A Fan-type condition for claw-free graphs to be Hamiltonian

Discrete Mathematics 219 (2000) 195–205www.elsevier.com/locate/disc

A Fan-type condition for claw-free graphsto be Hamiltonian

Rao Li1

Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, USA

Received 1 February 1996; revised 19 July 1999; accepted 26 July 1999

Abstract

Let G be a 2-connected claw-free graph of order n. If for each pair of vertices u and v

dist(u; v) = 3⇒ max(d(u); d(v))¿(n− 4)=2;

then G is Hamiltonian. c© 2000 Published by Elsevier Science B.V. All rights reserved.

1. Introduction

We use Bondy and Murty [2] for terminology and notation not de�ned here andconsider only simple graphs. A graph G is called claw-free if G has no inducedsubgraph which is isomorphic to K1;3. The distance between vertices x and y; dist(x; y),of a connected graph G is the least number of edges in a path connecting x and y.The diameter, diam(G), of a connected graph G is the maximum distance among allpossible pairs of vertices in the graph G. The circumference, c(G), of a graph G isthe length of the longest cycle of G.The following result is obtained by Fan [4].

Theorem 1 (Fan [4]). Let G be a 2-connected graph of order n. If for each pair ofvertices u and v

dist(u; v) = 2⇒ max(d(u); d(v))¿n=2;

then G is Hamiltonian.

E-mail address: [email protected] (R. Li).1 Current address: School of Computer and Information Sciences, Georgia Southwestern State University,Americus, GA 31709, USA.

0012-365X/00/$ - see front matter c© 2000 Published by Elsevier Science B.V. All rights reserved.PII: S0012 -365X(99)00341 -6

196 R. Li /Discrete Mathematics 219 (2000) 195–205

Fan’s theorem was generalized in several ways, one of them was obtained byBedrossian et al. They weakened the conditions in Fan’s theorem and proved strongerresults. The interested readers are refered to Bedrossian et al. [1].Gould established the following result in [5].

Theorem 2 (Gould [5]). Every 2-connected claw-free graph with diameter at most 2is Hamiltonian.

Thus it is reasonable to assume that the diam(G) is at least three when investigatingthe Hamiltonicity of a 2-connected claw-free graph G.The main result of this paper is the following theorem that can be roughly regarded

as a Fan-type theorem for claw-free graphs.

Theorem 3. Let G be a 2-connected claw-free graph of order n. If for each pair ofvertices u and v

dist(u; v) = 3⇒ max(d(u); d(v))¿(n− 4)=2;then G is Hamiltonian.

Let H denote the graph with n=2p+3 (p¿3) vertices obtained from three disjointcomplete graphs H1 =K3; H2 =Kp, and H3 =Kp by adding the edges of two trianglesbetween two disjoint triples of vertices, each triple containing one vertex of H1; H2,and H3. Then H is a 2-connected claw-free graph of order n such that max(d(u); d(v))is at least d(n − 4)=2e − 1 for each pair of vertices u; v with dist(u; v) = 3, but H isnot Hamiltonian. Thus the bound (n− 4)=2 in Theorem 3 is sharp.

2. Lemmas

We �rst introduce some additional notation. Let A; B be two disjoint subsets of vertexset V (G) of a graph G. A path P=av1v2 : : : vsb is called an (A; B)-path if a ∈ A; b ∈ Band vi’s 6∈A ∪ B; E(A; B) is de�ned as the set {ab : a ∈ A; b ∈ B; ab ∈ E(G)}. For acycle C of a graph G, we de�ne C+ a cycle with a given orientation. If u; v are inV (C), then C+[u; v] denotes the consecutive vertices on C from u to v in the directionspeci�ed by C+. The same vertices, in reverse order, are given by C−[v; u]. We willconsider C+[u; v] and C−[v; u] both as paths and as vertex sets. We use u+ to denotethe successor of u on C+ and u− its predecessor. u++:=(u+)+ and u−:=(u−)−. LetA be a subset of V (C), we de�ne A+:={a+ : a ∈ A}; A−:={a−− : a ∈ A}. We needthe following lemmas to prove the main theorem.

