Cycles of length 1 modulo 3 in graph

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1 Discrete Applied Mathematics 113 (2001) Note Cycles of length 1 modulo 3 in graph LU Mei, YU Zhengguang Department of Mathematical Sciences, Tsinghua University, Beijing , China Received 9 July 1999; received in revised form 19 October 2000; accepted 30 October 2000 Abstract We prove a conjecture of Saito that if a graph G with 3 has no cycle of length 1 (mod 3), then G has an induced subgraph which is isomorphic to the Petersen graph. The above result strengthened the result by Dean et al. that every 2-connected graph with 3hasa(1mod 3)-cycle if G is not isomorphic to the Petersen graph.? 2001 Elsevier Science B.V. All rights reserved. Keywords: Cycle; Modulo; Chord 1. Introduction We use Bondy and Murty [1] for terminology and notation not dened here and consider simple graphs only. Let G be a graph. If x V (G) and A V (G), then N A (x) is the neighborhood of x in A. When A = V (G), we denote N G (x)=n(x). Let C = v 1 v 2 v t v 1 be a cycle of G. For v i V (C), we use v i + ;vi to denote the successor and predecessor of v i on C, respectively. If v i ;v j V (C), then we use v i Cv j or v j Cv i to denote the v i ;v j -arc of C with the same or opposite orientation with respect to the orientation of C and we will consider v i Cv j and v j Cv i both as paths and as vertex sets. If e = v i v j E(G) \ E(C), then e is said to be a chord of C. Let f = v l v s be another chord of C. Ifv l = v i + and v s v j Cv i then e and f are said to be cross chords of C. Finally, we use l(c) and l(v i Cv j ) to denote the length of C and v i Cv j, respectively. For integers a and b a cycle is said to be an (a mod b)-cycle if its length is a (mod b). In [2], the following theorem is proved. Research supported by National Natural Science Foundation of Chain. Corresponding author. address: mlu@math.tsinghua.edu.cn (L. Mei) X/01/$ - see front matter? 2001 Elsevier Science B.V. All rights reserved. PII: S X(01)

2 330 L. Mei, Y. Zhengguang / Discrete Applied Mathematics 113 (2001) Theorem 1. (Dean et al. [2]). Let G be a 2-connected graph of minimum degree at least 3. (1) If G is not isomorphic to K 4 or K 3;n for some n 3, then G has a (2 mod 3)-cycle. (2) If G is not isomorphic to the Petersen graph, then G has a (1 mod 3)-cycle. In [3], Saito strengthened Theorem 1(1) in the following way. Theorem 2. (Saito [3]). Let G be a graph of minimum degree at least 3. If G has no (2 mod 3) -cycle, then G contains either K 4 or K 3;3 as an induced subgraph. He also conjectured that, under the same assumption, G contains (1 mod 3)-cycle unless G contains the Petersen graph as an induced subgraph. In this paper, we prove this conjecture. Theorem 3. Let G be a graph of 3. If G has no (1 mod 3)-cycle, then G contains the Petersen graph as an induced subgraph. From this theorem, we have: Corollary. Every 3-regular connected graph except for the Petersen graph has a (1 mod 3)-cycle. 2. Some lemma In this section, assume that G has no cycle of length 1 (mod 3), C a cycle in G and uv a chord on C. Lemma 1. (1) If l(c) 0 (mod 3), then l(ucv) 1 (mod 3) and l(vcu) 2 (mod 3) or l(ucv) 2 (mod 3) and l(vcu) 1 (mod 3). (2) If l(c) 2 (mod 3), then l(ucv) l(vcu) 1 (mod 3). Lemma 2. Let l(c) 2 (mod 3) and u 1 ucv. Ifl(uCu 1 ) 1 (mod 3) or l(ucu 1 ) 0( mod 3), then N C (u 1 ) ucv. Proof. Suppose there exists v 1 vcu s.t. u 1 v 1 E(G). Then by Lemma 1 and l(ucu 1 ) 1 (mod 3), l(u 1 Cv) l(v 1 Cu) 0 (mod 3). Thus l(vcv 1 ) 1 (mod 3). But the cycle uvcv 1 u 1 Cu would be a contradiction. The proof of l(ucu 1 ) 0 (mod 3) is similar. Lemma 3. Let l(c) 0 (mod 3), l(ucv) 2 (mod 3) and u 1 ucv. If l(ucu 1 ) 1 (mod 3), then N C (u 1 ) ucv. Specially, N C (u + ) ucv and N C (v ) ucv.

