition, we say that N is binary if every vertex has degree 1 or 3. In what follows, we will refer to unrooted phylogenetic networks as networks for short.
Let G and H be two partially labelled undirected multigraphs with the same label set, such that |V We say that f is an isomorphism between G and H if it is both label-preserving (that is, vertex a ∈ V (G) has label l if and only if f (a) has label l) and edge-preserving (that is, for any a, b ∈ V (G) the number of edges between a and b in G is equal to the number of edges between f (a) and f (b) in H ). We say G and H are equivalent, denoted G ∼ H , if there is an isomorphism between G and H .
Given an undirected multigraph G with no vertices of degree 2, and a vertex a ∈ V (G), we denote by G a the undirected multigraph derived from G by deleting a and all incident edges, and then suppressing any degree-2 vertices. We say G a is derived from G by removing the vertex a. For a label x, we may write G x to refer to G a , where a is the unique vertex in G with label x.
Given a network N on X , an X -reconstruction of N is a network N on X such that N x ∼ N x for all x ∈ X . We call a phylogenetic network N leaf-reconstructible if N ∼ N for every X -reconstruction N of N . That is, all X -reconstructions of N are isomorphic to each other.
It was conjectured in Van Iersel and Moulton (2018) that all unrooted phylogenetic networks with 5 or more leaves are leaf-reconstructible. (We note that phylogenetic trees on 5 or more leaves are leaf-reconstructible, as it is clearly possible to reconstruct every quartet in the tree.) In this paper, we show that the conjecture is false. More precisely, we will show that for each r ≥ 4, there exist binary unrooted phylogenetic networks N and N on X with |X | = r , such that N N , but N x ∼ N x for all x ∈ X . Thus, N and N are not leaf-reconstructible. 1 Finally, for an integer k, let [k] denote the set {1, 2, . . . , k}. Huber et al. (2014) showed that for any r ≥ 3, there exist rooted binary networks M and M on X with |X | = r , such that M M , but M| X ∼ M| X for any strict subset X of X . Here M| X denotes the subnet of M induced by X ; roughly speaking, M| X is derived from M by deleting any vertices not on a directed path from the root to an element of X , then suppressing any degree-2 vertices and parallel arcs [see Huber et al. (2014) for full details]. We note that one cannot create a counterexample to the leaf-reconstruction conjecture by simply taking the directed networks M, M given by Huber et al. and replacing them with their underlying undirected graphs G, G . A key observation here is that Fig. 2 The underlying undirected networks N and N of two rooted networks which, in Huber et al. (2014), were shown to have the same induced subnetworks for any strict subset of the leaf set X = {a, b, c, d}. We observe that N d N d , since the shortest path between a and b has length 7 in N d and length 6 in N d . Hence, these networks can not be used as counter-example for the leaf-reconstruction conjecture for any x ∈ X , the network M| X \{x} may have many fewer vertices and arcs than M, whereas the graph G x has at most two fewer edges and two fewer vertices than G. Indeed, Fig. 2 gives two networks N , N on X = {a, b, c, d} that correspond to the undirected versions (after suppressing degree-2 vertices) of the networks given by Huber et al. for r = 4. We observe that the distance between a and b is 7 in N d , and 6 in N d , and thus these networks do not have the same X -deck. Thus the approach of Huber et al. cannot be naively used to give our result. However, the two papers do use similar ideas, in particular the use of binary sequences in the construction of a network (see Sect. 2.2).

Binary sequences
Given an alphabet Σ, let w ∈ Σ * be a sequence of elements with elements drawn from Σ. If Σ = {0, 1} then we call w a binary sequence. The length of the sequence w, denoted l(w), is the number of elements in w. We write w i to denote the i'th element of w. We often write e 1 e 2 . . . e l to denote the sequence w such that l(w) = l and w i = e i for each i ∈ [l]. (Thus, for example, 1011 denotes the length-4 binary sequence whose second element is 0 and whose first, third and fourth elements are 1.) Given a binary sequence w, the weight of w is the number of 1's in w. For an integer l, we write B l to denote the set of binary sequences of length l. Given a sequence w ∈ B r and i ∈ [r ], let w i be the sequence derived from w by replacing the i'th element with 1 − w i (for example, if w = 1001 and i = 3, then w i = 1011).
