Long cycles in triangle-free graphs with prescribed independence number and connectivity

ARTICLE IN PRESS

Journal of Combinatorial Theory, Series B 91 (2004) 43–55 http://www.elsevier.com/locate/jctb

Long cycles in triangle-free graphs with prescribed independence number and connectivity Hikoe Enomoto,a Atsushi Kaneko,b Akira Saito,c,1 and Bing Weid,2 a

Department of Mathematics, Keio University, Hiyoshi 3-14-1, Kohoku-Ku, Yokohama 223-8522, Japan b Department of Electronic Engineering, Kogakuin University, Nishi-Shinjuku 1-24-2, Shinjuku-Ku, Tokyo 163-8677, Japan c Department of Applied Mathematics, Nihon University, Sakurajosui 3-25-40, Setagaya-Ku, Tokyo 156-8550, Japan d Institute of Systems Science, Academia Sinica, Beijing 100080, China Received 17 June 1999

Abstract + theorem says that a 2-connected graph with aðGÞpkðGÞ is hamiltonian. The Chva´tal-Erdos We extend this theorem for triangle-free graphs. We prove that if G is a 2-connected trianglefree graph of order n with aðGÞp2kðGÞ  2; then every longest cycle in G is dominating, and G has a cycle of length at least minfn  aðGÞ þ kðGÞ; ng: r 2003 Elsevier Inc. All rights reserved. Keywords: Longest cycle; Triangle-free graph; Independence number; Connectivity

For a graph G; we denote the independence number and the connectivity of G by + [4] proved the following theorem. aðGÞ and kðGÞ; respectively. Chva´tal and Erdos Theorem A (Chva´tal and Erdo¨s [4]). A 2-connected graph G with aðGÞpkðGÞ is hamiltonian.

E-mail addresses: [email protected] (A. Kaneko), [email protected] (A. Saito), [email protected] (B. Wei). 1 Partly supported by the Ministry of Education, Science, Sports and Culture of Japan, Grant-in-Aid for Scientific Research (A), 10304008, 1998. 2 Partly supported by National Natural Foundation of China. 0095-8956/$ - see front matter r 2003 Elsevier Inc. All rights reserved. doi:10.1016/j.jctb.2003.05.002

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In this paper, we study an extension of Theorem A for graphs G with aðGÞ4kðGÞ: The extension depends on the interpretation of a hamiltonian cycle. In [5], Kouider interpreted a hamiltonian cycle in a graph G as a covering of V ðGÞ by one cycle, and extended Theorem A in the form of the relation between the ratio aðGÞ=kðGÞ and the number of cycles that cover V ðGÞ: For a real number r; we denote by Jrn the least integer n with nXr: Theorem B (Kouider [5]). If kX2 and G is
a k-connected graph with independent number a; then G contains m mp ka cycles C1 ; y; Cm such that Sm V ðC Þ ¼ V ðGÞ: i i¼1 In this paper we interpret a hamiltonian cycle as a cycle of length jV ðGÞj; and try to extend Theorem A in terms of the circumference (the length of a longest cycle) of G: Let circðGÞ denote the circumference of G: As an immediate corollary of Theorem B, we have a bound circðGÞXm1 jV ðGÞj; where m ¼ JaðGÞ=kðGÞn: In this paper we consider the difference aðGÞ  kðGÞ instead of the ratio aðGÞ=kðGÞ: Let G be a 2-connected graph G with aðGÞpkðGÞ þ c; where c is a nonnegative integer. We hope to find a function f ðcÞ satisfying circðGÞXjV ðGÞj  f ðcÞ by restricting ourselves to a particular class of graphs. Note that we cannot obtain such a function for general graphs. For integers k; c and l with kX2; c40 and lX1; let H1 ; y; Hkþc be k þ c copies of Kl (the complete graph of S order l) and H0 be a copy of Kk : Then join each vertex in kþc i¼1 V ðHi Þ to every vertex in V ðH0 Þ: Let G be the resulting graph. (Using the notation in [3], G ¼ Kk þ ðk þ cÞKl :) Then aðGÞ ¼ k þ c; kðGÞ ¼ k; and hence aðGÞ  kðGÞ ¼ c: On the other hand, jV ðGÞj ¼ ðk þ cÞl þ k; circðGÞ ¼ kl þ k and hence jV ðGÞj  circðGÞ ¼ cl; which can be arbitrarily large. In the above example, however, we cannot replace Kl by a noncomplete graph since it would raise the independence number while the connectivity remains the same. Therefore, if we restrict ourselves to triangle-free graphs, we may be able to obtain a required function. Motivated by the above observation, we investigate the property of a longest cycle in a triangle-free graph G: A cycle C in a graph G is said to be dominating if G  V ðCÞ has no edge. First, we prove the following theorem. Theorem 1. Let G be a 2-connected triangle-free graph. If aðGÞp2kðGÞ  2; then every longest cycle is dominating. Then, using Theorem 1, we give a lower bound to the circumference. Theorem 2. Let G be a 2-connected triangle-free graph of order n: If aðGÞp2kðGÞ  2; then circðGÞXminfn  aðGÞ þ kðGÞ; ng: The bound in Theorem 2 is sharp. We do not know whether the condition aðGÞp2kðGÞ  2 is necessary in both theorems. But we know that the conclusions of