Lemma 1. Let G be a claw-free graph and C a longest oriented cycle in G. Ifv ∈V (G)\V (C) is adjacent to x ∈ C; then x+x− ∈ E(G).

R. Li /Discrete Mathematics 219 (2000) 195–205 197

Lemma 2. Let G be a 2-connected claw-free nonhamiltonian graph; C a longestoriented cycle in G and P=v1v2 : : : vk(k¿1) a path in V (G)\V (C) such that v1; vk areadjacent to distinct vertices x; y on C with N (vi)⊆V (C)∪V (P) for some i; 16i6k.Then

(1) |C+[x+; y−]|¿k + 2; |C+[y+; x−]|¿k + 2.(2) xy+; xy−; xy++; xy− 6∈ E(G);yx+; yx−; yx++; yx−− 6∈ E(G).(3) |V (C)|¿2d(vi) + 4.

Proof. Both (1) and (2) follow from the maximality of the cycle C. Proof of (3)goes as follows. Assume that |V (C) ∩ N (vi)| = t. Since N (vi)⊆V (C) ∪ V (P); k =|V (P)|¿d(vi)−t+1. If t62, then k¿d(vi)−1. By (1), |V (C)|¿2(k+2)+2¿2d(vi)+4.If t¿3, let |N (vi) ∩ C+[x+; y−]|= a and |N (vi) ∩ C+[y+; x−]|= b. Then a+ b¿t − 2and |V (C)|¿i+3a+ a− 1+ k − (i− 1)+ i+3b+ b− 1+ k − (i− 1)+ 6= 4a+4b+6 + 2k¿4t − 8 + 6 + 2d(vi)− 2t + 2 = 2d(vi) + 2t¿2d(vi) + 6.

Lemma 3. Let G be a 2-connected nonhamiltonian claw-free graph and C a longestcycle in G. If v∈V (G)\V (C) be a vertex such that vx ∈ E(G); where x ∈V (C); then|V (C)|¿2d(v) + 4.

Proof. By Lemma 2, it su�ces to show that there exists a (x; y)-path P containing vsuch that x; y ∈ V (C); x 6= y; V (C)∩ [V (P) \ {x; y}] = ∅ and N (v)⊆V (C)∪V (P). Let

A= {a ∈ V (G)\V (C) : av ∈ E; ax ∈ E};

B= {b ∈ V (G)\V (C) : bv ∈ E; bx 6∈ E}:Since G is claw-free and C is a longest cycle, A and B are complete.Case 1: A=∅. Since G is 2-connected, G\{v} is connected, there exists a path from

some vertex b ∈ B to the cycle C, say, P = b : : : rw. If x = w, note that rx ∈ E; A= ∅,then rv 6∈ E, thus G[x; x+; r; v] is isomorphic to a claw. Thus x 6= w and we can easily�nd a desired path.Case 2: B=∅. Since G is 2-connected, G\{x} is connected, there exists a path from

a ∈ A to the cycle C, say, P = a : : : rw with V (P) ∩ A= {a}; V (P) ∩ V (C) = {w} andw 6= x. Then the path starting at x, sweeping v and A, following P to w is a desiredpath.Case 3: A 6= ∅; B 6= ∅.Case 3.1: There exists a path P disjoint with V (C) ∪ {v} and joining A to B.Let P=au1u2 : : : ukb. Since G is 2-connected, there exists a path from v to the cycle

C not containing x, say, P1 = v : : : u : : : w, where w 6= x is the �rst vertex of C on P1,and u is the last vertex of A ∪ B ∪ {v} on P1. If u = v, then the path starting at x,sweeping A; P; B and reaching v, following P1 to w is a desired path. If u ∈ A, thenthe path starting at x, sweeping v; B; P; A\{u} and reaching u, following P1 to w is adesired path. If u ∈ B, then the path starting at x, sweeping A; v; B\{u} and reachingu, following P1 to w is a desired path.


