Heavy Cycles in 2-Connected Weighted Graphswith Large Weighted Degree Sums

Bing Chen1, Shenggui Zhang1,2, and T.C. Edwin Cheng2

1 Department of Applied Mathematics, Northwestern Polytechnical University,Xi’an, Shaanxi 710072, P.R. China

2 Department of Logistics, The Hong Kong Polytechnic University,Hung Hom, Kowloon, Hong Kong

Abstract. In this paper, we prove that a 2-connected weighted graphG contains either a Hamilton cycle or a cycle of weight at least 2m/3 ifit satisfies the following conditions: (1)

∑3i=1 dw(vi) ≥ m, where v1, v2

and v3 are three pairwise nonadjacent vertices of G, and two of them arenonadjacent vertices of an induced claw or an induced modified claw; (2)In each induced claw and each induced modified claw of G, all edges havethe same weight. This extends several previous results on the existenceof heavy cycles in weighted graphs.

Keywords: Hamilton cycle, weighted graph, induced claw (modifiedclaw.)

1 Terminology and Notation

We use Bondy and Murty [3] for terminology and notation not defined here andconsider finite simple graphs only. Let G be a graph. G is called a weighted graphif each edge e is assigned a nonnegative number w(e), called the weight of e. Fora subgraph H of G, V (H) and E(H) denote the sets of vertices and edges ofH , respectively. The weight of H is defined by w(H) =

∑e∈E(H) w(e). For a

vertex v ∈ V , NH(v) denotes the set, and dH(v) the number, of vertices in Hthat are adjacent to v. We define the weighted degree of v in H by dw

H(v) =∑h∈NH(v) w(vh). When no confusion occurs, we will denote NG(v), dG(v) and

dwG(v) by N(v), d(v) and dw(v), respectively.An unweighted graph can be regarded as a weighted graph in which each

edge is assigned weight 1. Thus, in an unweighted graph, dw(v) = d(v) for everyvertex v, and the weight of a subgraph is simply the number of its edges.

The number of vertices in a maximum independent set of G is denoted byα(G). If G is noncomplete, then for a positive integer k ≤ α(G) we denoteby σk(G) the minimum value of the degree sum of any k pairwise nonadjacentvertices, and by σw

k (G) the minimum value of the weighted degree sum of anyk pairwise nonadjacent vertices. If k > α(G), then both σk(G) and σw

k (G) aredefined as ∞.

An (x, y)-path is a path connecting the two vertices x and y. A y-path is apath which has y as one of its end-vertices. The distance between two vertices

Y. Shi et al. (Eds.): ICCS 2007, Part III, LNCS 4489, pp. 338–346, 2007.c© Springer-Verlag Berlin Heidelberg 2007

Heavy Cycles in 2-Connected Weighted Graphs 339

x and y, denoted by d(x, y), is the length of a shortest (x, y)-path. We call thegraph K1,3 a claw, and the graph K1,3+e (e is an edge between two nonadjacentvertices in the claw) a modified claw. For a graph G, if there exist three pairwisenonadjacent vertices in which two of them are nonadjacent in an induced clawor an induced modified claw, then by σw

3 (G) we denote the minimum value ofthe weighted degree sum of such three pairwise nonadjacent vertices. Otherwise,σw

3 (G) is defined as ∞. Clearly we have σw3 (G) ≥ σw

3 (G).

2 Results

In [8], Posa gave a degree sum condition for the existence of long cycles in graphs.This result was generalized by the following two theorems along different lines.

Theorem A (Fan [5]). Let G be a 2-connected graph such that max{d(x), d(y)|d(x, y) = 2} ≥ c/2. Then G contains either a Hamilton cycle or a cycle of lengthat least c.

Theorem B (Fournier & Fraisse [6]). Let G be a k-connected graph where2 ≤ k < α(G), such that σk+1(G) ≥ m. Then G contains either a Hamiltoncycle or a cycle of length at least 2m/(k + 1).

Bondy et al. [2] generalized the result of Posa to weighted graph. In [10], it wasshowed that if one wants to give a generalization of Theorem A to weightedgraphs, some extra conditions cannot be avoided. By adding two extra condi-tions, the authors gave a weighted generalization of Theorem A. Motivated bythis result, by adding two same extra conditions, Zhang et al. [9] gave a weightedgeneralization of Theorem B in the case k = 2.