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the theorems do not hold if aðGÞ is large. Detailed discussions are made at the end of the paper. There are several related works concerning longest cycles in triangle-free graphs. Aung [1] has studied properties of such cycles, and has proved the following theorem. Theorem C (Aung [1]). Let C be a longest cycle in a 2-connected triangle-free graph G with minimum degree d: If jGjp6d  6; then G  V ðCÞ has at most one edge. Brandt [2] has investigated the weak-pancyclicity of triangle-free graphs. Theorem D (Brandt [2]). Let G be a triangle-free nonbipartite graph of order n with independence number a: If G is not isomorphic to a cycle of order five and the minimum degree of G exceeds 13 n; then for each integer k with 4pkpminfn; 2ðn  aÞg; G has a cycle of length k: Note that while the above results assume large minimum degree, the result in this paper puts no assumption on minimum degree. Concerning a triangle-free graph G with aðGÞpkðGÞ; Lou has proved that such a graph is almost pancyclic. Theorem E (Lou [6]). A 2-connected triangle-free graph G with aðGÞpkðGÞ has a cycle of every length k for 4pkpjGj unless G is a balanced complete bipartite graph or a cycle of length five. For graph-theoretic terminology not explained in this paper, we refer the reader to [3]. Let G be a graph. We denote the minimum degree of G by dðGÞ: Let xAV ðGÞ and let H be a subgraph of G (possibly xeV ðHÞ). We define NH ðxÞ by NH ðxÞ ¼ fvAV ðHÞ: xvAEðGÞg: In particular, NG ðxÞ is the neighborhood of x in G: Let þþ  C ¼ x0 x1 ?xl1 x0 be a cycle. Then we define xþ ¼ xiþ2 i ¼ xiþ1 ; xi ¼ xi1 and xi (Suffices are counted modulo l). Furthermore, for a positive integer k; we define xðkÞþ ¼ xiþk : For X CV ðCÞ; let X þ ¼ fxþ : xAX g and X ðkÞþ ¼ fxðkÞþ : xAX g: We adopt the same notation for paths. In arguments dealing with two or more cycles and paths, we may have to specify which path or cycle we consider when we refer successors and predecessors. If we refer the successor of a vertex x along a path or a cycle T; we denote it by xþðTÞ : Similarly, we use notations such as xðTÞ and xþþðTÞ : In this paper, if no path or cycle is specified in this way, it is always assumed that the one we consider is the cycle C: The length of a path P (resp. a cycle C) is denoted by lðPÞ (resp. lðCÞ). For a path ~xj : The same path, P ¼ x1 x2 yxl ; the subpath xi xiþ1 yxj1 xj is denoted by xi P ’

traversed in the opposite direction, is denoted by xj P xi : When no fear of confusion ’

~xj and xj P xi as vertex sets. Thus, if arises, we sometimes consider these subpaths xi P ~xj ; we write uAxi P ~xj : For a cycle C and u; vAV ðCÞ; we use a vertex u appears in xi P

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~v and u C v in a similar way. For a path P ¼ x1 x2 ?xl ; we call x1 and xl the uC starting vertex and the terminal vertex, respectively. A path whose starting vertex and terminal vertex are u and v; respectively, is called a uv-path. For a positive integer k; a graph G is said to be k-path-connected if for every pair of distinct vertices u and v there exists a uv-path of length at least k: We consider a matching M in a graph G as a set of edges. Hence jMj is the number of edges in M: For e ¼ uvAEðGÞ; we denote V ðeÞ ¼ fx; ygCV ðGÞ; and for F CEðGÞ S we denote V ðF Þ by V ðF Þ ¼ eAF V ðeÞ: Thus, if F is a matching, then jV ðF Þj ¼ 2jF j: Before proving Theorem 1, we prove two lemmas. Lemma 3. Let G be a 2-connected triangle-free graph and let x0 AV ðGÞ: If degG ðxÞX3 for each xAV ðGÞ  fx0 g; then G is 4-path-connected. Proof. Let u and v be distinct vertices in G; and let P be a longest uv-path in G: Assume lðPÞp3: If possible, we choose P so that x0 eV ðPÞ  fu; vg: Since G is 2connected, lðPÞX2: Assume P ¼ ux0 v: By the choice of P; P is the only uv-path of length two. On the other hand, since G is 2-connected, there exists a uv-path Q in G  x0 : This is possible only if lðQÞ ¼ 1; or Q ¼ uv: However, this implies that G has a triangle uvx0 u: This contradicts the assumption. Thus, we have either x0 auþðPÞ or x0 avðPÞ : By symmetry, we may assume x0 auþðPÞ : Then degG ðuþðPÞ ÞX3 and hence NG ðuþðPÞ Þ  fu; uþþðPÞ ga|; say yANG ðuþðPÞ Þ  fu; uþþðPÞ g: Since lðPÞp3 and G is triangle-free, yeV ðPÞ: Since G is 2-connected, there exists a uy-path Q in G  uþðPÞ : If ~yuþðPÞ P ~v is a uv-path which is longer than P; a ðV ðQÞ  fugÞ-V ðPÞ ¼ |; then uQ contradiction. Therefore, ðV ðQÞ  fugÞ-V ðPÞa|: Take wAðV ðQÞ  fugÞ-V ðPÞ ’ ~y is as short as possible. Let P0 ¼ uuþðPÞ y Q wP ~v: Then P0 is a uv-path in G so that wQ with lðP0 ÞX3: Since P is a longest path and lðPÞp3; we have lðPÞ ¼ lðP0 Þ ¼ 3 and P0 ¼ uuþðPÞ yv: If vðPÞ ¼ uþþðPÞ ¼ x0 ; then P0 is a longest uv-path with x0 eV ðP0 Þ  fu; vg: This contradicts the choice of P: Thus, x0 auþþðPÞ : Then we can apply the same argument ’

as above to v P u and uþþðPÞ instead of P and uþðPÞ ; respectively, and we have that there exists a vertex y0 ANG ðuþþðPÞ Þ  V ðPÞ such that uy0 AEðGÞ: Note yay0 since G is triangle-free. Let P00 ¼ uy0 uþþðPÞ uþðPÞ yv: Then lðP00 ÞX44lðPÞ: This contradicts the assumption, and the lemma follows. & Lemma 4. Let C be a longest cycle of a graph G and let H be a 4-path-connected subgraph of G  V ðCÞ: Suppose x1 and x2 are distinct vertices of NC ðHÞ and ~x2 ÞX6 and xðiÞþ x ð jÞþ eEðGÞ for any i; j with jNH ðx1 Þ,NH ðx2 ÞjX2: Then lðx1 C 1 2 1pi; jp3:

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Proof. By the choice of x1 and x2 and the assumption, we can choose u1 ANH ðx1 Þ and u2 ANH ðx2 Þ so that u1 au2 : Then since H is 4-path-connected, there exists a u1 u2 path Q in H with lðQÞX4: ~u2 x2 C ~x1 : Then C 0 is a cycle with Let C 0 ¼ x1 u1 Q ~x2 Þ þ lðQÞ þ 2XlðCÞ  lðx1 C ~x2 Þ þ 6: lðC 0 Þ ¼ lðCÞ  lðx1 C ~x2 ÞX6: Since C is a longest cycle, we have lðx1 C ðiÞþ ð jÞþ ðiÞþ ~x2 and x ð jÞþ Ax2 C ~x1 : Let Assume x1 x2 AEðGÞ: Since lðQÞX4; x1 Ax1 C 2 ’

ð jÞþ ðiÞþ ~ x1 C x2 u2

C 0 ¼ x1 C x2

’

Q u1 x1 : Then lðC 0 ÞXlðCÞ  ði þ j  2Þ þ 5 ¼ lðCÞ þ 7 

ði þ jÞ: Since 0pi; jp3; lðC 0 Þ4lðCÞ: This is a contradiction.

&

Proof of Theorem 1. Let kðGÞ ¼ k; and assume G has a longest cycle C which is not dominating. Let H be a component of G  V ðCÞ with jHjX2: Claim 1. dðHÞX2: Proof. Let uAV ðHÞ: Since jHjX2; NH ðuÞa|: Let vANH ðuÞ: Then ðNC ðuÞ,NC ðvÞÞþ ,NH ðvÞ is an independent set since C is a longest cycle and G is triangle-free. Again since G is triangle-free, NC ðuÞ-NC ðvÞ ¼ | and jðNC ðuÞ,NC ðvÞÞÞþ ,NH ðvÞj ¼ jNC ðuÞ,NC ðvÞj þ jNH ðvÞj ¼ jNC ðuÞj þ jNC ðvÞj þ jNH ðvÞj ¼ degG v þ jNC ðuÞjpaðGÞp2k  2:

ð1Þ

On the other hand, degG vXk since G is k-connected. Therefore, jNC ðuÞjpk  2: This implies degH uXk  ðk  2Þ ¼ 2; and hence the claim follows. & Note that if degH u ¼ 2 in the proof of the above claim, then the equality holds in (1). This implies the following. ð Þ If uAV ðHÞ and degH u ¼ 2; then for each vANH ðuÞ; (1) ðNC ðuÞ,NC ðvÞÞÞþ ,NH ðvÞ is a maximum independent set, (2) degG v ¼ degG u ¼ k; and (3) jNC ðuÞj ¼ k  2: We consider two cases. Case 1: dðHÞX3: If H is not a block, let B be an endblock of H and let cB be the unique cutvertex of H contained in B: If H is a block, let B ¼ H and let cB be an arbitrarily chosen vertex in H: Since dðHÞX3; B is 2-connected and 4-pathconnected by Lemma 3. Let T ¼ NC ðB  fcB gÞ; and let T0 ¼ fxAT: NBfcB g ðxÞ ¼ NBfcB g ðx0 Þ ¼ fug for some x0 AT  fxg and uAB  fcB gÞg

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~x0 ; where x0 AT and xC ~x0 -T ¼ |: By and T1 ¼ T  T0 : For xAT1 ; let Ix ¼ xC ð3Þþ Lemma 4, lðIx ÞX6: In particular, x AIx for each xAT1 : Let S be a maximum independent set in B: ð3Þþ

Claim 2. S,T þ ,T1

is an independent set in G:

Proof. Since C is a longest cycle and SCV ðHÞ; S,T þ is independent. By Lemma 4, yxð3Þþ eEðGÞ for each yAS and xAT1 : (Note that this holds even if y ¼ cB :) Again by ð3Þþ ð3Þþ ð3Þþ Lemma 4, x1 x2 eEðGÞ for each x1 ; x2 AT1 and xþ eEðGÞ for each x1 AT 1 x2 þ ð3Þþ and x2 AT1 with x1 ax2 : Furthermore, x x eEðGÞ for each xAT1 since G is ð3Þþ triangle-free. Thus, S,T þ ,T1 is independent. & Since degB yX3 for yAB  fcB g; jBjX4: Furthermore, since G is triangle-free, NB ðyÞ is an independent set of order at least three. Hence jSjX3; and ð3Þþ jS,T þ ,T1 j ¼ jSj þ jTj þ jT1 jXjTj þ jT1 j þ 3: Suppose NBfcB g ðT0 ÞaB  fcB g: Then T1 ,NBfcB g ðT0 Þ,fcB g separates G; and hence jT1 ,NBfcB g ðT0 Þ,fcB gj ¼ jT1 j þ jNBfcB g ðT0 Þj þ 1Xk: By the definition of T0 ; jT0 jX2jNBfcB g ðT0 Þj: Hence we have jT1 j þ 12jT0 j þ 1Xk; ð3Þþ