Case 3.2: Any path from A to B must pass through some vertex in V (C) ∪ {v}.Since G is 2-connected, G\{v} is connected, there exists a path P from b ∈ B to

some vertex of V (C) in G\{v}, among all these paths, we choose the shortest one,say, P= b : : : rw. Then V (P)∩ A= ∅, otherwise there exists an (A; B)-path which doesnot pass through any vertex in V (C)∪ {v}. If w 6= x, it is easy to �nd a desired path;if w = x, then, as before, G[x; x+; r; v] is isomorphic to a claw.

The following closure operation was introduced by Ryj�a�cek [7]. Let G be a claw-freegraph. The Ryj�a�cek closure CR(G) of G is de�ned as the graph obtained from G byiteratively performing local completion at vertices with connected neighborhoods untilno more edges can be added.

Lemma 4 (Ryj�a�cek [7]). Let G be a claw-free graph. Then CR(G) is uniquelydetermined by G and c(G) = c(CR(G)).

We also use the combination of Theorem 5 and Proposition 7 in [3] as our lemma.

Lemma 5 (Broersma and Trommel [3]). Let G= (V; E) be a graph and let {s; t; y; z}be a subset of four vertices of V such that yz 6∈ E and {s; t}⊆N (y) ∩ N (z). If G isclaw-free and st ∈ E; then for every cycle C1 of G+ yz there exists a cycle C0 of Gsuch that V (C1)⊆V (C0).

3. Proof of Theorem 3

Proof of Theorem 3. It is easy to see that Theorem 3 is true if n68. So we assumethat n¿9. Suppose G is a graph satisfying the conditions in Theorem 3, but G isnon-hamiltonian. Let C be a longest oriented cycle in G. Since G is 2-connected, thereexists a path P in G\C connecting two distinct vertices in V (C). Among all these paths,choose P= x1x2 : : : xk (k¿1) to be the longest one and let x1u ∈ E; xkv ∈ E, where u; vare two distinct vertices in V (C). By Lemma 1, u+u−; v+v− ∈ E. If d(x1)¿(n−4)=2 ord(xk)¿(n−4)=2, then, by Lemma 3, |V (C)|¿n, a contradiction. Thus, we can assumethat d(x1); d(xk) are less than (n−4)=2. By (2) in Lemma 2, we have uv− 6∈ E; vu− 6∈ E.Let y be the �rst vertex on C+[u+; v−] which is not adjacent to u, and z be the �rstvertex on C+[v+; u−] which is not adjacent to v. Let

V1 = C+[u; y−]; V2 = C+[y; v−];

V3 = C+[v; z−]; V4 = C+[z; u−]; V5 = V (G)\V (C):For a vertex v ∈ V (G), we de�ne Ni(v)=N (v)∩Vi and di(v)= |Ni(v)|. Then we havethe following �ve claims, Claims 2, 4 are obvious, Claims 1, 3, 5 are followed bytheir proofs.


Claim 1. dist(y; x1) = 3; dist(z; xk) = 3.

Proof. Clearly, yx1 6∈ E and N5(y)∩N5(x1) = ∅. If w ∈ N (y)∩N (x1)∩V (C), then Ghas a cycle

x1wyC+[y+; w−]C+[w+; u−]C+[u+; y−]ux1

longer than C. Since x1uy−y is a path of length three, dist(y; x1) = 3. Similarly,dist(z; xk) = 3.

Claim 2. N (z) ∩ [V1 ∪ {y}] = ∅; N (y) ∩ [V3 ∪ {z}] = ∅.

Claim 3. N−2 (y) ∩ N2(z) = ∅; N−

4 (z) ∩ N4(y) = ∅.

Proof. If w ∈ N−2 (y) ∩ N2(z), then G has a cycle

x1uC−[y−; u+]C−[u−; z]C−[w; y]C+[w+; v−]C+[v+; z−]vxk : : : x1

longer than C, a contradiction. Similarly, N−4 (z) ∩ N4(y) = ∅.

Claim 4. N5(y) ∩ N5(z) = N5(y) ∩ V (P) = N5(z) ∩ V (P) = ∅.

Claim 5. n− 46d(y) + d(z)6|V1|+ |V2|+ |V3|+ |V4|+ |V5| − 2− k = n− 2− k.

