Domination and Cut Problems on Chordal Graphs with Bounded Leafage

The leafage of a chordal graph G is the minimum integer l such that G can be realized as an intersection graph of subtrees of a tree with l leaves. We consider structural parameterization by the leafage of classical domination and cut problems on chordal graphs. Fomin, Golovach, and Raymond [ESA 2018, Algorithmica 2020] proved, among other things, that Dominating Set on chordal graphs admits an algorithm running in time $2^{O(l^2)} n^{O(1)}$. We present a conceptually much simpler algorithm that runs in time $2^{O(l)} n^{O(1)}$. We extend our approach to obtain similar results for Connected Dominating Set and Steiner Tree. We then consider the two classical cut problems MultiCut with Undeletable Terminals and Multiway Cut with Undeletable Terminals. We prove that the former is W[1]-hard when parameterized by the leafage and complement this result by presenting a simple $n^{O(l)}$-time algorithm. To our surprise, we find that Multiway Cut with Undeletable Terminals on chordal graphs can be solved, in contrast, in $n^{O(1)}$-time.


Introduction
The intersection graph of a family F of nonempty sets is the graph whose vertices are the elements of F with two vertices being adjacent if and only if their corresponding sets intersect.The most natural and famous example of such intersection graphs are interval graphs where F is a collection of subpaths of a path.Due to their applicability in scheduling, interval graphs have received a considerable attention in the realm of algorithmic graph theory.One useful characterization of an interval graph is that its maximal cliques can be linearly ordered such that for every vertex, the maximal cliques containing that vertex occur consecutively [24].This property proves very useful for the design of polynomial-time dynamic programming based or greedy algorithms on interval graphs.
Consider the generalization where F is a collection of subtrees of a tree instead of subpaths of a path.In this case, the corresponding class of intersection graphs is exactly that of chordal graphs [9,23,45].Recall that a graph is chordal if every cycle of length at least 4 has a chord.Often, the algorithms of the types mentioned in the previous paragraph fail to generalize to this superclass as witnessed by the following problems that admit polynomialtime algorithms on interval graphs but are NP-complete on chordal graphs: Dominating Set [7,11], Connected Dominating Set [3,46], Steiner Tree [3,46], Multicut with Undeletable Terminals [27,42], Subset Feedback Vertex Set (Subset FVS) [20,43], Longest Cycle [26,32] 1 , Longest Path [31], Component Order Connectivity [18], s-Club Contraction [25], Independent Set Reconfiguration [5], Bandwidth [34], Cluster Vertex Deletion [33].Also, Graph Isomorphism on chordal graphs is polynomial-time equivalent to the problem on general graphs whereas it admits a linear-time algorithm on interval graphs [38].
The problems above remain hard even on split graphs, another well-studied subclass of chordal graphs.A graph is a split graph if its vertex set can be partitioned into a clique and an independent set.The collection of split graphs is a (proper) subset of the class of intersection graphs where F is a collection of substars of a star.As interval graphs are intersection graphs of subpaths of a path (a tree with two leaves) and split graphs are intersection graphs of substars of a star (a tree with arbitrary number of leaves), a natural question to consider is what happens to these problems on subclasses of chordal graphs that are intersection graphs of subtrees of a tree with a bounded number of leaves.Motivated by such questions, we consider the notion of leafage introduced by Lin et al. [37]: the leafage of a chordal graph G is the minimum integer ℓ such that G can be realized as an intersection graph of a collection F of subtrees of a tree that has ℓ leaves.Note that the leafage of interval graphs is at most 2 while split graphs have unbounded leafage.Thus the leafage measures, in some sense, how close a chordal graph is to an interval graph.Alternately, an FPT or XP algorithm parameterized by the leafage can be seen as a generalization of the algorithm on interval graphs.Related Work.Habib and Stacho [28] showed that we can compute the leafage of a connected chordal graph in polynomial time.Their algorithm also constructs a corresponding representation tree 2 T with the minimum number of leaves.In recent years, researchers have studied the structural parameterization of various graph problems on chordal graphs parameterized by the leafage.Fomin et al. [19] and Arvind et al. [2] proved, respectively, that the Dominating Set and Graph Isomorphism problems on chordal graphs are FPT parameterized by the leafage.Barnetson et al. [4] and Papadopoulos and Tzimas [44] presented XP-algorithms running in time n O(ℓ) for Fire Break and Subset FVS on chordal graphs, respectively.Papadopoulos and Tzimas [44] also proved that Subset FVS is W [1]hard when parameterized by the leafage.Hochstättler et al [30] showed that we can compute the neighborhood polynomial of a chordal graph in n O(ℓ) -time.
It is known that the size of asteroidal set in a chordal graph is upper bounded by its leafage [37].See [1,29] for the relationship between leafage and other structural properties of chordal graphs.Kratsch and Stewart [35] proved that we can effectively 2ℓ-approximate bandwidth of chordal graphs of leafage ℓ.Chaplick and Stacho [12] generalized the notion of leafage to vertex leafage and proved that, unlike leafage, it is hard to determine the optimal vertex leafage of a given chordal graph.Figueiredo et al. [16] proved that Dominating Set, Connected Dominating Set and Steiner Tree are FPT on chordal graphs when parameterized by the size of the solution plus the vertex leafage, provided that a tree representation with optimal vertex leafage is given as part of the input.
Our Results.We consider well-studied domination and cut problems on chordal graphs.As our first result, we prove that Dominating Set on chordal graphs of leafage at most ℓ admits an algorithm running in time 2 O(ℓ) • n O (1) .This improves upon the existing algorithm by Fomin et al. [19,Theorem 9] which runs in time 2 O(ℓ 2 ) • n O (1) .Despite being significantly simpler than the algorithm in [19], our algorithm in fact solves the Red-Blue Dominating Set problem, a well-known generalization of Dominating Set.In this generalized version, an input is a graph G with a partition (R, B) of its vertex set and an integer k, and the objective is to find a subset D of R that dominates every vertex in B, i.e., B ⊆ N(D).We further use this algorithm to solve other related domination problems.
Theorem 1.1.Dominating Set, Connected Dominating Set, and Steiner Tree can be solved in 2 O(ℓ) • n O (1) on chordal graphs of leafage at most ℓ.
The reductions in [7] and [46] used to prove that these problems are NP-complete on chordal graphs imply that these problems do not admit 2 o(n) , and hence 2 o(ℓ) •n O (1) , algorithms unless the ETH fails.
Arguably, the two most studied cut problems are MultiCut and Multiway Cut.In the MultiCut problem, an input is graph G, a set of terminal pairs P ⊆ V (G) × V (G) and an integer k, and the objective is to find a subset S ⊆ V (G) of size at most k such that no pair of vertices in P is connected in G − S. In the Multiway Cut problem, instead of terminal pairs, we are given a terminal set P and the objective is to find a subset S ⊆ V (G) of size at most k such that no two vertices in P are connected in G − S.These problems and variations of them have received a considerable attention which lead to the development of new techniques [8,13,14,39,40].Misra et al. [41] studied the parameterized complexity of these problems on chordal graphs.Guo et al. [27] proved that MultiCut with Deletable Terminals is NP-complete on interval graphs, thereby implying that this problem is paraNP-hard when parameterized by the leafage.We consider the MultiCut with Undeletable Terminals problem and prove the following result.Theorem 1.2.MultiCut with Undeletable Terminals on chordal graphs is W [1]hard when parameterized by the leafage ℓ and assuming the ETH, does not admit an algorithm running in time f (ℓ) • n o(ℓ) for any computable function f .However, it admits an XPalgorithm running in time n O(ℓ) .
Next, we focus on the Multiway Cut with Undeletable Terminals problem.We find it somewhat surprising that the classical complexity of this problem on chordal graphs was not known.Bergougnoux et al. [6], using the result in [19], proved that the problem admits an XP-algorithm when parameterized by the leafage 3 .Our next result significantly improves upon this and [41, Theorem 2] which states that the problem admits a polynomial kernel when parameterized by the solution size.
Theorem 1.3.Multiway Cut with Undeletable Terminals can be solved in n O (1) time on chordal graphs.
A well-known trick to convert an instance of Multiway Cut with Deletable Terminals into an instance of Multiway Cut with Undeletable Terminals is to add a pendant vertex to each terminal, remove that vertex from the set of terminals, and make the newly added vertex a terminal.As this reduction converts a chordal graph into another chordal graph, Theorem 1.3 implies that Multiway Cut with Deletable Terminals is also polynomial-time solvable on chordal graphs.Another closely related problem is Subset FVS which is NP-complete on split graphs [43].To the best of our knowledge, this is the first graph class on which the classical complexity of these two problems differ.
Next, we revisit the problems on chordal graphs with bounded leafage and examine how far we can generalize this class.An asteroidal triple of a graph G is a set of three vertices such that each pair is connected by some path that avoids the closed neighborhood of the third vertex.Lekkerkerker and Boland [36] showed that a graph is an interval graph if and only if it is chordal and does not contain an asteroidal triple.They also listed all minimal chordal graphs that contain an asteroidal triple (see, for instance, [10, Figure 1]).Among this list, we found the net graph to be the most natural to generalize.For a positive integer ℓ ≥ 3, we define H ℓ as a split graph on 2ℓ vertices with split partition (C, I) such that the only edges across C, I are a perfect matching.Note that H 3 is the net graph.As interval graphs are a proper subset of the collection of chordal graphs that do not contain a net graph as an induced subgraph, the collection of the chordal graph of leafage ℓ is a proper subset of the collection of chordal graphs that do not contain H ℓ+1 as an induced subgraph (see Observation 6.1).We show that, although the considered domination problems are polynomial-time solvable for constant ℓ, the fixed-parameter tractability results are unlikely to extend to this larger class.
Theorem 1.4.Dominating Set, Connected Dominating Set and Steiner Tree on H ℓ -induced-subgraph-free chordal graphs are W [1]-hard when parameterized by ℓ and assuming the ETH, do not admit an algorithm running in time f (ℓ) • n o(ℓ) for any computable function f .However, they all admit XP-algorithms running in time n O(ℓ) .
We observe a similar trend with respect to MultiCut with Undeletable Terminals as its parameterized complexity jumps from W[1]-hard on chordal graph of leafage ℓ to paraNP-hard on H ℓ -induced-subgraph-free chordal graphs when parameterized by ℓ.