Theorem 1 (Zhang et al. [9]). Let G be a 2-connected weighted graph whichsatisfies: (1) σw

3 (G) ≥ m; (2) w(xz) = w(yz) for every vertex z ∈ N(x) ∩N(y) with d(x, y) = 2; (3) In every triangle T of G, either all edges of T havedifferent weights or all edges of T have the same weight. Then G contains eithera Hamilton cycle or a cycle of weight at least 2m/3.

Theorem A was further extended by the following result.

Theorem C (Bedrossian et al.[1]). Let G be a 2-connected graph. If max{d(x), d(y)} ≥ c/2 for each pair of nonadjacent vertices x and y, which arevertices of an induced claw or an induced modified claw of G, then G containseither a Hamilton cycle or a cycle of length at least c.

By adding one extra condition, Fujisawa [7] gave a weighted generalization ofTheorem C.

Theorem 2 (Fujisawa [7]). Let G be a 2-connected weighted graph which sat-isfies: (1) For each induced claw and each induced modified claw of G, all itsnonadjacent pair of vertices x and y satisfy max{dw(x), dw(y)} ≥ s/2; (2) Foreach induced claw and each induced modified claw of G, all of its edges have thesame weight. Then G contains either a Hamilton cycle or a cycle of weight atleast s.

340 B. Chen, S. Zhang, and T.C.E. Cheng

Clearly, Conditions (2) and (3) of Theorem 1 imply Condition (2) of Theorem2. In a previous paper [4], the authors proved that Conditions (2) and (3) ofTheorem 1 can be replaced by Condition (2) of Theorem 2. Here we will furthershow that Condition (1) of Theorem 1 can also be replaced by a weaker one.

Theorem 3. Let G be a 2-connected weighted graph which satisfies the followingconditions: (1) σw

3 (G) ≥ m; (2) For each induced claw and each induced modifiedclaw of G, all of its edges have the same weight. Then G contains either aHamilton cycle or a cycle of weight at least 2m/3.

3 Proof of Theorem 3

We call a path P a heaviest longest path of a graph G if P has the followingproperties

• P is a longest path of G, and• w(P ) is maximum among all longest paths in G.

Proposition 1. Let G be a non-hamiltonian 2-connected weighted graph whichsatisfies conditions (1) and (2) of Theorem 3. Then G contains a heaviest longestpath P = v1v2 · · · vp, such that dw(v1) + dw(vp) ≥ 2m/3.

We prove this proposition in the next section. Theorem 3 can be proved bycombining Proposition 1 and the following lemma. The proof of Lemma 1 isimplicit in [2].

Lemma 1 (Bondy et al. [2]). Let G be a non-hamiltonian 2-connectedweighted graph and P = v1v2 · · · vp be a heaviest longest path in G. Then thereis a cycle C in G with w(C) ≥ dw(v1) + dw(vp).

Proof of Theorem 3. Suppose that G does not contain a Hamilton cycle. Thenby Proposition 1 and Lemma 1, we can find a cycle of weight at least 2m/3.

4 Proof of Proposition 1

In the proof of Proposition 1, we need the following two lemmas.

Lemma 2 (Fujisawa [7]). Let G be a weighted graph satisfying Condition (2)of Theorem 3. If x1yx2 is an induced path with w(x1y) �= w(x2y) in G, theneach vertex x ∈ N(y)\{x1, x2} is adjacent to both x1 and x2.

Lemma 3 (Fujisawa [7]). Let G be a weighted graph satisfying Condition (2)of Theorem 3. Suppose x1yx2 is an induced path such that w1 = w(x1y) andw2 = w(x2y) with w1 �= w2, and yz1z2 is a path such that {z1, z2}∩{x1, x2} = ∅and x2z2 /∈ E(G). Then the following (i) and (ii) hold:(i) {z1x1, z1x2, z2x1} ⊆ E(G), and yz2 /∈ E(G). Moreover, all edges in thesubgraph induced by {x1, y, x2, z1, z2}, other than x1y, have the same weight w2.

Heavy Cycles in 2-Connected Weighted Graphs 341

(ii) Let Y be the component of G − {x2, z1, z2} with y ∈ V (Y ). For each vertexv ∈ V (Y )\{x1, y}, v is adjacent to all of x1, x2, y and z2. Furthermore, w(vx1) =w(vx2) = w(vy) = w(vz2) = w2.

Proof of Proposition 1. Choose a heaviest longest path P = v1v2 · · · vp in Gsuch that dw(v1)+dw(vp) is as large as possible. It is clear that N(v1)∪N(vp) ⊆V (P ). And it is not difficult to prove that there exists no cycle of length p.