or 2jT1 j þ jT0 j ¼ jTj þ jT1 jX2k  2: Therefore, we have jS,T þ ,T1 jX 2k  2 þ 3 ¼ 2k þ 14aðGÞ: This contradicts Claim 2. Next, suppose NBfcB g ðT0 Þ ¼ B  fcB g: Choose y0 AB  fcB g so that jNC ðy0 Þ-T0 j is as small as possible. Let d0 ¼ jNC ðy0 Þ-T0 j and d1 ¼ jNC ðy0 Þ-T1 j: Then jNC ðy0 Þj ¼ d0 þ d1 ; d0 X2 and jT1 jXd1 : Since ðB  fy0 gÞ,NC ðy0 Þ separates G; jðB  fcB gÞ,NC ðy0 Þj ¼ jBj  1 þ d0 þ d1 Xk: Since jNC ðyÞ-T0 jXd0 for each yAB  fcB g; jT0 jXd0 jB  fcB gj ¼ d0 ðjBj  1Þ: By Claim 2, 2k  2XaðGÞX3 þ jTj þjT1 j ¼ 3 þ jT0 j þ 2jT1 j: Therefore, we have 2kXjT0 j þ 2jT1 j þ 5: Hence 2jBj þ 2d0 þ 2d1  2X2kXjT0 j þ 2jT1 j þ 5Xd0 ðjBj  1Þ þ 2d1 þ 5; which yields ðd0  2ÞðjBj  3Þ þ 1p0: Since d0 X2 and jBjX4; this is a contradiction. Therefore, the theorem follows in this case. Case 2: dðHÞ ¼ 2: Let y0 be a vertex in H with degH y0 ¼ 2: Let NH ðy0 Þ ¼ fy1 ; y2 g: Note that both NH ðy1 Þ and NH ðy2 Þ are independent sets since G is triangle-free. By the observation ð Þ after Claim 1, we have the following. (1) Both ðNC ðy0 Þ,NC ðy1 ÞÞþ ,NH ðy1 Þ and ðNC ðy0 Þ,NC ðy2 ÞÞþ ,NH ðy2 Þ are maximum independent sets, (2) degG y1 ¼ degG y2 ¼ k; and (3) jNC ðy0 Þj ¼ k  2: Claim 3. NC ðy1 Þ ¼ NC ðy2 Þa|:

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Proof. Since ðNC ðy0 Þ,NC ðy1 Þ,NC ðy2 ÞÞþ ,NH ðy1 Þ is an independent set while ðNC ðy0 Þ,NC ðy1 ÞÞþ ,NH ðy1 Þ is a maximum independent set, NC ðy2 ÞCNC ðy1 Þ: By symmetry, we have NC ðy1 ÞCNC ðy2 Þ and hence NC ðy1 Þ ¼ NC ðy2 Þ: If NC ðy1 Þ ¼ |; let S1 ¼ NC ðHÞþ ,NH ðy1 Þ: Then S1 is an independent set. Since G is k-connected, jNC ðHÞjXk: Furthermore, jNH ðy1 Þj ¼ k since NC ðy1 Þ ¼ |: Thus, jS1 j ¼ jNC ðHÞþ j þ jNH ðy1 Þj ¼ jNC ðHÞj þ jNH ðy1 Þj X 2k4aðGÞ: This is a contradiction. Therefore, NC ðy1 Þ ¼ NC ðy2 Þa|: & Let T ¼ NC ðy1 Þ,NC ðy0 Þ: ~x0 -T ¼ |: xþ C

For

xAT;

let

~x0 ; Ix ¼ xC

where

x0 AT

and

Claim 4. There exists no y1 y2 -path in H  y0 : In particular, NH ðy1 Þ-NH ðy2 Þ ¼ fy0 g: Proof. Assume there exists a y1 y2 -path Q in H  y0 : Since G is triangle-free, lðQÞX2: ~x0 : If x0 ANC ðy1 Þ ¼ NC ðy2 Þ; let C 0 ¼ xy1 y0 y2 x0 C ~x: If Let xANC ðy1 Þ and Ix ¼ xC ~y2 y0 x0 C ~x: Then in either case, C 0 is a cycle and hence x0 ANC ðy0 Þ; let C 0 ¼ xy1 Q lðC 0 ÞplðCÞ: This implies lðIx ÞX4 for each xANC ðy1 Þ: Since NC ðy1 Þa|; we can take a vertex x0 ANC ðy1 Þ: Let S2 ¼ ð3Þþ

ðNC ðy0 Þ,NC ðy1 ÞÞÞþ ,NH ðy1 Þ,fx0 independent. Since jIx jX4;

g: Since jS2 j ¼ aðGÞ þ 14aðGÞ; S2 is not

ð3Þþ NH ðx0 Þ

ð3Þþ

¼ |: Therefore, xþ x0 0

’

