The Complexity of Recognizing Tough Cubic Graphs

Transcription

1 The Complexity of Recognizing Tough Cubic Graphs D. auer 1 J. van en Heuvel 2 A. Morgana 3 E. Schmeichel 4 1 Department of Mathematical Sciences, Stevens Institute of Technology Hoboken, NJ 07030, U.S.A. 2 Department of Mathematics an Statistics, Simon Fraser University urnaby,.c., Canaa V5A 1S6 3 University of Rome Rome, Italy 4 Department of Mathematics & Computer Science, San Jose State University San Jose, CA 95192, U.S.A. June 29, 1998 Abstract We show that it is NP-har to etermine if a cubic graph G is 1-tough. We then use this result to show that for any integer t 1, it is NP-har to etermine if a 3 t-regular graph is t-tough. We conclue with some remarks concerning the complexity of recognizing certain subclasses of tough graphs. Keywors : toughness, cubic graphs, NP-completeness AMS Subject Classifications (1991) : 68R10, 05C38 Supporte in part by NATO Collaborative Research Grant CRG Supporte by a grant from the Natural Sciences an Engineering Council of Canaa. Current aress : Centre for Discrete an Applicable Mathematics, Department of Mathematics, Lonon School of Economics, Houghton Street, Lonon WC2A 2AE, Englan, U.K. Supporte in part by the National Science Founation uner Grant DMS

2 1 Introuction We begin with a few efinitions an some notation. A goo reference for any unefine terms is [7]. We consier only unirecte graphs with no loops or multiple eges. Let ω(g) enote the number of components of a graph G. A graph G is t-tough if S t ω(g S) for every subset S of the vertex set V (G) ofg with ω(g S) > 1. The toughness of G, enote τ(g), is the maximum value of t for which G is t-tough ( taking τ(k n )= for all n 1). A k-factor is a k-regular spanning subgraph. Of course, a hamiltonian cycle is a connecte 2-factor. We use δ(g) to enote the minimum vertex egree in G. Toughness was introuce by Chvátal in [9]. An obvious connection between toughness an hamiltonicity is that being 1-tough is a necessary conition for a graph to be hamiltonian. Chvátal raise a number of interesting questions concerning toughness. He conjecture that every k-tough graph on n vertices with n k +1 an kn even has a k-factor. It has been establishe that this is both true an best possible. Theorem 1.1 ( [15] ) Let k 1. (i) If G is a k-tough graph on n vertices with n k +1an kneven, then G has a k-factor. (ii) For any ε>0, there exists an infinite family of (k ε)-tough graphs with kn even an no k-factor. Chvátal also conjecture that there exists a constant t 0 such that every t 0 -tough graph is hamiltonian. This is still open. The smallest t 0 forwhichthismightbetrueist 0 =2. This follows by letting k = 2 in Theorem 1.1 (ii) above. The problem of etermining the complexity of recognizing t-tough graphs was first raise by Chvátal [8] an later appeare in [23] an [10, p. 429]. Consier the following ecision problem, where t 1 is a rational number, t-tough INSTANCE : Graph G. QUESTION : Is τ(g) t? The following was establishe in [3]. Theorem 1.2 ([3]) For any rational t 1, t-tough is NP-har. It seems natural to inquire whether the problem of recognizing t-tough graphs remains NPhar for various subclasses of graphs. For example, the following theorem of Dirac [14] might suggest that the problem of recognizing 1-tough graphs will be easier for ense graphs; i.e., graphs with many eges. 2

