Parameterized complexity of computing maximum minimal blocking and hitting sets
PParameterized complexity of computing maximumminimal blocking and hitting sets
Júlio Araújo
Departamento de Matemática, Universidade Federal do Ceará, Fortaleza, [email protected]
Marin Bougeret
LIRMM, Université de Montpellier, CNRS, Montpellier, [email protected]
Victor A. Campos
Departamento de Computação, Universidade Federal do Ceará, Fortaleza, [email protected]
Ignasi Sau
LIRMM, Université de Montpellier, CNRS, Montpellier, [email protected]
Abstract A blocking set in a graph G is a subset of vertices that intersects every maximum independent set of G .Let mmbs ( G ) be the size of a maximum (inclusion-wise) minimal blocking set of G . This parameterhas recently played an important role in the kernelization of Vertex Cover parameterized bythe distance to a graph class F . Indeed, it turns out that the existence of a polynomial kernel forthis problem is closely related to the property that mmbs ( F ) = sup G ∈F mmbs ( G ) is bounded by aconstant, and thus several recent results focused on determining mmbs ( F ) for different classes F . Weconsider the parameterized complexity of computing mmbs under various parameterizations, such asthe size of a maximum independent set of the input graph and the natural parameter. We provide apanorama of the complexity of computing both mmbs and mmhs , which is the size of a maximumminimal hitting set of a hypergraph, a closely related parameter. Finally, we consider the problemof computing mmbs parameterized by treewidth, especially relevant in the context of kernelization.Given the “counting” nature of mmbs , it does not seem to be expressible in monadic second-orderlogic, hence its tractability does not follow from Courcelle’s theorem. Our main technical contributionis a fixed-parameter tractable algorithm for this problem. Mathematics of computing → Graph algorithms.
Keywords and phrases maximum minimal blocking set, maximum minimal hitting set, parameterizedcomplexity, treewidth, kernelization, vertex cover, upper domination.
Funding
Júlio Araújo : CNPq-Pq 304478/2018-0, CAPES-PrInt 88887.466468/2019-00 and CAPES-STIC-AmSud 88881.569474/2020-01.
Victor A. Campos : FUNCAP - PNE-011200061.01.00/16.
Ignasi Sau : DEMOGRAPH (ANR-16-CE40-0028), ESIGMA (ANR-17-CE23-0010) and ELIT (ANR-20-CE48-0008-01).
Given a graph G , we denote by α ( G ) the maximum size of an independent set of G , thatis, of a set of pairwise non-adjacent vertices. For the sake of conciseness, we abbreviate“independent set” as is , and “maximum independent set” as mis . A set B ⊆ V ( G ) is a blockingset , abbreviated as bs , of G if α ( G \ B ) < α ( G ), where G \ B = G [ V ( G ) \ B ]. Equivalently, B is a blocking set of G if for every mis I ∗ ⊆ V ( G ), I ∗ ∩ B = ∅ . In this work we are interested in(inclusion-wise) minimal blocking sets, which we abbreviate as mbs . We denote by mmbs ( G )the maximum size of an mbs of G , and by Maximum Minimal Blocking Set ( MMBS a r X i v : . [ c s . D S ] F e b Parameterized complexity of computing maximum minimal blocking and hitting sets for short) the problem where, given a graph G and an integer β , the objective is to decidewhether mmbs ( G ) ≥ β . The main objective of this paper is to study the parameterizedcomplexity of MMBS . As discussed below, this problem is strongly related to the
MaximumMinimal Hitting Set ( MMHS ) problem, for which we also present several results.
Role of maximum minimal blocking sets in kernelization.
Given a graph G , a setof vertices S ⊆ V ( G ) is a vertex cover if it contains at least one endpoint of every edge.The Vertex Cover ( VC for short) problem consists in, given a graph G and an integer k , decide if there is a vertex cover S of G such that | S | ≤ k . For a fixed graph class F , the VC / dist - to - F parameterized problem is defined as follows. The input is a triple ( G, X, k )where G is a graph, X ⊆ V ( G ), and G \ X belongs to F . Set X is often referred to asa modulator to F , and | X | as the distance of G to F . The objective of the problem is todecide whether G admits a vertex cover of size at most k , and the parameter is | X | . A kernel of vertex size f for this problem is a polynomial-time algorithm that, given an input( G, X, k ), outputs an equivalent instance ( G , X , k ) with | V ( G ) | ≤ f ( | X | ). Informally, sucha kernel compresses the input graph G to a smaller graph G whose size is bounded by afunction f depending only on | X | . If f is a polynomial (resp. linear) function, we speak of a polynomial (resp. linear ) kernel. We refer the reader to Section 2 for formal definitions. The VC / dist - to - F problem has been defined by Jansen and Bodlaender [22] for F being theclass of forests as a way to improve the linear kernel for Vertex Cover parameterized by thestandard parameter k , and the main result of [22] is a polynomial kernel for VC / dist - to - F (for F being the forests).This result triggered a long line of follow-up research, which aimed to find the mostgeneral graph families F such that VC / dist - to - F admits a polynomial kernelization [18].Several results where proved for specific families F such as those of degree at most two,of bounded treedepth, pseudo-forests (see [8, 18] for a complete list of references), and amajor open question in this area is to find a characterization of the families F for which VC / dist - to - F admits a polynomial kernel [8]. This is where parameter β comes into play.Kernelization algorithms for VC / dist - to - F usually proceed in two steps. In step 1,they reduce the number of connected components of G \ X to a polynomial in | X | , and instep 2 they reduce the size of each connected component of G \ X to a polynomial in | X | as well. Minimal blocking sets have been introduced in the seminal paper of Jansen andBodlaender [22] for the case of F being the class of forests as a handy tool to achieve step 1.After that, this notion has been generalized and reused for example in [8,9,20], finally leadingto the following black box tool for step 1, where mmbs ( F ) = sup G ∈F mmbs ( G ). (cid:73) Theorem 1 (Hols et al. [20]) . Let F be a hereditary graph class on which VC can be solvedin polynomial time. There is a polynomial-time algorithm that, given an instance ( G, X, k ) of VC / dist - to - F , returns an equivalent instance ( G , X, k ) of VC / dist - to - F such that G \ X ∈ F and has at most O ( | X | mmbs ( F ) ) connected components. Informally, Theorem 1 states that, if mmbs ( F ) is bounded, then “half” of the kernelizationalgorithm can be done automatically. Moreover, it has been shown that mmbs ( F ) beingbounded by a constant is necessary in order to obtain a polynomial kernel: (cid:73) Theorem 2 (Hols et al. [20]) . Unless NP ⊆ coNP/poly , VC / dist - to - F does not admit akernel of size O ( | X | mmbs ( F ) − ε ) for any ε > . These two theorems might suggest that mmbs might be the right candidate to characterizegraph classes F for which VC / dist - to - F admits a polynomial kernel. However, it turns out úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 3 that there exists a class F where mmbs ( F ) is bounded but for which there is no polynomialkernel for VC / dist - to - F under standard complexity assumptions [20]. Nevertheless, forminor-closed families , the following theorem shows that mmbs is indeed the correct parameterin order to characterize the existence of polynomial kernels for VC / dist - to - F . (cid:73) Theorem 3 (Bougeret et al. [8]) . If F is a minor-closed graph class, then VC / dist - to - F admits a polynomial kernel if and only if mmbs ( F ) is bounded by a constant. To summarize this discussion, for general graph classes F , having bounded mmbs ( F ) isnecessary but not sufficient, although having bounded mmbs ( F ) yields “half” of the kernel.For minor-closed classes F , having bounded mmbs ( F ) is indeed the correct characterization.These results explain the recent interest in computing mmbs ( F ) for different classes F [20],and thus our motivation to study the complexity of the MMBS problem.Let us also mention that computing mmbs can in addition be useful when implementingany of the previously mentioned kernels. Indeed, given an instance (
G, X, k ) of VC / dist - to - F with | V ( G ) | = n , the algorithm behind Theorem 1 takes as additional input the value mmbs ( F ) and outputs the claimed equivalent instance in time n mmbs ( F )+ O (1) . However,when implementing this algorithm, we can rather first compute mmbs ( G \ X ), and usethe algorithm of Theorem 1 with additional input mmbs ( G \ X ) instead of mmbs ( F ) (notethat mmbs ( G \ X ) ≤ mmbs ( F ), potentially much smaller), and thus obtain a running time n mmbs ( G \ X )+ O (1) , and an equivalent instance ( G , X, k ), where G \ X ∈ F , with at most O ( | X | mmbs ( G \ X ) ) connected components. Contribution and related work.
In what follows we present our contribution and relate itto previous work, by considering each parameterization of the considered problems separately.
Choice of the parameters.
As the VC / dist - to - F problem has only been considered for graphclasses F where Maximum Independent Set ( IS for short) can be solved in polynomialtime, we also incorporate this assumption in this work, hence motivating the parameterizationof MMBS by combinations of α (i.e., the size of a mis of the input graph) and the threshold β (the natural parameter). Moreover, since in all the previously mentioned cases where VC / dist - to - F has a polynomial kernel [8, 18] the graphs in the class F have bounded treewidth , we also consider the MMBS problem parameterized by the treewidth of the inputgraph G . Problems related to computing mmbs . We denote by
Maximum Minimal Hitting Set ( MMHS for short) the problem where, given a hypergraph H and an integer β , the objectiveis to decide whether mmhs ( H ) ≥ β , where mmhs ( H ) is the size of a largest minimal hittingset of H , that is, an inclusion-wise minimal set of vertices of H containing at least one vertexof every hyperedge. A dominating set in a graph G is a subset of vertices S ⊆ V ( G ) suchthat every vertex in V ( G ) \ S has a neighbor in S . We denote by Upper DominatingSet ( Up-Dom for short) the problem of computing a maximum inclusion-wise minimaldominating set in an input graph G . As pointed out by Bazgan et al. [2], Up-Dom is aspecial case of
MMHS , as we can create a hyperedge for each closed neighborhood of thevertices in G , implying that the negative results stated below for Up-Dom transfer directlyto
MMHS . As mentioned before, we only consider graph classes F where IS can be solvedin polynomial time. A natural special case of such classes is when α is constant, and in thiscase MMBS reduces to
MMHS by simply generating in time n α + O (1) a hyperedge for each A graph class is minor-closed if any minor of a member in the class also belongs to it, and a graph H isa minor of a graph G is H can be obtained from a subgraph of G by contracting edges. Parameterized complexity of computing maximum minimal blocking and hitting sets mis of G . Let us now define parameterizations for MMBS and
MMHS . A parameterizationis, in a nutshell, a function mapping instances of a parameterized problem to non-negativeintegers (see Section 2 for the details). For
MMBS , we define, with slight abuse of notation,the parameters α as α ( G, β ) = α ( G ) and β as β ( G, β ) = β . Similarly, for MMHS , we definethe parameters α as α ( H , β ) = max H ∈ E ( H ) | H | and β as β ( H , β ) = β .The first objective of this paper is to obtain a complete landscape of the parameterizedcomplexity of MMBS and
MMHS under different combinations of α and β to compare thebehavior of these two problems. To the best of our knowledge, the parameterized complexityof MMBS has not been considered before in the literature. On the other hand, the
MMHS problem has received considerable attention, especially concerning the enumeration of minimalhitting sets, as it allows to construct the so-called dual hypergraph, that is, the hypergraphhaving a hyperedge for every minimal hitting set of the original hypergraph. Let us nowpresent our contributions together with the related work about the parameterized complexityof
MMHS , for each different parameterization that we consider.
Parameterization by α or β separately. When parameterizing by α only, both MMBS and
MMHS are para- NP -hard, meaning NP -hard for fixed values of the parameter. Indeed, theparticular case α = 2 of MMHS corresponds to the
Maximum Minimal Vertex Cover ( MMVC for short) problem, which is NP -hard [7]. When parameterizing by β only, we showin Proposition 6 that MMBS is para- NP -hard, whereas MMHS is W[1] -hard [2] and XP [4].As discussed in Section 3, the W[1] -hardness proof of [2] implies that, unless the ExponentialTime Hypothesis (
ETH for short) fails,
Up-Dom cannot be solved in time f ( k ) · n o ( √ k ) on n -vertex graphs for any computable function f . We improve this lower bound by showing inTheorem 7 that Up-Dom cannot be solved in time f ( k ) · n o ( k ) for any computable function f , implying the same result (replacing k by β ) for MMHS . We point out that very recentlyand independently from our work, this improved lower bound for
Up-Dom has also beenproved by Dublois et al. [17].
Parameterization by one parameter while fixing the other.
When fixing α and parameterizingby β , MMBS reduces to
MMHS , which was known to be
FPT [14], and for which we providein Proposition 16 a polynomial kernel with O ( β α ) vertices, generalizing the known quadratickernel for MMVC [7, 19]. When fixing β and parameterizing by α , we show in Proposition 6that MMBS is W[1] -hard, whereas
MMHS is FPT (at it is even
FPT parameterized by thesum as explained in the next paragraph).
Parameterization by the sum.
Finally, when parameterizing by α + β , the hardness resultgiven in Proposition 6 (i.e., parameterizing by α for fixed β ) implies that MMBS is W[1] -hard,whereas
MMHS is FPT for the following reasons. We first provide in Corollary 18 a simple
FPT algorithm for
MMHS that reduces to an extension problem considered by Bläsius etal. [4], and then design in Theorem 24 a more involved an ad-hoc algorithm to improve therunning time to O ∗ (2 αβ ), where the O ∗ -notation hides multiplicative polynomial terms (seeSection 2).Our results considering parameters α and β are summarized in Table 1, where Π /κ denotes problem Π parameterized by function κ (see again Section 2). Parameterization by treewidth.
Let us now turn to the second objective of this paper, whichis the parameterization by treewidth. It is known that both
Up-Dom [2] and
MMVC [7]are
FPT parameterized by treewidth, but none of these results implies the same resultfor
MMBS . We also mention that that the problem of finding a maximum minimal setintersecting all maximum cliques of a graph is
FPT parameterized by treewidth [24], implying úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 5
Parameter
MMBS MMHS β para- NP -hard (Prop. 6) XP ([4], Corollary 18) W[1] -hard [2] (cid:64) f ( β ) · ( | V ( H ) | + | E ( H ) | ) o ( β ) (Cor. 11) α para- NP -hard (Prop. 6) para- NP -hard ( MMVC for α = 2) β ( α fixed) Reducible to MMHS in time n α + O (1) Kernel with O ( β α ) vertices (Prop. 16) FPT in O ∗ ( α β ) [14] ( O ∗ hides n f ( α ) ) α ( β fixed) W[1] -hard (Prop. 6) XP (trivial) FPT (as
FPT by α + β ) α + β W[1] -hard (Prop. 6)
FPT (Cor. 18 and Thm. 24)
Figure 1
Parameterized complexity of
MMBS and
MMHS with parameters α and β . The maindifferences between MMBS and
MMHS show up in the parameterizations by β and α + β . that MMBS is FPT parameterized by treewidth of the complement of the input graph. Weprove in Theorem 51 that
MMBS / tw is FPT , which is the main technical result of this paper.Let us mention that
MMBS does not seem to be (at least, easily) expressible in monadicsecond-order logic, due to the fact that the a blocking set in a graph G is defined so that itintersects every maximum-sized independent set of G , and properties involving counting the sizes of arbitrarily large sets are typically non-expressible in monadic second-order logic [6,12].This is why, in order to deduce that MMBS / tw is FPT , we cannot directly apply Courcelle’stheorem [11], and we need to design an ad-hoc algorithm that is quite involved, needing anumber of technical lemmas. In we discuss the list of parameters used in the tables of ourdynamic programming algorithm, along with several examples to illustrate the difficultiesthat motivate their choice.
Organization.
In Section 2 we provide preliminaries about graphs, parameterized com-plexity, the formal statement of the considered problems, and treewidth, and we state inLemma 4 several useful properties of minimal blocking and hitting sets. Section 3 is devotedto parameterizations by α and β , and Section 4 to the algorithm for MMBS parameterizedby treewidth. We conclude the article in Section 5 with some directions for further research.
Graphs and functions.
We only provide here basic definitions and refer the reader to [15]for any missing definitions about graphs. We only consider finite simple graphs with no loopsnor multiple edges. For a graph G and a vertex v ∈ V ( G ), we denote by N G ( v ) the set ofvertices of G adjacent to v and, for a subset S ⊆ V ( G ), we let N S ( v ) = N G ( v ) ∩ S . When thegraph G is clear from the context, we may omit the subscript. Given B ⊆ V ( G ), we denote G \ B = G [ V ( G ) \ B ] where G [ X ] denotes the graph induced by X ⊆ V ( G ). We denote a triangle , that is, a complete graph on three vertices, on vertices u, v, w by ( u, v, w ). Givena graph G , we say that X ⊆ V ( G ) is a vertex cover if for any edge e ∈ E ( G ), e ∩ X = ∅ ,and that X is a dominating set if for every v ∈ V ( G ) \ X , there exists u ∈ X such that { u, v } ∈ E . Given a hypergraph H , we say that I ⊆ V ( H ) is an independent set if for every H ∈ E ( H ), H (cid:42) I , and that X ⊆ V ( H ) is a hitting set if for every S ∈ E ( H ), S ∩ X = ∅ . Agraph class is hereditary if it is closed under induced subgraphs.If a set A is partitioned into pairwise disjoint subsets A , . . . , A k , we denote it by A = A ] · · · ] A k . If A is a set, we denote by 2 A the collection containing all the subsets of A . Given a function f : A → B and a subset A ⊆ A , we denote by f | A the restriction of Parameterized complexity of computing maximum minimal blocking and hitting sets f to A . For a positive integer k , we let [ k ] be the set containing every integer i such that1 ≤ i ≤ k .For a function f mapping graphs to integers (such as the parameterizations α and β discussed before) and a class of graphs F , we define f ( F ) = sup G ∈F f ( G ). Given twofunctions f and f , mapping instances I of a problem to N ), if there exists a polynomial p such that for every instance I , f ( I ) ≤ f ( I ) · p ( | I | ), where | I | is the size of I , then we write f = O ∗ ( f ). Parameterized complexity.