Central to the proof of our result is the idea that for a binary sequence w, one needs to know all elements of w in order to decide whether w has odd or even weight. (Note that here and in the rest of the paper, we consider a sequence of weight 0 to have even weight.) For some integer r , consider the set B even r of all length-r binary sequences of even weight, and the set B odd r of all length-r binary sequences of odd weight. Given a length-r binary sequence w and integer i ∈ [r ], let w −i denote the sequence on {0, 1, * } derived from w by replacing the i'th element with * . Then for each w ∈ B even r , there exists a sequence w ∈ B odd r such that (w ) −i = w −i (indeed, w i is such a sequence). For a set of sequences S and i ∈ [r ], let B −i r = {w −i ; w ∈ S}. Then it follows that for each i ∈ [r ], the sets (B odd r ) −i and (B even r ) −i are the same. We will use this concept to guide our construction of two networks N even and N odd on a set X = {x 1 , . . . , x r }. Roughly speaking, N even can be thought of as a representation of B even r , and N odd can be thought of as a representation of B odd r . Then we will be able to show that (N even ) x i and (N odd ) x i are equivalent, while originally N even and N odd are different.

A non-binary example
In order to demonstrate the main concepts of our construction, we first give a construction using non-binary graphs. In the next section, we will construct an example with binary phylogenetic networks, using these non-binary graphs as a guide.
For some integer r ≥ 4, let X denote the set of labels {x 1 , . . . , x r }. We will construct two graphs M even and M odd , in which the leaves are bijectively labelled by the elements of X . As in the previous section, let B even r denote the set of all length-r binary sequences of even weight, and let B odd r denote the set of all length-r binary sequences of odd weight.
The graph M even is constructed as follows. For each i ∈ [r ], let M even contain vertices v i,0 and v i,1 , and a leaf labelled with x i , such that x i is adjacent to v i,0 . 2 For each w ∈ B even r , let M even contain a vertex u w . For each w ∈ B even r and i ∈ [r ], let u w be adjacent to v i,0 if w i = 0, and let u w be adjacent to v i,1 if w i = 1. This completes the construction of M even (see Fig. 3a).
The construction of M odd is identical to that of M even , except that we have a vertex u w for each w ∈ B odd r rather than each w ∈ B even r . For completeness, the full construction is as follows: For each i ∈ [r ], let M odd contain vertices v i,0 and v i,1 , and a leaf labelled with This completes the construction of M odd (see Fig. 3b).

Lemma 1 M even and M odd are not equivalent.
Proof Suppose for a contradiction that M even and M odd are equivalent, and let f : be an isomorphism between M even and M odd . Let 0 denote the all-0 sequence from B even r . Observe that for each i ∈ [r ], the distance between u 0 and x i is 2 (as both u 0 and Observe that by construction of M odd (in particular, the fact that it is a bipartite graph with one side consisting of vertices v j,0 or v j,1 ), the distance between any leaf x i and any vertex v j,0 or v j,1 is odd. It follows that f (u 0 ) must be the vertex u w , for some w ∈ B odd r (any other vertex is either a leaf, which has distance 0 from itself, or has odd distance from any leaf). However, for any w ∈ B odd r there exists i ∈ [r ] such that w i = 1, and so u w is not adjacent to v i,0 . As v i,0 is the only vertex adjacent to x i , it follows that the distance between u w and x i is greater than 2, and so f (u 0 ) = u w .
As there is no choice for f (u 0 ) that satisfies the conditions of an isomorphism, we have that there is no possible isomorphism between M even and M odd , and so M even and M odd are not equivalent.
