When Three Trees Go to War

How many reticulations are needed for a phylogenetic network to display a given set of k phylogenetic trees on n leaves? For k = 2, Baroni et al. [Ann. Comb. 8, 391-408 (2005)] showed that the answer is n − 2. Here, we show that, for k ≥ 3 the answer is at least (3 /2 − ε )n. Concretely, we prove that, for each ε > 0, there is some n ∈ N such that three n-leaf caterpillar trees can be constructed in such a way that any network displaying these caterpillars contains at least (3 /2 − ε )n reticulations. The case of three trees is interesting since it is the easiest case that cannot be equivalently formulated in terms of agreement forests. Instead, we base the result on a surprising lower bound for multilabelled trees (MUL-trees) displaying the caterpillars. Indeed, we show that one cannot do (more than an ε ) better than the trivial MUL-tree resulting from a simple concatenation of the given caterpillars. The results are relevant for the development of methods for the Hybridization Number problem on more than two trees. This fundamental problem asks to construct a binary phylogenetic network with a minimum number of reticulations displaying a given set of phylogenetic trees


Introduction
A fundamental task of evolutionary analysis is to construct a phylogeny for a set of taxa depicting their ancestral relations.While many biological studies are content with the simplification that this phylogeny is a tree, there are circumstances, such as the presence of horizontal gene transfer (observed in many bacteria; Dagan et al., 2008) or hybridization (common among plant species and also observed among animals; Mallet, 2005), that require constructing phylogenetic networks which, in contrast to trees, allow modeling such "reticulate" evolution.Mathematically, a rooted phylogenetic network is a directed acyclic graph (DAG) with a single root and leaves bijectively labelled by the elements of a set X , modeling the considered taxa.In this paper, we will only consider phylogenetic networks that are rooted and binary.In such networks, the indegree-2 nodes represent reticulate evolutionary events and are called reticulations.
The construction of the most parsimonious (that is, containing the least amount of reticulations) network that is still "compatible" with a given set of trees is modeled by the Hybridization Number problem, which is well understood for the special case of having exactly two input trees (Baroni et al., 2005a,b;Bordewich and Semple, 2007;van Iersel and Kelk, 2011;Kelk et al., 2012).To formalize this problem, we say that a network displays a tree if this tree can be obtained from a subgraph of the network by suppressing nodes with exactly one incoming and exactly one outgoing arc.The Hy b r i d i z at i o n Nu m b e r problem then asks for a smallest network (in terms of the number of reticulations) displaying all input trees.Throughout this paper, we will focus on the simplified version of this problem where all input trees and the output network are required to be binary and all have the same set of leaf labels.
Unfortunately, many observations made for this case do not generalize to more than two input trees.One such observation is that for any two trees with n leaves, there is always a network with n − 2 reticulations displaying the two trees.This bound is tight because two "inverse" caterpillar trees need exactly n − 2 reticulations (Baroni et al., 2005b).In this work, we show that, for three or more trees, at least ( 3 /2 − ϵ)n reticulations may be required, even if the trees are caterpillars.See Fig. 1 for an example of our construction.This result represents a first lower bound for more than two trees that improves upon the n − 2 bound.In particular, it refutes the tempting conjecture that n reticulations are sufficient to display any set of three phylogenetic trees.If the bound of n had held, it would have had positive consequences for the development of methods for the Hy b r i d i z at i o n Nu m b e r problem, by bounding the worst-case complexity of subnetworks that need to be considered inside an algorithm.
To prove the ( 3 /2 − ϵ)n bound, we fist derive a corresponding bound for "multilabelled trees" (MUL-trees), that is, trees in which each leaf has one label, but each label may be used more than once.Again, the goal is to find a smallest (in terms of the number of leaves) MUL-tree displaying all input trees.Surprisingly, we show that, given at most three caterpillars, one cannot do better (up to an ϵ) than the trivial MUL-tree that simply concatenates the given caterpillars.More precisely, we show that, for each ϵ > 0 and t ≤ 3, there is some n ∈ N, and t caterpillars with n leaves, such that any MUL-tree displaying the caterpillars has at least (t − ϵ)n leaves.This is very close to the upper bound of t • n for any set of t trees, which holds because the t trees can simply be concatenated into a single MUL-caterpillar with t • n leaves.