We refer the reader to [13, 16] for basic background onparameterized complexity, and we recall here only some basic definitions. A parameterizedproblem is a language L ⊆ Σ ∗ × N , where Σ is some fixed alphabet. For an instance I = ( x, k ) ∈ Σ ∗ × N , k is called the parameter . Given a classical (non-parameterized) decisionproblem L c ⊆ Σ ∗ and a function κ : Σ ∗ → N , we denote by L c /κ = { ( x, κ ( x ) } | x ∈ L c } theassociated parameterized problem.A parameterized problem L is fixed-parameter tractable ( FPT ) if there exists an algorithm A , a computable function f , and a constant c such that given an instance I = ( x, k ), A (called an FPT algorithm ) correctly decides whether I ∈ L in time bounded by f ( k ) · | I | c .For instance, the Vertex Cover problem parameterized by the size of the solution is
FPT .A parameterized problem L is XP if there exists an algorithm A (called an XP algorithm )and two computable functions f and g such that given an instance I = ( x, k ), A (calledan XP algorithm ) correctly decides whether I ∈ L in time bounded by f ( k ) · | I | g ( k ) . Forinstance, the Independent Set problem parameterized by the size of the solution is XP .Within parameterized problems, the W -hierarchy may be seen as the parameterizedequivalent to the class NP of classical decision problems. Without entering into details(see [13, 16] for the formal definitions), a parameterized problem being W [1]- hard can beseen as a strong evidence that this problem is not FPT . The canonical example of W [1]-hardproblem is Independent Set parameterized by the size of the solution.The most common way to transfer
W[1] -hardness is via parameterized reductions. A parameterized reduction from a parameterized problem L to a parameterized problem L isan algorithm that, given an instance ( x, k ) of L , outputs an instance ( x , k ) of L such that( x, k ) is a yes-instance of L if and only if ( x , k ) is a yes-instance of L , k ≤ g ( k ) for some computable function g , andthe running time is bounded by f ( k ) · | x | O (1) for some computable function f If L is W [1]-hard and there is a parameterized reduction from L to L , then L is W [1]-hardas well.A parameterized problem is para- NP -hard if it is NP -hard for a fixed value of the parameter,implying in particular that the problem cannot be in XP unless P = NP .A kernelization algorithm for a parameterized problem L is an algorithm A that, givenan instance ( x, k ) of L , generates in polynomial time an equivalent instance ( x , k ) of Q such that | x | + k ≤ f ( k ), for some computable function f : N → N . If f ( k ) is bounded fromabove by a polynomial function, we say that L admits a polynomial kernel. In particular, if f ( k ) is bounded by a linear (resp. quadratic) function, then we say that L admits a linear (resp. quadratic ) kernel.The Exponential Time Hypothesis ( ETH for short) of Impagliazzo et al. [21] is a complexityassumption implying that the 3-SAT problem cannot be solved in time 2 o ( n ) restricted toformulas with n variables. List of considered problems.
We denote by IS the Maximum Independent Set problemwhere, given a graph G and an integer k , the objective is to decide whether α ( G ) ≥ k , and by CIS the
Multicolored Independent Set problem, where given graph G and an integer úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 7 k such that V ( G ) is partitioned into k cliques { V i | i ∈ [ k ] } , the goal is to decide whether α ( G ) ≥ k .In the two following problems, recall that mmbs ( G ) (resp. mmhs ( H )) denotes the size ofa largest minimal blocking set of G (resp. largest minimal hitting set of H ). Maximum Minimal Blocking Set ( MMBS ) Input:
A graph G and a positive integer β . Question: mmbs ( G ) ≥ β ? Maximum Minimal Hitting Set ( MMHS ) Input:
A hypergraph H and a positive integer β . Question: mmhs ( H ) ≥ β ?The problem Maximum Minimal Vertex Cover ( MMVC ) corresponds to the restric-tion of
MMHS to instances where all hyperedges have size two, i.e., graphs. For a fixedpositive integer α , we also define α - MMHS as the
MMHS problem restricted to instanceswhose hypergraph H is such that | H | ≤ α for every H ∈ E ( H ), and α - MMBS as the
MMBS problem restricted to instances whose graph G is such that α ( G ) ≤ α . Notice that in any FPT or kernel algorithm for α - MMHS or for α - MMBS , as α is fixed, the running timegiven using the O ∗ -notation might typically hide a term n f ( α ) , where n is the number ofvertices of the graph or hypergraph under consideration. Finally, we define MMBS = (resp. MMBS ≤ ) as the MMBS problem where the objective is to decide whether mmbs ( G ) = β (resp. mmbs ( G ) ≤ β ). Extension-MMHS ( Ext-MMHS ) Input:
A hypergraph H and a two subsets X, Y ⊆ V ( H ) such that X ∩ Y = ∅ . Question:
Does there exist a minimal hitting set S of H such that X ⊆ S ⊆ V ( H ) \ Y ?Problem Ext-MMHS was defined by Bläsius et al. [4]. We also define
Simple-Ext-MMHS as the special case of the
Ext-MMHS where Y = ∅ .The last problem we define here is the “max-min” version of Dominating Set . Upper Dominating Set ( Up-Dom ) Input:
A graph G and an integer k . Question:
Does G contain a minimal dominating set of size at least k ? Tree decompositions and treewidth. A tree decomposition of a graph G is a pair D = ( T, B ), where T is a tree and B = { X w | w ∈ V ( T ) } is a collection of subsets of V ( G ),called bags , such that: S w ∈ V ( T ) X w = V ( G ),for every edge { u, v } ∈ E , there is a w ∈ V ( T ) such that { u, v } ⊆ X w , andfor every { x, y, z } ⊆ V ( T ) such that z lies on the unique path between x and y in T , X x ∩ X y ⊆ X z .We call the vertices of T nodes of D and the sets in B bags of D . The width of a treedecomposition D = ( T, B ) is max w ∈ V ( T ) | X w | −
1. The treewidth of a graph G , denoted by tw ( G ), is the smallest integer t such that there exists a tree decomposition of G of width atmost t . We need to introduce nice tree decompositions, which will make the presentation of Parameterized complexity of computing maximum minimal blocking and hitting sets the algorithm of Section 4 much simpler.
Nice tree decompositions.
Let D = ( T, B ) be a rooted tree decomposition of G (meaningthat T has a special vertex r called the root ). As T is rooted, we naturally define an ancestorrelation among bags, and say that X w is a descendant of X w if the vertex set of the uniquesimple path in T from r to w contains w . In particular, every node w is a descendant ofitself. For every w ∈ V ( T ), we define G X w = G [ S { X w | X w is a descendant of X w in T } ].Such a rooted decomposition is called a nice tree decomposition of G if the followingconditions hold: X r = ∅ ,every node of T has at most two children in T ,for every leaf ‘ ∈ V ( T ), X ‘ = ∅ . Each such a node ‘ is called a leaf node ,if w ∈ V ( T ) has exactly one child w , then either X w = X w ∪ { v } for some v X w . Each such a node is called an introduce node , X w = X w \ { v } for some v ∈ X w . Each uch a node is called a forget node , andif w ∈ V ( T ) has exactly two children w L and w R , then X w = X w L = X w R . Notice thatthere is no edge in G X w between V ( G X wL ) \ X w and V ( G X wR ) \ X w . Each such a nodeis called a join node .Given a tree decomposition of a graph G , it is possible to transform it in polynomial timeinto a nice one of the same width [23].For the sake of simplicity of the (already quite heavy) notation used in the dynamicprogramming algorithm of Section 4, we will drop the vertices of V ( T ) from the notation ofbags defined above. Therefore, in the case of an introduce or forget node, the bag X w andits child X w will be denoted X and X C , respectively, and in the case of a join node, thebag X w and its children X w L and X w R will be denoted X, X L , and X R respectively. Basic properties.
We now state some basic properties of the considered problems that willbe used later. The ones concerning minimal blocking sets have been already (explicitly orimplicitly) observed in [20], but for the sake of completeness we prove all of them here. (cid:73)
Lemma 4.
The following properties hold. For every graph G , B ⊆ V ( G ) is an mbs of G if and only if B is a bs of G and, for every v ∈ B , there is a mis I v of G such that I v ∩ B = { v } . For every hypergraph H , B ⊆ V ( H ) is an minimal hitting set of H if and only if B is a hitting set of H and, for every v ∈ B , there is a hyperedge H v of H such that H v ∩ B = { v } . For every graph G , there exists a unique mis in G if and only if mmbs ( G ) = 1 . Given an instance ( H , β ) of MMHS , we can obtain in polynomial time an equivalentinstance ( H , β ) such that V ( H ) = V ( H ) , E ( H ) ⊆ E ( H ) , and no hyperedge of H iscontained in another hyperedge of H . Given an instance ( H , β ) of MMHS , we can obtain in polynomial time an equivalentinstance ( H , β ) such that V ( H ) ⊆ V ( H ) , and every vertex of H belongs to at least onehyperedge of H . Proof:
Property 1 . For the forward implication, consider an mbs B of G . As B is minimal,for every v ∈ B , there exists a mis I v such that I v ∩ { B \ { v }} = ∅ . As B is a bs , I v ∩ B = ∅ ,implying I v ∩ B = { v } . The backward implication is immediate. The proof of Property 2 isalmost the same. Property 3 . For the forward implication, let I be the unique mis of G , B be a bs of G ,and v ∈ B ∩ I . If | B | ≥ B is not minimal as { v } is still a bs . Let us now prove the úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 9 contrapositive of the backward implication. Suppose that G contains two distinct mis I and I . Let us consider the reduction mapping G to a hypergraph H on the same vertex set, andhaving one hyperedge for each mis of G . Let H and H be the hyperedges corresponding to I and I , respectively. Observe that { H , H } is a sunflower (see Definition 14 in page 15).By Lemma 15 (page 15), we get that mmhs ( H ) ≥
2, and as mmhs ( H ) = mmbs ( G ), we obtainthe desired result. Property 4 . Suppose that there exist two distinct hyperedges H and H of H such that H (cid:40) H . Let ˜ H be the hypergraph obtained from H by removing H . Let us prove thatfor every B ⊆ V ( H ), B is a minimal hitting set in H if and only if B is a minimal hittingset in ˜ H . The equivalence of being a hitting set is immediate, and the minimality in ˜ H directly implies the minimality in H . Finally, suppose that B if minimal in H . Consideran arbitrary vertex v ∈ B and let H v ∈ E ( H ) such that B ∩ H v = { v } , which exists byProperty 2. Let H be an inclusion-wise minimal hyperedge of H such that v ∈ H ⊆ H v .We get that H ∈ E ( ˜ H ), and H ∩ B = { v } . We now repeat this operation of removing ahyperedge containing another until no hyperedge is included in another, and define H asthe obtained hypergraph. By Property 1, it follows that B is a minimal hitting set in ˜ H . Property 5 . Suppose that there exists v ∈ V ( H ) such that no hyperedge of H contains v .Let ˜ H be the hypergraph obtained from H by removing v . We immediately have that forevery B ⊆ V ( H ), B is a minimal hitting set in H if and only if B is a minimal hitting set in˜ H , as no minimal hitting set in ˜ H can contain vertex v . We now repeat this operation untilwe get the claimed hypergraph H . (cid:50) α and β In this section we establish the results summarized in Table 1 about the parameterizedcomplexity of
MMBS and
MMHS under several parameterizations depending on α and β .We present the negative and the positive results in Section 3.1 and Section 3.2, respectively. It is natural to ask, for a graph G , whether computing α ( G ) can help toward computing mmbs ( G ), and vice-versa. In fact, the parameters α and mmbs are linked by the dualityrelation discussed in what follows.Given a ground set S , a clutter is a family A of subsets of S such that no set A ∈ A contains another set A ∈ A . Given a clutter A , the family of blocking sets of A , denoted by b ( A ), is the set of minimal subsets B of S such that B intersects every set A ∈ A . Noticethat b ( A ) is a clutter, and thus b ( b ( A )) is well-defined. The following theorem provides aduality relation and can be found, for instance, in [3]. (cid:73) Theorem 5. b ( b ( A )) = A . If we apply Theorem 5 to our setting, namely with A being the set of all mis of a graph G , we get that b ( A ) is the set of all minimal blocking sets, and that the set of minimal setsintersecting all the sets in b ( A ) is the set of all mis . Even if this theorem gives a relationbetween mbs and mis , it seems, to the best of our knowledge, that it does not provide a wayto compute α ( G ) from mmbs ( G ), or mmbs ( G ) from α ( G ).Let us start with the easy direction. (cid:73) Property 1.
Let F be a hereditary graph class. If the problem of computing an mbs (ofany size) is polynomial on F , then IS is polynomial on F . This implies that if MMBS ispolynomial on F , then IS is polynomial on F . Proof:
Suppose that we have an algorithm that, given a graph G ∈ F , outputs in polynomialtime an mbs B of G . According to Lemma 4 (Property 1), for every v ∈ B there exists a mis I v such that I v ∩ B = { v } , implying that α ( G \ B ) = α ( G ) −
1. As G \ B ∈ F because F ishereditary, we can repeat the same argument to G \ B , stopping when we obtain an emptygraph. It follows that α ( G ) is equal to the number of iterations of this procedure. (cid:50) Let us now show that there is no hope to get the same kind of property in the backwarddirection. We point out a related result in [20] showing that there is a graph class F where mmbs ( F ) = 1 (as there is a unique mis for any G ∈ F , see Property 3 of Lemma 4), but IS is not polynomial unless NP = RP . This result is obtained through a reduction from Unique-SAT and guarantees that if the original instance is a yes -instance, then there is aunique mis of size k , and otherwise a unique mis of size k −
1. In the following property, thesituation is different as we target a complexity result for
MMBS , and not IS . For a graphclass F , let α ( F ) = sup G ∈F α ( G ). (cid:73) Proposition 6.
There exista (hereditary) graph class F where α ( F ) ≤ and on which MMBS is NP -hard (implyingthat - MMBS is NP -hard), anda graph class F where IS is polynomial, and MMBS / α is W[1] -hard, even the particularcase of deciding, given an input graph G , whether mmbs ( G ) > . This implies that MMBS / β is para- NP -hard, and that MMBS /( α + β ) is W[1] -hard.
Proof:
We consider the reduction of Boria et al. [7, Theorem 1] from IS to MMVC showingan inapproximability result for
MMVC . Given an input graph G of the IS problem, thisreduction produces a graph G by starting from G , and adding for every v ∈ V ( G ) a newprivate vertex v p with N G ( v p ) = { v } . As it is known that IS remains NP -hard on triangle-freegraphs (just by subdividing every edge twice), and as this reduction does not create triangles,we get that MMVC is NP -hard on triangle-free graphs.To prove our first statement, we reduce from MMVC on triangle-free graphs, and givenan input G of MMVC , we define our input G of MMBS as the complement of G (that is,the graph obtained from G by swapping edges and non-edges). We may also assume that G contains at least one edge. Observe that as G is triangle-free and G contains at least oneedge, α ( G ) = 2. Moreover, there is a bijection between edges of G and mis of G . Thisimplies that for every subset B ⊆ V ( G ), B is a vertex cover of G if and only if B is a bs of G , and thus that B is a minimal vertex cover of G if and only if B is an mbs of G .To prove our second statement, we reduce from the Multicolored Independent Set ( CIS ) problem. It is known [13] that
CIS / k is W[1] -hard. Given an input (
G, k ) of
CIS ,where V ( G ) is partitioned into k cliques { V i | i ∈ [ k ] } , let G be a copy of G and let G bethe graph composed of an is of size k . We define G as the graph obtained by taking thedisjoint union of G and G , and adding all edges between V ( G ) and V ( G ). Observe that | V ( G ) | = | V ( G ) | + k , α ( G ) ≤ k as α ( G i ) ≤ k , and α ( G ) = k as V ( G ) is an is .If ( G, k ) is a yes -instance, there are two distinct (even disjoint) mis I i in G : we can define I ⊆ V ( G ) as an is of size k in G , and I = V ( G ). This implies by Lemma 4 (Property 3)that mmbs ( G ) ≥ G, k ) is a no -instance, the unique mis of G is V ( G ), implying by Lemma 4(Property 3) that mmbs ( G ) = 1. Finally, this is a parameterized reduction as α ( G ) ≤ k ,and IS is clearly polynomial restricted to the family of graphs produced by the reduction. (cid:50) Let us now turn to lower bounds for
MMHS / β . It is known that MMHS / β is W[1] -hard [2] and that, unless the
ETH fails,
Simple-Ext-MMHS cannot be solved in time f ( | X | ) · ( n + m ) o ( | X | ) for any computable function f : N → N , where n and m are the number úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 11 of vertices and hyperedges of the input hypergraph, respectively [4]. In our next theorem weprove the same lower bound for Up-Dom , transferring the result to
MMHS as well (recallthat
Up-Dom is a special case of
MMHS ).Let us mention that the reduction for
MMHS / β of Bazgan et al. [2] is a reduction from Multicolored Independent Set parameterized by k , showing that, in fact, Up-Dom is W[1] -hard parameterized by the solution size, where the parameter of the
Up-Dom instanceis O ( k ). While being indeed a parameterized reduction, it only implies that, unless the ETH fails,
Up-Dom cannot be solved in time f ( k ) · ( n + m ) o ( √ k ) for any computable function f : N → N . We also mention that very recently and independently from our work, Theorem 7has also been proved by Dublois et al. [17], by using a reduction quite similar to ours. (cid:73) Theorem 7.
