maximal k3's and hamiltonicity of 4-connected claw-free graphs

Maximal K3’s andHamiltonicity of4-Connected Claw-FreeGraphs

Jun Fujisawa1 and Katsuhiro Ota2

1DEPARTMENT OF APPLIED SCIENCE, KOCHI UNIVERSITY2-5-1 AKEBONO-CHO, KOCHI, 780-8520 JAPAN

E-mail: [email protected]

2DEPARTMENT OF MATHEMATICS, KEIO UNIVERSITYYOKOHAMA, 223-8522 JAPAN

E-mail: [email protected]

Received February 10, 2009; Revised July 13, 2010

Published online in Wiley Online Library (wileyonlinelibrary.com).DOI 10.1002/jgt.20599

Abstract: Let cl(G) denote Ryjacek’s closure of a claw-free graph G. Inthis article, we prove the following result. LetG be a 4-connected claw-freegraph. Assume that G[NG(T )] is cyclically 3-connected if T is a maximal K3in G which is also maximal in cl(G). Then G is hamiltonian. This result isa common generalization of Kaiser et al.’s theorem [J Graph Theory 48(4)(2005), 267–276] and Pfender’s theorem [J Graph Theory 49(4) (2005),262–272]. � 2011 Wiley Periodicals, Inc. J Graph Theory

MSC 2010: 05C45, 05C38, 05C75

Keywords: maximal K3; Hamiltonian cycle; claw-free graph

1. INTRODUCTION

In this article, a graph or a simple graph means a finite undirected graph without loopsor multiple edges. A multigraph may contain multiple edges but no loops. For a graph

Journal of Graph Theory� 2011 Wiley Periodicals, Inc.

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2 JOURNAL OF GRAPH THEORY

G and U⊆V(G), G[U] denotes the induced subgraph of G induced by U. Moreover,we define NG(U)=⋃

v∈U NG(v). We often identify a subgraph H of G with its vertexset V(H).

In 1984, Matthews and Sumner [5] conjectured that every 4-connected claw-freegraph is hamiltonian, and this conjecture is still wide open. Among a lot of studies onthis conjecture, one of the remarkable results is Ryjacek’s closure concept. Let G be aclaw-free graph. We call a vertex v of G locally connected (resp. locally disconnected)if G[NG(v)] is connected (resp. disconnected). For a locally connected vertex v of G,the operation of adding all possible edges between vertices in NG(v) is called localcompletion at v. In [7], it is shown that this operation preserves the claw-freeness of theoriginal graph. Iterating local completions, we obtain a graph G∗ in which G∗[NG∗ (v)]is a complete graph for every locally connected vertex v. We call this graph the closureof G, and denote it cl(G).

Theorem 1 (Ryjacek [7]). Let G be a claw-free graph. Then

• cl(G) is uniquely defined,• cl(G) is the line graph of some triangle-free simple graph,• G is hamiltonian if and only if cl(G) is hamiltonian.

It follows from Theorem 1 that Matthews and Sumner’s conjecture is equivalent toThomassen’s conjecture, which states that every 4-connected line graph is hamiltonian.In the following, we investigate hamiltonian cycles in line graphs.

In order to find a hamiltonian cycle in a line graph, the following theorem is useful.A dominating closed trail of a graph G is a closed trail T such that G−V(T) has noedge.

Theorem 2 (Harary and Nash-Williams [3]). Let H be a graph with |E(H)|≥3. Thenthe line graph L(H) is hamiltonian if and only if H has a dominating closed trail.

For a graph G and F⊆E(G), F is called an essential k-edge cut if |F|=k and G−Fhas at least two components which contain an edge. A graph is called essentiallyk-edge-connected if there is no essential k′-edge cut for every k′<k. For k≥2, we calla graph G cyclically k-connected if G is 2-connected and G contains no vertex cut Dof at most k−1 vertices such that at least two components of G−D contain a cycle.It is easy to see that a graph H is essentially k-edge-connected if and only if L(H) isk-connected or complete. Therefore, if every essentially 4-edge-connected graph hasa dominating closed trail, then Thomassen’s conjecture holds, and so does Matthewsand Sumner’s conjecture.

As for dominating closed trails in essentially 4-edge-connected graphs, the followingproposition is well known from Theorem 2 of [1].

Proposition 3. Let H be an essentially 4-edge-connected simple graph which has novertex of degree 3. Then H has a dominating closed trail.

We call a complete subgraph X of a graphGmaximal if there is no complete subgraphX′ such that V(X)�V(X′) . Since three edges of a graph H which is adjacent to a vertexof degree 3 induces a maximal K3 in L(H), the following proposition immediatelyfollows from Proposition 3.

Journal of Graph Theory DOI 10.1002/jgt

THE 4-CONNECTED CLAW-FREE GRAPHS 3

FIGURE 1. Hourglass.

FIGURE 2. T3.

Proposition 4. Every 4-connected line graph which has no maximal K3 ishamiltonian.