3 Theorem 1.3 ( [14] ) Let G be a graph on n 3 vertices with δ(g) 1 2 n.thengis hamiltonian. In [4], a relatively satisfying solution for the complexity of recognizing t-toughness in ense graphs was obtaine. Let Ω(r) enote the class of all graphs G satisfying δ(g) r V (G). Theorem 1.4 ([4]) Let t 1 be any rational number. (i) (ii) Every graph in Ω( t t+1 ) is t-tough. For any ε>0, t-tough remains NP-har for graphs in Ω( t t+1 ε). Of course, Theorem 1.4 (i) for t = 1 is an immeiate consequence of Theorem 1.3. Another interesting class of graphs is the class of bipartite graphs. Obviously τ(g) 1 for any bipartite graph G. The complexity of recognizing 1-tough, bipartite graphs has been raise by several authors; see, e.g., [6, p. 119], an we have the following recent result. Theorem 1.5 ( [19] ) 1-TOUGH remains NP-har for bipartite graphs. On the other han, there exist classes of graphs for which it is NP-har to etermine whether a graph in the class is hamiltonian, but polynomial to etermine if it is 1-tough. One such class is the class of split graphs. A graph G is calle a split graph if V (G) can be partitione into an inepenent set an a clique. Determining if a split graph is hamiltonian was shown to be NP-har in [11]. On the other han, the following was shown in [19]. Theorem 1.6 ( [19] ) The class of 1-tough split graphs G(V,E) can be recognize in O( E V ) time. Our main interest in this paper is the complexity of toughness for the class of cubic graphs. Chvátal [9] showe that a necessary an sufficient conition on the integer n for the existence of an n-vertex, 3/2-tough cubic graph is that n =4orn 0 ( mo 6 ). Later, Jackson an Katerinis [18] strengthene this by characterizing 3/2-tough, cubic graphs. Theorem 1.7 ( [18] ) Let G be a cubic graph. Then G is 3/2-tough if an only if G = K 4, G = K 2 K 3,orG is obtaine from a 3-connecte cubic graph by replacing all the vertices of this graph by triangles. The above characterizationallows3/2-tough, cubic graphs to be recognize in polynomial time. However, the arguments use in [18] seem very epenent on the particular constant 3/2, an it seems unlikely that a polynomial algorithm exists to recognize t-tough, cubic graphs for any t such that 1 t<3/2. In fact, the following theorem in support of this assertion is our main result. 3

4 Theorem TOUGH remains NP-har for cubic graphs. Using Theorem 1.8 we also prove the following more general result. Theorem 1.9 For any integer t 1, t-tough remains NP-har for 3 t-regular graphs. The following conjecture of Goar an Swart [17] is relate to Theorem 1.9. Conjecture 1.10 ( [17] ) Let G be a k-regular graph. Then G is 1 2 k-tough if an only if G is k-connecte an K 1,3-free. The if irection of Conjecture 1.10 was establishe by the following result of Matthews an Sumner [20]. Theorem 1.11 ( [20] ) Let G be a noncomplete K 1,3 -free graph. Then τ(g) is equal to one half of the connectivity of G. More evience for the truth of Conjecture 1.10 was given in Theorem 1.7; in particular, Theorem 1.7 clearly implies the truth of Conjecture 1.10 for k = 3. If Conjecture 1.10 is true, then t-tough, 2 t-regular graphs can be recognize in polynomial time. y contrast, we now make the following conjecture. Conjecture 1.12 For any integer t 1, t-tough remains NP-har for (2 t +1)-regular graphs. We present the proofs of Theorems 1.8 an 1.9 an a possible approach to Conjecture 1.12 in the following section. We then conclue the paper by iscussing several open problems on the complexity of recognizing certain subclasses of tough graphs. 2 Proofs of Theorems 1.8 an 1.9 ProofofTheorem1.8 We will prove that 1-TOUGH is NP-har for cubic graphs by reucing 1-TOUGH for general connecte graphs with at least two vertices. Let G K 1 be any connecte graph. Construct the corresponing cubic graph H = H(G) as follows. Each vertex v V (G) will correspon to the graph H v in Figure 1 below, in which there are G (v) black vertices of egree 2 on each sie of the inicate ege e. Arbitrarily esignate the black vertices in H v on one sie of e by A v, an those on the other sie by v. Note that every vertex in H v (A v v ) belongs to a triangle in H v. 4