Unless the
ETH fails, the
Up-Dom problem cannot be solved in time f ( k ) ·| V ( G ) | o ( k ) for any computable function f : N → N . Proof:
Chen et al. [10] proved that, unless the
ETH fails, the k -Clique problem (wherewe have to decide if a given graph have a clique of size at least k ) cannot be solved in time f ( k ) · n o ( k ) on n -vertex graphs for any computable function f : N → N . Since there is asimple parameterized reduction from k -Clique to te Multicolored Independent Set problem parameterized by k , namely CIS / k , with linear dependency on the parameter (seefor instance [13]), the result of Chen et al. [10] implies that the CIS cannot be solved in time f ( k ) · n o ( k ) on n -vertex graphs for any computable function f : N → N .We present a parameterized reduction from CIS to Up-Dom such that, given an instance(
G, k ) of
CIS , creates in polynomial time a graph G that contains a minimal dominating ofsize at least 3 k if and only if G contains a multicolored is of size k . By the above discussion,such a reduction concludes the proof of the theorem.Given G , with V ( G ) = V ] · · · ] V k , for every i ∈ [ k ] we add to G three copies A i , B i , C i of V i , and let U i = A i ∪ B i ∪ C i be their union. We denote A = S i ∈ [ k ] A i , B = S i ∈ [ k ] B i , C = S i ∈ [ k ] C i , and, for a vertex v ∈ V ( G ), we denote by v A , v B , v C its corresponding copyin A, B, C , respectively. For every i ∈ [ k ], the set U i induces a clique minus the triangles { ( v A , v B , v C ) | v ∈ V i } . That is, within the same color i , every vertex is adjacent to all othervertices except for its two other copies. For every edge { u, v } ∈ E ( G ) such that u ∈ V i and v ∈ V j with i = j , we add to G the edges { u A , v B } and { u B , v A } . This concludes theconstruction of G . We claim that G contains a multicolored is of size k if and only if G that contains a minimal dominating of size at least 3 k .Let first S ⊆ V ( G ) be a multicolored is of size k . Let D ⊆ V ( G ) contain, for everyvertex v ∈ S , its three copies v A , v B , v C . Note that | D | = 3 k . We claim that D is a minimaldominating set of G . Since D contains a vertex in each of the 3 k cliques into which V ( G ) ispartitioned, D is clearly a dominating set. Consider a vertex v A ∈ D ∩ A (the case v B ∈ D ∩ B is symmetric). Then D \ { v A } is not a dominating set, since by the hypothesis that S is an is , no vertex in D \ { v A } is adjacent to v A . Consider now a vertex v C ∈ D ∩ C , with v ∈ V i .Then D \ { v C } is not a dominating set either, as D ∩ ( A i ∪ B i ) = { v A , v B } , and none of v A and v B is adjacent to v C . Hence, D is a minimal dominating set of G and we are done.Conversely, let D ⊆ V ( G ) be a minimal dominating set with | D | ≥ k . (cid:66) Claim 8.
For every i ∈ [ k ], | D ∩ A i | ≤ | D ∩ B i | ≤ Proof of the claim:
We say that an index i ∈ [ k ] is abnormal if | D ∩ A i | ≥ | D ∩ B i | ≥ normal otherwise. We will construct a set D ⊆ V ( G ) with | D | = | D | suchthat if i is normal, | D ∩ U i | ≤
3, and if i is abnormal, | D ∩ U i | ≤
2. Hence, if there exists anabnormal index, it holds that | D | = | D | < k , contradicting the hypothesis that | D | ≥ k . We now proceed to the construction of D , which is not required to be a dominating set of G . We start with D = D , and we update D as follows.Let i be an abnormal index. Since | D ∩ A i | ≥ | D ∩ B i | ≥
2, by construction of G wehave that D ∩ ( A i ∪ B i ) dominates U i , and since N ( C i ) = A i ∪ B i , necessarily | D ∩ C i | = 0,as otherwise D would not be minimal. If | D ∩ U i | ≤ | D ∩ U i | = | D ∩ U i | ≤
2. Assume henceforth that | D ∩ U i | = | D ∩ ( A i ∪ B i ) | ≥ | D ∩ A i | ≥ | D ∩ B i | (the other case is symmetric),so we have that | D ∩ A i | ≥
2. Note that any two vertices in the set D ∩ A i , say u A and v A ,dominate the whole set U i . Hence, since D is a minimal dominating set of G , for every othervertex w A ∈ ( D ∩ A i ) \ { u A , v A } (resp. w B ∈ D ∩ B i ) there must exist an index j = i (resp. ‘ = i ) and a vertex z B ∈ B j (resp. z A ∈ A ‘ ) not in D and dominated only by w A (resp. w B ), that is, with z B / ∈ D (resp. z A / ∈ D ) and N D ( z B ) = { w A } (resp. N D ( z A ) = { w B } );see Figure 2(a) for an illustration, where the vertices in D are depicted in red. Note thatsuch an index j (resp. ‘ ) is necessarily normal, as otherwise vertex z B (resp. z A ) would bealready dominated within U j (resp. U ‘ ). Note also that, for the same reason, D ∩ B j = ∅ (resp. D ∩ A ‘ = ∅ ). For each such a vertex w A ∈ D (resp. w B ∈ D ), we remove vertex w A (resp. w B ) from D and we add vertex z B (resp. z A ) to D ; see Figure 2(b) for an illustration,where the vertices in D are depicted in red. We say that vertex z B ∈ B j (resp. z A ∈ A ‘ )is a sink . This concludes the construction of D . It just remains to verify that the claimedproperties of D are satisfied. A i B i C i A j B j C j A (cid:96) B (cid:96) C (cid:96) u A v A w A z B w (cid:48) B z (cid:48) A (b) A i B i C i A j B j C j A (cid:96) B (cid:96) C (cid:96) u A v A w A z B w (cid:48) B z (cid:48) A (a) Figure 2
Configuration in the proof of Claim 8. Index i is abnormal, while indices j and ‘ arenormal. Vertices z B and z A are sinks. (a) The vertices in D are depicted in red. (b) The vertices in D are depicted in red. By construction, we clearly have that | D | = | D | . Note that, if i is an abnormal index asin the above paragraph, it cannot contain any sink since all the vertices of U i are alreadydominated by D ∩ U i . Hence, no vertex is added to D ∩ U i and it holds that D ∩ U i = { u A , v A } ,so we indeed have that | D ∩ U i | ≤ j is a normal index, then | D ∩ U j | ≤
3. Since the vertices in C j have neighbors only in U j , necessarily | D ∩ U j | ≥
1. Hence, at most one vertex in A j andat most one vertex in B j are not dominated by the vertices in D ∩ U j . Thus, each of A j and B j contains at most one sink. We distinguish three cases according to the number of sinksin A j ∪ B j .Suppose first that A j ∪ B j contains no sink. Then | D ∩ U j | = | D ∩ U j | ≤
3, where thelast inequality follows easily by using that, since j is normal, | D ∩ A j | ≤ | D ∩ B j | ≤ A j ∪ B j contains exactly one sink, so we have that | D ∩ U i | = | D ∩ U i | +1.Suppose without loss of generality that the sink is a vertex z A ∈ A j , so we have | D ∩ A j | = 0. úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 13 Since z A is not dominated by D ∩ U i , necessarily | D ∩ B j | ≤ | D ∩ C j | ≤
1, so | D ∩ U i | ≤ | D ∩ U i | ≤ A j ∪ B j contains two sinks z A ∈ A j and z B ∈ B j , so we have that | D ∩ U i | = | D ∩ U i | + 2, | D ∩ A j | = 0, and | D ∩ B j | = 0. Also, since none of z A and z B can be dominated by D ∩ U j , necessarily z = z and D ∩ C j = { z C } , so | D ∩ C j | = 1. Thus, | D ∩ U i | ≤ | D ∩ U i | ≤
3, and the claim follows. (cid:5) (cid:66)
Claim 9.
For every i ∈ [ k ], | D ∩ U i | ≤ Proof of the claim:
Suppose for contradiction that there exists i ∈ [ k ] such that | D ∩ U i | ≥ | D ∩ C i | ≥
2. If | D ∩ C i | ≥
3, then deleting all but any two verticesin D ∩ C i results in a proper subset of D that is still a dominating set of G , contradictingthe minimality of D . Hence | D ∩ C i | = 2, | D ∩ A i | = 1, and | D ∩ B i | = 1. Let u A ∈ D ∩ A i and let v C , w C ∈ D ∩ C i . At least one among v and w , say v , is not equal to u . Then the set D \ { w C } is still a dominating set of G , contradicting again the minimality of D . (cid:5) (cid:66) Claim 10.
For every i ∈ [ k ], | D ∩ A i | = | D ∩ B i | = | D ∩ C i | = 1. Proof of the claim:
Since by hypothesis we have that | D | ≥ k and by Claim 9 it holdsthat | D ∩ U i | ≤ i ∈ [ k ], necessarily | D ∩ U i | = 3 for every i ∈ [ k ]. Since byClaim 8 we have that | D ∩ A i | ≤ | D ∩ B i | ≤ i ∈ [ k ], we conclude that | D ∩ A i | = | D ∩ B i | = | D ∩ C i | = 1 for every i ∈ [ k ]. (cid:5) We proceed to define from D a multicolored is S ⊆ V ( G ) with | S | = k . Consider anarbitrary index i ∈ [ k ]. By Claim 10, D ∩ A i = { u A } , D ∩ B i = { v B } , and D ∩ C i = { w C } .Note that if u = v , then u A and v B would dominate the whole set U i and w C could beremoved from D , contradicting its minimality. Thus, we have that u = v , which in turnimplies that w = u as well. We define S ∩ V i = { v } . It remains to verify that S is indeed an is of G . Consider u, v ∈ S with u ∈ V i and v ∈ V j . If { u, v } ∈ E ( G ) then D \ { u A } wouldstill be a dominating set of G . Indeed, vertex u A would be dominated by v B , and the othervertices in A i would still be dominated by u B . Thus, { u, v } / ∈ E ( G ) and S is an is in G . (cid:50) Theorem 7 immediately yields the following corollary for
MMHS . (cid:73) Corollary 11.
Unless the
ETH fails,
MMHS cannot be solved in time f ( β ) · ( | V ( H ) | + | E ( H ) | ) o ( β ) for any computable function f : N → N . Let us now turn to positive results, and consider a class where IS is polynomial. Asaccording to Proposition 6 we cannot hope for solving MMBS in polynomial time, weconsider the parameterized complexity of the
MMBS problem. The first results show thecrucial difference between the problems of, given a graph G and a positive integer β , decidingwhether mmbs ( G ) = β and deciding whether mmbs ( G ) ≥ β . In any maximization problem,the first property implies the second one, but the backward implication is not always true. Inparticular, for MMBS , mmbs ( G ) ≥ β does not imply that G contains an mbs of size exactly β , and this is informally what makes the inequality version harder.As observed in [19] or in [14, Proposition 1], deciding whether there exists a minimalhitting set of size exactly β , or at most β , in a hypergraph with hyperedges of size at most α can be trivially decided by a search-tree in time O ∗ ( α β ). However, we cannot use directlythis result, as a reduction to MMHS would require time n α ( G ) to generate the hyperedges,and thus we have to define an ad-hoc algorithm, which is also based on branching. (cid:73) Proposition 12.
Let F be a hereditary graph class on which IS is polynomial. Then MMBS = / ( α + β ) and MMBS ≤ / ( α + β ) are FPT restricted to input graphs in F . Moreprecisely, they can both be solved in time O ∗ ( α ( G ) β ) . Proof:
We only prove the result for
MMBS = , as it directly implies the result for MMBS ≤ .Consider an input graph G ∈ F . Let us define an algorithm A ( X ) that, given a set X ⊆ V ( G )with | X | ≤ β , answers “ yes ” if and only if there exists an mbs B of G such that X ⊆ B and | B | = β , in which case we say that X is a yes -set . The algorithm starts with X = ∅ ,and calls itself recursively for a larger set X obtained from branching on vertices of a mis of G \ X , as detailed below. Note that if | X | = β , then A ( X ) answers “ yes ” if and only if X isan mbs . This can be checked in polynomial time, as it is equivalent to the properties that α ( G \ X ) < α ( G ) and α ( G \ ( X \ { v } )) = α ( G ) for every v ∈ X . Let us now consider thecases where 0 ≤ | X | < β . If α ( G \ X ) < α ( G ), then we answer “ no ” as X is already a bs , andthus no superset X (cid:41) X can be an mbs . Otherwise, we have that α ( G \ X ) = α ( G ). Since G \ X ∈ F as F is hereditary and IS is polynomial on F , we can compute in polynomialtime a mis I of G \ X (notice that beeing able to solve IS implies that we can construct sucha mis ). Observe that I is also a mis of G , and that I ∩ X = ∅ . In this case, A ( X ) returns W v ∈ I A ( X ∪ { v } ).Let us now prove the correctness of this latter case. Suppose first that X is a yes -set, andlet B be an mbs in G of size β such that X ⊆ B . As B is a bs , there exists v ∈ I ∩ B . As I ∩ X = ∅ , we get v / ∈ X , and thus X ∪ { v } is a yes -set and A ( X ) returns “ yes ”. We provethe other direction by reverse induction on | X | , the case | X | = β being correct as discussedabove. Consider a set X with | X | < β , and suppose inductively that the claimed property iscorrect for sets of size | X | + 1. Thus, if A ( X ∪ { v } ) returns “ yes ” for some v ∈ I , then byinduction ( X ∪ { v } ) is a yes -set, implying by definition of A that X is also a yes -set.Let us finally discuss the running time of the algorithm, given an input graph G . Startingwith X = ∅ , for every set X the algorithm performs a polynomial number of operations, andthen branches on a set I of size at most α ( G ), as such a set I is always a mis of a subgraphof G ). As the depth of the branching tree corresponding to the algorithm has depth at most β , the claimed running time follows. (cid:50) Observe that, unless
FPT = W[1] , we cannot obtain results similar to Proposition 12 todecide whether mmbs ( G ) ≥ β , and even in time f ( α ( G ) , β ) for any computable function f ,as it would imply that MMBS /( α + β ) is FPT , contradicting the fact that
MMBS / β ispara- NP -hard by Proposition 6. Thus, we need consider a stronger assumption than assumingthat IS is polynomial on F . Namely, in what follows we consider the α - MMBS problem,that is, the case where α is fixed. Recall that, according to Proposition 6, even 2- MMBS remains NP -hard, motivating the study of the parameterized complexity of α - MMBS . (cid:73) Proposition 13.
For every fixed positive integer α , α - MMBS / β and α - MMHS / β are FPT . More precisely, both problems can be solved in time O ∗ ( α β ) . Proof:
Given an input graph G of α - MMBS , we compute in time O ∗ ( n α ) the hypergraph H where V ( H ) = V ( G ), and H is a hyperedge in H if and only if H is a mis in G . By definitionof α - MMBS , all hyperedges of H have size exactly α , and for every B ⊆ V ( G ), B is an mbs in G if and only if B is a minimal hitting set in H . Then, according to [14, Lemma 6], as α is fixed we can decide whether there is a minimal hitting set of H of size at least β in time O ∗ ( α β ). (cid:50) We point out that in both Proposition 13 and [14, Lemma 6], in order to decide whetherthere is a minimal hitting set of H of size at least β , there is term | V ( H ) | f ( α ) hidden inside úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 15 the O ∗ -notation. This means that [14, Lemma 6] does not imply that MMHS /( α + β ) is FPT (recall that function α in the parameterization of MMHS denotes the size of a largesthyperedge of H ). However, according to the two following propositions, it turns out that MMHS /( α + β ) is indeed FPT . This highlights a difference between
MMHS and
MMBS , asaccording to Proposition 6
MMBS /( α + β ) is unlikely to be FPT .Let us start with a kernelization result, using the well-known notion of sunflower. (cid:73)
Definition 14.
Let β ∈ N . Given a hypergraph H , a sunflower in H with β petals and core C ⊆ V ( H ) is a collection of β hyperedges H , . . . , H β of H such that H i ∩ H j = C forall i = j , and for every i ∈ [ β ] the so-called petal H i \ C is not empty. We say that a function s : N → N is a sunflower function if, for every hypergraph H whose hyperedges have size atmost α , if | E ( H ) | > s ( α, β ) then H admits a sunflower with β petals. It is known (see for instance [13]) that s ( α, β ) := ( α )! α ( β − α is a sunflower function.Even if this is not relevant for our next proposition, where α is fixed, we point out that thisbound has been recently been improved by Rao [25] to s ( α, β ) := ( cβ · log( αβ )) α . (1) (cid:73) Lemma 15.
Let β ∈ N , and let H be a hypergraph such that no hyperedge is included inanother hyperedge. If H has a sunflower with β petals, then mmhs ( H ) ≥ β . Proof:
Let { H i | i ∈ [ β ] } be a sunflower of H with β petals. Let C = T i ∈ [ β ] H i and S = V ( H ) \ C . As there is no H ∈ E ( H ) such that H ⊆ C , it follows that S is a hitting setof H . Let S ⊆ S be a minimal hitting set of H . For every i ∈ [ β ], H i \ C = ∅ by definition ofa sunflower, hence S must contain at least one vertex in H i \ C . This implies that | S | ≥ β ,and thus that mmhs ( H ) ≥ β . (cid:50) (cid:73) Proposition 16.
Let s ( α, β ) be a sunflower function that is polynomial in β for fixed α .For every fixed integer α , α - MMBS / β admits a polynomial kernel with at most α · s ( α, β ) vertices, which can beconstructed in time O ∗ ( | V ( G ) | α ) , and α - MMHS / β admits a polynomial kernel with at most α · s ( α, β ) vertices, which can beconstructed in time O ∗ ( | V ( H ) | ) (that is, not depending on α ). Proof:
Let us start with α - MMHS . Consider an instance ( H , β ) of α - MMHS . By Lemma 4(Property 5), we can compute in polynomial time an equivalent instance ( H , β ) where everyvertex belongs to a hyperedge. If | E ( H ) | ≤ s ( α, β ), then we get | V ( H ) | ≤ α · | E ( H ) | ≤ α · s ( α, β ) and we are done. Otherwise, as s is a sunflower function, it follows that H containsa sunflower with β petals. According to Lemma 15, we get that ( H , β ) is a yes-instance.The result for α - MMBS is now straightforward, but we provide the details as we cannotdirectly say that any kernel for α - MMHS implies a kernel for
MMBS . For example, removinga hyperedge in α - MMHS cannot necessarily be translated to α - MMBS . Let (
G, β ) be aninstance of α - MMBS . In time O ∗ ( | V ( G ) | α ) we can compute an equivalent instance ( H , β ) of α - MMHS by creating a hyperedge for every mis of G . As all hyperedges have size exactly α ( G ), no hyperedge can be included in another hyperedge. Now, if the previous kernel detectsa yes -instance, we are done, and otherwise we output ( G , β ) where G = G [ V ( H )]. As allvertices of V ( G ) \ V ( G ) are vertices that do not belong to any mis of G , by the argumentsof the proof of Lemma 4 (Property 5), we get an equivalent instance. (cid:50) Even if Proposition 16 implies that
MMHS /( α + β ) is FPT , the running obtaining byapplying brute force to the kernelized instance is O ∗ (2 α · s ( α,β ) ), and thus doubly exponential in α . This motivates the question of obtaining a faster FPT algorithm for
MMHS /( α + β ).We point out that trying to improve the running time by considering separated parameters,instead of the aggregated parameter α + β , is not possible as MMHS / β is W[1] -hard [2], and
MMHS / α is already NP -hard for α = 2, as it corresponds to MMVC . A first way to get afaster
FPT algorithm is to reduce to the
Simple-Ext-MMHS problem.Bläsius et al. [4] proved that
Ext-MMHS (and thus
Simple-Ext-MMHS ) can be solvedin time O ∗ ( λ | X | ), where λ = min (cid:16) | E ( H ) || X | , ∆( H ) (cid:17) and ∆( H ) = max v ∈ V ( H ) |{ H ∈ E ( H ) | v ∈ H }| is the maximum degree of H . Informally, this algorithm, in the simplified settingof Simple-Ext-MMHS , just guesses for each x ∈ X its “private” hyperedge H x such that H x ∩ X = { x } , and checks that there is no H ∈ E ( H ) such that H ⊆ ( S x ∈ X H x ) \ X .Thus, guessing a hyperedge for every x ∈ X yields the claimed running time. In the nextproposition we formalize these ideas, using ideas similar to the proof of Proposition 12. (cid:73) Proposition 17.
