The Complexity of Toughness in Regular Graphs

Transcription

1 The Complexity of Toughness in Regular Graphs D Bauer 1 J van den Heuvel 2 A Morgana 3 E Schmeichel 4 1 Department of Mathematical Sciences Stevens Institute of Technology, Hoboken, NJ 07030, USA 2 Centre for Discrete and Applicable Mathematics London School of Economics, Houghton Street, London WC2A 2AE, UK 3 University of Rome Rome, Italy 4 Department of Mathematics & Computer Science San Jose State University, San Jose, CA 95192, USA Abstract Let t 1 be an integer We sho that it is NP-hard to determine if an r-regular graph is t-tough for any fixed integer r 3 t We also discuss the complexity of recognizing if an r-regular graph is t-tough, for any rational t 1 Keyords : toughness, regular graph, NP-completeness AMS Subject Classifications (1991) : 68R10, 05C38 Supported in part by the National Science Foundation under Grant DMS

2 1 Introduction We begin ith a fe definitions and some notation A good reference for any undefined terms in graph theory is [7] and for computational complexity is [12] We consider only undirected graphs ith no loops or multiple edges Let G be a graph Then G is hamiltonian if it has a Hamilton cycle, ie, a cycle containing all of its vertices It is traceable if it has a path containing all of its vertices Let ω(g) denote the number of components of G We say G is t-tough if S t ω(g S) for every subset S of the vertex set V (G) ofg ith ω(g S) > 1 The toughness of G, denoted τ(g), is the maximum value of t for hich G is t-tough ( taking τ(k n )= n 1 2 for all n 1) A k-factor is a k-regular spanning subgraph Of course, a Hamilton cycle is a 2-factor Toughness as introduced by Chvátal in [9] In recent years much ork has been done to understand the relationship beteen the toughness of a graph and its cycle structure For survey papers in this area, see [4, 5, 6] An obvious connection beteen toughness and hamiltonicity is that being 1-tough is a necessary condition for a graph to be hamiltonian Chvátal conjectured that there exists a finite constant t 0 such that every t 0 -tough graph is hamiltonian This conjecture is still open Until recently it as believed that the smallest value of t 0 for hich this conjecture ould be true as t 0 = 2 This belief, knon as the 2-tough conjecture, stemmed in large part by letting k = 2 in the to results belo The first of these results shoed that Chvátalascorrect in hisconjecture that every k-tough graph on n vertices ith n k +1 and kneven has a k-factor Theorem 11 [11] LetGbeak-tough graph on n vertices ith n k +1 and kn even Then G has a k-factor Theorem 12 [11] Letk 1 For any ε>0, thereexistsa(k ε)-tough graph G on n vertices ith n k +1 and kn even hich has no k-factor We no kno that the 2-tough conjecture is false Theorem 13 [1] For every ε>0, thereexistsa( 9 4 ε)-tough nontraceable graph 2

3 In this note e continue to study the complexity of recognizing t-tough graphs This problem as first raised by Chvátal [8] and later appeared in [19] and [10, p 429] Consider the folloing decision problem, here t>0 is a rational number t-tough INSTANCE : Graph G QUESTION : Is τ(g) t? The folloing as established in [2] Theorem 14 [2] For any rational t>0, t-tough is NP-hard It seems natural to inquire hether the problem of recognizing t-tough graphs remains NP-hard for various subclasses of graphs A number of results in this direction are discussed in [3] In this note e focus on the complexity of recognizing tough regular graphs, and begin ith the class of cubic graphs In [9], Chvátal shoed that a necessary and sufficient condition on an integer n for the existence of an n-vertex, 3/2-tough cubic graph is that n =4orn 0 ( mod 6 ) Later, Jackson and Katerinis [15] strengthened this by characterizing 3/2-tough, cubic graphs Theorem 15 [15] Let G be a cubic graph Then G is 3/2-tough if and only if G = K 4, G = K 2 K 3,orG is obtained from a 3-connected cubic graph by replacing all the vertices of this graph by triangles The above characterization allos 3/2-tough, cubic graphs to be recognized in polynomial time Hoever, the arguments used in [15] seem very dependent on the particular constant 3/2, and it seems unlikely that a polynomial time algorithm exists to recognize t-tough, cubic graphs for any t such that 1 t<3/2 In [3] e established this for t =1 Theorem 16 [3] 1-TOUGH remains NP-hard for cubic graphs In [3] e ere also able to use Theorem 16 to establish the folloing result Theorem 17 [3] For any integer t 1, t-tough remains NP-hard for 3 t-regular graphs 3