By the above observation, in order to prove Matthews and Sumner’s conjecturein special cases, it is natural to focus on the structure around maximal K3s in linegraphs. Considering the connectivity of the graphs induced by maximal K3s and theirneighbors, we obtain the following result.

Theorem 5. Let G be a 4-connected claw-free graph. Assume that G[NG(T)] iscyclically 3-connected if T is a maximal K3 in G which is also maximal in cl(G). ThenG is hamiltonian.

Readers may ask the following question: What kind of maximal K3 in G will alsobe a maximal K3 in cl(G)? As little is known about properties of locally disconnectedvertices of a graph G which turn to locally connected vertices in cl(G), it seems difficultto give a tidy answer for this question. However, some structural information is givenin Lemma 4 of Section 2.

Proof of Theorem 5 is given in Sections 2–5. In Section 6, we show that Theorem 5generalizes the following two theorems. The graphs in Figures 1 and 2 are called anhourglass and T3, respectively.

Theorem 6 (Kaiser et al. [4]). Let G be a 4-connected claw-free graph. Assume that,in every induced hourglass S, there are two non-adjacent vertices which have a commonneighbor in V(G)\V(S). Then G is hamiltonian.

Theorem 7 (Pfender [6]). Let G be a 4-connected claw-free graph. If G has noinduced subgraph isomorphic to T3, then G is hamiltonian.

2. PRELIMINARIES

For terminology and notation not defined in this article, we refer the readers to [2].For a claw-free graph G, the set of locally connected vertices and locally disconnected



vertices are denoted by LC(G) and LD(G), respectively. If v∈LC(G) andG[NG(v)] is nota complete graph, then we call v eligible vertex, and EL(G) denotes the set of eligiblevertices of G. Since G is claw-free, G[NG(v)] consists of two complete graphs if v∈LD(G). For p and q with p≤q, we define LDp,q(G)={v∈LD(G)|G[NG(v)]�Kp∪Kq}.LDi(G) denotes the set

⋃j≥i LD

i,j(G). For x∈EL(G),Gx denotes the graph obtained fromG by local completion at x. For a multigraph H, we define Vi(H)={v∈V(H)|dH(v)= i}and V≥i(H)={v∈V(H)|dH(v)≥ i}. A cycle of length l is called an l-cycle.

In the rest of this section, we prepare some lemmas on properties of claw-free graphsand its closures.

Lemma 1 (Pfender [6]). Let G be a claw-free graph. Then LD(Gx)⊆LD(G) for everyx∈EL(G).

Since V(G)\LD(G)=LC(G) holds for every claw-free graph G, Lemma 1 impliesthe following corollary.

Corollary 1. Let G be a claw-free graph and let v∈LC(G). Then v∈LC(Gx) for everyx∈EL(G).

Lemma 2. Let G be a claw-free graph, let v∈LD(G) and let a,b∈NG(v) such thatab /∈E(G). If ab∈E(cl(G)), then v∈LC(cl(G)).

Proof. Since ab /∈E(G) and ab∈E(cl(G)), there exist a graph G∗ and a vertexx∈EL(G∗) such that G∗ is obtained from G by repeating local completions, x∈EL(G∗),ab /∈E(G∗) and ab∈E(G∗

x). If v∈LC(G∗), then it follows from Corollary 1 that v∈LC(cl(G)). If v∈LD(G∗), then since G∗[NG(v)] consists of two complete graphs, v isa vertex of LC(G∗

x ). Thus we obtain v∈LC(cl(G)) by Corollary 1. �

Lemma 3. Let v be a vertex of G such that v∈LD(cl(G)). Assume that there existsa 4-cycle C which is an induced cycle of G containing v. Then, for every x∈EL(G), Cis an induced 4-cycle in Gx.

Proof. Let vacbv be an induced 4-cycle in G. Since a,b∈NG(v), ab /∈E(G) andv /∈LC(cl(G)), we obtain ab /∈E(cl(G)) by Lemma 2. Hence ab /∈E(Gx). On the otherhand, if vc∈E(Gx), thenGx[NGx (v)] contains a path acb. This implies that ab∈E(cl(G)),a contradiction. Thus vc /∈E(Gx), and hence the assertion holds. �

Lemma 4. Let G be a 4-connected claw-free graph and let T be a maximal K3 in Gas well as a maximal K3 in cl(G). Let v1,v2 and v3 be the three vertices of T and letUi=NG(vi)\V(T). Then the following holds:

(a) Each vi is in LD2(G),(b) Ui∩Uj=∅ for every 1≤ i<j≤3 and(c) if xy∈E(G) for x∈Ui and y∈Uj with i �= j, then xy′ /∈E(G) for every y′ ∈Uj\{y}.

Proof. (a) Since T is a maximal K3 in cl(G), each vi is in LD(cl(G))⊆LD(G). Also,since dG(vi)≥4 and T is maximal, we have vi∈LD2(G). This proves (a).

(b). Assume that there exists x∈Ui∩Uj. Let V(T)\{vi,vj}={vk}, then the path vkvixis contained in G[NG(vj)]. Hence {vk,vi,x,vj} induces a complete graph in cl(G). Thiscontradicts the fact that T is a maximal K3 in cl(G), and hence (b) holds.



