Katona, Gyula Y.

Kovács, István

Varga, Kitti

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The complexity of recognizing minimally toughgraphsGyula Y Katona1Department of Computer Science andInformation TheoryBudapest University of Technology andEconomics1111 Budapest, Mu˝egyetem rkpt. 3, HungaryandMTA-ELTE Numerical Analysis and LargeNetworks Research Groupkiskat@cs.bme.huIstva´n Kova´csDepartment of Control Engineering andInformation TechnologyBudapest University of Technology andEconomics1111 Budapest, Mu˝egyetem rkpt. 3, Hungarykovika@iit.bme.huKitti Varga1Department of Computer Science andInformation TheoryBudapest University of Technology andEconomics1111 Budapest, Mu˝egyetem rkpt. 3, Hungaryvkitti@cs.bme.huAbstract: Let t be a positive real number. A graph is called t-tough, if the removal of anycutset S leaves at most |S|/t components. The toughness of a graph is the largest t for whichthe graph is t-tough. A graph is minimally t-tough, if the toughness of the graph is t andthe deletion of any edge from the graph decreases the toughness. The complexity class DPis the set of all languages that can be expressed as the intersection of a language in NP anda language in coNP. We prove that recognizing minimally t-tough graphs is DP-complete forany positive integer t and for any positive rational number t ≤ 1/2.Keywords: 3–6 keywords toughness, complexity, DP-complete1 IntroductionAll graphs considered in this paper are finite, simple and undirected. Let ω(G) denote the number ofcomponents and α(G) denote the independence number. For a graph G and a vertex set V ⊆ V (G), letG[V ] denote the subgraph of G induced by V .The complexity class DP was introduced by C. H. Papadimitriou and M. Yannakakis [4].Definition 1 A language L is in the class DP if there exist two languages L1 ∈ NP and L2 ∈ coNP suchthat L = L1 ∩ L2.We mention that DP 6= NP∩ coNP, if NP 6= coNP. Moreover, NP∪ coNP ⊆ DP. A language is calledDP-hard if all problems in DP can be reduced to it in polynomial time. A language is DP-complete if itis in DP and it is DP-hard.1Research is supported by by the National Research, Development and Innovation Office NKFIH (grant number K108947)1A critical-type DP-complete problem is CriticalClique [5], in our proofs we use an equivalent formof it, α-Critical.CriticalCliqueInstance: a graph G and a positive integer k.Question: is it true that G has no clique of size k, but adding any missing edge e to G, the resultinggraph G+ e has a clique of size k?By taking the complement of the graph, we can obtain α-Critical from CriticalClique.Definition 2 A graph G is called α-critical, if α(G− e) > α(G) for all e ∈ E(G).α-CriticalInstance: a graph G and a positive integer k.Question: is it true that α(G) < k, but α(G− e) ≥ k for any edge e ∈ E(G)?Since a graph is clique-critical if and only if its complement is α-critical, α-Critical is also DP-complete.Corollary 3 α-Critical is DP-complete.The notion of toughness was introduced by Chva´tal [2].Definition 4 Let t be a positive real number. A graph G is called t-tough, ifω(G− S) ≤ |S|tfor any cutset S of G (i.e. for any S with ω(G− S) > 1). The toughness of G, denoted by τ(G), is thelargest t for which G is t-tough, taking τ(Kn) =∞ for all n ≥ 1.We say that a cutset S ⊆ V (G) is a tough set if ω(G− S) = |S|/τ(G).For all positive rational number t we can define a separate problem:t-ToughInstance: a graph G,Question: is it true that τ(G) ≥ t?Bauer et al. proved the following.Theorem 5 ([1]) For any positive rational number t, t-Tough is coNP-complete.The critical form of this problem is minimally toughness.Definition 6 A graph G is minimally t-tough, if τ(G) = t and τ(G− e) < t for all e ∈ E(G).Given t we define:Min-t-ToughInstance: a graph G,Question: is it true that G is minimally t-tough?Our main result is the following.Theorem 7 Min-t-Tough is DP-complete for any positive integer t and for any positive rational numbert ≤ 1/2.First we prove this theorem for t = 1, then we