3 L. Mei, Y. Zhengguang / Discrete Applied Mathematics 113 (2001) Proof. Obviously, l(u 1 Cv) 1 (mod 3). If there exists v 1 vcu such that u 1 v 1 E(G), then we can assume that l(v 1 Cu 1 ) 2 (mod 3) by Lemma 1. Thus l(v 1 Cu) 1 (mod 3). But the cycle v 1 Cuv Cu 1 v 1 would be a contradiction. Lemma 4. Let l(c) 0 (mod 3) and l(ucv) 1 (mod 3). If there exists a chord u 1 v 1 in ucv, then N C (u + 1 ) u 1Cv 1 and N C (v 1 ) u 1Cv 1. Proof. Since ucvu is a cycle of length 2 (mod 3), N C (u 1 + ) (ucu 1 v+ 1 Cv)=N C(v 1 ) (ucu 1 v+ 1 Cv)= by Lemma 2. Suppose there exist y 1;y 2 vcu so that y 1 u 1 + ;y 2v 1 E(G). We would consider three cases. (1) l(ucu 1 ) 0 (mod 3) and then l(v 1 Cv) 0 (mod 3). Then the cycles y 1 Cu 1 + y 1 and vcy 1 u 1 + Cuv derive l(y 1 Cu) 2 (mod 3) and l(vcy 1 ) 1 (mod 3). And the cycles v 1 Cy 2v 1 and y 2Cu 1 v 1 v 1 y 2 derive l(vcy 2 ) 2 (mod 3) and l(y 2 Cu) 1 (mod3). Since l(vcu) 2 (mod 3), we have that l(y 1 Cu) 0 (mod 3) or l(vcy 2 ) 0 (mod 3). But the cycles y 1 Cuv Cv 1 u 1 u 1 + y 1 or vcy 2 v 1 v 1u 1 Cuv would be a contradiction. (2) l(ucu 1 ) 1 (mod 3) and then l(v 1 Cv) 2 (mod 3). Then the cycle y 1 Cu 1 + y 1 and v 1 Cy 1 u 1 + u 1v 1 derive l(y 1 Cu) 1 (mod 3) and l(vcy 1 ) 2 (mod 3). The cycles v 1 Cy 2v 1 and y 2Cu 1 v 1 v 1 y 2 derive l(vcy 2 ) 0 (mod 3) and l(y 2 Cu) 0 (mod 3). Since l(vcu) 2 (mod 3), we have l(vcy 1 ) 0 (mod 3) or l(vcy 2 ) 1 (mod 3). But the cycles vcy 1 u 1 + u 1Cuv or vcy 2 v 1 Cuv would be a contradiction. (3) l(ucu 1 ) 2 (mod 3) and then l(v 1 Cv) 1 (mod 3). This situation is similar to (2). From (1) (3), N C (u 1 + ) u 1Cv 1 and N C (v 1 ) u 1Cv Proof of Theorem 3 Assume that G has no cycle of length 1 (mod 3). Let P be the set of longest paths in G. For P = x 1 x 2 x n P, let m(p)=max{l : x l N(x 1 ) V (P)}. Choose P P so that m(p) is as large as possible. Let m(p)=m, then m 4 and C = x 1 x m x 1 is a cycle. Choose x t such that x t is the rst vertex in x + 2 Px m with x 1 x t E(G). Then x 1 x t is a chord of C. In the following proof, we would use the Search Procedure S 1 (x; y) on C. Search Procedure S 1 (x; y) (x; y V(C)): Step 1. Let u 0 :=x and v 0 :=y; i:=1. Step 2. u i :=v i 1 (i 1 (mod 2)) or u i:=v + i 1 (i 0 (mod 2)). If d C(u i )=2 or N(u i ) \ u i 1 Cv i 1 (i 1 (mod 2)) or N(u i ) \ v i 1 Cu i 1 (i 0 (mod 2)), then stop; else v i N (u i ) \{u + i ;u i }, i:=i + 1 and goto Step 2.