Upper bounds on the number of reticulations needed to display a set of trees follow from results on "universal tree-based" networks (Bordewich and Semple, 2018), which are, roughly speaking, networks that can be obtained from any tree on the same set of leaves by subdividing Figure 1 -Three caterpillar trees with 9 leaves, resulting from Construction 1 for n = 3 i = 9.Lemma 2 implies that any MUL-tree displaying these caterpillars needs at least 19 leaves.Lemma 3 further implies that any network displaying these caterpillars needs at least 2 reticulations.While this second number does not seem particularly surprising, the strength of Lemma 3 is in the asymptotic bound it provides for growing n.
arcs of the tree and adding arcs between subdividing nodes (Francis and Steel, 2015;Hayamizu, 2016;Zhang, 2016).Bordewich and Semple (2018) showed that such a network has Θ(n log n) reticulations.The lower bound does not (directly) carry over to our question, but the upper bound does.Concretely, Bordewich and Semple (2018) proved that any network displaying all phylogenetic trees on n leaves needs at least O(n log n) reticulations.
The structure of this paper is as follows.After the preliminaries in Section 2, we prove the bound for MUL-trees in Section 3 and the bound for networks in Section 4, concluding with some open problems in Section 5.

Preliminaries
We will deal with sequences of letters over an unspecified alphabet.To differentiate such letters from variable names (even those referring to letters), we will typeset them in typewriter font such as a, 3, B. For all sequences s ′ that can be produced from s by removing zero or more letters, we say that s ′ is a subsequence of s and we write s ′ ⊴ s.We use • to denote the usual concatenation operator on sequences, where s • s ′ denotes the result of writing out s ′ after s.For letters a and b, we write a ≤ s b if the last occurence of a preceeds the first occurence of b in s, that is, some prefix of s contains all occurences of a but no occurence of b.
We also deal with (binary, phylogenetic) MUL-networks, which are directed, acyclic graphs (DAGs) with only the following types of nodes: (1) a unique source (called the root) with outdegree zero or two; (2) sinks (called leaves) with in-degree at most one, labeled using a function L from the set of leaves to some set of labels; (3) in-degree one and out-degree two nodes (tree nodes); and (4) in-degree two, out-degree one nodes (reticulation nodes or reticulations).A MUL-tree is a MUL-network without reticulations.A network is a MUL-network whose labelling function L is injective.A tree is a network without reticulations.A caterpillar tree (or just a caterpillar) is a tree in which each node is either a leaf or the parent of a leaf.If X is a MUL-network, a subset of nodes in a MUL-network, or a sequence, then L(X ) is the set (not the multiset) of labels/letters occurring in X .If X is a MUL-network, then n(X ) denotes the number of leaves in X .If X is a sequence, then n(X ) denotes its length.
We use the following correspondence between sequences and caterpillars.We say that sequence s corresponds to caterpillar P if the elements of s are exactly the leaf labels of P and these labels are ordered in s by decreasing distance from the root in P. Observe that each caterpillar has exactly two corresponding sequences since it has exactly two leaves with the same distance from the root.Conversely, each sequence has exactly one corresponding caterpillar.See Fig. 1 for an example of three caterpillars with corresponding sequences abc123ABC, CABcab312 and 231BCAbca.