Suppose dw(v1)+ dw(vp) < 2m/3. Without loss of generality, we may assumedw(v1) < m/3.

Claim 1. Let P1 and P2 be two heaviest longest paths such that P1 has v′ andvp as its end-vertices, and P2 has v′′ and vp as its end-vertices. Then v′ and v′′

cannot be nonadjacent vertices of an induced claw or an induced modified claw.

Proof. Suppose v′ and v′′ are nonadjacent vertices of an induced claw or aninduced modified claw. Since P1 and P2 are heaviest longest paths, v′vp /∈ E(G)and v′′vp /∈ E(G). By the choice of the path P in (b), dw(v′) ≤ dw(v1) anddw(v′′)+dw(vp) < 2m/3. So we have dw(v′)+dw(v′′)+dw(vp) < m, contradictingCondition (1) of Theorem 3. ��

Since G is 2-connected, v1 is adjacent to at least one vertex on P other than v2.Choose vk ∈ N(v1) such that k is as large as possible. It is clear that 3 ≤ k ≤p − 1.

Case 1. N(v1) = {v2, · · · , vk}.

Since P is longest and N(v1) = {v2, · · · , vk}, N(vi) ⊆ V (P ) for every i withi = 2, · · · , k − 1; Since G is a non-hamiltonian 2-connected graph, k + 2 ≤ p;Since G − vk is connected, there must be an edge vjvs ∈ E(G) with j < k < s.

We assume that such an edge vjvs was chosen so that: (i) s is as large aspossible; (ii) j is as large as possible, subject to (i). Clearly we have s ≤ p − 1.

Claim 2. w(v1vj+1) = w(vjvj+1).

Proof. Suppose j ≤ k − 2. By the choices of vk, vj and vs, we have v1vs /∈ E(G)and vj+1vs /∈ E(G). So {vj, v1, vj+1, vs} induces a modified claw, which impliesthat w(v1vj+1) = w(vjvj+1).

Suppose j = k − 1 and s ≥ k + 2. By the choices of vk and vs, we havev1vs /∈ E(G), v1vs+1 /∈ E(G) and vjvs+1 /∈ E(G). If vkvs+1 ∈ E(G), then{vk, v1, vk−1, vs+1} induces a modified claw. So w(v1vk) = w(vk−1vk). If vkvs /∈E(G), then {vk−1, v1, vk, vs} induces a modified claw. So w(v1vk) = w(vk−1vk).If vkvs+1 /∈ E(G) and vkvs ∈ E(G), then {vk, v1, vk+1, vs} induces a claw or amodified claw. So w(v1vk) = w(vkvs); On the other hand, {vs, vk−1, vk, vs+1}induces a modified claw. So w(vkvs) = w(vk−1vk); Then w(v1vk) = w(vk−1vk).

Suppose j = k − 1 and s = k + 1. Since P is longest, N(vk) ⊂ V (P ); SinceG − vk+1 is connected, there exists an edge vkvt ∈ E(G) with t ≥ k + 2. Choosevkvt such that t is as small as possible. So {vk, v1, vk−1, vt} induces a modifiedclaw. Then w(v1vk) = w(vk−1vk). ��

342 B. Chen, S. Zhang, and T.C.E. Cheng

Claim 3. s = k + 1.

Proof. Suppose s ≥ k + 2. Let P ′ = vs−1vs−2 · · · vj+1v1 v2 · · · vjvsvs+1 · · · vp. Ifvjvs−1 ∈ E(G), then {vj , v1, vs−1, vs} induces a modified claw, so w(vs−1vs) =w(vjvs). By Claim 2, now P ′ is a heaviest longest vp-path different from P ,contradicting Claim 1. If vjvs−1 /∈ E(G), then {vs, vj , vs−1, vs+1} induces a clawor a modified claw. Thus w(vs−1vs) = w(vjvs). By Claim 2, both vjvj−1 · · · v1vj+1vj+2 · · · vp and P ′ are heaviest longest vp-paths, contradicting Claim 1. ��As in the proof of Claim 2, since s = k + 1 and G − vk+1 is connected, thereexists an edge vkvt ∈ E(G) with k + 2 ≤ t ≤ p − 1. Choose vkvt such that t is assmall as possible.

Case 1.1. t ≥ k + 3.

