On the Tractability of Optimization Problems on H-Graphs

For a graph H, a graph G is an H-graph if it is an intersection graph of connected subgraphs of some subdivision of H. H-graphs naturally generalize several important graph classes like interval graphs or circular-arc graph. This class was introduced in the early 1990s by Bíró, Hujter, and Tuza. Recently, Chaplick et al. initiated the algorithmic study of H-graphs by showing that a number of fundamental optimization problems like Maximum Clique, Maximum Independent Set, or Minimum Dominating Set are solvable in polynomial time on H-graphs. We extend and complement these algorithmic findings in several directions. First we show that for every fixed H, the class of H-graphs is of logarithmically-bounded boolean-width (via mim-width). Pipelined with the plethora of known algorithms on graphs of bounded boolean-width, this describes a large class of problems solvable in polynomial time on H-graphs. We also observe that H-graphs are graphs with polynomially many minimal separators. Combined with the work of Fomin, Todinca and Villanger on algorithmic properties of such classes of graphs, this identify another wide class of problems solvable in polynomial time on H-graphs. The most fundamental optimization problems among the problems solvable in polynomial time on H-graphs are Maximum Clique, Maximum Independent Set, and Minimum Dominating Set. We provide a more refined complexity analysis of these problems from the perspective of parameterized complexity. We show that Maximum Independent Set and Minimum Dominating Set are W[1]-hard being parameterized by the size of H plus the size of the solution. On the other hand, we prove that when H is a tree, then Minimum Dominating Set is fixed-parameter tractable parameterized by the size of H. For Maximum Clique we show that it admits a polynomial kernel parameterized by H and the solution size.


Introduction
The notion of H-graph was introduced in the work of Bíró, Hujter, and Tuza [BHT92] on precoloring extensions of graphs.H-graphs nicely generalize several popular and widely studied classes of graphs.For example, the classical definition of an interval graph is as a graph which is an intersection graph of intervals of a line.Equivalently, a graph is interval if it is an intersection graph of some subpaths of a path.Or, equivalently, if it is an intersection graph of some subgraphs of some subdivision (which is a graph obtained by placing vertices of degree 2 on the edges) of P 2 , the graph with two adjacent vertices.Similarly, every chordal graph is an intersection graph of subtrees of some tree.More generally, for a fixed graph H, a graph G is an H-graph if it is an intersection graph of some connected subgraphs of some subdivision of H. Thus for example, an interval graph is a P 2 -graph, a circular-arc graph is a C 2 -graph, where C 2 is a double-edge with two endpoints, a split graph is a K 1,d -graph for some d ≥ 0, where K 1,d is a star with d leaves, etc..
The main motivation behind the study of H-graphs is the following.It is well-known that on interval, chordal, circular-arc, and other graphs with "simple' intersection models many NP-hard optimization problems are solvable in polynomial time, see e.g. the book of Golumbic [Gol04] for an overview.It is a natural question whether at least some of these algorithmic results can be extended to more general classes of intersection graphs.Chaplick et al. [CTVZ16] and Chaplick and Zeman [CZ17] initiated the systematic study of algorithmic properties H-graphs.It appeared that a number of fundamental optimization problems like Maximum Independent Set, Minimum Dominating Set, or Maximum Clique are solvable in polynomial time on H-graphs for a fixed H. Most of the algorithms developed on H-graphs in [CTVZ16,CZ17] run in time n f (H) , where n is the number of vertices in the input graph and f is some function.In other words, being parameterized by H most of the problems are known to be in the class XP.
Our work is driven by the following question.
• Are there generic explanations why many problems admit polynomial time algorithms on H-graphs?
We address the first question by proving the following combinatorial results.We show first that every n-vertex H-graph has boolean-width at most 2|E(H)| • log n.Moreover, a decomposition of boolean-width 2|E(H)| • log n can be found in polynomial time.This combinatorial result extends the results of Belmonte and Vatshelle [BV13] on the boolean-width of interval (resp.circular interval) graphs to H-graphs.Together with the algorithms for a vast class of problems called LC-VSP problems [BXTV11,BV13], which are solvable on n-vertex graphs of boolean-width b in time 2 b n O(1) , this implies immediately that all these problems are solvable in polynomial time on H-graphs.The illustrative problems solvable in polynomial time on H-graphs by making use of this approach are Maximum Weight Independent Set, Minimum Weight Dominating Set, Total Dominating Set, Induced Matching, and many others.Then we prove that every n-vertex H-graph has at most (2n + 1) |E(H)| + |E(H)| • (2n) 2 minimal separators. 1 Pipelining the bound on the number of minimal separators in H-graphs with the meta-algorithmic theorem about induced subgraphs of bounded treewidth which properties are expressible in Counting Monadic Second Order Logic (CMSOL) of Fomin, Todinca and Villanger [FTV15], we obtained another wide class of problems solvable in polynomial time on H-graphs.Examples of such problems are Treewidth, Minimum Feedback Vertex Set, Maximum Induced Subgraph excluding a planar minor, and various packing problems.
All these generic algorithmic results provide XP algorithms when parameterized by the size of H.This brings us immediately to the second question defining the direction of our research.
• What is the parameterized complexity of the fundamental optimization problems being parameterized by the size of H?
The first steps in this direction were done by Chaplick et al. in [CTVZ16] who showed that Minimum Dominating Set is fixed-parameter tractable (FPT) on K 1,d -graphs parameterized by d.In this paper we show that Minimum Dominating Set is W[1]-hard parameterized by H and the size of the solution.Thus the existence of an FPT algorithms for a general graph H is very unlikely.(We refer to books [DF13, CFK + 15] for definitions from parameterized complexity and algorithms.)We also prove a similar lower bound for Maximum Independent Set parameterized by H and the solution size.On the positive side, we show that when H is a tree, then Minimum Dominating Set is FPT parameterized by H.This significantly extends the result from [CTVZ16] for stars to arbitrary trees.We actually prove a slightly more general result, namely that Minimum Dominating Set is FPT on chordal graphs parameterized by the leafage of the graph.
Finally we show that Clique is W[1]-hard parameterized by H.This nicely complement the result of Chaplick and Zeman who showed that Clique is FPT parameterized by H and size the solution size.We also strengthen the algorithmic result of Chaplick and Zeman by showing that for such a parameterization Clique admits a polynomial kernel.Algorithmic results about H-graphs are summarized in Table 1.
Organization of the paper.Section 2 contains the necessary definitions.In Section 3, we upper-bound the boolean-width of H-graphs and provide algorithmic applications.Section 4 is devoted to the study of minimal separators in H-graphs, again with algorithmic consequences.Last, Section 5 contains our results on the parameterized complexity of some classic optimization problems on H-graphs.

Definitions
Basics.All graphs in this paper are finite, undirected, loopless, and may have multiple edges.If G is a graph, we denote by |G| its number of vertices and by G its number of  1: Summary of algorithmic results on H-graphs.We always assume that a representation of the input as an H-graph is given.For each of the mentioned problems, k denotes the solution size.See Sections 3 and 4, respectively, for details about the two first problem.edges (counting multiplicities).For every graph G, we denote by ω(G) the largest order of a complete subgraph of G.If X, Y ⊆ V (G), X is the complement of X in V (G) (i.e.X = V (G) \ X), G[X] is the subgraph of G induced by the vertices of X, and G[X, Y ] is the bipartite subgraph of G induced by those edges that have one endpoint in X and the other in Y .Unless otherwise specified, logarithms are binary.
H-graphs.Let H be a (multi) graph.We say that a graph G is an H-graph if there is a subdivision H of H and a collection M = {M v } v∈V (G) (called an H-representation or, simply, representation) of subsets of V (H ), each inducing a connected subgraph, such that G is isomorphic to the intersection graph of M. To avoid confusion, we refer to the vertices of H as nodes.We also say that the nodes of H are branching vertices of H and the other nodes are subdivision nodes.If v is a vertex of G, then M v is the model of v in the representation M.
For every set A ⊆ V (G), we define M A = v∈A M v .For every node u of H , we denote by V u the set of vertices of G whose model contains u, that is, contains only subdivision nodes of H from the path in H corresponding to e in H.The set of e-vertices is denoted by V G (e).

H-graphs have logarithmic boolean-width
Boolean-width is a graph invariant that has been introduced in [BXTV11] and which is related to the number of different neighborhoods along a cut.Belmonte and Vatshelle showed in [BV13] that n-vertex interval graphs and circular interval graphs have boolean-width O(log n).In this section, we generalize their result by proving that, for H a fixed graph, n-vertex H-graphs have boolean-width O(log n).Using the results of [BXTV13], we obtain polynomial algorithms for a vast class of optimization problems on H-graphs.Before we proceed with the proofs, we need to introduce some notions specific to this section.