Proof Observe that v i,0 and v i,1 each have 2 r −2 ≥ 4 neighbors in M even not including x i (as |B even r | = 2 r −1 and exactly half of the sequences in B even r have 1 as their i'th element). Also any vertex u w has r ≥ 4 neighbors in M even . It follows that if x i is deleted from M even , the remaining graph has no vertices of degree 2, and thus (M even ) x i is exactly M even with x i deleted. By a similar argument, By construction, f is a bijective function from It remains to show that f is label-preserving and edge-preserving. As f is the identity on all labelled vertices, f is label-preserving. As

A binary example
In this section, we show how to construct two binary networks on X that are Xreconstructions of each other but are not equivalent, for |X | ≥ 4. (An example of two such networks for the case when |X | = 4 is given in Fig. 7.) This is enough to show that networks on r ≥ 4 leaves are not leaf-reconstructible. Given the non-binary networks M even and M odd constructed in the previous section, we proceed to construct two graphs G even and G odd in the following way. For each binary sequence w ∈ B r , u w will be expanded into a caterpillar Cat(w) (details of the construction are given below). Each vertex v i,h will be expanded into a lexicographic tree Lex(i, h) even or Lex(i, h) odd (defined below). These subgraphs contain leaves denoted z w,i , for w ∈ B r and i ∈ [r ]. Two subgraphs Cat(w) and Lex(i, h) even (or Cat(w) and Lex(i, h) odd ) will share a vertex z w,i if and only if w i = h (analogous to how in M even and M odd , the vertices u w and v i,h are adjacent if and only if w i = h).
Similarly to M even and M odd , we will show that G even and G odd are not equivalent, but that they become equivalent if a single leaf x i is deleted.
We note that G even and G odd are not technically networks, because while they have maximum degree 3, they contain some vertices of degree 2 (in particular, every vertex z w,i has degree 2). In the last part of this section, we will produce two networks N even and N odd from G even and G odd .
We now define the two types of tree that will be used in our construction.
Definition 1 For any sequence w ∈ B r , the caterpillar Cat(w) is the tree with internal vertices u w and y w,i for each i ∈ [r − 3], leaves z w,i for each i ∈ [r ], and edges u w z w,1 , u w z w,2 , u w y w,1 , y w,r −3 z w,r −1 , y w,r −3 z w,r , and y w,i z w,i+2 , y w,i y w,i+1 for each 1 ≤ i ≤ r − 4.
See Fig. 4 for an example. Observe that all internal vertices of Cat(w) have degree 3.

Observation 1 Given sequences w, w ∈ B r , the trees Cat(w) and Cat(w ) are equivalent. In particular, there exists an isomorphism f between Cat(w) and Cat(w ) such that f (u w ) = u w and f (z w,i ) = z w ,i for all i ∈ [r ].
Definition 2 Given a set S of binary sequences such that |S| = 2 t for some positive integer t, and i ∈ [r ], the lexicographic tree Lex(i, S) is a fully balanced binary tree with leaves z w,i for w ∈ S. All non-leaf vertices have degree 3 except for a single vertex, called the root, of degree 2, and all leaves are of distance exactly t from the root. Moreover, the leaves are arranged in such a way that there exists a depth-first search of the vertices of Lex(i, S) that traverses the leaves z w,i in lexicographic order with respect to w. (Note that this uniquely determines Lex(i, S).)

Also, for any h ∈ {0, 1} there exists an isomorphism f between Lex(i, h) even and Lex
Proof Observe that the root of a lexicographic tree is unique, as it is the only vertex of degree 2. Then for any integer l and leaf z w, j in a lexicographic tree, we may define the depth-l ancestor of z w, j as follows. The depth-l ancestor of z w, j is the unique vertex on a path between z w, j and the root, that has distance l from z w, j . Note that we count the root itself as a depth-(r − 2) ancestor of every leaf, and each leaf is the depth 0 ancestor of itself. Moreover, because a lexicographic tree is fully balanced, if a vertex a is the depth-l ancestor of one leaf and the depth-l ancestor of another leaf then l = l .
In order to prove the first claim, we first show that for any two sequences w, w ∈ (B even r ) j:h and integer l, the leaves z w, j , z w , j share a depth-l ancestor in Lex( j, h) even if and only if z w i , j , z w i , j share a depth-l ancestor in Lex( j, h) odd . Indeed, it is easy to see that z w, j , z w , j share a depth-l ancestor if and only if w, w agree on the first r − 2 − l elements not including j. But if w, w agree on these elements then so do w i , w i , and so z w i , j , z w i , j also share a depth-l ancestor.
Thus, we may define a bijective function f : V (Lex( j, h) even ) → V (Lex( j, h) odd ) as follows. For any vertex a ∈ V (Lex( j, h) even ) with distance r − 2 − l from the root, choose any sequence w ∈ (B even r ) j:h such that a is a depth-l ancestor of z w, j , and let f (a) be the depth-l ancestor of z w i , j in Lex( j, h) odd . Observe that f is well-defined, since we have just shown that if two leaves z w, j , z w , j share a as a depth-l ancestor, then z w i , j , z w i , j also have the same depth-l ancestor.

