International audienceUne instance du problème est un graphe, un ensemble F de sommets interdits et un ensemble R de sommets obligatoires. Nous montrons que construire un vertex cover, connexe ou pas, de taille minimale, contenant tous les sommets de R et aucun sommet de F, peut être 2-approché (s'il existe). Nous montrons aussi que décider s'il existe ou pas un ensemble dominant indépendant contenant tout R et aucun sommet de F est NP-complet

Delbot, Francois

Laforest, Christian

Phan, Raksmey

HAL Clermont Université

Graphs with Forbidden and Required VerticesFrancois Delbot, Christian Laforest, Raksmey PhanTo cite this version:Francois Delbot, Christian Laforest, Raksmey Phan. Graphs with Forbidden and RequiredVertices. ALGOTEL 2015 - 17e`mes Rencontres Francophones sur les Aspects Algorithmiquesdes Te´le´communications, Jun 2015, Beaune, France. <hal-01148233>HAL Id: hal-01148233https://hal.archives-ouvertes.fr/hal-01148233Submitted on 4 May 2015HAL is a multi-disciplinary open accessarchive for the deposit and dissemination of sci-entific research documents, whether they are pub-lished or not. The documents may come fromteaching and research institutions in France orabroad, or from public or private research centers.L’archive ouverte pluridisciplinaire HAL, estdestine´e au de´poˆt et a` la diffusion de documentsscientifiques de niveau recherche, publie´s ou non,e´manant des e´tablissements d’enseignement et derecherche franc¸ais ou e´trangers, des laboratoirespublics ou prive´s.Sommets interdits et obligatoiresFrancois Delbot1 and Christian Laforest2 and Raksmey Phan21Universite´ Paris Ouest Nanterre / LIP6, CNRS UMR7606 - 4 place Jussieu, 750005 Paris, Francefrancois.delbot@u-paris10.fr2Universite´ Blaise Pascal / LIMOS, CNRS UMR 6158 - 24 avenue des Landais, 63173 Aubibe`re Cedex, France{laforest,phan}@isima.frUne instance du proble`me est un graphe, un ensemble F de sommets interdits et un ensemble R de sommets obligatoires.Nous montrons que construire un vertex cover, connexe ou pas, de taille minimale, contenant tous les sommets de R etaucun sommet de F , peut eˆtre 2-approche´ (s’il existe). Nous montrons aussi que de´cider s’il existe ou pas un ensembledominant inde´pendant contenant tout R et aucun sommet de F est N P -complet.Keywords: Graphs, Vertex Cover, Minimum Independent Dominating Set, Approximation Algorithms.1 IntroductionIn this paper we consider undirected, unweigted graphs G = (V ,E) where V is its set of vertices and E itsset of edges. We also distinguish two subsets of vertices: F ⊆ V the set of Forbidden vertices and R ⊆ Vthe set of Required vertices.The generic problem addressed in this paper is to construct a subset S of vertices of G having somegiven properties (for example S can be a vertex cover of G or a dominating set of G , etc) and such that no(forbidden) vertex of F is in S (F ∩S = /0) and every (required) vertex of R must be in S (R⊆ S ).An instance is given by the triplet (G ,F,R). Once the properties of the solution are defined, the mainobjectives are of twofold: (1) Decide whether a solution for the instance (G ,F,R) exists; if it is the case (2)try to minimize its size. Of course R∩F = /0 otherwise no solution S is possible.Notations. We give here some notations and definitions that will be useful throughout the paper. LetG = (V ,E) be any undirected, unweighted graph. An edge between vertices u and v of G is noted uv; inthis case u and v are neighbors. Let S ⊆ V a set of vertices of G . We note G [S] the graph induced by S inG : its set of vertices is S and its edges are the ones of G between vertices of S: {uv : u ∈ S,v ∈ S,uv ∈ E}.Set S is a vertex cover of G if each edge e= uv of G is covered by (i.e. contains) at least a vertex of S: u∈ Sor v ∈ S or both. S is a connected vertex cover of G if S is a vertex cover of G and if G [S] is a connectedgraph. Set S is an independent set of G if G [S] contains non edges. S is a dominating set of G if for allu ∈V −S, u has at least one neighbor in S: ∀u ∈V −S,∃v ∈ S,uv ∈ E . S is an independent dominating setof G if S is a dominating set of G and also an independent