Definition 1 (Graph decompositions).A tree decomposition of a graph G is a pair (T, δ) where T is a full binary rooted tree (that is, every non-leaf vertex has degree 3) and δ is a bijection from the leaves of T to the vertices of G.A tree decomposition (T, δ) is a caterpillar decomposition if T can be obtained from a path by adding a vertex of degree one adjacent to every internal vertex.If w ∈ V (T ), let us denote by V w the set of vertices of G in bijection with the leaves of the subtree of T rooted at w. Definition 2 (Maximum induced matching over a cut).A set of vertices of a graph G is an induced matching if it induces a disjoint union of edges.If X ⊆ V (G), we denote by mim(X) the maximum number of edges in an induced matching of G[X, X].If (T, δ) is a tree decomposition of G, we denote by mim(T, δ) the maximum of mim(V w ) taken over all w ∈ V (T ).
Definition 3 (Neighborhood equivalence).Let G be a graph, let d ∈ N and and let A ⊆ V (G).We say that two subsets X, Y ⊆ A are d-neighborhood equivalent, what we denote by We write nec d (A) for the number of equivalence classes of the relation ≡ d A .If (T, δ) is a tree decomposition of G, we denote by nec d (T, δ) the maximum of nec d (V w ) and nec d (V w ) over all w ∈ V (T ).
The following lemma relates maximum induced matchings to neighborhood equivalence.
Lemma 1 ( [BV13, Lemma 1 and Lemma 2]).For every n-vertex graph G and A ⊆ V (G), Definition 4 (Boolean-width).If (T, δ) is a tree decomposition of a graph G, the booleanwidth of (T, δ), that we write boolw(T, δ), is defined as the maximum of log(nec(≡ 1 Vw )) over all w ∈ V (T ).The boolean-width of G, denoted by boolw(G), is the minimum boolean-width of a tree decomposition of G.
Our results on the boolean-width of H-graphs will follow from the next result.
Theorem 1.Let H be a graph.For every H-graph G on n vertices whose intersection model is given, we can compute in polynomial time a caterpillar decomposition (T, δ) with mim(T, δ) ≤ 2 H .
Proof.Let F be the subdivision of H in which G can be realized and let {M v } v∈V (G) be the intersection model of G. Let us arbitrarily fix a vertex branching vertex r of F .Let v 1 , . . ., v n be an ordering of V (G) by increasing distance of M v i 's to r.
Claim 1.For every prefix A of v 1 , . . ., v n and every S ⊆ A, there is a set R ⊆ S of size at most 2 H such that N (R) ∩ A = N (S) ∩ A.
Proof.Let A be a prefix of v 1 , . . ., v n and let S ⊆ A. Let M A = v∈A M v and similarly for M A and M S .Let us consider the path P e corresponding to some edge e ∈ E(H).Let x 1 , . . ., x p be the vertices of P e in the same order.
Let v ∈ A and notice that since, by definition, G[M v ] is connected, the vertex set M v ∩ V (P e ) induces at most two connected components in P e .Let us assume that it induces at least one connected component and let x i and x j be the first and last vertices (wrt.the ordering x 1 , . . ., x p ) of this component.If {x 1 , . . ., x i−1 } is disjoint from M A , we say that v is a left-protector of P e .If j is maximum among all vertices that protects the left of P e , then v is a rightmost left-protector.(Informally, it extends the most to the right.)Similarly, v is a right-protector the right of P e if {x j+1 , . . ., x p } is disjoint from M A and is a leftmost right-protector if i is minimal.
Let Z e be a set containing one (arbitrarily chosen) rightmost left-protector and one leftmost right-protector of e if some exist, and let R = e∈E(H) Z e .Clearly |R| ≤ 2 H . Let us now show that N (S) ∩ A ⊆ N (R) ∩ A. We consider a vertex u ∈ N (S) ∩ A and we show that it also belongs to N (R).Let v be a neighbor of u in S. As u and v are adjacent, M u and M v have non-empty intersection.Let e be an edge of H such that M u and M v meet on P e , i.e.M u ∩ M v ∩ V (P e ) = ∅.Again, we denote by x 1 , . . ., x p the vertices of P e .
Proof.If there are vertices u, u of A such that M u and M u respectively intersect {x 1 , . . .x i−1 } and {x j+1 , . . ., x p }, then one of dist F (M u , r) and dist F (M u , r) is smaller than dist F (M w , r).This contradicts the fact that w ∈ A whereas u, u ∈ A.
Claim 3. Let w ∈ A. If M w intersects V (P e ) then it is a right-protector or a left-protector.
Proof.By definition, if x 1 ∈ M w then M w is a left-protector of P e (and symmetrically for the right).The case where M w contains none of x 1 and x p follows from Claim 2.
As M u intersects M v on P e , it intersects the vertex set C of one component induced by M v on P e (recall that there are either one or two such components).In the case where there are two components, we assume without loss of generality that this is the "left" one (i.e. that with smallest indices).In the case where there is one component, we assume that v is a left-protector of P e (according to Claim 3, v is a left-protector or a right-protector of P e ).
Observe that in both cases, v is a left-protector of P e .Let z be the rightmost left-protector of P e that belongs to R and let x k , . . .x k be the vertices of the corresponding component of P e [M z ∩ V (P e )] (that is, the component used in the definition of left-protector).
Observe that when H is a tree, G is chordal and so the intersection model can be computed in linear time.The next result follows from the application to the decomposition provided by Theorem 1 of Lemma 1.(ii), with the fact that mim(A) = mim(A) for every A ⊆ V (G).
Corollary 1.Let H be a graph.For every H-graph G on n vertices whose intersection model is given, we can compute in polynomial time a caterpillar decomposition (T, δ) with For the definition of boolean-width, we also get: Corollary 2. Let H be a graph.Every n-vertex H-graph has boolean-width at most 2 H • log n.
By choosing H to be a single or double edge, we recover the results of [BV13] on the boolean-width of interval and circular interval graphs, respectively, as special cases of Corollary 2. As proved in the same paper, there is a infinite family of interval graphs with boolean-width Ω(log n).Observe that unless H is edgeless (in which case H-graphs are disjoint unions of cliques), every interval graph is an H-graph.This shows that the bound in Corollary 2 is tight up to a constant factor.
Let us describe provide algorithmic applications of our results.Boolean-width has been used [BXTV11] to design parameterized algorithms for the problems Maximum Weight Independent Set and Minimum Weight Dominating Set.Later, invariants of the form nec d were used in [BXTV13] as parameters of FPT algorithms for a vast class of problems, called LC-VSP problems, that include fundamental problems as Independent Set, Independent Dominating Set, Total Dominating Set, Induced Matching, and many others (see [BXTV13]).The main result of [BXTV13] is the following.

Theorem 2 ( [BXTV13]
).For every LC-VSP problem Π, there are constants d and q such that Π can be solved in time O(n 4 • q • nec d (T, δ) 3q ) if a decomposition (T, δ) of the input is given.
As Corollary 1 provides, in polynomial time, a tree decomposition where the value of nec d is polynomial in the number of vertices of the graph (for fixed d), we get the following meta-algorithmic consequence from Theorem 2. Theorem 3. Let H be a graph.Every LC-VSP problem can be solved in polynomial time on H-graphs.

Minimal separators in H-graphs
The study of minimal separators is an active line of research that found many algorithmic applications (see e.g.[KBMK93, BBC99, BT01, FTV15]).In general, the number of minimal separators of a graph may be as large as exponential in its number of vertices.We prove in this section that in an H-graph, this number is upper-bounded by a polynomial (Theorem 4).By combining this finding with meta-algorithmic results of Fomin, Todinca and Villanger [FTV15], we deduce that a wide class of optimization problems can be solved in polynomial time on H-graphs (Corollary 3).
Proof.Let G be a H-graph and let F be a subdivision of H where G can be represented as the intersection graph of {M v , v ∈ V (G)}.For every subset V ⊆ V (G), the border edges of V are the edges of F with one endpoint in M V and one endpoint in V (F ) \ M V .Let R be the union of border edges over {M v , v ∈ V (G)}.Observe that for every V ⊆ V (G), the set of border edges of V is a subset of R. For every edge e ∈ E(F ), we set and extend this notation to sets S ⊆ E(F ) as follows: Informally, V S is the set of all vertices of G whose models contain some edge of S.
Claim 4. For every minimal separator X in G, there is a S ⊆ R such that X = V S .
Proof.Let A, B be two connected components of G \ X such that N (A) = N (B) = X.As X is an (A, B)-separator (i.e.A and B are included in the vertex sets of distinct connected components of G \ X), M A ∩ M B = ∅.Let S be the set of border edges of M A met on all inclusion-wise minimal paths that start in A and end in B. As noted above, S ⊆ R. First we show V S ⊆ X.Let v ∈ V S .That is, v is a vertex G such that M v ⊇ s for some s ∈ S. Then M v contains both endpoints of s, one of which belongs to M A .The vertex v is adjacent to A but does not belong to A (as M v contains a vertex of A), so it has to belong to the separator X.Therefore, V s ⊆ X.Now we show X ⊆ V S .Let x ∈ X.By definition, x has a neighbor in both A and B. Therefore, M x meets both M A and M B .As M x induces a connected subgraph of F and M A is disjoint from M B , it contains an edge s ∈ E(F ) with one endpoint in M A and the other in V (F ) \ M A .Then s is a border edge of M A in a minimal path from A to B: x ∈ V S .Hence X = V S .