By the choices of vs and vt, vjvk+2 /∈ E(G) and vkvk+2 /∈ E(G). So {vk+1, vj ,vk, vk+2} induces a claw or a modified claw. Thus w(vkvk+1) = w(vjvk+1).

Suppose vkvt+1 /∈ E(G). By the choice of vt, vkvt−1 /∈ E(G). So {vt, vk, vt−1,vt+1} induces a claw or a modified claw. Thus w(vt−1vt) = w(vkvt). By Claim 2,both vkvk−1 · · · vj+1v1 · · · vjvk+1 · · · vp and vt−1vt−2 · · · vk+1vjvj−1 · · · v1vj+1 · · ·vkvtvt+1 · · · vp are heaviest longest vp-paths, contradicting Claim 1. Supposevkvt+1 ∈ E(G). By the choice of vk, v1vt /∈ E(G) and v1vt+1 /∈ E(G). So{vk, v1, vt, vt+1} induces a modified claw. Thus w(vkvt+1) = w(vtvt+1). By Claim2, vtvt−1 · · · vk+1vjvj−1 · · · v1vj+1 · · · vkvt+1vt+2 · · · vp is a heaviest longest vp-path different from P , contradicting Claim 1.

Case 1.2. t = k + 2.

By the choice of vk, now {vk, v1, vk+1, vk+2} induces a modified claw. So we getw(vkvk+1) = w(vkvk+2) = w(vk+1vk+2) = w(v1vk).

Since G−vt is connected, there exists an edge vk+1vt′ ∈ E(G) with t′ ≥ k+3.By the choice of vs, vjvt /∈ E(G) and vjvt′ /∈ E(G). So {vs, vj , vt, vt′} inducesa claw or a modified claw. In this case s = k + 1 and t = k + 2, thus we havew(vjvk+1) = w(vk+1vk+2). By Claim 2, now vk+1vjvj−1 · · · v1vj+1vj+2 · · · vkvk+2· · · vp is a heaviest longest vp-path different from P . At the same time, v1 andvk+1 are nonadjacent vertices in the modified claw induced by{vk, v1, vk+1, vk+2},contradicting Claim 1.

This completes the proof of Case 1.

Case 2. N(v1) �= {v2, · · · , vk}.Choose vr /∈ N(v1) with 2 < r < k such that r is as large as possible. Thenv1vi ∈ E(G) for every i with r < i ≤ k. Let j be the smallest index such thatj > r and vj /∈ N(v1) ∩ N(vr). Since vr+1 ∈ N(v1) ∩ N(vr), we have j ≥ r + 2.On the other hand, it is obvious that j ≤ k + 1. By the choice of vr and vj ,now v1 and vr are nonadjacent vertices of a claw or a modified claw induced by{vj−1, v1, vr, vj}. By Claim 1 and the choice of P , we have the following claim.

Claim 4. There exists no heaviest longest vp-path with vr as it’s another end-vertex.

Claim 5. w(v1vr+1) �= w(vrvr+1).

Heavy Cycles in 2-Connected Weighted Graphs 343

Proof. If w(v1vr+1) = w(vrvr+1), then vrvr−1 · · · v1vr+1vr+2 · · · vp is a heaviestlongest vp-path different from P , contradicting Claim 4. ��

Claim 6. vr+1vj /∈ E(G).

Proof. By the choice of vr, v1vr /∈ E(G). So, if vr+1vj ∈ E(G), then by thechoice of vj , we know that {vr+1, v1, vr, vj} induces a claw or a modified claw.Thus w(v1vr+1) = w(vrvr+1), contradicting Claim 5. ��

By Claim 6, we have r + 3 ≤ j ≤ k + 1. Let n be the largest index such thatr + 1 ≤ n ≤ j − 2 and vnvj /∈ E(G). Then vivj ∈ E(G) for every i withn + 1 ≤ i ≤ j − 1.

Claim 7. If r + 3 ≤ j ≤ k, then w(v1vn+1) = w(vnvn+1).

Proof. By the choice of vn, vn+1vj ∈ E(G). So {vn+1, v1, vr, vj} induces a mod-ified claw. Thus w(v1vn+1) = w(vn+1vj). Since vrvj /∈ E(G) and vnvj /∈ E(G),{vr, vn, vn+1, vj} induces a modified claw. Thus w(vnvn+1) = w(vn+1vj). Sow(v1vn+1) = w(vnvn+1). ��

Case 2.1. r + 3 ≤ j ≤ k − 1.

