Parameterized Complexity of the Sparsest -Subgraph Problem in …bougeret/publis/ToCS-revised1-16-04-15.pdf · 2019-09-25 · Presentation of the problem. Given a simple undirected

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Parameterized Complexity of the Sparsestk-Subgraph Problem in Chordal Graphs

Marin Bougeret · Nicolas Bousquet ·Rodolphe Giroudeau · Remi Watrigant

Received: date / Accepted: date

Abstract In this paper we tackle the Sparsest k-Subgraph problem whichconsists in finding a subset of k vertices in a graph which induces the minimumnumber of edges. The Sparsest k-Subgraph problem is a natural general-ization of the Independent Set problem, and thus is NP-hard (and evenW [1]-hard) in general graphs. In this paper we investigate the parameterizedcomplexity of both Sparsest k-Subgraph and Densest k-Subgraph inchordal graphs. We first provide simple proofs that Densest k-Subgraphin chordal graphs is FPT and does not admit a polynomial kernel unlessNP ⊆ coNP/poly (both parameterized by k). More involved proofs ensure thesame behavior for Sparsest k-Subgraph in the same graph class. We lastlyprovide an FPT algorithm in interval graphs for Sparsest k-Subgraph, butparameterized by the number of edges of the solution (a stronger parameteri-zation than by k).

Keywords parameterized algorithm · kernel lower bound · chordal graphs ·sparsest k subgraph

1 Introduction

Presentation of the problem. Given a simple undirected graph G = (V,E) andan integer k, the Sparsest k-Subgraph problem asks to find k vertices in Ginducing the minimum number of edges (such a set will be called an optimalsolution), where an edge {u, v} is induced by a set S whenever {u, v} ⊆ S. The

This work has been funded by grant ANR 2010 BLAN 021902.A preliminary version of this work appeared in the proceedings of SOFSEM 2014 [4].

Universite Montpellier 2LIRMM, 161 rue Ada, 34095 Montpellier, FranceTel.: +33 467 41 85 85Fax: +33 467 41 85 00E-mail: {bougeret, bousquet, rgirou, watrigant}@lirmm.fr

2 Marin Bougeret et al.

decision version asks if there exists a k-subgraph inducing at most C edges,where C ∈ N is given in the input of the problem. As a generalization of theclassical independent set problem (for C = 0), Sparsest k-Subgraph isNP-hard in general graphs, as well as W [1]-hard when parameterized by k(as independent set is W [1]-hard [13]). In addition, there is an obvious XPalgorithm for Sparsest k-Subgraph when parameterized by k, as all subsetsof size k can be enumerated in O(nk) time, where n is the number of verticesin the graph.

Related problems. Several problems closely related to Sparsest k-Subgraphhave been extensively studied in the past decades. Among them, one can men-tion the maximization version of Sparsest k-Subgraph, namely the Dens-est k-Subgraph, for which several results have been obtained in general orrestricted graphs. Corneil and Perl [10], showed that Densest k-Subgraphremains NP-hard in bipartite, comparability and chordal graphs, and is poly-nomial time solvable in trees, cographs, and split graphs. The complexity sta-tus of Densest k-Subgraph in interval graphs, proper interval graphs andplanar graphs is left as an open problem, and is still not answered yet. Morerecently, Broersma et al. [6] improved some of these results by showing thatboth Densest k-Subgraph and Sparsest k-Subgraph are polynomial-timesolvable in bounded cliquewidth graphs, and Bonnet et al. [3] developed ex-act algorithms for Sparsest k-Subgraph, Densest k-Subgraph and othersimilar problems in general graphs parameterized by k and the maximum de-gree ∆ of the graph. During the past two decades, a large amount of work hasbeen dedicated to the approximability of Densest k-Subgraph in generalgraphs. So far, the best approximation ratio is O(nδ) for some δ < 1/3 [12],while the only negative result is due to Khot [17] ruling out a PT AS undersome complexity assumptions. Still concerning Densest k-Subgraph but inrestricted graph classes, Nonner [19] developed a PT AS in interval graphs,while constant approximation algorithms were proved for chordal graphs [9,18]. In a previous work [20], we recently proved that Sparsest k-Subgraphremains NP-hard in chordal graphs and admits a 2-appoximation algorithm.We can also mention the dual version of Sparsest k-Subgraph, namely themaximum partial vertex cover problem, for which we are looking for kvertices in the input graph which cover the maximum number of edges (anedge {u, v} is said to be covered by a set S whenever {u, v} ∩ S 6= ∅). Veryrecently, the NP-hardness of maximum partial vertex cover in bipartitegraphs was proven by three independent works [1,8,16], which directly trans-fers to Sparsest k-Subgraph (since finding k vertices covering the maximumnumber of edges is equivalent to find (n − k) vertices inducing the minimumnumber of edges).

More generally, Sparsest k-Subgraph, Densest k-Subgraph and max-imum partial vertex cover fall into the family of cardinality constrainedoptimization problems introduced by Cai [7]. In its survey, the author provedthat these three problems are W [1]-hard in regular graphs, and gives an XP

Parameterized Complexity of Sparsest k-Subgraph in Chordal Graphs 3

algorithm for general graphs with a better running time than the trivial algo-rithm.

As said previously, Sparsest k-Subgraph and Densest k-Subgraph arenatural generalizations of k-independent set and k-clique, and are thusimportant both from a theoretical and practical point of view. Our motivationis to study their computational (parameterized) complexity in graph classeswhere they remain NP-hard whereas k-independent set and k-clique arepolynomial-time solvable, such as the well-known class of perfect graphs andsome of its subclasses. To that end, we study their parameterized complexityin the class of chordal graphs, an important subclass of perfect graphs whicharises in many practical situations [15]. More precisely, we prove that bothSparsest k-Subgraph and Densest k-Subgraph in chordal graphs arefixed-parameter tractable and do not admit a polynomial kernel under someclassical complexity assumptions. As we will see, the results are quite easyto obtain for Densest k-Subgraph, but require some efforts for Sparsestk-Subgraph.

Organization of the paper. The paper is organized as follows: in the followingsection (Section 2), we recall the classical definitions of parameterized com-plexity and chordal graphs. Our two main results, namely the FPT algorithmand kernel lower bound for Sparsest k-Subgraph in chordal graphs, arepresented respectively in Sections 4 and 5. Lastly in Section 6, we developan FPT algorithm for Sparsest k-Subgraph in interval graphs but param-eterized by the number of edges in a solution, a stronger parameter than k(since it is always at most

(k2

)). Before all these, we study as prolegonema

the parameterized complexity of Densest k-Subgraph in chordal graphs inSection 3.

2 Parameterized Algorithms, Chordal Graphs, Interval Graphs

Parameterized algorithms. An interesting way to tackle NP-hard problems isparameterized complexity. A parameterized problem Q is a subset of Σ∗ ×N,where the second component is called the parameter of the instance. A fixed-parameter tractable (FPT for short) problem is a problem for which thereexists an algorithm which, given (x, k) ∈ Σ∗ × N, decides whether (x, k) ∈Q in time f(k)|x|O(1) for some computable function f . Such an algorithmbecomes efficient with an hopefully small parameter. A kernel is a polynomialalgorithm which, given (x, k) ∈ Σ∗ × N, outputs an instance (x′, k′) suchthat (x, k) ∈ Q ⇔ (x′, k′) ∈ Q and |x′| + k′ ≤ f(k) for some computablefunction f . For decidable problems, the existence of a kernel is equivalent tothe existence of an FPT -algorithm. Nevertheless one can ask the function fto be a polynomial. If so, then the kernel is called a polynomial kernel. If aproblem admits a polynomial kernel, then it roughly means that we can, inpolynomial time, compress the initial instance into an instance of size poly(k)


which contains all the hardness of the instance. In order to rule out polynomialkernels, we will use the recent technique of cross-composition [2].