5 t @,, t @, P t, Figure 1 The graph H v. An ege vw in G will be represente in H by joining any previously unuse vertex in A v ( i.e., one whose egree is then 2 ) to any previously unuse vertex in w, an any previously unuse vertex in A w toanypreviouslyunusevertexin v. It is immeiate that the resulting graph H(G) is 2-connecte cubic. To complete the proof, it now suffices to show the following. Claim G is 1-tough if an only if H(G) is 1-tough. Proof of the Claim Suppose first that G is not 1-tough. Then there exists a nonempty set X V (G) withω(g X) > X. LetY V (H) begivenbyy = (A v v ). It is easily verifie that ω(h Y ) > Y, an thus H is also not 1-tough. Conversely, suppose H is not 1-tough. Then there exists a nonempty set Y V (H) with ω(h Y ) > Y. We will now establish a series of properties ( Lemmas 2.1 to 2.4 ) which we may assume Y satisfies, since otherwise we may select a nonempty set Y V (H) with ω(h Y ) > Y satisfying the esire properties. Lemma 2.1 We may assume, for each y Y,thatN H (y) is an inepenent set. Proof Suppose N H (y) is not inepenent for some y Y. Set Y = Y {y}, sothat Y = Y 1 an since H is cubic, ω(h Y ) ω(h Y ) 1. In particular, ω(h Y ) > Y. If Y =Ø,thenY is a singleton an thus H is not 2-connecte, a contraiction. Otherwise we have a nonempty Y V (H) such that ω(h Y ) > Y an we can simply iterate this moification to Y until the esire conition hols. 2 v X 5

6 Since every vertex in H v (A v v ) belongs to a triangle, Lemma 2.1 implies that we may assume Y (A v v ). v V (G) Lemma 2.2 We may assume, for all v V (G), thata v Y (resp., v Y )iseithera v (resp., v )orø. Proof Suppose A v Y is neither A v nor Ø. Then there are two consecutive vertices a, a A v (see Figure 2) with a Y, a / Y. Set Y = Y {a }. It is immeiate P e e a PP e, Figure 2 Consecutive vertices in A v. e Y = Y +1 an ω(h Y ) ω(h Y ) + 1, an thus we have a nonempty Y V (H) such that ω(h Y ) > Y. We simply iterate this moification to Y until A v Y = A v. The proof for v is ientical. 2 We will call v V (G) asplit vertex if exactly one of A v, v belongs to Y. In a moment we will show we may assume there are no split vertices. First we nee the following result. Lemma 2.3 Let v V (G). IfA v Y an v Y, then we may assume, for all w N G (v), that w Y. Proof Suppose A v Y, v Y, but w Y for some w N G (v). Set Y = Y A v.we fin Y = Y + G (v) anω(h Y ) ω(h Y )+ G (v),anthuswehaveanonempty Y V (H) such that ω(h Y ) > Y. Now simply iterate the moification of Y until the esire conition hols. 2 Lemma 2.4 We may assume there are no split vertices. Proof Suppose S = { v V (G) v is a split vertex } = Ø. Consier any component C of S in G, anletv (C) ={v 1,v 2,...,v m }. y Lemma 2.3 we may assume, without loss of generality, that A vi Y, vi Y,fori =1, 2,...,m.Notethatifwv i E(G) for some i, 1 i m, anw/ V (C), then by Lemma 2.3, A w, w Y.Alsonotethat m (H vi vi ) inuces i=1 a subgraph in H with exactly G (v 1 )+ + G (v m )components. NowsetY = Y \ ( m vi ). 6 i=1

7 Suppose Y =Ø. IfV (C) V (G), then G is isconnecte by the above observation on eges wv i. On the other han, if V (C) =V (G), then ω(h Y )= G (v) = Y, a contraiction. Hence Y Ø. Thuswehave m (1) Y = Y G (v i ) 1. i=1 y Lemma 2.3, we also have ( ω(h Y m ) (2) ) = ω(h Y ) G (v i ) 1. i=1 v V (G) Using (1), (2), an ω(h Y ) > Y, we fin that Y is a nonempty subset of V (H) such that m ω(h Y ) > Y G (v i )+1 = Y +1 > Y. i=1 Now simply iterate the above moification to Y for every component C of S. 2 Finally, let X = { v V (G) A v, v Y }. Since there are no split vertices, it is easy to check that X is a nonempty subset of V (G) such that ω(g X) > X an so G is not 1-tough. This proves the Claim, an completes the proof of Theorem 1.8. Proof of Theorem 1.9 We will reuce 1-TOUGH for 2-connecte, cubic graphs to t- TOUGH for 3 t-regular graphs, where t 1 is an integer. Let G be any 1-tough, 2-connecte cubic graph. y a well-known theorem of Petersen [21], G can be ege-partitione into a 1-factor an a 2-factor. Construct H = H(G) as follows. Each vertex in G is replace by a K t in H. Each ege in the 1-factor in G is replace by a matching ( henceforth calle an m-join ) between the corresponing K t s in H, while each ege in the 2-factor in G is replace by a complete bipartite join ( henceforth calle a c-join ) between the corresponing K t s. It is immeiate that H is 3 t-regular. We next want to show that H is 2t-connecte. Since G is 2-connecte, isconnecting G requires removing at least two vertices, one vertex an a nonincient ege, or two inepenent eges. Thus isconnecting H requires removing at least two K t s, removing one K t an breaking a nonincient m-join, or breaking two inepenent m-joins. Note that breaking an m-join means removing enough vertices in the m-join to eliminate all eges in the join without completely removing either K t ; obviously this requires removing a total of at least t vertices in the two K t s. It follows that any cutset in H must contain at least 2 t vertices, an thus H is 2 t-connecte. To complete the proof, we now show G is 1-tough if an only if H is t-tough. If G is not 1-tough, there exists a cutset X V (G) withω(g X) > X. Let Y V (H) consist 7