We can decide an instance ( H , β ) of MMHS in time O ∗ (( α ( H ) · λ ) β ) ,where λ = min (cid:16) | E ( H ) | β , ∆( H ) (cid:17) and ∆( H ) = max v ∈ V ( H ) |{ H ∈ E ( H ) | v ∈ H }| . Proof:
Let ( H , β ) be an instance of MMHS . Let us define an algorithm A ( H , β, X ) suchthat, given a set X ⊆ V such that | X | ≤ β , decides whether there exists a minimal hittingset S of H such that X ⊆ S and | S | ≥ β . As in the proof of Proposition 12, the algorithmstarts with X = ∅ . If | X | = β then we return B ( H , X ), where B is the algorithm of Bläsiuset al. [4]. Let us now consider the cases where 0 ≤ | X | < β . If there is no H ∈ E ( H ) suchthat H ∩ X = ∅ then we answer “ no ”. Otherwise, let H ∈ E ( H ) such that H ∩ X = ∅ , andin this case A ( H , β, X ) returns W v ∈ H A ( H , β, X ∪ { v } ).Let us prove that A is correct by induction on β − | X | . The only non-trivial case iswhen A returns W v ∈ H A ( H , β, X ∪ { v } ). If A ( H , β, X ) returns “ yes ” then there exists v ∈ H such that A ( H , β, X ∪ { v } ) returns “ yes ”, implying by induction that ( H , β, X ∪ { v } ) is a yes -instance, and thus that ( H , β, X ) is a yes -instance as well. Conversely, if ( H , β, X ) is a yes -instance certified by a solution S ∗ , then there exists v ∗ ∈ S ∗ ∩ H since S ∗ is a hitting set,implying that ( H , β, X ∪ { v ∗ } ) is a yes -instance.As at each step we branch on all vertices of a hyperedge H , the running time is boundedby O ∗ ( α β · f ( α, β )) where O ∗ ( f ( α, β )) is the running time of algorithm B , implying f ( α, β ) = O ∗ ( λ β ) and λ = min (cid:16) | E ( H ) | β , ∆( H ) (cid:17) . (cid:50) Proposition 17 implies the following corollary, where the XP algorithm follows directly fromthe algorithm of Proposition 17. On the other hand, the FPT algorithm is obtained by firstapplying the kernel for
MMHS of Proposition 16 to ensure that ∆( H ) ≤ | E ( H ) | ≤ s ( α, β ),where s is the sunflower function of Equation (1). (cid:73) Corollary 18.
The following claims hold:
MMHS / β is XP . MMHS / ( α + β ) is FPT . More precisely, it can be solved in time O ∗ ( α β ( cβ · log( αβ )) αβ ) ,where c is the constant in the s sunflower function of Equation (1). Even if the algorithm of Proposition 17 gives a running time matching the lower boundof Corollary 11 for the dependency on | E ( H ) | , we can get a faster FPT algorithm for
MMHS /( α + β ) using an ad-hoc algorithm that does not reduce to the extension problem.Namely, we present in Theorem 24 an algorithm for MMHS /( α + β ) running in time O ∗ (2 αβ ).We first need some preliminaries. (cid:73) Definition 19.
Let H be a hypergraph, let I ⊆ V ( H ) be an is in H , and let X ⊆ V ( H ) . Let úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 17 H I such that V ( H I ) = V ( H ) \ I , and E ( H I ) = { H \ I | H ∈ E ( H ) } , E ¯ X = { H ∈ E ( H ) | H ∩ X = ∅} , and H ¯ X such that V ( H ¯ X ) = V ( H ) and E ( H ¯ X ) = E ¯ X . (cid:73) Lemma 20.
Let H be a hypergraph and let I ⊆ V ( H ) be an is in H . For every minimal hitting set S of H I , S is also a minimal hitting set of H . This implies mmhs ( H ) ≥ mmhs ( H I ) . For every minimal hitting set S ∗ of H such that S ∗ ∩ I = ∅ , S ∗ is also a minimal hittingset of H I . Proof:
For the first property, let S be a minimal hitting set of H I . Let us first prove that S is a hitting set of H . Consider an arbitrary hyperedge H ∈ E ( H ). As I is an is , H \ I = ∅ ,and as S is a hitting set of H I and H \ I ∈ E ( H I ), we get S ∩ ( H \ I ) = ∅ . Let us now provethat S is minimal. Consider an arbitrary vertex v ∈ S . By the minimality in H I , there exists H ∈ E ( H I ) such that ( S \ { v } ) ∩ H = ∅ , implying that ( S \ { v } ) ∩ ( H ∪ I ) = ∅ as S ∩ I = ∅ ,where H ∪ I ∈ E ( H ).For the second property, let H ∈ E ( H I ), where H = H \ I , H ∈ E ( H ). As S ∗ is ahitting set of H , S ∗ ∩ H = ∅ , and as S ∗ ∩ I = ∅ , we get S ∗ ∩ H = ∅ . Let us now verifythe minimality. Consider an arbitrary vertex v ∈ S ∗ . As S ∗ is minimal in H , there exists H ∈ E ( H ) such that ( S ∗ \ { v } ) ∩ H = ∅ , implying ( S ∗ \ { v } ) ∩ ( H \ I ) = ∅ . (cid:50) (cid:73) Lemma 21.
Let H be a hypergraph, let X ⊆ V ( H ) , and let S be a minimal hitting set of H ¯ X . There exists a minimal hitting set S of H such that S ⊆ S . Proof:
Let S = S ∪ X . Observe that S is a hitting set of H . Now, as far as there exists v ∈ S ∩ X such that S \{ v } is still a hitting set of H , remove v from S . Let S ∗ be the obtainedset, which satisfies S ⊆ S ∗ ⊆ S , and let us verify that S ∗ is minimal. For every v ∈ S ∗ ∩ X ,by definition of S ∗ we have that S ∗ \ { v } is not a hitting set of H . For every v ∈ S ∗ ∩ S , as S is minimal in H ¯ X , it follows that there exists H ∈ E ¯ X such that ( S \ { v } ) ∩ H = ∅ . As H ∩ X = ∅ , we get ( S ∗ \ { v } ) ∩ H = ∅ as well. (cid:50) We are now ready to present our
FPT algorithm. (cid:73)
Definition 22.
For a positive integer β , we define algorithm A β ( H , X ) , where H is ahypergraph without empty hyperedges and X ⊆ V ( H ) , as follows:If E ¯ X = ∅ ,if | X | ≥ β and X is minimal hitting set of H , return “ yes ”.Otherwise, return “ no ”.Otherwise, let S be a minimal hitting set of H ¯ X .If | S | ≥ β , return “ yes ”.Otherwise, return W S ∈L A β ( H S , X ∪ ( S \ S )) , where L = { S ⊆ S | S is an is of H ¯ X } . In order to analyze the algorithm, given an input ( H , X ) of A β , we define the measure m ( H , X ) = ( max {| H | | H ∈ E ( H ¯ X ) } , if E ( H ¯ X ) = ∅ S is a hitting set of H ¯ X , for every S ∈ L we have m ( H S , X ∪ ( S \ S )) < m ( H , X ). Indeed, a hyperedge of H ¯ X either intersects S \ S and is not taken into account in the ‘max’, or intersects S and thus its correspondinghyperedge in the hypergraph H S has smaller size. Observe also that as H does not containan empty hyperedge, m ( H , X ) = 0 is equivalent to E ( H ¯ X ) = ∅ . (cid:73) Lemma 23.
The following statement hold: If A β ( H , X ) returns “ yes ” then mmhs ( H ) ≥ β . If there exists a minimal hitting set S ∗ of H such that X ⊆ S ∗ and | S ∗ | ≥ β , then A β ( H , X ) returns “ yes ”.The above properties imply that, given an instance ( H , β ) of MMHS , A β ( H , ∅ ) returns “ yes ”if and only if mmhs ( H ) ≥ β . Proof:
We use the notation introduced in Definition 22. Let us first argue that the prerequisitethat the input hypergraph H does not contain an empty hyperedge is always satisfied. Tothat end, let us consider such an input H , and we shall prove that for any S ∈ L , H S doesnot contain an empty hyperedge either. Observe that as S is minimal and hyperedges of H ¯ X do not intersect X , we have S ∩ X = ∅ , implying S ∩ X = ∅ . Together with the fact that S is an is of H ¯ X , this implies that S is an is of H . Thus, H S does not contain an emptyhyperedge.We prove both properties by induction on m ( H , X ). Let us start with the first property.If m ( H , X ) = 0, then E ( H ¯ X ) = ∅ , and the claimed property is true. Let is now assume that m ( H , X ) >
0. Suppose that A β ( H , X ) returns “ yes ”. As E ( H ¯ X ) = ∅ , the algorithm goes tothe second case and chooses S . If | S | ≥ β , then by Lemma 21 we get that mmhs ( H ) ≥ | S | ≥ β .Otherwise, there exists an is S of H ¯ X such that A β ( H S , X ∪ ( S \ S )) returns “ yes ”. As m ( H S , X ∪ ( S \ S )) < m ( H , X ), by induction we get mmhs ( H S ) ≥ β , implying by Lemma 20that mmhs ( H ) ≥ β .Let us now turn to the second property, and assume that there exists a minimal hittingset S ∗ of H such that X ⊆ S ∗ and | S ∗ | ≥ β . Suppose first that m ( H , X ) = 0, implying that E ( H ¯ X ) = ∅ . In this case, we have that X is already a hitting set of H , and thus, as S ∗ isminimal and X ⊆ S ∗ , we get that S ∗ = X , implying that | X | ≥ β and that the algorithmreturns “ yes ”. Suppose now that m ( H , X ) >
0, implying E ( H ¯ X ) = ∅ , and thus that thealgorithm goes to the second case and chooses S . If | S | ≥ β then we are done. Otherwise,let S ∗ = S ∩ S ∗ and S ∗ = S \ S ∗ . As S ∗ is a hitting set of H , there is no H ∈ E ( H ) suchthat H ⊆ S ∗ , implying that S ∗ is an is and thus that S ∗ ∈ L . As by Lemma 20, S ∗ is also aminimal hitting set of H S ∗ , as X ∪ ( S \ S ∗ ) ⊆ S ∗ , and as m ( H S ∗ , X ∪ ( S \ S ∗ )) < m ( H , X ),by induction we get that A β ( H S ∗ , X ∪ ( S \ S ∗ )) returns “ yes ”, and thus that A ( H , X ) returns“ yes ” as well. (cid:50) (cid:73) Theorem 24.
MMHS / ( α + β ) can be solved in time O ∗ (2 αβ ) . Proof:
Given an instance ( H , β ) of MMHS , we simply call A β ( H , ∅ ). According to Lemma 23,this algorithm correctly decides whether mmhs ( H ) ≥ β . Let us now analyze the running time.Let f ( β, α, n ) we the worst case running time of the algorithm A β ( H , X ) when m ( H , X ) ≤ α and | V ( H ) | = n . We get that there exists a polynomial p such that f ( β, , n ) ≤ p ( n )(as when m ( H , X ) = 0 we have E ( H ¯ X ) = ∅ and the algorithm only checks that X is aminimal hitting set of size at least β ), and f ( β, α, n ) ≤ p ( n ) + 2 β − f ( β, α − , n ). Tosimplify the notation, let b = 2 β − . Using a straightforward induction on α it follows that f ( β, α, n ) ≤ p ( n ) · (cid:16) b α − + b α − − b − (cid:17) , implying the claimed running time. (cid:50) úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 19 MMBS parameterized by treewidth
In this section we prove that
MMBS/ tw is FPT . The algorithm requires a long case analysis.As one may expect, we present a dynamic programming (DP) algorithm using a nice treedecomposition of the input graph G . In Section 4.1, we present the notation we need and weprovide some intuition about the parameters that we store in the tables of the algorithm. InSections 4.2, 4.3, and 4.4 we present how to compute the table entries for a join, introduce,and forget node, respectively. In Section 4.5 we combine the previous ingredients to completethe algorithm. Consider a graph G and subsets X ⊆ V ( G ), B ⊆ V ( G ), and Z ⊆ X such that Z is an is of G . We say that a set I ⊆ V ( G ) is an ( X, Z ) - is if I is an is of G such that I ∩ X = Z . Wedenote by α ( X,Z ) ( G ) the size of a largest ( X, Z )- is in G , and by α B ( X,Z ) ( G ) the size of a largest ( X, Z )- is I in G such that I ∩ B = ∅ .In both cases, if such a set does not exist, we set the corresponding parameter to −∞ . Wesay that a set B ⊆ V ( G ) is an ( X, Z ) - bs in G if α B ( X,Z ) ( G ) < α ( X,Z ) ( G ), and we say that( X, Z ) is blocked by B in G . These concepts are illustrated in Figure 3. Observe that if B ∩ Z = ∅ then B is an ( X, Z )- bs (as α B ( X,Z ) ( G ) = −∞ ), but the backward implication isnot necessarily true as B may contain one vertex in X \ Z of each maximum ( X, Z )-is of G . Observe also that an ( ∅ , ∅ )- is of G is simply an is of G , implying that α ( ∅ , ∅ ) ( G ) = α ( G ).Similarly, an ( ∅ , ∅ )- bs is a bs of G . Z u XZ u Figure 3
In this example there is only one maximum (
X, Z )- is which is I = Z ∪ { u , u } . Notethat B = { u } is an ( X, Z )- bs , but B is not an ( X, Z )- bs . In what follows we assume that we are given a nice tree decomposition D = ( T, B ) of theinput graph G as defined in Section 2. In particular, recall thatevery node of T has at most two children,if a bag X corresponds to a node of T having two children with bags X L and X R , then X = X L = X R (the node corresponding to X is a join node);if a bag X corresponds to a node of T having one children with bag X C , theneither X (cid:40) X C and | X C | = | X | + 1 (the node corresponding to X is a forget node), or X C (cid:40) X and | X | = | X C | + 1 (the node corresponding to X is an introduce node). Discussion on the list of parameters used in the DP algorithm
As usual, our dynamic programming algorithm performs a leaf-to-root traversal of a nice treedecomposition D = ( T, B ) of an input graph G computing, for each node of the correspondingtree T , a set of tuples from the corresponding tuples of its children. Let us first explain the intuition behind each parameter of such tuples we shall compute and why they are needed.The formal details are presented in Definition 26 (page 23).To simplify the presentation, we call a bag of the tree decomposition join bag (resp. forgetbag , introduce bag ) its corresponding node is a join node (resp. forget node, introduce node).We also speak about the children of a bag, meaning the bags corresponding to the childrenof the considered node.Consider a join bag X with children X L = X R = X , and suppose we look for a maximum mbs B of G X . First, finding separately a maximum mbs B L in G X L and B R in G X R willnot guarantee that the size of B L ∪ B R (assuming B L ∪ B R is an mbs in G X ) is maximum,and thus we introduce a parameter B ⊆ X and look for an mbs B of the graph G X suchthat B ∩ X = B .Let I be a mis of G X , I L = I ∩ V ( G X L ), and I R = I ∩ V ( G X R ). Observe that I L (resp. I R ) is not necessarily a mis of G X L (resp. G X R ), and thus it is pointless to find an mbs B L (resp. B R ) in G X L (resp. G X R ), as blocking maximum independent sets of G X L and G X R may not imply that we block maximum independent sets of G X . This motivates the abovenotion of ( X, Z )- is in G X . More precisely, let L = { Z ⊆ X | there exists a mis I of G X suchthat I ∩ X = Z } . Then, B is an mbs of G X if and only if: (blocking condition) for every Z ∈ L , B is an ( X, Z )- bs in G X , and (minimality condition) for every v ∈ B , there must exist Z ∈ L such that B \ { v } is notan ( X, Z )- bs in G X .This explains why we have, in our list of parameters of our dynamic programming (and alsothe input of our auxiliary problem in Definition 27), a list L of subsets of X , in additionto the set B . Notice that there may exist Z ∈ L with Z = ∅ . Toward the correct notionof the operator ‘ ‘ ’ given in Definition 26, let us introduce some intermediate ones that wedenote by ‘ ‘ ’, ‘ ‘ ’, ‘ ‘ ’ and whose scope is limited to this preliminary discussion (as theywill not be used in the eventual DP algorithm). Given ( X, B , L ) and a set B , we say that B ‘ ( X, B , L ) if and only if B ∩ X = B , and B satisfies the two properties above (blocking and minimality conditions).Such a set B will be called a solution to ( X, B , L ) (instead of bs ).Let us now argue that these three parameters are still not sufficient to design our algorithm,by exhibiting two situations where we suppose that we computed “small solutions”, butextending these small solutions to the current bag creates a solution which no longer respectsthe minimality condition. v v Z Z Z G X L G X R u w w X Figure 4 B ‘ ( X, ∅ , L ) where L = { Z , Z , Z } , B = { u, v , v } , and B = { u, w , w } . Let us start with the first situation. Suppose first that we have a solution B ‘ ( X, B , L )for some L = { Z , Z , Z } , as depicted in Figure 4. Recall that X = X L = X R and let úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 21 B L = B ∩ V ( G X L ) and B R = B ∩ V ( G X R ). We prove in Lemma 31 (page 24) that, forevery Z ∈ L , B is an ( X, Z )- bs in G X if and only if B L is an ( X L , Z )- bs in G X L or B R is an ( X R , Z )- bs in G X R . Thus, it may be the case, as in Figure 4, that B L = { u } isan ( X L , Z )- bs in G X L for Z ∈ L L = { Z , Z } , and B R = { v , v } is an ( X R , Z )- bs in G X R for Z ∈ L R = { Z , Z } . Suppose now that we compute B L and B R such that B L ‘ ( X L , B , L L ) and B R ‘ ( X R , B , L R ), and let B = B L ∪ B R . It may be the casethat B does not verify the previous minimality condition 2. Indeed, let u ∈ B L \ X andsuppose that B L \ { u } is not an ( X L , Z )- bs in G X L . Unfortunately, if B R is an ( X R , Z )- bs in G X R (even if Z / ∈ L R ), we will have that B \ { u } is still a ( X, Z )- bs in G X , and thusmaybe still an ( X, Z )- bs for any Z ∈ L . We overcome this problem by forcing B R not to bean ( X R , Z )- bs in G X R . This explains why we have in the input a list S of subsets of X ,and we now impose that for any Z ∈ S , B must not be an ( X, Z )- bs in G X .Thus, now we denote by B ‘ ( X, B , L , S ) the property that B ‘ ( X, B , L ) and, forany Z ∈ S , B is not an ( X, Z )- bs in G X . u Z Z u vG X L G X R XZ Z Figure 5 B = { v } , B L = { u , v } , and B R = { u , v } . Let us now turn to the second situation, which is depicted Figure 5, where L = { Z , Z , Z , Z } and B = { v } . Suppose that we compute B L and B R such that B L ‘ ( X L , B , L L , S L ) where L L = { Z , Z , Z } , S L = { Z } and B R ‘ ( X R , B , L R , S R ) where L R = { Z , Z , Z } and S R = { Z } . Let B = B L ∪ B R . Let v ∈ B . By minimalitycondition 2, there exists Z ∈ L L such that B L \ { v } is not an ( X L , Z )- bs in G X L (where Z = Z in Figure 5). In the same way, there exists Z ∈ L R such that B R \ { v } is not an( X R , Z )- bs in G X R (where Z = Z in Figure 5). If Z = Z , we may not be able to concludethat there exists a Z ∈ L such that B \ { v } is not an ( X, Z )- bs in G X . In the exampledepicted in Figure 5, B \ { v } is still an ( X, Z )- bs in G X for every Z ∈ L . Thus, for v ∈ B ,we will keep control of the minimality condition in a more precise way by • introducing another list L of subsets of X , and still ask that B is an ( X, Z )- bs in G X for any Z ∈ L , • introducing a function f : B → L , and • (minimality condition in B ) requiring that for every v ∈ B , B \{ v } is not an ( X, f ( v ))- bs in G X .In the previous example, we would have to set either f L ( v ) = f R ( v ) = Z or f L ( v ) = f R ( v ) = Z , but none of these choices leads to a feasible solution on both the left and the right handsides. This is not surprising, as in fact there is no set B such that B ‘ ( X, B , L , S ) where L = { Z , Z , Z , Z } , B = { v } , and S = ∅ . Indeed, being an ( X, Z )- bs in G X forces any B to contain u (as B cannot contain the vertex of Z ), and being a ( X, Z )- bs in G X forcesany B to contain u . This means that we necessarily have { u , u , v } ⊆ B . Then, observe that B is not minimal as { u , u } is still a ( X, Z )- bs in G X for any Z ∈ L . The conclusion isthat in this situation, forcing v to be in any solution leads to an infeasible instance.Finally, even when using function f , and defining accordingly B ‘ ( X, B , L , L , f, S )if B ‘ ( X, B , L , S ) and Y respects the previous minimality condition in B , there is alast important detail. Suppose B L ‘ ( X L , B , L L , L L , f L , S L ) where for example that L L = { Z } , L L = { Z } , and consider v ∈ B L \ B . We know that there exists Z such that B L \ { v } is not an ( X L , Z )- bs in G X L , but we must even impose that Z ∈ L L , as otherwiseif Z = Z then B \ { v } would still be an ( X, Z )- bs in G X . Thus, the minimality conditionis finally as follows: (minimality condition outside B , forcing Z ∈ L ) ∀ v ∈ B \ B , ∃ Z ∈ L such that B \ { v } is not an ( X, Z )- bs in G X . (minimality condition in B ) ∀ v ∈ B , B \ { v } is not an ( X, f ( v ))- bs in G X , where f ( v ) ∈ L .Even if we only discussed here the case where X is a join node, it turns out that this listof parameters is also enough for the introduce and forget nodes. Defining the auxiliary problem
Let us now define the auxiliary problem that will be solved by our DP algorithm. (cid:73)
Definition 25.