Roughly speaking, a cross-composition is a polynomial reduction from tinstances of a (non-parameterized) problem A to a single instance of a param-eterized problem B such that the constructed instance is positive iff one of theinput instances is positive. In addition, the parameter of the constructed in-stance must be of size polynomial in the maximum size of the input instancesand the logarithm of t. It is known that if A is NP-hard and A cross-composesinto B, then B cannot admit a polynomial kernel unless NP ⊆ coNP/polysome complexity assumptions. For a stronger background concerning the pa-rameterized complexity, we refer the reader to the monograph of Downey andFellows [11]. Formal definitions of cross-composition and related notions areavailable in Appendix A.

Chordal graphs. A graph G = (V,E) is a chordal graph if it does not containan induced cycle of length at least four. As said previously, chordal graphsform an important subclass of perfect graphs. One can also equivalently definechordal graphs in terms of a special tree decomposition. Indeed, it is known [14]that a graph G = (V,E) is a chordal graph iff one can find a tree T = (X , A)with X ⊆ 2V such that for all v ∈ V , the set of nodes of T containing v, thatis Xv = {X ∈ X : v ∈ X}, induces a (connected) tree, and such that for allu, v ∈ V we have {u, v} ∈ E iff Xu ∩Xv 6= ∅. Moreover, given a chordal graph,this corresponding tree, called tree decomposition into cliques, can be found inpolynomial time. From this definition, it is clear that each X ∈ X induces aclique in G.

Interval Graphs. An interesting subclass of chordal graphs is the class of in-terval graphs. A graph G = (V,E) is a interval graph if there exists an intervalmodel of G, i.e. if each vertex v ∈ V can be mapped to an interval Iv of thereal line, such that two intervals overlap iff their corresponding edges are ad-jacent. A graph is an interval graph if and only if it has a tree decompositionof the above type that is a path and where every bag is a clique.

3 The Case of Densest k-Subgraph

Recall that the goal of Densest k-Subgraph is, given a graph G = (V,E)and k ∈ N, to find a set of k vertices which induces the maximum numberof edges (such a set of vertices will be called an optimal solution). The goalof this section is to prove that Densest k-Subgraph in chordal graph, pa-rameterized by k, is FPT but does not admit a polynomial kernel unlessNP ⊆ coNP/poly.

Theorem 1 Densest k-Subgraph in chordal graphs is FPT parameterizedby k.


Proof Notice that if the input graph G contains a clique of size k or more(which can be tested in polynomial time in chordal graphs), then this set isobviously an optimal solution. Otherwise, it implies that the treewidth of G isupper bounded by k (since the treewidth equals the maximum clique number inchordal graphs), and we can apply a classical dynamic programming algorithmover a tree decomposition ofG in order to compute an optimal solution in FPTtime. For instance, we may use the following result [5].

Theorem 2 ([5]) Densest k-Subgraph can be solved in1 O∗(2ω) on graphsof treewidth at most ω.

Which concludes the proof. utTheorem 3 Densest k-Subgraph in chordal graph parameterized by k doesnot admit a polynomial kernel, unless NP ⊆ coNP/poly.

Proof We show that Densest k-Subgraph cross-composes into itself (pa-rameterized by k). Indeed, suppose that we are given a sequence (G1, . . . , Gt)of chordal graphs on n vertices together with k ≤ n, C ≤

(k2

)(i.e. t instances of

Densest k-Subgraph in chordal graphs sharing the same number of verticesand values for k and C, which is the polynomial equivalence relation), whereGi = (Vi, Ei) for all i ∈ {1, . . . , t}. Let us consider the disjoint union of theset graphs, and for all i ∈ {1, . . . , t}, add a clique Ki on n2 vertices connectedto all vertices of Gi (and not connected to the others). Clearly, the resultinggraph G∗ is a chordal graph on t · (n+n2) vertices (recall that Gi is a chordalgraph for all i ∈ {1, . . . , t}, and we cannot create any cycle of length four ormore with the clique Ki). Then, if there exists i ∈ {1, . . . , t} such that Gi con-tains a set X of k vertices inducing at least C edges, then observe that X ∪Ki

is a set of k′ = n2+k vertices in G∗ which induces at least C ′ =(n2

2

)+kn2+C

edges. Conversely, let S be a set of k′ vertices in G∗ which induces at leastC ′ edges. Suppose first that there exists i ∈ {1, . . . , t} such that there existsu ∈ S ∩ Vi and v ∈ Ki \ S. Now, observe that (S \ {u}) ∪ {v} does not induceless edges than S. By iteratively applying this exchange argument, we mustfall into one of the following case:

– if there exists i ∈ {1, . . . , t} such that Ki ⊆ S, then S ∩ Ki induces(n2

2

)

edges, and thus one can observe that all vertices of S \Ki must have anaverage degree of at least n2 in S, implying that S \ Ki ⊆ Vi, i.e. thereexists a set of k vertices in Gi inducing at least C edges.

– for all i ∈ {1, . . . , t} we have Vi ∩ S = ∅, in which case we claim thatS cannot induce at least C ′ edges. Indeed, observe that by an exchangeargument, we may assume that Ki ⊆ S, and S \Ki ⊆ Kj for some i, j ∈{1, . . . , t}, and thus the number of induced edges would be

(n2

2

)+(k2

)< C ′.

This concludes the equivalence of solutions. Hence, since Densest k-Subgraphin chordal graphs is NP-hard [10], it does not admit a polynomial kernel un-less NP ⊆ coNP/poly. ut

1 The O∗(.) notation supresses polynomial factors and terms.


An interesting fact about the previous cross-composition is that it alsoholds for interval graphs as long as the input graphs are interval graphs.Unfortunately, since the complexity (NP-hard versus Polynomial) of Dens-est k-Subgraph in interval graphs is still unknown, we cannot conclude anynegative result under the classical complexity assumptions. However, it stillshows that the NP-hardness of Densest k-Subgraph in interval graphswould imply that it does not even admit a polynomial kernel (unless NP ⊆coNP/poly). Taking the contrapositive, and still under the same complexityassumption, we have that it is sufficient2 to derive a polynomial kernel in or-der to show that Densest k-Subgraph in interval graphs is not NP-hard.Observe however that the above does not hold for proper interval graphs.

4 FPT Algorithm for Sparsest k-Subgraph in Chordal Graphs

4.1 Notation

Throughout this section, G = (V,E) and k ∈ N will denote the input ofSparsest k-Subgraph. Let T = (X , A) be a tree decomposition into cliquesof G as defined in Section 2, and assume that T is rooted at some arbitraryleaf Xr ∈ X , which thus allows us to define, for each X ∈ X , the uniqueancestor anc(X) of X in T (with the convention that anc(Xr) = ∅). It isalso possible to define the depth of a node in the tree: depth(Xr) = 0, andfor any X 6= Xr, depth(X) = depth(anc(X)) + 1. For any subset X ′ ⊆ X ,we define V (X ′) =

⋃X∈X ′ X. For a vertex u ∈ V , let top(u) be the node

of T with the smallest depth among those containing u, or, more formally,top(u) = arg minX∈Xu depth(X).

In the following, let L and B denote respectively the set of leaves andbranching nodes (nodes of degree at least 3) of T . We now partition X intopaths in the following way: for any X ∈ (L∪B)\{Xr}, let YX be the branchingnode of maximum depth in the path between X and Xr (excluding X), orYX := Xr if there is no such branching node, and define PX to be the set ofall nodes in the path between X and YX , including X, but excluding YX ifYX 6= Xr. Observe that P = {PX : X ∈ (L ∪ B) \ {Xr}} is indeed a partitionof X , and that |P| = |L|+ |B| − 1.