4 332 L. Mei, Y. Zhengguang / Discrete Applied Mathematics 113 (2001) Claim 1. l(c) 0 (mod 3) and then l(x 1 Px m ) 2 (mod 3). Proof. Suppose l(c) 2 (mod 3). Then by Lemma 2 and the choice of P, S 1 (x 1 ;x t ) would stop at a vertex u i such that d(u i ) = 2, a contradiction. Claim 2. Assume, without loss of generality, that l(x 1 Cx t ) 1 (mod 3). Proof. If l(x 1 Cx t ) 2 (mod 3) then replace P by P = x m 1 Px 1 x m Px n and assume x is the rst vertex in x m 3 Px m with x x m 1 E(G). If x x t Px m 3 then l(x m 1 P x ) 1 (mod 3) by Lemma 1. If x x 1 Px t and l(x m 1 P x ) 2 (mod 3), then l(x Px t ) 0 (mod 3) by l(x t Px m 1 ) 2 (mod 3). But the cycle x Px t x 1 x m x m 1 x would be a contradiction. Hence l(x m 1 P x ) 1 (mod 3). Claim 3. There exists x s x t+1 Cx m such that x s x t 1 E(G) and l(x t Cx s ) 0 (mod 3). Proof. Obviously, N (x t 1 ) V (C). If N(x t 1 ) x 1 Cx t, assume y N(x t 1 ) x 1 Cx t, then by Lemma 4 and the choice of P, S 1 (x t 1 ;y) would stop at u i with d(u i )=2, a contradiction. Hence there exists x s x t+1 Cx m such that x s x t 1 E(G). Obviously, l(x s Cx 1 ) 1 (mod 3) and l(x t Cx s ) 2 (mod 3). But l(x t Cx 1 ) 2 (mod 3), so l(x t Cx s ) 0 (mod 3) and l(x s Cx 1 ) 2 (mod 3). Claim 4. There exists x l x 2 Cx t 1 such that x l x s 1 E(G) and l(x 1 Cx l ) 1 (mod 3). Proof. By the choice of P, N (x s 1 ) V (C). Since l(x t 1 Cx s ) 1 (mod 3), there exists x l x t 2 Cx s+1 such that x l x s 1 E(G) by Claim 3. If x l x s+1 Cx 1, then l(x s Cx l ) 2 (mod 3) and the cycle x l Cx t 1 x s x s 1 x l derives l(x l Cx 1 ) 1 (mod 3). Hence l(x s Cx l ) 0 (mod 3), l(x l Cx 1 ) 2 (mod 3) and x l x 1. But the cycle x l Cx 1 x t x t 1 x s x s 1 x l would be a contradiction. Therefore x l x 2 Cx t 1. Let x l be the rst vertex in x 2 Cx t 1 with x l x s 1 E(G). By Lemma 3, l(x s 1 Cx l ) 1 (mod 3) and l(x 1 Cx l ) 1 (mod 3): Claim 5. s = m 1. Proof. Suppose s m 1, then N(x m 1 ) V (C) by the choices of P. Denote y N(x m 1 )\{x m ;x m 2 }.Ify x t Cx s 1 thenthe cycles x t Cyx m 1 x m x 1 x t and x t 1 Cyx m 1 Cx s x t 1 derive l(x t Cy) 0 (mod 3) and l(x t Cy) 1 (mod 3). So l(x t Cy) 2 (mod 3) and then l(ycx s ) 1 (mod 3). But the cycle ycx s x t 1 x t x 1 x m x m 1 y would be a contradiction. If y x 1 Cx l, then l(x m 1 Cy) 1 (mod 3) by the cycle x s 1 Cx l x s 1 with length 2 (mod 3). By Claim 4, l(x 1 Cx l ) 1 (mod 3) and then l(ycx l ) 2 (mod 3). But the cycle ycx l x s 1 x s x t 1 x t x 1 x m x m 1 y would be a contradiction. If y x + l Cx t 1, then the cycles x l Cyx m 1 Cx s 1 x l and ycx t 1 x s Cx m 1 y derive l(x l Cy) 1 (mod 3) and l(ycx t 1 ) 2 (mod 3), respectively. Since l(x l Cx t 1 )