Let N be a MUL-network and let u be a node of N. If u has a descendant v in N (that is, N contains a directed u-v -path), then we write v ≤ N u and we call u an ancestor of v .Note that the "≤ X "-relations for MUL-networks and sequences X naturally extend to sets of nodes/letters, that is, P < X Q if p < X q for all p ∈ P and q ∈ Q.For a set L of nodes of N, the MUL-network of N induced by L (written N[L]) is the result of removing all nodes u of N that have no descendant in L followed by the exhaustive contraction of indegree-one outdegree-one nodes onto their respective parents (this last operation is also called "supression").If L is a set of labels, then N[L] is the MUL-network of N induced by all nodes with a label in L. The result of removing all nodes x from N with x ≰ N u is denoted by N u and, if u is a reticulation and N u does not contain any reticulations of N, then u is called a lowest reticulation.If N is a MUL-tree and x and y are nodes in N, then the lowest common ancestor (LCA) of x and y in N is the unique minimum with respect to "≤ N " of all nodes u of N such that N u contains both x and y .
An embedding of a MUL-network T into a MUL-network N is a function ϕ that maps the nodes of T to nodes of N and the arcs of T to directed paths in N such that (1) the paths in the image of ϕ are arc-disjoint; (2) for each arc uv of T , ϕ(uv ) starts in ϕ(u) and ends in ϕ(v ) in N.
We say that MUL-network N displays MUL-network T if there is an embedding of T into N.
The backbone of a caterpillar is the path containing all edges not incident to a leaf.The backbone of an embedding ϕ of a caterpillar P in a MUL-network N is the path obtained by merging the paths ϕ(e) for all edges e on the backbone of P.

Lower Bound on MUL-Trees
In this section, we construct a family C of triples of caterpillars such that for any ϵ > 0, the familiy C contains a triple (C 1 , C 2 , C 3 ) of n-leaf caterpillars (where n depends on the choice of ϵ) such that any MUL-tree displaying all three caterpillars has at least (3 − ϵ)n leaves.As a byproduct, we show the existance of a family of pairs of caterpillars with a (2 − ϵ)n lower-bound on the leaf-number in any displaying MUL-tree.
A relabeling is a function mapping a label to another label and we allow applying relabelings to sets of labels, sequences and (MUL-)trees in the natural way.
sequences of length 3 i as follows: Let r 1 , r 2 , and r 3 be relabelings defined on the labels of C i−1 with disjoint images.Then, (1) ) are identical and we refer to this set by A i .Similarly, we abbreviate It turns out that sequences constructed by Construction 1 have very short common subsequences.
Case 1: k = 3.The proof is by induction on i.For i = 0, all three sequences contain a single label λ so the claim is trivially true.Suppose in the following that the claim holds for i − 1.Let s 3 be a common subsequence of X i , Y i , and Z i .Since all labels in the image of r 1 preceed all labels in the image of r 3 in X i and all labels in the image of r 3 preceed all labels in the image of r 1 in Y i , we know that s 3 does not contain labels of the images of both r 1 and r 3 .Similarly, it can be seen that s 3 cannot contain labels of any two of r 1 , r 2 , and r 3 .Thus, without loss of generality, s 3 consists only of labels of r 1 , implying that s 3 is a common subsequence of the result of removing all labels of r 2 and r 3 from X i , Y i , and and, by induction hypothesis, the length of s 3 is 1.