Dom Set, ConnDom Set, Steiner Tree
MultiCut with Un-Del Term

Interval Graphs
Poly-time [3,11] Poly-time [27] Poly-time [6] Chordal graphs of leafage ℓ Table 1: Overview of the known results and our contributions.Every graph class mentioned in the first column is a proper subset of the graph class mentioned below.
Table 1 summarises our results.
Our Methods.We briefly discuss the methods used in our two main algorithms, namely the algorithm for Dominating Set and the one for Multiway Cut.
Red-Blue Dominating Set in Chordal Graphs.As mentioned earlier, the linear ordering of cliques in interval graphs is particularly useful for the design of polynomial-time algorithms.Such an ordering is not possible even if G is a chordal graph whose representation tree T is a star.Consider the case where the model of every red vertex in G includes the center of the star T (and possibly some leaves) and the model of every blue vertex is (only) a leaf.We can solve this instance by converting it to an instance of Set Cover and solving it using the FPT algorithm parameterized by the size of the universe.In this case, the size of the universe is at most the number of leaves which is upper bounded by the leafage.In the other case where the properties of red vertices and blue vertices are reversed, we obtain a similar result by creating an equivalent instance of Hitting Set.
These ideas can be used in a more general setting as long as the following two properties are satisfied: (1) the model of each vertex is local, that is, it contains at most one branching node, and (2) each branching node is contained only in models of either red vertices or blue vertices.Based on this observation, we introduce a restricted version of the problem in which the input graph is required to satisfy these two conditions.We then show that the general case reduces to this restricted version: indeed, we prove that there is a branching algorithm that constructs 2 O(ℓ) many instances (where ℓ is the leafage of the input graph) of the restricted version of the problem such that the input instance is a Yes-instance if and only if one of these newly created instances is a Yes-instance.These two properties ensure that the graph induced by the red and blue vertices whose model intersect the subtree rooted at a farthest branching node (from some fixed root) satisfies the premise of at least one of the cases mentioned in the previous paragraph.We then present a greedy procedure, based on solving the Set Cover and Hitting Set problems, that identifies some part of an optimum solution.Apart from this greedy selection procedure, all other steps of the algorithm run in polynomial time.
Multiway Cut in Chordal Graphs.We give a polynomial-time algorithm for Multiway Cut on chordal graphs by solving several instances of the (s, t)-Cut problem (not necessarily with unit capacities).Our strategy is based on a bottom-up dynamic programming (DP) on a tree representation of a chordal graph.An interesting aspect of our DP is that we need to look-up all DP table values that are already computed to compute a new entry.This is in contrast to typical DP-based algorithms that do computations only based on local entries.
We remark that we do not expect to design an algorithm for Multiway Cut on chordal graphs using much simpler arguments (like a simple dynamic programming procedure etc.) as the problem generalizes some well-studied cut-flow based problems.As an example, recall the Vertex Cover problem on bipartite graphs where given a bipartite graph G with bipartition (A, B), the goal is to find The Vertex Cover problem on bipartite graphs reduces to the Multiway Cut problem on chordal graphs: indeed, let G ′ be the graph obtained from G by making B a clique, adding new pendant vertex t a to each vertex a ∈ A, and further adding another new vertex t that is adjacent to all vertices of B. Then G ′ is a chordal graph and letting T = t ∪ {t a | a ∈ A}, it is easy to see that S ⊆ V (G) is a vertex cover of G if and only if S is a T -multiway-cut in G ′ .As mentioned earlier, our algorithm solves several instances of the (s, t)-Cut problem, which also sits at the heart of some algorithms for Vertex Cover on bipartite graphs.The above reduction suggests that an algorithm for Multiway Cut on chordal graphs using much simpler techniques, would imply an algorithm for Vertex Cover on bipartite graphs that uses much simpler techniques as well.
Note that a similar reduction would work from the weighted variant of the Vertex Cover problem on bipartite graphs.This can be achieved by further replacing each vertex of the graph G by a clique of size proportional to the weight of this vertex and making each vertex of the clique adjacent to all the neighbors of this vertex.This reduction still preserves the chordality of the resulting graph.
Organization of the Paper.In Section 2, we define the notations and terminology used throughout the paper.In Section 3, we present the FPT algorithm for the generalized Red-Blue Dominating Set problem parameterized by the leafage.In Section 4, we consider the Multicut problem and provide the proof of Theorem 1.2.We present the polynomialtime algorithm for Multiway Cut on chordal graphs in Section 5.In Section 6, we revisit the aforementioned problems by restricting the input to H ℓ -induced-subgraph-free chordal graphs and prove Theorem 1.4 for Dominating Set as well as Theorem 1.5.Finally in Section 7, we consider the Connected Dominating Set and the Steiner Tree problems and complete the proofs of Theorem 1.1 and Theorem 1.4.

Preliminaries
For a positive integer q, we denote the set {1, 2, . . ., q} by [q] and for any 0 ≤ p ≤ q, we denote the set {p, . . ., q} by [p, q].We use N to denote the set of all non-negative integers.Given a function f : Graph Theory.For a graph G, we denote by V (G) and E(G) the set of vertices and edges of G, respectively.Unless specified otherwise, we use n to denote the number of vertices in G.We denote the edge with endpoints u, v by uv.
When the graph G is clear from the context, we omit the subscript G.For any S ⊆ V (G), G − S denotes the graph obtained from G by deleting vertices in S. We denote the subgraph of G induced by S, i.e., the graph G − (V (G) \ S), by G[S].We say graph G contains graph H as in induced subgraph if H can be obtained from G by series of vertex-deletions.Recall that for a directed graph H, we denote by N + H (v) the out-neighbors of v ∈ V (H) and by N − H (v) the in-neighbors of v ∈ V (H).If H is clear from the context, we omit the subscript H.Given a (directed) path P in a graph G and two vertices u, v ∈ V (P ), we denote by P [u, v] the subpath of P from u to v.For any further notation from basic graph theory, we refer the reader to [17].
Trees.A tree T is a connected acyclic graph.Consider a tree T rooted at r.We define function parent(t, T ) : V (T ) \ {r} → V (T ) to specify unique parent of the nodes in rooted tree T .For any node t ∈ T , we denote by T t the subtree rooted at t.A subdivided star is a tree with at most one vertex of degree at least 3 (in other words, it is a tree obtained by repeatedly subdividing the edges of a star graph).The sets V ≥3 (T ) and V =1 (T ) denote the set of vertices of degree at least 3, and of degree equal to 1, respectively.The set V ≥3 (T ) is also called the set of branching vertices of T and the set Chordal graphs and Tree representations.A graph is called a chordal graph if it contains no induced cycle of length at least four.It is well-known that chordal graphs can be represented as intersection graphs of subtrees in a tree, that is, for every chordal graph G, there exists a tree T and a collection M of subtrees of T in one-to-one correspondence with V (G) such that two vertices in G are adjacent if and only if their corresponding subtrees intersect.The pair (T, M) is called a tree representation of G.For every v ∈ V (G), we denote by M(v) the subtree corresponding to v and refer to M(v) as the model of v in T .Throughout this article, we use nodes to refer to the vertices of the tree T to avoid confusion with the vertices of the graph G. Furthermore, we use the greek alphabet to denote nodes of T and the latin alphabet to denote vertices of G.For notational convenience, for any node α ∈ V (T ) and edge e ∈ E(T ), we may abuse notation and write α ∈ M(v) in place of α ∈ V (M(v)) as well as e ∈ M(v) in place of e ∈ E(M(v)).
For every node α ∈ V (T ), we let ver(α is the set of vertices in G that contain the node α is their model.A vertex v ∈ V (G) whose model contains α may also be referred to as an α-vertex.Similarly, for every edge e ∈ E(T ), we let ver(e) = {v ∈ V (G) | e ⊆ M(v)}, that is, ver(e) is the set of vertices of G that contain the edge e in their model.Given a subtree T ′ of T , we denote by G |T ′ the subgraph of G induced by those vertices x ∈ V (G) such that V (M(x)) ⊆ V (T ′ ).If T is rooted, then for each vertex v ∈ V (G), we call the node in M(v) that is closest to the root of T , the topmost node of M(v) and denoted it by top M (v).
The leafage of chordal graph G, denoted by lf(G), is defined as the minimum number of leaves in the tree of a tree representation of G.A tree representation (T, M) for G such that the number of leaves in T is lf(G), can be computed in time O(|V (G)| 3 ) [28].Furthermore, the number of nodes in T is at most O(|V (G)|).
Parameterized Complexity.The input of a parameterized problem comprises an instance I, which is an input of the classical instance of the problem, and an integer k, which is called the parameter.A parameterized problem Π is said to be fixed-parameter tractable (FPT for short) if given an instance (I, k) of Π, we can decide whether (I, k) is a Yes-instance of Π in time f (k) • |I| O(1) for some computable function f depending only on k.We say that an instance (I, k) of a parameterized problem Π and an instance (I ′ , k ′ ) of a parameterized problem Π ′ (possibly Π = Π ′ ) are equivalent if (I, k) ∈ Π if and only if (I ′ , k ′ ) ∈ Π ′ .A reduction rule, for a parameterized problem Π, is a polynomial-time algorithm that takes as input an instance (I, k) of Π and outputs an instance (I ′ , k ′ ) of Π.If (I, k) and (I ′ , k ′ ) are equivalent then we say that the reduction rule is safe.For more details on parameterized algorithms, and in particular parameterized branching algorithms, we refer the reader to the book by Cygan et al. [15].

Dominating Set
In the Dominating Set problem (DomSet for short), the input is a graph G and an integer k, and the objective is to decide whether G has a dominating set of size at most k.We assume that the leafage of the input graph is given as part of the input.If not, recall that it can be computed in polynomial time [28].We consider a generalized version of this problem as defined below.
Red-Blue Dominating Set (Red-Blue-DomSet) Input: A graph G, a partition (R, B) of V (G), and an integer k.Question: Does there exist a set X ⊆ R of size at most k such that B ⊆ N(X)?
We first prove that to solve DomSet, it is sufficient to solve Red-Blue-DomSet even when the input is restricted to chordal graphs of leafage ℓ.There is indeed a simple reduction from the former problem to the latter that preserves the properties in which we are interested.Lemma 3.1.There is a polynomial-time algorithm that given an instance (G, k) of DomSet constructs an equivalent instance (G ′ , (R ′ , B ′ ), k) of Red-Blue-DomSet such that if G has leafage at most ℓ, then so does G ′ .
Proof.We construct G ′ from G as follows.For every vertex v ∈ V (G), add two copies v R and v B to V (G ′ ) and add an edge Suppose that the DomSet instance has a solution S ⊆ V (G).Then the set S R = {v R | v ∈ S}, i.e., S R contains the red version of each vertex in S, is a solution for the Red-Blue-DomSet instance: indeed, the blue vertices v B such that v / ∈ S are dominated since S is a solution, and if v ∈ S, then v B is dominated because of the newly added edges.Conversely, if S R ⊆ R ′ is a solution for the Red-Blue-DomSet instance, then it is easy to see that Finally note that a tree representation for G ′ can be obtained from a tree representation for G by duplicating the model of each vertex, and making the original model a model for the blue version of the vertex, and the copy a model for its red version.In particular, the leafage of G ′ is at most that of G.
In the remainder of this section, we present an FPT algorithm for Red-Blue-DomSet when parameterized by the leafage ℓ of the input graph.The algorithm consists of two parts.In the first part, the algorithm constructs 2 O(ℓ) many instances of a "restricted version" of the problem such that the input instance is a Yes-instance if and only if one of these newly created instances is a Yes-instance.Moreover, the graphs in the newly created instances satisfy certain properties that allow us to design a fast algorithm.See Lemma 3.2 for the formal statement.In the second part (cf.Lemma 3.3), the algorithm solves the restricted version of Red-Blue-DomSet which is defined as follows.
Restricted-Red-Blue Dominating Set (Rest-Red-Blue-DomSet) Input: A chordal graph G, a partition (R, B) of V (G), an integer k and tree representation (T, M) of G such that • for every vertex in G, its model contains at most one branching node of T , and • for all branching nodes γ ∈ V (T ), there are either only red γ-vertices or only blue γ-vertices.Question: Does there exist a set D ⊆ R of size at most k such that B ⊆ N(D)?