(c). Assume that xy′ ∈E(G) for some y′ ∈Uj\{y}. Since the path xyvj is containedin G[NG(y′)], {x,y,vj} induces a complete graph in cl(G). Thus, xvj∈E(cl(G)). Thiscontradicts (b) in cl(G), and hence (c) holds. �

Lemma 5. Let G be a 4-connected claw-free graph and let T be a maximal K3 ofcl(G). Then T is also a maximal K3 of G.

Proof. Assume that T is a maximal K3 in cl(G). If V(T)={v1,v2,v3} also inducesa K3 in G, then it is clear that T is maximal in G. Thus, for the sake of a contradiction,we assume that an edge in T is not an edge of G. Without loss of generality, we mayassume that v1v2 /∈E(G). Since v1v2∈E(cl(G)), there exist a graph G∗ and a vertex x inEL(G∗) such that G∗ is obtained from G by repeating the local completion operation,v1v2 /∈E(G∗) and v1v2∈E(G∗

x ). This implies that v1,v2∈NG∗(x)⊆Ncl(G)(x). If x=v3,then it follows from Corollary 1 that x∈LC(G∗)⊆LC(cl(G)), which contradicts (a) ofLemma 4. Thus we have x /∈V(T). However, this yields x∈NG∗

x(v1)∩NG∗

x(v2), which

contradicts (b) of Lemma 4. �

3. STABILITY AND LINE GRAPHS

First, we investigate the structure around T which is a maximal K3 such that G[NG(T)]is cyclically 3-connected. For a claw-free graph G, let T (G) be the set of maximalK3s of G each of which is also maximal in cl(G), and let T∗(G)=⋃

T∈T (G)V(T).A claw-free graph G satisfies the LD2-property if G satisfies the following:

(i) For every v∈LD2,2(G)∩T∗(G), there exists a vertex x∈V(G)\NG(v) such thatxy,xz∈E(G), where y,z∈NG(v) with yz /∈E(G), and

(ii) for every v∈LD2,r(G)∩T∗(G) with r≥3, there exist two vertices x1,x2∈V(G)\NG(v) such that xiyi,xizi∈E(G) for i=1,2, where yi,zi∈NG(v) with yizi /∈E(G).

Lemma 6. Let G be a graph which satisfies the conditions of Theorem 5. Then Gsatisfies the LD2-property.

Proof. Let T ∈T , let v1,v2 and v3 be the three vertices of T and let NG(vi)\V(T)=Ui for 1≤ i≤3. Then by (a) of Lemma 4, each vi is in LD2(G). By symmetry, it sufficesto prove that (i) and (ii) hold for v1. It follows from (b) of Lemma 4 that u1vi /∈E(G)for every u1∈U1 and i∈{2,3}. Note that V(T) is contained in NG(T). Since G[NG(T)]is cyclically 3-connected (and thus 2-connected), G[NG(T)]−{v1} is connected. Hencethere exists an edge between U1 and U2∪U3. This implies (i) in case of v1∈LD2,2(G),and hence we assume that v1∈LD2,r(G) for r≥3. Without loss of generality, we mayassume that there exists an edge between U1 and U2. Let u1∈U1 and u2∈U2, whereu1u2∈E(G).

Since v1∈LD2,r(G) for r≥3 and T is a maximal K3, |U1|≥3 holds. If there is no edgebetween U1 and U2∪U3\{u2}, then G[NG(T)]−{v1,u2} is disconnected. Moreover,each of G[U1] and G[U3∪{v3}] contains a cycle, and these two cycles belong todifferent components of G[NG(T)]−{v1,u2}. This contradicts the fact that G[NG(T)]is cyclically 3-connected. Thus, there exists an edge between U1 and U2∪U3\{u2}.Consequently, (ii) holds. �



Proposition 8. If G satisfies the LD2-property, then cl(G) also satisfies it.

Proof. Assume not. Then there exists a vertex v in cl(G) such that v is containedin T which is a maximal K3 of cl(G) and (i) or (ii) does not hold for v in cl(G). ByLemma 5, T is a maximal K3 of G. Thus (a) of Lemma 4 implies that v∈LD2,r(G) andv∈LD2,r′ (cl(G)) for some r,r′ ≥2 with r′ ≥r. Since E(G)⊆E(cl(G)), (i) holds in caseof r= r′ =2. In case of r,r′ ≥3, it follows from Lemma 3 and the fact v∈LD(cl(G))that an induced 4-cycle which contains v in G is also induced in cl(G). Therefore (ii)holds. Hence we may assume that r=2 and r′ ≥3. Let NG(v)={a1,a2,b1,b2}, wherea1a2,b1b2∈E(G).

Since G satisfies (i), there exist c∈V(G)\{v} such that aic,bjc∈E(G). Without lossof generality, we may assume that i= j=1, and let C=va1cb1v. Then it follows fromLemma 3 that vc /∈E(cl(G)). Since v,c∈Ncl(G)(a1) and v,c∈Ncl(G)(b1), we obtain a1,b1∈LD(cl(G)).