By construction, it is clear that
To see that f is an isomorphism it remains to show that f is edgepreserving. To see this, observe that two vertices a, b ∈ V (Lex( j, h) even ) are adjacent if and only if one is the depth-l ancestor and the other the depth(l + 1) ancestor of some leaf, and that this holds if and only if f (a), f (b) are also adjacent.
The proof of the second claim is similar.
We can now describe the structure of G even and G odd . For each w ∈ B even r , let G even contain the caterpillar Cat(w). For each i ∈ [r ] and h ∈ {0, 1}, let G even contain the lexicographic tree Lex(i, h) even . Finally, for each i ∈ [r ] let G even contain the labelled leaf x i adjacent to v even i,0 . The construction of G odd is similar: For each w ∈ B odd r , let G odd contain the caterpillar Cat(w). For each i ∈ [r ] and h ∈ {0, 1}, let G odd contain the lexicographic tree Lex(i, h) odd . Finally, for each i ∈ [r ] let G odd contain the labelled leaf x i adjacent to v even i,0 . Observe that in both G even and G odd , the vertices z w,i have degree 2 (as they appear as a leaf in the caterpillar Cat(w) and in the lexicographic tree Lex(i, w i ) even or Lex(i, w i ) odd ). The vertices v i,1 also have degree 2, and all other non-leaf vertices have degree 3.
We will later show that G even and G odd are not equivalent. First though, we will show that the multigraphs derived from G even and G odd by deleting (not removing) the same leaf are in fact equivalent. (Recall that the difference between deleting and removing a vertex v is that removing v involves the extra step of suppressing any degree-2 vertices left after deleting v.) Lemma 4 For i ∈ [r ], let G even − x i be the graph derived from G even by deleting x i and its incident edge, and similarly let G odd − x i be the graph derived from G odd by deleting x i and its incident edge. Then G even − x i and G odd − x i are equivalent.
Proof We will describe a set of isomorphisms between subgraphs of G even − x i and G odd − x i , then combine them to produce an isomorphism between G even − x i and G odd − x i . Each isomorphism will be one that maps vertex z w, j to z w i , j .
For each w ∈ B even r , Observation 1 implies that there exists an isomorphism f between Cat(w) and Cat(w i ) such that f (z w, j ) = z w i , j for each j ∈ [r ]. For each j ∈ [r ] \ {i} and h ∈ {0, 1}, Lemma 3 implies that there exists an isomorphism f between Lex( j, h) even and Lex( j, h) odd , such that f (v j,h ) = v j,h and f (z w, j , j) = z w i ; j for each leaf z w, j . Finally, for each h ∈ {0, 1}, Lemma 3 implies that there exists an isomorphism f between Lex(i, h) even and Observe that all of these isomorphisms agree on z w, j for any w ∈ B even r , j ∈ [r ] (that is, they each map this vertex to z w i , j ), and such vertices are the only vertices that are shared between caterpillars and lexicographic trees. Thus we can combine these isomorphisms into a single edge-preserving function f that maps every non-leaf vertex of G even − x i to a non-leaf vertex of G odd − x i . Moreover, as each caterpillar and lexicographic tree in G even −x i is mapped to a different caterpillar or lexicographic tree in G odd − x i , this function is a bijection. Finally, set f (x j ) = x j for every j ∈ [r ] \ i.
Then f is now a bijective function from V (G even − x i ) to V (G odd − x i ) that is both edge-preserving and label-preserving.
We note that we cannot extend the above graph isomorphism between G even − x i and G odd − x i to an isomorphism between G even and G odd by setting f ( , and so there would be no edge between In fact, the next lemma shows that there is no isomorphism between G even and G odd . Lemma 5 Let 0 denote the all-0 sequence from B r . For two vertices a, b in G even , let dist even (a, b) denote the distance between a and b in G even . Similarly for two vertices a, b in G odd , let dist odd (a, b) denote the distance between a and b in G odd . Then for any vertex a in G odd : This holds even if we suppress all degree-2 vertices in G even and G odd .
Proof We consider the two parts of the claim separately.