From Claim 4 we can already deduce that the number of minimal separators of G is at most the number of subsets of R. In order to obtain better bounds, we need other observations.Claim 5.For every V ⊆ V (G) such that M V induces a connected subgraph of F , and every e ∈ E(H), the set M V has at most two border edges in E(P e ).Hence, |R| ≤ 2|G| • H . Proof.Follows from the fact that F [M V ] is connected.Claim 6.For every minimal separator X of G, if S ⊆ R is the subset of edges of F defined in the proof of Claim 4, then • either |S ∩ E(P e )| ≤ 1 for every e ∈ E(H); • or |S| = 2 and S ⊆ E(P e ) for some e ∈ E(H).
Proof.Let A and B be as in the proof of Claim 4. According to Claim 5 and as S is a subset of the border edges of M A , we deduce |S ∩ E(P e )| ≤ 2 for every e ∈ E(H).Let us assume that |S ∩ E(P e )| = 2 for some e ∈ E(H).Let u, u and v, v be the endpoints of the two edges shared by S and E(P e ), respectively and in this order on the path.Then the model of one of A and B has its vertices in the subpath Q of P e delimited by u and v. Indeed, both {u, u } and {v, v } and have an endpoint that does not belong to M A .As M A induces a connected subgraph of F , either these endpoints are u and v (intuitively, the exterior endpoints) or they are u, v (the interior endpoints).In the first case M A ⊆ E(Q) and in the second one, M B ⊆ E(Q).From the definition of R, we can then conclude that S ⊆ E(P e ) and we get |S| = 2. Therefore, for every minimal separator X of G, there is a set S ⊆ R such that: 1. either |S ∩ E(P e )| ≤ 1 for every e ∈ E(H); 2. or |S| = 2 and S ⊆ E(P e ) for some e ∈ E(H); In order to upper-bound the number of possible minimal separators of G, we can consequently upper-bound the number of sets S ⊆ R that satisfy one of the two conditions above.As noted in Claim 5, for every e ∈ E(H) we have R ∩ E(P e ) ≤ 2|G|.Hence there are at most (2|G|) 2 possible choices of set S that satisfy (2) for each e ∈ E(H).We deduce that there are at most H • (2|G|) 2 distinct sets S ⊆ R satisfying (2).Let us now consider sets S ⊆ R that satisfy (1).For every e ∈ E(H), either S contains one of the 2|G| edges of R ∩ E(P e ) or it does not contain any of them.This makes 2|G| + 1 possible choices for each e ∈ E(H), and (2|G| + 1) H in total.Consequently, G has at most (2|G| + 1) H + H • (2|G|) 2 minimal separators.
For every r ∈ N, let θ r be the graph with 2 vertices and r parallel edges.The following shows that the exponential contribution of H in Theorem 4 cannot be avoided.
Lemma 2. For every r ∈ N, there is a θ r -graph G with at least |G|−2 r r minimal separators.
Proof.Let G be the graph obtained from θ r by subdividing k times each edge.Then G is a θ r -graph and |G| = kr + 2. Notice that any choice of r subdivision vertices, each corresponding to a different edge of θ r , gives a distinct minimal separator of G. Hence G has at least k r = |G|−2 r r minimal separators.
Our results on minimal separators have algorithmic consequences.We consider the following generic problem described in [FTV15].
Optimal Induced Subgraph for P and t Input: For various choices of P and t, this generic problem corresponds to natural families of optimization meta-problems like F-minor-deletion (where F is a class of graphs containing at least one planar graph) whose goal is to delete a minimum number of vertices in order to get an F-minor free graph 2 and Independent F-packing (where F is a class of connected graphs), which asks for a maximum number of disjoint copies of graphs in F as pairwise independent subgraphs of the input.Fomin, Todinca, and Villanger proved that when the property P can be expressed in Counting Monadic Second Order logic (CM-SOL, see [FTV15]), the above problem can be easily solved on classes of graphs that have a polynomial number of minimal separators.

Theorem 5 ( [FTV15]
).For any fixed t and CSMO property P, problem Optimal Induced Subgraph for P and t can be solved in polynomial time for classes of graphs with a polynomial number of minimal separators.
We deduce the following result about H-graphs.
Corollary 3. Let H be a graph.For any fixed t and CSMO property P, the problem Optimal Induced Subgraph for P and t can be solved in polynomial time for H-graphs.

Parameterized complexity of basic problems for Hgraphs
In this section we investigate the parameterized complexity of some basic graph problems for H-graphs: Dominating Set, Independent Set and Clique.First, in Subsection 5.1, we show that Dominating Set and Independent Set are W[1]-hard when parameterized by the solution size and the size of H and then we prove that Clique is W[1]-hard when parameterized by the size of H even if an H-representation of the input graph is given.In Subsection 5.2, we show that Dominating Set is FPT when parameterized the solution size and by the number of vertices of H if H is a tree.In fact, we show a more general result by proving that Dominating Set is FPT for chordal graphs if the problem is parameterized by the solution size and the leafage of the input graph, that is, by the minimum number of leaves in a clique tree for the input graph.This result is somehow tight since Dominating Set is well-known to be W[2]-hard for split graphs when parameterized by the solution size [RS08].Recall also that Independent Set and Clique are polynomial for chordal graphs [Gav72,Gol04] and, therefore, for H-graphs if H is a tree.Finally, in Subsection 5.3, we show that Clique admits a polynomial kernel when parameterized by the solution size and the size of H.

Hardness
In this section we prove W[1]-hardness of Dominating Set, Independent Set and Clique for H-graphs.To show hardness, we reduce from the Multicolored Clique problem.This problem, given a graph G with a k-partition of its vertex set V 1 , . . ., V k , asks whether G has a k-clique with exactly one vertex in each V i for i ∈ {1, . . ., k}.The problem is well-known to be W[1]-complete when parameterized by k [FHRV09,Pie03].First, we consider Dominating Set and Independent Set.Recall that Dominating Set and Independent Set, given a graph G and a positive integer k, ask whether G has a dominating set of size at most k and independent set of size at least k respectively.Theorem 6. Dominating Set and Independent Set are W[1]-hard for H-graphs when parameterized by k + ||H|| and the hardness holds even if an H-representation of G is given.
Proof.First, we show the W[1]-hardness for Independent Set and then explain how to modify the reduction for Dominating Set.The reduction is from Multicolored Clique.
Let (G, V 1 , . . ., V k ) be an instance of Multicolored Clique.We assume that k ≥ 2 and |V i | = p for i ∈ {1, . . ., k}.The second assumption can be made without loss of generality because we always can add isolated vertices to the sets V 1 , . . ., V k to ensure that they have the same size.Denote by v i 1 , . . ., v i p the vertices of V i for i ∈ {1, . . ., k}.We construct the multigraph H as follows (see Figure 1 (ii) For every 1 ≤ i < j ≤ k, construct a node w i,j and two pairs of parallel edges u i w i,j and u j w i,j .
Note that Then we construct the subdivision H of H obtained by subdividing each edge p times.We denote the subdivision nodes for 4 edges of H constructed for each pair 0 as it is shown in Figure 1 b).To simplify notations, we assume that Figure 1: The construction of H for k = 3 and the subdivision of the edges of H.
M r (i,j) where the model of each vertex is a connected subset of V (H ) (see Figure 2).Recall that G is the graph of the original instance of Multicolored Clique.
(i) For each i ∈ {1, . . ., k} and s ∈ {1, . . ., p}, construct a vertex z i s with the model (ii) For each edge v i s v j t ∈ E(G) for s, t ∈ {1, . . ., p} and 1 Finally, we define k = k(k + 1)/2.We claim that (G, V 1 , . . ., V k ) is a yes-instance of Multicolored Clique if and only if G has an independent set of size k .The proof is based on the following crucial property of our construction, that can be easily checked.
of vertices of G .It is straightforward to verify using Claim 7 that I is an independent set of size k in G .Suppose now that G has an independent set I of size k .For each i ∈ {1, . . ., k}, the set cliques form a partition of V (G ), we have that for each i ∈ {1, . . ., k}, there is a unique z i h i ∈ Z i ∩ I, and for every 1 ≤ i < j ≤ k, there is a unique r (i,j) s i ,s j is not adjacent to z i h i and z j h j , we obtain that s i = h i and s j = h j by Claim 7. It implies that This completes the W[1]-hardness proof for Independent Set.Now we explain how we modify our proof to show the W[1]-hardness of Dominating Set.This time we reduce not from Multicolored Clique but from the Multicolored Independent Set problem that, given a graph G with a k-partition of its vertex set V 1 , . . ., V k , asks whether G has am independent set of size k with exactly one vertex in each V i for i ∈ {1, . . ., k}.Clearly, the W[1]-completeness of Multicolored Clique parameterized by k [FHRV09, Pie03] immediately implies the same for Multicolored Independent Set.