set of G .Related works. Recently several papers have been published concerning the construction of structures ingraphs under constraints like conflicts of edges: if two edges e and e′ are in conflict they cannot be partof the same structure (at most one can be in the structure, not both). This has been investigated for paths,trees and Hamiltonian paths in [DPSW11, KLM13a, KLM13b, Sze03]. The same kind of study has beencarried out when the conflicts concern pairs of vertices (see [Kov13] for example). All these constraints areconditional: if an edge (or a vertex) is in the structure then the other one cannot be part of it. This can beused in applications to take into account incompatible devices in a network for example.In this paper, there is no such conditional exclusion. If a vertex is in the forbidden set F then it cannot bein the structure. If it is in the required set R then it must be in the structure. This changes the nature of theconstraints and type of applications.Francois Delbot and Christian Laforest and Raksmey PhanMotivations for network monitoring. Some monitoring devices or software can be installed on somenodes of a network; these equipped nodes will monitor all their incident links (vertex cover or connectedvertex cover) or neighbors (dominating set). However, some nodes are incompatible with the device orsoftware and cannot be equipped (forbidden nodes). On the contrary, some nodes must be in the set becausethey have particular properties (required nodes).2 The Vertex Cover Problem in (G ,F,R)Given (G ,F,R) a Vertex Cover with Forbidden and Required Vertices (VCwFaRV) S is a vertex cover of Gsuch that F ∩S = /0 and R⊆ S . We first make the following remark, easy to prove.Remark 1 Let (G ,F,R) be any instance. If F ∩R 6= /0 or if F is not an independent set of G then (G ,F,R)has no VCwFaRV. Otherwise S = V −F is a VCwFaRV of (G ,F,R).This remark induces the fact that deciding whether there is a VCwFaRV in (G ,F,R) is polynomial.Optimizing the size of a VCwFaRV is hard, since the very particular case R = F = /0 is the classical N P -complete vertex cover problem.Theorem 1 When (G ,F,R) has at least a VCwFaRV then a minimum size one can be 2-approximated witha polynomial approximation algorithm.Proof. Let (G ,F,R) be any instance such that F ∩R = /0 and F is an independent set of G . Let N(F) bethe set of neighbors of the vertices of F in G : N(F) = {v : uv ∈ E ,u ∈ F} (as F is an independent set, atmost one extremity of any edge can be in F). Any VCwFaRV must contain N(F) otherwise edges incidentto F cannot be covered. By definition any VCwFaRV must also contain R, thus it must contain R∪N(F).Let V ′ =V −(F ∪R∪N(F)) and G ′ =G [V ′] be the graph induced by the vertices of V ′ in G . Let S ′ beany vertex cover of G ′ obtained by applying any polynomial 2-approximation algorithm (see [DLP13] forrecent new ones) and let S = S ′ ∪R∪N(F) be the final result. By polynomial construction, S contains Rand contains no vertex of F . S covers all edges of E : edges incident to F are covered by vertices of N(F),edges incident to R are covered by R (a part of the solution S ) and all the other edges are covered by S ’. Sis then a VCwFaRV of (G ,F,R). Let us prove now the approximation ratio.Let S∗ be a minimum size VCwFaRV of (G ,F,R) and OPT = |S∗| be its size. Let T = S∗− (R∪N(F))(thus we have OPT = |T |+ |R∪N(F)|). We prove that T is an optimal vertex cover ofG ′=G [V ′]. Supposethat it is false: Thus a vertex cover T2 of G ′ with |T2|< |T | exists; Hence S2 = T2∪R∪N(F) is a VCwFaRVof size |S2| ≤ |T2|+ |R∪N(F)| < |T |+ |R∪N(F)| = |S∗|. This is in contradiction with the optimality ofS∗.As S ′ is a 2-approximation of an optimal vertex cover of G ′ we obtain: |S ′| ≤ 2|T |. Moreover, as T isonly a part of S∗ we also have: |T | ≤ OPT . Combining all these points we get: |S | ≤ |S ′|+ |R∪N(F)| ≤2|T |+ |R∪N(F)| ≤ OPT + |T |+ |R∪N(F)|= 2OPT . 