Constructing Rest-Red-Blue-DomSet Instances
In this section, we prove the following result.Lemma 3.2.Let I = (G, (R, B), k) be an instance of Red-Blue-DomSet where G is a chordal graph of leafage at most ℓ.We can construct, in time 2 O(ℓ) • n O (1) , a collection ℓ) ], G i is a chordal graph of leafage at most 2ℓ, and • I is a Yes-instance of Red-Blue-DomSet if and only if at least one of the instances in the collection is a Yes-instance of Rest-Red-Blue-DomSet.
Proof.Let G be a chordal graph and let (T, M) be a tree representation of G.We define the following functions.
• Let f T (G) denote the number of branching nodes γ ∈ V (T ) such that there exist both a red vertex and a blue vertex whose models contain γ.
• Let f r (G) denote the number of pairs of consecutive branching nodes α, β in T (that is, no node on the unique path in T from α to β is a branching node) such that there is red vertex whose model contains both α and β.
• Similarly, let f b (G) denote the number of pairs of consecutive branching nodes α, β in T such that there is blue vertex whose model contains both α and β.We further define µ(G) Note that, by definition, µ(G) ≥ lf(G).We design a polynomial-time branching algorithm whose measure µ decreases in each branch.We first show that if µ(G) = lf(G) then (G, (R, B), k) is in fact an instance of Rest-Red-Blue-DomSet and then show how the branching algorithm proceeds.
Assume therefore that µ(G) = lf(G).Then f T (G) = f r (G) = f b (G) = 0 by definition.However, when f T (G) = 0, then, by definition, for every branching node γ ∈ V (T ), all the vertices containing γ in their model are either red or blue; and when f r (G) = f b (G) = 0 then, considering the fact that every model is a subtree in T , for every vertex in G, its model contains at most one branching node in T .Therefore if µ(G) = lf(G), then (G, (R, B), k) is also an instance of Rest-Red-Blue-DomSet. Now assume that µ(G) > lf(G).Then f T (G) + f r (G) + f b (G) > 0. We consider the following three exhaustive cases.
Case-I.f T (G) > 0. Let γ be a branching node in T such that there is both a red-vertex and a blue-vertex whose models contain γ.Suppose that I is a Yes-instance of Red-Blue-DomSet and let D be a solution.Consider first the case where D includes a red vertex whose model contains γ.In this case, we return the instance • Let T 1 be the tree obtained from T by adding a node δ and making it adjacent to γ only.Note that V (T 1 ) \ {δ} ⊆ V (T ).
It is easy to verify that (T 1 , M 1 ) is a tree representation of G 1 and that T 1 has exactly one more leaf than T , i.e., lf(G 1 ) ≤ lf(G) + 1.However, since we have deleted all the blue vertices whose models contained γ, f T (G 1 ) = f T (G) − 1.As the other parts of the measure do not change, µ(G 1 ) < µ(G).
In the second case where no vertex in D contains γ in its model, we return the instance If I is a Yes-instance, then at least one of I 1 or I 2 is a Yes-instance as these two branches are exhaustive.If I 1 is a Yes-instance, then any optimum solution must include a red γ-vertex because of the newly added vertex x.As R 2 ⊆ R, if I 2 is a Yes-instance, then I is a Yes-instance.Hence, this branching step is correct.
Case-II.f T (G) = 0 and f r (G) > 0. Let α, β be two consecutive branching nodes in T such that there is a red vertex whose model contains both α and β.Suppose that I is a Yes-instance of Red-Blue-DomSet and let D be a solution.Consider the case where D includes a red vertex whose model contains both α and β.In this case, we return the instance • Let T 1 be the tree obtained from T by contracting the unique path P αβ from α to β in T and let γ αβ be the node resulting from this contraction.Add a node δ to T 1 and make it adjacent to γ αβ only.Note that V (T 1 ) \ {γ αβ , δ} ⊆ V (T ).
• For every (red or blue It is easy to verify that (T 1 , M 1 ) is a tree representation of G 1 and that T 1 has one more leaf than T which implies lf(G 1 ) ≤ lf(G) + 1.Since we have contracted the path P αβ to obtain the node γ αβ , f r (G 1 ) < f r (G).As the other parts of the measure do not change, µ(G 1 ) < µ(G).
In the second case where no vertex in D contains both α and β in its model, we return an instance R, respectively, by deleting red vertices whose model contains both α and β.It is easy to verify that µ(G 2 ) < µ(G).We argue as in the previous case for the correctness of this branching steps.
Case-III.f T (G) = 0 and f b (G) > 0. Let α, β be two consecutive branching nodes in T such that there is a blue vertex whose model contains both α and β.Note that since f T (G) = 0, for every red vertex Suppose that I is a Yes-instance of Red-Blue-DomSet and let D be a solution.Consider first the case where D includes a red vertex whose model is in V (P αβ ) \ {α, β}.In this case, we return the instance I 1 = (G 1 , (R, B 1 ), k) where G 1 , B 1 are obtained from G and B as follows.
• Delete all the blue vertices whose model contains both α and β.
• Add a blue vertex x to V (G 1 ) (and to B 1 ) with M(x) = V (P αβ ) \ {α, β}.It is easy to verify that (T, M) is a tree representation of G 1 and f b (G 1 ) < f b (G).As the other parts of the measure do not change, µ(G 1 ) < µ(G).
In the second case where there is no vertex in D whose model is in V (P αβ ) \ {α, β}, we consider the following two subcases.If there is a blue vertex v such that M(v) ⊆ V (P αβ ), then we return a trivial No-instance.Otherwise, we return the instance • Let T 2 be the tree obtained from T by contracting the path P αβ from α to β in T and let γ αβ be the node resulting from this contraction.Note that V (T 2 ) \ {γ αβ } ⊆ V (T ).
• For every (red or blue The correctness of this branching step follows from the same arguments as in the previous cases and the fact that in the second case, since there is no red vertex whose model intersects V (P αβ ), it is safe to contract that path.
Finishing the Proof.The correctness of the overall algorithm follows from the correctness of branching steps in the above three cases.To bound its running time and the number of instances it outputs, note that f as these functions either count the number of branching nodes or the unique paths containing exactly two (consecutive) branching nodes.

Solving an instance of Rest-Red-Blue-DomSet
In this section, we present an algorithm to solve Rest-Red-Blue-DomSet. Formally, we prove the following lemma.We first state some easy reduction rules before we handle two cases based on whether the farthest branching node4 is contained only in the models of red vertices or blue vertices.We present Greedy Select 3.9 and Greedy Select 3.12 to handle these cases.The proof of the lemma follows from Lemma 3.10, Lemma 3.13 and the fact that each application of the greedy selection procedure deletes some vertices in the graph.
We first introduce some notations.Recall that an instance of Rest-Red-Blue-DomSet contains a chordal graph G, a partition (R, B) of V (G), an integer k and tree representation (T, M) of G such that for every vertex in G, its model contains at most one branching node of T , and for all branching nodes γ ∈ V (T ), there are either only red γ-vertices or only blue γ-vertices.We assume, without loss of generality, that the tree T is rooted at node r.Unless mentioned otherwise, α denotes the farthest branching node in T from the root, that is, each proper subtree of T α is a path.If there are more than one branching node that satisfy the property, we arbitrarily select one of them.Let β be the closest branching ancestor of α, that is, no internal node in the unique path from α to β is a branching node in T . 5Recall that for a vertex v ∈ V (G), we define top M (v) as the node η ∈ M(v) that is closest to the root.Likewise if a leaf λ is fixed, we define bot λ M (v) as the node η ∈ M(v) that is closest to λ.For ease of notation, we omit λ as it is always clear from the context.Definition 3.4.Let γ be a node of the tree T .We define the following sets of vertices in G.
where the model is completely contained inside the tree rooted at γ but does not contain γ, respectively, i.e.
Simplifications.We first apply the following easy reduction rules whose correctness readily follows from the definition of the problem.It is also easy to see that the reduction rules can be applied in polynomial time and the reduced instance is also a valid instance of Rest-Red-Blue-DomSet. Reduction Rule 3.5.If there is a blue vertex, which is not adjacent to a red vertex, or if k < 0, then return a trivial No-instance.
• If there are two blue vertices u, v such that M(u) ⊆ M(v), then delete v.
• If there are two red vertices u, v such that M(u) ⊆ M(v), then delete u.
Consider a blue vertex v in G whose model is contained in the subtree rooted at α.Moreover, let v be such a vertex for which top M (v) is farthest from the root and v is not adjacent to a red vertex whose model contains α.Hence, there is a natural ordering amongst the red neighbors of v.Note that such an ordering is not possible if some of its neighbors contain α in their models.As any solution contains a red neighbor of v, it is safe to include its neighbor v r for which top M (v r ) is closest to α.
Reduction Rule 3.7.Suppose that there is a blue vertex v ∈ B ⊆ † α such that top M (v) is farthest from the root and v is not adjacent to any red α-vertices.Moreover, amongst all the red neighbors of v, let v r be the node such that top M (v r ) is closest to α.Then, remove v r and all of its blue neighbors and decrease k by 1.
We remark that the above reduction rule is applicable irrespective of the fact whether either all α-vertices are red or all α-vertices are blue.
Case-1: All the vertices that contain α in their models are red vertices.Let β be the closest branching ancestor of α.Consider the blue vertices whose model intersect the path from α to β.Note that there may not be any such blue vertex; however, we find it convenient to present an uniform argument.With a slight abuse of notation, let b 1 , . . ., b d be these blue vertices ordered according to their endpoint in the direction of α, that is, for i < j we have either bot we compute an optimal solution for dominating the vertices whose model is in the tree rooted at α (i.e., the vertices of B ⊆ † α ) and the vertex b i while only using red α-vertices.Formally, we want to compute an optimal solution for the following instance: We also define instance to handle the cases when there are no blue vertices whose model intersects the path from α to β or when b 1 (and hence, the other blue vertices mentioned above) are not dominated by red α-vertices in an optimum solution.To simplify notation we set OPT i := OPT(I i ) in the following.If I i is not defined, then we set OPT i = ∞.Note that the solution OPT i also dominates the blue vertices b 1 , . . ., b i−1 due to the ordering of the b i s.Hence, for any i, j ∈ [0, d] such that i < j, we have |OPT i | ≤ |OPT j |.We use this monotonicity to prove the following structural lemma.Lemma 3.8.Let q ∈ [0, d] be the largest value such that |OPT q | = |OPT 0 |.If there is a solution, then there is an optimum solution containing OPT q .
Proof.Let OPT be an optimum solution of (G, (R, B), k).Let S denote the collection of vertices in OPT whose model contains nodes in the subtree rooted at α, i.e., S := OPT∩R ∩ α .We claim that we can replace S by a super-set S ′ of OPT q of equal size to obtain another solution.
Let j ∈ [0, d] be the largest integer such that b j is dominated by some vertex in S. If j ≤ q, then by our choice of q, |S| = |OPT q |.By the definition of the I i s, we get that OPT q is also a solution for I i .Hence, we can replace S by OPT q to get another optimal solution.Suppose therefore that j > q.By our choice of q, we have |S| > |OPT q |.Let r j be the red α-vertex with top M (r j ) closest to β such that b j is a neighbor of r j .Such a vertex exists, as by assumption, S contains one of these vertices which dominates b j .Then we replace S by S ′ = OPT q ∪ {r j }.As |S| > |OPT q |, we have |S ′ | ≤ |S|.Moreover, observe that S ′ ∪ OPT \ S is still a solution as all vertices in B ⊆ † α and the vertices b 1 , . . ., b q are dominated by some vertex in OPT q , vertex r j dominates the vertices b q+1 , . . ., b j and, by the choice of j, the vertices b j+1 , . . ., b d are dominated by some vertex not contained in S.
We devise a greedy selection step based on the above lemma.
Greedy Select 3.9.Let q ∈ [0, d] be the largest value such that |OPT q | = |OPT 0 |.Include the vertices of OPT q in the solution, i.e., delete the red vertices in OPT q , the blue vertices that are adjacent to vertices in OPT q , and decrease k by |OPT q |.Lemma 3.10.Greedy Select 3.9 step is correct and can be completed it time 2 O(ℓ) • n O (1) .
Proof.The correctness of the step follows directly from Lemma 3.8.In the remaining proof, we show how to compute, for every i by constructing an instance of Set Cover.Before constructing such an instance, we justify that only one blue vertex (which is farthest from α) is critical while constructing this Set Cover instance.
Let α ′ be a child of α.As α is a farthest branching node of T from the root, the tree rooted at α ′ is a path.Let λ be the another endpoint of this path.Consider a blue vertex v α ′ whose model is contained in As Reduction Rule 3.7 is not applicable, there exists at least one red neighbor of v α ′ which is an α-vertex.Hence, an optimum solution can always include a red neighbor of v α ′ which is also an α-vertex.This red α-vertex also dominates all the blue vertices in B ⊆ α ′ .We now explain how to construct an instance (U, F ) of Set Cover.For every child α ′ of α, if the vertex v α ′ mentioned in the previous paragraph exists, then add an element u α ′ corresponding to it to U. When i = 0, add another element u i corresponding to b i to U. For every red α-vertex v, we define set S v ⊆ U as the collection of elements corresponding to the blue vertices in I i that are adjacent to v.This completes the construction of the instance.
It is easy to see the one-to-one correspondence between the optimum solutions of these two instances.The running time of the algorithm follows from the known algorithms for Set Cover (see, for instance, [21]) and the fact that α has at most ℓ many children.
Case-2: All the vertices that contain α in their models are blue vertices.Let β be the closest branching ancestor of α.We consider two cases depending on whether there is a red vertex whose model intersects the path from α to β.If there is no such red vertex, then we consider the graph induced by all the red vertices whose model is (properly) contained in the subtree rooted at α and the blue vertices whose model intersects the subtree rooted at α. Formally, we define . Consider the other case and suppose that there are d ≥ 1 many red vertices whose model intersects the path from α to β.Let r 1 , . . ., r d be these vertices ordered according to their endpoints in the direction of α, that is, for i < j, we have either bot M (r i ) = bot M (r j ) or bot M (r i ) is closer to α than bot M (r j ).For each such red vertex v i , we compute the optimal solution to dominate the vertices in B ∩ α by vertices in R ⊆ α assuming that v i is already selected.Note that we only have to focus on the blue vertices in B ∩ α which are not adjacent to v i .Formally we define . It is possible that the optimum solution does not include any of the vertices in {r 1 , r 2 , . . ., r d }.To handle this case, we define To simplify notation, we set OPT i := OPT(I i ) in the following.Note that for the instance defined above, R i is same for every instance whereas B i ⊆ B i+1 because of the ordering.Hence, for any i, j ∈ [d + 1] such that i < j, we have We use this monotonicity to prove the following structural lemma.Lemma 3.11.If there is a red vertex whose model intersects the path from α to β, let q ∈ [d + 1] be the largest value such that If there is a solution for the instance, then there is an optimum solution OPT such that Proof.If there is no red vertices whose model intersects the path from α to β, then all the red vertices in G that are adjacent to blue vertices in I 0 are the red vertices in I 0 .Hence, the statement of the lemma follows.
We now consider the case where there are red vertices whose model intersects the path from α to β.Let OPT be an optimum solution of (G, (R, B), k).Let S denote the collection of vertices in OPT whose model is (properly) contained in the subtree rooted at α, i.e., S := OPT ∩ R ⊆ † α .We claim that we can replace S by a super-set S ′ of OPT q of equal size to obtain another optimum solution.
Let j ∈ [d] be the smallest index such that v j is contained in OPT.Note that, by definition, j = d + 1 as there are only d red vertices with the said property.If j ≤ q, then by our choice of q, |S| ≥ |OPT j |.By the definition of I j and the fact blue vertices in I j are subset of blue vertices in I q , OPT q is also a solution for I j .Hence, we can replace S by OPT q to get another optimal solution.Suppose therefore that j > q.By our choice of q, we have |OPT j | > |OPT q |.As OPT is a solution, all vertices in B ∩ α must be covered by OPT.Hence, we can replace S by S ′ = OPT q ∪ {r q } and get a solution of not larger size which still dominates all vertices in B ∩ α .Indeed, the vertices which are not dominated by OPT q are dominated by r q .
We devise a greedy selection step based on the above lemma.
Greedy Select 3.12.If there is a red vertex whose model intersects the path from α to β, let q ∈ [d + 1] be the largest value such that |OPT q | = |OPT 1 |.Otherwise, define OPT q = OPT 0 .Include OPT q in the solution, i.e., delete the red vertices in OPT q , the blue vertices that are adjacent to vertices in OPT q , and decrease k by |OPT ℓ |.Lemma 3.13.Greedy Select 3.12 step is correct and can be completed in time 2 O(ℓ) • n O (1) .
Proof.The correctness of the step follows directly from Lemma 3.11.In the remaining proof, we show how to compute OPT i for every i ∈ [0, d + 1], by constructing an instance of Hitting Set.As in Lemma 3.10, we first argue that only one red vertex (which is closest to α) is critical while constructing a Hitting Set instance.
Recall that, by assumption, none of the previous reduction rules are applicable.As in the previous case, let α ′ be a child of α.We first argue that there are no blue vertices whose path is completely contained in the path rooted at α ′ .Assume, for the sake of contradiction, that there exists such a blue vertex v.As Reduction Rule 3.5 is not applicable, v is adjacent to at least one red vertices.However, since all α-vertices are blue, by the property of the instance, there are no red α-vertices.In particular, v is not adjacent to any red α-vertex.This contradicts the fact that Reduction Rule 3.7 is not applicable.Hence, there is no blue vertex whose model is contained in the path rooted at α ′ .Since this is true for any child of α, there are no blue vertices in Since all blue vertices contain α in their model, the only critical red vertex in this leg is v α ′ .
We now explain how to construct an instance (U, F ) of Hitting Set.For every child α ′ of α, let v α ′ be the vertex as mentioned above.Add an element u α ′ corresponding to v α ′ in U.For every blue α-vertex v, we define S v ⊆ U as the collection of elements corresponding to the red vertices in I i that are adjacent to v.This completes the construction of the instance.
It is easy to see the one-to-one correspondence between the optimum solutions of these two instances.The running time of the algorithm follows from the simple brute-force algorithm for Hitting Set parameterized by the size of the universe and the fact that α has at most ℓ many children.