Proof. By Claim 1, d(y); d(z) ¿(n− 4)=2. By Claim 2, d1(y) + d1(z)6|V1| − 1 andd3(y) + d3(z)6|V3| − 1. By Claim 3, d2(y) + d2(z) = |N−

2 (y) ∪ N2(z)|6|V2| andd4(y)+d4(z)= |N−

4 (z)∪N4(y)|6|V4|. By Claim 4, d5(y)+d5(z)6|V5|−k. ThereforeClaim 5 follows from the foregoing inequalities.

Hence d(y)=(n−2)=2 and d(z)=(n−4)=2, or d(y)=(n−4)=2 and d(z)=(n−2)=2,or d(y) = d(z) = (n − 3)=2, or d(y) = d(z) = (n − 4)=2. The remainder of this paperis divided into two main cases.Case 1: d(y)=(n−2)=2 and d(z)=(n−4)=2, or d(y)=(n−4)=2 and d(z)=(n−2)=2,

or d(y) = d(z) = (n− 3)=2.Then, by the proof of Claim 5, we have the following Claim 1.1.

Claim 1.1. (1) k = 1.(2) d1(y) = |V1| − 1; d1(z) = 0;d3(y) = 0; d3(z) = |V3| − 1.(3) V2 = N−

2 (y) ∪ N2(z) which implies that zv−∈E.(4) V4 = N−

4 (z) ∪ N4(y) which implies that yu−∈E.(5) V5 = N5(y) ∪ N5(z) ∪ {x1} which implies that d5(y) + d5(z) = |V5| − 1.

Claim 1.2. |N (y) ∩ N (z)|= 2.

Proof. Notice that u; v; x1; y; z 6∈ N (y) ∪ N (z), thus n¿5 + |N (y) ∪ N (z)| = n + 2 −|N (y)∩N (z)| and |N (y)∩N (z)|¿2. If |N (y)∩N (z)|¿3, choose a; b; c ∈ N (y)∩N (z).


Since G[y; a; b; c] is not isomorphic to a claw, we have at least one of ab; bc; ca belongsto E. Note that G + yz has a cycle of length c(G) + 1, then by Lemma 5 we have acontradiction. Hence |N (y) ∩ N (z)|= 2.

It is not di�cult to see that there exist vertices s ∈ N2(y)∩N2(z); t ∈ N4(y)∩N4(z)such that

N (y) = C+[u+; y−] ∪ C+[y+; s] ∪ C+[t; u−] ∪ N5(y);N (z) = C+[v+; z−] ∪ C+[z+; t] ∪ C+[s; v−] ∪ N5(z):

Clearly, s 6= v−; t 6= u−, otherwise G[a−; a; y; z] is isomorphic to a claw, where a= uor v. Also s 6= y+; t 6= z+, otherwise we can �nd a longest cycle C1 by replacing x1by y (resp. z) such that y (resp. z) belongs to G\C1. Note that d(y); d(z)¿(n− 3)=2,then we can derive contradictions by Lemma 3.

Claim 1.3. N5(y) = N5(z) = ∅.

Proof. Suppose that w ∈ N5(y) 6= ∅. Since G is claw-free, st or ws or wt∈E. If st∈E,then by a similar argument as in the proof of Claim 1.2 we can derive a contradiction.If ws ∈ E, then G has a cycle

x1uC−[y−; u+]C−[u−; z]sC+[y; s−]C+[s+; v−]C+[v+; z−]vx1

longer than C, a contradiction. Similarly, we have a contradiction if wt ∈ E. ThusN5(y) = ∅. Similarly, N5(z) = ∅.