generalize that proof for positive integers, and finallywe prove it for any positive rational number t ≤ 1/2.22 PreliminariesIn this section we prove some useful lemmas.Proposition 8 Let G be a connected noncomplete graph on n vertices. Then τ(G) ∈ Q+, and if τ(G) =a/b, where a, b are positive integers and (a, b) = 1, then 1 ≤ a, b ≤ n− 1.Proof: By definition,τ(G) = minS⊆V (G)cutset|S|ω(G− S)for a noncomplete graph G. Since G is connected and noncomplete, 1 ≤ |S| ≤ n − 2 and since S is acutset, 2 ≤ ω(G− S) ≤ n− 1. Corollary 9 Let G and H be two connected noncomplete graphs on n vertices. If τ(G) 6= τ(H), then|τ(G)− τ(H)| > 1n2.Claim 10 For every positive rational number t, Min-t-Tough ∈ DP.Proof: For any positive rational number t,Min-t-Tough = {G graph | τ(G) = t and τ(G− e) < t for all e ∈ E(G)} == {G graph | τ(G) ≥ t} ∩ {G graph | τ(G) ≤ t}∩∩{G graph | τ(G− e) < t for all e ∈ E(G)}.LetL1,1 = {G graph | τ(G− e) < t for all e ∈ E(G)},L1,2 = {G graph | τ(G) ≤ t}andL2 = {G graph | τ(G) ≥ t}.L2 ∈ coNP, a witness is a cutset S ⊆ V (G) whose removal leaves more than |S|/t components. L1,1 ∈ NP,the witness is a set of cutsets: Se ⊆ V (G) for each edge e whose removal leaves more than |Se|/tcomponents.Now we show that L1,2 ∈ NP, i.e. we can express L1,2 in a form ofL1,2 = {G graph | τ(G) < t+ ε},which belongs to NP. Let a, b be positive integers such that t = a/b and (a, b) = 1, and let G be anarbitrary graph on n vertices. If G is disconnected, then τ(G) = 0, and if G is complete, then τ(G) =∞,so in both cases G is not minimally t-tough. By Proposition 8, if 1 ≤ a, b ≤ n− 1 does not hold, then Gis also not minimally t-tough. So we can assume that t = a/b, where a, b are positive integers, (a, b) = 1and 1 ≤ a, b ≤ n− 1. With this assumptionL1,2 = {G graph | τ(G) ≤ t} ={G graph∣∣∣∣ τ(G) < t+ 1|V (G)|2},so L1,2 ∈ NP.Since L1,1 ∩ L1,2 ∈ NP, L2 ∈ coNP and Min-t-Tough = (L1,1 ∩ L1,2) ∩ L2, we can conclude thatMin-t-Tough ∈ DP. 3Claim 11 Let t be a positive rational number and G a minimally t-tough graph. For every edge e of G,1. the edge e is a bridge in G, or2. there exists a vertex set S = S(e) ⊆ V (G) withω(G− S) ≤ |S|tand ω((G− e)− S) > |S|t,and the edge e is a bridge in G− S.In the first case, we define S = S(e) = ∅.Proof: Let e be an arbitrary edge of G, which is not a bridge. Since G is minimally t-tough, τ(G− e) < t.So there exists a cutset S = S(e) ⊆ V (G − e) = V (G) in G − e satisfying ω((G − e) − S) > |S|/t. Onthe other hand, τ(G) = t, so ω(G − S) ≤ |S|/t. This is only possible if e connects two components of(G− e)− S. Finally we cite a Lemma that our proof relies on.Lemma 12 (Problem 14 of 8 in [3]) If we replace a vertex of an α-critical graph with a clique, andconnect every neighbor of the original vertex with every vertex in the clique, then the resulting graph isstill α-critical.3 Recognizing minimally 1-tough graphsTo show that Min-1-Tough is DP-hard, we reduce α-Critical to it.Theorem 13 Min-1-Tough is DP-complete.Proof: In Claim 10 we have already proved that Min-1-Tough ∈ DP.Let G be an arbitrary connected graph on the vertices v1, . . . , vn. Let Gα be defined as follows. Itwill be easy to see that it can be constructed from G in polynomial time. For all i ∈ [n], letVi = {vi,1, vi,2, . . . , vi,α}and place a clique on the vertices of Vi. For all i, j ∈ [n], if vivj ∈ E(G), then place a complete bipartitegraph on (Vi;Vj). For all i ∈ [n] and for all j ∈ [α] add the vertex ui,j to the graph and connect it tovi,j . LetV =n⋃i=1ViandU = {ui,j | i ∈ [n], j ∈ [α]}.Add the vertex setW = {w1, . . . , wα}to the graph and for all j ∈ [α] connect wj to v1,j , . . . , vn,j .4GV1V2Vnu1,1u1,αu2,1u2,αun,1un,αUw1wαWFigure 1: The graph Gα.We need to prove that G is α-critical with α(G) = α if and only if Gα is minimally 1-tough. First weprove the following lemma.Lemma 14 Let G be a graph with α(G) ≤ α. Then Gα is 1-tough.Proof: Let S ⊆ V (Gα) be a cutset. We show that ω(Gα − S) ≤ |S|.Case 1: W ⊆ S. If a vertex of U has only one neighbor in V (Gα) \ S, then we can assume that thisvertex is not in S. Then there are two types of components in Gα − S: isolated vertices from U andcomponents containing at least one vertex from V . There are at most α(G) components of the secondtype and (exactly) |V ∩S| = |S|−α components of the first type. Thus ω(Gα−S) ≤ |S|−α+α(G) ≤ |S|.Case 2: W 6⊆ S. First, we make two convenient assumptions for S.