Case 2: k = 2. Again, the proof is by induction on i and the induction base case i = 0 is trivially true, so we will suppose that the claim holds for i − 1.By symmetry, we can further suppose without loss of generality that s 2 is a common subsequence of X i and Y i .If s 2 only uses labels from the image of one r ∈ {r 1 , r 2 , r 3 }, then r −1 (s 2 ) is a common subsequence of X i−1 and Y i−1 so the claim holds by induction hypothesis.Otherwise, s 2 uses labels of at least two of r 1 , r 2 , and r 3 .Since all labels of r 3 succeed all labels of r 1 and r 2 in X i but preceed them in Y i , we know that s 2 uses labels of r 1 and r 2 but not of r 3 .Thus, s 2 admits two subsequences s ′ and s ′′ such that s ′ and s ′′ contain only labels of r 1 and r 2 , respectively, and |s ′ | + |s ′′ | = |s 2 |.Then, however, s ′ is a subsequence of the result of removing all labels of r 2 and r 3 from In the following, we prove lower bounds on the number of leaves in any MUL-tree displaying k of the caterpillars in C i for k ∈ {1, 2, 3} and i ∈ N. We denote these bounds by N (k) i and we note that, by the concatenation argument, N Proof.The proof is by induction on i.For the induction base, observe that all of X 0 , Y 0 and T consist of a single leaf and n(T ) = 2 − 1 = 1.For the induction step, suppose that the lemma holds for all j < i.Let T A , T B and T C denote the subtrees of T induced by labels in A i , B i and C i , Leo van Iersel et al. 5 Peer Community Journal, Vol. 4 (2024), article e54 https://doi.org/10.24072/pcjournal.419respectively, and observe that their label multisets are a partition of the label multiset of T since A i , B i and C i are disjoint.In the following, we show that at least one of T A , T B and T C contains 2N (1) i−1 leaves.Since, by definition, the other two contain at least N (2) i−1 leaves, we have Towards a contradiction, assume that T A , T B and T C contain less than 2N (1) i−1 leaves.For all F ∈ {A i , B i , C i }, let χ F and ψ F be respective embeddings of X i [F ] and Y i [F ] into T F and note that, by assumption, T F contains strictly less than 2N (1) i−1 = n(X i ) + n(Y i ) leaves, implying that some leaf ℓ F in T F is mapped-to by both χ F and ψ F .In particular, the label of each ℓ F occurs only once in T F and, thus, in T , which allows us to use ℓ F and its label interchangeably.Now, since Then, there is some n ∈ N and two caterpillar trees of the same set of n labels, such that any MUL-tree displaying them has at least (2 − ϵ)n leaves.
Proof.Let i ∈ N such that ( 2 /3) i ≤ ϵ and, hence, Proof.The proof is by induction on i.For the induction base, observe that all of X 0 , Y 0 , Z 0 , and T consist of a single leaf and n(T ) = 3 − 2 = 1.For the induction step, suppose that the lemma holds for all j < i.Let T A , T B and T C denote the subtrees of T induced by labels in A i , B i and C i , respectively, and observe that their label multisets are a partition of the label multiset of T since A i , B i and C i are disjoint.If any of T A , T B , and T C contains 3N (1) Further, if any two of T A , T B , and Thus, in the following, suppose that neither of the two cases holds.In particular, at least two trees among T A , T B , and T C contain strictly less than N (2) i−1 leaves.By symmetry, suppose these are T A and T B .For each F ∈ {A, B, C }, let χ F , ψ F , and ϕ F denote the respective embeddings of and let us say that two among them overlap if T F has a leaf that is assigned-to by both.Note that removing the n(X i−1 leaves of T A that are mapped-to by χ results in a MUL-tree with strictly less than N (2) i−1 leaves and, by Lemma 1, this MUL-tree cannot display both Y i [A i ] and Z i [A i ]).Thus, χ A overlaps one of ψ A and ϕ A and the analog holds for ψ A and ϕ A .By pigeonhole principle, one among the three embeddings overlaps both others (while the other two may not necessarily overlap).Let ℓ A and k A denote the corresponding leaves (possibly ℓ A = k A if all three embeddings assign to ℓ A ). Since the same argument holds for B, we define ℓ B and k B analogously (see Fig. 3 for an illustration).Now, since T C contains strictly less than 3N (1) i−1 leaves by assumption, we know that two of χ C , 6 Leo van Iersel et al.
Peer Community Journal, Vol. 4 (2024), article e54 https://doi.org/10.24072/pcjournal.419ψ C , and ϕ C overlap in a leaf ℓ C of T C .Next, we consider the relative positions of these five leaves in T .