Let (G, V 1 , . . ., V k ) be an instance of Multicolored Independent Set.We assume without loss of generality that k ≥ 2 and |V i | = p for i ∈ {1, . . ., k}.As before, denote by v i 1 , . . ., v i p the vertices of V i for i ∈ {1, . . ., k}.We construct the same multigraph H and its subdivision H as above.We construct the H-graph G from the graph G constructed above by adding k new vertices d 1 , . . ., d k with the models We show that (G, V 1 , . . ., V k ) is a yes-instance of Multicolored Independent Set if and only if G has a dominating set of size k.
Suppose that {v 1 h 1 , . . ., v k h k } is an independent set of G. Consider the set D = {z 1 h 1 , . . ., z k h k }.By Claim 7 and the construction of G , we obtain that D is a dominating set of G .
Let now D be a dominating set of G with |D| = k.Note that each vertex d i is adjacent only to the vertices of the set Since Z i is a clique, we can assume without loss of generality that D ∩ Z i = ∅ as, otherwise, we can replace d i in D by an arbitrary vertex of Z i .Since Z i ∩ Z j = ∅ if i = j, we conclude that D contains a unique vertex from each Z i and no other vertices.
h i ,h j is not adjacent to z s hs for any s ∈ {1, . . ., k} such that s = i, j, we have that r Proof.We reduce from Multicolored Clique.Let (G, V 1 , . . ., V k ) be an instance of Multicolored Clique.We assume without loss of generality that k ≥ 3 and The construction of H for k = 3; the edges of S are not shown.
We construct the multigraph H as follows (see Figure 3).
(iii) For every 1 ≤ i < j ≤ k, construct a node w i,j and two pairs of parallel edges u i w i,j and u j w i,j .
(iv Denote by S the set of the edges added in (iv); note that some of them are parallel to the edges constructed in (ii) and (iii).Observe that Our next aim is to construct the subdivision H of H.We do it by subdividing p times some edges of E(H) \ S.
The subdivision of H.
(i) For every i ∈ {1, . . ., k}, subdivide the edge a 2 i b 2 i and denote the subdivision nodes by c i 1 , c i 2 , . . ., c i p as it is shown in Figure 4 a).(ii) For every i ∈ {1, . . ., k}, subdivide the edges u i a 1 i and u i b 1 i p times, and denote the subdivision nodes byf i 1 , . . ., f i p and g i 1 , . . ., g i p respectively as it is shown in Figure 4 a).(iii) For every pair 0 ≤ i < j ≤ k, subdivide two pairs of parallel edges u i w i,j and u j w i,j p times and denote the subdivision nodes by x Note that the edges of S are not subdivided.To define the weight function ω, we introduce two constants α = 3p 3 k 3 and β = p 2 k 2 (recall that k is the number of parts in the partition of V (G) and p is the size of each of them).Now we construct the weighted H-graph G by defining its H-representation M = {M v } v∈V (G ) where the model of each vertex is a connected subset of V (H ) defined hereafter and depicted in Figure 5. (i) For each i ∈ {1, . . ., k} and s ∈ {1, . . ., p − 1}, construct a vertex d i s with the model and set ω(d i s ) = α (see Figure 5 a)).(ii) For each i ∈ {1, . . ., k} and s ∈ {2, . . ., p}, construct a vertex di s with the model 5 a)).(iii) For each i ∈ {1, . . ., k} and s ∈ {1, . . ., p}, construct a vertex z i s with the model (iv) For each i ∈ {1, . . ., k} and s ∈ {1, . . ., p}, construct a vertex zi s with the model and set ω(z i s ) = β (see Figure 5 a)).
(v) For each edge v i s v j t ∈ E(G) for s, t ∈ {1, . . ., p} and 1 ≤ i < j ≤ k, construct a vertex r (i,j) s,t with the model M r (i,j) s,t =({x (i,j) s , . . ., x (i,j)  p }) ∪ ({y and set ω(r We claim that (G, V 1 , . . ., V k ) is a yes-instance of Multicolored Clique if and only if (G , ω, k ) is a yes-instance of Weighted Clique.
For i ∈ {1, . . ., k} and h ∈ {1, . . ., p}, let and of vertices of G .We show that K is a clique of G .We already observed that each Because k ≥ 3, it follows that any two vertices of D i h i and D j h j for 1 ≤ i < j ≤ k are adjacent.For i ∈ {1, . . ., k}, the vertices of Z i h i are adjacent to the vertices of , and either . ., k} and h ∈ {1, . . ., p}, we obtain that the vertices of Z i h i are adjacent to the vertices of Z j h j and D j h j if i, j ∈ {1, . . ., k} and i . It follows, that r (i,j) h i ,h j is adjacent to the vertices of D s hs for s ∈ {1, . . ., k} and, moreover, the vertices r (i,j) h i ,h j are pairwise adjacent for all 1 ≤ i < j ≤ k.Since {u 1 , . . ., u k } \ {u i , u j } ⊆ M r (i,j) h i h j , r (i,j) h i ,h j is adjacent to the vertices of Z s hs for s ∈ {1, . . ., k} if s = i, j.Finally, notice that x (i,j) and this implies that r (i,j) h i ,h j is adjacent to the vertices of Z i h i and Z j h j .Combining all these observations together, we obtain that the vertices of K are pairwise adjacent, that is, K is a clique of G . Since . ., k} and h ∈ {1, . . ., p}, and ω(r Therefore, (G , ω, k ) is a yes-instance of Weighted Clique.
We now show the reverse direction.Suppose that (G , ω, k ) is a yes-instance of Weighted Clique.Let K be a clique in G with ω(K) ≥ k .Consider the vertices d i s and di t for i ∈ {1, . . ., k} and s, t ∈ {1, . . ., p}.Observe that these vertices are adjacent if and only if s ≥ t.It implies that the sets D i h for h ∈ {1, . . ., p} form the set of inclusion maximal cliques in G[D i ], where for some h i ∈ {1, . . ., p} for each i ∈ {1, . . ., k}.
We show that K ∩ D i = D i h i for i ∈ {1, . . ., k}.To obtain a contradiction, assume that (1) Recall that α = 3p 3 k 3 and β = p 2 k 2 .Then Combining (1) and (2), we have that Notice that only the vertices z i h i , zi h i ∈ Z i are adjacent to all the vertices of D i h i .Hence, K can contain only these vertices.We claim that K contains both of them, that is, (3) Notice that β = p 2 k 2 > p 2 k(k − 1)/2 and, therefore, by (4), We obtained that k(p − 1) vertices of K of weight α belong to ∪ k i=1 D i and 2k vertices of weight β belong to ∪ k i=1 Z i .The total weight of these vertices is ) is adjacent to all the vertices of the sets Z h i and Z h j if and only if s = h i and t = h j .Hence, K contains at most one vertex from R (i,j) .Since |R ∩ K| ≥ k(k − 1)/2, K contains exactly one vertex r (i,j) ) is a yes-instance of Multicolored Clique and this completes the proof.
Corollary 4. Clique is W[1]-hard for H-graphs when parameterized by ||H|| and the hardness holds even if an H-representation of G is given.
Proof.Given an instance (G, ω, k) of Weighted Clique, we construct the instance (G , k) of Clique where G is obtained from G by replacing each vertex v ∈ V (G) by ω(v) true twins (i.e.ω(v) pairwise adjacent vertices, each of which having in V (G) \ v the same neighborhood as v).Clearly, (G, ω, k) is a yes instance of Weighted Clique if and only if (G , k) is a yes-instance of Clique.By Theorem 7, Weighted Clique parameterized by ||H|| is W[1]-hard for H-graphs with a given H-representation even if the weight of every vertex is encoded in unary.Then the claim of the corollary follows immediately.

Dominating Set for T -graphs
In this section we show that Dominating Set is FPT for chordal graphs if the problem is parameterized by the solution size and the leafage (hereafter defined) of the input graph.
Let G be a graph.As it is standard, we say that Let G be a graph.Let K be the set of (inclusion-wise) maximal cliques of G and let Note that a clique tree of a chordal graph is not necessarily unique.For a connected chordal graph G, the leafage (G) of G is the minimum number of leaves in its clique tree.In was shown by Habib and Stacho in [HS09] that the leafage of a connected chordal graph G can be found in polynomial time.Their algorithm also constructs the corresponding clique tree T with the minimum number of leaves.In other words, given a connected chordal graph G, we can construct in polynomial time its clique tree T with (G) leaves and a T -representation of G.
Let T be a tree and let G be a connected T -graph with its T -representation M = {M v } v∈V (G) with respect to a subdivision T of T .
Recall that for nonempty U ⊆ V (T ), we say that we write u-vertex instead of {u}-vertex.Recall also that we denote the set of U -vertices by V G (U ) and V G (u) if U = {u}.We also denote by V G (T ) the set of all U -vertices of G for all nonempty U ⊆ V (T ).For e ∈ E(T ), v ∈ V (G) is an e-vertex if M v contains only subdivision nodes of T from the path in T corresponding to e in T .The set of e-vertices is denoted by V G (e).