We say N (z) is disconnected, otherwise tv− ∈ E(CR(G)) and CR(G) has a cyclex1C+[v; t]C−[v−; y]C+[t+; u−]C+[u+; y−]ux1

of length c(G) + 1 and by Lemma 4 we have a contradiction. Thus N (z) is dis-connected, namely, E(C+[z+; t]; C+[s; v−] ∪ C+[v+; z−]) = ∅. Since G is claw-free,G[C+[z+; t]], G[C+[s; v−] ∪ C+[v+; z−]] are complete in G. Similarly, N (y) is dis-connected, E(C+[y+; s]; C+[t; u−] ∪ C+[u+; y−]) = ∅ and G[C+[y+; s]], G[C+[t; u−] ∪C+[u+; y−]] are complete in G.Note that V5={x1}, then N5(z+)=N5(s−)=∅ otherwise we can �nd longer cycles than

C. Moreover, we have N (z+)⊆C+[z; t] \ {z+} and N (s−)⊆C+[y; s] \ {s−}, otherwiseit can be easily seen that G has longer cycles than C. Therefore dist(z+; s−) = 3;max(d(z+); d(s−))6(n− 2)=2− 2 = (n− 6)=2, a contradiction.Case 2: d(y)=d(z)=(n−4)=2. By Claim 5, k62. By Lemma 5 again we can show

that |N (y)∩N (z)|62. When k=2, Claim 1.1 is still true except k=2. Note that u, v,x1, x2, y, z 6∈ N (y)∪N (z), we have n¿6+ |N (y)∪N (z)|= n+2− |N (y)∩N (z)| andtherefore |N (y)∩N (z)|¿2. Thus |N (y)∩N (z)|=2. Repeating the remaining arguments(almost the same) in Case 1, we can also derive a contradiction. Thus we can assumek = 1.


Clearly, |N (y) ∩ N (z)| 6= 0, otherwise n¿|N (y)|+ |N (z)|+ |{x1; u; v; y; z}|= n+ 1.Therefore, |N (y) ∩ N (z)|= 2 or 1 and we have the following two subcases.Case 2.1: |N (y) ∩ N (z)| = 1. Note that n¿|N (y) ∪ N (z)| + |{x1; u; v; y; z}| = n, we

have V (G) = N (y) ∪ N (z) ∪ {x1; u; v; y; z}. Without loss of generality, we assume thatthe vertex s ∈ N (y) ∩ N (z) is in C+[y+; v−]. By the proof of Claim 5, we have atleast one of yu−, zv− belongs to E. We �rst assume that both yu− and zv− are in E.Then it is not di�cult to see there exists a vertex t ∈ N4(z) such that

N (y) = C+[u+; y−] ∪ C+[y+; s] ∪ C+[t+; u−] ∪ N5(y);N (z) = C+[v+; z−] ∪ C+[z+; t] ∪ C+[s; v−] ∪ N5(z):

Clearly, t 6= u−, t+ 6= z+. As before, s 6= y+, v−.We say N5(z) = ∅. Suppose w ∈ N5(z) 6= ∅, since G is claw-free, ws or wt or

st ∈ E. For all three cases here we can easily �nd cycles in G that are longer than C,contradictions. Similarly, N5(y) = ∅.Furthermore, we say N (z) is disconnected, otherwise tv− ∈ E(CR(G)) and CR(G)

has a cycle

x1C+[v; t]C−[v−; y]C+[t+; u−]C+[u+; y−]ux1

of length c(G) + 1, we have a contradiction by Lemma 4. Thus N (z) is disconnected,namely, E(C+[z+; t]; C+[s; v−] ∪ C+[v+; z−]) = ∅. Since G is claw-free, G[C+[z+; t]],G[C+[s; v−] ∪ C+[v+; z−]] are complete in G. Similarly, N (y) is disconnected,E(C+[y+; s]; C+[t+; u−]∪C+[u+; y−])=∅ and G[C+[y+; s]], G[C+[t+; u−]∪C+[u+; y−]]are complete in G.Obviously, N5(z+)=N5(s−)=∅. Thus N (z+)⊆C+[z; t]\{z+} and N (s−)⊆C+[y; s]\

{s−}, otherwise it can be easily seen that G has longer cycles than C. Thereforedist(z+; s−) = 3;max(d(z+); d(s−))6(n− 4)=2− 2 = (n− 8)=2, a contradiction.Now we assume that only one of yu−, zv− belongs to E. Without loss of generality,

we assume that zv− ∈ E, yu− 6∈ E. Then it is not di�cult to see that there exists avertex s ∈ N2(y) ∩ N2(z) such that