23 The Connected Vertex Cover Problem in (G ,F,R)Given (G ,F,R) a Connected Vertex Cover with Forbidden and Required Vertices (CVCwFaRV) S is a vertexcover of G such that G [S ] is connected, F ∩S = /0 and R⊆ S . We suppose that G is connected.Proposition 1 Let G be a connected graph. (G ,F,R) contains a CVCwFaRV if and only if F is an inde-pendent set of G , F ∩R = /0 and G [V −F ] is connected.Proof. If (G ,F,R) contains a CVCwFaRV noted S , this means that F ∩R= /0 and F is an independent setof G (otherwise edges of G [F ] could not be covered). Moreover, as S is a vertex cover of G , any edge uvof G has at least one extremity in S . Let us consider any vertex u ∈ V −F : u is in S or has a neighbor in S(since G is connected, u has at least a neighbor) or both. Thus for any u,v ∈ V −F there is a path betweenu and v in G [V −F ] (through the connected graph G [S ]) which is so connected.Now suppose that G [V −F ] is connected, that F ∩R = /0 and that F is an independent set of G . Thismeans that S = V −F contains no vertex of F and contains all vertices of R. Moreover, let uv be any edgeProblmes de graphes avec sommets interdits et obligatoiresof G . As F is an independent set of G , u ∈ S or v ∈ S (or both). This means that S is a vertex cover of G .In conclusion, S is a CVCwFaRV of (G ,F,R). 2By the previous result it is polynomial to decide whether there exists a CVCwFaRV of (G ,F,R) and, if itis the case, to construct one (namely V −F). Like in Section 2, if F = R = /0, we get the classic connectedvertex cover N P -complete problem.Theorem 2 When (G ,F,R) has at least a CVCwFaRV then a minimum size one can be 2-approximatedwith a polynomial algorithm.Proof. Proposition 1 gives conditions that can be checked in polynomial time under which (G ,F,R)contains a CVCwFaRV . We suppose now that these conditions are verified.Let us consider the problem where F = /0 (no forbidden vertices). Let G = (V ,E) be the graph and Rbe the set of required vertices. Let us construct G+ = (V+,E+) the graph G in which we add for eachvertex r of R a new proper neighbor r+ (of degree 1, that is a leaf of G), and the associated new edgerr+. G+ is also connected. Now, use any polynomial approximation algorithm of ratio 2 to construct a2-approximation connected vertex cover of G+ (the most widely known is based on DFS [Sav82]). Anynew edge rr+ is covered by this solution but if r is not in the solution and r+ is in the solution, then justreverse: put r into the solution and extract the leaf r+ from it. Let us note S+ this new solution that is alwaysa connected vertex cover of G+, with the same size and that verifies now R ⊆ S+. We construct the finalsolution, returned by our algorithm: S = S+∩V ; S is then S+ except the new vertices of G+ that could bein S+. The set S always contains R. Moreover, S is a vertex cover of G . In addition, G [S ] is connected sinceG+[S+] is connected and the only potential deleted vertices are of degree one, useless to ensure connectivityfor the remaining vertices. S is then a CVCwFaRV of (G , /0,R) that can be constructed with a polynomialalgorithm.Let us study its approximation ratio. Let S∗ be an optimal CVCwFaRV of (G ,R) and OPT be its size.Let OPT+ be the size of an optimal connected vertex cover of G+. By definition, S∗ contains all R, coversevery edge of G and G [S∗] is connected. Hence, S∗ is a connected vertex cover of G+ and its size is thenlarger than the optimal one: OPT+ ≤ OPT . Now, by construction of S and with previous properties wehave: |S | ≤ |S+| ≤ 2OPT+ ≤ 2OPT .The last part of this proof is just to take into account a set F of forbidden vertices. We have now aninstance (G ,F,R). We suppose that it satisfies the conditions of Proposition 1 that ensures the existenceof at least a solution. Consider N(F) the set of neighbors of vertices of F in G . All the vertices of N(F)must be in any solution to cover the edges incident to F . So now consider R′ = R∪N(F) as the new set ofrequired vertices and solve the problem as previously in the graph G [V −F ] by considering that the set offorbidden vertices is empty. This is a polynomial 2-approximation algorithm for the instance (G ,F,R). 