5 L. Mei, Y. Zhengguang / Discrete Applied Mathematics 113 (2001) (mod 3), l(x l Cy) 2 (mod 3) and l(ycx t 1 ) 0 (mod 3). But the cycle ycx t 1 x s x s 1 x l Cx m 1 y would be a contradiction. Therefore y x s Cx m. Since l(x s 1 Cx l ) 1 (mod 3), N(y + ) ycx m 1 by Lemma 4 and the choice of P. Thus S 1 (x m 1 ;y) would stop at u i with d(u i ) = 2, a contradiction. Claim 6. x l = x 2. Proof. Suppose x l x 2, then N(x 2 ) V (C). Let y N(x 2 ) \{x 1 ;x 3 }. Obviously, y {x m 1 ;x t ;x t 1 } and y x m by Lemma 2. If y x + l Cx t 1, then l(x 2 Cy) 1 (mod 3) by Lemma 1. Thus l(x l Cy) 1 (mod 3). But the cycle x l Cyx 2 x 1 x m x m 1 x m 2 x l would be a contradiction. If y x t Cx m 2, then the cycles yx 2 Cy and x t Cyx 2 x 1 x t derive l(ycx m 2 ) 2 (mod 3) and l(x t Cy) 1 (mod 3). Since l(x t Cx m 2 ) 2 (mod 3), l(x t Cy) 2 (mod 3) and l(ycx m 2 ) 0 (mod 3). But the cycle x l Cx 2 y Cx t 1 x m 1 x m 2 x l would be a contradiction. Hence y x 3 Cx l. So S 1 (x 2 ;y) would stop at u i with d(u i ) = 2 by Lemma 4, a contradiction. Claim 7. d(x 1 )=d(x m 1 )=d(x 2 )=d(x m 2 )=3. Proof. Suppose there exists y V (C) \{x m ;x 2 } such that yx 1 E(G), then y x t+1 Cx m 1 by the choice of x t. Obviously, y x m 2 and y x m 1. On the other hand, the cycles ycx 1 y and x t Cyx 1 x 2 x m 2 x m 1 x t 1 x t derive l(ycx m 2 ) 0 (mod 3) and l(x t Cy) 1 (mod 3). Since l(x t Cx m 2 ) 2 (mod 3), l(ycx m 2 ) 2 (mod 3). But the cycle ycx m 1 x t 1 x t x 1 y would be a contradiction. Hence d(x 1 )=3. Suppose there exists y V (C) \{x 1 ;x 3 } such that y x 2 E(G), then y {x m 1 ;x m }.Ify x 3 Cx t, then S 1 (x 2 ;y ) would stop at u i with d(u i ) = 2 by the choice of P and Lemma 4, a contradiction. So y x t Cx m 3. Since the cycles y Cx m 2 x 2 y and x t Cy x 2 x 1 x t derive l(y Cx m 2 ) 2 (mod 3) and l(x t Cy ) 1 (mod 3), l(x t Cy ) 2 (mod 3). But the cycle x t 1 Cy x 2 x m 2 x m 1 x t 1 would be a contradiction. Hence d(x 2 )=3. By the proof of Claims 5 and 6, d(x m 1 )=d(x m 2 )=3. Claim 8. N (x 3 ) V (P). Proof. Suppose there exists y 0 V (P) s.t. x 3 y 0 E(G), then N(y 0 ) V (C) bythe choice of P. Let