Multicut with Undeletable Terminals
This section considers the MultiCut with Undeletable Terminals problem formally defined as follows.
MultiCut with Undeletable Terminals (MultiCut with UnDel Term) Input: An undirected graph G, a set P ⊆ V (G) × V (G), and an integer k.Question: Is there a set S ⊆ V (G) \ V (P ) such that |S| ≤ k and for all (p, p ′ ) ∈ P , there is no path between p and p ′ in G − S?
In the following, a set S ⊆ V (G) \ V (P ) such that for all (p, p ′ ) ∈ P , there is no path between p and p ′ in G − S is called a P -multicut in G.We first prove that when the input is restricted to chordal graphs, the problem is unlikely to admit an FPT algorithm when parameterized by the leafage.We then complement this result with an XP-algorithm parameterized by the leafage.We restate the theorem with the precise statement for the reader's convenience.
Theorem 1.2.MultiCut with Undeletable Terminals on chordal graphs is W[1]hard when parameterized by the leafage ℓ and assuming the ETH, does not admit an algorithm running in time f (ℓ) • n o(ℓ) for any computable function f .However, it admits an XPalgorithm running in time n O(ℓ) .
To prove that the problem is W[1]-hard, we present a parameter preserving reduction from Multicolored Clique.An instance of this problem consists of a simple graph G, an integer q, and a partition (V 1 , V 2 , . . ., V q ) of V (G).The objective is to determine whether there is a clique in G that contains exactly one vertex from each part V i .Such a clique is called a multicolored clique.We assume, without loss of generality, that each V i is an independent set and that |V 1 | = . . .= |V q | = n. 6This implies, in particular, that |E(G)| < n 2 • q 2 .For every i ∈ [q], we denote by v i 1 , . . ., v i n the vertex set of V i and for every i = j ∈ [q], we denote by E i,j ⊆ E(G) the set of edges between V i and V j .We define Reduction.The reduction takes as input an instance (G, q, (V 1 , . . ., V q )) of Multicolored Clique and outputs an instance (H, P, k) of MultiCut with UnDel Term which is constructed as follows. .Then, it adds edges so that p 1 is complete to K 1 , p n+1 complete to K n , and p a complete to K a−1 ∪ K a for every a ∈ [n] \ {1}.This completes the construction of B (see Figure 1).
• For each i ∈ [q], the reduction introduces two vertex-disjoint copies B i,α and B i,β of B. For every i ∈ [q], let p i,α 1 , . . ., p i,α n+1 denote the copies of p 1 , . . ., p n+1 in B i,α and K i,α 1 , . . ., K i,α n denote the copies of K 1 , . . ., K n in B i,α .Moreover, for every 1 ≤ a 1 ≤ a 2 ≤ n + 1, we define, for notational convenience, We define p i,β a , K i,β a , p i,β [a 1 , a 2 ], and K i,β [a 1 , a 2 ] in a similar way.• For i ∈ [q] and a ∈ [n], the reduction uses p i,α a , p i,β n+1−a , K i,α a , and K i,β n+1−a to encode vertex v i a .
• For every edge e = v i a i v j a j ∈ E(G), the reduction introduces an edge-vertex v e and adds edges so that v e is complete to the following sets. - • The reduction introduces a central clique K of size 2M 2 and makes it complete to {p i,α n+1 , p i,β n+1 | i ∈ [q]} and V E where V E = {v e | e ∈ E(G)} is the set of edge-vertices.This completes the construction of H.
• The reduction further defines The reduction returns (H, P, k) as the instance of MultiCut with UnDel Term.This completes the reduction.It is easy to see that H is chordal and has leafage at most 2q.See Figure 2 for a tree representation of H.
Intuition.We first provide the intuition behind the reduction.Recall that the reduction uses p i,α a , p i,β n+1−a , K i,α a , and K i,β n+1−a to encode vertex v i a where i ∈ [q] and a ∈ a and p i,β b correspond to the same vertex.Note that the pairs in P do not correspond to the vertices associated with v i a .Rather, p i,α a+1 is paired with p By the construction of H and P , for a P -multicut S of H, if there is a path from p i,α a to p i,β b in H − S, then a + b ≥ n + 3. Now, consider the terminal pairs (p i,α 1 , p i,β n+1 ) in P for some i ∈ [q].Because of the size constraints, S cannot contain all the vertices of the central clique K. Since S cannot contain a terminal, it needs to include one clique from B i,α .Let a i ∈ [n] be the largest index such that K i,α a i ⊆ S. Using similar arguments, there must also exist b i ∈ [n] such that K i,β b i ⊆ S and b i is largest such index.By definition of a i , b i and construction of H, there is a path from p i,α The discussion in the previous paragraph implies that a i +1+b i +1 ≥ n+3, i.e., a i + b i ≥ n + 1.However, by definition of the solution size k and the size of the cliques, we have a i + b i ≤ n + 1.Hence, the structure of the auxiliary graphs and the terminal pairs ensure that the selected cliques in S ∩ V (B i,α ) and S ∩ V (B i,β ) encode selecting a vertex in V i in G.
Suppose that {v 1 a 1 , v 2 a 2 , . . ., v q aq } are the vertices in G that are selected by S. Recall that V E is the collection of edge-vertices in H. Considering the remaining budget, a solution S can include at most |E(G)| − q(q − 1)/2 many vertices in V E .We argue that q(q − 1)/2 edges in G corresponding to vertices in V E \ S should have their endpoints in {v 1 a 1 , v 2 a 2 , . . ., v q aq } as otherwise some terminal pair is connected in H −S. Hence, a P -multicut S of H corresponds to a multicolored clique in G.We formalize this intuition in the following two lemmas.
. We construct a P -multicut S of H as follows.First, we add V E \ {v e | e ∈ {v i a i v j a j | i, j ∈ [q]}} to S. For every i ∈ [q], we further add K i,α a i and K i,β n+1−a i to S. It is easy to verify that |S| = q(n + 1)M + |E(G)| − q(q − 1)/2 = k.
Let us show that S is indeed a P -multicut.Fix indices i ∈ [q] and a ∈ [n], and consider the terminal pair (p i,α a , p i,β n+2−a ) in P .Suppose first that a ≤ a i .By construction of H, any path from p i,α a to p i,β n+2−a in H contains a vertex of K i,α a i or of ); in particular, it is not adjacent to p i,α a i .As S only excludes edge-vertices in V E that encode edges incident on v i a i , it contains every vertex in N(p i,α a i ) ∩ V E .Since S also contains every vertex in K i,α a i , we conclude that there is no path from p i,α a to p i,β n+2−a in H − S. Now, consider the case where a i < a, i.e., n+2−a < n+2−a i .In this case, it is convenient to consider a path from p i,β n+2−a to p i,α a .Once again, by construction of H, any path from n+1−a i .As S only excludes edgevertices in V E that encode edges incident on v i a i , it contains every vertex in N(p i,β n+1−a i ) ∩ V E .Since S also contains every vertex in K i,β n+1−a i , we conclude that there is no path from p i,β n+2−a to p i,α a .This implies that no terminal pair in P is connected in H − S which concludes the proof.Lemma 4.2.If (H, P, k) is a Yes-instance of MultiCut with UnDel Term, then (G, q, (V 1 , V 2 , . . ., V q )) is a Yes-instance of Multicolored Clique.
Proof.Assume that (H, P, k) is a Yes-instance of MultiCut with UnDel Term and let S be a P -multicut of H of size at most k.Recall that, by definition of the problem, S ∩ V (P ) = ∅.Also, recall that the reduction adds the clique K of size 2M 2 and makes it complete to {p i,α n+1 , p i,β n+1 | i ∈ [q]} and V E .Note that K \ S = ∅ as k < 2M 2 .Consider an index i ∈ [q].It is easy to see that there exists a ∈ [n] such that K i,α a ⊆ S as otherwise, there is a path from p i,α Note that by definition of a i , b i and the fact that K \ S = ∅, there is path from p i,α a i +1 to p i,β b i +1 in H − S. Now suppose for a contradiction that a i +1+b i +1 ≤ n+2.Then there exists a ′ i ∈ [n] such that a ′ i ≥ a i and a ′ i +1+b i +1 = n+2 and so, by definition of P , (p i,α a ′ i +1 , p i,β b i +1 ) ∈ P .Moreover, by construction of H, the existence of a path from p i,α a i +1 to p i,β b i +1 in H − S implies that there is path from p i,α a ′ i +1 to p i,β b i +1 in H − S; this however, contradicts the fact that S is a P -multicut of H. Therefore a i + 1 + b i + 1 ≥ n + 3, i.e., a i + b i ≥ n + 1.Since this holds for any i ∈ [q], we have that Since |E(G)| − q(q − 1)/2 < M and S has size at most k = q(n + 1)M + |E(G)| − q(q − 1)/2, it follows that, in fact, In the remaining proof, we argue that there are at least q(q − 1)/2 edges with endpoints in {v 1 a 1 , v 2 a 2 , . . ., v q aq }.Since |E(G)| − q(q − 1)/2 < M, and every clique in B i,α is of size at least M, for any a ∈ [n] such that a < a i , we have K i,α a \ S = ∅.In other words, there is at least one vertex in H − S from each clique K i,α a where a < a i .Since a i is the largest index such that K i,α a i ⊆ S, this also holds for every a > a i .As S intersects every path from p i,α 1 to p i,β n+1 , it contains every vertex in N(p i,α a i ) ∩ V E .Using similar arguments, we conclude that S also contains every vertex in In particular, v e is not adjacent to p i,α a i and p i,β n+1−a i .This implies that only edges-vertices that correspond to edges incident on v i a i can be excluded from S. As this holds for any i ∈ [q], every vertex in V E \ S has its endpoints in {v 1 a 1 , v 2 a 2 , . . ., v q aq }.
Since G is a simple graph, it follows that {v 1 a 1 , v 2 a 2 , . . ., v q aq } is a multicolored clique in G.This concludes the proof of the lemma.
Finally, it is known that, assuming the ETH, there is no algorithm that can solve Multicolored Clique on instance (G, q, (V 1 , V 2 , . . .V q )) in time f (q) • |V (G)| o(q) for any computable function f (see, e.g., [15,Corollary 14.23]).Thus, together with the fact that the reduction takes polynomial time in the size of the input, Lemmas 4.1 and 4.2, and arguments that are standard for parameter preserving reductions, we conclude that the following holds.Proof.Let (G, P ) be an instance of MultiCut with UnDel Term where G is a chordal graph of leafage at most ℓ.Let (T, M) be a tree representation of G of leafage at most ℓ.We say that a path in T is a maximal degree-2 path if it contains no branching nodes, except for possibly the first and last node of the path, and it cannot be extended without violating this property (that is, it is maximal).A P -multicut S of G is said to destroy an edge e ∈ E(T ) if ver(e) ⊆ S.
Let us root T at an arbitrary node r ∈ V (T ).Since the number of leaves of T is at most ℓ, T has at most 2ℓ − 2 maximal degree-2 paths, one starting at each each leaf or branching node (except the root) and ending at the first ancestor in T which is a branching node.Now for each maximal degree-2 path Q from α to β in T , guess the first (i.e., closest to α) and last (i.e., closest to β) edge of Q, say e Q 1 and e Q 2 , respectively, such that S destroys e Q 1 and e Q 2 .Note that, it might be the case that an optimal solution does not destroy an edge of Q or only destroys one edge of Q (i.e., e Q 1 = e Q 2 ).Since the length of any maximal degree-2 path is O(n), this creates at most (n + 1) 2ℓ branches.
In each such branch, let D ⊆ E(T ) be the set of guessed edges of T .Pick V D = {ver(e) | e ∈ D} in the solution and delete V D from G: let (G ′ , P ′ ) be the resulting instances and further let T ′ be obtained from T by deleting the edges in D and set M ′ = M| V (G ′ ) .Observe that the tree representation of each connected component of G ′ is given by some tree of the forest T ′ together with M ′ restricted to the vertices of the corresponding connected component.Note that it is enough to solve the problem independently on each connected component of G ′ .
Thus, without loss of generality, assume that G ′ is connected and let (T ′ , M ′ ) be a tree representation of G ′ as defined above.Suppose that G ′ has at least one terminal pair in P ′ , we may assume that for every leaf η of T , ver(η) ∩ P = ∅.Note that, consequently, for every internal node α ∈ V (T ), the truncation of T α w.r.t.P |Tα exists.Now let T 0 be the tree obtained by adding a new node r 0 and connecting it to an arbitrary node r ∈ V (T ).Observe that (T 0 , M) is also a tree representation of G.In the following, we root T 0 at r 0 .To prove Theorem 1.3, we design a dynamic program that computes, in a bottom-up traversal of T 0 , the entries of a table A whose content is defined as follows.The table A is indexed over the edges of E(T 0 ).For each node α ∈ V (T ), A[α parent T 0 (α)] stores the size of a minimum P |Tα -multiway-cut in G |Tα .The size of a minimum P -multiway-cut in G may then be found in A[rr 0 ].We describe below how to compute the entries of A.
Update Procedure.For every leaf η of T , we set A[η parent T 0 (η)] = 0. Consider now an internal node α of T .We show how to compute A[α parent T 0 (α)] assuming that for every edge e ∈ E(T α ), the entry A[e] is correctly filled.
Let T be the truncation of T α w.r.t.P |Tα and let G = G | T .Denote by η 1 , . . ., η q the leaves of T .Recall that, by construction, for every i ∈ [q], there exists p i ∈ P |Tα such that }. Furthermore, it may be that P |Tα ∩ ver(r) is nonempty: we let P r = P |Tα ∩ ver(r).Note that | P r | ≤ 1: if P r = ∅ then we refer to the terminal in P r as the root terminal.Observe that V ( G) ∩ P |Tα = V ( G) ∩ P = P ∪ P r by construction.To compute A[α parent T 0 (α)], we distinguish two cases: (1) if P r = ∅ then we construct a unique instance (H 0 , s, t, wt 0 ) of (s, t)-Cut; (2) otherwise, for every i ∈ [0, q], we construct an instance (H i , s, t, wt i ) of (s, t)-Cut.We describe below how such instances are constructed.First, recall that an instance of the (s, t)-Cut problem consists of a digraph D, vertices s, t ∈ V (D), a weight function wt : E(D) → N ∪ {∞}, and the goal is to find a set X ⊆ E(D) such that D − X has no (s, t)-path and wt(X) is minimum with this property, where wt(X) = u∈X wt(u).
Construction of the (s, t)-Cut Instances.For every i ∈ [q], let us denote by P i = P \ {p i } and let P 0 = P .Consider i ∈ [0, q].Before turning to the formal construction of the instance (H i , s, t, wt i ), let us first give an intuitive idea of the construction.The digraph H i is obtained from T by orienting all edges of T towards its root r = α and further adding vertices and weighted arcs to encode the graph G |Tα .The arcs in H i corresponding to the edges of T are called the tree arcs and the nodes in H i corresponding to the nodes of T are called the tree nodes.The idea is that we separate, for each terminal p ∈ P i , the node top M (p) from the root r.To achieve this, we add a source node s and source arcs from s to top M (p) (of infinite weight) and look for an (s,r)-cut in H i .Since the edges of T can presumably not be independently destroyed in a P -multiway-cut, we need some additional vertices to encode these dependencies.For each vertex v ∈ V ( G) \ P i , we introduce a node γ(v) in H i which is reachable via connection arcs (with infinite weight) from all the tree nodes that are contained in the model of v.This node γ(v) is further connected via a sink arc (of weight one) to top M (v) which ensures that if we want to cut a tree arc, we also have to cut all the sink arcs associated to vertices containing the corresponding edge in their model.The index i is then used to specify which root-to-leaf path of T is uncut: if i = 0 then every such path is cut, otherwise the (η i ,r)-path is uncut.To encode the rest of the solution, we associate with each tree arc (β, δ) a weight wt i ((β, δ)) corresponding to the size of a minimum P |β -multiway-cut in G |β .
We proceed with the formal construction of H i .The vertex set of that is, Γ contains a node of every non-terminal vertex in G.For every z ∈ Γ, we denote by γ −1 (z) the corresponding vertex in V ( G) \ P .The arc set of H i is partitioned into four sets: • the set E T of tree arcs containing all the edges of T oriented towards the root r, Furthermore, if P r = ∅, then we let E rterm ⊆ E T be the set of tree arcs (β, δ) ∈ E T such that the edge βδ is contained in the model of the root terminal; otherwise, we let E rterm = ∅.The weight function wt i : E(H i ) → N ∪ {∞} is defined as follows.For every j ∈ [q], let ρ j be the path in T from η j to r and let − → ρ j be the corresponding directed path in H i (that is, − → ρ j is the path in H i from η j to r consisting only of tree arcs).Then for every arc e of H i , Note, in particular, that every arc in E rterm (if any) has infinite weight.Similarly, if i = 0, then every arc of the path − → ρ i has infinite weight.This completes the construction of the instance (H i , s, t = r, wt i ) (see Figure 3).It is easy to see that such an instance can be constructed in O(n 2 )-time.Now let X 0 be an (s,r)-cut in H 0 such that wt 0 (X 0 ) is minimum; and if P r = ∅, then for every i ∈ [q], further let X i be an (s,r)-cut in H i such that wt i (X i ) is minimum.For each i ∈ [q], let us denote by cost i = A[η i parent T 0 (η i )] and let cost 0 = 0. Then we set In the following, for convenience, we let I = [0, q] if P r = ∅, and I = {0} otherwise.We next show that the entry A[α parent T 0 (α)] is updated correctly.To this end, we show that G |Tα has a P |Tα -multiway-cut of size at most k if and only if there exists i ∈ I such that H i has an (s,r)-cut of weight at most k − cost i w.r.t.wt i .
The instance (H 2 , s,r, wt 2 ) (thick arcs have infinite weight).Proof.Assume that there exists i ∈ I such that H i has an (s,r)-cut Y where wt i (Y ) ≤ k − cost i .For every j ∈ [q] \ {i}, let A j be the set of tree arcs on the path − → ρ j belonging to Y (recall that − → ρ j is the path in H i from η j to r consisting only of tree arcs).Note that since Y is an (s,r)-cut, A j = ∅ for every j ∈ [q] \ {i}.
Claim 5.2.For every terminal j ∈ [q] \ {i}, there exists an arc (x, y) ∈ A j such that for every , the sink arc with tail z belongs to Y .Proof.Suppose for a contradiction that this does not hold for some index j ∈ [q] \ {i}, that is, for every arc (x, y) ∈ A j , there exists z ∈ N + H i (x) \ (N − H i (x) ∪ {y}) such that the sink arc with tail z does not belong to Y .Let (x 1 , y 1 ), . . ., (x a , y a ) be the arcs of A j ordered according to their order of appearance when traversing the path − → ρ j .We show that, in this case, there is a path from s to r in H − Y .For every b ∈ [a], denote by ) the set of vertices z such that the sink arc with tail z does not belong to Y .Let b 1 , . . ., b w ∈ [a] be the longest sequence defined as follows: is a path from s to r in H − Y , a contradiction which proves our claim.
For every j ∈ [q] \ {i}, let e j = (x j , y j ) ∈ A j be the arc closest to r such that for every z ∈ N + H i (x j ) \ (N − H i (x j ) ∪ {y j }), the sink arc with tail z belongs to Y (note that we may have e j = e j ′ for two distinct j, j ′ ∈ [q] \ {i}).Denote by E = {e j | j ∈ [q] \ {i}} ∪ {e * } where e * = (η i , parent(η i )).For every e = (x, y) ∈ E, let P e ⊆ P i be the set of terminals in P i which are also terminals in the instance restricted to T x .Note that {P e | e ∈ E \ {e * }} is a partition of P i : indeed, by construction, every p ∈ P i belongs to at least one such set and if there exist e, e ′ ∈ E \ {e * } such that P e ∩ P e ′ = ∅, then for any j ∈ [q] \ {i} such that p j ∈ P e ∩ P e ′ , e, e ′ ∈ A j ; in particular, both e and e ′ lie on the path − → ρ j , a contradiction to the choice of the arc in A j .Now for every e = (x, y) ∈ E, let S e be a minimum P |Tx -multiway-cut in G |Tx and denote by Proof.Since for every e = (x, y) ∈ E, S e is a P |Tx -multiway-cut in G |Tx , it is in fact enough to show that for every e, e ′ ∈ E, p ∈ P e and p ′ ∈ P e ′ , there is no path from p to p ′ in G |Tα −S.Consider therefore j, j ′ ∈ [q]\{i} such that p j ∈ P e and p j ′ ∈ P e ′ for two distinct e, e ′ ∈ E. Since, as shown above, ∈ P e and p j / ∈ P e ′ ; in particular, e ′ does not lie on the path − → ρ j and e does not lie on the path − → ρ j ′ .It follows that any path in G |Tα from p j to p j ′ contains at least one vertex x whose model contains the edge corresponding to e; but then, γ(x) ∈ N e and so, x ∈ S by construction.Thus, there is no path from p j to p j ′ in G |Tα − S.
Finally, note that, by construction, Proof.Recall that for every j ∈ [q], ρ j is the unique (η j ,r)-path in T .To prove the lemma, we first show the following.Claim 5.5.If there exists i ∈ [q] such that G |Tα has a P |Tα -multiway-cut X of size at most k where (1) X does not destroy any edge of ρ i and (2) for every j ∈ [q] \ {i}, X destroys an edge of ρ j , then Proof.Assume that such an index i ∈ [q] exists and let X be a P |Tα -multiway-cut X of size at most k satisfying item (1) and (2).Note that since X does not destroy any edge of ρ i , P r = ∅ for, otherwise, p i and the root terminal would be in the same connected component of G |Tα −X thereby contradicting the fact that X is a P |Tα -multiway-cut.For every j ∈ [q]\{i}, let e j ∈ E( T ) be the closest edge to η j on ρ j such that ver(e j ) ⊆ X (note that the edges e 1 , . . ., e q are not necessarily pairwise distinct).Denote by E = {e j | j ∈ [q] \ {i}}.We construct an (s,r)-cut Y in H i as follows: Y contains the tree arcs of H i corresponding to the edges in E and for each v ∈ X such that M(v) contains at least one edge of E (that is, v ∈ ver(e) for some edge e ∈ E), we include in , respectively) be the set of nodes of the subpath of ρ j from η j to the tail of e j (the head of e j to r, respectively).We contend that for every j ∈ [q] \ {i}, there is no (V j − , V j + )-path in H i − Y .Note that if true, this would prove that Y is indeed an (s,r)-cut in H i .For the sake of contradiction, suppose that, for some in particular, the sink arc e with tail z must belong to L. By construction of H i , it must then be that M(γ −1 (z)) contains the edge e j , that is, γ −1 (z) ∈ ver(e j ); but then, γ −1 (z) ∈ X and so, e ∈ Y by construction, a contradiction which proves our claim.
Let us finally show that wt i (Y ) ≤ k − cost i .To this end, for every e ∈ E, let X e ⊆ X be the restriction of X to T te where t e is the endpoint of e the furthest from r (note that for any two distinct e, e ′ ∈ E, X e ∩ X e ′ = ∅).Then, for every e ∈ E, X e is a P |Tt e -multiway-cut in G |Tt e and so, wt i (e) ≤ |X e |.Similarly, the restriction X i of X to T η i is a P |Tη i -multiway-cut in G |Tη i and so, |X i | ≥ cost i (note that, by construction, X i ∩ X e = ∅ for every e ∈ E).Letting X ′ = e∈E ver(e), it then follows from the definition of Y that Using similar arguments, we can also prove the following.Claim 5.6.If G |Tα has a P |Tα -multiway-cut X of size at most k such that for every i ∈ [q], X destroys an edge of ρ i , then H 0 has an (s,r)-cut Y such that wt i (Y ) ≤ k.
To conclude the proof of Lemma 5.4, let us show that for any P |Tα -multiway-cut S in G |Tα , S destroys an edge of every root-to-leaf path of T , except for at most one when P r = ∅.Note that if the claim is true, the lemma would then follow from Claims 5.5 and 5.6.
Let S be a P |Tα -multiway-cut in G |Tα .Observe first that if P r = ∅ then for every i ∈ [q], S must destroy an edge of ρ i for, otherwise, p i and the root terminal are in the same connected component of G |Tα −S, thereby contradicting the fact that S is a P |Tα -multiway-cut.Assume therefore that P r = ∅ and suppose, for the sake of contradiction, that there exist two distinct indices i, j ∈ [q] such that S destroys no edge of ρ i and no edge of ρ j .Then for every edge e of ρ i ∪ ρ j , ver(e) \ S = ∅: for each such edge e, let α e ∈ ver(e) \ S. It is now not difficult to see that there is a path in G |Tα − S from p i to p j using only vertices from {α e | e is an edge of ρ i ∪ ρ j }, a contradiction to the fact that S be a P |Tα -multiway-cut in G |Tα .
We now conclude by Lemmas 5.1 and 5.4, that A[α parent T 0 (α)] indeed stores the size of a minimum P |Tα -multiway-cut in G |Tα .Since the construction of each H i takes polynomialtime, an (s, t)-cut in H i can be computed in polynomial time (see, for instance, [22]) and the number of H i s is at most n, it takes plynomial-time to update A[α parent T 0 (α)].Finally, since the number of edges of T is linear in n, the overall running time is polynomial in n, which proves Theorem 1.3.We remark that a more careful analysis of the running time of the algorithm leads to an upper bound of O(n 4 ).