4.2 First Observations

It is well known that a maximum independent set can be computed in poly-nomial time in chordal graphs (since chordal graphs are perfect). Hence, wecan first determine in polynomial time if there exists an independent set in G

2 A polynomial kernel is theoretically easier to find than a polynomial algorithm, sincea polynomial (and even a constant size) kernel can be easily derived from a polynomialalgorithm.


of size k. In this case, we return this set which is naturally an optimal solu-tion. Thus, we assume in the following that the graph G does not contain anindependent set of size k.

Notice that we may assume that for every leaf L ∈ L we do not haveL ⊆ anc(L) (otherwise we can contract the two nodes to obtain another treedecomposition into cliques of G). Therefore, for each leaf L of the tree, there isa vertex xL ∈ L of G such that xL /∈ anc(L), i.e. such that XxL = {L}. Since,for two distinct leaves L,L′ we have {xL, xL′} /∈ E, the set {xL : L ∈ L} is anindependent set of G, which implies:

Observation 1 We may assume that:

(i) |L| ≤ k − 1,(ii) |B| ≤ k − 3,

(iii) |P| ≤ 2k − 5.

We may also assume that depth(top(u)) is different for every u ∈ V , byeventually duplicating nodes in some paths between two branching nodes ofT (notice that such a modification does not change |L|, |B| or |P|, and thusObservation 1 still holds).

Finally, another important property is the following, whose proof comesdirectly from the above definitions.

Observation 2 For all X ∈ (L ∪ B) \ {Xr} and any x, y, z ∈ V (PX) suchthat top(y), top(z) ∈ PX and depth(top(x)) < depth(top(y)) < depth(top(z)),it holds that {x, z} ∈ E =⇒ {x, y} ∈ E.

4.3 The Algorithm

We fix an optimal solution S of Sparsest k-Subgraph. We also assumethat S = {s1, . . . , sk} is sorted in a non-increasing order with respect todepth(top(si)) for each i ∈ [k], or, informally, from the bottom to the topof the tree. For each i ∈ [k] we note Si = {s1, . . . , si}, and for each u ∈ V ,we note Ni(u) = N(u) ∩ Si, where N(u) = {v|{u, v} ∈ E}. The idea of thealgorithm consists in finding each si iteratively for i = 1, . . . , k. To do so, ateach step i the algorithm guesses the node Xi ∈ (L ∪ B) \ {Xr} such thattop(si) ∈ V (PXi) together with the set N∗i ⊆ Si−1 such that Ni−1(si) = N∗i .Then, as we show in Lemma 1 that without loss of optimality we can alwayssuppose that si is the vertex of maximum depth among all vertices u such thattop(u) ∈ V (PXi) \Si−1 and Ni−1(u) = N∗i , knowing Xi and N∗i allows the al-gorithm to find and add si to the solution. Since, for each iteration 1 ≤ i ≤ k,Xi is chosen among a set of size 2k − 4, and Ni is chosen among a set of size2i−1, the running time of this algorithm is O∗(2O(k2)). It now remains to showits correctness:

Lemma 1 For all i ∈ [k], without loss of optimality we may assume thatsi is the vertex of maximum depth among all vertices u such that top(u) ∈V (PXi) \ Si−1 and Ni−1(u) = N∗i .


Proof Let u∗ be the vertex of maximum depth among all vertices such thattop(u) ∈ V (PXi) \ Si−1 and Ni−1(u) = N∗i , and assume that u∗ 6= si. Letj ∈ [k]. If j < i, then since Ni−1(u∗) = Ni−1(si), we have {sj , u∗} ∈ E iff{sj , si} ∈ E. Now, if i < j and {sj , u∗} ∈ E, then, since depth(top(sj)) <depth(top(si)) < depth(top(u∗)), we must have {sj , si} ∈ E (using Observa-tion 2), which proves that S′ := (S \ {si}) ∪ {u∗} does not induce more edgesthan S. ut

We thus obtain the following result:

Theorem 4 Sparsest k-Subgraph can be solved in O∗(2O(k2)) in chordalgraphs.

5 Kernel Lower Bound of Sparsest k-Subgraph in Chordal Graphs

Intuition of the proof. The following kernel lower bound is obtained usinga cross-composition. It is an extension of our previous work [20], showingthe NP-hardness of Sparsest k-Subgraph in chordal graphs. Let us firstgive the intuition of this result, and then explain the modification we applywhich leads to the kernel lower bound. We then explain in details the wholeconstruction of the cross-composition and give a formal proof of the result.

The NP-hardness proof is a reduction from the classical k-clique problemin general graphs and roughly works as follows. Given an input instance G =(V,E), k ∈ N of k-clique, we first build a clique A representing the verticesof G. We also represent each edge ej = {u, v} ∈ E by a gadget Fj , and connectthe representative vertices of u and v in A to some vertices of Fj (see Figure1). The reduction will force the solution to take in A the representatives of(n− k) vertices of G (corresponding to the complement of a solution S of sizek in G), and also to take the same number of vertices among each gadget. Thekey idea is that the cost of a gadget Fj increases by one if it is adjacent to oneof the selected vertices of A. Thus, since the goal is to minimize the cost, wewill try to maximize the number of gadgets adjacent to the representatives ofS (i.e. vertices we did not pick in A), the maximum being reached when S isa clique in G.

To adapt this reduction into a cross-composition, we add an instance selec-tor composed of 2 log t gadgets adjacent to A (where t is the number of inputinstances of the cross-composition) which encodes the binary representation ofeach instance index. These gadgets have the same structure as the Fj . For tech-nical reasons, this instance selector has to be duplicated many times, as wellas the clique A which we must duplicate t times in order to encode the vertexset of each instance. The Figure 2 represents the construction in a simplifiedway. Let us now define formally the gadgets and state their properties.

Definition of a gadget. Let T ∈ N (we will set the value of T later). The vertexset of each gadget is composed of three sets of T vertices X,Y and Z, with


X = {x1, . . . , xT }, Y = {y1, . . . , yT } and Z = {z1, . . . , zT }. The set X inducesan independent set, the set Z induces a clique, and there is a (T −1)-clique on{y2, . . . , yT }. In addition, for all i ∈ {1, . . . , T}, we connect yi to all vertices ofZ and to xi. The Figure 1 summarizes the construction.In the following cross-composition, we will force the solution to take 2T verticesamong each gadget F . It is easy to see that the sparsest 2T -subgraph of Fis composed of the sets X and Z, which induces

(T2

)edges. In addition, if

we forbid the set Z to be in the solution (if the gadget is adjacent to somepicked vertices of A), then the remaining 2T vertices (namely X and Y ) induce((T2

)+ 1) edges.

Xj

Yj

Zj

Ai

gadget Fj for ej = {u, v} ! Ei

Ai1

Ain

T

n

n(clique)

1

Fig. 1: Schema of the gadget and its adjacencies with a clique Ai.

Theorem 5 Sparsest k-Subgraph does not admit a polynomial kernel inchordal graphs unless NP ⊆ coNP/poly (parameterized by k).

Proof Let (G1, k1), . . . , (Gt, kt) be a sequence of t instances of k-clique,with Gi = (Vi, Ei) for all i ∈ {1, . . . , t}. For all i ∈ {1, . . . , t}, let Ei ={ei1, . . . , ei|Ei|}. W.l.o.g. we suppose that t = 2q for some q ∈ N, and define

T = n(n− k) as well as M = n6.