1. We first calculate the value of dist even (u 0 , x 1 ). Recall that in G even , x 1 is adjacent to the root v 1,0 of Lex(1, 0) even , and (by definition) every leaf of Lex(1, 0) even has distance r − 2 from v 1,0 . As u 0 is adjacent to a leaf z 0,1 of Lex(1, 0) even , it follows that dist even (u 0 , x 1 ) = 1 + r − 2 + 1 = r (there is no shorter path from u 0 to x 1 , as any path must pass through z w,1 for some w). As all leaves in Lex(1, 0) odd have distance r − 2 from v 1,0 in G odd , and therefore distance r − 1 = dist even (u 0 , x 1 ) − 1 from x 1 , it follows that the only vertices in G odd of distance dist even (u 0 , x 1 ) from x 1 are those which are not in Lex(1, 0) odd but adjacent to a leaf z w,1 of Lex(1, 0) odd . By construction, all such vertices are u w for some w ∈ B odd r such that w 1 = 0. When degree-2 vertices are suppressed, a similar argument holds, except that dist even (u 0 , x 1 ) is reduced by 1 (as we suppress z 0,1 ). It remains the case that the vertices in G odd of distance dist even (u 0 , x 1 ) from x 1 are those which are incident to a vertex from Lex(1, 0) odd but not in Lex(1, 0) odd themselves, and again all such vertices are u w for some w ∈ B odd r . 2. For any w ∈ B odd r , there exists i ∈ [r ] such that w i = 1. Any path from u w to x i must pass through a vertex z w ,i where w i = 0, and all such vertices have equal distance from x i . Thus, it is enough to show that the distance in G odd between u w and any such z w ,i is greater than the distance between u 0 and z 0,i in G even . To see this, consider a path P between u w and z w ,i . As w i = 0, we note that w = w and so P must traverse at least one lexicographic tree. We construct a mapping g : V (P) → V (Cat(0)), as follows. For any a ∈ V (P), if a is in Cat(w ) for any w ∈ B odd r (including w or w ), set g(a) = f (a), where f is the isomorphism between Cat(w ) and Cat(0) such that f (u w ) = u 0 and f (z w , j ) = z 0, j for all j ∈ [r ] (such an isomorphism exists by Observation 1).
Otherwise, it must be the case that a ∈ Lex( j, h) odd for some j ∈ [r ], h ∈ {0, 1}. In this case, set g(a) = z 0, j . Let Q be the set of all g(a) for any vertex a in P. Observe that for any vertices a, b in P, if a and b are adjacent then either g(a) = g(b) or g(a) and g(b) are adjacent. It follows that Q forms a connected set of vertices in Cat(0), and thus Q contains a path between g(u w ) = u 0 and g(z w ,i ) = z 0,i . Moreover, as P must traverse at least one lexicographic tree, there are consecutive vertices in P that are mapped to the same vertex by g. It follows that the path in Q is shorter than the path P, as required. It follows that the distance between u w and x i is greater than dist even (u 0 , x i ). We note that a similar argument applies even when vertices of degree 2 are suppressed.
Corollary 1 G even and G odd are not equivalent.
The next lemma will be used to show that when we suppress degree-2 vertices in G even and G odd , the resulting graphs N even and N odd are networks.

Lemma 6 In both G even and G odd , there exists a single blob containing all non-leaf vertices.
Proof Observe that any non-leaf vertex is part of a path between u w and v i,h for some w ∈ B r , i ∈ [r ], h ∈ {0, 1}. Furthermore every vertex v i,h appears on a path between u w and u w for some w, w . Therefore it is enough to show that for any w = w , u w and u w appear in the same blob.
Here the path between z 00 * ,1 and z 01 * ,1 passes through Lex(1, 0) even , the path between z 01 * ,2 and z 11 * ,2 passes through Lex(2, 1) even , the path between z 11 * ,1 and z 10 * ,1 passes through Lex(1, 1) even , and the path between z 10 * ,2 and z 00 * ,2 passes through Lex(2, 0) even . See Fig. 6 for an example when 00 * = 0000, 01 * = 0101, 11 * = 1100 and 10 * = 1001. As 00 * , 01 * , 11 * , 10 * appear on a cycle, they appear in the same blob of G even . Moreover as any vertex u w could fill the role of one of 00 * , 01 * , 11 * , 10 * , we have that all u w appear in the same blob. A similar argument holds for G odd . Now we are ready to construct the networks N even and N odd : Let N even be derived from G even by suppressing all vertices of degree 2. Similarly, let N odd be derived from G odd by suppressing all vertices of degree 2 (see Fig. 7 for the networks when r = 4).