N (y) = C+[u+; y−] ∪ C+[y+; s] ∪ N5(y);N (z) = C+[s; v−] ∪ C+[v+; z−] ∪ C+[z+; u−] ∪ N5(z):

As before again, s 6= y+, v−. It is clear N5(z)=∅, otherwise let w be a vertex in N5(z)and then G[z; s; u−; w] would be isomorphic to a claw.As before, N (z) is disconnected, E(C+[z+; u−]; C+[s; v−]∪C+[v+; z−])=∅ and both

G[C+[z+; u−]] and G[C+[s; v−]∪C+[v+; z−]] are complete in G. We say N5(z+) = ∅,otherwise let w ∈ N5(z+), then w ∈ N5(y) ∪ {x1} and we can easily �nd a cycle in Gthat is longer than C. It can also be veri�ed that N (z+)⊆(C+[z; u−]\{z+}). Therefored(z+)6(n−4)=2−2=(n−8)=2. Since z+u−ux1 is a path of length three, z+x1 6∈ E andN (z+)∩N (x1)=∅, dist(z+, x1)=3. However max(d(z+); d(x1))6(n−4)=2−2=(n−8)=2,a contradiction


Case 2.2: |N (y) ∩ N (z)| = 2. Then there exists exactly one vertex p ∈ V (G) \{x1; u; v; y; z} such that p 6∈ N (y)∪N (z). Let N (y)∩N (z)={a; b}, then we claim that

a ∈ C+[y+; v−]; b ∈ C+[z+; u−] or a ∈ C+[z+; u−]; b ∈ C+[y+; v−]:Suppose to the contrary, without loss of generality, we assume that a, b ∈ C+[y+; v−]and a+ ∈ C+[a; b]. As before, a 6= y+, b 6= v−. Since G[a; y; a+; z] is not isomorphicto a claw and ya+, yz are not E, za+ ∈ E. Similarly, zb+ ∈ E. It is not di�cult tosee that p ∈ C+[a++; b−−]. Note that V4 \ {z}=N4(y)∪N4(z) and N4(y)∩N4(z)= ∅,then there exists a vertex w ∈ C+[z+; u−] such that

N4(y) = C+[w+; u−]; N4(z) = C+[z+; w] or N4(y) = ∅; N4(z) = C+[z+; u−]:

Clearly, ab 6∈ E, otherwise we can derive a contradiction by using Lemma 5. If N4(y)=C+[w+; u−], N4(z) = C+[z+; w], since G[z; w; a; b] is not isomorphic to a claw, wa orwb ∈ E. Let s= a or b, then G has a cycle

x1uC−[y−; u+]C−[u−; w+]C+[y; s]C−[w; z]C+[s+; v−]C+[v+; z−]vx1

longer than C, a contradiction. If N4(y)= ∅, N4(z)=C+[z+; u−], since G[z; u−; a; b] isnot isomorphic to a claw, u−a or u−b ∈ E. Let s= a or b, then G has a cycle

x1C+[u; s]C−[u−; z]C+[s+; v−]C+[v+; z−]vx1

longer than C, a contradiction.We claim that p must be in V (C). Otherwise there exist vertices s ∈ N2(y)∩N2(z),

t ∈ N4(y) ∩ N4(z) such that s 6= v−, y+; t 6= u−, z+ and

N (y) = C+[u+; y−] ∪ C+[y+; s] ∪ C+[t; u−] ∪ N5(y);N (z) = C+[v+; z−] ∪ C+[z+; t] ∪ C+[s; v−] ∪ N5(z):

Repeating the arguments in Case 1, we �nally have N (z+)⊆(C+[z+; t] \ {z+}) ∪ {p}and N (s−)⊆(C+[y+; s]\{s−})∪{p}. It is clear that z+ and s− cannot have a commonneighbor in V (G)\V (C). Therefore dist(z+, s−) = 3;max(d(z+); d(s−))6(n− 4)=2−2 + 1 = (n− 6)=2, a contradiction.By the proof of Claim 5, we have at least one of yu−, zv− belongs to E. We

�rst assume that yu− ∈ E, zv− ∈ E. Then there exists vertices s ∈ N2(y) ∩ N2(z),t ∈ N4(y) ∩ N4(z) such that