Restricting to H ℓ -induced-subgraph-free chordal graphs
In this section, we consider problems restricted to H ℓ -induced-subgraph-free chordal graphs.Recall that H ℓ is the split graph on 2ℓ vertices such that if V (H ℓ ) = C ⊎ I is a split partition then (i) |C| = |I| = ℓ, (ii) every vertex in C is adjacent to exactly one vertex in I, and (iii) every vertex in I is adjacent to exactly one vertex in C. As mentioned in the Introduction, the class of H ℓ -induced-subgraph-free chordal graphs is a natural generalization of the class of chordal graphs of leafage at most ℓ.In fact, denoting by C ℓ the collection of all chordal graphs that have leafage at most ℓ and by C is ℓ the collection of all chordal graphs that do not contain H ℓ as a induced subgraph (that is, the collection of H ℓ -induced-subgraph-free chordal graphs), the following holds.Observation 6.1.C ℓ C is ℓ+1 .Let us briefly explain why Observation 6.1 holds true.Walter generalized the concept of asteroidal triple in order to characterize other subclasses of chordal graphs [45] as follows.A subset of nonadjacent vertices of G is an asteroidal set if the removal of the closed neighborhood of any one of its elements does not disconnect the remaining ones.Formally, a set of vertices A of a graph G is asteroidal if for each a ∈ A, the vertices in A \ {a} belong to a common connected component of G − N[a].The asteroidal number of G, denoted by at(G), is then the size of a largest asteroidal set of G.Note that in the graph H ℓ+1 , I is an asteroidal set of size ℓ + 1 and thus, at(H ℓ+1 ) ≥ ℓ + 1.By definition, if H is a subgraph of G and H is connected, then at(H) ≤ at(G).Lin et al. [37,Therorem 1] proved that for a connected chordal graph G, at(G) ≤ lf(G).Hence, if lf(G) ≤ ℓ, then it cannot contain H ℓ+1 as an induced subgraph.This implies that C ℓ ⊆ C is ℓ+1 .To see that C ℓ is proper subset of C is ℓ+1 , consider a graph obtained from a star by subdividing every edge once.Then it is easy to see that this graph does not contain H 3 as induced subgraph but can have unbounded leafage.
The remainder of this section is organized as follows.In Subsection 6.1, we argue that the FPT algorithms for domination problems cannot be generalized to this larger graph class.We complement this with an XP-algorithm, which is optimal under the ETH.In Subsection 6.2, we present a simple algorithm to prove that MultiCut with UnDel Term is paraNP-hard on this graph class.This implies that the XP-algorithm presented in Section 4 cannot be generalized for this larger class.