Let G be a graph and let D = ( T, B ) be a nice tree decomposition of G . Let E ( G, D ) be the set containing all tuples ( X, B , L , L , f, S ) such that: • X ∈ B , • B ⊆ X , • L , L , S ⊆ X such that for every Z ∈ L ∪ L ∪ S , Z is an is of G , and • f : B → L . (cid:73) Definition 26.
Let G be a graph and let D = ( T, B ) be a nice tree decomposition of G . Forevery ( X, B , L , L , f, S ) ∈ E ( G, D ) and B ⊆ V ( G X ) , we write B ‘ ( X, B , L , L , f, S ) ifand only if i) B ∩ X = B , ii) ∀ Z ∈ L ∪ L , B is an ( X, Z ) - bs in G X , iii) ∀ Z ∈ S , B is not an ( X, Z ) - bs in G X , iv) and the following two minimality conditions are satisfied: a) ∀ v ∈ B \ B , ∃ Z ∈ L such that B \ { v } is not an ( X, Z ) - bs in G X , and b) ∀ v ∈ B , B \ { v } is not an ( X, f ( v )) - bs in G X . Let us point out that there may exist Z ∈ L ∪ L ∪ S with Z = ∅ , and that if ∃ Z ∈ S such that B ∩ Z = ∅ , then there is no solution (because of Property iii). (cid:73) Definition 27.
We define the optimization problem Π as follows, where we consider thatthe input graph G and a nice tree decomposition D of G are fixed: Input : A tuple (
X, B , L , L , f, S ) ∈ E ( G, D ). Output : A set B ⊆ V ( G X ) such that B ‘ ( X, B , L , L , f, S ). Objective : Maximize | B | .We say that an instance I of Π is feasible if there exists a set B such that B ‘ ( X, B , L , L , f, S ). Let us now show that being able to solve optimally problem Π issufficient for computing the parameter mmbs ( G ), for a given graph G . úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 23 (cid:73) Proposition 28.
Let G be a graph and D = ( T, B ) be a nice tree decomposition of G suchthat T is rooted at X = {∅} . For every B ⊆ V ( G ) , B ‘ ( ∅ , ∅ , {∅} , ∅ , ∅ , ∅ ) if and only if B is an mbs of G. Proof:
Let (
X, B , L , L , f, S ) = ( ∅ , ∅ , {∅} , ∅ , ∅ , ∅ ). Recall that being an ( ∅ , ∅ )- bs in G X isequivalent to being a bs in G .Suppose first that B ‘ ( ∅ , ∅ , {∅} , ∅ , ∅ , ∅ ). By Property ii, B is an ( ∅ , ∅ )- bs in G X , implyingthat B is a bs of G . Let us now prove that B is minimal. Let v ∈ B . As v ∈ B \ B , byProperty iva, there exists Z ∈ L such that B \ { v } is not an ( X , Z )- bs in G . As L = {∅} ,we obtain that B \ { v } is not an ( ∅ , ∅ )- bs in G , and thus not a bs in G .Suppose now that B is an mbs of G . Property ii is satisfied as B is a bs in G . Let us nowprove Property iva. Let v ∈ B \ B . As B is minimal, B \ { v } is not a bs in G , and thus notan ( ∅ , ∅ )- bs in G X = G , where ∅ ∈ L . (cid:50) The following proposition is now immediate. (cid:73)
Proposition 29.
Given an n -vertex graph G with treewidth tw ( G ) = t , if • t ( n, t ) is the time to compute a nice tree decomposition D of G of width t , and • t A ( n, t ) is the time to compute an optimal solution of problem Π ,then one can compute mmbs ( G ) in time O ( t ( n, t ) + t A ( n, t )) . In what follows, namely in Sections 4.2, 4.3, and 4.4, we fix an input graph G and a nicetree decomposition D = ( B , T ) of G of width t . Before proving Lemma 32 corresponding to the join case, let us first prove the followingtechnical lemmas. (cid:73)
Lemma 30.
For every I ⊆ V ( G X ) and every Z ⊆ X where Z is an is of G X , I is amaximum ( X, Z ) - is in G X if and only if I L = I ∩ V ( G X L ) is a maximum ( X L , Z ) - is in G X L and I R = I ∩ V ( G X R ) is a maximum ( X R , Z ) - is in G X R . Proof:
For the forward implication, suppose I is a maximum ( X, Z )- is in G X . Let I L ∗ be amaximum ( X L , Z )- is in G X L . Then, ( I \ I L ) ∪ I L ∗ is still an is as there is no edge between I L ∗ ∩ X and I R (as I L ∗ ∩ X = I L ∩ X = Z ), and there is no edge between I L ∗ \ X and I R \ X as, by the properties of a tree decomposition, there is no edge even between V ( G X L ) \ X and V ( G X R ) \ X . As (( I \ I L ) ∪ I L ∗ ) ∩ X = Z , this implies that ( I \ I L ) ∪ I L ∗ is an ( X, Z )- is in G X and that | I | ≥ ( I \ I L ) ∪ I L ∗ . As I ∩ I L = I ∩ I L ∗ = Z , we get | I L | ≥ | I L ∗ | . As I L ∩ X L = Z ,we obtain that I L is a maximum ( X L , Z )- is in G X L . The same arguments hold for I R .For the backward implication, suppose I L is a maximum ( X L , Z )- is in G X L and I R is a maximum ( X R , Z )- is in G X R . Observe first that I L ∪ I R is an ( X, Z )- is in G X , asthere is no edge between V ( G X L ) \ X and V ( G X R ) \ X . Let I ∗ be a maximum ( X, Z )- is in G X . Note that I L ∗ := I ∗ ∩ V ( G X L ) is an ( X L , Z )- is in G X L , and symmetrically that I R ∗ := I ∗ ∩ V ( G X R ) is an ( X R , Z )- is in G X R . This implies that | I L ∗ | ≤ | I L | and | I R ∗ | ≤ | I R | .As I L ∗ ∩ I R ∗ = I L ∩ I R = Z , the previous inequalities imply | I ∗ | ≤ | I | , meaning that I is amaximum ( X, Z )- is in G X . (cid:50) (cid:73) Lemma 31.
Let Z ⊆ X . For every B ⊆ V ( G X ) , B is an ( X, Z ) - bs in G X if and onlyif B L = B ∩ V ( G X L ) is an ( X L , Z ) - bs in G X L or B R = B ∩ V ( G X R ) is an ( X R , Z ) - bs in G X R . Proof:
For the forward implication, suppose B is an ( X, Z )- bs in G X . Suppose by contra-diction that there exists a maximum ( X L , Z )- is I L in G X L such that I L ∩ B L = ∅ , and amaximum ( X R , Z )- is I R in G X R such that I R ∩ B R = ∅ . Let I = I L ∪ I R . By Lemma 30, I is a maximum ( X, Z )- is in G X . As I L ∩ B = I L ∩ B L = ∅ , and also I R ∩ B = I R ∩ B R = ∅ ,we get I ∩ ( B L ∪ B R ) = I ∩ B = ∅ , a contradiction to the hypothesis that B is an ( X, Z )- bs in G X .For the backward implication, suppose B L is an ( X L , Z )- bs in G X L or B R is an ( X R , Z )- bs in G X R . Suppose by contradiction that there exists a maximum ( X, Z )- is I in G X suchthat I ∩ B = ∅ . By Lemma 30, I L = I ∩ V ( G X L ) is a maximum ( X L , Z )- is in G X L and I R = I ∩ V ( G X R ) is a maximum ( X R , Z )- is in G X R . As I L ∩ B = I R ∩ B = ∅ , we obtain that B L is not an ( X L , Z )- bs in G X L and that B R is not an ( X R , Z )- bs in G X R , a contradiction. (cid:50) We are now ready to state the main lemma of this section. (cid:73)
Lemma 32.
Let ( X, B , L , L , f, S ) ∈ E ( G, D ) where X ∈ B is a join node and X L , X R are the children of X (with X = X L = X R ). For every B ⊆ V ( G X ) , it holds that B ‘ ( X, B , L , L , f, S ) if and only if there exist sets B L , B R , L A , L B , L C , L A , L B , L C suchthat the following properties hold: B = B L ∪ B R , L = L A ] L B ] L C and L = L A ] L B ] L C , for every v ∈ B , f ( v ) ∈ L B , B L ‘ ( X L , B , L L , L L , f L , S L ) , where • L L = L A , • L L = L B ∪ L B ∪ L A , • f L = f , and • S L = S ∪ L C ∪ L C ;and B R ‘ ( X R , B , L R , L R , f R , S R ) , where • L R = L C , • L R = L B ∪ L B ∪ L C , • f R = f , and • S R = S ∪ L A ∪ L A . Proof:
For the forward implication, suppose first that B ‘ ( X, B , L , L , f, S ). Let B L = B ∩ V ( G X L ) and B R = B ∩ V ( G X R ), satisfying Property 1 of the lemma. For i ∈ [2], let • L Ai = { Z ∈ L i | B L is an ( X L , Z )- bs in G X L and B R is not an ( X R , Z )- bs in G X R } , • L Ci = { Z ∈ L i | B L is not an ( X L , Z )- bs in G X R and B R is an ( X R , Z )- bs in G X R } , and • L Bi = { Z ∈ L i | B L is an ( X L , Z )- bs in G X L and B R is an ( X R , Z )- bs in G X R } .By Definition 26, for every Z ∈ L i , as B is an ( X, Z )- bs in G X . By Lemma 31, weobtain that B L is an ( X L , Z )- bs in G X L or B R is an ( X R , Z )- bs in G X R . This implies that L i = L Ai ] L Bi ] L Ci , and thus Property 2 is satisfied.For Property 4, let us only prove that B L ‘ ( X L , B , L L , L L , f L , S L ), as the proof for B R follows the same arguments. We verify that each of the (non-trivial) properties ofDefinition 26 is satisfied. Property ii . We need to prove that B L is an ( X L , Z )- bs in G X L for every Z ∈ L L ∪ L L .This follows from the definition of the sets L Li , and by the hypothesis that B is an ( X, Z )- bs in G X for every Z ∈ L ∪ L (since B ‘ ( X, B , L , L , f, S )). úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 25 Property iii . Let us prove that B L is not an ( X L , Z )- bs in G X L , for every Z ∈ S L = S ∪ L C ∪ L C . Let Z ∈ S L . If Z ∈ S , then since B is not an ( X, Z )- bs in G X , because B ‘ ( X, B , L , L , f, S ), we have by Lemma 31 that B L is not an ( X L , Z )- bs in G X L . If Z ∈ L C ∪ L C , then the result follows from definition of L Ci . Property iva . We have to prove that ∀ v ∈ B L \ B , ∃ Z ∈ L L = L A such that B L \ { v } is not an ( X L , Z )- bs in G X L . If B L = B , the statement trivially holds. Otherwise, let v ∈ B L \ B . As B ‘ ( X, B , L , L , f, S ), there exists Z ∈ L such that B \ { v } is not an( X, Z )- bs in G X , by Definition 26. This implies, by Lemma 31, that B L \ { v } is not an( X L , Z )- bs in G X L . As B \ { v } ⊇ B R , we get that B R is not an ( X, Z )- bs in G X , and thusby Lemma 31 that B R is not an ( X R , Z )- bs in G X R , implying that Z ∈ L A . Property ivb . We finally have to prove that ∀ v ∈ B , B L \ { v } is not an ( X L , f L ( v ))- bs in G X L . Let v ∈ B . Let Z = f L ( v ) = f ( v ), where Z ∈ L . As B ‘ ( X, B , L , L , f, S ), B \ { v } is not an ( X, Z )- bs in G X , implying by Lemma 31 that B L \ { v } is not an ( X L , Z )- bs in G X L . Moreover, as B is an ( X, Z )- bs in G X and B \ { v } is not an ( X, Z )- bs in G X , wededuce that v ∈ Z , implying that Z ∈ L B ⊆ L L . Note also that f L ( v ) ∈ L B , as required byProperty 3.For the backward implication, suppose that there exist B L , B R , L A , L B , L C , L A , L B , L C satisfying the lemma’s conditions. Let us prove that B ‘ ( X, B , L , L , f, S ), by verifyingagain that each of the (non-trivial) properties of Definition 26 is satisfied. Property ii . We have to prove that B = B L ∪ B R is an ( X, Z )- bs in G X , for every Z ∈ L ∪ L . By hypothesis, we know that: B L ‘ ( X L , B , L L , L L , f L , S L ) and B R ‘ ( X R , B , L R , L R , f R , S R ). By Definition 26, we deduce that B L is an ( X L , Z )- bs in G X L ,for every Z ∈ L L ∪ L L = L A ∪ L A ∪ L B ∪ L B . By Lemma 31, B is an ( X, Z )- bs in G X , forevery Z ∈ L A ∪ L A ∪ L B ∪ L B . Analogously, one may deduce that B is an ( X, Z )- bs in G X ,for every Z ∈ L C ∪ L C ∪ L B ∪ L B . Thus, B is an ( X, Z )- bs in G X , for every Z ∈ L ∪ L . Property iii . We have to prove that B is not an ( X, Z )- bs in G X , for every Z ∈ S .Let Z ∈ S . Since B L ‘ ( X L , B , L L , L L , f L , S L ) and S ⊆ S L , we have that B L is not an( X L , Z )- bs in G X L , by Definition 26. As B L is not an ( X L , Z )- bs in G X L and, analogously, B R is not an ( X R , Z )- bs in G X R , it implies by Lemma 31 that B is not an ( X, Z )- bs in G X . Property iva . Let us now prove that for every v ∈ B \ B , there is Z ∈ L = L A ] L B ] L C such that B \ { v } is not an ( X, Z )- bs in G X . If B = B , then there is nothing to prove.Otherwise, let v ∈ B \ B , and suppose without loss of generality that v ∈ B L \ B . As B L ‘ ( X L , B , L L , L L , f L , S L ), and as L L = L A , there exists Z ∈ L A such that B L \ { v } isnot an ( X L , Z )- bs in G X L . As Z ∈ L A , L A ⊆ S R , and B R ‘ ( X R , B , L R , L R , f R , S R ), B R is not an ( X R , Z )- bs in G X R . Thus, by Lemma 31, B \ { v } is not an ( X, Z )- bs in G X . Property ivb . Let us finally prove that for each v ∈ B , B \ { v } is not an ( X, f ( v ))- bs in G X . Let v ∈ B and let Z = f ( v ). By Property 3, we know that Z ∈ L B , i.e. Z is bothin L L and L R . Then, as B L ‘ ( X L , B , L L , L L , f L , S L ) and f L = f , by Property ivb weget that B L \ { v } is not an ( X L , Z )- bs in G X L . Using the same arguments for B R , we getthat B R \ { v } is not an ( X R , Z )- bs in G X R . By Lemma 31, we obtain that B \ { v } is not an( X, Z )- bs in G X . (cid:50) (cid:73) Definition 33.