N (y)⊆C+[u+; y−] ∪ C+[y+; s] ∪ C+[t; u−] ∪ N5(y);

N (z)⊆C+[v+; z−] ∪ C+[z+; t] ∪ C+[s; v−] ∪ N5(z):Without loss of generality, we can assume that p ∈ C+[v+; z−]∪C+[z+; t]∪C+[s; v−]=(N (z) ∪ {p})\N5(z). Then the proof of Claim 5 implies that N (y) = C+[u+; y−] ∪C+[y+; s] ∪ C+[t; u−] ∪ N5(y).As before, s 6= y+, v−; t 6= z+, u−. By a similar argument as in the Case

1, we still can prove that N5(y) = N5(z) = ∅; N (y) and N (z) are disconnected;E([C+[y+; s]; C+[u+; y−] ∪ [C+[t; u−]) = ∅ and G[C+[y+; s]] and G[C+[u+; y−] ∪


[C+[t; u−]] are complete; N (z) is a disjoint union of two cliques in G and there existno edges between these two cliques.We observed that st 6∈ E, v−t 6∈ E, and z+z− 6∈ E. Both st 6∈ E and v−t 6∈ E are

obvious. If z+z− ∈ E, then G has a cycle

x1uC−[y−; u+]C−[u−; t−]C+[y; v−]zC−[t; z+]C−[z−; v]x1

longer than C, a contradictionIf p ∈ C+[z++; t−], then G[C+[z+; t]\{p}] and G[C+[s; v−]∪C+[v+; z−]] are com-

plete. It can be easily veri�ed that N (z+)⊆C+[z; t]\{z+} and N (s−)⊆C+[y; s]\{s−}.Therefore dist(z+; s−)=3;max(d(z+); d(s−))6(n−4)=2−1=(n−6)=2, a contradiction.If p ∈ C+[s+; v−−], then G[C+[z+; t]] and G[(C+[s; v−] \ {p}) ∪C+[v+; z−]] are

complete. It can be easily veri�ed that N (z+)⊆C+[z; t] \ {z+} and N (s−)⊆C+[y; s] \{s−}. Therefore dist(z+; s−)= 3;max(d(z+); d(s−))6(n− 4)=2− 2= (n− 8)=2, a con-tradiction.If p ∈ C+[v+; z−−], then G[C+[z+; t]] and G[C+[s; v−] ∪ (C+[v+; z−] \ {p})] are

complete. It can be easily veri�ed that N (z+)⊆C+[z; t] \ {z+} and N (s−)⊆C+[y; s] \{s−}. Therefore dist(z+; s−)= 3;max(d(z+); d(s−))6(n− 4)=2− 2= (n− 8)=2, a con-tradiction.Now we assume that only on of zv−, yu− belongs to E. Without loss of generality,

we let zv− ∈ E, yu− 6∈ E and p ∈ C+[v+; u−].If p= u−, Then there exist vertices s ∈ N2(y) ∩ N2(z), t ∈ N4(y) ∩ N4(z) such that

N (y) = C+[u+; y−] ∪ C+[y+; s] ∪ C+[t; u−−] ∪ N5(y);N (z) = C+[s; v−] ∪ C+[v+; z−] ∪ C+[z+; t] ∪ N5(z):

It is obvious that t 6= u−. As before, s 6= y+, v−; t 6= z+. By a similar argu-ment as in the Case 1, we can again prove that N5(z) = ∅; N (z) is disconnected;E(C+[z+; t]; C+[v+; z−]∪C+[s; v−])=∅ and G[C+[z+; t]], G[C+[v+; z−]∪C+[s; v−]] arecomplete. We say N5(y)=∅, otherwise let w ∈ N5(y), since G[y; w; s; t] is not isomor-phic to a claw and st 6∈ E, ws ∈ E or wt ∈ E. For both cases we can easily �nd longercycles than C in G. We say N (y) is disconnected, otherwise st ∈ E(CR(G)) and CR(G)has a cycle of c(G) + 1, by Lemma 4 we have a contradiction. Therefore C+[y+; s],C+[u+; y−] and C+[t; u−] induce two complete subgraphs in G. By a similar argumentfor z+z− 6∈ E above, we can prove that y+y− 6∈ E. Thus E(C+[y+; s]; C+[u+; y−] ∪C+[t; u−]) = ∅ and G[C+[y+; s]], G[C+[u+; y−] ∪ C+[t; u−−]] are complete. It canbe easily veri�ed that N (z+)⊆C+[z; t] \ {z+} and N (s−)⊆C+[y; s] \ {s−}. Thereforedist(z+; s−) = 3;max(d(z+); d(s−))6(n− 4)=2− 2 = (n− 8)=2, a contradiction.Now we assume that p 6= u−. Then it is not di�cult to see there exists vertices