In the following, a leaf a in T A with parent t is said to be above a leaf b in T B (written a ⇝ b) if b < T t, and analogously for any pair chosen from the combined leaf-set of T A , T B and T C .A third leaf ℓ is said to be between a and b if a ⇝ ℓ ⇝ b.For all leaves a whose parent is an ancestor of b in T A , we have that a is above both a and b and, likewise, for T B and T C .By symmetry, suppose that ℓ F ⇝ k F for all F ∈ {A, B} Now, as C < Y i A and C < Z i A, all leaves of ψ C and ϕ C are below all leaves of ψ A and ϕ A .Since ℓ C is contained in the former and both ℓ A and k A are contained in the latter, we have that ℓ A and k A are both above ℓ C .Further, as B < X i C and B < Z i C , we have ℓ C ⇝ ℓ B .Thus, in particular both of ℓ A , k A are above both of ℓ B , k B .However, the caterpillar X i contains at least one leaf mapped to ℓ A or k A , and at least one leaf mapped to ℓ B or k B .But since A < X i B, this implies that at least one of ℓ B and k B is above one of ℓ A and k A , contra- Then, there is some n ∈ N and three caterpillar trees of the same set of n labels, such that any MUL-tree displaying them has at least (3 − ϵ)n leaves.

Lower Bound on Networks
In this section, we build on the lower bound developed for MUL-trees in Section 3 to prove that, for any ϵ > 0 and large enough n, any single-labeled phylogenetic network displaying the three n-leaf caterpillars constructed in Construction 1 has at least ( 3 /2 − ϵ)n reticulations.To this end, we will give an algorithm that transforms any network displaying the caterpillars into a MULtree displaying the caterpillars by "unzipping" (or "duplicating") subtrees.Then, we show that, if the network had fewer than ( 3 /2 − ϵ)n = ( 3 /2 − ϵ)3 i reticulations, then the resulting MUL-tree has fewer than 3 i+1 − 2 i+1 leaves, contradicting Lemma 2. The embedding of the three caterpillars X i , Y i and Z i is depicted as green solid, red dashed and blue dotted lines, respectively, within the network outlined in gray.Note that all leaves of N z into which ϕ embeds leaves of X i as well as at least one of Y i and Z i , are duplicated in the process.Note also that not all three caterpillars are necessarily embedded in N z , as previous unzip operations may have split a caterpillar off N z .

Transforming the Network into a MUL-tree
In the following, we present a transformation acting on a given embedding of the three caterpillars into a multi-labeled network.Each application of our transformation rule will reduce the number of reticulations by one at the cost of creating new leaves.Hence, exhaustive application will result in an embedding of the three caterpillars into a MUL-tree.The rule acts on the subtree below a lowest reticulation and also manipulates a reservoir of virtual "tokens" which will help in the amortized analysis of how many new leaves are created in the process.
In the following, we work with pairs (N, ϕ), where N is a MUL-network with 3 i distinct labels and ϕ is an embedding of the caterpillars X i , Y i and Z i (as constructed by Construction 1) into N such that all arcs of N are used by the embedding ϕ.We call such pairs caterpillar embeddings.Note that the assumptions that all arcs of N are used is satisfied by any embedding of the caterpillars into a network with smallest reticulation number.We make use of the fact that no embedding of any caterpillar can use both arcs incoming to any reticulation r of N, so the caterpillars with leaves embedded below r can be divided into two groups, depending on which incoming arc of r is used in their embedding.We call this the parity of a caterpillar with respect to r .We say that a caterpillar that does not have leaves below r has parity ⊥ with respect to r .Note that, since all arcs of N are used by ϕ, there are two caterpillars with different non-⊥ parity with respect to r .Let N r be the subnetwork of N rooted at r .We say that the backbone of a caterpillar Q is embedded in N r (or below r ) if ϕ maps a non-leaf of Q into N r .Note that this is the case if and only if at least two leaves of N r are used by the embedding of Q into N.The central operation in the transformations "unzips" (N, ϕ) at r (see Fig. 4).
Definition 1.Let (N, ϕ) be a caterpillar-embedding, let r be a reticulation in N with child z such that the subnetwork N z of N rooted at z does not contain reticulations.Let xr and yr denote the incoming arcs of r with x ̸ = y .The operation of unzipping N at r consists in the following steps: (1) Remove the node r from N.