4 In this note e extend both Theorem 16 and Theorem 17 We first extend Theorem 16 by proving Theorem 18 belo The proof is modelled after the proof of Theorem 16 given in [3] Theorem 18 For any fixed integer r 3, 1-TOUGH remains NP-hard for r-regular graphs We then extend Theorem 17 by proving Theorem 19 belo The proof, hich uses Theorem 18, is modelled after the proof of Theorem 17 given in [3] Theorem 19 For any integer t 1 and any fixed integer r 3 t, t-tough remains NP-hard for r-regular graphs The truth of Theorems 18 and 19 is not surprising in light of Theorems 16 and 17, respectively Nevertheless, it is not alays easy to establish this type of intuitively obvious result For example, after Holyer [14] established that it is NP-hard to decide if a cubic graph is class one ( relative to edgecoloring ), it required a surprisingly difficult proof by Levin and Galil [16] to sho the problem is also NP-hard for r-regular graphs for fixed r>3 We no consider the problem of recognizing t-tough r-regular graphs for any rational t 1 Note that if r < 2 t, anr-regular graph cannot be t-tough since removing the neighborhood of any vertex yields at least to components If r = 2t, thent-tough ould be polynomial if the folloing conjecture of Goddard and Sart [13] is true Conjecture 110 [13] LetG be an r-regular graph Then G is r -tough if 2 and only if G is r-connected and K 1,3 -free Note that Conjecture 110 is true for r = 3 by Theorem 15 In addition, the if direction of Conjecture 110 as established by a result of Matthes and Sumner [17] Theorem 111 [17] LetG be a noncomplete K 1,3 -free graph Then τ(g) is equal to one half of the connectivity of G We no make the folloing conjecture 4

5 Conjecture 112 For any rational t 1 and any fixed integer r 1, t-tough remains NP-hard for the class of r-regular graphs if and only if r>2 t In the next to sections e present the proofs of Theorems 18 and 19 Finally, in the last section e conclude ith a discussion of Conjecture Proof of Theorem 18 We already kno that recognizing 1-tough cubic graphs is NP-hard [3] For fixed r 4, e no prove that 1-TOUGH is NP-hard for r-regular graphs by reducing 1-TOUGH for cubic graphs Let G be any cubic graph For r 4, construct the corresponding r-regular graph H r = H r (G) as follos Each vertex v V (G) ill correspond to a graph H v,r as in Figures 1 and 2 belo The structure of H v,r ill depend on the parity of r In each case the three black vertices on each side of the connection graph ill have degree r 1inH v,r Case 1 : r =2k, k 2 In this case there are four basic graphs, B a 1, B b 1, B a 2, B b 2, and one connection graph C 3 For j {a, b}, the graph B j 1 is a complete graph on 2 k vertices Let V (B j 1 )=Rj 1 Sj 1,hereRj 1 = {rj 1,1,rj 1,2,,rj 1,k } and Sj 1 = {sj 1,1,sj 1,2,,s j 1,k } For j {a, b}, the graph B j 2 is a complete graph on 2 k + 1 vertices ith certain edges removed Let V (B j 2 )=Rj 2 Sj 2 {}, hererj 2 = {r j 2,1,rj 2,2,,rj 2,k } and Sj 2 = {sj 2,1,sj 2,2,,sj 2,k } No remove the edges {r j 2,1 sj 2,1,rj 2,2 sj 2,2,,rj 2,k 1 sj 2,k 1 } The graph C 3 is a complete graph on 2 k vertices ith certain edges removed First let V (C 3 )=R 3 S 3,hereR 3 = {r 3,1,r 3,2,,r 3,k } and S 3 = {s 3,1,s 3,2,,s 3,k } No remove the edges {r 3,1 s 3,1,r 3,2 s 3,2,, r 3,k 1 s 3,k 1 } 5