Dominating Set and related problems
In this subsection, we prove Theorem 1.4.We first show the hardness results of the theorem and provide afterwards the XP-algorithms for the problems.Lemma 6.2.Dominating Set, Connected Dominating Set and Steiner Tree on H ℓ -induced-subgraph-free chordal graphs are W[1]-hard when parameterized by ℓ and assuming the ETH, do not admit an algorithm running in time f (ℓ) • n o(ℓ) for any computable function f .Proof.We present a parameter preserving reduction from Multicolored Independent Set.An instance of this problem consists of a graph G, an integer q, and a partition (V 1 , . . ., V q ) of V (G).The objective is to determine whether G has an independent set which contains exactly one vertex from every part V i .We assume, without loss of generality, that each V i is an independent set.We present a slight modification of a known reduction (see [15,Theorem 13.9]).
Reduction.The reduction takes as input an instance (G, q, (V 1 , . . ., V q )) of Multicolored Independent Set and constructs a graph G ′ as follows.
• For every vertex v ∈ V (G), the reduction introduces a vertex v into G ′ : we denote by C the set of all these vertices in G ′ .Note that the sets V i carry over directly to G ′ .
• The reduction turns the set C into a clique in G ′ by adding edges between any two distinct vertices of C.
• For every i ∈ [q], the reduction introduces two new vertices x i , y i into G ′ and makes them adjacent to every vertex of V i .
• For every edge e = uv ∈ E(G) with endpoints u ∈ V i and v ∈ V j , the reduction introduces a vertex w e into G ′ and makes it adjacent to every vertex of (V i ∪V j )\{u, v}.For DomSet and connDomSet the reduction returns the instance (G ′ , q).For Steiner Tree, it sets all the vertices in V (G ′ ) \ C as terminals and returns the instance (G ′ , V (G ′ ) \ C, q).Since by construction, u 1 is not adjacent to v 2 and v 3 , and v 2 , v 3 are in C, it must be that u 1 ∈ I.But then, u 1 is adjacent to one vertex in V i , namely v 1 , and nonadjacent to two vertices in V i , namely v 2 and v 3 , a contradiction to the fact that vertices in I can be partitioned into the three sets described above.Therefore, G does not contain H 2q+2 as an induced subgraph.
It is known that, assuming the ETH, there is no algorithm that can solve Multicolored Independent Set on instance (G, q, (V 1 , V 2 , . . .V q )) in time f (q) • |V (G)| o(q) for any computable function f (see, e.g., [15,Corollary 14.23]).Note finally, that and G ′ is an H 2q+2 induced-subgraph-free split graph.These facts, together with arguments that are standard for parameter preserving reductions, concludes the proof of the lemma.
In the following, we give the XP-algorithms for the three problems.Instead of giving the algorithm for DomSet, we give an algorithm for the more general Red-Blue-DomSet.Recall that, from Lemma 3.1, there is a reduction from the former to the latter problem.There remains to argue that this reduction preserves the property of being H ℓ -inducedsubgraph-free.Lemma 6.5.There is a polynomial-time algorithm that given an instance (G, k) of DomSet constructs an equivalent instance Proof.As in Lemma 3.1, we construct G ′ from G as follows.For every vertex v ∈ V (G), add two copies v R and v B to V (G ′ ) and add an edge v R v B to E(G ′ ).For every edge uv ∈ E(G), add edges v R u R , v R u B , v B u R , and v B u B to E(G ′ ).This completes the construction of G ′ .By the proof of Lemma 3.1, it is known that these two instances are equivalent.In the following, we let Now assume that G is H ℓ -induced-subgraph-free and suppose, for the sake of contradiction, that G ′ contains H ℓ as an induced subgraph.Let I be the vertices forming the independent set and C the vertices forming the clique of H ℓ .We claim that for no vertex v ∈ V (G), we have that v B , v R ∈ C ∪ I.Note that if the claim holds, then using the original version of each vertex would give an induced H ℓ in G and thus contradict our assumption.
There remains to prove the claim.To this end, consider v ∈ V (G).Since I is an independent set, v B and v R cannot both be contained in I.Moreover, it can also not be the case that v B ∈ I and v R ∈ C (or vice-versa) as then v B would also be adjacent to all vertices in C. Hence, assume that v B , v R ∈ C. Assume, without loss of generality, that u B ∈ I is the unique vertex adjacent to v B in C (the case where u R ∈ I is the unique adjacent vertex is symmetric).Since there is an edge from v B to u B , we know that u and v are adjacent in G. Hence, by construction, there must also be an edge from v R to u B which contradicts the fact that we have an H ℓ graph.
We are now ready to show that Red-Blue-DomSet on chordal graphs admits an XPalgorithm if the input graph does not contain H ℓ as induced subgraph.Lemma 6.6.Red-Blue Dominating Set restricted to H ℓ -induced-subgraph-free chordal graphs admits an algorithm running in time n O(ℓ) .
Proof.Let (G, (R, B), k) be an instance of Red-Blue-DomSet where G is an H ℓ -inducedsubgraph-free chordal graph, and let (T, M) be a tree representation of G. First, we add a node r to T by connecting it to an arbitrary node of T and root T at r (note that, by construction, no model in M contains r).We use dynamic programming to compute the entries of two tables T 1 and T 2 in a bottom-up traversal of T .The contents of T 1 and T 2 are defined as follows.For every node α ∈ V (T ) and every nonempty set X ⊆ R ∈ α of size at most ℓ, Intuitively, this stores the (size of the) smallest set of red vertices containing X such that all blue vertices in T α are dominated.
For every node α ∈ V (T ) and every set Y ⊆ R ⊆ † α of size at most ℓ, Intuitively, this stores the (size of the) smallest set of red vertices intersecting with T α but not α which dominate all blue vertices below α and the α-blues that are neighbors of the red vertices in Y .Initially, every entry of T 1 and T 2 is set to +∞.The output is Yes if and only if T 2 [r, ∅] ≤ k.We next show how to update the entries of T 1 and T 2 .
Updating the Leaves.Let α ∈ V (T ) be a leaf of T .Then set For every i ∈ I, further let X i = X ∩ R ∈ β i .We update T 1 [α, X] according to the following procedure.
1.For every i ∈ I, set m i = min 2. For every i ∈ I, let and set To update the entries of T 1 [β i , Z] and m 2 i = min We next show that the entries of T 1 [α, •] and T 2 [α, •] are updated correctly.To this end, we first introduce some useful notation.Given a set X ⊆ B, a set S ⊆ R minimally dominates X if X ⊆ N(S) and for every x ∈ S, X ⊆ N(S \ {x}).Additionally, we prove the following.Claim 6.7.For every node α ∈ V (T ), the following hold.
(i) For every minimum red-blue dominating set S of G, |S ∩ R ∈ α | ≤ ℓ. (ii) For every set X ⊆ B ∈ α and every set Y ⊆ R \ R ∈ α minimally dominating X, |Y | ≤ ℓ.Proof.To prove item (i), let S be a minimum red-blue dominating set of G. Since S is minimum, for every x ∈ S ∩ R ∈ α , there exists p x ∈ N(x) ∩ B such that p x / ∈ y∈S\{x} N(y), i.e., the blue vertex p x is only dominated by x.Then {p x | x ∈ S ∩ R ∈ α } is an independent set: indeed, if there exist x, y ∈ S ∩ R ∈ α such that p x p y ∈ E(G) then x, p x , p y , y induces a C 4 , a contradiction as α and let Y ⊆ R\R ∈ α be a set minimally dominating X.Since Y is minimal, for every x ∈ Y , there exists p x ∈ N(x) ∩ X such that p x / ∈ y∈Y \{x} N(y), i.e., the blue vertex p x is only dominated by x.This implies that Y is an independent set: indeed, if there exist x, y ∈ Y such that xy ∈ E(G) then x, p x , p y , y induces a C 4 , a contradiction as We now move towards proving the correctness of the update procedure.We start with the first table.For every i ∈ I, further let X i = X ∩ R ∈ β i .For every i ∈ I, let m i be as defined in Step 6.1 and let Then, since for every i ∈ I, T 1 [β i , •] has been correctly filled, there exists a set and S i dominates every vertex in B ∩ β i .Similarly, for every i ∈ I, let m 1 i and m 2 i be as defined in Step 6.1.Further let I 1 ⊆ I be the set of indices i ∈ I such that min{m 1 i , m 2 i } = m 1 i and set . Thus, we conclude by the update step and the above that as claimed.Now by observing that S ∩R ∩ α is a minimum-sized set dominating every vertex in B ∩ α and whose intersection with R ∈ α is X (S would otherwise not be minimum), we conclude by the above that T 1 [α, •] is updated correctly.
Finally, it is not difficult to see that it takes n O(ℓ) -time to update one entry of T 1 [α, •] and since there are n O(ℓ) entries, the claim follows.
We next show the correctness of the update procedure for the second table.Claim 6.9.For every internal node α ∈ V (T ), the entries of T 2 [α, •] are updated correctly.Furthermore, T 2 [α, •] can be updated in n O(ℓ) -time.
Proof.Let α ∈ V (T ) be an internal node of T with children β 1 , . . ., β p and assume that for every i and m 2 i be as defined in Step 6.1.Further let I 1 ⊆ I be the set of indices i ∈ I such that min{m 1 i , m 2 i } = m 1 i and set α as considered in Step 6.1 such that the final value of Int N is minimum among all such final values taken over every partition of N(Y ) ∩ B ∈ α as considered in Step 6.1.For every i ∈ I, let Y N i , m 1 i and m 2 i be as defined in Step 6.1.Let I 1 ⊆ I be the set of indices i ∈ I such that min{m 1 i , m 2 i } = m 1 i and set I 2 = I \ I 1 .
For every i ∈ I ).We contend that the set M = i∈I S i is the desired S. Indeed, observe first that, by the update step, Similarly, let . Thus, we conclude by the update step and the above that ) (S would otherwise not be minimum), we conclude by the above that T 2 [α, •] is updated correctly.
Finally, it is not difficult to see that Step 6.1 can be done in n O(ℓ) -time and that, similarly, for a fixed partition, Steps 6.1-6.1 can be done in n O(ℓ) -time.Now observe that |I| ≤ ℓ since |Y | ≤ ℓ and thus, there are at most n O(ℓ) partitions to consider in Step 6.1.
The lemma now follows from Claims 6.8 and 6.9.