(2) Add a copy N ′ z of N z with root z ′ to N and add the arcs xz and yz ′ .(3) For each caterpillar Q such that ϕ embeds Q using the arc yr , replace all nodes u of N z by their copy u ′ in N ′ z in the embedding of Q. (4) repeatedly remove all leaves of the resulting MUL-network that are not used by the embedding, and suppress indegree-one outdegree-one nodes.
Observation 1.Let (N, ϕ) be a caterpillar embedding and let (N ′ , ϕ ′ ) be the result of unzipping N at a reticulation r .Then, (N ′ , ϕ ′ ) is a caterpillar embedding.In particular, all arcs of N ′ are used by ϕ ′ .
The rest of this section depends on an arbitrary number q ∈ N, which we will pick "sufficiently large" in the proof of the main theorem (we assume q ∈ N for ease of presentation, but all proofs also work for q ∈ R).Further, we also suppose that our input caterpillars are "sufficiently large" with respect to q, that is, their length n satisfies (1) n > 12qn log 3 2 .
Transformation Rule 1.Let (N, ϕ) be a caterpillar-embedding, let r be a lowest reticulation in N, and let Q denote the set of caterpillars whose backbone is embedded in N r .Then, (1) unzip N at r , (2) create three tokens in the token reservoir, and (3) for each leaf ℓ below r in N and each pair of different-parity caterpillars in Q whose embedding uses ℓ, remove 2q tokens from the token reservoir.
As we will see, we never need to remove more tokens than are contained in the reservoir.
Lemma 3. Let N be a network with n leaves, let (N, ϕ) be a caterpillar embedding, let k be the number of reticulations of N, and let (T , ϕ ′ ) be the result of applying Transformation Rule 1 exhaustively to (N, ϕ).Then, T has at most n + 4(q+1)k /3q leaves.
Proof.Intuitively, the proof is based on the observation that, whenever the transformation creates many new leaves for only a single reticulation it removes, then we can use half of these leaves to construct a common subsequence of two caterpillars.Then, Proposition 1 implies that this cannot happen too often.
Formally, we consider a series of "configurations" C 0 , C 1 , ... , C Ω , each consisting of a caterpillar embedding and a token reservoir where C 0 := ((N, ϕ), t 0 = 0) and C Ω = ((T , ϕ ′ ), t Ω ) for some t Ω and each C j results from an application of Transformation Rule 1 to the previous configuration C j−1 .To show Lemma 3, we assign a "weight" ω to each C i .We prove that ω is monotonically non-increasing with respect to Transformation Rule 1.This implies an upper bound on the number of leaves of the MUL-tree T displaying all three caterpillars.For a configuration C := ((Γ, ψ), t), (1) let # Γ r denote the number of reticulations in Γ, (2) let # Γ i denote the number of leaves of Γ that are used by the embedding of exactly i caterpillars, and define (2) ω((Γ, ψ), t) := i∈{1,2,3,r} where c 1 := c 3 := 3q, c 2 := 4q, c r := 4(q + 1), and c t := 1.We will omit the superscript Γ when it is clear from the context and we abbreviate the total number of leaves as # Γ := i∈{1,2,3} # Γ i .
Claim 1. ω is monotonically non-increasing with respect to Transformation Rule 1. Suppose that there are at least two r i -bad leaves since, otherwise, s i is already a subsequence of Q.Let u denote the lowest node of the tree N j r i that still has the backbones of both P and Q embedded in it.Clearly, all r i -bad leaves that are not below u form a suffix of s i that is also a subsequence of Q.By definition of u, it is not a leaf of N j and u has children v P and v Q in N j such that at most one leaf ℓ Q below v P is used in the embedding of Q and at most one leaf ℓ P below v Q is used in the embedding of P. But then, removing ℓ P from s i yields a subsequence of Q.Note that the removal of ℓ P is necessary to form a subseqence of Q since the sequences corresponding to P and Q may disagree on the relative ordering of ℓ P and ℓ Q .Now, the concatenation s of all subsequences of Q corresponding to the s i is a common subsequence of P and Q and it contains at least half of all bad leaves (recall that we only remove a leaf from s i if it contains at least two leaves).Then, by Proposition 1, the number of bad leaves is at most 2 • 2 i = 2 • 2 3 n = 2 • n log 3 2 , where n = 3 i is the number of leaves in N. Summing over all three caterpillar pairs, we get an upper bound of 6n log 3 2 bad leaves overall.