Ai

X1

Y1

Z1

Xm

Ym

Zm

Z!11

F!11

F"11

clique ontn2 vertices A1 At

k n ! k

!k2

"gadgets m !

!k2

"gadgets

T

bin. representationof i : 1 1 0

2Mqgadgets

T

F1 Fm

Y!11

X!11

Z"Mq

Y"Mq

X"MqF!M

jF"M

j

F!1q

F"1q

1

Fig. 2: Schema of the cross-composition. Grey rectangles represent vertices ofthe solution, supposing that Gi contains a clique of size k. Notice that gadgetsof the bottom have been drawn in the reverse direction (e.g. Xβ1

1is below

Yβ11). Edges of the clique A have not been drawn for sake of clarity.

Our polynomial equivalence relation is the following: for 1 ≤ i, j ≤ t,(Gi, ki) is equivalent to (Gj , kj) if |Vi| = |Vj | = n, |Ei| = |Ej | = m andki = kj = k. One can verify that this relation is a polynomial equivalencerelation. In what follows we suppose that all instances of the sequence are inthe same equivalence class. The output instance G′ = (V ′, E′), k′, C ′ is definedas follows (see Figure 2):

– For each i ∈ {1, . . . , t} we construct a clique Ai on n2 vertices, where Ai

is composed of n subcliques Ai1, . . . , Ain. We also add all possible edges

between all cliques (Ai)i=1..n. Hence, A =⋃ti=1A

i is a clique of size tn2.– Since all instances have the same number of edges, we construct m gad-

gets (Fj)j=1..m, where each Fj is composed of Xj , Yj and Zj as describedpreviously. For all i ∈ {1, . . . , t}, if there is an edge eij = {u, v} ∈ Ei, then


we connect all vertices of Zj to all vertices of Aiu and Aiv. Let us defineF =

⋃mj=1 Fj the subgraph of all gadgets of the ”edge selector”.

– We add 2qM gadgets (Fαhj )h=1..Mj=1..q and (Fβhj )h=1..M

j=1..q , where all gadgets are

isomorphic to the edge gadgets, and thus composed of Xαhj, Yαhj and Zαhj

(resp. Xβhj, Yβhj and Zβhj ) for all h ∈ {1, . . . ,M} and all j ∈ {1, . . . , q}.

Let i ∈ {1, . . . , t}, and consider its binary representation b ∈ {0, 1}q (forconvenience, and because our indices start from 1, the binary representationof t will always be the 0 vector on q bits) . For all j ∈ {1, . . . , q}, ifthe jth bit of b equals 0, then connect all vertices of Ai to all vertices of⋃Mh=1 Zαhj . Otherwise, connect all vertices of Ai to all vertices of

⋃Mh=1 Zβhj .

Let us define B =⋃Mh=1

⋃qj=1(Fαhj ∪ Fβhj ) the subgraph of all gadgets of

the ”instance selector”.– We set k′ = T +2Tm+4TqM and C ′ =

(T2

)+(T2

)(m+2Mq)+(m−

(k2

))+

Mq.

It is clear that G′, k′ and C ′ can be constructed in time polynomial in∑ti=1 |Gi| + ki. Then, one can verify that G′ is a chordal graph. Indeed, it

is known [15] that a graph is chordal iff one can repeatedly find a simplicialvertex (a vertex whose neighborhood is a clique) and delete it from the graphuntil it becomes empty. Such an ordering is called a simplicial eliminationorder. It is easily seen that for each gadget, X,Y and then Z is a simplicialelimination order (each gadget is only adjacent to the clique A via its set Z).The remaining vertices are A which is a clique, and can thus be eliminated.

In addition, notice that the parameter k′ is a polynomial in n, k and log tonly and thus respects the definition of a cross-composition. We finally provethat there exists i ∈ {1, . . . , t} such that Gi contains a clique K of size k iffG′ contains a set K ′ of k′ vertices inducing C ′ edges or less.

Lemma 2 If there exists i ∈ {1, . . . , t} such that Gi contains a k-clique, thenG′ contains k′ vertices inducing at most C ′ edges.

Proof Suppose that K ⊆ Vi is a clique of size k in Gi. W.l.o.g. suppose thatK = {v1, . . . , vk}, and that {{u, v}, u, v ∈ K} = {e1, . . . , e(k2)}. Let b ∈ {0, 1}qbe the binary representation of i. We build K ′ as follows (see Figure 2).

– For all j ∈ {1, . . . ,(k2

)}, K ′ contains Xj and Zj (2T vertices inducing

(T2

)

edges for each gadget Fj).

– For all j ∈ {(k2

)+ 1, . . . ,m}, K ′ contains Xj and Yj . (2T vertices inducing

((T2

)+ 1) edges for each gadget Fj).

– For all u /∈ {1, . . . , k}, K ′ contains Aiu (T = n(n− k) vertices inducing(T2

)

edges).– For all h ∈ {1, . . . ,M}, and all j ∈ {1, . . . , q}, K ′ contains Xαhj

and Xβhj.

Moreover, if the jth bit of b equals 1, then K ′ contains Yβhj and Zαhj ,

otherwise K ′ contains Zβjjand Yαhj (4T vertices inducing (2

(T2

)+ 1) edges

for each pair of gadgets Fαhj and Fβhj ).


One can verify that K ′ is a set of k′ vertices inducing C ′ edges. ut

The following lemma concludes the proof.

Lemma 3 If G′ contains k′ vertices inducing at most C ′ edges, then thereexists i ∈ {1, . . . , t} such that Gi contains a k-clique.

Proof Suppose now that K ′ is a set of k′ vertices inducing C ′ edges. For a setS ⊆ V ′, we denote by tr(S) = S ∩K ′ the trace of the solution on S. For allv ∈ V ′, let µ(v) = |tr(N(v))| be the number of neighbors of v belonging toK ′.Let I = {1, . . . ,m}∪{αhj }h=1..M

j=1..q ∪{βhj }h=1..Mj=1..q be the set of all indices of gad-

gets. As in the definition of the gadgets given above, we define for all γ ∈ Ithe sets Xγ = {xγ1 , . . . , xγT }, Yγ = {yγ1 , . . . , yγT } and Zγ = {zγ1 , . . . , zγT }.We define E0 = {γ ∈ I such that ∀x ∈ tr(A), no vertex of Zγ is adjacent tox}, i.e. E0 represents the indices of all gadgets Fγ which do not induce edgesin the solution with vertices in the clique A. Then, define E1 = I \E0, whichrepresents indices of gadgets which induce at least one edge in the solutionwith a vertex of tr(A).

In the three following Claims (1, 2 and 3), we show that we can restructurethe solution inside each gadget in order to encode a solution for the k-cliqueinstance. To do so, we define the notion of safe replacement :

Let u ∈ K ′ and v ∈ V ′\K ′. We say that (K ′\{u})∪{v} is a safe replacementif we have µ(v) ≤ µ(u) if {u, v} /∈ E′ and µ(v)− 1 ≤ µ(u) if {u, v} ∈ E′. It iseasily seen that in this case (K ′\{u}) ∪ {v} does not induce more edges thanK ′. For the sake of readability, we will keep the same notations and updatethe set K ′ when applying replacements, as well as the sets E0 and E1 whenreplacing vertices of A (e.g. if there exists γ ∈ E1 such that Fγ is adjacent toa unique vertex u ∈ tr(A), and if a replacement removes u from the solution,then γ now belongs to E0).

Claim 1 K ′ can be safely modified such that for all γ ∈ I we have Xγ ⊆ K ′.