Suppose vr+1vj+1 /∈ E(G). By Claim 6, vr+1vj /∈ E(G), so {v1, vj , vj+1, vr+1}induces a modified claw. Thus w(v1vj+1) = w(vjvj+1). Suppose vr+1vj+1 ∈E(G). By Claim 5 and the choice of vj , applying Lemma 3 (i) to {v1, vr+1, vr,vj+1, vj}, we get w(v1vj+1) = w(vjvj+1).

By Claim 7, both vnvn−1 · · · v1vn+1vn+2 · · · vp and vjvj−1 · · · v1vj+1vj+2 · · · vp

are heaviest longest vp-paths. Furthermore, vn and vj are nonadjacent verticesin the modified claw induced by {vr, vn, vn+1, vj}, contradicting Claim 1.

Case 2.2 j = k.

Since {vj−1, v1, vr, vj} induces a claw or a modified claw and j = k, we have

Claim 8. w(v1vk−1) = w(v1vk) = w(vk−1vk).

Claim 9. vr+2vk /∈ E(G).

Proof. Suppose vr+2vk ∈ E(G). Applying Lemma 3 (ii) to {v1, vr+1, vr, vr+2, vk}and the vertex v2 (which is adjacent to v1), we get v2vr+1 ∈ E(G) and w(v1v2) =w(v2vr+1) = w(v1vk) = w(vrvr+1). By Claim 8, vrvr−1 · · · v2vr+1vr+2 · · · vk−1v1vkvk+1 · · · vp is a heaviest longest vp-path, contradicting Claim 4. ��

Now, we have vr+1vk−1 /∈ E(G). Since otherwise, applying Lemma 3 (ii) to{v1, vr+1, vr, vk−1, vk} and the vertex vr+2 (which is adjacent to vr+1), we getvr+2vk ∈ E(G), contradicting Claim 9. By Claim 6, vk−1 and vr+1 are nonadja-cent vertices in the modified claw induced by {v1, vr+1, vk−1, vk}.

By Claims 6 and 9, now {v1, vr+1, vr+2, vk} induces a modified claw. Thenw(v1vr+2) = w(vr+1vr+2). By Claim 8, both vk−1vk−2 · · · v1vkvk+1 · · · vp andvr+1vr · · · v1vr+2 · · · vp are heaviest longest vp-paths, contradicting Claim 1.

344 B. Chen, S. Zhang, and T.C.E. Cheng

Case 2.3 j = k + 1.

Since {vj−1, v1, vr, vj} induces a claw or a modified claw, we have

Claim 10. w(v1vk) = w(vkvk+1) = w(vrvk).

Claim 11. vr+1vk /∈ E(G).

Proof. Suppose vr+1vk ∈ E(G). Applying Lemma 3 (ii) to {v1, vr+1, vr, vk, vk+1}and the vertex vr−1 (which is adjacent to vr), we get vr−1vk+1 ∈ E(G) andw(vr−1vr) = w(vr−1vk+1). By Claim 10, we get a heaviest longest vp-pathvrvr+1 · · · vkv1v2 · · · vr−1vk+1vk+2 · · · vp, contradicting Claim 4. ��

Claim 12. vivk+1 /∈ E(G) for every vi ∈ N(v1) ∩ N(vr) \ {vk}.

Proof. Suppose there exists a vertex vi ∈ N(v1)∩N(vr)\{vk} such that vivk+1 ∈E(G). If w(v1vr+1) �= w(v1vk), then applying Lemma 3 (i) to {vr+1, v1, vk, vi,vk+1}, we get w(v1vr+1) = w(vkvk+1). So w(v1vk) �= w(vkvk+1), contradict-ing Claim 10. So we get w(v1vr+1) = w(v1vk). Similarly, we can prove thatw(vrvr+1) = w(vrvk). By Claim 10, we get w(v1vr+1) = w(vrvr+1), contradict-ing Claim 5. ��

Claim 13. w(v1vk−1) = w(vk−1vk).

Proof. By the choice of vk and Claim 12, we get v1vk+1 /∈ E(G) and vk−1vk+1 /∈E(G). So{vk, v1, vk−1, vk+1}induces a modified claw and w(v1vk−1)=w(vk−1vk).

��

Case 2.3.1. w(vrvr+1) �= w(vrvk).