We need the following lemma that allows to upper bound the number of vertices in a minimum dominating set whose models contain given nodes of T .Lemma 3. Let D be a minimum dominating set of G. Let X ⊆ V (T ) be an inclusion maximal set of nodes of T such that i) for every x ∈ X, there is u ∈ D with x ∈ M u and ii) for every xy ∈ E(T ) with x, y ∈ X, there is u ∈ D with x, y ∈ M u .Then the set Proof.Denote by v 1 , . . ., v s the vertices of N T (X).To obtain a contradiction, assume that that |U | > |N T [X]| − 1 = s + |X| − 1.For each i ∈ {1, . . ., s}, let u i ∈ U be a vertex such that the distance between M u i and v i in T is minimum.For each xy ∈ E(T ) with x, y ∈ X, we select arbitrary In particular, since |N T (X)| is at most the number of leaves , we have that |U | ≤ |X| + − 1.
The next lemma gives an upper bound for the number of vertices in a minimal dominating set whose models contain nodes of T .
Lemma 4. Let D be a minimum dominating set of G.
Proof.Let D be a minimum dominating set of G. Consider the set W of nodes of T that are included in the models of the vertices of D. Let X 1 , . . ., X r be the partition of W into inclusion maximal subsets such that for each i ∈ {1, . . ., r} and adjacent x, y ∈ X i , there is u ∈ D with x, y ∈ M u .By Lemma 3, For en edge e ∈ E(T ), we say that G is obtained by contracting e in T if G is the (T /e)-graph with the model obtained as follows: (i) contract xy in T and, respectively, the (x, y)-path P in T , and denote the node obtained from x and y by z, (ii) delete all e-vertices of G, (iii) for each remaining vertex u ∈ V (G), delete from M u the subdivision nodes of P and replace x and y by z if at least one of these nodes is in M u .
Note that We say that M is a nice representation if |M v ∩ V (H)| ≤ 1 for v ∈ V (G), i.e., each set M v contains at most one branching node of T .
We say that a T -representation M = {M v } v∈V (G) of G with respect to a subdivision T is an r-rooted representation if a node r of T is chosen to be a root.The root defines the parent-child relation on V (T ) and V (T ).For x ∈ V (T ) (x ∈ V (T )), we denote by T x (T x respectively) the subtree of T (T respectively) induced by x and its descendants.For x ∈ V (T ) and its child y, we denote by T xy the subtree of T induced by x, y and the descendants of y.
it holds that the number of nodes x ∈ V (T ) such that D contains an x-vertex is maximum and for each x ∈ V (T ), Task: Decide whether there is a dominating set D of G of size at most k containing at most d x-vertices for x ∈ V (T ) such that for each x ∈ V (T ), Note that Dominating Set Extension is a promise problem: we are promised that there is D with the described properties but D itself is not given.Moreover, the promise could be false but we are not asked to verify it.
Lemma 7. Dominating Set Extension can be solved in time 2 O((d+ ) log d) n O(1) .Moreover, it can be done by an algorithm that either return a correct yes-answer or (possible incorrect) no-answer even if the promise is false.
Proof.Let G be a T -graph for a tree T rooted in r that has a nice r-rooted T -representation M with respect to a subdivision T of T .If G is disconnected, then we reduce the problem to solving Dominating Set Extension for the components of G. Assume from now that G is connected.Let also k and d be positive integers, c : ∪ x∈V (T ) V G (x) → N is a labeling function, and {C x } x∈V (T ) is a collection of sets where each C x ⊆ c(V G (x)).Since G is connected, we can assume without loss of generality that for every x ∈ V (T ), there is v ∈ V (G) with x ∈ M v .Otherwise, we can replace T by its subtree without increasing the number of leaves.We also assume that |C x | ≤ d ≤ |V G (x)| as, otherwise, we have a trivial no-instance of Dominating Set Extension.We construct a dynamic programming algorithm for the problem that finds the minimum size of a dominating set D of G containing at most d x-vertices for x ∈ V (T ) such that for each x ∈ V (T ), C x = c(D ∩ V G (x)).Our algorithm assumes that the promise is fulfilled for the considered instance of Dominating Set Extension.The algorithm uses the properties that every node of T has at most children, i.e., the number of the children is bounded by the parameter, and for each edge e of T , the set of e-edges of G composes an interval graphs for which the domination problem can be solved efficiently.First, we construct a subroutine that solves the following auxiliary problem for each e = xy ∈ E(T ).Let P e be the (x, y)-path in T corresponding to e. Let U e = {v ∈ V (G) | M v ∩ V (P ) = ∅}.For X ⊆ U e , α e (X) is the minimum size of a set S of e-vertices of G that dominates X; we assume that α e (X) = 0 if X = ∅ and α e (X) = +∞ if such a dominating set of e-vertices S does not exist.
Claim A. For every e ∈ E(T ) and X ⊆ U e , α e (X) can be computed in time n O(1) .Proof of Claim A. If X = ∅, then α e (X) = 0 by the definition.Assume that X = ∅.If there is a vertex in X that is not dominated by any e-vertex of G, then we set α e (X) = +∞.Otherwise, it is straightforward to see that S exists, and we construct S using the well-known greedy approach for constructing a minimum dominating set in an interval graph.
Initially, we set S = ∅ and then increase it iteratively until all the vertices of X are dominated.Denote by Y ⊆ X the set of vertices that are not dominated by the the current S. Then we do the following: 1. Find a vertex w in Y that is maximum with respect to the ordering e .
2. Find a minimum with respect to e e-vertex v that dominates w, set S = S ∪ {v} and recompute Y .

If Y = ∅, then return to
Step 1.
It is straightforward to verify that the algorithm correctly computes α e (X) in polynomial time.
We say that a node x ∈ V (T ) is loaded of C x = ∅ and x is unloaded otherwise.We say that a set of vertices S is extendable if there is a dominating set D such that (a) D has at most d x-vertices for x ∈ V (T ), that has the minimum size and contains S and the conditions of the promise is fulfilled: the number of nodes x ∈ V (T ) such that D contains an x-vertex is maximum and for each We are ready to explain our dynamic programming algorithm for Dominating Set Extension.It works on T starting from the leaves and moving towards the root.To avoid dealing with the root that has no parent separately, we add an artificial node r to T and T and make r the parent of r and the new root.Observe that r is the unique node of the tree that is not included in M v for some v ∈ V (G).
We start with defining the tables of data that the algorithm stores for each x = r of T .Let y be a parent of x and let e = yx.Consider the set W = {v 1 , . . ., v p } of x-vertices of G and assume that v 1 e . . .e v p (note that if x = r and y = r , then the ordering is arbitrary).
If x loaded, then for x and i ∈ {1, . . ., p}, the algorithm stores the value β(x, i) that is either the minimum size of a set S ⊆ V x (G) such that (iv) S dominates all the vertices of V x (G), or we may set β(x, i) = +∞ if we detect that there is no S ⊆ V x (G) satisfying (i)-(iv) such that S is extendable.In particular, it can happen if every set S satisfying (i)-(iv) contradicts the promise of Dominating Set Extension.
Similarly, if x is unloaded, then for x and i ∈ {0, . . ., p} the algorithm stores the value γ(x, i) that is either the minimum size of a set S ⊆ V x (G) such that (v) v i+1 , . . ., v p are dominated by S, (viii) S dominates all the vertices of V x (G) \ {v 1 , . . ., v p }, or we may set γ(x, i) = +∞ if we detect that there is no S ⊆ V x (G) satisfying (v)-(viii) such that S is extendable.Similarly to β(x, i), we do it if we detect that every set S satisfying (v)-(viii) contradicts the promise of Dominating Set Extension.In particular, it happens when we gain by including an x-vertex of G into a partial solution.
It also is assumed that β(x, i) = +∞ and γ(x, i) = +∞ if there is no any S that satisfy the conditions (i)-(iv) or (v)-(viii) respectively.Now we explain how we compute the values β(x, i) and γ(x, i).First, we do it for leaves.
Computing β(x, i) and γ(x, i) for leaves.Let x be a loaded leaf of T .We set Let x be unloaded leaf.We set Now we compute the values β(x, i) and γ(x, i) for non-leaves.Suppose that x = r is a non-leaf node of T .Let z 1 , . . ., z s be the children of x in T and let e j = xz i for j ∈ {1, . . ., s}.Assume that the functions β and γ are computed for the children of x depending on whether they are loaded or not.
Computing β(x, i) for loaded x.Consider z j for j ∈ {1, . . ., s}.Let W j = {u 1 , . . ., u q } be the set of z j -vertices of G.We assume that u 1 e j . . .e j u q .For i ∈ {1, . . ., p}, we compute the values of δ j (x, i).To do it, we consider two cases.