Since NG(v) �=Ncl(G)(v), there exists a graph G∗ and a vertex y in EL(G∗) such thatG∗ is obtained from G by repeating the local completion operation, NG(v)=NG∗(v) andNG(v) �=NG∗

y(v). By the definition of the local completion, it follows that y∈NG∗(v). The

fact y∈EL(G∗) and Corollary 1 yields y∈LC(cl(G)), and hence y=a2 or b2. Withoutloss of generality, we may assume that y=a2.

It holds that a2b1,a2b2 /∈E(G∗), since otherwise v∈LC(cl(G)) follows fromLemma 2. Thus, NG∗(v)∩NG∗ (a2)={a1}. Let y′ be a vertex in NG∗

y(v)\NG∗ (v).

Then, since G∗[NG∗(a2)] is connected, there exists a path P which joins y′ and v inG∗[NG∗(a2)]. Since NG∗(v)∩NG∗ (a2)={a1}, P passes through a1. Let z be the neighborof a1 in P which is not v, then z∈NG∗(a1). Thus z �=b1,b2 holds, which impliesz /∈NG∗(v).

Since local completion keeps the claw-freeness of the original graph, G∗ is claw-free.Recall that v,z∈NG∗(a1) and vz /∈E(G∗). Since vz∈cl(G), it follows from Lemma 2that a1∈LC(cl(G)), a contradiction. �

Next, we investigate the property of a graph H such that L(H) satisfies the LD2-property.

Proposition 9. Let H be a triangle-free simple graph. If L(H) is a 4-connected graphsatisfying the LD2-property, then H is an essentially 4-edge-connected graph thatsatisfies the following:

(I) For every xy∈E(H) with dH(x)=dH(y)=3, x and y are contained in a 4-cycle,and

(II) For every xy∈E(H) with dH(x)≥4 and dH(y)=3, x and y are contained in twodistinct 4-cycles.

Proof. Since L(H) is 4-connected, H is essentially 4-edge-connected. Let xy∈E(H) such that dH(x)=dH(y)=3. Let NH(x)\{y}={x1,x2} and NH(y)\{x}={y1,y2}.Then, xx1,xx2,xy induce a maximal K3 in L(G). Since NL(H)(xy)={xx1,xx2,yy1,yy2},it follows from (a) of Lemma 4 that xy∈LD2,2(L(H)). Therefore, by (i), there existse∈V(L(H))\NL(H)(xy) such that ee1,ee2∈E(L(H)), where e1,e2∈NL(H)(xy) with e1e2 /∈E(L(H)). Then xy,e1,e,e2 induce a 4-cycle in H, and hence (I) holds. By a similarargument, we obtain (II) for xy∈E(H) with dH(x)≥4 and dH(y)=3. �



Now we can deduce that the following proposition implies Theorem 5. The proofof this proposition will be given in the next section.

Proposition 10. Let H be an essentially 4-edge-connected simple graph. If H istriangle-free and H satisfies (I) and (II) of Proposition 9, then H has a dominatingclosed trail.

Proof of Theorem 5. Let G be a graph which satisfies the conditions ofTheorem 5. Then by Lemma 6 and Proposition 8, cl(G) satisfies the LD2-property. ByTheorem 1, there exists a triangle-free simple graph H such that L(H)=cl(G). Then byPropositions 9 and 10, H contains a dominating closed trail. It follows from Theorem 2that cl(G) is hamiltonian, and hence Theorem 1 implies that G is hamiltonian. �

4. CORE

The core of a graph G, denoted by core(G), is a graph obtained by recursively deletingall the vertices of degree 1, recursively contracting exactly one edge xz or zy for eachpath xzy with dG(z)=2 (we call this operation suppressing), and recursively deletingthe created loops. It is easy to see that core(H) is uniquely defined. Some propertiesof cores of graphs are investigated in [8]. The following lemmas follow immediatelyfrom the definitions, but useful in our proof.

Lemma 7. Let H be an essentially 4-edge-connected multigraph such that L(H) is nota complete graph. Then core(H) is also essentially 4-edge-connected, V≥4(core(H))=V≥4(H), V3(core(H))=V3(H) and �(core(H))≥3.

In our proof of Proposition 10, the following proposition is essential.

Proposition 11. Let H be an essentially 4-edge-connected multigraph which satisfiesthe following:

(I′) For every xy∈E(H) with dH(x)=dH(y)=3, xy is contained in a cycle of lengthat most 4, and

(II′) for every xy∈E(H) with dH(x)≥4 and dH(y)=3, xy is contained in a cycle oflength at most 3, or x, y is contained in two distinct 4-cycles.

If �(H)≥3, then H has a closed trail that contains all vertices of V≥4(H).

The proof of Proposition 11 will be given in Section 5. Here we prove Proposition 10by using Proposition 11.