6 Finally, consider the 6 black vertices {u 1,u 2,,u 6 } in Figure 1 vertex u 1 is joined to (R a 1 Ra 2 ) {ra 2,k }, u 2 is joined to (S2 a R 3 ) {s a 2,k }, u 3 is joined to (S1 a R 3 ) {r 3,k }, u 4 is joined to (R b 2 S 3) {r2,k b }, u 5 is joined to (R b 1 S 3 ) {s 3,k } and u 6 is joined to (S1 b S2) b {s b 2,k } The The graph H v,r for r even is shon in Figure 1 B a 2 B b 2 u 2 u 4 u 1 C 3 u 6 u 3 u 5 B a 1 B b 1 Fig 1 : H v,r for r even Case 2 : r =2k +1,k 2 In this case there are four basic graphs, B a 4, B b 4, B c 4, B d 4, and one connection graph C 5 For j {a, b, c, d}, the graph B j 4 is a complete graph on 2 k +2 vertices ith certain edges removed Let V (B j 4 ) = Rj 4 Sj 4, here Rj 4 = {r j 4,1,rj 4,2,,rj 4,k+1 } and Sj 4 = {sj 4,1,sj 4,2,,sj 4,k+1 } No remove the edges {r j 4,1 sj 4,1,rj 4,2 sj 4,2,,rj 4,k sj 4,k } The graph C 5 is a complete graph on 2 k + 2 vertices ith certain edges removed Let V (C 5 )=R 5 S 5,hereR 5 = {r 5,1,r 5,2,,r 5,k+1 } and 6

7 S 5 = {s 5,1,s 5,2,,s 5,k+1 } No remove the edges {r 5,1 s 5,2,r 5,2 s 5,3,, r 5,k s 5,k+1 } and {r 5,2 s 5,1,r 5,3 s 5,2,,r 5,k+1 s 5,k } Finally, consider the 6 black vertices { 1, 2,, 6 } in Figure 2 vertex 1 is joined to (R a 4 R c 4) {r4,k+1 a,rc 4,k+1 }, 2 is joined to (S4 a R 5) {s a 4,k+1,r 5,k+1}, 3 is joined to (S4 c R 5) {s c 4,k+1,r 5,1}, 4 is joined to (R b 4 S 5 ) {r4,k+1 b,s 5,k+1}, 5 is joined to (R d 4 S 5 ) {r4,k+1 d,s 5,1} and 6 is joined to (S4 b Sd 4 ) {sb 4,k+1,sd 4,k+1 } The The graph H v,r for r odd is shon in Figure 2 B a 4 B b C B c 4 B d 4 Fig 2 : H v,r for r odd For r even, arbitrarily designate the black vertices in H v,r on one side of C 3 by A v,r, and those on the other side by B v,r Similarly, for r odd arbitrarily designate the black vertices in H v,r on one side of C 5 by A v,r, and those on the other side by B v,r Thus the set of black vertices in H r is B = v V (G) A v,r B v,r LetW be the remaining set of hite vertices in H r Anedgev in G ill be represented in H r by joining any previously unused vertex in A v,r to any previously unused vertex in B,r, and any previously unused vertex in A,r to any previously unused vertex in B v,r NotethatH r is an r-regular 2-connected graph 7