We consider all possible partitions P = {J 1 , . . ., J t } for |C x | ≤ t ≤ d of {0, 1, . . ., s} where 0 ∈ J 1 and some other sets could be empty.We then consider all possible surjections ϕ : {1, . . ., t} → C x .For x and i ∈ {1, . . ., p}, we set where we assume to simplify notations that Computing γ(x, i) for unloaded x.Consider z j for j ∈ {1, . . ., s}.Let W j = {u 1 , . . ., u q } be the set of z j -vertices of G.We assume that u 1 e j . . .e j u q .Consider also the ordering v i j 1 , . . ., v i j p of the vertices of W such that v i j 1 e j . . .e j v i j p .We compute the functions η(j) and ψ(j) as follows depending on whether z j is loaded or not.
Case 1.The vertex z j is loaded.For h ∈ {1, . . ., q}, let For t ∈ {1, . . ., p} and h ∈ {1, . . ., q + 1}, denote and let Case 2. The vertex z j is unloaded.For h ∈ {0, . . ., q}, let For t ∈ {1, . . ., p} and h ∈ {1, . . ., q + 1}, denote and let Now for i ∈ {0, . . ., p}, we set We compute β and γ for all nodes of T except the artificial root r .The algorithm is based on the following properties of these values.Claim B. If x is a loaded node, then for x and i ∈ {1, . . ., p}, the following is fulfilled: if β(x, i) < +∞, then there is a set S ⊆ V x (G) satisfying the conditions (i)-(iv) of size at most β(x, i), and if there is a set S ⊆ V x (G) satisfying the conditions (i)-(iv) that is extendable, then |S| = β(x, i).Similarly, if x is unloaded, then for x and i ∈ {0, . . ., p}, the following is fulfilled: if γ(x, i) < +∞, then there is a set S ⊆ V x (G) satisfying the conditions (v)-(viii) of size at most γ(x, i), and if there is a set S ⊆ V x (G) of minimum size satisfying the conditions (v)-(viii) that is extendable, then |S| = γ(x, i).
Proof of Claim B. It is straightforward to verify these properties for the leaves of T by the definition of β and γ.
We use standard approach for proving correctness of dynamic programming algorithms.We assume inductively that the properties of the values of β and γ are fulfilled for the children of x using as the base of the induction the fact that we already verified the properties of β and γ for the leaves of T .
As in the description of the algorithm, we assume that W = {v 1 , . . ., v p } is the set of x-vertices and assume that v 1 e . . .e v p for e = yx where y is the parent of x.In the same way, z 1 , . . ., z s are the children of x.
First, we prove the claim for β.Let x be a loaded node and let i ∈ {1, . . ., p}.Let β(x, i) < +∞.We show that there is a set (iv) S dominates all the vertices of V x (G).
Consider a partition P = {J 1 , . . ., J t } and a mapping ϕ : {1, . . ., t} → C x for which the minimum in the right part of (4) is achieved.Further, for each j ∈ {1, . . ., s}, let i j be a value of i for which min{δ We define S x = {v i 1 , . . ., v it }; observe that some vertices could be repeated and in this case we remove the duplicates.We have that |S x | ≤ t ≤ d and v i ∈ S x .Since ϕ is a surjection, C x = c(S x ∩ V G (x)).We obtain (i)-(iii) are fulfilled for z = x.Clearly, all x-vertices of G are dominated by S x .
Consider z j for j ∈ {1, . . ., s}.As in the description of the algorithm, we assume that W j = {u 1 , . . ., u q } is the set of z j -vertices of G.We also assume that u 1 e j . . .e j u q .Let j ∈ J i .We consider two cases depending on whether z j is loaded or not.
Case 1.The vertex z j is loaded.Let h ∈ {1, . . ., q} be such that β(z j , h) + α e j (X i ,h ) has the minimum value.
By the inductive assumption, there is S z j ⊆ V z j (G) of size at most β(z j , h) such that (i ) u h ∈ S j , (ii ) S z j contains at most d z-vertices for each z ∈ V (T z j ), (iii ) for each z ∈ V (T z j ), C z = c(S z j ∩ V G (z)), and (iv ) S z j dominates all the vertices of V z j (G).
Notice that if w is a e j -vertex and w / ∈ X i ,h , then w is dominated either by v i ∈ S x or u h ∈ S z j , i.e., w is dominated by S x ∪ S z j .By the definition of α e j (X i ,h ), there is a set S xz j of e j -vertices of size α e j (X i ,h ) that dominates X i ,h .
Case 2. The vertex z j is unloaded.Let h ∈ {1, . . ., q} be such that γ(z j , h) + α e j (Y i ,h ) has the minimum value.
By the inductive assumption, there is S z j ⊆ V z j (G) of size at most γ(z j , h) such that (v ) u h+1 , . . ., u q are dominated by S z j , (vi ) S z j contains at most d z-vertices for each z ∈ V (T z j ), (vii ) for each z ∈ V (T z j ), C z = c(S ∩ V G (z)), and (viii ) S z j dominates all the vertices of V z j (G) \ {u 1 , . . ., u q }.
Recall that we defined If w is a e j -vertex and w / ∈ X i ,h , then w is dominated by v i ∈ S x , i.e., w is dominated by S x ∪ S z j .By the definition of α e j (Y i ,h ), there is a set S xz j of e jvertices of size α e j (Y i ,h ) that dominates Y i ,h .This set dominates the e j -vertices that are not dominated by S x and the z j -vertices u 1 , . . ., u h if h ≥ 1 that are the only vertices of V z j (G) that (possibly) are not dominated by S z j .

Now we let
We have that (i)-(iv) are fulfilled for S. It remains to notice that |S| ≤ β(x, i) by the definition.
Assume now that S ⊆ V x (G) is a set of minimum size satisfying (i)-(iv) and S is extendable.Since S is a set of minimum size satisfying these conditions, as we already proved, |S| ≤ β(x, i).We prove that |S| ≥ β(x, i).
We consider the partition (S x , S z 1 , . . ., S zs , S xz 1 , . . ., S xzs ) of S, where S z j ⊆ V z j (G) and S xz j are e j -vertices for j ∈ {1, . . ., s}; some sets in the partition could be empty.
Let t = |S x |.For each j ∈ {1, . . ., s}, select v i(j) ∈ S x to be a maximum with respect to the relation e j element of S x .Observe that the selection is not necessarily unique.We consider a partition P = {J 1 , . . ., J t } of {0, 1, . . ., s} such that 0 ∈ J 1 and it holds that j, j ∈ {1, . . ., s} are in the same set of P if and only if i(j) = i(j ).We define ϕ : {1, . . ., t} → C x as follows.If J h = ∅, then ϕ(h) = c(v i(j) ) for j ∈ J h .Then we extent ϕ on h ∈ {1, . . ., t} with J t = ∅ greedily to ensure that ϕ is a surjection.Such a mapping always exists because Consider z j for j ∈ {1, . . ., s}.It is assumed again W j = {u 1 , . . ., u q } is the set of z jvertices of G.We also assume that u 1 e j . . .e j u q .Let j ∈ J i .We consider two cases depending on whether z j is loaded or not.
Case 1.The vertex z j is loaded.Let h ∈ {1, . . ., q} be such that u h is a minimum with respect to e j element of S z j .By our inductive assumption, we have that By the definition of i(j) and h, the vertices of X i(j),h are not dominated by S x ∪ S z j .Therefore, they are dominated by S xz j .By the definition of α e j , we have that α e j (X i(j),h ) ≤ |S xz j |.Then Case 2. The vertex z j is unloaded.Let h ∈ {0, . . ., q} be the minimum integer such that u h+1 , . . ., u q are dominated by S z j .Clearly, if h > 0, then u h is not nominated by S z j .By the inductive assumption, we have that β(z j , h) ≤ |S z j | as S z j is extendable.Consider the set Y i(j),h .By the definition of this set, we obtain that the vertices of X i(j),h are not dominated by S x ∪ S z j .Therefore, they are dominated by S xz j .By the definition of α e j , we have that α e j (Y i(j),h ) ≤ |S xz j |.Then Now we combine (10) and (11) and conclude that for each h ∈ {1, . . ., t}, where i = i(j) for j ∈ J h .
By the definition of α e j (X h,t ), there is a set of e j vertices S xz j of size α e j (X h,t ) that dominates X h,t .Note that |S z j | + |S xz j | ≤ η(j).Observe also that S z j ∪ S xz j dominates all vertices of V z j (G) and the e j -vertices.
Case 2. The vertex z j is unloaded.We select t ∈ {1, . . ., p + 1} for which the minimum achieved in (8), that is, ψ(j) = t.In particular, we have that it holds where and Let h ∈ {1, . . ., q} be such that the minimum in the right part of ( 14) is achieved for this value.