Proof of Proposition 10. We may assume that L(H) is not a complete graph.Let H′ =core(H). First, we show that H′ satisfies the conditions of Proposition 11.By Lemma 7, H′ is essentially 4-edge-connected, �(H′)≥3, V≥4(H′)=V≥4(H) andV3(H′)=V3(H). Let v1v2 be an edge of H′, where dH′(v2)=3. Then v1v2 is also anedge of H, for otherwise there exists an essential 3-edge-cut of H consisting of twoedges adjacent to v2 and an edge adjacent to v1. Assume that dH′(v1)≥4. Then, sincev1v2∈E(H), dH(v1)≥4 and dH(v2)=3, there exist two distinct 4-cycles v1v2w1u1v1and v1v2w2u2v1 in H. Each of the two cycles corresponds to a cycle of length at most 4



that contains v1v2 in H′, and thus v1v2 satisfies the condition (II′) of Proposition 11. Bya similar argument, in case of dH′(v1)=3, we can show that v1v2 satisfies the condition(I′) of Proposition 11. Consequently, H′ satisfies the conditions of Proposition 11, andhence H′ has a closed trail which contains all the vertices of V≥4(H′).

Next, we show that there exists a dominating closed trail in H′ which contains all thevertices of V≥4(H′). Take a closed trail T in H′ so that V≥4(H′)⊆V(T) and |T∩V3(H′)|is as large as possible. Assume, for the sake of a contradiction, that there exists x∈V3(H′)\V(T) such that |NH′(x)∩V(T)|≤2. Since V≥4(H′)⊆V(T) and �(H′)≥3, we cantake x with NH′(x)∩V(T) �=∅. Let y∈NH′(x)∩V(T). Since H′ satisfies the conditionsof Proposition 11, there exists a cycle C1 of length at most 4 which contains xy. If|E(T)∩E(C1)|≤1, then let T∗ be the graph induced by E(T) E(C1), where denotesthe symmetric difference. Then T∗ is a closed trail which contains x and all the verticesin V(T). This contradicts the maximality of |V(T)∩V3(H′)|. Therefore, we obtain|E(T)∩E(C1)|≥2. It follows from x /∈V(T) that C1 has length 4. Let C1=xyz1z2x,then it holds that yz1,z1z2∈E(T). Let w be the vertex in NH′(x)\{y,z2}. Since H′satisfies the conditions of Proposition 11, there exists a cycle C2 of length at most 4which contains xw. By a similar argument as above, we obtain |E(T)∩E(C2)|≥2. Itfollows from x /∈V(T) that w∈V(T), which implies |NH′ (x)∩V(T)|=3, a contradiction.Consequently, it holds that |NH′ (x)∩V(T)|=3 for every x∈V3(H′)\V(T). Hence T isa dominating closed trail in H′ which contains all the vertices of V≥4(H′).

Finally, we show that there exists a dominating closed trail of H. Let F=E(T)\E(H).Then by the construction of H′ from H, for every e=xy∈F, there exists a path Pe whichjoins x and y in H such that E(Pe)∩E(H′)=∅. Let T0 be the graph which is obtainedfrom T by deleting all the edges in F and adding the path Pe for every e∈F. Then T0is a closed trail in H such that V(T)⊆V(T0). Since V≥4(H)=V≥4(H′)⊆V(T)⊆V(T0)and H is essentially 4-edge-connected, T0 dominates every edge which is adjacent toa vertex in V1(H)∪V2(H)∪V≥4(H). Moreover, since an edge e of H which joins twovertices of V3(H) is also an edge of H′, e is dominated by T . Thus e is dominatedby T0. Therefore, T0 is a dominating closed trail of H. �

5. PROOF OF PROPOSITION 11

Let G be a multigraph and H be a subgraph of G. Then G /H denotes the multigraphobtained from G by identifying the vertices of H with a new vertex, which we denotevH , and by deleting the created loops. If F is a subgraph ofG, F /H stands for F / (H∩F).Before proving Proposition 11, we prepare some lemmas.

Lemma 8. Let G be an essentially 4-edge-connected multigraph with �(G)≥3 whichsatisfies (I′) and (II′) of Proposition 11. Then for every subgraph H of G with �(H)≥2,G /H is also an essentially 4-edge-connected multigraph which satisfies (I′) and (II′).

Proof. Let G′ =G /H, then it is easy to see that G′ is essentially 4-edge-connected. In case of E(G−V(H))=∅, the assertion is obvious, and thus we assumeE(G−V(H)) �=∅. Let v1v2∈E(G′) such that dG′(v2)=3. Since G is essentially4-edge-connected and E(G−V(H)) �=∅, dG′(vH)≥4. This implies v2 �=vH, and hencev2∈V(G)\V(H) and dG(v2)=3 holds.



First consider the case dG(v1)=3. Since G satisfies (I′) of Proposition 11 anddG(v2)=3, v1v2 is contained in a cycle of length at most 4, say C, in G. Since v2 /∈V(H),C′ =H /C is a cycle of length at most 4 which contains v1v2 in G′. This proves that (I′)holds for G′ in case of dG′(v1)=3. In case of dG′(v1)≥4, it follows from dG(v1)=3and �(H)≥2 that E(H)∩E(C) �=∅. Thus the length of C′ is at most 3. This proves that(II′) holds for G′.