(1) U ∩ S = ∅.It is easy to see that if ui,j ∈ S, then we can assume that vi,j 6∈ S. Now there are two cases.Case 2.1: vi,j is not isolated in Gα − S. Then we can consider S′ = (S \ {ui,j})∪ {vi,j} instead of S.Case 2.2: vi,j is isolated in Gα−S. Since there are no isolated vertices in G, there exists k ∈ [n] suchthat vivk ∈ E(G). Then vk,j ∈ S, so uk,j 6∈ S, which means that wj is not isolated in Gα − S, so we canconsider S′ = (S \ {ui,j}) ∪ {wj} instead of S.(2) For all i ∈ [n], either Vi ⊆ S or Vi ∩ S = ∅.After the assumption (1), assume that only a proper subset of Vi is contained in S. Let v be an elementof this subset. We can consider the cutset S \ {v} instead of S, since this decreases the number ofcomponents by at most one. So we can repeat this procedure until Vi ∩ S = ∅.So in Gα − S there are isolated vertices from U and one more component containing the remainingvertices of W and V . So there are less than |V ∩ S| isolated vertices, thusω(Gα − S) ≤ |V ∩ S| ≤ |S|.So Gα is 1-tough. We show that G is α-critical with α(G) = α if and only if Gα is minimally 1-tough.5Let us assume that G is α-critical with α(G) = α. So by Lemma 14 Gα is 1-tough. Let e ∈ E(Gα)be an arbitrary edge. If e has an endpoint in U , then this endpoint has degree 2, so τ(Gα − e) < 1. If edoes not have an endpoint in U , then it connects two vertices of V . By Lemma 12 Gα[V ] is α-critical,so in Gα[V ] − e there exists an independent vertex set I of size α(G) + 1. Let S = (V \ I) ∪W . Then|S| = (|V | − α(G)− 1)+ α = |V | − 1 and ω((Gα − e)− S) = |V |, so τ(Gα − e) < 1.Let us assume that G is not α-critical with α(G) = α.Case 1: α(G) > α. Let I be an independent vertex set of size α(G) in Gα[V ] and let S = (V \ I)∪W .Then |S| = (|V | − α(G)) + α < |V | and ω(Gα − S) = |V |, so τ(Gα) < 1, which means that Gα is notminimally 1-tough.Case 2: α(G) ≤ α. Since G is not α-critical there exists an edge e ∈ E(G) such that α(G − e) ≤ α.By Lemma 14 (G− e)α is 1-tough, but we can obtain (G− e)α from Gα by edge-deletion, which meansthat Gα is not minimally 1-tough. 4 Further resultsTheorem 15 For every positive integer t, Min-t-Tough is DP-complete.To prove this more general theorem, first we generalize the construction on Figure 1. We follow asimilar argument to show that this construction has the required properties. However, due to the morecomplicated construction, the proof is harder.The case when t ≤ 1/2 is also covered in the paper.Theorem 16 For every positive rational number t = a/b ≤ 1/2, Min-t-Tough is DP-complete.It is shown that Min-1-Tough can be reduced to this problem. The construction and the proof usesdifferent ideas than the previous proofs.We were not able to prove the DP-completeness for the remaining t values, but we make the followingconjecture.Conjecture 17 Min-t-Tough is DP-complete for any positive rational number t.References[1] D. Bauer, S. L. Hakimi, and E. Schmeichel, Recognizing tough graphs is NP-hard, Discrete Applied Mathematics 28(1990), 191–195.[2] V. Chva´tal, Tough graphs and hamiltonian circuits, Discrete Mathematics 5 (1973), 215–228.[3] L. Lovsz, Combinatorial problems and exercises, AMS Chelsea Publishing, Providence, Rhode Island, 2007.[4] C. H. Papadimitriou and M. Yannakakis, The Complexity of Facets (and Some Facets of Complexity), Journal ofComputer and System Sciences 28 (1984), 244–259.[5] C. H. Papadimitriou and D. Wolfe, The Complexity of Facets Resolved, Journal of Computer and System Sciences 37(1988), 2–13.6

The complexity of recognizing minimally tough graphs

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