Let G be a graph, X ⊆ V ( G ) , v ∈ X , and R ⊆ X . We denote • R ( v ) = { Z ∈ R | v ∈ Z } , • R (¯ v ) = { Z ∈ R | v / ∈ Z } , and • r v ( R ) = { Z \ { v } | Z ∈ R} . Before proving Lemma 38 corresponding to the introduce case, let us first prove thefollowing lemmas where we assume that X ∈ B is an introduce node and that X C is thechild of X with X C = X \ { v } for some vertex v ∈ X . (cid:73) Lemma 34.
Let Z ⊆ X such that Z is an is with v ∈ Z . For every I ⊆ V ( G ) such that v ∈ I , I is a maximum ( X, Z ) - is in G X if and only if I \ { v } is a maximum ( X C , Z \ { v } ) - is in G X C . Proof:
For the forward implication, suppose that I is a maximum ( X, Z )- is in G X such that v ∈ I . Note that I \ { v } is an ( X C , Z \ { v } )- is in G X C . Let I be a maximum ( X C , Z \ { v } )- is in G X C . As Z is an is of G , and N G X ( v ) ⊆ X by the properties of a tree decomposition, I ∪ { v } is an ( X, Z )- is in G X , implying | I ∪ { v }| ≤ | I | . Therefore | I \ { v }| ≥ | I | , hence I \ { v } is a maximum ( X C , Z \ { v } )- is in G X C .For the backward implication, suppose that I \ { v } is a maximum ( X C , Z \ { v } )- is in G X C . Note that I is an ( X, Z )- is in G X . Let I be a maximum ( X, Z )- is in G X . As I \ { v } is an ( X C , Z \ { v } )- is in G X C , we get | I \ { v }| ≤ | I \ { v }| and therefore, since both I and I contain v , | I | ≤ | I | and the lemma follows. (cid:50) (cid:73) Lemma 35.
Let Z ⊆ X such that Z is an is with v / ∈ Z . For every I ⊆ V ( G ) such that v / ∈ I , I is a maximum ( X, Z ) - is in G X if and only if I is a maximum ( X C , Z ) - is in G X C . Proof:
For the forward implication, suppose that I is a maximum ( X, Z )- is in G X such that v / ∈ I . Note that I is an ( X C , Z )- is in G X C . Let I be a maximum ( X C , Z )- is in G X C . As I ’is an ( X, Z )- is in G X , | I | ≤ | I | , leading to the desired result.For the backward implication, consider that I is a maximum ( X C , Z )- is in G X C . Notethat I is an ( X, Z )- is in G X . Let I be a maximum ( X, Z )- is in G X . As v / ∈ Z , v / ∈ I , and I is an ( X C , Z )- is in G X C , implying | I | ≤ | I | . (cid:50) (cid:73) Lemma 36.
Let Z ⊆ X such that Z is an is . For every B ⊆ V ( G X ) such that v / ∈ B , B is an ( X, Z ) - bs in G X if and only if B is an ( X C , Z \ { v } ) - bs in G X C . Proof:
For the forward implication, assume that B is an ( X, Z )- bs in G X such that v / ∈ B .Let I be a maximum ( X C , Z \ { v } )- is in G X C . Suppose first that v ∈ Z . By Lemma 34, weget that I ∪ { v } is a maximum ( X, Z )- is in G X , implying that B ∩ ( I ∪ { v } ) = ∅ . As v / ∈ B ,we get B ∩ I = ∅ . Suppose now that v / ∈ Z . By Lemma 35, we get that I is a maximum( X, Z )- is in G X , implying B ∩ I = ∅ .For the backward implication, suppose that B is an ( X C , Z \ { v } )- bs in G X C . Let I be amaximum ( X, Z )- is in G X . Suppose first that v ∈ Z . By Lemma 34, we get that I \ { v } isa maximum ( X C , Z \ { v } )- is in G X C , implying that B ∩ ( I \ { v } ) = ∅ . Suppose now that v / ∈ Z . By Lemma 35, we get that I is a maximum ( X C , Z )- is in G X C , implying B ∩ I = ∅ . (cid:50) (cid:73) Lemma 37.
Let Z ⊆ X such that Z is an is with v / ∈ Z . For every B ⊆ V ( G X ) such that v ∈ B , B is an ( X, Z ) - bs in G X if and only if B \ { v } is an ( X C , Z ) - bs in G X C . Proof:
For the forward implication, suppose that B is an ( X, Z )- bs in G X such that v ∈ B .Let I be a maximum ( X C , Z )- is in G X C . By Lemma 35, I is a maximum ( X, Z )- is in G X ,implying that B ∩ I = ∅ . As v / ∈ I , we get ( B \ { v } ) ∩ I = ∅ .For the backward implication, suppose that B \ { v } is an ( X C , Z )- bs in G X C . Let I be amaximum ( X, Z )- is in G X . By Lemma 35, I is a maximum ( X C , Z )- is in G X C , implyingthat ( B \ { v } ) ∩ I = ∅ . (cid:50) úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 27 We are now ready to state the main lemma of this section. Let us recall that given afunction f : A → B and a subset A ⊆ A , we denote by f | A the restriction of f to A . (cid:73) Lemma 38.
Let ( X, B , L , L , f, S ) ∈ E ( G, D ) where X ∈ B is an introduce node and X C is the child of X with X C = X \ { v } . For every B ⊆ V ( G X ) , B ‘ ( X, B , L , L , f, S ) if and only if one of the following two cases holds: Case 1 : v ∈ B and there exist L A , L B such that L ( v ) = L A ] L B , f ( v ) ∈ L A , for every Z ∈ S , v / ∈ Z , and B \ { v } ‘ ( X \ { v } , B \ { v } , L C , L C , f C , S C ) , where • L C = L (¯ v ) , • L C = L (¯ v ) , • f C = f | B \{ v } , and • S C = S ∪ r v ( L A ) . Case 2 : v / ∈ B and B ‘ ( X \ { v } , B , L C , L C , f C , S C ) , where • L C = r v ( L ) , • L C = r v ( L ) , • f C ( v ) = f ( v ) \ { v } for every v ∈ B , and • S C = r v ( S ) . Proof:
For the forward implication, suppose that B ⊆ V ( G X ) is such that B ‘ ( X, B , L , L , f, S ).We distinguish the two cases considered in Lemma 38. In both cases, we verify that each ofthe corresponding properties is satisfied. Case 1 . Suppose that v ∈ B , and thus v ∈ B , as v ∈ X and B = B ∩ X . Let L A = { Z ∈ L | B \ { v } is not an ( X, Z )- bs in G X } . By Property ivb applied to v , we getthat f ( v ) ∈ L A , implying Property 2. Moreover, for every Z ∈ L A , there exists a maximum( X, Z )- is I in G X such that I ∩ ( B \ { v } ) = ∅ , and thus if we had v / ∈ Z , then v / ∈ I and I ∩ B = ∅ , a contradiction. This implies that L A ⊆ L ( v ), and we define L B = L ( v ) \ L A ,implying Property 1. By Property iii, as B is not an ( X, Z )- bs in G X for every Z ∈ S and v ∈ B , we get Property 3. Let us now prove Property 4, by verifying each of the non-trivialproperties of Definition 26 applied to B \ { v } . Property ii . Recall that X C = X \ { v } . We need to prove that B \ { v } is an ( X C , Z )- bs in G X C , for every Z ∈ L C ∪ L C = L (¯ v ) ∪ L (¯ v ). Let Z ∈ L C ∪ L C . As v / ∈ Z , Lemma 37implies that B \ { v } is an ( X C , Z )- bs in G X C . Property iii . We must prove that B \ { v } is not an ( X C , Z )- bs in G X C , for every Z ∈ S C = S ∪ r v ( L A ). Let Z ∈ S C . If Z ∈ S , as v / ∈ Z (which we know from Property 3)and B is not an ( X, Z )- bs in G X (since B ‘ ( X, B , L , L , f, S )), Lemma 37 implies that B \ { v } is not an ( X C , Z )- bs in G X C . If Z ∈ r v ( L A ), then let Z be such that Z = Z \ { v } .We know that B \ { v } is not an ( X, Z )- bs in G X . By Lemma 36, we get that B \ { v } is notan ( X C , Z )- bs in G X C . Property iva . Let us now prove that, for every v ∈ ( B \ { v } ) \ ( B \ { v } ) there is Z ∈ L C = L (¯ v ) such that ( B \ { v } ) \ { v } is not an ( X C , Z )- bs in G X C . Since v ∈ B , let v ∈ B \ B . Since B ‘ ( X, B , L , L , f, S ), there exists Z ∈ L such that B \ { v } is not an( X, Z )- bs in G X . As v ∈ B \ { v } , this implies that Z ∈ L (¯ v ). As v / ∈ Z , from Lemma 37we get that ( B \ { v } ) \ { v } is not an ( X C , Z )- bs in G X C . Property ivb . We now have to prove that for every v ∈ B \ { v } , ( B \ { v } ) \ { v } is not an( X C , f C ( v ))- bs in G X C . If B \{ v } = ∅ , we have nothing to prove. Otherwise, let v ∈ B \{ v } . Since B ‘ ( X, B , L , L , f, S ), there exists Z ∈ L such that Z = f C ( v ) = f ( v ) and B \{ v } is not an ( X, Z )- bs in G X . As v ∈ B \ { v } , this implies that Z ∈ L (¯ v ). As v / ∈ Z , Lemma 37implies that ( B \ { v } ) \ { v } is not an ( X C , Z )- bs in G X C . Case 2 . Suppose that v / ∈ B . Let us prove that B ‘ ( X \{ v } , B , r v ( L ) , r v ( L ) , f C , r v ( S )),where f C ( v ) = f ( v ) \ { v } for every v ∈ B . Property ii . We first prove that B is an ( X C , Z )-set in G X C , for every Z ∈ L C ∪ L C = r v ( L ) ∪ r v ( L ). Let Z ∈ r v ( L ) ∪ r v ( L ) where Z = Z \ { v } and Z ∈ L ∪ L . As v / ∈ B and B is an ( X, Z )- bs in G X , Lemma 36 implies that B is an ( X C , Z )- bs in G X C . Property iii . Let us prove that B is not an ( X C , Z )- bs in G X C , for every Z ∈ S C = r v ( S ).Let Z ∈ r v ( S ), where Z = Z \ { v } and Z ∈ S . As v / ∈ B , and as B is not an ( X, Z )- bs in G X , Lemma 36 implies that B is not an ( X C , Z )- bs in G X C . Property iva . We now prove that, for every v ∈ B \ B , there exists Z ∈ L C = r v ( L )such that B \ { v } is not an ( X C , Z )- bs in G X C . Suppose that B \ B = ∅ , as otherwisethe statement trivially holds. Let v ∈ B \ B . Since B ‘ ( X, B , L , L , f, S ), there exists Z ∈ L such that B \ { v } is not an ( X, Z )- bs in G X . As v / ∈ B \ { v } , Lemma 36 impliesthat B \ { v } is not an ( X C , Z \ { v } )- bs in G X C , and Z \ { v } ∈ L C . Property ivb . We must finally prove that, for every v ∈ B , B \{ v } is not an ( X C , f C ( v ))- bs in G X C . Let v ∈ B . Since B ‘ ( X, B , L , L , f, S ), let Z = f ( v ), where Z ∈ L , suchthat B \ { v } is not an ( X, Z )- bs in G X (by Property ivb). As v / ∈ B \ { v } , Lemma 36implies that B \ { v } is not an ( X C , Z \ { v } )- bs in G X C , and Z \ { v } = f C ( v ).We now focus on the backward implication, and we distinguish again the two casesaccording to the possible hypothesis. In both cases, remind that our goal is to prove that B ‘ ( X, B , L , L , f, S ). Case 1 . Let B with v ∈ B and suppose that there exist L A , L B satisfying the statementof the lemma. Property ii . Let us prove that B is an ( X, Z )- bs in G X , for every Z ∈ L ∪ L . Let Z ∈ L ∪ L . If v ∈ Z , then as v ∈ B it follows that B is an ( X, Z )- bs in G X . Otherwise, byProperty ii, we get that B \ { v } is an ( X C , Z )- bs in G X C . As v / ∈ Z , Lemma 37 implies that B is an ( X, Z )- bs in G X . Property iii . We now prove that B is not an ( X, Z )- bs in G X , for every Z ∈ S . Let Z ∈ S . We know that B \ { v } is not an ( X C , Z )- bs in G X C . By hypothesis, v / ∈ Z , and thusLemma 37 implies that B is not an ( X, Z )- bs in G X . Property iva . We have to show that, for every v ∈ B \ B , there exists Z ∈ L such that B \ { v } is not an ( X, Z )- bs in G X . Assume that B \ B = ∅ . Let v ∈ B \ B . By hypothesis,there exists Z ∈ L C = L (¯ v ) such that ( B \ { v } ) \ { v } is not an ( X C , Z )- bs in G X C . As v / ∈ Z , Lemma 37 implies that ( B \ { v } ) is not an ( X, Z )- bs in G X . Property ivb . We finally prove that, for every v ∈ B , B \ { v } is not an ( X, f ( v ))- bs in G X . Let v ∈ B and let Z = f ( v ). Suppose first that v = v . By Property 2 we knowthat Z ∈ L A , and by Property 1 we know that v ∈ Z , implying then that there exists Z such that Z = Z ∪ { v } . As r v ( L A ) ⊆ S C , it follows that B \ { v } is not an ( X C , Z )- bs in G X C . By Lemma 36, B \ { v } is not an ( X, Z )- bs in G X . Suppose now that v = v . Byhypothesis, ( B \ { v } ) \ { v } is not an ( X C , Z )- bs in G X C , and Z ∈ L (¯ v ). As v / ∈ Z , byLemma 37 B \ { v } is not an ( X, Z )- bs in G X . Case 2 . Let B with v / ∈ B and B ‘ ( X \ { v } , B , r v ( L ) , r v ( L ) , f C , r v ( S )), where f C ( v ) = f ( v ) \ { v } for every v ∈ B . Property ii . Let us prove that B is an ( X, Z )- bs in G X , for every Z ∈ L ∪ L . Let Z ∈ L ∪ L . As v / ∈ B and as B is an ( X C , Z \ { v } )- bs in G X C , Lemma 36 implies that B úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 29 is an ( X, Z )- bs in G X . Property iii . We now prove that B is not an ( X, Z )- bs in G X , for every Z ∈ S . Let Z ∈ S . As v / ∈ B and as B is not an ( X C , Z \ { v } )- bs in G X C , Lemma 36 implies that B isnot an ( X, Z )- bs in G X . Property iva . We have to show that, for every v ∈ B \ B , there exists Z ∈ L suchthat B \ { v } is not an ( X, Z )- bs in G X . If B \ B = ∅ , then we have nothing to prove. Let v ∈ B \ B . By hypothesis, there exists Z ∈ L C such that B \ { v } is not an ( X C , Z )- bs in G X C . Let Z ∈ L such that Z = Z \ { v } . As v / ∈ B \ { v } , Lemma 36 implies that B \ { v } is not an ( X, Z )- bs in G X . Property ivb . We finally prove that, for every v ∈ B , B \ { v } is not an ( X, f ( x ))- bs in G X . Let v ∈ B and let Z = f ( v ), where Z ∈ L . Recall that f C ( v ) = Z \ { v } . Byhypothesis, B \ { v } is not an ( X C , Z \ { v } )- bs in G X C . As v / ∈ B \ { v } , Lemma 36 impliesthat B \ { v } is not an ( X, Z )- bs in G X . (cid:50) Let us start with some preliminaries related to the notion of criticality . (cid:73) Definition 39.
Let G be a graph, X ⊆ V ( G ) , Z ⊆ X such that Z is an is , and v ∈ V ( G ) .We say that ( X, Z ) is • v -critical in G if for every maximum ( X, Z ) - is I in G , v ∈ I , • ¯ v -critical in G if for every maximum ( X, Z ) - is I in G , v / ∈ I , and • v -mixed in G if there exists a maximum ( X, Z ) - is I in G with v ∈ I and there exists amaximum ( X, Z ) - is I in G with v / ∈ I .Given v ∈ V ( G ) and a set R ⊆ X such that for each Z ∈ R , Z is an is , we denote • R ( v,X ) = { Z ∈ R | ( X, Z ) is v -critical in G X } , • R (¯ v,X ) = { Z ∈ R | ( X, Z ) is ¯ v -critical in G X } , • R ( ∗ v,X ) = { Z ∈ R | ( X, Z ) is v -mixed in G X } , and • a v ( R ) = { Z ∪ { v } | Z ∈ R} . (cid:73) Lemma 40.
Let G be a graph, X ⊆ V ( G ) , and Z ⊆ X such that Z is an is and v ∈ V ( G ) .Deciding whether ( X, Z ) is v -critical in G , ¯ v -critical in G or v -mixed in G can be done intime O ∗ (2 tw ( G ) ) . Proof: If v ∈ Z , then ( X, Z ) is by definition v -critical in G , and if v ∈ X \ Z then ( X, Z ) isby definition ¯ v -critical G . Suppose now that v / ∈ X . Then, observe that • ( X, Z ) is v -critical in G if and only if α ( X ∪{ v } ,Z ) ( G ) < α ( X,Z ) ( G ), • ( X, Z ) is ¯ v -critical in G if and only if α ( X ∪{ v } ,Z ∪{ v } ) ( G ) < α ( X,Z ) ( G ), and • ( X, Z ) is v -mixed in G if and only if α ( X ∪{ v } ,Z ∪{ v } ) ( G ) = α ( X ∪{ v } ,Z ) ( G ) = α ( X,Z ) ( G ).Let us now prove that, for every ( X , Z ) such that Z ⊆ X and Z is an is , α ( X ,Z ) ( G ) can becomputed in time O ∗ (2 tw ( G ) ). Indeed, observe that α ( X ,Z ) ( G ) = | Z | + α ( G \ ( X ∪ N ( Z ))).As tw ( G \ ( X ∪ N ( Z ))) ≤ tw ( G ) and α ( G ) can be computed in O ∗ (2 tw ( G ) ) [13], we get thedesired result. (cid:50) Before proving Lemma 45 corresponding to the forget case, let us first prove the followinglemmas, where we assume that X ∈ B is a forget node and X C is the child of X with X C = X ∪ { v } , for some vertex v ∈ X C . (cid:73) Lemma 41.