8 To complete the proof, it no suffices to sho the folloing Claim G is 1-tough if and only if H r (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) ithω(g X) > X LetY V (H r )be given by Y = (A v,r B v,r ) It is easily verified that ω(h r Y ) > Y, v X and thus H r is also not 1-tough Conversely, suppose H r is not 1-tough Then there exists a nonempty set Y V (H r )ithω(h r Y ) > Y We ill no establish a series of properties (Lemmas 21 to 24 ) hich e may assume Y satisfies, since otherise e may select a nonempty set Y V (H r )ithω(h r Y ) > Y satisfying the desired properties Lemma 21 We may assume that Y W = Proof :Foragivenv V (G), let F be any basic or connection subgraph of H v,r H r, together ith the black neighbors of this subgraph ( see Figures 1 and 2 ) Let Y F (W )(resp,y F (B) ) denote Y V (F ) W (resp, Y V (F ) B ), ith Y F (W ) We first prove Fact 1 : ω((f Y F (B)) Y F (W )) Y F (W ) +1 ProofofFact1:Whenr 5, F Y F (B) is hamiltonian, and hence 1-tough, regardless of hich vertices are in Y F (B) Thus ω((f Y F (B)) Y F (W )) Y F (W ) < Y F (W ) +1 If r = 4, it is easy to verify the inequality directly for each F 2 Let Y = Y Y F (W ) We next prove Fact 2 : ω(h r Y ) > Y Proof of Fact 2 :ByFact1ehave ω(h r Y ) ω(h r Y ) [ ω((f Y F (B)) Y F (W )) 1] > Y [( Y F (W ) +1) 1] = Y Y F (W ) = Y 2 8

9 To complete the proof of Lemma 21, simply iterate the above procedure for each basic or connection graph F in H r having Y F (W ) We finally obtain a set Y B ith ω(h r Y ) > Y ThissetY must be nonempty; otherise the previous set Y consists solely of hite vertices in a single basic or connection graph F and satisfies ω(h r Y ) > Y This clearly is impossible 2 Lemma 22 We may assume, for all v V (G), that A v,r Y (resp, B v,r Y )iseithera v,r (resp, B v,r )or Proof : Suppose A v,r Y is neither A v,r nor Then there are to consecutive vertices a, b A v,r ( see Figure 3 ) ith a Y, b/ Y Set Y = Y {b} B j i a Fig 3 : Consecutive vertices in A v,r b It is immediate that Y = Y +1 andω(h r Y ) ω(h r Y ) + 1, and thus e have a nonempty Y V (H r ) such that ω(h Y ) > Y We simply iterate this modification to Y until A v,r Y = A v,r The proof for B v,r is identical 2 We ill call v V (G) asplit vertex if exactly one of A v,r,b v,r belongs to Y In a moment e ill sho e may assume there are no split vertices First e need the folloing result Lemma 23 Let v V (G) If A v,r Y and B v,r Y, then e may assume, for all N G (v), that B,r Y Proof : Suppose A v,r Y, B v,r Y, but B,r Y for some N G (v) Set Y = Y A v,r We find Y = Y +3 and ω(h r Y ) ω(h r Y )+3, 9

10 and thus e have a nonempty Y V (H r ) such that ω(h r Y ) > Y No simply iterate the modification of Y until the desired condition holds 2 Lemma 24 We may assume there are no split vertices Proof : Suppose S = { v V (G) v is a split vertex } Consider any component K of S in G, andletv (K) ={v 1,v 2,,v m } By Lemma 23 e may assume, ithout loss of generality, that A vi,r Y, B vi,r Y, for i =1, 2,,m Note that if v i E(G) forsomei, 1 i m, and / V (K), then by Lemma 23, A,r,B,r Y Also note that m (H vi,r B vi,r) induces a subgraph in H r ith exactly 3 m components i=1 No set Y = Y \ ( m B vi,r ) Suppose Y = IfV (K) V (G), then G i=1 is disconnected by the above observation on edges v i On the other hand, if V (K) =V (G), then ω(h r Y )=3 V(G) = Y, a contradiction Hence Y Thusehave (1) Y = Y 3 m 1 By Lemma 23, e also have (2) ω(h r Y )=ω(h r Y ) (3 m 1) Using (1), (2), and ω(h r Y ) > Y, e find that Y is a nonempty subset of V (H r ) such that ω(h r Y ) > Y 3 m +1 = Y +1 > Y No simply iterate the above modification to Y for every component K of S 2 Finally, let X = { v V (G) A v,r,b v,r 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 and so G is not 1-tough This proves the Claim, and completes the proof of Theorem 18 2 Note that for every r 4, H v,r ( and thus H r ) contains a 1-factor Thus the proof of Theorem 18 actually proves the folloing result : For every fixed integer r 3, 1-TOUGH remains NP-hard for the class of 2-connected, r-regular graphs hich contain a 1-factor 10