y 1 N (y 0 ) \{x 3 }, then y 1 x t.ify 1 x 4 Cx t 1, then the cycles x 3 Cy 1 y 0 x 3 and x 3 y 0 y 1 Cx t x 1 x 2 x 3 force l(x 3 Cy 1 ) 1 (mod 3). But when l(x 3 Cy 1 ) 1 (mod 3), we have l(y 1 Cx 3 ) 2 (mod 3) and then the cycle y 1 Cx 3 y 0 y 1 would be a contradiction. If there exist y 1 ;y 2 x t+1 Cx m 3, say y 2 y + 1 Cx m 3, such that y 1 ;y 2 N(y 0 ), then the cycles y 0 y 1 Cy 2 y 0 and y 0 y 2 Cy 1 y 0 derive l(y 1 Cy 2 ) 2 (mod 3) and

6 334 L. Mei, Y. Zhengguang / Discrete Applied Mathematics 113 (2001) l(y 2 Cy 1 ) 2 (mod 3). Since l(c) 0 (mod 3), l(y 2 Cy 1 ) 0 (mod 3). But the cycle y 0 y 2 Cy 1 y 0 has a chords x 1 x t, x 2 x m 2, a contradiction with Lemma 2. Hence N(y 0 ) x t+1 Cx m 3 1. Since 3, x m y 0 E(G). Obviously, t 5. Then the path x 4 Px t 1 x m 1 Px t x 1 x 2 x 3 y 0 x m Px n would be longer than P, a contradiction. Claim 9. There exists x k x m+1 Px n such that x k x 3 E(G) and l(x m Px k ) 1 (mod 3). Proof. Suppose that N (x 3 ) V (C). Since l(x m 1 Cx t 1 ) 2 (mod3), N(x 3 ) x m 1 Cx t 1 by Lemma 3. By Claim 7 and the fact x 3 x m E(G), N(x 3 ) x 2 Cx t 1. Denote x 3 N(x 3) \{x 2 ;x 4 }, then l(x 3 Cx 3 ) 1 (mod 3). Thus S 1(x 3 ;x 3 ) would stop at N(u 2i+1 ) \ (u 2i Cv 2i ) or N (u 2i ) \ v 2i 1 Cu 2i 1. Assume N(u 2i+1 ) \{u 2i Cv 2i }, then by Lemma 4, we just have N (u 2i+1 ) x m+1 Px n or N(u 2i+1 ) \ V (P). We use induction to prove l(x 3 Cu k ) 0 (mod 3) for k =1; 2;:::. Since l(x 3 Cx 3 ) 1 (mod 3), l(x 3 Cu 1 ) 0 (mod 3). Suppose it is true for all integers less than k. Assume k is an even number. Since l(x 3 Cx 3 x 3) 2 (mod 3), l(v k 1 Cu k 1 ) 1 (mod 3) by Lemma 1. Hence l(u k Cu k 1 ) 0 (mod 3) and then l(x 3 Cu k ) 0 (mod 3). Specially, l(x 3 Cu 2i+1 ) 0 (mod 3). If there exists v N (u 2i+1 ) x m+1 Px n, then the cycles x m Cu 2i+1 v Px m and u 2i+1 Cx 3 x 3x 2 x 1 x m Pvu 2i+1 derive l(x m Pv) 0 (mod 3) and l(x m Pv) 1 (mod 3). But when l(x m Pv) 2 (mod 3), the cycle u 2i+1 Cx 3 x 2 x m 2 x m 1 x m Pvu 2i+1 would be a contradiction. If there exists y 0 V (P) such that y 0 u 2i+1 E(G), then by the choice of P, N(y 0 ) V (C). If there exists y 1 N (y 0 ) x 2 Cx t 1 and y 1 u 2i+1, assume y 1 u 2i+1 Cx t 1, then the cycles u 2i+1 Cy 1 y 0 u 2i+1 and y 1 Cu 2i+1 y 0 y 1 derive l(u 2i+1 Cy 1 ) 2 (mod 3) and l(y 1 Cu 2i+1 ) 2 (mod 3). Since l(c) 0 (mod 3), l(y 1 Cu 2i+1 ) 0 (mod 3). But the cycle y 1 