8 of the K t s corresponing to the vertices in X. It is easy to see that Y is a cutset an ω(h Y )=ω(g X) > X = Y /t, an thus H is not t-tough. Conversely, suppose H is not t-tough. Then there exists a cutset Y V (H) with ω(h Y ) > Y /t. We now establish the following. Claim We may assume each K t in H is entirely containe in Y or entirely isjoint from Y ( i.e., no K t in H is split by Y.) Assuming we have establishe the Claim, let X V (G) enote the vertices in G corresponing to the K t s in Y. Then X is a cutset in G an ω(g X) =ω(h Y ) > Y /t = X, an thus G is not 1-tough as esire. To prove the claim consier a cutset Y V (H) such that ω(h Y ) > Y /t an Y splits as few K t s in H as possible. If Y splits no K t s there is nothing to prove, so suppose A is a K t split by Y,anlet enote the K t whichism-joinetoa. We now consier several cases. Case 1. is not split by Y Let Y = Y (A Y ). Then Y < Y while ω(h Y ) = ω(h Y ), since A is still Y c-joine to the same K t s as was A Y. Thus we have ω(h Y ) < Y ω(h Y ) <t,or ω(h Y ) > Y /t. SinceY is a cutset an ω(h Y )=ω(h Y), Y is also a cutset in H. Since Y splits fewer K t s than Y, this violates the optimality of Y. Case 2.1. is split by Y an A Y + Y <t Set Y = Y (A Y ) ( Y ). Then ω(h Y )=ω(h Y), since A Y an Y belong to the same component of H Y,anA (resp, ) is c-joine to its ajacent K t s besies Y (resp, A ). Thus we get ω(h Y ) < Y ω(h Y ) <t,orω(h Y ) > Y /t. Since Y is a cutset an ω(h Y )=ω(h Y), Y is also a cutset in H. Again, Y splits fewer K t s than Y, an this violates the optimality of Y. Case 2.2. is split by Y an A Y + Y t Let Y = Y (A Y ) Z, wherez Y is any subset with Z = t A Y > 0. Since H is 2t-connecte an Y is a cutset in H, wehave Y = Y t 2 t t = t. Note that ω(h Y ) ω(h Y ) 1, since we might still lose one component by pulling together the two components containing A Y an Y, but nothing more. Since ω(h Y ) > Y t,we Y get ω(h Y ) Y t ω(h Y ) 1 <t,orω(h Y ) > Y 1. Thus Y is a cutset in H. t Since Y splits fewer K t s than Y, this violates the optimality of Y. This proves the Claim, an thereby proves Theorem 1.9. We conclue this section by remarking that a possible approach towar proving Conjecture 1.12 is to interchange the roles of the m-joins an c-joins in the construction of H(G) in the proof 8