Next consider the case dG(v1)≥4. Since G satisfies (II′) of Proposition 11 anddG(v2)=3, v1v2 is contained in a cycle of length at most 3, or v1,v2 is contained in twodistinct 4-cycles. In both cases, since v2 /∈V(H), any cycle of length l which containsv2 in G is also a cycle of length at most l which contains v2 in G′. Thus (II′) holdsfor G′, and hence this lemma holds. �

A multigraph H is called collapsible if for every even set S⊆V(H), there existsa spanning connected subgraph F of H such that v∈S if and only if dF(v) is odd.This concept was introduced by Catlin [1]. Collapsible subgraphs have the followingproperty.

Lemma 9. Let G be a multigraph and H⊂G be a collapsible subgraph. If G /H hasa closed trail T which contains vH , then G has a closed trail T∗ which contains all thevertices in V(T)∪V(H)\{vH}.

Proof. Let T ′ be a subgraph of G such that T ′ /H=T and E(T ′)∩E(H)=∅, andlet S={v∈V(G)||NG(v)∩T ′| is odd}. Since T ′ is a closed trail in G /H, it holds that|S| is even and S⊆V(H). Since H is collapsible, there exists a spanning connectedsubgraph F of H such that v∈S if and only if dF(v) is odd. Now T ′∪F is the requiredclosed trail. �

We call a multigraph G reduced if G has no collapsible subgraph of order at leasttwo. We use the following two theorems in our proof of Proposition 11.

Theorem 12 (Catlin [1]). Let G be a reduced multigraph. Then G has no cycle oflength at most 3.

Theorem 13 (Catlin [1]). Any 4-edge-connected multigraph is collapsible.

Proof of Proposition 11. Assume not, and letH be a counterexample of least order.It is easy to see that |V(H)|≥3.

Claim 1. H is reduced. In particular, the girth of H is at least 4.

Proof. Suppose, for the sake of a contradiction, thatH has a collapsible subgraphH′of order at least two. Then by definition, �(H′)≥2 holds. In case of E(H−V(H′))=∅,by Lemma 9, it is easy to find a spanning closed trail in H, a contradiction. Thereforewe assume that V(H) �=V(H′) and E(H−V(H′)) �=∅. Then it follows from Lemma 8that H /H′ is an essentially 4-edge-connected graph which satisfies (I′) and (II′) ofProposition 11. Since H is essentially 4-edge-connected, the degree of vH′ in H /H′is at least 4, thus the minimum degree of H /H′ is at least 3. By the minimality ofH, there exists a closed trail in H /H′ that contains all the vertices of V≥4(H /H′).Note that this trail contains the vertex vH′ . By Lemma 9, we can find a closedtrail in H which contain every vertex in V(H′)∪V≥4(H /H′)\{vH}, a contradiction.



Therefore H is reduced. It follows from Theorem 12 that the girth of H is atleast 4. �

If V3(H)=∅, then H is 4-edge-connected. It follows from Theorem 13 that H iscollapsible, contradicting the fact that H is reduced. Hence there exists a vertex x inV3(H). Let x1,x2,x3 be the neighbors of x in H. By applying (II′) of Proposition 11 tothe three edges xx1,xx2 and x3, we obtain two cycles C1 and C2 of length 4 containingx, since the girth of H is at least 4. Without loss of generality, we may assume that xx2 iscontained in both C1 and C2. Let C1=xx1yx2x and C2=xx3zx2x, and let C=H[V(C1)∪V(C2)]. It follows from Lemma 8 that H /C is an essentially 4-edge-connected graphwith minimum degree at least 3 which satisfies (I′) and (II′) of Proposition 11. By theminimality of H, there exists a closed trail T ′ in H /C that contains all the verticesof V≥4(H /C). Let T be a subgraph of H such that T /C=T ′ and E(T)∩E(C)=∅, andlet O(T)={v|v∈V(C), dT (v) is odd}. Then since every vertex of T ′ has even degree,|O(T)| is even. Moreover, since x has no neighbor in V(H)\V(C), x /∈O(T). Now weconsider two cases.

Case 1. y �= z.

Here, we assume that T was chosen so that O(T) �= {y,z}, if possible. In case ofO(T) �= {y,z}, all the cases are divided into the following, up to symmetry. In each casewe define T∗ as follows:

• If O(T)=∅, let T∗ =T∪E(C1)∪E(C2)\{xx2},• if O(T)={x1,y}, let T∗ =T∪E(C1)∪E(C2)\{x1y,xx2},• if O(T)={x1,x2}, let T∗ =T∪E(C1)∪E(C2)\{x1x},• if O(T)={x1,z}, let T∗=T∪E(C1)∪E(C2)\{xx1,zx2},• if O(T)={x1,x3}, let T∗ =T∪E(C1)∪E(C2)\{xx1,xx2,xx3},• if O(T)={y,x2}, let T∗ =T∪E(C1)∪E(C2)\{yx2,xx2},• if O(T)={x1,y,x2,z}, then either T∪E(C1)∪E(C2)\{xx1,yx2,x2z} or T∪E(C1)∪E(C2)\{x1y,xx2,x2z} is connected, and let T∗ be the connected one,