s ∈ N2(y) ∩ N2(z), t ∈ N4(y) ∩ N4(z) such that

N (y) = C+[u+; y−] ∪ C+[y+; s] ∪ C+[t; p−] ∪ N5(y);N (z) = C+[s; v−] ∪ C+[v+; z−] ∪ C+[z+; t] ∪ C+[p+; u−] ∪ N5(z):


As before, s 6= y+, v−; t 6= z+. t 6= u− is obvious in this case. Next we willshow dist(x1, u−) = 3. Since V (G) = N (y) ∪ N (z) ∪ {u; v; x1; y; z; p} and dist(x1, y) =dist(x1, z) = 3, it su�ces to show that vu−− 6∈ E, uu−− 6∈ E, and px1 6∈ E. vu−− 6∈ Efollows from (2) in Lemma 2. If uu− ∈ E, then G has a cycle

x1uC−[u−−; z]u−C+[u+; v−]C+[v+; z−]vx1

longer than C, a contradiction. If px1 ∈ E, then G has a cycle

x1C−[p; z]C+[z+; v−]C+[v+; z−]vx1

longer than C. Since x1uu−u−− is a path of length three, dist(x1, u−−)=3. Since d(x1)is less than (n−4)=2, we have d(u−−)¿(n−4)=2. We claim that p 6= u−−, otherwisewe can �nd a longest cycle C1 in G by replacing x1 by u− such that p=u− ∈ G \C1,and we can derive a contradiction by Lemma 3. Since G[z; t−; u−; v−] is not isomorphicto a claw and u−v− 6∈ E, u−t− or v−t− ∈ E, therefore G has cycle

x1uC−[y−; u+]u−C−[t−; z]C−[u−−; t]C+[y; v−]C+[v+; z−]vx1

when u−t− ∈ E or G has a cycle

x1vC+[v+; t−]C−[v−; y]C+[t; u−]C+[u+; y−]ux1

when v−t− ∈ E. Both these cycles are longer than C. This contradiction completes theproof of Theorem 2.

4. For further reading

The following reference is also of interest to the reader: [6]

Acknowledgements

The author thanks Dr. Richard H. Schelp for his help in the preparation of this paper.The author also thanks one of anonymous referees for his/her idea of using closureconcept in the proof of the main theorem. This idea plays a vital role in shorteningthe initial lengthy proof.

References

[1] P. Bedrossian, G. Chen, R.H. Schelp, A generalization of Fan’s condition for Hamiltonicity, pancyclicity,and Hamiltonian connectedness, Discrete Math. 115 (1993) 39–50.

[2] J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, Macmillan, London and Elsevier,New York, 1976.

[3] H.J. Broersma, H. Trommel, Closure concepts for claw-free graphs, Discrete Math. 185 (1998)231–238.


[4] Geng-Hua Fan, New su�cient conditions for cycles in graphs, J. Combin. Theory Ser. B 37 (1984)221–227.

[5] R.J. Gould, Traceablity in Graphs, Doctoral Thesis, Western Michigan University, 1979.[6] M.M. Matthews, D.P. Sumner, Longest paths and cycles in K1;3-free graphs, J. Graph Theory 9 (1985)

269–277.[7] Z. Ryj�a�cek, On a closurs concept in claw-free graphs, J. Combin. Theory Ser. B 70 (1997) 217–224.

Documents

A Fan-type condition for claw-free graphs to be Hamiltonian