Proof Let S =⋃γ∈I Xγ . Since we have k′ > |S|, we have K ′\S 6= ∅. Suppose

that there exists γ ∈ I and i ∈ {1, . . . , T} such that xγi /∈ K ′. Recall thatyγi is the only neighbor of xγi . If yγi /∈ K ′, then we have µ(xγi ) = 0 and wecan thus safely replace any other vertex of K ′\S by xγi . Now, if yγi ∈ K ′,then µ(xγi ) = 1. Since xγi and yγi are adjacent, (K ′\{yγi }) ∪ {xγi } is a safereplacement. ut

In the following, we suppose that for all γ ∈ I we have Xγ ⊆ K ′. InClaim 2, we will only modify gadgets related to E0, while in Claim 3, we willonly modify gadgets related to E1.

Claim 2 K ′ can be safely modified such that one of the two following casesmust happen (see Figure 3):

– case A1: for all γ ∈ E0 we have tr(Zγ) = Zγ .


– case A2: for all γ ∈ E0 we have tr(Yγ) = ∅.

Proof Let us first restructure each gadget of E0 separately. For all γ ∈ E0 suchthat tr(Yγ) 6= ∅ and tr(Zγ) 6= Zγ , let j0 = max{j ∈ {1, . . . , T} : yγj ∈ tr(Yγ)}and let j1 be such that zγj1 /∈ tr(Zγ). Recall that Claim 1 ensures that xγj0 isin K ′. If j0 6= 1, then µ(yγj0) = y + z + 1, where y = |N(yγj0) ∩ tr(Yγ)| andz = |N(yγj0) ∩ tr(Zγ)|. On the other side, we have µ(zγj1) ≤ y + z + 1 (moreprecisely, µ(zγj1) = y+z+1 if yγ1 ∈ K ′, and µ(zγj1) = y+z if yγ1 /∈ K ′). Roughlyspeaking, when replacing yγj0 by zγj1 in K ′, one edge will be “lost” due to thevertex of Xγ , but one edge will be “won” due to yγ1 . Hence µ(zγj1) ≤ µ(yγj0)and (K ′\{yγj0}) ∪ {z

γj1} is a safe replacement. If j0 = 1, then it means that

tr(Yγ) = {yγ1 }. Suppose that there exists j1 such that zγj1 /∈ tr(Zγ). We haveµ(yγ1 ) = z + 1 where z = |N(yγ1 ) ∩ tr(Zγ)|, and µ(zγj1) = z + 1. Here again(K ′\{yγ1 }) ∪ {zγj1} is a safe replacement. After all these replacements, givenany γ ∈ E0, tr(Yγ) 6= ∅ implies that tr(Zγ) = Zγ .Then, we proceed to replacements between gadgets Fγ , γ ∈ E0. If one canfind a, b ∈ E0 such that tr(Ya) 6= ∅ and tr(Zb) 6= Zb, then let j0 be such thatyaj0 ∈ tr(Ya) and let j1 be such that zbj1 /∈ tr(Zb). We have µ(yaj0) ≥ T + 1 and

µ(zbj1) ≤ T − 1. Thus, (K ′\{yaj0}) ∪ {zbj1} is a safe replacement.These replacements end either when tr(Yγ) = ∅ for all γ ∈ E0 or whentr(Zγ) = Zγ for all γ ∈ E0, which achieves the proof of Claim 1. ut

Claim 3 K ′ can be safely modified such that one of the two following casesmust happen (see Figure 3):

– case B1: for all γ ∈ E1 we have tr(Yγ) = Yγ .– case B2: for all γ ∈ E1 we have tr(Zγ) = ∅.

Proof The proof is roughly based on the fact that replacing a vertex of Zγby a vertex of Yγ will remove from K ′ at least one edge with one enpoint inA and add to K ′ one edge with one endpoint in Xγ . Let us formally proveClaim 3. Similarly to the proof of Claim 2, we first restructure each gadgetof E1 separately: for all γ ∈ E1 such that tr(Zγ) 6= ∅ and tr(Yγ) 6= Yγ , let j0be such that yγj0 /∈ K ′ and let j1 be such that zγj1 ∈ tr(Zγ). Recall that by

definition of E1, there exists i, j ∈ {1, . . . , n} such that zγj1 is adjacent to aji .We have µ(zγj1) ≥ y + z + 1, where y = |N(zγj1) ∩ Yγ | and z = |N(zγj1) ∩ Zγ |.On the other side, we have µ(yγj0) ≤ z + y + 2 (indeed, |N(yγj0) ∩ Zγ | = z + 1,|N(yγj0) ∩ Yγ | ≤ y and |N(yγj0) ∩Xγ | = 1). Since {yγj0 , z

γj1} ∈ E′, it holds that

(K ′\{zγj1}) ∪ {yγj0} is a safe replacement. After all these replacements, given

any γ ∈ E1, tr(Zγ) 6= ∅ implies that tr(Yγ) = Yγ .We now proceed to replacements between gadgets Fγ , γ ∈ E1. If one can finda, b ∈ E1 such that tr(Za) 6= ∅ and tr(Yb) 6= Yb, then let j0 be such thatybj0 /∈ tr(Yb) and let j1 be such that zaj1 ∈ tr(Za). We have µ(zaj1) ≥ T + 1 and

µ(ybj0) ≤ T − 1. Thus (K ′ \ {zaj1}) ∪ {ybj0} is a safe replacement.As previously, the replacements ends either when tr(Yγ) = Yγ for all γ ∈ E1

or when tr(Zγ) = ∅ for all γ ∈ E1. ut


γ ∈ E0

Dγ

case A1

γ ∈ E0

Dγ

case A2

γ ∈ E1

Dγcase B1

γ ∈ E1

Dγ

case B2

Xγ

Yγ

Zγ

1

Fig. 3: Schema of different cases. Shaded rectangles represent part of K ′.

We now define each γ ∈ I the set of vertices Dγ ⊆ Yγ ∪ Zγ that have tobe replaced, depending on the case we are in.

– case A1: for all γ ∈ E0, Dγ = Yγ ∩K ′– case A2: for all γ ∈ E0, Dγ = Zγ \K ′– case B1: for all γ ∈ E1, Dγ = Zγ ∩K ′– case B2: for all γ ∈ E1, Dγ = Yγ \K ′

Notice that if Dγ = ∅ for all γ ∈ E0 (resp. for all γ ∈ E1), then cases A1 andA2 (resp. B1 and B2) collapse. If such a case happens for all γ ∈ I, we will beable to conclude, as we will see in Claim 6. To terminate the proof, it is nowsufficient to combine cases A1, A2 with cases B1, B2, leading to four differentsituations. We first settle cases (A2, B2), (A2, B1), and (A1, B2), by provingthat they either reduce to (A1, B1) or directly to the case where Dγ = ∅ forall γ ∈ I.

– Case A2 and B2: let ∆0 =∑γ∈E0

|Dγ |, ∆1 =∑γ∈E1

|Dγ |, and ∆ =∆0 + ∆1, and suppose that ∆ > 0 (otherwise we conclude by Claim 6).Since tr(A) = T + ∆ > T , we have µ(u) ≥ T for all u ∈ tr(A). On theother hand, for all γ ∈ I such that there exists v ∈ Dγ , we claim thatµ(v) ≤ T . Indeed, if γ ∈ E1, then since v ∈ Yγ , tr(N(v)∩Yγ) ≤ T − 1, andtr(N(v) ∩Xγ) = 1. Now, if γ ∈ E0, then observe that v is not adjacent totr(A) by definition of E0, and thus tr(N(v)) ( Zγ which implies µ(v) < T .Thus (K ′\{u}) ∪ {v} is a safe replacement. Since before this replacementwe had tr(A) = T +∆, it is clear that we can repeat this replacement (i.e.K ′\{u}∪ {v} where u ∈ tr(A) and v ∈ Dγ for some γ ∈ I) ∆ times safely.At this point, the updated value of ∆ is 0, i.e. Dγ = ∅ for all γ ∈ I. Wethen conclude by Claim 6.