By the inductive assumption, there is a set S z j ⊆ V z j (G) of size at most γ(z j , h) such that (v ) u h+1 , . . ., u q are dominated by S z j , (vi ) S z j contains at most d z-vertices for each z ∈ V (T z j ), (vii ) for each z ∈ V (T z j ), C z = c(S z j ∩ V G (z)), and (viii ) S z j dominates all the vertices of V x (G) \ {u 1 , . . ., u q }.By the definition of α e j (Y h,t ), there is a set of e j vertices S xz j of size α e j (Y h,t ) that dominates Y h,t .We have that |S z j | + |S xz j | ≤ η(j).Notice that S z j ∪ S xz j dominates all vertices of V z j (G) and the e j -vertices.Now we define S = ∪ s j=1 (S z j ∪ S xz j ).We have that By the definition of S, we have that (vi)-(viii) are fulfilled.To show (v), recall that γ(x, i) < +∞.Then for each h ∈ {i + 1, . . ., p}, there is j ∈ {1, . . ., s} such that v i j ψ(j) e j v h and, therefore, v h is dominated by S xz j .
Assume now that S ⊆ V x (G) is a set of minimum size satisfying (v)-(viii) for x and i ∈ {1, . . ., p} and S is extendable.Because S is a set of minimum size satisfying these conditions, |S| ≤ γ(x, i).We show that |S| ≥ γ(x, i).
(ii ) S z j contains at most d z-vertices for each z ∈ V (T z j ), (iii ) for each z ∈ V (T z j ), C z = c(S z j ∩ V G (z)), and (iv ) S z j dominates all the vertices of V z j (G).
By the definition of α e j (X h * ), there is a set of e j vertices S xz j of size α e j (X h * ) that dominates X h * .Note that |S z j | + |S xz j | ≤ η(j).Observe also that S z j ∪ S xz j dominates all vertices of V z j (G) and the e j -vertices.Let S = S z j ∪ S xz j ∪ {v 1 }.This set dominates all the vertices of V Case 2. The vertex z j is unloaded.
By (8), η(j) < min 1≤h≤q (γ(z j , h) + α e j (Y h,t )), where Let h ∈ {0, . . ., q} be the minimum index such that u h+1 , . . ., u q are dominated by S z j .By induction, |S z j | ≥ γ(z j , h).Since the vertices of G that are in Y h,t are not dominated by S z j , they are dominated by S xz j .By the definition of α e j (X h,t ), Let h * ∈ {1, . . ., q} be such that the minimum in the right part of ( 14) is achieved for this value, that is, η(j) = γ(z j , h * ) + α e j (Y h * ).By the inductive assumption, there is a set S z j ⊆ V z j (G) of size at most γ(z j , h * ) such that (v ) u h * +1 , . . ., u q are dominated by S z j , (vi ) S z j contains at most d z-vertices for each z ∈ V (T z j ), (vii ) for each z ∈ V (T z j ), C z = c(S z j ∩ V G (z)), and (viii ) S z j dominates all the vertices of V x (G) \ {u 1 , . . ., u q }.By the definition of α e j (X h ), there is a set of e j vertices S xz j of size α e j (Y h * ) that dominates Observe also that S z j ∪ S xz j dominates all vertices of V z j (G) and the e j -vertices.Let S = S z j ∪ S xz j ∪ {v 1 }.This set dominates all the vertices of V x (G).Note that |S | ≤ |S z j ∪ S xz j |.Now we use the obtained in both cases set S to obtain a contradiction with the extendability of S. The set S is extendable, that is, there is a dominating set D of G such that (a) D has at most d x-vertices for x ∈ V (T ), that has the minimum size and contains C and the conditions of the promise is fulfilled: the number of nodes z ∈ V (T ) such that D contains an z-vertex is maximum and for each It is straightforward to see that D is a dominating set and |D | ≤ |D|.We also have that (a) and (b) are fulfilled for D , because of (ii ), (iii ), (vi ) and (vii ), but this contradicts the condition that the number of nodes x ∈ V (T ) such that D contains an x-vertex is maximum because D contains the x-vertex v 1 .Hence, γ(x, i) = s j=1 η(j) and this completes the proof of the claim that |S| ≥ γ(x, i).Now we are ready to complete the description of our algorithm for Dominating Set Extension.The algorithm computes the table of values of β(r, i) if r is loaded and the table of values of γ(r, i) if r is unloaded.If r is loaded, we find the minimum value β(r, i * ) in the table for i.By Claim B, if β(r, i * ) < +∞, then G has a dominating set S of size at most β(r, i * ) containing at most d x-vertices for x ∈ V (T ) such that for each x ∈ V (T ), C x = c(S ∩ V G (x)).Moreover, if the promise is true, then β(r, i * ) < +∞ and the minimum size of S is β(r, i * ).If r is unloaded, then we consider γ(r, 0) in the table for r.By Claim B, if γ(r, 0) < +∞, then G has a dominating set S of size at most γ(r, 0) containing at most d x-vertices for x ∈ V (T ) such that for each x ∈ V (T ), C x = c(S ∩ V G (x)), and if the promise is true, then γ(r, 0) < +∞ and the minimum size of S is γ(r, 0).It remains to check whether β(r, i * ) ≤ k or γ(r, 0) ≤ k respectively and return the answer.
To evaluate the running time, observe that to compute β(x, i), we consider all possible partitions P = {J 1 , . . ., J t } for 1 ≤ t ≤ d of {0, 1, . . ., s} into non-empty sets such that 0 ∈ J 1 where s is the number of children of x.Since s ≤ , we have that the number of partitions is 2 O( log d) .Then for each partition P = {J 1 , . . ., J t }, we consider all possible surjections ϕ : {1, . . ., t} → C x .Since t ≤ d and each |C x | ≤ d, there are 2 O(d log d) choices of ϕ.Because each value of α and γ can be computed in polynomial time, it implies that the total running time of the algorithms is 2 O(( +d) log d) n O(1) .Now we are ready to prove the main theorem of the section.
Proof.Let (G, k) be an instance of Dominating Set where G is a connected chordal graph.
We use the algorithm of Habib and Stacho [HS09] to compute its leafage (G).If (G) > , we stop and return a no-answer.Otherwise, we consider the clique tree T of G constructed by the algorithm.We construct the tree T from T by dissolving nodes of degree two, that is, for a node x of degree two with the neighbors y and z, we delete x and make y and z adjacent.Observe that since T is a tree with at most leaves that has no node of degree two, |V (T )| ≤ 2 − 2. We have that G is a T -graph.Note also that the algorithm of Habib and Stacho [HS09] gives us a T -representation We consider 2 |V (T )| − 1 ≤ 2 2 −2 − 1 nonempty subsets of V (T ) and construct a coloring c : V G (T ) → {1, . . ., 2 |V (T )| } such that for u, v ∈ V G (T ), c(u) = c(v) if and only if u and v are U -vertices for the same U ⊆ V (T ).
By Lemma 4, a minimum dominating set of G contains at most 3|V (T )| − 2 ≤ 6 − 8 vertices of V G (T ).Clearly, these vertices can have at most 6 − 8 distinct colors.We consider all set C ⊆ {1, . . ., 2 |V (T )| } of distinct colors of size at most 6 − 8 and for each C, we aim to find a minimum dominating set of G whose vertices in V G (T ) are colored by the maximum number of distinct colors and are colored exactly by the colors of C. Since we consider all possible choices of C, it holds for some C.
Toward this aim, we apply the following rule.
To see that the rule is safe, observe that if D is minimum dominating set of G whose vertices in V G (T ) are colored exactly by the colors of C, then w is dominated by some xyvertex w .We have that v / ∈ D, because c(v) / ∈ C. Then it is straightforward to see that D = (D \ {w }) ∪ {v} is a minimum dominating set of G whose vertices in V G (T ) are colored by |C| + 1 colors.Now we are looking for a dominating set D of minimum size such that c(D ∩ V G (T )) = C.We use the following rule.
Rule 2. If there is a U -vertex u of G for nonempty U ⊆ V (T ) such that i) c(u) / ∈ C and ii) there is c ∈ C such that for every v ∈ V G (T ) with c(v) = c, v dominates u, then delete u.
To see that the rule is safe, observe that u cannot be included in a dominating set D of minimum size such that c(D ∩ V G (T )) = C and u is dominated by any set D such that c Observe that because of Rule 1, there is no e-vertex for e ∈ A .We contract the edges e ∈ A ∪ A .Denote by T the obtained from T tree and let T be the tree obtained from T by contracting the paths that correspond to the contracted edge.We also construct the graph Ĝ that is obtained from G by contracting these edges of T and we also construct its T -representation M = { Mv } v∈V ( Ĝ) where Mv ∈ V ( T ) for v ∈ V ( Ĝ).We set ĉ = c| V ( Ĝ) and let C x = {c | there is u ∈ V Ĝ( T ) with ĉ(u) = c} for x ∈ V ( Ĝ). Observe that M is a nice T -representation of Ĝ.Indeed, for every xy ∈ E(T ) such that x, y ∈ M u for u ∈ V G (T ) we have that xy To see that the rule is safe, assume that the input graph G has a minimum dominating set D whose vertices in V G (T ) are colored exactly by the colors of C. By Lemma 3, we have that if a set of nodes X of T is contracted into a single vertex x of T , then D has at most |X|+ −1 vertices whose models contain a vertex of X and, therefore, the number of vertices colored by the colors of C x in D is at most d.