xANC ðy0 Þ,NC ðy1 Þ: If xANC ðy0 Þ; then xax0 : Let C ¼ x C

AEðGÞ for some

ð3Þþ ~x0 y1 Q ~y2 y0 x: x0 xþ C

Then C 0 is a cycle, which is longer than C: If xANC ðy1 Þ; then xax0 since G is ’ ð3Þþ ~x0 y1 y0 y2 x: Then again C 0 is a cycle, which is triangle-free. Let C 0 ¼ x C x0 xþ C longer than C; a contradiction. Therefore, there does not exist a y1 y2 -path in H  y0 : & Since degG y1 ¼ degG y2 ¼ k and NC ðy1 Þ ¼ NC ðy2 Þ; degH y1 ¼ degH y2 : Furthermore, since degH y1 ¼ degH y2 XdðHÞX2 and NH ðy1 Þ-NH ðy2 Þ ¼ fy0 g; we have NH ðy1 Þ  NH ðy2 Þa| and NH ðy2 Þ  NH ðy1 Þa|: Take yANH ðy2 Þ  NH ðy1 Þ and let S3 ¼ ðNC ðy0 Þ,NC ðy1 ÞÞþ ,NH ðy1 Þ,fyg: Then S3 is not independent. Since NG ðyÞ-ðNC ðy0 Þ,NC ðy1 ÞÞþ ¼ |; NH ðy1 Þ-NH ðyÞa|; say y0 ANH ðy1 Þ-NH ðyÞ: If y0 ay0 ; then y1 y0 yy2 is a y1 y2 -path in H  fy0 g; which contradicts Claim 4. If y0 ¼ y0 ; then yy0 y2 y is a triangle in G; again a contradiction. This completes the proof of Case 2, and the theorem follows. & A matching F ¼ fx1 y1 ; x2 y2 ; y; xk yk g in a graph G is said to be a pendant matching in G if for every i; 1pipk; either degG xi ¼ 1 or degG yi ¼ 1 holds.

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Lemma 5. Let G be a 2-connected triangle-free graph of order n: If aðGÞp2kðGÞ  2; then G  V ðCÞ has no pendant matching F with jF jXcircðGÞ  jV ðCÞj þ 1 for any cycle C in G: Proof. Assume, to the contrary, that there exists a cycle C such that G  V ðCÞ has a pendant matching F ¼ fx1 y1 ; y; xk yk g with kXcircðGÞ  jV ðCÞj þ 1: Choose such C so that C is as long as possible. We may assume degGV ðCÞ yi ¼ 1 for 1pipk: By Theorem 1, C is not a longest cycle, which implies kX2: Let S ¼ NGV ðCÞ ðx1 Þ,ðNC ðx1 Þ,NC ðy1 ÞÞþ : Since G is triangle-free, NG ðx1 Þ-NG ðy1 Þ ¼ | and hence jSj ¼ jNGV ðCÞ ðx1 Þj þ jðNC ðx1 Þ,NC ðy1 ÞÞþ j ¼ jNGV ðCÞ ðx1 Þj þ jNC ðx1 Þj þ jNC ðy1 Þj ¼ jNG ðx1 Þj þ jNC ðy1 ÞjXkðGÞ þ kðGÞ  1 ¼ 2kðGÞ  14aðGÞ: Therefore, S is not an independent set. Let u and v be a pair of adjacent vertices in S: Since G is triangle-free, fu; vggNGV ðCÞ ðx1 Þ: ~v x1 uv: If If uANGV ðCÞ ðx1 Þ and vANC ðx1 Þþ (possibly u ¼ y1 ), let C 0 ¼ vC ueV ðF Þ or u ¼ y1 ; let F 0 ¼ F  fx1 y1 g: If uAV ðF Þ  fy1 g; then u ¼ xi for some i; 2pipk since F is a pendant matching and degGV ðCÞ yi ¼ 1: In this case let F 0 ¼ F  fx1 y1 ; xi yi g: Since V ðCÞCV ðC 0 Þ; F 0 is a pendant matching in G  V ðC 0 Þ: Moreover, jV ðC 0 Þj ¼ jV ðCÞj þ 2: Therefore, jF 0 jX jF j  2XcircðGÞ  jV ðCÞj þ 1  2 ¼ circðGÞ  jV ðC 0 Þj þ 1: If uANGV ðCÞ ðx1 Þ and vANC ðy1 Þþ ; then uay1 since G is triangle-free. Let ~v y1 x1 uv; and C 0 ¼ vC  F  fx1 y1 g if ueV ðF Þ; 0 F ¼ F  fx1 y1 ; xi yi g if u ¼ xi for some i; 2pipk: Then F 0 is a pendant matching in G  V ðC 0 Þ; jV ðC 0 Þj ¼ jV ðCÞj þ 3 and jF 0 jXjF j  2XcircðGÞ  jV ðCÞj þ 1  24circðGÞ  jV ðC 0 Þj þ 1: ’

~u x1 v C uv and F 0 ¼ The other cases are similar. If fu; vgCNC ðx1 Þþ ; let C 0 ¼ vC ’

~u x1 y1 v C uv and F 0 ¼ F  fx1 y1 g: If uANC ðx1 Þþ and vANC ðy1 Þþ ; let C 0 ¼ vC ’

~u y1 v C uv and F 0 ¼ F  fx1 y1 g: F  fx1 y1 g: And if fu; vgCNC ðy1 Þþ ; let C 0 ¼ vC Then in every case, F 0 is a pendent matching in G  V ðC 0 Þ with jF 0 jXcircðGÞ  jV ðC 0 Þj þ 1: Now in every case we have a cycle C 0 and a pendant matching F 0 in G  V ðC 0 Þ with jF 0 jXcircðGÞ  jV ðC 0 Þj þ 1 and jV ðC 0 Þj4jV ðCÞj: This contradicts the maximality of jV ðCÞj: Therefore, the lemma follows. &