Next, we show that the first token retraction is preceeded by the creation of enough tokens to compensate for all retractions.To this end, consider the first configuration C j = ((N j , ϕ j ), t j ) such that applying Transformation Rule 1 to a reticulation r in N j incurs a withdrawal from the token reservoir.In particular, this implies the existance of two different-parity caterpillars P and Q such that their backbone is embedded below r and both their embeddings use a common leaf ℓ below r .However, by construction of P and Q, none of the labels occuring in the lowest third of P occurs in the lowest third of Q and, thus, ℓ is preceeded by at least n /3 leaves in either P or Q; without loss of generality, suppose Q.Since the backbone of Q is embedded below r , there are at least n /3 leaves below r used by the embedding of Q, but not that of P.However, since all leaves in N are used by all three caterpillars, all these n /3 leaves were "unzipped" in previous operations.
Since C j is the first configuration in which P and Q have different parity and "share" a leaf, we know that in all previous "unzip" operations, either (a) P and Q have the same parity or (b) the embeddings of P and Q share no leaves below the corresponding reticulation or (c) the embedding of at least one of P and Q uses only one leaf below the corresponding reticulation.Clearly, in cases (a) and (b), the unzip operation does not "separate P from Q" in any leaf, that is, the unzip operation does not reduce the number of leaves used by the embeddings of both P and Q.In case (c), each unzip operation can "separate P from Q" in at most one leaf, implying that C j is preceeded by at least n /3 unzip operations, each creating 3 tokens in the reservoir.Thus, by the time the first withdrawal is made from the reservoir, it contains at least n tokens which is sufficient to cover all withdrawals (recall that we withdraw 2q tokens for each of the at most 6n log 3 2 bad leaves) since n > 12qn log 3 2 by (1).■ With Claim 1 and Claim 2 we can now prove the bound on the number of leaves # T in T in the number n of leaves in N and the number k of reticulations in N. To this end, note and recall that (a) t 0 = 0, (b) all leaves in N are used by the embeddings of all three caterpillars, and (c) T has no reticulations.Then, Lemma 3 tells us that if we can construct a network with few reticulations that displays our caterpillars, then we can also construct a MUL-tree with few leaves that displays our caterpillars.Since Corollary 1 says that such MUL-trees do not exist, we conclude that such networks do not exist.
Theorem 1.Let ϵ > 0.Then, there are three caterpillar trees, each with n ∈ N leaves, such that any network displaying all three caterpillars has at least ( 3 /2 − ϵ)n reticulations.
Proof.Let δ := 2ϵ /3 and choose q large enough so that β := δ − 1 /q+1 > 0. Finally, choose i such that β ≥ (1 − 1 /q+1)( 2 /3) i .Let N be a network displaying X i , Y i and Z i , that is, there is a caterpillar embedding (N, ϕ).Let (T , ϕ) be the result of applying Transformation Rule 1 exhausively to (N, ϕ) and note that, by Observation 1, (T , ϕ) is a caterpillar embedding, that is, T displays X i , Y i , and Z i .Further, by Lemma 2, T has at least 3 i+1 − 2 i+1 leaves.Let k denote the number of reticulations in N.Then,

Discussion and Open Problems
We have shown that, for each ϵ > 0 and t ≤ 3, there is some n ∈ N, and t caterpillars with n leaves, such that any MUL-tree displaying the caterpillars has at least (t − ϵ)n leaves.Whether this result can be generalized to t ≥ 4 remains an interesting open question, even more so for networks, where the question would be to generalize our lower bound of ( 3 /2 − ϵ)n reticulations required to display three caterpillars to t ≥ 4 caterpillars.