• if O(T)={x1,y,x2,x3}, let T∗ =T∪E(C1)∪E(C2)\{x1y,xx3},• if O(T)={x1,y,z,x3}, then either T∪E(C1)∪E(C2)\{x1y,xx2,x3z} or T∪E(C1)∪E(C2)\{x1y,xx3,x2z} is connected, and let T∗ be the connected one

(Fig. 3). In either case T∗ is a closed trail which contains V≥4(H). Such a T∗ contra-dicts the assumption. Therefore we obtain O(T)={y,z}. Moreover, if x1∈V3(H), thenT∗ =T∪E(C1)∪E(C2)\{yx1,xx1,zx2} is a closed trail which contains V≥4(H), sincedT (x1)=0 (Fig. 4I). Such a T∗ contradicts the assumption, and hence x1∈V≥4(H) holds.

In case of yx3∈E(H), let T∗=T∪E(C1)∪E(C2)∪{yx3}\{xx2,x3z} (Fig. 4II). ThenT∗ is a closed trail which contains V≥4(H), a contradiction. Thus yx3 /∈E(H), and bysymmetry, x1z /∈E(H). Since x1∈V≥4(H) and H satisfies (II′) of Proposition 11, thereexists an edge y′x′ �=yx2, where y′ ∈NH(x1)\{x} and x′ ∈NH(x)\{x1}. If y′ =y, then sincethe girth of H is at least 4, it follows that x′ =x3, contradicting the fact that yx3 /∈E(H).Hence y′ �=y. Moreover, since the girth ofH is at least 4, y′ �=x2,x3. Therefore, y′ /∈V(C).It follows from x′ ∈NH(x)\{x1} that x′ =xi, where i=2 or 3.

In case of {y′x1,y′xi}�E(T), let T∗ be the graph induced by E(T) {y′x1,y′xi}.Then T∗ /C is a closed trail in H /C that contains all the vertices of V≥4(H /C) and



FIGURE 3. Circled vertices are inO(T ), and dotted linesstand for the edges in E(T ∗)\E(T ).

I II III IV

FIGURE 4. Circled vertices are in O(T ), dotted lines stand for the edges inE(T ∗)\E(T ) and broken lines stand for the edges in E(T ).

O(T∗)={x1,xi,y,z}. This contradicts the choice of T . Therefore, both y′x1 and y′xi arein E(T). Let T∗=T∪E(C1)∪E(C2)\{yx2,xxi,zxi} (Figs. 4III and IV). Then, whetheri=2 or 3, T∗ is a closed trail which contains V≥4(H), a contradiction.

Case 2. y= z.

We assume that T was chosen so that O(T) �=∅, if possible. In case of O(T) �=∅, allthe cases are divided into the following, up to symmetry. In each case we define T∗



FIGURE 5. Circled vertices are in O(T ) and dotted linesstand for the edges in E(T ∗)\E(T ).

I II

FIGURE 6. Dotted lines stand for the edges in E(T ∗)\E(T ) and brokenlines stand for the edges in E(T ).

as follows:

• If O(T)={x1,y}, let T∗ =T∪E(C1)∪E(C2)\{xx1};• if O(T)={x1,x2}, let T∗ =T∪E(C1)∪E(C2)\{x1y,xx2};• if O(T)={x1,x2,x3,y}, let T∗ =T∪{yx1,yx2,yx3}

(Fig. 5). Then in either case T∗ is a closed trail which contains V≥4(H). Such a T∗contradicts the assumption. Therefore we obtain O(T)=∅. Moreover, if x1∈V3(H),then T∗ =T∪E(C1)∪E(C2)\{x1y,x1x} is a closed trail which contains V≥4(H), sincedT (x1)=0 (Fig. 6I). Such a T∗ contradicts the assumption, and hence x1∈V≥4(H)holds. Since H satisfies (II′) of Proposition 11, there exists an edge y′x′ �=yx2 wherey′ ∈NH(x1)\{x} and x′ ∈NH(x)\{x1}. We have y′ /∈V(C), because the girth of H is atleast 4. It follows from x′ ∈NH(x)\{x1} that x′ =xi, where i=2 or 3.

In case of {y′x1,y′xi}�E(T), let T∗ be the graph induced by E(T) {y′x1,y′xi}.Then T∗ /C is a closed trail in H /C that contains all the vertices of V≥4(H /C) andO(T∗)={x1,xi}. This contradicts the choice of T . Therefore, both y′x1 and y′xi are inE(T). Let T∗=T∪E(C1)∪E(C2)\{yx1,xx1} (Fig. 6II). Then, whether i=2 or 3, T∗is a closed trail which contains V≥4(H), a contradiction. This completes the proof ofProposition 11. �

6. REMARKS

We conclude this article by showing the following proposition.



Proposition 14. Theorem 5 generalizes Theorems 6 and 7.