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Proof of Theorem 2. Assume circðGÞominfn  aðGÞ þ kðGÞ; ng; and let C be a longest cycle in G: By the assumption, C is not a hamiltonian cycle. Let V ðGÞ  V ðCÞ ¼ fx0 ; x1 ; y; xm g ðmXaðGÞ  kðGÞÞ: By Theorem 1, fx0 ; x1 ; y; xm g is an independent set. Claim 1. For each xi AV ðGÞ  V ðCÞ and for each y1 ; y2 ANC ðxi Þ with y1 ay2 ; there ~  exists a vertex zAyþ 1 C y2 such that NG ðzÞCV ðCÞ: Proof. Assume, to the contrary, that there exist a vertex xi AV ðGÞ  V ðCÞ and ~  distinct neighbors y1 ; y2 ANC ðxi Þ such that each zAyþ 1 C y2 satisfies NGV ðCÞ ðzÞa|: ~y2 is as short as possible. Since C is a longest We choose such ðxi ; y1 ; y2 Þ so that y1 C cycle, y2 ayþ 1: ~  Assume NGV ðCÞ ðz1 Þ-NGV ðCÞ ðz2 Þa| for some distinct vertices z1 z2 Ayþ 1 C y2 : Let ~y : Then NGV ðCÞ ðzÞa| for xj ANGV ðCÞ ðz1 Þ-NGV ðCÞ ðz2 Þ: We may assume z2 Azþ C 1

2

~  ~ ~ each zAzþ 1 C z2 ; and z1 C z2 is shorter than y1 C y2 : This contradicts the choice of ~  ðxi ; y1 ; y2 Þ: Therefore, NGV ðCÞ ðz1 Þ-NGV ðCÞ ðz2 Þ ¼ | for each z1 ; z2 Ayþ 1 C y2 with z1 az2 : ~  For each zAyþ 1 C y2 ; take a neighbor xjðzÞ ANGV ðCÞ ðzÞ: By the choice of ~  ðxi ; y1 ; y2 Þ; jðzÞai; and j : yþ 1 C y2 -f0; 1; y; mg  fig is an injection. Further~y1 xi y2 and F ¼ more, for each xjðzÞ ; N þ ~  ðxjðzÞ ÞÞ ¼ fzg: Let C 0 ¼ y2 C y1 C y2

þ~  0 ~  fzxjðzÞ : zAyþ 1 C y2 g: Then V ðC Þ ¼ V ðCÞ  y1 C y2 ,fxi g and hence þ~  ~  jV ðC 0 Þj ¼ jV ðCÞj  jyþ 1 C y2 j þ 1 ¼ circðGÞ  jy1 C y2 j þ 1:

~  On the other hand, V ðGÞ  V ðC 0 Þ ¼ ðfx0 ; y; xm g  fxi gÞ,yþ 1 C y2 ; and hence F is 0 ~  a pendant matching in G  V ðC 0 Þ of order jF j ¼ jyþ 1 C y2 j ¼ circðGÞ  jV ðC Þj þ 1: This contradicts Lemma 5. Therefore, the claim follows. & Let NG ðx0 Þ ¼ fy1 ; y2 ; y; yk gCV ðCÞ: Then kXkðGÞX2: We may assume ~yiþ1 ð1pipkÞ; y1 ; y; yk appear in the consecutive order along C: Let Ii ¼ yi C where we consider ykþ1 ¼ y1 : By Claim 1, for each i; 1pipk; there exists a vertex þ~ ~  zi Ayþ i C yiþ1 such that NG ðzi ÞCV ðCÞ: Choose such zi so that yi C zi is as short as possible. Let T ¼ fx0 ; x1 ; y; xm ; z1 ; y; zk g: Since jTj ¼ m þ 1 þ kXaðGÞ  kðGÞ þ kðGÞ þ 1 ¼ aðGÞ þ 1; T is not independent. On the other hand, fx0 ; y; xm g is independent, and there does not exist an edge between fx0 ; y; xm g and fz1 ; y; zk g: Thus, zi zj AEðGÞ for some i; j with 1piojpk: þ~ ~ Choose vi Ayþ i C zi and vj Ayj C zj which satisfy vi vj AEðGÞ or

NGV ðCÞ ðvi Þ-NGV ðCÞ ðvj Þa|

ð

Þ



þ~ ~ so that jyþ i C vi j þ jyj C vj j is as small as possible. Since ðzi ; zj Þ satisfies ( ), vi and vj are well-defined.

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’

~yj x0 : If NGV ðCÞ ðvi Þ-NGV ðCÞ ðvj Þa|; we may If vi vj AEðGÞ; let C0 ¼ x0 yi C vj vi C ’

~yj x0 : Since C is a longest assume xm ANG ðvi Þ-NG ðvj Þ and let C0 ¼ x0 yi C vj xm xi C þ~  ~  cycle, yþ i C vi ,yj C vj a|: By the choices of zi and zj ; NG ðuÞ-fx1 ; y; xm ga| þ~  ~  for each uAyþ i C vi ,yj C vj : Furthermore, by Claim 1 and the choices of vi ~v ,yþ C ~v : Take and vj ; NG ðuÞ-NG ðu0 Þ-fx1 ; y; xm g ¼ | for each u; u0 Ayþ C i

i

j

j

þ~  ~  Then vertex xjðuÞ ANG ðuÞ-fx1 ; y; xm g for each uAyþ i C vi ,yj C vj : þ~  þ~  j : yi C vi ,yj C vj -f1; y; mg is an injection. Furthermore, by the choices ~v ,yþ C ~v Þ if and vj ; mejðyþ C xm AV ðC0 Þ: Hence F¼ of vi

a

i

i

j

j

þ~  ~  fuxjðuÞ : uAyþ i C vi ,yj C vj g is a matching in G  V ðC0 Þ: þ~  ~  Claim 2. degGV ðC0 Þ xjðuÞ ¼ 1 for every uAyþ i C vi ,yj C vj :