Cu 2i+1 y 0 y 1 has a cross chords x 1 x t and x 2 x m 2, a contradiction with Lemma 2. Hence N (y 0 ) x 2 Cx t 1 = {u 2i+1 } and similarly, N(y 0 ) x t Cx m 2 1. Hence x m u 2i+1 E(G) and the cycle u 2i+1 Cx 3 x 3x 2 x 1 x m y 0 u 2i+1 would be a contradiction. Hence there exists x k x m+1 Px n s.t. x 3 x k E(G). Obviously, the cycles x m Px k x 3 x 2 x 1 x m and x m Px k x 3 Cx t x 1 x m derive l(x m Px k ) 1 (mod 3). Claim 10. t =5. Proof. Suppose t 5, then x t 2 exists. If v 0 V (P) s.t. x t 2 v 0 E(G), then N(v 0 ) V (C) by the choice of P. Obviously, N (v 0 ) x 2 Cx t 1 ={x t 2 } and N(v 0 ) x t Cx m 2 1. Hence x m v 0 E(G). But the cycle x 3 Cx t 2 v 0 x m Px k x 3 would be a contradiction. Thus N(x t 2 ) V (P). Let y N (x t 2 ) \{x t 1 ;x t 3 }, then y x m 1. Since l(x m 1 Cx t 1 ) 2 (mod 3), N C (x t 2 ) x m Cx t 1 by Lemma 3. If y x m Px n then the cycles x 1 x m Pyx t 2 x t 1 x t x 1 and x 1 x m Pyx t 2 Cx 1 derive l(x m Py) 1 (mod 3). But the cycles x t 2 y Px k x 3 x 2 x m 2 x m 1 x t 1 x t 2 (x k x m Py) and x t 2 ypx k x 3 x 2 x m 2 x m 1 x t 1 x t 2 (y x m Px k ) would be a contradiction. Therefore y x 2 Cx t 2. Then by the same proof as that of Claim 9, we can derive a contradiction.

7 L. Mei, Y. Zhengguang / Discrete Applied Mathematics 113 (2001) Claim 11. m =9. Proof. By Claim 10 and replacing P with x m 1 Px 1 x m Px n, we have m = 9 and there exists x h x m Px n such that x 6 x h E(G) and l(x m Px h ) 1 (mod 3). By Claims 7, 10, 11 and the choice of P, we have Note 1 d(x 1 )=d(x 2 )=d(x 4 )=d(x 5 )=d(x 7 )=d(x 8 )=3. By the proofs of Claims 9 and 11, we have Note 2 For any x 9 x k path P 1 x m Px n and x 9 x h path P 2 x m Px n, l(p 1 ) l(p 2 ) 1 (mod 3). Claim 12. k = 10. Proof. Suppose that k 10. We rst use the following Search Procedure (2) to get a x 9 x k path. Search Procedure S 2 (x; y) (x; y V (P)): Step 1. Let a 0 :=x; b 0 :=y; i:=0. Step 2. Let a i :=b i 1 (on P). If N (a i) b + i 1 Px n =, then stop; else let b i :=N(a i ) b + i 1 Px n, i:=i + 1 and goto step 2. Suppose that S 2 (x 9 ;x k ) stop at i 0 +1 (assume, i 0 1 (mod 2)), then we have a x 9 x k path Q =x 9 Pa 1 b 1 P a i0 b i0 Pb i0 1a i0 1 b 2 a 2 Px k. By Note 2, l(q) 1 (mod 3). Thus we have a cycle C = x 1 x 9 Qx k x 3 x 2 x 7 x 6 x 5 x 1 with length 2 (mod 3): P(1) For any x V (Q), if there exists y V (Q) so that xy E(G), then N(y) (V (C) V (Q)) 2. Let