By Claim 11, vr+1vk /∈ E(G). Furthermore, vrvk+1 /∈ E(G), since otherwise itfollows from Lemma 2 that vr+1vk+1 ∈ E(G), which contradicts Claim 6.

Applying Lemma 2 to the induced path vr+1vrvk and the vertex vr−1 ∈N(vr)\{vr+1, vk}, we get vr−1vr+1 ∈ E(G) and vr−1vk ∈ E(G). By Claim5, w(v1vr+1) �= w(vrvr+1). Applying Lemma 2 to the induced path v1vr+1vr

and the vertex vr−1 ∈ N(vr+1)\ {v1, vr}, we get v1vr−1 ∈ E(G). This impliesthat vr−1 ∈ N(v1) ∩ N(vr). By Claim 12, we have vr−1vk+1 /∈ E(G). Then{vk, vr−1, vr, vk+1} induces a modified claw, so w(vr−1vr) = w(vr−1vk). There-fore, by Claim 13, vrvr+1 · · · vk−1v1v2 · · · vr−1vkvk+1 · · · vp is a heaviest longestvp-path, contradicting Claim 4.

Case 2.3.2. w(vrvr+1) = w(vrvk).

By Claim 5 and Claim 10, we can get that w(v1vr+1) �= w(v1vk).

Claim 14. v2vk, v2vr+1 and v2vr ∈ E(G).

Proof. By Claim 11, vr+1vk /∈ E(G). Applying Lemma 2 to the induced pathvr+1v1vk and the vertex v2 ∈ N(v1)\{vr+1, vk}, we get v2vk ∈ E(G) andv2vr+1 ∈ E(G). By Claim 5, w(v1vr+1) �= w(vrvr+1). Applying Lemma 2 tothe induced path v1vr+1vr and the vertex v2 ∈ N(vr+1)\{v1, vr}, we get v2vr ∈E(G). ��

Heavy Cycles in 2-Connected Weighted Graphs 345

Claim 15. vrvk+1 /∈ E(G).

Proof. Suppose vrvk+1 ∈ E(G). By Claim 14, we get v2 ∈ N(v1) ∩ N(vr),v2vk and v2vr+1 ∈ E(G). By Claims 6 and 12, we get vr+1vk+1 /∈ E(G) andv2vk+1 /∈ E(G). Thus both {vr, v2, vr+1, vk+1} and {vk, v1, v2, vk+1} induce mod-ified claws. Then we have w(v2vr+1) = w(vrvr+1) = w(v1v2) = w(v1vk). ByClaim 13, we get w(v1vk−1) = w(vk−1vk). So, vrvr−1 · · · v2vr+1vr+2 · · · vk−1v1vk

vk+1 · · · vp is a heaviest longest vp-path, contradicting Claim 4. ��

Suppose v1vr−1 ∈ E(G). Then by applying Lemma 2 to the induced pathvkv1vr+1 and vr−1 ∈ N(v1)\{vk, vr+1}, we get vr−1vk ∈ E(G) and vr−1vr+1 ∈E(G). By Claims 12 and 15, we get vr−1vk+1 /∈ E(G) and vrvk+1 /∈ E(G).Then {vk, vr−1, vr, vk+1} induces a modified claw. Thus w(vr−1vr) = w(vr−1vk).Therefore, by Claim 13, we have vrvr+1 · · · vk−1v1v2 · · · vr−1vkvk+1 · · · vp is aheaviest longest vp-path, contradicting Claim 4.

Suppose v1vr−1 /∈ E(G). Then by Claim 5 and Lemma 2, we have vr−1vr+1 /∈E(G). Furthermore, v2vr−1 /∈ E(G), since otherwise, applying Lemma 3 (ii)to {vr, vr+1, v1, v2, vr−1} and the vertex vk (which is adjacent to vr), we getw(v1vr+1) = w(v1vk), a contradiction. By Claims 12 and 14, v2vk+1 /∈ E(G).Then, both {vr, vr+1, v2, vr−1} and {vk, v1, v2, vk+1} induce modified claws. Thusw(v2vr+1) = w(vrvr+1) and w(v1v2) = w(v1vk). By Claim 13, we know thatvrvr−1 · · · v2vr+1vr+2 · · · vk−1v1vk vk+1 · · · vp is a heaviest longest vp-path, con-tradicting Claim 4.

The proof of the theorem is complete.

Acknowledgements

This work was supported by NSFC (No. 60642002). The second and the thirdauthors were supported by The Hong Kong Polytechnic University under grantnumber G-YX42.

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