Proof. Let G be a graph satisfying the hypothesis of Theorem 6 or 7. Then G is a4-connected claw-free graph. Let T be a maximal K3 of G which is also maximal incl(G). Moreover, let v1,v2,v3 be the three vertices of T , let Ui=NG(vi)\V(T) and letU=⋃3

i=1Ui. Then, it follows from (b) of Lemma 4 that uivj /∈E(G) for every ui∈Uiand j �= i.

Claim 2. Assume that the following holds for each i:

(A) If |Ui|=2, then there exists ui∈NG(vi) such that ui has a neighbor in U\Ui and(B) if |Ui|≥3, then there exist ui,wi∈NG(vi) with ui �=wi such that ui and wi have

a neighbor in U\Ui.

Then G[NG(T)] is cyclically 3-connected.

Proof. Assume, for the sake of a contradiction, that G[NG(T)] is not cyclically3-connected. By (A) and (B), it holds that

there exists an edge joining a vertex of Ui and a vertex of U\Ui for each i. (1)

Moreover, by (a) of Lemma 4, each vi is in LD2(G). Hence G[Ui] is a completegraph for 1≤ i≤3. This implies that G[NG(T)] is 2-connected, and hence there existx1,x2∈NG(T) such that at least two components of G[NG(T)]−{x1,x2} contain a cycle.

If {x1,x2}⊆V(T) or {x1,x2}∩V(T)=∅, then G[NG(T)]−{x1,x2} is connected, acontradiction. Hence one vertex of {x1,x2} is in T and the other one is not in T . Withoutloss of generality, we may assume that x1=v1. By (1), there exist u1∈U1 and u2∈U2∪U3 such that u1u2∈E(G). Without loss of generality, we may assume that u2∈U2.If there exist u′

1∈U1\{u1} and u3∈U3 such that u′1u3∈E(G), then G[NG(T)]−{x1} is

2-connected, contradicting the fact that G[NG(T)]−{x1,x2} is disconnected. Hencethere is no edge between U1\{u1} and U3. (2)

Suppose that |U1|=2. By (1), there exists an edge between Uj and U3, where j=1or 2. If j=1, then it follows from (2) that u1 is the only cut-vertex of G[NG(T)]−{x1}.Hence x2=u1. If j=2, then u1 and u2 are the only cut-vertices of G[NG(T)]−{x1}.Hence x2=u1 or u2. In either case, G[NG(T)]−{x1,x2} contains only one componentwhich contain a cycle, a contradiction. Thus we have |U1|≥3.

It follows from (B) and (2) that there exists an edge u′1u

′2, where u′

1∈U1, u′2∈U2

and u′1 �=u1. By (c) of Lemma 4, u2 �=u′

2 holds. Moreover, by (1), there exists anedge between U1∪U2 and U3. This implies that G[NG(T)]−{x1} is 2-connected, acontradiction. �

Here we assume thatG satisfies the conditions of Theorem 6, and prove thatG[NG(T)]is cyclically 3-connected. By Claim 2 and the symmetry, it suffices to show that (A)and (B) of Claim 2 holds for i=1. Let {a1,b1}⊆U1. Since {a1,b1,v1,v2,v3} inducesan hourglass in G, there exists an edge e between {a1,b1} and U2∪U3. This implies(A) in case of |U1|=2, and hence we assume |U1|≥3. Without loss of generality,we may assume that e is incident with a1. Let c1∈U1, where c1 �=a1,b1. Then, since{b1,c1,v1,v2,v3} induces an hourglass in G, there exists an edge e between {b1,c1}



and U2∪U3, This implies (B). Consequently, G[NG(T)] is cyclically 3-connected. Thisimplies that Theorem 5 generalizes Theorem 6.

Finally, we assume that G satisfies the conditions of Theorem 7, and prove thatG[NG(T)] is cyclically 3-connected. By a similar argument as above, it suffices to showthat (A) and (B) of Claim 2 holds for i=1. Let {ai,bi}⊆Ui for every i with i=1,2and 3. Then by (b) of Lemma 4 and the fact that G is T3-free, the graph induced by{a1,b1,v1,v3,v2,a2,b2} contains a T3 which is not induced. Since uivj /∈E(G) for everyui∈Ui and j �= i, there exists no edge between {a1,b1} and {v2,v3} and between {a2,b2}and {v1,v3}. Thus there exists an edge e between {a1,b1} and {a2,b2}. This implies(A) in case of |U1|=2, and hence we assume |U1|≥3. Without loss of generality, wemay assume that e is incident with a1. Let c1∈U1, where c1 �=a1,b1. Then by (b) ofLemma 4 and the fact that G is T3-free, the graph induced by {b1,c1,v1,v3,v2,a2,b2}contains a T3 which is not induced. By a similar argument as above, there exists an edgebetween {b1,c1} and {a2,b2}, which implies (B). Consequently, G[NG(T)] is cyclically3-connected, and this implies that Theorem 5 generalizes Theorem 7. �

ACKNOWLEDGMENTS

The authors thank anonymous referees for their careful reading and many helpfulcomments.

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maximal k3's and hamiltonicity of 4-connected claw-free graphs

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