y 0 N (y) (V (C) Q), then y 0 {x 6 } Q by Claim 8 and Note 1. If y 0 Q\{x}, say y 0 x 9 Qx, then l(y 0 Qx) 0 (mod 3) and l(y 0 Qx) 2 (mod 3) by l(c ) 2 (mod 3). Thus l(y 0 Qx) 1 (mod 3) and then we have a x 9 x k path x 9 Qy 0 yxqx k which has length 2 (mod 3), a contradiction with Note 2. So N(y) (V (C) Q) {x 6 ;x}. Next we would consider a i0+1 and would show N(a i0+1) b i0 C b i0 1, that is, N(a i0+1) b i0 1Pb i0. If there exists y V (P) such that ya i0+1 E(G), then N(y) x 1 Pb i0 by the choice of P. Thus d(y) 2 by P(1), a contradiction. Hence N(a i0+1) x 1 Pb i0. When i 0 3, we have a parallel chords a i0 b i0 1, b i0 2a i0 1 in C and l(c ) 2 (mod 3) and l(b i0 2C a i0+1) 1 (mod 3), then N(a i0+1) b i0 2C a i0 1 by Lemma 2. Let b i0+1 N(a i0+1) and b i0+1 is not the successor or predecessor of a i0+1 on C. Obviously, b i0+1 a i0 1;a i0. If b i0+1 b i0 2C a i0, then the cycles b i0+1c a i0+1b i0+1 and b i0+1a i0+1c a i0 1b i0 2C b i0+1 derive l(b i0+1c a i0 ) 1 (mod3) and l(b i0 2C b i0+1) 2 (mod3). By l(b i0 2C a i0 ) 2 (mod3), l(b i0 2C b i0+1) 0 (mod 3). Since l(x 9 C b i0 2) +l(a i0 1C x k ) 0 (mod 3), the x 9 x k path x 9 C b i0+1a i0+1b i0 a i0 b i0 1C x k would contradict with Note 2. Therefore N (a i0+1) b i0 C b i0 1. If i 0 = 1, we similarly have N(a 2 ) x k Pb 1.

8 336 L. Mei, Y. Zhengguang / Discrete Applied Mathematics 113 (2001) By S 2 (a i0+1;b i0+1), we would have a vertex u j0+1 with N(u j0+1) \ V (C ) by l(c ) 2 (mod 3). By P(1) and the choice of P, N(u j0+1) V (P), thus N(u j0+1) b i0 Px n by Note 1. Let v j0+1 N (u j0+1) b i0 Px n, then we have a x 9 x k path Q = x 9 a 1 b 1 P a i0 b i0 Pv j0+1u j0+1pu j0 v j0 P u 0 v 0 Pb i0 a i0 P b 2 a 2 Px k (assume, j 0 1 (mod2)) with length 1 (mod 3) by Note 2. Thus, the cycle x 1 x 9 Q x k x 3 x 2 x 7 x 6 x 5 x 1 with length 2 (mod 3) but has a cross chords b i0 u 0 ;v 0 u 1, a contradiction with Lemma 2. Hence k = 10. Similarly, we have: Claim 13. h = 10. Thus, G[{x 1 ; ;x 10 }] is a Petersen graph. Acknowledgements The authors express their deep thanks to the referees and editors for their helpful suggestions for the improvement of this work. References [1] J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, Macmillan Press, London, [2] N. Dean, A. Kaneko, K. Ota, B. Toft, Cycle modulo 3, preprint. [3] A. Saito, Cycle of length 2 modulo 3 in graph, Discrete Math. 101 (1992)