MultiCut with Undeletable Terminals
We present a simple reduction from Vertex Cover to Multicut with UnDel Term to prove Theorem 1.5.Consider an instance (G, q) of Vertex Cover where G has n vertices.Let G ′ be a graph obtained from a star with center r and n + 1 leaves by subdividing each of its edge once.Fix an injective mapping f : V (G) → V (G ′ ) such that f (v) is a leaf for every v ∈ V (G).Let w be the unique leaf which is not in the range of f .Then, the set of terminal pairs P is defined as follows: P = {(f (u), f (v) | uv ∈ E(G)} ∪ {(r, w)}.It is easy to see that (G, q) is a yes-instance of Vertex Cover if and only if (G ′ , P, q) has a multicut of size at most q.As G ′ is acyclic, it is H 3 -induced free.

Other domination-related problems
The aim of this section is to complete the proofs of Theorem 1.1 and Theorem 1.4.More precisely, we show that Connected Red-Blue-DomSet and Steiner Tree are FPT parameterized by leafage and admit a n O(ℓ) -algorithm on H ℓ -induced-subgraph-free chordal graphs.The two problems are considered in two separate subsections.

Connected Red-Blue Dominating Set
In this subsection, we aim to prove that Connected Dominating Set is FPT parameterized by the leafage and admits a n O(ℓ) -algorithm on H ℓ -induced-subgraph-free chordal graphs.Formally, we prove the following.intersection is nonempty).Further let β ∈ V (T ) be the neighbor of α i on the path in T from α i to α.Then since T is minimal, I = {x ∈ V (G) | α i ∈ M(x)}\{x ∈ V (G) | β ∈ M(x)} = ∅; and since α i , β ∈ M(v i ), v i / ∈ I. Thus, we may set x i = x where x ∈ I.

Steiner Tree
The aim of this section is to prove that Steiner Tree is FPT parameterized by the leafage and admits an n O(ℓ) -algorithm on H ℓ -induced-subgraph-free chordal graphs.To obtain these results, we give two parameter preserving reductions to Red-Blue-DomSet.We first present a general reduction rule for Steiner Tree instances.) is a Yes-instance for Steiner Tree and let S be a solution.By construction, for every neighbor y of v C in S, there exists x ∈ V (C) such that y ∈ N(x): for every y ∈ N(v C ) ∩ S, let x y ∈ V (C) be an arbitrary vertex such that y ∈ N(x y ).Set V = {x y | y ∈ N(v C ) ∩ S} and for every x ∈ V , denote by N x = {y ∈ N(v C ) ∩ S | x y = x}.Now let x 1 , . . ., x p be an arbitrary ordering of V and let y 1 , . . ., y q be an arbitrary ordering of V (C) \ V .Then the tree obtained from S by removing the vertex v C to replace it with the path x 1 . . .x p y 1 . . .y q and adding the edges {x i z | i ∈ [p] and z ∈ N x i } is readily seen to be a solution for (G, T , k).