11 3 Proof of Theorem 19 Before proving Theorem 19, e require the folloing lemma The proof is essentially the same as the proof of Theorem 17 given in [3] Lemma 31 Let t 1 and k 3 be integers Then t-tough remains NP-hard for the class of kt-regular graphs Proof of Lemma 31 : We ill reduce 1-TOUGH for 2-connected, k-regular graphs containing a 1-factor ( see the remark at the end of the previous section ) Let G be any such graph, and construct H = H(G) as follos Each vertex in G is replaced by a K t in H Eachedgeinthe 1-factor in G is replaced by a matching ( henceforth called an m-join )beteen the corresponding K t s in H, hile each edge not in the 1-factor in G is replaced by a complete bipartite join ( henceforth called a c-join ) beteen the corresponding K t s It is immediate that H is kt-regular We next sho that H is 2 t-connected Since G is 2-connected, disconnecting G requires removing at least to vertices, one vertex and a nonincident edge, or to independent edges Thus disconnecting H requires removing at least to K t s, removing one K t and breaking a nonincident m-join, or breaking to independent m-joins Note that breaking an m-join means removing enough vertices in the m-join to eliminate all edges in the join ithout completely removing either K t ; obviously this requires removing a total of at least t vertices in the to K t s It follos that any cutset in H must contain at least 2 t vertices, and thus H is 2 t-connected To complete the proof, e no sho G is 1-tough if and only if H is t-tough If G is not 1-tough, there exists a cutset X V (G) ithω(g X) > X Let Y V (H) consist of the K t s corresponding to the vertices in X Itis easy to see that Y is a cutset and ω(h Y )=ω(g X) > X = Y /t, and thus H is not t-tough Conversely, suppose H isnot t-tough Then there exists a cutset Y V (H) ith ω(h Y ) > Y /t We no establish the folloing Claim We may assume each K t in H is entirely contained in Y or entirely disjoint from Y ( ie, no K t in H is split by Y ) Assuming e have established the Claim, let X V (G) denotethevertices in G corresponding to the K t s in Y Then X is a cutset in G and 11

12 ω(g X) =ω(h Y ) > Y /t = X, and thus G is not 1-tough as desired To prove the claim consider a cutset Y V (H) such that ω(h Y ) > Y /t and Y splitsasfek 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,andletB denote the K t hich is m-joined to A We no consider several cases Case 1 : B is not split by Y Let Y = Y (A Y ) Then Y < Y hile ω(h Y )=ω(h Y), since A is still c-joined to the same K t s as as A Y Thus e have Y ω(h Y ) < Y ω(h Y ) <t,orω(h Y ) > Y /t Since Y is a cutset and ω(h Y )=ω(h Y), Y isalsoacutsetinh Since Y splits feer K t s than Y, this violates the optimality of Y Case 21 : B is split by Y and A Y + B Y <t Set Y = Y (A Y ) (B Y ) Then ω(h Y )=ω(h Y), since A Y and B Y belong to the same component of H Y,andA(resp, B )is Y c-joined to its adjacent K t s besides B (resp,a) Thus e get ω(h Y ) < Y ω(h Y ) <t,orω(h Y ) > Y /t SinceY is a cutset and ω(h Y )= ω(h Y ), Y isalsoacutsetinh Again, Y splits feer K t s than Y, and this violates the optimality of Y Case 22 : B is split by Y and A Y + B Y t Let Y = Y (A Y ) Z, herez B Y is any subset ith Z = t A Y > 0 Since H is 2 t-connected and Y is a cutset in H, ehave Y = Y t 2 t t = t Note that ω(h Y ) ω(h Y ) 1, since e might still lose one component by pulling together the to components containing A Y and B Y, but nothing more Since ω(h Y ) > Y t,e Y get ω(h Y ) Y t ω(h Y ) 1 <t,orω(h Y ) > Y 1 Thus Y is t acutsetinh SinceY splits feer K t s than Y, this violates the optimality of Y This proves the Claim, and thereby proves Lemma