– Case A2 and B1: if there exists γ ∈ E0 such that there exists u ∈ Dγ , thenµ(u) < T (as in the case A2 and B2). If such a vertex exists, then either|tr(A)| > T or there exists γ′ ∈ E1 such that there exists v ∈ Dγ′ ⊆ Zγ . Inthe first case, for all x ∈ tr(A) we have µ(x) ≥ T (as in the case A2 andB2), and (K ′ \{x})∪{u} is a safe replacement. In the second case we haveµ(v) > T since tr(N(v) ∩ Yγ′) = Yγ′ , and N(v) ∩ tr(A) 6= ∅ by definitionof E1. This, here again (K ′ \ {v}) ∪ {u} is a safe replacement.After these replacements we must have Dγ = ∅ for all γ ∈ E0, and we can


apply the case A1 and B1.

– Case A1 and B2: if there exists γ ∈ E1 such that there exists u ∈ Dγ , thenµ(u) < T (as in the case A2 and B2). If such a vertex exists, then either|tr(A)| > T or there exists γ′ ∈ E0 such that there exists v ∈ Dγ′ . In thefirst case, for all x ∈ tr(A) we have µ(x) ≥ T , and (K ′ \ {x}) ∪ {u} isa safe replacement. In the second case we have µ(v) > T and here again(K ′ \ {v}) ∪ {u} is a safe replacement.After these replacements we must have Dγ = ∅ for all γ ∈ E1, and we canapply the case A1 and B1.

Concerning the remaining case (A1, B1), we need an intermediate result(Claim 4) saying that in this case, the solution must have a particular shapewithin A. We then prove (Claim 5) that (A1, B1) reduces to the case whereDγ = ∅ for all γ ∈ I, which is finally handled in Claim 6.

Claim 4 If cases A1 and B1 happen, then there exists i ∈ {1, . . . , t} such thattr(A) ⊆ Ai, i.e. the solution K ′ intersects A in exactly one clique Ai.

Proof Let ∆ =∑γ∈I |Dγ |, and suppose by contradiction that there exists

i, j ∈ {1, . . . , t} with i 6= j such that K ′ ∩Ai 6= ∅ and K ′ ∩Aj 6= ∅. First, sincewe are in case A1 and B1, the number of edges induced by the solution in eachgadget is at least

(T2

). Then, let S (resp. S) be the number of pairs of gadgets

(Fαhp , Fβhp ), for some p ∈ {1, . . . , q} and h ∈ {1, . . . ,M}, corresponding to anindex for which the binary representations of i and j have the same value(resp. different values). Recall that S+ S = Mq. Then, since i 6= j, the binaryrepresentations of i and j must differ on at least one bit, which implies S ≥M .Let us count the number of edges induced by the solution by analyzing eachpair of gadget, depending whether they correspond to a bit value shared bythe binary representation of i and j or not. In the following, we will say thata given pair of gadget F and F ′ adds at least x edges if the trace of K ′ insideF ∪ F ′ covers at least x edges (we recall that an edge {u, v} is covered by aset S if S ∩ {u, v} 6= ∅). By considering the edges covered by every mutuallydisjoint pairs of gadgets together with the edges induces by tr(A), we willobtain a lower bound on the number of edges induced by K ′.Let p ∈ {1, . . . , q} be an index such that the binary representations of i and jcoincide at index p. Then, for all h ∈ {1, . . . ,M}, three cases may happen:

– Yαhp ⊆ K ′ and Yβhp ⊆ K ′. In this case the pair of gadgets adds at least

2(T2

)+ 2 edges.

– Yαhp ⊆ K ′ and Zβhp ⊆ K ′ (or the contrary). In this case the pair of gadgets

adds at least 2(T2

)+ 1 edges.

– Zαhp ⊆ K ′ and Zβhp ⊆ K ′. In this case the pair of gadgets adds at least

2(T2

)+ T edges, since either Zαhp or Zβhp is adjacent to at least one vertex

of tr(Ai).

Hence, in all three cases the solution in each pair of such gadgets covers atleast (2

(T2

)+ 1) edges.


Let us now focus on some index p ∈ {1, . . . , q} for which the binary rep-resentations of i and j differ. Then, for all h ∈ {1, . . . ,M}, notice that bothZαhp and Zβhp are adjacent to at least one vertex in tr(Ai)∪ tr(Aj). Here againthree cases may happen:

– Yαhp ⊆ K ′ and Yβhp ⊆ K ′. In this case the pair of gadgets adds at least

2(T2

)+ 2 edges.

– Yαhp ⊆ K ′ and Zβhp ⊆ K ′ (or the contrary). In this case the pair of gadgets

adds at least 2(T2

)+ T + 1 edges.

– Zαhp ⊆ K ′ and Zβhp ⊆ K ′. In this case the pair of gadgets adds at least

2(T2

)+ 2T edges.

Hence, in all three cases the solution in each pair of such gadgets covers atleast (2

(T2

)+ 2) edges. In addition, it is easily seen that the number of edges

induced by tr(A) is(T2

)−(∆2

)−∆(T −∆), since it is a clique of size (T −∆).

To resume:

– tr(A) induces ((T2

)−(∆2

)−∆(T −∆)) edges.

– Each gadget (both from the edge or the instance selector) adds at least(T2

)

edges (there are (m+ 2Mq) gadgets), and more precisely:– Each pair of gadgets corresponding to a shared bit value of the binary

representation of i and j adds (2(T2

)+ 1) edge (i.e. one more than the

”normal” ones). There are S such pairs of gadgets.– Each pair of gadgets corresponding to a different bit value of the binary

representation of i and j adds (2(T2

)+ 2) edge (i.e. two more than the

”normal” ones). There are S such pairs of gadgets.

Thus we have:

E(K ′) ≥(T

2

)−(∆

2

)−∆(T −∆) +

(T

2

)(m+ 2Mq) + S + 2S

=

(T

2

)−(∆

2

)−∆(T −∆) +

(T

2

)(m+ 2Mq) +Mq + S

≥(T

2

)−(∆

2

)−∆(T −∆) +

(T

2

)(m+ 2Mq) +Mq +M

And thus

E(K ′)− C ′ ≥M −m+

(k

2

)−(∆

2

)−∆(T −∆)

Since M = n6, we have E(K ′) > C ′ for some large value of n (we may assumethat the value of n is above some given constant in the input instances), whichis impossible. ut

We are now ready to deal with the last case:

Claim 5 If cases A1 and B1 happen, then Dγ = ∅ for all γ ∈ I.


Proof Recall that for γ ∈ E0, Dγ = tr(Yγ), while for γ ∈ E1, Dγ = tr(Zγ) let∆ =

∑γ∈I |Dγ |, and suppose that ∆ > 0 (otherwise we conclude by Claim 6).