Proof. Assume, to the contrary, degGV ðC0 Þ xjðuÞ X2: Then since V ðGÞ  V ðC0 ÞC þ~  ~  and fx1 ; y; xm g is independent, NG ðxjðuÞ Þ; y; xm g,yþ i C vi ,yj C vj þ~  þ~  ~v ,yþ C ~v Þ  fug: By symðy C v ,y C v Þ  fuga|; say u0 ANG ðxjðuÞ Þ-ðyþ C i

i

j

j

~  uAyþ i C vi :

0

i þ~  Ayj C vj ;

i

j

j

If u then xjðuÞ ANG ðuÞ-NG ðu0 Þ; metry we may assume which contradicts the choice of vi and vj : On the other hand, by Claim 1, ~  u0 eyþ i C vi : Therefore, we have degGV ðC0 Þ xjðuÞ ¼ 1: & By Claim 2 F is a pendant matching in G  V ðC0 Þ: However, þ~  ~  jF j ¼ jyþ i C vi j þ jyj C vj jXjV ðCÞj  jV ðC0 Þj þ 1

¼ circðGÞ  jV ðC0 Þj þ 1: This contradicts Lemma 5, and the theorem follows. & The lower bound minfn  aðGÞ þ kðGÞ; ng in Theorem 2 is sharp. Let G ¼ Km;mþc with mX2 and cX0: Then G is a 2-connected triangle-free graph of order 2m þ c with aðGÞ ¼ m þ c and kðGÞ ¼ m: Hence jV ðGÞj  aðGÞ þ kðGÞ ¼ 2m: On the other hand, circðGÞ ¼ 2m: Hence we have infinitely many graphs which attain the equality. We do not know whether the upper bound 2kðGÞ  2 of aðGÞ in Theorems 1 and 2 can be relaxed. However, we know that there exists a limit within which the conclusion of each theorem remains true. For i ¼ 1; 2; let Gi be the complete ðiÞ ðiÞ ðiÞ ðiÞ bipartite graph with partite sets fx1 ; y; xr1 g and fy1 ; y; ysrþ1 g; where rX2 and sX2r  1: Let G be the graph with V ðGÞ ¼ V ðG1 Þ,V ðG2 Þ and EðGÞ ¼ ð1Þ ð2Þ

EðG1 Þ,EðG2 Þ,fyj yj : 1pjps  r þ 1g: In other words, G is the graph obtained ð1Þ

ð1Þ

ð2Þ

ð2Þ

from G1 and G2 by joining fy1 ; y; ysrþ1 g and fy1 ; y; ysrþ1 g by a matching. Then G is a 2-connected triangle-free graph with aðGÞ ¼ s; kðGÞ ¼ r and hence

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jV ðGÞj  aðGÞ þ kðGÞ ¼ 2s  s þ r ¼ s þ r: On the other hand, it is not difficult to see circðGÞ ¼ 6r  6: Therefore, if s45r  6; circðGÞojV ðGÞj  aðGÞ þ kðGÞ: Hence the conclusion of Theorem 2 is valid only if aðGÞp5kðGÞ  6: Though the upper bound 2kðGÞ  2 of aðGÞ in Theorem 1 may not be best possible, it is almost best possible. To see this, for a positive integer s we construct a graph Gs in the following way. Let X ¼ fx1 ; x2 ; y; xsþ1 ; x01 ; x02 ; y; x0s g; Y ¼ fy1 ; y2 ; y; ysþ1 ; y01 ; y02 ; y; y0s g; and W ¼ fw1 ; y; w2s g be mutually disjoint sets. Define Gs by V ðGs Þ ¼ X ,Y ,W and (see Fig. 1) EðGs Þ ¼ fxi yi : 1pips þ 1g,fx0i y0i : 1pipsg ,fxi wj : 1pips þ 1; ipjpi þ s  1g ,fyi wj : 1pips þ 1; 1pjpi  1g ,fyi wj : 1pips þ 1; i þ spjp2sg ,fx0i wj : 1pips; ipjpi þ s  1g ,fy0i wj : 1pips; 1pjpi  1g,fy0i wj : 1pips; i þ spjp2sg: Then Gs is a triangle-free graph of order 6s þ 2: Furthermore, X ; Y and Z are all independent sets. Since jW j ¼ 2s and Gs  W ¼ ð2s þ 1ÞK2 ; no cycle in Gs is dominating. We investigate the connectivity and the independence number of Gs :

W w1

y1

yi

ys+1

x1

xi

xs+1

wi-1

wi

ws-1

ws+1

ws

x′1

x′i

y′1

y′i Fig. 1. Gs :

wi+s-1 wi+s

x′s

y′s

w2s-1

w2s

ARTICLE IN PRESS 54

H. Enomoto et al. / Journal of Combinatorial Theory, Series B 91 (2004) 43–55

Lemma 6. kðGs Þ ¼ s þ 1: Proof. By the construction of Gs ; we have jNGs ðwk Þ-fxi ; yi gj ¼ 1 and jNGs ðwk Þ-fx0j ; y0j gj ¼ 1ð1pips þ 1; 1pjps; 1pkp2sÞ:

ð Þ

Since dðGs Þ ¼ s þ 1; we have kðGs Þps þ 1: We claim that Gs is ðs þ 1Þ-connected. First, consider wk and wl ð1pkolp2sÞ: By ( ), for each i with 1pips þ 1; there exists a path from wk and wl which only passes through vertices in fxi ; yi g: (Therefore, the length of this path is two or three.) Similarly, for each j with 1pjps; there exists a path from wk to wl which only passes through vertices in fx0j ; y0j g: These 2s þ 1 paths are openly disjoint. Now take TCV ðGs Þ with jTj ¼ s and consider Gs  T: By the argument in the previous paragraph, vertices in W  T lie in the same component of Gs  T: If fxi ; yi ggT (1pips þ 1), then again by ( ), there exists an edge between fxi ; yi g  T and W  T: Hence fxi ; yi g  T and W  T are contained in the same component of Gs  T: Similarly, if fx0j ; y0j ggT (1pjps), then fx0j ; y0j g  T and W  T are contained in the same component (1pjps). Therefore, Gs  T is connected, and hence Gs is ðs þ 1Þ-connected. & Lemma 7. aðGs Þ ¼ 2s þ 2: Proof. Let W1 ¼ fw1 ; y; ws g and W2 ¼ fwsþ1 ; y; w2s g: By the construction of Gs ; if wi AW1 ; then 1pips and NGs ðwi Þ ¼ fx1 ; y; xi ; yiþ1 ; y; ysþ1 g,fx01 ; y; x0i ; y0iþ1 ; y; y0s g; and if wi AW2 ; then s þ 1pip2s and NGs ðwi Þ ¼ fy1 ; y; yis ; xisþ1 ; y; xsþ1 g,fy01 ; y; y0is ; x0isþ1 ; y; x0s g: Since fw1 g,fy1 ; x2 ; y; xsþ1 g,fy01 ; x02 ; y; x0s g is independent, aðGs ÞX2s þ 2: Assume aðGs ÞX2s þ 3: Let S be a maximum independent set of Gs ; and let k ¼ jS-W j: Since Gs  W ¼ ð2s þ 1ÞK2 ; jS-ðX ,Y Þjp2s þ 1; and hence kX2: Since jS-W j ¼ k; jS-W1 jX12 k or jS-W2 jX12 k: We claim jS-W1 j ¼ jS-W2 j ¼ 12 k: If jS-W1 jX12 k; then since kX2; S-W1 a|: Choose wi1 ; wj1 AS-W1 so that j1  i1 is as large as possible. Then j1  i1 X12 k  1 (possibly k ¼ 2 and j1 ¼ i1 ). Let t1 ¼ j1  i1 : Since NGs ðwi1 Þ,NGs ðwj1 Þ ¼ fx1 ; yxj1 ; yi1 þ1 ; y; ysþ1 g,fx01 ; y; x0j1 ; y0i1 þ1 ; y; y0s g; S-ðX ,Y ÞCfy1 ; y; yi1 ; xj1 þ1 ; y; xsþ1 g,fy01 ; y; y0i1 ; x0j1 þ1 ; y; x0s g and hence jS-ðX ,Y Þjp2s þ 1 þ 2ð j1  i1 Þ ¼ 2s þ 1  2t1 : This implies jSjp2s þ 1  2t1 þ k: Since jSjX2s þ 3; we have t1 p12 k  1: Therefore, t1 ¼ 12 k  1: This implies k  0 ðmod 2Þ; jS-W1 j ¼ 12 k and jS-W2 j ¼ 12 k; and the claim follows in this case. If jS-W2 jX12 k; we can prove the claim by a similar argument.

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Now, since jS-W1 j ¼ jS-W2 j ¼ 12 k; we have S-ðX ,Y Þ ¼ fy1 ; y; yi1 ; xj1 þ1 ; y; xsþ1 g,fy01 ; y; y0i1 ; x0j1 þ1 ; y; x0s g for some wi1 ; wj1 AS-W1 with j1  i1 ¼ 12 k  1: In particular, we have xsþ1 AS since j1 ps: Since jS-W2 j ¼ 12 k; there exist wi2 ; wj2 AS-W2 with j2  i2 X12 k  1: Then NGs ðwi2 Þ,NGs ðwj2 Þ ¼ fy1 ; y; yj2 s ; xi2 sþ1 ; y; xsþ1 g ,fy01 ; y; y0j2 s ; x0i2 sþ1 ; y; x0s g and hence S-ðX ,Y ÞCfx1 ; y; xi2 s ; yj2 sþ1 ; y; ysþ1 g,fx01 ; y; x0i2 s ; y0j2 sþ1 ; y; y0s g: This implies jS-ðX ,Y Þjp2s þ 1  2ð j2  i2 Þp2s þ 3  k and jSjp2s þ 3: Thus, the equality holds, and hence we have S-ðX ,Y Þ ¼ fx1 ; y; xi2 s ; yj2 sþ1 ; y; ysþ1 g,fx01 ; y; x0i2 s ; y0j2 sþ1 ; y; y0s g: In particular, we have ysþ1 AS since j2 p2s: Now we have fxsþ1 ; ysþ1 gCS; but since xsþ1 ysþ1 AEðGs Þ; this is a contradiction. Therefore, we have aðGs Þ ¼ 2s þ 2: & By Lemma 7, we have aðGs Þ ¼ 2kðGs Þ: Therefore, the conclusion of Theorem 1 holds only if aðGÞp2kðGÞ  1:

References [1] M. Aung, Longest cycles in triangle-free graphs, J. Combin. Theory Ser. B 47 (1989) 171–186. + II, Springer, [2] S. Brandt, Cycles and paths in triangle-free graphs, The Mathematics of Paul Erdos Berlin, 1997, pp. 32–42. [3] G. Chartrand, L. Lesniak, Graphs & Digraphs, 3rd Edition, Wadsworth & Brooks/Cole, Monterey, CA, 1996. + A note on Hamiltonian circuits, Discrete Math. 2 (1972) 111–113. [4] V. Chva´tal, P. Erdos, [5] M. Kouider, Cycles in graphs with prescribed stability number and connectivity, J. Combin. Theory Ser. B 60 (1994) 315–318. + condition for cycles in triangle-free graphs, Discrete Math. 152 (1996) [6] D. Lou, The Chva´tal–Erdos 253–257.