Reduction
Finally, let us remark that a tree representation for G/V (C) can be obtained from a tree representation (T, M) of G by merging the models in {M(x) | x ∈ V (C)} into a single model representing v C ; in particular, the leafage of G/V (C) is at most that of G. Lemma 7.6.Let (G, T , k) be an instance of Steiner Tree and let (G R , T R , k) be the instance resulting from an exhaustive application of Reduction Rule 7.4 to (G, T , k).If G is H ℓ -induced-subgraph-free then G R is H ℓ+1 -induced-subgraph-free.
Proof.Assume that G[T ] contains at least one connected component of size greater than 1 (the lemma is trivial otherwise) and let C 1 , . . ., C p be all such connected components of G[T ].For every i ∈ [p], denote by v C i ∈ V (G R ) the vertex resulting from the contraction of C i .Now assume that G is H ℓ -induced-subgraph-free and suppose for a contradiction that G R contains an induced H ℓ+1 .Let v 1 , u 1 , . . ., v ℓ+1 , u ℓ+1 ∈ V (G R ) be 2(ℓ+1) vertices inducing an H ℓ+1 in G R where {v i | i ∈ [ℓ + 1]} is the clique and for every i ∈ Proof.As mentioned above, we reduce to Connected Red-Blue-DomSet: given an instance (G, T , k) of Steiner Tree, we construct an instance (H, (R, B), k H ) of Connected Red-Blue-DomSet as follows.First, we assume that Reduction Rule 7.4 has been exhaustively applied to (G, T , k).This implies, in particular, that T is an independent set of G. Let us further assume that |T | > 1 (the problem is trivial otherwise).Now let G * be the supergraph of G obtained by making each terminal simplicial, that is, for every t ∈ T , the neighborhood N G (t) (= N G * (t)) of t induces a clique in G * .Observe that the leafage of G * is at most that of G: indeed, a tree representation for G * can be obtained from a tree representation (T, M) of G as follows.For every terminal t ∈ T , let α t ∈ V (T ) be a node of T contained the model M(t) of t.If there exists a neighbor x ∈ N G * (t) such that M(x) does not contain α t , then we extend M(x) by adding to it the path in T from α t to α x where α x ∈ M(x) is the closest node to α t in T .By iterating this process and leaving all the other models intact, we obtain a tree representation (T * , M * ) for G * where T * has the same number of leaves as T .
Reduction.We may now construct the graph (H, (R, B)): the set R = {r x | x ∈ V (G)} of red vertices contains a copy of each vertex in V (G) and the set B = {b t | t ∈ T } of blue vertices contains a copy of each terminal.The graph H[R] is then isomorphic to G * and for every t ∈ T , b t is a true twin to r t .Finally, we set k H = k − |T |.We next show that the instances (G, T , k) and (H, (R, B), k H ) are equivalent.Proof.Assume that (G, T , k) is a Yes-instance for Steiner Tree and let S be a solution.Note that since |T | > 1 by assumption, necessarily V (S) \ T = ∅.We contend that the set D = {r x | x ∈ V (S) \ T } is a solution for (H, (R, B), k H ). Indeed, it is clear that for every t ∈ T , b t has a neighbor in D. To see that D is connected, observe that if a terminal t ∈ T is not a leaf of S, then t has at least two neighbors in S; but the neighborhood of r t (and in time 2 O(ℓ) • n O(1) -time for the Dominating Set problem on chordal graphs, and used it to obtain similar results for the Connected Dominating Set and Steiner Tree problems.Regarding cut problems, we proved that MultiCut with Undeletable Terminals on chordal graphs is W[1]-hard when parameterized by the leafage.We also presented a polynomial-time algorithm for Multiway Cut with Undeletable Terminals on chordal graphs.We find it surprising that the complexity of this problem was not known before.Finally, we examined these problems on H ℓ -induced-subgraph-free chordal graphs to check the extent of our approach.
In the case of chordal graphs, we believe the leafage to be a more natural parameter than other popular parameters such as vertex cover, feedback vertex set or treewidth.It would be interesting to examine the structural parameterized complexity of problems such as Longest Cycle, Longest Path, Component Order Connectivity, s-Club Contraction, Independent Set Reconfiguration, Bandwidth, or Cluster Vertex Deletion.These problems are known to be NP-complete on split graphs and admit polynomial-time algorithms on interval graphs.Hence it is plausible that they admit an FPT or XP algorithm on chordal graphs parameterized by the leafage.We believe it is a representative list, though not exhaustive, of problems that exhibit this behavior.In fact, it would be fascinating to find a natural problem that does not exhibit this behavior, i.e., a problem that is NP-complete on interval graphs but admits a polynomial-time algorithm on split graphs.

Figure 1 :
Figure 1: The auxiliary graph B. Rectangles represent cliques and thick edges indicate that the corresponding vertex is complete to the corresponding cliques.

•
The reduction starts by constructing an auxiliary graph B. The vertex set of B consists of n + 1 vertices p 1 , . . ., p n+1 and n vertex-disjoint cliques K 1 , . . ., K n such that |K a | = a • M for every a ∈ [n]

Lemma 4 . 3 .
MultiCut with Undeletable Terminals on chordal graphs is W[1]-hard when parameterized by leafage ℓ and assuming the ETH, does not admit an algorithm running in time f (ℓ) • n o(ℓ) for any computable function f .The remainder of this section is devoted to the proof of the following lemma, which together with Lemma 4.3 proves Theorem 1.2.

Lemma 4 . 4 .
MultiCut with Undeletable Terminals on chordal graph of leafage at most ℓ admits an XP-algorithm running in time n O(ℓ) .

Figure 3 :
Figure 3: An illustration of the construction of the (s, t)-Cut instances.
by construction, and so, h b l necessarily lies on − → ρ j [y b l , x b l +1 ].Now observe that, by maximality of the sequence, b w = a: indeed, if b w < a then the sequence could be extended as Z bw+1 = ∅ by assumption.Since z bw / ∈ N − H i (x bw ), this implies, in particular, that h bw lies on the path − → ρ j [y bw ,r].It follows that

T 2
[α, ∅] = 0 and for every nonempty set X ⊆ R ∈ α of size at most ℓ, set T 1 [α, X] = |X|.Updating Internal Nodes.Let α ∈ V (T ) be an internal node of T and let β 1 , . . ., β p be the children of α.To update the entries of T 1 [α, •], we proceed as follows.Let X ⊆ R ∈ α be a nonempty set of size at most ℓ.Denote by I ⊆ [p] the set of indices i ∈ [p] such that X ∩ R ∈ β i = ∅ and set I = [p] \ I.
T 2 [α, •], we proceed as follows.Let Y ⊆ R ⊆ † α be a set of size at most ℓ.Denote by I ⊆ [p] the set of indices i ∈ [p] such that Y ∩ R ∩ β i = ∅ and set I = [p] \ I.We update T 2 [α, Y ] according to the following procedure.1. Initialise OPT I = 0 and OPT I = +∞.2. For every i ∈ I do: 2.a.Let

2
.b. Set OPT I = OPT I + min{m 1 i , m 2 i }. 3.For every partition N = {N i | i ∈ I} of N(Y ) ∩ B ∈ α where for every i ∈ I, N i ⊆ N(Y i ) ∩ B ∈ α do: 3.a.Initialise Int N = 0. 3.b.For every i ∈ I do: 3.b.i.Let Y

Claim 6 . 8 .
For every internal node α ∈ V (T ), the entries ofT 1 [α, •] are updated correctly.Furthermore, T 1 [α, •] can be updated in n O(ℓ) -time.Proof.Let α ∈ V (T ) be an internal node of T with children β 1 , . . ., β p and assume that for everyi ∈ [p], T 1 [β i , •] and T 2 [β i , •] have been correctly filled.Let us first show that for every nonempty set X ⊆ R ∈ α of size at most ℓ, there exists a set S ⊆ R ∩ α of size T 1 [α, X] such that S ∩ R ∈ α = Xand S dominates every vertex in B ∈ α .Consider a nonempty set X ⊆ R ∈ α of size at most ℓ.Let I ⊆ [p] be the set of indices i ∈ [p] such that X ∩ R ∈ β i = ∅ and set I = [p] \ I.

Lemma 7 . 1 .
Connected Red-Blue Dominating Set is FPT parameterized by the leafage and admits n O(ℓ) -algorithm on H ℓ -induced-subgraph-free chordal graphs.

Rule 7 . 4 .
Let (G, T , k) be an instance of Steiner Tree.If G[T ] has a connected component C of size greater than 1, then return the instance (G/V (C), (T \V (C))∪ {v C }, k − |V (C)| + 1) where v C is the vertex resulting from the contraction of C in G. Lemma 7.5.Reduction Rule 7.4 is safe.Furthermore, the leafage of G/V (C) is at most that of G. Proof.Suppose that such a connected component C exists.Assume first that (G, T , k) is a Yes-instance for Steiner Tree and let S be a solution for (G, T , k) such that the number of connected component in S[V (C)] is minimum amongst all solutions for (G, T , k).We claim that S[V (C)] has only one connected component.Indeed, suppose to the contrary that S[V (C)] has at least two connected components.Since C is connected, there exist two connected components C 1 and C 2 of S[V (C)] such that C 1 and C 2 are adjacent, that is, there is an edge xy ∈ E(G) where x ∈ V (C 1 ) and y ∈ V (C 2 ).Let L = z 1 . . .z p be a shortest path in S from C 1 to C 2 .Then the tree S ′ = S − {z 1 z 2 } + {xy} is a solution for (G, T , k) such that S ′ [V (C)] contains fewer connected component than S[V (C)], a contradiction to the choice of S. Thus, S[V (C)] has only one connected component and it is easy to see that S/V (C) is a solution for (G/V (C), (T \ V (C)) ∪ {v C }, k − |V (C)| + 1).Conversely, assume that (G/V (C), (T \ V (C)) ∪ {v C }, k − |V (C)| + 1 ≤ 1: let us assume without loss of generality that{v i | i ∈ [ℓ]}∩{v C i | i ∈ [p]} = ∅.On the other hand, if v C i = u j i for some i ∈ [p] and j i ∈ [ℓ], then, by construction, there existsx i ∈ V (C i ) such that x i v j i ∈ E(G): let I ⊆ [p] be the set of such indices.Then X = {x i | i ∈ I}∪{u i | i ∈ [ℓ]\I} is an independent set in G where each vertex in X has exactly one neighbor in K = {v i | i ∈ [ℓ]}, that is, K ∪ X induces an H ℓ in G, a contradiction.Lemma 7.7.Steiner Tree parameterized by the leafage is FPT.

Claim 7 . 8 .
If (G, T , k) is a Yes-instance for Steiner Tree, then (H, (R, B), k H ) is a Yes-instance for Connected Red-Blue-DomSet.
as claimed.Let us finally show that M dominates every vertex in B ∩ α .First observe that since X = ∅, every vertex in B ∈ α is dominated by M. Consider therefore a vertex x ∈ B ⊆ † α .Then there exists i ∈ [p] such that x ∈ B ∩ β i .If i ∈ I then x is dominated by S i by definition; similarly, if i ∈ I 1 then x is dominated by S i by definition.Thus, suppose that i ∈ I 2 .Then either x ∈ B ⊆ † β i in which case x is dominated by S i by definition; or x ∈ B ∈ β i and since x / ∈ B ∈ α by assumption, x ∈ N(Z i ) by construction and thus, x is dominated by S i by definition.Therefore, M dominates every vertex in B ∩ α and so, M is indeed the desired S. Consider now a minimum red-blue dominating set S of G such that S ∩ R ∈ α = ∅.Then by Claim 6.7(i), |S ∩R ∈ α | ≤ ℓ.Let us show that |S ∩R ∩ α | ≥ T 1 [α, S ∩R ∈ α ].Denote by X = S ∩R ∈ α and for every i ∈ [p], let S i = S ∩ R ∩ β i .Further let I ⊆ [p] be the set of indices i ∈ [p] such that S i ∩ R ∈ β i = ∅ and set I = [p] \ I.By Claim 6.7(i), for every and let us show that x is dominated by M. Suppose first that x / ∈ B ∈ α .Then there exists i ∈I such that x ∈ B ∩ β i .If i ∈ I 1 ∪ I 1 then x is dominated by S i by definition.Suppose therefore that i ∈ I 2 ∪ I 2 .If x ∈ B ⊆ † β i then x is dominated by S i by definition; otherwise, x ∈ B ∈ β i and since x / ∈ B ∈ α by assumption, x ∈ N(Z i) by construction and so, x is dominated by S i by definition.Suppose second that x ∈ N(Y ) ∩ B ∈ α .Then there exists i ∈ I such that x ∈ N i .If i ∈ I 1 then x is dominated by S i by definition.Suppose therefore that i ∈ I 2 .If x ∈ B ⊆ † β i then x is dominated by S i by definition; otherwise, x ∈ B ∈ β i and since x ∈ N i ⊆ N(Z i ) by construction, x is dominated by S i by definition.Therefore, M dominates every vertex B ⊆ † α ∪ (N(Y ) ∩ B ∈ α ) and so, M is indeed the desired S. Consider now a minimum red-blue dominating set S of G such that S ∩ R ∈ α = ∅ and let Y ⊆ S ∩ R ⊆ † α be a set minimally dominating N(S ∩ R ⊆ † α ) ∩ B ∈ α .Then by Claim 6.7(ii), |Y | ≤ ℓ.Let us show that |S