13 Proof of Theorem 19 : Given any integer r 3 t, choose the integer k 3 such that kt r (k +1)t 1 In the construction of H(G) given in the proof of Lemma 31, replace each m-join beteen a pair of K t s by r kt+ 1 edge-disjoint 1-factors beteen the pair of K t s The resulting graph H (G)isthenr-regular The remainder of the proof that G is 1-tough if and only if H (G) ist-tough is essentially the same as in the proof of Lemma 31 In particular, if H (G) isnott-tough, then clearly H(G) isnot t-tough Thus G is not 1-tough as before 2 4 Concluding Remarks Conjecture 112 appears quite difficult, particularly hen r is slightly larger than 2 t In [3] e considered a special case of Conjecture 112; namely, that for any integer t 1, t-tough remains NP-hard for (2 t +1)-regular graphs We thought that a reasonable approach toard attacking this special case as to use the same approach as in the proof of Theorem 17 given in [3], except to interchange the roles of the m-joins and c-joins in the construction of H(G) Note that this yields a graph Ĥ(G) thatis(2t +1)- regular Unfortunately, e recently found a 1-tough graph G for hich the graph Ĥ(G ) constructed in this ay is not t-tough References [1] D Bauer, H J Broersma, and H J Veldman Not every 2-tough graph is hamiltonian To appear in Discrete Appl Math [2] D Bauer, S L Hakimi, and E Schmeichel Recognizing tough graphs is NP-hard Discrete Appl Math, 28 : , 1990 [3] D Bauer, J van den Heuvel, A Morgana, and E Schmeichel The complexity of recognizing tough cubic graphs Discrete Appl Math, 79 : 35 44, 1997 [4] D Bauer, E Schmeichel, and H J Veldman Some recent results on long cycles in tough graphs In Y Alavi, G Chartrand, O R Oellermann, and A J Schenk, editors, Graph Theory, Combinatorics, and 13

14 Applications - Proceedings of the Sixth Quadrennial International Conference on the Theory and Applications of Graphs, pages John Wiley & Sons, Inc, Ne York, 1991 [5] D Bauer, E Schmeichel, and H J Veldman Cycles in tough graphs - Updating the last four years In Y Alavi and A J Schenk, editors, Graph Theory, Combinatorics, and Applications - Proceedings of the Seventh Quadrennial International Conference on the Theory and Applications of Graphs, pages John Wiley & Sons, Inc, Ne York, 1995 [6] D Bauer, E Schmeichel, and H J Veldman Progress on tough graphs - Another four years In Y Alavi, D Jones, H Gavlas, A Schenk, and M Schultz, editors, The Proceedings of the Eighth Quadrennial International Conference on Graph Theory, Combinatorics, Algorithms and Applications John Wiley & Sons, Inc To appear [7] G Chartrand and L Lesniak Graphs and Digraphs Chapman and Hall, London, 1996 [8] V Chvátal Private communication [9] V Chvátal Tough graphs and hamiltonian circuits Discrete Math, 5 : , 1973 [10] V Chvátal Hamiltonian cycles In E L Laler, J K Lenstra, A H G Rinnooy Kan and D B Shmoys, editors, The Traveling Salesman Problem, A Guided Tour of Combinatorial Optimization, Ch 11, pp John Wiley & Sons, Chichester, 1985 [11] H Enomoto, B Jackson, P Katerinis, and A Saito Toughness and the existence of k-factors J Graph Theory, 9 : 87 95, 1985 [12] M R Garey and D S Johnson Computers and Intractability Freeman, San Francisco, CA, 1979 [13] W Goddard and H Sart On some extremal problems in connectivity In Y Alavi, G Chartrand, O R Oellermann, and A J Schenk, editors, Graph Theory, Combinatorics, and Applications - Proceedings of the Sixth Quadrennial International Conference on the Theory and Applications of Graphs, pages John Wiley & Sons, Inc, Ne York,

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