9 of Theorem 1.9. However at present, we have not been able to verify the Claim in the proof of Theorem 1.9 for H(G) constructe in this new way. 3 Concluing Remarks There remain a number of interesting questions concerning the complexity of recognizing special classes of tough graphs. Recall the following well-known conjecture. arnette s Conjecture Every 3-connecte, cubic, planar, bipartite graph is hamiltonian. It is well-known that if any of the hypotheses in this conjecture are roppe, the conclusion that the graph is hamiltonian nee not follow. Thus it seems interesting to consier the complexity of recognizing 1-tough graphs when one or more of the hypotheses in arnette s Conjecture are roppe. It is easy to see that every 3-connecte cubic graph is 1-tough. On the other han, there are 2-connecte, cubic, planar, bipartite graphs which are not 1-tough ( see, e.g., [2] ). The complexity of recognizing 1-tough graphs remains open for the following classes of graphs. 2-connecte, cubic, planar, bipartite graphs; 2-connecte, cubic, planar graphs; 2-connecte, cubic, bipartite graphs; 2-connecte, planar, bipartite graphs; 2-connecte, planar graphs. Tutte [25] has shown that every 4-connecte planar graph is hamiltonian an Thomassen [24] has shown that every such graph graph is hamiltonian connecte. On the other han, there exist 3-connecte, planar, bipartite graphs which are not 1-tough ( e.g., the Herschel graph [5, p. 53] ). The complexity of recognizing 1-tough graphs remains open for the following classes. 3-connecte, planar, bipartite graphs; 3-connecte, planar graphs; 3-connecte, bipartite graphs. It is interesting to note that the complexity of recognizing hamiltonian graphs is known to be NP-har for all of the above classes except possibly 3-connecte, planar, bipartite graphs [1, 16]. Finally, let us focus on the class of planar graphs. As inicate above, we o not know the complexity of recognizing 1-tough, planar graphs. However, the next result might yiel a clue. It follows from theorems in [13, 22]. 9

10 Theorem 3.1 ( [13, 22] ) Let G be a planar graph on at least 5 vertices. Then G is 4-connecte if an only if ω(g X) X 2, for all cutsets X V (G) with X 3. Since 4-connecte graphs can be recognize in polynomial time, it follows that for planar graphs G, it can be etermine in polynomial time whether ω(g X) X 2, for all cutsets X V (G) with X 3. To etermine if G is 1-tough, one nees to ecie the superficially similar inequality ω(g X) X, for all cutsets X V (G). Perhaps this suggests that recognizing 1-tough, planar graphs can be one in polynomial time. Dillencourt [12] has also inquire about the complexity of recognizing 1-tough, maximal planar graphs, noting that recognizing hamiltonian, maximal planar graphs is NP-har. All we know is that there exist maximal planar graphs which are not 1-tough. References [1] T.Akiyama,T.Nishizeki,anN.Saito,NP-completeness of the hamiltonian cycle problem for bipartite graphs. J.Infor.Proc.3 (1980) [2]T.Asano,N.Saito,G.Exoo,anF.Harary,The smallest 2-connecte cubic bipartite planar nonhamiltonian graph. Discrete Math. 38 (1982) 1 6. [3] D. auer, S.L. Hakimi, an E. Schmeichel, Recognizing tough graphs is NP-har. Discrete Appl. Math. 28 (1990) [4] D. auer, A. Morgana, an E. Schmeichel, On the complexity of recognizing tough graphs. Discrete Math. 124 (1994) [5] J.A. ony an U.S.R. Murty, Graph Theory with Applications. Macmillan, Lonon an Elsevier, New York (1976). [6] H.J.roersma,J.vanenHeuvel,anH.J.Velman,eitors,Upate contributions to the Twente Workshop on Hamiltonian Graph Theory. Technical report, University of Twente, The Netherlans (1992). [7] G. Chartran an L. Lesniak, Graphs an Digraphs. Wasworth Inc., elmont, CA (1986). [8] V. Chvátal. Private communication. [9] V. Chvátal, Tough graphs an hamiltonian circuits. Discrete Math. 5 (1973) [10] V. Chvátal, Hamiltonian cycles. In : E.L. Lawler, J.K. Lenstra, A.H.G. Rinnooy Kan, an D.. Shmoys, eitors, The Traveling Salesman Problem, A Guie Tour of Combinatorial Optimization, chapter 11, pages John Wiley & Sons, Chichester (1985). [11] C.J. Colburn an L.K. Stewart, Dominating cycles in series-parallel graphs. ArsCombi- natoria 19A (1985)

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