Let us count the number of edges induced by such a solution. To do so, wecount separately the number of edges of the solution induced by vertices of A,and the number of edges of the solution with at least one endpoint in somegadget. By abuse of language, we will say in the following that a gadget F(or the trace of the solution in F ) covers an edge {u, v} if {u, v} ⊆ K ′ and{u, v} ∩ tr(F ) 6= ∅ (in particular, cover will never refer to an edge that is notinduced by K ′) First, it is clear that |tr(A)| = T −∆, and thus the numberof edges induced by tr(A) is (

(T2

)−(∆2

)− ∆(T − ∆)) since A is a clique. In

addition, since we are in case A1 and B1, the trace of the solution in all gadgets(both from the edge or the instance selector) covers at least

(T2

)edges. More

precisely, for each gadget γ ∈ I, we have:

– if γ ∈ E0, then since each vertex of Yγ is connected to all vertices of Zγ and

to one vertex of Xγ , we have that tr(Fγ) covers at least ((T2

)+ |Dγ |(T +1))

edges (see Figure 3).– if γ ∈ E1, then since each vertex of Zγ is connected to all vertices of Yγ ,

and to at least one vertex of tr(A), we have that tr(Fγ) covers at least

((T2

)+ 1 + |Dγ |(T + 1)) edges (recall that if Dγ = ∅ the gadget covers

exactly ((T2

)+ 1) edges).

Summing up all gadgets, the solution among all gadgets covers at least ((T2

)(m+

2Mq) + |E1|+∆(T + 1)) edges.We define Ee0 = E0 ∩ {1, . . . ,m} the restriction of E0 to the edge selector andEb0 = E0 \ Ee0 the restriction of E0 to the instance selector, as well as thecorresponding sets Ee1 = E1 ∩ {1, . . . ,m} and Eb1 = E1 \ Ee1 .By Claim 4, there exist i ∈ {1, . . . , t} such that tr(A) ⊆ Ai. This implies that|Eb0| = |Eb1| = Mq (roughly speaking, for each pair of gadgets of the instanceselector, only one of the two is connected to Ai and thus to tr(A), depending onthe corresponding bit value). Thus, we have |E1| = Mq+|Ee1 | = Mq+m−|Ee0 |.Combining all these, we obtain:

E(K ′) ≥(T

2

)−(∆

2

)−∆(T−∆)+

(T

2

)(m+2Mq)+∆(T+1)+Mq+m−|Ee0 |

And thus:

E(K ′)− C ′ ≥(k

2

)+∆(T + 1)− |Ee0 | −

(∆

2

)−∆(T −∆)

=∆(∆+ 3)

2+

(k

2

)− |Ee0 |

Thus, since we supposed E(K ′)− C ′ ≤ 0 it implies

|Ee0 | ≥∆(∆+ 3)

2+

(k

2

)(1)


On the other hand, let us deduce the maximum number of vertices of Gi whichcan induce all edges of Ee0 . Informally, for a vertex u ∈ Vi to be able to inducean edge of Ee0 , its corresponding clique Aiu must not intersect K ′. Since the sizeof all cliques Aiu is n, and since |Ai \tr(A)| = nk+∆, the number of vertices of

Gi which can induce all edges of Eb0 is at most b |Ai\tr(A)|n c = k + b∆n c. Hence

we have |Ee0 | ≤(k+b∆n c

2

), and:(k + b∆n c

2

)≥ ∆(∆+ 3)

2+

(k

2

)(2)

If ∆ < n, then b∆n c = 0 and by Equation 2 we have ∆ = 0. If ∆ ≥ n, thenit contradicts inequality (1) since we have by definition |Ee0 | ≤ m. Hence, wemust have ∆ = 0 and the result follows by Claim 6.

Finally, the following result terminates the proof of Lemma 3 and thus theproof of Theorem 5.

Claim 6 If Dγ = ∅ for all γ ∈ I, then there exists i ∈ {1, . . . , t} such that Gcontains a clique of size k.

Proof By construction, we have |tr(A)| = T and |tr(Fγ)| = 2T for all γ ∈ I.

Thus, E(tr(A)) =(T2

)and E(tr(Fγ)) =

(T2

)+1 if γ ∈ E1, and E(tr(Fγ)) =

(T2

)

if γ ∈ E0. Hence, we have E(K ′) ≥(T2

)+(T2

)(m+ 2Mq) + |E1|.

By Claim 4 (recall that Dγ = ∅ for all γ ∈ I is a special case of cases A1and B1), there exists i ∈ {1, . . . , t} such that tr(A) ⊆ Ai. Thus, there are atmost Mq gadgets among the instance selector whose respective traces are notadjacent to tr(A), and which can belong to E0. This implies that there are atleast Mq gadgets among the instance selector which must belong to E1. LetEe0 = E0∩{1, . . . ,m} be the restriction of E0 to the edge selector, and similarlyEe1 = E1 ∩ {1, . . . ,m}. Hence, by the foregoing, we have |E1 \ Ee1 | ≥ Mq. Bydefinition of K ′ we have E(K ′) ≤ C ′, we have that |E1| ≤ m−

(k2

)+Mq, and,

since |E1 \ Ee1 | ≥ Mq, we have |Ee1 | ≤ m −(k2

), which implies |Ee0 | ≥

(k2

). In

addition, each gadget j ∈ Ee0 corresponding to the edge eij = {u, v} of Gi is

adjacent to the cliques Aiu and Aiv, which must be such that tr(Aiu) = tr(Aiv) =∅ by definition of E0. However, since |tr(A)| = |tr(Ai)| = T , the number ofcliques Aix, 1 ≤ x ≤ n, such that tr(Aix) = ∅ is at most n − bTn c = k. This

proves that |Ee0 | =(k), corresponding to

(k2

)edges of Gi induced by exactly k

vertices of Gi, i.e. Gi contains a clique of size k. ututut

6 Dynamic Programming Algorithm in Interval Graphs forSparsest k-Subgraph

In this section we provide a dynamic programming for Sparsest k-Subgraphin interval graphs, leading to an FPT algorithm parameterized by C, the num-ber of edges in the solution. Remark that C is a smaller parameterization than


the natural parameterization by k, in the sense that we may always assumethat C ≤

(k2

). In addition, for C = 0, we obtain the known polynomial algo-

rithm for Independent Set in interval graphs. Notice that the complexity(NP-hard versus Polynomial) is still unknown for Sparsest k-Subgraph ininterval graphs. However, we believe that this question is a tough open prob-lem, similar to the corresponding one for Densest k-Subgraph in the samegraph class [10].

6.1 Preliminaries

In the following, we will assume that the interval model of the input graphis given. In addition, we will make no distinction between a vertex and itscorresponding interval, as well as we will make no distinction between edgesin the graph and overlaps in the corresponding interval model.

For the rest of the section, G = (V,E) denotes the input graph of theproblem, with V = {v1, . . . , vn}. For all u ∈ V , we denote by Iu the intervalcorresponding to vertex u, and I = {Iu : u ∈ V }. Without loss of generality,we may assume that all endpoints are pairwise distinct. Given I ∈ I, wedenote by r(I) ∈ R (resp. `(I) ∈ R) the right (resp. left) endpoint of I. Byextension, for any set S ⊆ I, we define `(S) = minI∈S `(I) (resp. r(S) =maxI∈S r(I)), where, by convention, `(∅) = −∞, r(∅) = +∞. Then, for a non-empty subset S ⊆ I, we define first(S) = arg minI∈S `(I) (resp. last(S) =arg maxI∈S r(S)). Finally, for a non-empty subset S ⊆ I, we define next(S) tobe the interval having the leftmost right endpoint among all intervals I suchthat r(S) < `(I) (informally, next(S) is the first interval after S in the orderingof I not being adjacent to S). Unless otherwise stated, we assume that V issorted by increasing right endpoints of the intervals (i.e. for all i ∈ {2, . . . , n}we have r(Ivi−1) < r(Ivi).

Definition 1 For u ∈ V , we say that a subset of vertices Z is left-optimal foru if the following holds:

(i) Iu = first(Z) (i.e. Iu is the interval of Z with the leftmost left endpoint);(ii) the sequence of the right endpoints of intervals of the vertices of Z, sorted

decreasingly, is lexicographically minimum possible among subsets Z ′

satisfying (i) and such that |Z ′| = |Z| and |E(G[Z ′])| ≤ |E(G[Z])|.