Note that Theorem 1 can be stated more precisely as "there is some function α(n) ∈ o(1) such that, for each n, there are three caterpillars with n leaves that cannot be displayed by any network with fewer than ( 3 /2 − α(n))n reticulations".This raises the question how the "smallest" functions α for which this statement still holds, may look like (other than being in o(1)).In the MUL-tree case, a closer inspection of our proofs gives α(n) ≤ 2 • ( 2 /3) log 3 n .If we force the MUL-tree displaying the t trees to be a caterpillar, the question is equivalent to the question of the shortest supersequence that any collection of t permutations over {1, ... , n} may have.This can be shown (Hunter, 2023) to be at least (t − O(n −1/2 ))n, implying α(n) ∈ O(n −1/2 ) ⊂ o(( 2 /3) log 3 n ).If there is always a MUL-caterpillar among the optimal MUL-trees displaying the t caterpillars, then this would imply a stricter bound for both MUL-trees and networks than what we showed here.Apart from improving the function α(n), it is interesting whether the factor of 3 /2 can be improved.In other words, is there a family of triples of phylogenetic trees for which more than 3n /2 reticulations are needed?We do know that all triples of phylogenetic trees can have a network with 2(n − 2) reticulations that display them: the network can be obtained from a tree on two leaves by inserting each remaining leaf using two reticulations.However, the same three trees might also be displayed by a network with strictly fewer reticulations.The best Leo van Iersel et al. 13 Peer Community Journal, Vol. 4 (2024), article e54 https://doi.org/10.24072/pcjournal.419possible bound for three trees is therefore between ( 3 /2 − ϵ)n and 2(n − 2).More generally, the best possible bound for t ≥ 3 trees is between ( 3 /2 − ϵ)n and (t − 1)(n − 2).Can this gap be closed or narrowed?As our results are for caterpillars, we expect that trees of varying topology can be used to prove such tighter bounds.The last two questions beg the, somewhat philosophical question of whether reticulations are strictly more powerful than multiple leaves?For MUL-trees, we know that we cannot do much better than the trivial upper bound of t • n.Is the same true for networks, can we not do much better than the trivial upper bound of (t − 1)(n − 2), or are networks really more powerful than MUL-trees in this sense?
Finally, our original motivation for considering this problem came from the Hy b r i d i z at i o n Number problem.However, the bounds proven in this paper do not have direct formal consequences for that problem.Hence, another interesting direction for future research is to see if our bounds can be used to prove a negative result regarding exact algorithms for Hy b r i d i z at i o n Number, e.g.parameterized by treewidth (van Iersel et al., 2022).

Figure 4 -
Figure4-Illustration of the operation of "unzipping" (N, ϕ) at a lowest reticulation.The embedding of the three caterpillars X i , Y i and Z i is depicted as green solid, red dashed and blue dotted lines, respectively, within the network outlined in gray.Note that all leaves of N z into which ϕ embeds leaves of X i as well as at least one of Y i and Z i , are duplicated in the process.Note also that not all three caterpillars are necessarily embedded in N z , as previous unzip operations may have split a caterpillar off N z .

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Leo van Iersel et al.Peer Community Journal, Vol. 4 (2024), article e54 https://doi.org/10.24072/pcjournal.419 Illustration of the five leaves handled in the proof of Lemma 2. The three parts depict possible embeddings ofX i [F ], Y i [F ], and Z i [F ] for all F ∈ {A i , B i , C i }.We assume that the embedding ψ A of Y i [A i ] overlaps χ A in k A and ϕ A in ℓ A (left part) and, likewise for B. For C , only χ C and ϕ C overlap, in ℓ C (right part).While T is not necessarily a caterpillar, drawing T linearly like that may help understand the situation.