We select arbitrarily a node r be the root of T and T respectively.Then we apply Lemma 7 for the instance ( T , k, d, ĉ, {C x } x∈V ( T ) ) of Dominating Set Extension.
Recall that Dominating Set Extension is a promise problem.If the algorithm from Lemma 7 returns a yes-answer, it means that there is a dominating set D of Ĝ of size at most k such that for each x ∈ V ( T ), C x = c(D ∩ V Ĝ(x)).It means that the input graph G has a dominating set of size at most k.Still, if the promise is false, the algorithm can return an incorrect no-answer.Recall that the promise of Dominating Set Extension is the following: for a dominating set D of Ĝ of minimum size with the properties that (a) D has at most d x-vertices for x ∈ V ( T ), (b) for each x ∈ V ( T ), C x ⊆ c(D ∩ V G (x)), it holds that the number of nodes x ∈ V ( T ) such that D contains an x-vertex is maximum and for each x ∈ V ( T ), C x = c(D ∩ V G (x)).By Lemmas 5 and 6, we have that if C is chosen in such a way that G has a minimum dominating set D that has the maximum number of vertices of V G (T ) and whose vertices in V G (T ) are colored exactly by the colors of C, then this promise holds for the corresponding instance of Dominating Set Extension constructed for this choice of C. Therefore, if (G, k) is a yes-instance of Dominating Set, then for some choice of C, we obtain a yes-answer.
To evaluate the running time of the algorithm, observe that T , T and the representation M are constructed in polynomial time by the algorithm of Habib and Stacho [HS09].The coloring function c : V G (T ) → {1, . . ., 2 |V (T )| } can be constructed in time 2 O( ) • n O(1) .Then we construct 2 O( 2 ) sets C ⊆ {1, . . ., 2 |V (T )| } of size at most 6 − 8 and it can be done in time 2 O( 2 ) .For each C, the Rules 1 and 2 can be applied in polynomial time.Similarly, the construction of the instance ( T , k, d, ĉ, {C x } x∈V ( T ) ) of Dominating Set Extension for a given C can be done in polynomial time.Clearly, Rule 3 can be applied in polynomial time.Then we solve the constructed instance of Dominating Set Extension in time 2 O( log ) • n O(1) .Hence, the total running time of the algorithm is 2 O( 2 ) • n O(1) .
The theorem immediately gives the following corollary for T -graphs.
Corollary 5. Dominating Set can be solved in time 2 O(|V (T )| 2 ) • n O(1) for T -graphs if T is a tree.

A polynomial kernel for Clique
It was observed in [CZ17] that the Clique problem is FPT for H-graphs when parameterized by the solution size k and ||H|| if an H-representation of G is given.We show that Clique admits a polynomial kernel.
Let G be an H-graph with an H-representation M = {M v } v∈V (G) where M v ⊆ V (H ) for the corresponding subdivision H of H. Recall that for e ∈ E(T ), v ∈ V (G) is an e-vertex if M v contains only subdivision nodes of H from the path in H corresponding to e in H.We claim that we can find a maximum clique in G that contains some e-vertex in polynomial time.
Lemma 8. Let G be an H-graph given together with its H-representation.Then a clique of maximum size that contains at least one e-vertex for some e ∈ E(G) can be found in time O(n 3/2 m).
Proof.Let M = {M v } v∈V (G) be an H-representation of G.For each e-vertex u of G, we find a maximum clique K such that M u is inclusion minimal for K, that is, there is no v ∈ K with M v ⊂ M u .Let e = xy for x, y ∈ V (H) and denote by P the (x, y)-path corresponding to e in the subdivision H of G. Since u is an e-vertex, M u ⊆ V (P ), that is, the vertices of M u form a subpath of P .Denote by x and y the end-vertices of the subpath.Note that it can happen that x = y .Because M u is inclusion maximal model of a vertex of K, for every Notice that U can be partitioned into two cliques It means that G is a cobipartite graph.A maximum clique in a cobipartite graph can be found in time O( √ nm) by the algorithm of Hopcroft and Karp [HK73] as finding a maximum clique in G is equivalent to finding a maximum independent set in the complement of G that is a bipartite graph.Note that a maximum clique in G always contains u, because u is adjacent to every other vertex of G .
Since we consider all e-vertices to find a maximum clique containing some e-vertex for some e ∈ E(H), the total running time is O(n 3/2 m).Now we a ready to construct our kernel.
Theorem 9.The Clique problem for H-graphs admits a kernel with at most (k − 1)|V (H)| vertices if an H-representation of the input graph is given.
Proof.Let G be an H-graph with an H-representation M = {M v } v∈V (G) where M v ⊆ V (H ) for the corresponding subdivision H of H.
First, we use Lemma 8 to check whether G has a clique of size at least k that contains at least one e-vertex for some e ∈ E(G).If we find such a clique we return a yes-answer.Assume that this is not the case.Let G be the graph obtained from G by the deletion of all e-vertices for e ∈ E(H).We have that G has a clique of size at least k if and only if G has a clique of size at least k.
If there is x ∈ V (H) such that V x = {v ∈ V (G ) | x ∈ M v } has size at least k, then it is a clique of size at least k and we return a yes-answer.Otherwise, we return G .Clearly, |V (G )| ≤ (k − 1)|V (H)| in this case.
j is not dominated by D. This contraction shows the claim.Now we consider the Clique problem.Recall that Clique, given a graph G and a positive integer k, ask whether G has a clique of size at least k.Note that it was shown by Chaplick and Zeman[CZ17] that Clique is FPT when parameterized by k + ||H||.We show that the problem is W[1]-hard when parameterized by the size of H only.To do it, we first prove W[1]-hardness for the Weighted Clique problem that, given a graph G, a weight function ω : V (G) → N and a positive integer k, asks whether G has a clique K of weight ω(K) = v∈K ω(v) ≥ k.Theorem 7. Weighted Clique is W[1]-hard in the strong sense (i.e., if the weights are encoded in unary) for H-graphs when parameterized by ||H|| and the hardness holds even if an H-representation of G is given.
and for every pair x, y of distinct nodes of U , add a new edge xy.
shown in Figure 4 b).

Figure 5 :
Figure 5: The construction of G .
and only if u and v are U -vertices for the same U ⊆ V (T ).The next lemmas are used to simplify the models of vertices of G by contracting edges of T .Lemma 5. Let C ⊆ c(V G (T )) and let A = {xy ∈ E(T ) | x, y ∈ M u for u ∈ V G (T ), c(u) ∈ C}.Let also G be the graph obtained from G by consecutive contracting edges of A in T .Then for a set D ⊆ V (G), D is a dominating of G of minimum size satisfying the condition C = c(D ∩ V G (T )) if and only if D is a dominating set of G of minimum size satisfying the same condition C = c(D ∩ V G (T )).
be a nice r-rooted T -representation of G with respect to T .For a vertex x ∈ V (T ), we denote by dist(x) the distance between r and x in T .For a set X ⊆ V (T ), we denote by d min (X) = min{dist(x) | x ∈ X} and by d max (X) = max{dist(x) | x ∈ X}.Consider an edge e = xy ∈ E(T ), where y is a child of x, and denote by P e the (x, y)-path in T corresponding to e.Let u, v ∈ V (G) such that M e u = M u ∩ V (P e ) = ∅ and M e v = M v ∩ V (P e ) = ∅.We write u e v if either d min (M e u ) < d min (M e v ) or d min (M e u ) = d min (M e u ) and d max (M e u ) ≤ d max (M e v ).Respectively, u ≺ e v if either d min (M e u ) < d min (M e v ) or d min (M e u ) = d min (M e v ) and d max (M e u ) < d max (M e v ).We consider the following auxiliary problem for T -graphs with nice representations.Dominating Set ExtensionInput: A tree T with leaves and a graph G with a nice r-rooted T -representation of G with respect to a subdivision T of T , positive integers k and d, a labeling function c : ∪ x∈V (T ) V G (x) → N, and a collection of sets {C x } x∈V (T ) of size at most d where each C x ⊆ c(V G (x)) (some sets could be empty) such that for a dominating set D of G of minimum size with the properties that (a) D has at most d x-vertices for x ∈ V (T ), (b) for each ∈ A if c(u) ∈ C and xy ∈ A if c(u) / ∈ C because of Rule 2, and all such edges xy are contracted.Combining Lemmas 5 and 6 we obtain that D is a dominating set of minimum size withC = c(D ∩ V G (T )) if and only if D is a dominating set of G of minimum size such that C = ĉ(D ∩ V Ĝ( T )).Note that the condition C = ĉ(D ∩ V Ĝ( T )) is equivalent to the condition that for every x ∈ V ( T ), C x = ĉ(D ∩ V Ĝ(x)), because the T -representation of Ĝ is nice and C x ∩ C y = ∅ for distinct x, y ∈ V ( T ).We set d = |V (T )| + − 1 ≤ 3 − 3 and apply the next rule.Rule 3.If there is x ∈ V ( T ) with |C x | > d, then discard the current choice of C.