Lemma 7 N even and N odd are networks on X .
Proof We show that N even is a network on X (the proof for N odd is similar). By construction, all vertices in N even have degree 1 or 3 and the leaves are bijectively labelled with the elements of X . It remains to show that contracting each blob into a single vertex gives a tree with no degree-2 vertices, which we will do by showing that N even has only one blob. By Lemma 6, all non-leaf vertices in G even are part of the same blob in G even . Observe that if two degree-3 vertices are in the same blob, then they are still in the same blob after contracting degree-2 vertices. Thus, all non-leaf vertices in N even are part of the same blob, and thus N even has a single blob, as required.
Lemma 8 N even and N odd are not equivalent.
Proof As N even and N odd are derived from G even and G odd by suppressing degree-2 vertices, Lemma 5 implies that there is no vertex in N odd that has the same distance from each leaf x i as u 0 has from x i in N even .
This implies that there is no isomorphism between N even and N odd , as if f is edgepreserving and label-preserving then the distance between u 0 and x i is equal to the distance between f (u 0 ) and f (x i ) = x i .

Lemma 9
For each i ∈ [r ], (N even ) x i and (N odd ) x i are equivalent.
Proof Recall the definitions of G even − x i and G odd − x i , and observe that (N even ) x i (respectively, (N odd ) x i ) can be derived from G even − x i (G odd − x i ) by suppressing degree-2 vertices. By Lemma 4, there exists an isomorphism f between G even − x i and G odd − x i . So define a bijective function f : for all a ∈ V ((N even ) x i ). Note that if a does not have degree 2 in G even − x i , f (a) also does not have degree 2 in G odd − x i . Thus if a ∈ V ((N even ) x i ) then f (a) = f (a) ∈ V ((N odd ) x i ), and so f is well-defined.
By construction, f is label-preserving. To see that f is edge-preserving, consider some a, b ∈ V ((N even ) x i ). Observe that the number of edges between a and b in (N even ) x i is equal to the number of paths between a and b in G even − x i whose internal vertices have degree 2. As f is an isomorphism, this is equal to the number of paths between f (a) and f (b) in G odd − x i whose internal vertices have degree 2, which in turn is equal to the number of edges between f (a) and f (b) in (N odd ) x i . Thus, f is edge-preserving, and so f is an isomorphism.
Lemmas 7, 8 and 9 imply the following theorem: Binary example of N even and N odd for the case when when r = 4. The vertex u 0000 in N even has distance d 1 = 3 from x 1 , d 2 = 3 from x 2 , d 3 = 4 from x 3 , and d 4 = 4 from x 4 . Moreover there is no vertex in N odd with distance d i from x i for each i ∈ [4]. Thus N even and N odd are not equivalent. However, for each i ∈ [4] the multigraphs (N even ) x i and (N odd ) x i are equivalent, using an isomorphism that maps each vertex u w to u w i Theorem 2 For any r ≥ 4, there exist networks N even , N odd on X with |X | = r, such that N odd is a leaf-reconstruction of N even , but N even and N odd are not equivalent. Thus, N even is not leaf-reconstructible.

Concluding remarks
Although we have shown that not all phylogenetic networks with five or more leaves are leaf-reconstructible, this does not mean that reconstructing networks from subnet-works is completely hopeless. There are already some positive results for interesting restricted network classes (Van Iersel and Moulton 2018). Moreover, since the presented counter-examples are very complex, it is certainly possible that other reasonable network classes are also leaf-reconstructible. For example, while it is known that all networks with at least five leaves and |E| − |V | ≤ 3 are leaf-reconstructible, the counter-examples presented in this paper have |E| − |V | = 2 r −1 (r − 1) − 2r , with r the number of leaves. Hence, whether networks with 3 < |E| − |V | < 2 r −1 (r − 1) − 2r are leaf-reconstructible is still open. In particular, is it possible to construct counter-examples where |E| − |V | is bounded by a linear function of the number of leaves?
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