We have the following lemma, whose proof is straighforward:

Lemma 4 Suppose there exists a set S of k vertices inducing at most C edges,and let S1, . . . , Sq be its connected components. Then there exists a set S′ of kvertices inducing at most C edges, having q connected components S′1, . . . , S

′q

such that for all i ∈ [q], S′i is left-optimal for first(Si).

We are now ready to prove the main result of this section:

Theorem 6 Sparsest k-Subgraph in interval graphs can be solved in O∗(2C).


Proof Let us first define, for every u ∈ V , the set Fu of all subsets S satisfyingthe following properties:

(i) first(S) = Iu;(ii) S induces a connected subgraph;(iii) |S| ≤ C + 1 and |E(G[S])| ≤ C;(iv) S is left-optimal for u.

The first step of the algorithm consists in applying, for each u ∈ V , a branchingprocedure which outputs a set Su of subsets containing Fu. This proceduretakes as parameter a set S of intervals (initially S = {u}), and which, at eachstep, consider the set of candidates

cand = {I ∈ I \ S : `(S) < `(I)}If cand = ∅, |S| = C + 1 or |E(G[S])| = C, then we output S. Otherwise,let η = maxI∈cand |N(I)∩ S|, and, for all t = {1, . . . ,max{η, C − |E(G[S])|}},make a recursive call of the algorithm with parameter S ∪ {It}, where It isthe interval of cand with leftmost right endpoint among those having exactlyt neighbors in S. Denote by Su the set of all sets returned by the algorithm,and let us prove Fu ⊆ Su. For any S ∈ Fu if we sort S = {s1, . . . , s|S|} so thats1 = Iu and {Is2 , . . . , Is|S|} have increasing right endpoints, we can prove byinduction that for every i ∈ {1, . . . , |S|}, there is a call of the algorithm withinput {s1, . . . , si}. The base case corresponds to the first call with parameter{Iu}, and, for input {s1, . . . , si−1}, one can observe that Isi is the intervalwith leftmost right endpoint among those of cand having the same numberof neighbors in {Is1 , . . . , Isi−1

}. Indeed, if Is∗ 6= Isi was this interval, thenS′ = (S \ {si}) ∪ {s∗} would clearly be smaller than S (w.r.t. a lexigraphicorder on the right endpoints ordered decreasingly). Then, since r(Is∗) < r(Isi),for any j > i we have {s∗, sj} ∈ E =⇒ {si, sj} ∈ E. Using the fact that|Ni−1(s∗)| = |Ni−1(si)|, it proves that |E(G[S′])| ≤ |E(G[S])|, and S wouldnot be left-optimal for u, a contradiction.

Let us analyze the size of Su: at each step, the procedure makes a recursivecall for each t ∈ {1, . . . ,max{η, C−|E(G[S])|}} with as parameter a set havingt more edges. Since it stops whenever the input set has C edges, we obtain|Su| = O(2C) (remark that we cap t by C − |E(G[S])|, so that the input setnever induces more than C edges).

Now, for each ke ≤ k and Ce ≤ C, if Su contains a set of ke intervals in-ducing Ce edges, then let Suke,Ce be the one with the lexicographically smallestsequence of right endpoints (ordered by decreasing right endpoints). Observethat by definition of Fu and since Fu ⊆ Su, we have Fu = {Suke,Ce , ke ≤k and Ce ≤ C}. Moreover, this construction can be carried in O∗(2C) time.

We now perform a dynamic programming running through I sorted by in-creasing right endpoints, which, given i ∈ [n], ke ≤ k, Ce ≤ C, outputs a subsetof ke intervals inducing at most Ce edges included in the set of all intervals Isuch that r(Ivi) ≤ r(I) if such a set exists, or “No” otherwise. In Algorithm 1,we describe a function called DP to achieve this, whose correctness is providedby the following lemma:


Algorithm 1 DP (i, ke, Ce)if ke = 0 then

return ∅if i = n then

return “No”for j ← i to n do

for all k′ ≤ ke and C′ ≤ Ce doif S

ujk′,C′ exists then

ui∗ ← next(Sujk′,C′ )

S ← DP (i∗, ke − k′, Ce − C′)if S 6=“No” then

return Sujk′,C′ ∪ S

return “No”

Lemma 5 DP (i, ke, Ce) is correct.

Proof Suppose first that there exists a set S∗ of ke interval inducing at mostCe edges included in the subset of all intervals I such that r(Ivi) ≤ r(I),and let S∗1 , . . . , S

∗q be the connected components induced by this set. Assume

w.l.o.g. that for all i ∈ [q], S∗i is left-optimal for first(S∗i ). Then observe that:

(i) S∗1 ⊆ Ffirst(S∗1 );(ii) S∗2 ∪ · · · ∪ S∗q is a set of ke − |S∗1 | intervals inducing Ce −E(G[S∗1 ]) edges

included in the set of all intervals I such that r(Inext(S∗1 )) ≤ r(I).

Conversely, ifDP (i, ke, Ce) outputs a set S, then by definition of the algorithm,S = S

ujk′,C′ ∪ DP (i∗, ke − k′, Ce − C ′) for some j ∈ {i, . . . , n} , with ui∗ =

next(Sujk′,C′). By definition, S

ujk′,C′ is a set of k′ intervals among intervals I

such that r(Ivi) ≤ r(Ivj ) ≤ r(I), and which induces at most C ′ edges. On theother hand, DP (i∗, ke − k′, Ce − C ′) returns a set of ke − k′ intervals amongintervals I such that r(Ivi∗ ) ≤ r(I) and which induces at most Ce −C ′ edges.Thus, S

ujk′,C′ ∩ DP (i∗, ke − k′, Ce − C ′) = ∅, which implies that S is a set of

ke intervals among all intervals I such that r(Ivi) ≤ r(I) and which inducesat most Ce edges, as desired. ut

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A Formal Definitions for Kernel Lower Bounds

In order to establish kernel lower bounds, we use the technique of cross-composition ofBodlaender et al. [2]:

Definition 2 (Polynomial equivalence relation [2]) An equivalence relation R on Σ∗

is called a polynomial equivalence relation if the two following conditions hold:

– There is an algorithm that given two strings x, y ∈ Σ∗, decides whether x and y belongto the same equivalence class in (|x|+ |y|)O(1) time.

– For any finite set S ⊆ Σ∗, the equivalence relation R partitions the elements of S intoat most (maxx∈S |x|)O(1) classes.

Definition 3 (OR-cross-composition [2]) Let L ⊆ Σ∗ be a set and let Q ⊆ Σ∗ ×N be a parameterized problem. We say that L OR-cross-composes into Q if there is apolynomial equivalence relation R and an algorithm which, given t strings belonging to thesame equivalence class of R, computes an instance (x∗, k∗) ∈ Σ∗ ×N in time polynomial in∑t

i=1 |xi| such that:

– (x∗, k∗) ∈ Q⇔ xi ∈ L for some 1 ≤ i ≤ t– k∗ is bounded by a polynomial in maxti=1|xi|+ log t

Theorem 7 ([2]) If some set L ⊆ Σ∗ is NP-hard and L OR-cross-composes into the pa-rameterized problem Q, then there is no polynomial kernel for Q unless NP ⊆ coNP/poly.

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Parameterized Complexity of the Sparsest -Subgraph Problem in …bougeret/publis/ToCS-revised1-16-04-15.pdf · 2019-09-25 · Presentation of the problem. Given a simple undirected