Let I ⊆ V ( G ) such that v ∈ I and let Z ⊆ X such that Z is an is . Thefollowing claims hold: • If I is a maximum ( X, Z ) - is in G X , then I is a maximum ( X C , Z ∪ { v } ) - is in G X C . • If ( X, Z ) is not ¯ v -critical in G X , then I is a maximum ( X, Z ) - is in G X if and only if I is a maximum ( X C , Z ∪ { v } ) - is in G X C . Proof:
For the first item, let I be a maximum ( X, Z )- is in G X . As v ∈ I , I ∩ X C = Z ∪ { v } ,and I is also an ( X C , Z ∪ { v } )- is in G X C . Let I be a maximum ( X C , Z ∪ { v } )- is in G X C . As I is also an ( X, Z )- is in G X , | I | ≥ | I | , and thus I is a maximum ( X C , Z ∪ { v } )- is in G X C .For the second item, the sufficiency is already proved in the first item. For the backwardimplication, let I be a maximum ( X C , Z ∪ { v } )- is in G X C . Observe first that I is an ( X, Z )- is in G X . Let I be a maximum ( X, Z )- is in G X such that v ∈ I , which exists as ( X, Z ) isnot ¯ v -critical in G X . By the previous property, I is a (maximum) ( X C , Z ∪ { v } )- is in G X C ,implying | I | ≥ | I | and the desired result. (cid:50) (cid:73) Lemma 42.
Let I ⊆ V ( G ) such that v / ∈ I and let Z ⊆ X such that Z is an is . Thefollowing claims hold: • If I is a maximum ( X, Z ) - is in G X , then I is a maximum ( X C , Z ) - is in G X C . • If ( X, Z ) is not v -critical in G X , then I is a maximum ( X, Z ) - is in G X if and only if I is a maximum ( X C , Z ) - is in G X C . Proof:
For the first item, let I be a maximum ( X, Z )- is in G X . As v / ∈ I , I ∩ X C = Z ,hence I is also an ( X C , Z )- is in G X C . Let I be a maximum ( X C , Z )- is in G X C . As I isalso an ( X, Z )- is in G X , | I | ≥ | I | , and thus I is a maximum ( X C , Z )- is in G X C .For the second item, again we only need to prove the backward implication. Let I bea maximum ( X C , Z )- is in G X C . Observe first that I is an ( X, Z )- is in G X . Let I be amaximum ( X, Z )- is in G X such that v / ∈ I , which exists as ( X, Z ) is not v -critical in G X .By the first item, I is a (maximum) ( X C , Z )- is in G X C , implying | I | ≥ | I | and the desiredresult. (cid:50) (cid:73) Lemma 43.
Let Z ⊆ X where Z is an is . The following claims hold: • If ( X, Z ) is v -critical in G X , then for each B ⊆ V ( G X ) , B is an ( X, Z ) - bs in G X if andonly if B is an ( X C , Z ∪ { v } ) - bs in G X C . • If ( X, Z ) is ¯ v -critical in G X , then for each B ⊆ V ( G X ) , B is an ( X, Z ) - bs in G X if andonly if B is an ( X C , Z ) - bs in G X C . • If ( X, Z ) is v -mixed in G X , then for each B ⊆ V ( G X ) , B is an ( X, Z ) - bs in G X if andonly if B is an ( X C , Z ∪ { v } ) - bs in G X C and B is an ( X C , Z ) - bs in G X C . Proof:
For the first item, let Z ⊆ X such that ( X, Z ) is v -critical in G X .For the forward implication, suppose that B is an ( X, Z )- bs in G X . Let I be a maximum( X C , Z ∪ { v } )- is in G X C . As v ∈ I and as ( X, Z ) is not ¯ v -critical in G X , by Lemma 41, I isalso a maximum ( X, Z )- is in G X , implying that I ∩ B = ∅ .For the backward implication, let B be an ( X C , Z ∪ { v } )- bs G X C . Let I be a maximum( X, Z )- is in G X . As ( X, Z ) is v -critical in G X , we know that v ∈ I . By Lemma 41, I is alsoa maximum ( X C , Z ∪ { v } )- is in G X C , implying that I ∩ B = ∅ .For the second item, let Z ⊆ X such that ( X, Z ) is ¯ v -critical in G X .For the forward implication, suppose that B is an ( X, Z )- bs in G X . Let I be a maximum( X C , Z )- is in G X C . As v / ∈ I and as ( X, Z ) is not v -critical in G X , by Lemma 42, I is also amaximum ( X, Z )- is in G X , implying that I ∩ B = ∅ .For the backward implication, let B be an ( X C , Z )- bs G X C . Let I be a maximum( X, Z )- is in G X . As ( X, Z ) is ¯ v -critical in G X , we know that v / ∈ I . By Lemma 42, I is alsoa maximum ( X C , Z )- is in G X C , implying that I ∩ B = ∅ . úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 31 For the third item, let Z ⊆ X such that ( X, Z ) is v -mixed in G X .For the forward implication, assume that B is an ( X, Z )- bs in G X . Let I be a maximum( X C , Z ∪ { v } )- is in G X C and I be a maximum ( X C , Z )- is in G X C . As ( X, Z ) is both not v -critical and not ¯ v -critical in G X , by Lemmas 41 and 42 we now that both I and I aremaximum ( X, Z )- is in G X , implying B ∩ I = ∅ and B ∩ I = ∅ .For the backward implication, let finally B be an ( X C , Z ∪ { v } )- bs in G X C and an( X C , Z )- bs in G X C . Let I be a maximum ( X C , Z )- is in G X C . If v ∈ I , by Lemma 41, I isalso a maximum ( X C , Z ∪ { v } )- is in G X C , implying I ∩ B = ∅ , and if v / ∈ I , by Lemma 42, I is also a maximum ( X C , Z )- is in G X C , implying I ∩ B = ∅ as well. (cid:50) (cid:73) Lemma 44.
Let
L ⊆ X where for each Z ∈ L , Z is an is . For every B ⊆ G X , B isan ( X, Z ) - bs in G X for every Z ∈ L if and only if B is an ( X C , Z ) - bs in G X C for every Z ∈ a v ( L ( v,X ) ) ∪ L (¯ v,X ) ∪ a v ( L ( ∗ v,X ) ) ∪ L ( ∗ v,X ) . Proof:
For the forward implication, suppose that B is an ( X, Z )- bs in G X , for every Z ∈ L .Let Z ∈ a v ( L ( v,X ) ) (resp. Z ∈ a v ( L ( ∗ v,X ) )), implying that Z = Z ∪ { v } with Z ∈ L ( v,X ) (resp. Z ∈ L ( ∗ v,X ) ). Observe that for every Z ∈ L , v / ∈ Z , implying that v / ∈ Z and thusthat we also have Z = Z \ { v } . This implies that ( X, Z ) is v -critical (resp. v -mixed)in G X . By hypothesis, B is an ( X, Z )- bs in G X , implying, as ( X, Z ) is v -critical (resp. v -mixed) in G X , that B is an ( X C , Z )- bs in G X C by Lemma 43. Let now Z ∈ L (¯ v,X ) (resp. Z ∈ L ( ∗ v,X ) ), implying that ( X, Z ) is ¯ v -critical (resp. v -mixed) in G X . By hypothesis, B isan ( X, Z )- bs in G X , implying, as ( X, Z ) is ¯ v -critical (resp. v -mixed) in G X , that B is an( X C , Z )- bs in G X C by Lemma 43.For the backward implication, suppose that B is an ( X C , Z )- bs in G X C for every Z ∈ a v ( L ( v,X ) ) ∪ L (¯ v,X ) ∪ a v ( L ( ∗ v,X ) ) ∪ L ( ∗ v,X ) . Let Z ∈ L . If Z ∈ L ( v,X ) , then there exists Z ∈ a v ( L ( v,X ) ) such that Z = Z ∪ { v } . By hypothesis, B is an ( X C , Z )- bs in G X C ,implying, as ( X, Z ) is v -critical in G X , that B is an ( X, Z )- bs in G X by Lemma 43. If Z ∈ L (¯ v,X ) , then by hypothesis, B is an ( X C , Z )- bs in G X C , implying, as ( X, Z ) is ¯ v -criticalin G X , that B is an ( X, Z )- bs in G X by Lemma 43. If Z ∈ L ( ∗ v,X ) , then by hypothesis B isan ( X C , Z ∪ { v } )- bs in G X C and B is an ( X C , Z )- bs in G X C , implying, as ( X, Z ) is v -mixedin G X , that B is an ( X, Z )- bs in G X by Lemma 43. (cid:50) We are now ready to state the main lemma of this section. (cid:73)
Lemma 45.
Let ( X, B , L , L , f, S ) ∈ E where X ∈ B is a forget node and X C is thechild of X with X C = X ∪ { v } . For each B ⊆ V ( G X ) , B ‘ ( X, B , L , L , f, S ) if and onlyif one of the following two cases holds: Case 1 : there exists Z ∗ ∈ L such that ( X, Z ∗ ) is not ¯ v -critical in G X , for each Z ∈ S , ( X, Z ) is not v -critical in G X , for each v ∈ B , ( X, f ( v )) is not v -critical in G X , and B ‘ ( X ∪ { v } , B ∪ { v } , L C , L C , f C , S C ) , where • L C = a v ( L ( v,X )1 ) ∪ L (¯ v,X )1 ∪ a v ( L ( ∗ v,X )1 ) ∪ L ( ∗ v,X )1 , • L C = a v ( L ( v,X )2 ) ∪ L (¯ v,X )2 ∪ a v ( L ( ∗ v,X )2 ) ∪ L ( ∗ v,X )2 ∪ { Z ∗ ∪ { v }} , • f C : B ∪ { v } → L C is such that f C ( v ) = ( Z ∗ ∪ { v } , if v = v,f ( v ) , otherwise, and • S C = S . Case 2 : there exist S A , S B , and f C such that S ( ∗ v,X ) = S A ] S B and B ‘ ( X ∪ { v } , B , L C , L C , f C , S C ) , where • L C = a v ( L ( v,X )1 ) ∪ L (¯ v,X )1 ∪ a v ( L ( ∗ v,X )1 ) ∪ L ( ∗ v,X )1 , • L C = a v ( L ( v,X )2 ) ∪ L (¯ v,X )2 ∪ a v ( L ( ∗ v,X )2 ) ∪ L ( ∗ v,X )2 , • f C : B → L C is such that f C ( v ) = ( f ( v ) ∪ { v } , if f ( v ) ∈ L ( v,X )2 ,f ( v ) , if f ( v ) ∈ L (¯ v,X )2 , otherwise f C ( v ) ∈ { f ( v ) , f ( v ) ∪ { v }} , and • S C = a v ( S ( v,X ) ) ∪ S (¯ v,X ) ∪ a v ( S A ) ∪ S B . Proof:
Observe first that in both cases, for every Z ∈ L C ∪ L C ∪ S C , Z is an is as required inthe definition of E . Indeed, for each Z ∈ L ∪ L ∪ S , we only add Z ∪ { v } to L C ∪ L C ∪ S C when Z is not ¯ v -critical in G X , implying that Z ∪ { v } is an is .For the forward implication, suppose that B ⊆ V ( G X ) is such that B ‘ ( X, B , L , L , f, S ),and let us distinguish two cases.Suppose first that v ∈ B . In this case, we will prove that all statements correspondingto Case 1 hold. Recall that X = X C \ { v } and B = B ∩ X . Thus v ∈ B \ B . Since B ‘ ( X, B , L , L , f, S ), Property iva in Definition 26 implies that there exists Z ∗ ∈ L suchthat B \ { v } is not an ( X, Z ∗ )- bs in G X , implying that there exists a maximum ( X, Z ∗ )- is I ∗ in G X such that I ∗ ∩ ( B \ { v } ) = ∅ . Moreover, v ∈ I ∗ as otherwise, I ∗ ∩ B = ∅ , contradictingthe fact that B is an ( X, Z ∗ )- bs in G X . This implies Property 1 of Case 1, i.e. ( X, Z ∗ ) isnot ¯ v -critical in G X .Let v ∈ B , Z = f ( v ), and B = B \ { v } . As B ‘ ( X, B , L , L , f, S ), Property ivbimplies that B is not an ( X, Z )- bs in G x . Thus, there exists a maximum ( X, Z )- is I in G X such that I ∩ B = ∅ . As v ∈ B , we deduce that ( X, Z ) is not v -critical in G X , implyingProperty 3 of Case 1. Let us now prove Property 4 and along the proof we will verify thatProperty 2 is also satisfied.Thus, let us now check all properties of Definition 26 to prove that B ‘ ( X ∪ { v } , B ∪{ v } , L C , L C , f C , S C ), where L C , L C , f C , and S C are defined as in Case 1. Property ii . Let us prove that B is an ( X C , Z )- bs in G X C , for each Z ∈ L C ∪ L C . ByLemma 44, for each Z ∈ L C ∪ ( L C \ { Z ∗ ∪ { v }} ), we know that B is an ( X C , Z )- bs in G X C .Moreover, recall that Z ∗ ∈ L and that B ‘ ( X, B , L , L , f, S ). Thus, B is an ( X, Z ∗ )- bs in G X and, as ( X, Z ∗ ) is either v -critical or v -mixed in G X , this implies by Lemma 43 that B is an ( X C , Z ∗ ∪ { v } )- bs in G X C . Property iii . We now prove that B is not an ( X C , Z )- bs in G X C , for each Z ∈ S C = S .Let Z ∈ S . As B ‘ ( X, B , L , L , f, S ), B is not an ( X, Z )- bs in G X . Since v ∈ B , it followsthat ( X, Z ) is not v -critical in G X , implying Property 2 of Case 1. If ( X, Z ) is ¯ v -critical in G X , then by Lemma 43, we get that B is not an ( X C , Z )- bs in G X C . If ( X, Z ) is v -mixed in G X , since B is not an ( X, Z )- bs in G X , by Lemma 43 we get that B is not an ( X C , Z )- bs in G X C or B is not an ( X C , Z ∪ { v } )- bs in G X C . The latter case is not possible as v ∈ B , andthus we get the desired property. úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 33 Property iva . Let us prove that, for each v ∈ B \ ( B ∪ { v } ), there is Z ∈ L C such that B \ { v } is not an ( X C , Z )- bs in G X C . Let v ∈ B \ ( B ∪ { v } ). As B ‘ ( X, B , L , L , f, S ),there exists Z ∈ L such that B = B \ { v } is not an ( X, Z )- bs in G X . Notice that as v ∈ B , ( X, Z ) is not v -critical in G X . If ( X, Z ) is ¯ v -critical in G X , then by Lemma 43, B is not an ( X C , Z )- bs in G X C , and we are done as Z ∈ L (¯ v,X )1 ⊆ L C . If ( X, Z ) is v -mixed in G X , then by Lemma 43, B is not an ( X C , Z )- bs in G X C or B is not an ( X C , Z ∪ { v } )- bs in G X C . Again, this latter case is not possible as v ∈ B . Thus, we conclude the proof as Z ∈ L ( ∗ v,X )1 ⊆ L C . Property ivb . To finish this case, let us prove that for each v ∈ B ∪ { v } , B \ { v } isnot an ( X C , f C ( v ))- bs in G X C . Let v ∈ B C = B ∪ { v } . Let B = B \ { v } . Let us firstconsider the case v = v . In this case, remind that f C ( v ) = Z ∗ ∪ { v } , where, as chosen above, Z ∗ ∈ L such that B \ { v } is not an ( X, Z ∗ )- bs in G X . Then let us consider I ∗ defined above,i.e. a maximum ( X, Z ∗ )- is in G X such that I ∗ ∩ ( B \ { v } ) = ∅ . As v ∈ I ∗ , according toLemma 41, I ∗ is a maximum ( X C , Z ∗ ∪ { v } )- is in G X C , and B ∩ I ∗ = ∅ . Suppose now that v = v and remind that, in this case, f C ( v ) = f ( v ). Then, since B ‘ ( X, B , L , L , f, S )and v ∈ B , we have that B is not an ( X, Z )- bs in G X , where Z = f ( v ) ∈ L . As v ∈ B ,we deduce that ( X, Z ) is not v -critical in G x . If ( X, Z ) is v -mixed in G X , then by Lemma 43, B is not an ( X C , Z )- bs in G X C or B is not an ( X C , Z ∪ { v } )- bs in G X C . This last case isagain not possible as v ∈ B . Thus, we deduce that B is not an ( X C , Z )- bs . If ( X, Z ) is¯ v -critical in G X , then by Lemma 43, we also get that B is not an ( X C , Z )- bs in G X C .Suppose now v / ∈ B . We now prove that Case 2 of lemma’s statement holds. Since B ‘ ( X, B , L , L , f, S ), we have that, for each Z ∈ S ( ∗ v,X ) , B is not an ( X, Z )- bs in G X .By Lemma 43 we get that either B is not an ( X C , Z )- bs in G X C , in which case we add Z to S B , and if it is not the case then B is not an ( X C , Z ∪ { v } )- bs in G X C , in which case we add Z to S A . It remains to define the function f C for v = v such that f ( v ) ∈ L ( ∗ v,X )2 . Let v = v such that f ( v ) ∈ L ( ∗ v,X )2 , and let Z = f ( v ). Since B ‘ ( X, B , L , L , f, S ), B \ { v } is notan ( X, Z )- bs in G X . By Lemma 43, as ( X, Z ) is mixed in G X , we get that either B \ { v } is not an ( X C , Z )- bs in G X C , in which case we define f C ( v ) = Z , and if it not the casethen B \ { v } is not an ( X C , Z ∪ { v } )- bs in G X C , in which case we define f C ( v ) = Z ∪ { v } .Let us now prove that B ‘ ( X C , B , L C , L C , f C , S C ) where L C , L C , and S C are definedas in the Case 2 of the lemma’s statement, by verifying that the required properties inDefinition 26 are satisfied. To prove Property ii, one should argue that B is an ( X C , Z )- bs in G X C , for each Z ∈ L C ∪ L C where L C = a v ( L ( v,X )1 ) ∪ L (¯ v,X )1 ∪ a v ( L ( ∗ v,X )1 ) ∪ L ( ∗ v,X )1 and L C = a v ( L ( v,X )2 ) ∪ L (¯ v,X )2 ∪ a v ( L ( ∗ v,X )2 ) ∪ L ( ∗ v,X )2 . It is immediate using Lemma 44 and thehypothesis that B ‘ ( X, B , L , L , f, S ). Property iii . Let us now prove that B is not an ( X C , Z )- bs in G X C , for each Z ∈ S C = a v ( S ( v,X ) ) ∪ S (¯ v,X ) ∪ a v ( S A ) ∪ a v ( S B ). Let Z ∈ S C . If Z ∈ a v ( S ( v,X ) ) then Z = Z ∪ { v } where Z ∈ S ( v,X ) . Since B ‘ ( X C , B , L C , L C , f C , S C ), we know that B is not an ( X, Z )- bs in G X . As ( X, Z ) is v -critical in G X , by Lemma 43 we know that B is not an ( X C , Z ∪ { v } )- bs in G X C . If Z ∈ S (¯ v,X ) then by the hypothesis B ‘ ( X C , B , L C , L C , f C , S C ), we knowthat B is not an ( X, Z )- bs in G X . As ( X, Z ) is ¯ v -critical in G X , by Lemma 43 we know that B is not an ( X C , Z )- bs in G X C . If Z ∈ a v ( S A ) ∪ S B , then by definition of S A and S B , B isnot an ( X C , Z )- bs in G X C . Property iva . We must now prove that, for each v ∈ B \ B , there is Z ∈ L C = a v ( L ( v,X )1 ) ∪ L (¯ v,X )1 ∪ a v ( L ( ∗ v,X )1 ) ∪ L ( ∗ v,X )1 such that B \ { v } is not an ( X C , Z )- bs in G X C .Recall that v / ∈ B and let v ∈ B \ ( B ∪ { v } ). As B ‘ ( X, B , L , L , f, S ), there exists Z ∈ L such that B = B \ { v } is not an ( X, Z )- bs in G X . By Lemma 44 with the list L = { Z } , there exists Z ∈ L C such that B is not an ( X C , Z )- bs in G X C . Property ivb . Let us finally prove that, for each v ∈ B C = B , B \ { v } is not an( X C , f C ( v ))- bs in G X C . Let v ∈ B C . As v ∈ B , by the hypothesis B ‘ ( X, B , L , L , f, S )we know that B = B \ { v } is not an ( X, Z )- bs in G X , where Z = f ( v ) with Z ∈ L . If( X, Z ) is ¯ v -critical in G X , then by Lemma 43, we also get that B is not an ( X C , Z )- bs in G X C , and we are done as Z = f C ( v ). If ( X, Z ) is v -critical in G X , then by Lemma 43, wealso get that B is not an ( X C , Z ∪ { v } )- bs in G X C , and we are done as Z ∪ { v } = f C ( v ).Finally, ( X, Z ) is v -mixed then Z ∈ L ( ∗ v,X )2 and by the definition of f C ( v ) we get that B isnot an ( X C , f ( v ))- bs in G X C .For the backward implication, let us prove that B ‘ ( X, B , L , L , f, S ), by distinguishingagain both cases in the statement of the lemma. Case 1 . Suppose that there exist Z ∗ ∈ L as required in Case 1. Property ii follows directlyfrom Lemma 44. Let us verify that the other properties of Definition 26 are also verified. Property iii . Let us prove that B is not ( X, Z )- bs in G X , for each Z ∈ S = S C . Let Z ∈ S . By hypothesis, B ‘ ( X C , B C , L C , L C , f C , S C ) and thus B is not an ( X C , Z )- bs in G X C . Consequently, there exists a maximum ( X C , Z )- is I in G X C such that I ∩ B = ∅ . Inaddition, since B C = B ∪ { v } ⊆ B , we have that v / ∈ I . As ( X, Z ) is not v -critical in G X ,we get by Lemma 42 that I is a maximum ( X, Z )- is in G X . Property iva . We now argue that, for each v ∈ B \ B , there is Z ∈ L such that B \ { v } is not an ( X, Z )- bs in G X . Let v ∈ B \ B . If v = v , then as v ∈ B C , by definition of f C we get that B \ { v } is not an ( X C , Z ∗ ∪ { v } )- bs in G X C . As ( X, Z ∗ ) is either v -criticalor v -mixed in G X , in both cases by Lemma 43 we get that B \ { v } is not an ( X, Z ∗ )- bs in G X . As by hypothesis Z ∗ ∈ L , this implies Property iva. Suppose now v = v . Since B ‘ ( X C , B C , L C , L C , f C , S C ), we know that there exists Z ∈ L C such that B \ { v } is notan ( X C , Z )- bs in G X C . By Lemma 44, there exists Z ∈ L such that B \ { v } is not an( X, Z )- bs in G X . Property ivb . Finally, we prove that for each v ∈ B , B \ { v } is not an ( X, f ( v ))- bs in G X . Let v ∈ B and let Z = f ( v ). Recall that B = B ∩ X and thus v / ∈ B , implying that v = v . By definition, we thus have f C ( v ) = f ( v ). Since B ‘ ( X C , B C , L C , L C , f C , S C ),we know that B = B \ { v } is not an ( X C , Z )- bs in G X C . By Property 3, ( X, Z ) is not v -critical in G X . As ( X, Z ) is ¯ v -critical or v -mixed in G X , then by Lemma 43, B is not an( X, Z )- bs in G X . Case 2 . Suppose that there exist S A , S B , and f C as required in Case 2. Property ii followsagain directly from Lemma 44. Property iii . Let us prove that B is not ( X, Z )- bs in G X , for each Z ∈ S . Let Z ∈ S .Recall that S C = a v ( S ( v,X ) ) ∪ S (¯ v,X ) ∪ a v ( S A ) ∪ S B and that B ‘ ( X C , B , L C , L C , f C , S C )as claimed in Case 2. If Z ∈ S ( v,X ) , then B is not an ( X C , Z ∪ { v } )- bs in G X C , and byLemma 43, as ( X, Z ) is v -critical in G X , B is not an ( X, Z )- bs in G X . If Z ∈ S (¯ v,X ) , then B is not an ( X C , Z )- bs in G X C , and by Lemma 43, as ( X, Z ) is ¯ v -critical in G X , B is not an( X, Z )- bs in G X . It remains to treat the case where Z ∈ S ( ∗ v,X ) = S A ∪ S B . In this case, if Z ∈ S A then B is not an ( X C , Z ∪ { v } )- bs in G X C , or ( Z ∈ S B ) B is not an ( X C , Z )- bs in G X C . In both cases, as ( X, Z ) is v -mixed in G X , by Lemma 43 B is not an ( X, Z )- bs in G X . Property iva . We now show that, for each v ∈ B \ B , there is Z ∈ L such that B \ { v } is not an ( X, Z )- bs in G X . Let v ∈ B \ B . Recall that, since B ‘ ( X C , B , L C , L C , f C , S C ), B ∩ X C = B ⊆ X and { v } = X C \ X . As v / ∈ B , we deduce v = v . By hypothesis, we knowthat there exists Z ∈ L C such that B \ { v } is not an ( X C , Z )- bs in G X C . By Lemma 44,there exists Z ∈ L such that B \ { v } is not an ( X, Z )- bs in G X . úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 35 Property ivb . Finally, we prove that for each v ∈ B , B \ { v } is not an ( X, f ( v ))- bs in G X . Let v ∈ B , Z = f ( v ) and Z = f C ( v ). By hypothesis, we know that B = B \ { v } isnot an ( X C , Z )- bs in G X C . If Z ∈ L ( v,X )2 , then by definition of f C we have Z = Z ∪ { v } ,and by Lemma 43 we get that B is not an ( X, Z )- bs in G X . If Z ∈ L (¯ v,X )2 , then by definitionof f C we have Z = Z , and by Lemma 43 we get that B is not an ( X, Z )- bs in G X . Finally,if Z ∈ L ( ∗ v,X )2 , then by definition of f C we have Z ∈ { Z, Z ∪ { v }} , and again by Lemma 43we get that B is not an ( X, Z )- bs in G X . (cid:50) Let us now assume once again that the input graph G and the nice tree decomposition D of G are provided, and let us describe a recursive algorithm A that solves problem Π. Wedistinguish several cases as follows. In each case, namely join, introduce, or forget, we usethe notations introduced in the corresponding lemma, namely Lemma 32, Lemma 38, orLemma 45, respectively, and define Algorithm A as follows. (cid:73) Definition 46.
Suppose we are given an instance I = ( X, B , L , L , f, S ) ∈ E of problem Π such that X is a join node with children X L = X R = X . For each collection P = {L A , L B , L C , L A , L B , L C } such that L = L A ] L B ] L C and L = L A ] L B ] L C ,we denote by I L ( P ) = ( X L , B , L L , L L , f L , S L ) and I R ( P ) = ( X R , B , L R , L R , f R , S R ) asdefined by Lemma 32.In the join case, Algorithm A enumerates all such collections, and returns A ( I L ( P )) ∪A ( I R ( P )) , where P maximizes | A ( I L ( P )) | + | A ( I R ( P )) | . (cid:73) Definition 47.
Suppose we are given an instance I = ( X, B , L , L , f, S ) ∈ E of problem Π such that X is an introduce node with child X C = X \ { v } . For each collection P = {L A , L B } of L as required in Case of Lemma 38, we denote by I ( P ) the instance defined in Case of Lemma 38, and we denote by I the instance defined in Case of Lemma 38.In the introduce case, if v ∈ B then Algorithm A enumerates all such collections andreturns { v } ∪ A ( I ( P )) , where P maximizes |A ( I ( P )) | , and if v / ∈ B then Algorithm A returns A ( I ) . (cid:73) Definition 48.
Suppose we are given an instance I = ( X, B , L , L , f, S ) ∈ E of problem Π such that X is a forget node with child X C = X ∪ { v } . For each Z ∗ ∈ L as requiredin Case of Lemma 45, we denote by I ( Z ∗ ) the instance defined in Case of Lemma 45,and for each partition P = {S A , S B } of S ( ∗ v,X ) and function f C as required in Case ofLemma 45, we denote by I ( P , f C ) the instance defined in Case of Lemma 45.In the forget case, Algorithm A enumerates all sets Z ∗ ∈ L as required in Case , andcomputes B = A ( I ( Z ∗ )) where Z ∗ maximizes | I ( Z ∗ ) | . Then, Algorithm A enumeratesall sets S A , S B , and functions f C as required in Case , and computes B = A ( I ( P , f C )) where P , f C maximizes | I ( P , f C ) | . Finally, Algorithm A returns the largest solution amongall B and B . In any of the three cases (join, forget, introduce), A returns max x ∈ E A ( x ) for someappropriate set E , and we point out that it may be the case that E = ∅ , when none ofthe enumerated parameters respect the required conditions of Lemma 32, Lemma 38, andLemma 45, and in this case Algorithm A returns −∞ instead of a solution. Concerning thebase case, we can always assume that the underlying tree decomposition has leaves where X = ∅ . On such a leaf, ∅ is the only candidate solution, and thus A returns ∅ if it is a validsolution, or −∞ otherwise. (cid:73) Lemma 49. A solves Π optimally: for every instance I ∈ E , if I is feasible then A ( I ) returns an optimal solution. Otherwise, it returns −∞ . Proof:
The proof is by induction on the number of remaining bags in G X in the providednice tree decomposition D . Let B be the solution returned by A ( I ), and let B ∗ be an optimalsolution of I . We distinguish the different types of nodes in the nice tree decomposition D of G . In the three types of nodes, if I is not feasible, then by Lemma 32, Lemma 38, andLemma 45, and by the inductive hypothesis, any of the recursive calls will output −∞ , andthus A ( I ) will return −∞ as well. We suppose now that I is feasible, and let B = A ( I ) andlet B ∗ an optimal solution. Join node . By Lemma 32, there exists a collection P ∗ as defined in Definition 46 andsets B ∗ L , B ∗ R such that B ∗ L ‘ I L ( P ∗ ) and B ∗ R ‘ I R ( P ∗ ). Let P be the collection chosenby A . We have | B | = | A ( I L ( P )) | + | A ( I L ( P )) | − | B | ≥ | A ( I L ( P ∗ )) | + | A ( I L ( P ∗ )) | − | B | ≥| B ∗ L | + | B ∗ R | − | B | = | B ∗ | . Moreover, by Lemma 32, B ‘ I . Introduce node . If v ∈ B ∗ , then according to Case 1 of Lemma 38 there existsa collection P ∗ as defined in Definition 47 such that B ∗ \ { v } ‘ I ( P ∗ ). Let P be thecollection chosen by A . As in this case we have v ∈ B , we have | B | = 1 + | A ( I ( P )) | ≥ | A ( I ( P ∗ )) | ≥ | B ∗ \ { v }| = | B ∗ | . Moreover, by Lemma 38, B ‘ I . If v / ∈ B ∗ , thenaccording to Case 2 of Lemma 38, B ∗ ‘ I . As in this case we have v / ∈ B , we have | B | = | A ( I ) | ≥ | B ∗ | . Moreover, according to Lemma 38, B ‘ I . Forget node . If v ∈ B ∗ , then according to Case 1 of Lemma 45 there exist Z ∗∗ ∈ L such that B ∗ ‘ I ( Z ∗∗ ). Let Z ∗ be the element chosen by A for the first case, and ( P , f C )the elements chosen for the second case. We have | B | ≥ | A ( I ( Z ∗ )) | ≥ | A ( I ( Z ∗∗ )) | ≥ | B ∗ | .If v / ∈ B ∗ , then according to Case 2 of Lemma 45 there exist P ∗ and f ∗ C such that B ∗ ‘ I ( P ∗ , f ∗ C ). We have | B | ≥ | A ( I ( P , f C )) | ≥ | A ( I ( P ∗ , f ∗ C )) | ≥ | B ∗ | . (cid:50) (cid:73) Lemma 50.
Algorithm A runs in time O ∗ (2 O (2 t ) ) , where t is the width of the given nicetree decomposition of the input graph. Proof:
The time complexity of Algorithm A is O ∗ ( x · x ), where x = |E| is the number ofpossible inputs of Π and x is the maximum time necessary to compute A ( I ) for each I ∈ E .We denote n = | V ( G ) | . From the definitions of the corresponding objects, it can be routinelyverified that x ≤ n · t · t · t · t · (2 t ) t · t ≤ t +2 t +1 + t = 2 O (2 t ) .Let us now bound x . To that end, let θ be an upper bound on the number ofenumerated subinstances made in any of the three cases (join, introduce, or forget) and letby θ be an upper bound on the time complexity related to all operations like taking theminimum, and verifying that each enumerated subinstance verifies required properties (likefor example, in Case 1 of Lemma 45). In the join case, the number of subinstances is at most3 |L | · |L | ≤ t +1 (to consider all L A , L B , L C , L A , L B , L C ), and for each subinstance thecomplexity is polynomial in n . In the introduce case, the number of subinstances is 2 t + 1(to consider all L A ), and for each subinstance the complexity is polynomial in n . In the forgetcase, the number of subinstances is at most 2 t + 2 t · t (to consider all Z ∗ ∈ L in Case 1and all S A and f C in Case 2), and for each subinstance the complexity is in O ∗ (2 t ) as inCase 1, for every Z ∈ S (resp. every v ∈ B ,) we must verify if ( X, Z ) (resp. (
X, f ( v ))) isnot v -critical in G X , and this verification can be done in time O ∗ (2 t ) according to Lemma 40.All in all, we can choose θ = 3 t +2 = 2 O (2 t ) , θ = O ∗ (2 t ), and the lemma follows. (cid:50) úlio Araújo, Marin Bougeret, Victor A. Campos, and Ignasi Sau 37 As according to [5] we can determine whether tw ( G ) ≤ t in time O ∗ ( t O ( t ) ), and con-struct the corresponding (nice) tree decomposition of G in case of a positive answer, fromProposition 29 and Lemma 50 the following theorem is now immediate. (cid:73) Theorem 51.
The
MMBS / tw problem is FPT . More precisely, it can we solved in time O ∗ (2 O (2 tw ( G ) ) ) . We presented a number of negative and positive results for the
MMBS and
MMHS problems.Several interesting questions remain open. Concerning
MMBS , even if seems implausiblethat the problem could be expressed in monadic second-order logic, it would be nice to proveit formally. For that, one may try to use the framework introduced by van Bevern et al. [26].Simplifying the dynamic programming algorithm behind Theorem 51 is also worth trying.As for
MMHS , we believe that the main challenge is trying to get a algorithm para-meterized by α + β running faster than O ∗ (2 αβ ) (Theorem 24). Let us consider the case α = 2, corresponding to the MMVC problem. The parameterized complexity of
MMVC hasreceived some attention recently, with results concerning
FPT algorithms in time O ∗ (2 β ) [19],and even in time O ∗ ( c β ) for c < .
54 [7],
FPT algorithms for structural parameterizations [27],and kernelization [1]. This motivates the problem of trying to improve the running time O ∗ (2 αβ ) of Theorem 24, for example by solving MMHS in time O ∗ ( α β ), or even O ∗ ( α O ( β ) ).Recall that the algorithm of Proposition 13 runs in time O ∗ ( α β ) for fixed α , and that ithides a term | V ( H ) | f ( α ) for some function f .Achieving a running time of O ∗ ( α β ) might be typically done by guessing, at each step,only which vertex of a given hyperedge should be added to the solution. However, guessingonly a vertex v and applying recursion on a remaining instance ( H , β − H is definedby removing v and all hyperedges containing v , is not correct. Indeed, ( H , β −
1) being a yes -instance, certified by a solution S , does not imply that ( H , β ) is also a yes -instance, as S ∪ { v } may not be minimal anymore. Thus, we believe that, in order to solve MMHS intime O ∗ ( α β ) or even O ∗ ( α O ( β ) ), a significantly new approach should be devised. Acknowledgement . We would like to thank Mamadou Moustapha Kanté for helpfulsuggestions concerning the non-expressibility of problems in monadic second-order logic.
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