24 Independent Dominating Set in (G ,F,R)An Independent Dominating Set with Forbidden and Required Vertices (IDwFaRV) in (G ,F,R) is an inde-pendent dominating set of G containing all vertices of R and no vertices of F . Note that when R = F = /0the problem is polynomial.Theorem 3 Given (G ,F,R), deciding whether an IDwFaRV exists is N P -complete, even if R = /0.Proof. The problem is clearly in N P . We reduce it to the X3C (Exact Cover by Sets of size 3) N P -complete problem that we recall now. Let X = {u1,u2, . . . ,u3q} a set of 3q elements and F a family of ksets Ci ⊆ X such that |Ci|= 3 (i ∈ {1, . . . ,k}) and ∪ki=1Ci = X . Given the instance (X ,F ), the X3C problemis to determine whether an exact cover SF ⊆ F of X exists: ∀Ci ∈ SF ,∀C j ∈ SF −{Ci},Ci ∩C j = /0 and∪Ci∈SF Ci = X . Let (X ,F ) be any X3C instance. We construct an instance (G ,F,R) as follows. Each u j ∈ Xbecomes a vertex (also noted u j) of G . Each Ci ∈ F becomes a vertex (also noted Ci) of G . In G eachvertex Ci is connected to the 3 vertices ua,ub,uc if, in F , Ci = {ua,ub,uc}. Two distinct vertices Ci and C jare connected in G if, in F , the two sets are non-disjoint: Ci∩C j 6= /0. The set F of forbidden vertices is X :Francois Delbot and Christian Laforest and Raksmey PhanF = X . The set R of required vertices is empty: R = /0. This construction can be done in polynomial time.For any X3C instance (X ,F ) we note (G ,F,R) the associated instance of our problem.Suppose first that there is a solution SF for the X3C problem. Let S be the set composed of all thecorresponding vertices of SF . S is an independent set of G : indeed, as SF is an exact cover of X , all thesets Ci ∈ SF are pairwise disjoint and then the corresponding vertices of S are non connected in G . S isalso a dominating set of G because each vertex ui is dominated by exactly one C j ∈ SF (the one containingit) and each C j that is not in S is dominated by at least a vertex in S ; Indeed, as SF is a cover of X , the3 elements of set C j are covered by sets of SF meaning that there is Ci ∈ SF such that Ci∩C j 6= /0; vertexC j 6∈ S is then dominated by vertex Ci ∈ S in G by the edge CiC j. To finish, as S contains no forbiddenvertices (because F = X ), S is a IDwFaRV of (G ,F,R).Suppose now that there is an IDwFaRV noted S , of (G ,F,R). In this case, S contains no vertex of X(because these vertices are forbidden, F = X ). Hence S must contain some vertices Ci. Note SF the familyof sets corresponding to the vertices of S . As S is a dominating set of G , ∪Ci∈SF Ci = X . Moreover, as S isan independent set of G , for all Ci ∈ SF and all C j ∈ SF , there is no edge connecting them, i.e. Ci∩C j = /0.Thus SF is a solution for the X3C problem. 2References[DLP13] Franc¸ois Delbot, Christian Laforest, and Raksmey Phan. New approximation algorithms forthe vertex cover problem. In IWOCA, volume 8288 of Lecture Notes in Computer Science,pages 438–442. Springer, 2013.[DPSW11] Andreas Darmann, Ulrich Pferschy, Joachim Schauer, and Gerhard J. Woeginger. Paths, treesand matchings under disjunctive constraints. Disc. Appl. Math., 159(16):1726–1735, 2011.[KLM13a] Mamadou Moustapha Kante´, Christian Laforest, and Benjamin Mome`ge. An exact algorithmto check the existence of (elementary) paths and a generalisation of the cut problem in graphswith forbidden transitions. In SOFSEM, volume 7741 of Lecture Notes in Computer Science,pages 257–267. Springer, 2013.[KLM13b] Mamadou Moustapha Kante´, Christian Laforest, and Benjamin Mome`ge. Trees in graphs withconflict edges or forbidden transitions. In TAMC (Theory and Applications of Models of Com-putation), volume 7876 of Lecture Notes in Computer Science, pages 343–354. Springer, 2013.[Kov13] Jakub Kova´cˇ. Complexity of the path avoiding forbidden pairs problem revisited. DiscreteApplied Mathematics, 161(10–11):1506–1512, 7 2013.[Sav82] Carla Savage. Depth-first search and the vertex cover problem. Information Processing Letters,14(5):233–235, July 1982.[Sze03] Stefan Szeider. Finding paths in graphs avoiding forbidden transitions. Discrete Applied Math-ematics, 126(2-3):261–273, 2003.

Sommets interdits et obligatoires

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Sommets interdits et obligatoires

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