Closure, 2-factors, and cycle coverings in claw-free graphs

Closure, 2-Factors, andCycle Coverings inClaw-Free Graphs

Zdenek Ryjacek,1 Akira Saito,2 and R. H. Schelp3

1DEPARTMENT OF MATHEMATICSUNIVERSITY OF WEST BOHEMIA

UNIVERZITNI 22, 306 14 PLZENCZECH REPUBLIC

E-mail: [email protected] OF MATHEMATICS

NIHON UNIVERSITYSAKURAJOSUI 3-25-40

SETAGAYA-KU, TOKYO 156-8550JAPAN

E-mail: [email protected] OF MATHEMATICAL SCIENCES

THE UNIVERSITY OF MEMPHISMEMPHIS, TN 38152

E-mail: [email protected]

Received January 16, 1997; revised January 31, 1999

Abstract: In this article, we study cycle coverings and 2-factors of a claw-freegraph and those of its closure, which has been defined by the first author (On aclosure concept in claw-free graphs, J Combin Theory Ser B 70 (1997), 217–224).For a claw-free graph G and its closure cl(G), we prove: (1) V (G) is covered byk cycles in G if and only if V (cl(G)) is covered by k cycles of cl(G); and (2) Ghas a 2-factor with at most k components if and only if cl(G) has a 2-factor withat most k components. c© 1999 John Wiley & Sons, Inc. J Graph Theory 32: 109–117, 1999

Keywords: closure; claw-free graph; 2-factor; cycle covering

Contract grant sponsor: NSFContract grant number: DMS-9400530Correspondence to: Akira Saito

c© 1999 John Wiley & Sons, Inc. CCC 0364-9024/99/020109-09

110 JOURNAL OF GRAPH THEORY

1. INTRODUCTION

For graph theoretic notation not defined in this article, we refer the reader to [2]. Avertex x of a graph G is said to be locally connected, if the neighborhood NG(x)of x in G induces a connected graph. A locally connected vertex x is said to beeligible, if NG(x) induces a noncomplete graph. Let x be an eligible vertex of agraph G. Consider the operation of joining every pair of nonadjacent vertices inNG(x) by an edge, so that NG(x) induces a complete graph in the resulting graph.This operation is called local completion ofG at x. For a graphG, letG0 = G. Fori ≥ 0, if Gi is defined and it has an eligible vertex xi, then apply local completionofGi at xi to obtain a new graphGi+1. IfGi has no eligible vertex, let cl(G) = Giand call it the closure ofG. The above operation was introduced, and the followingtheorems were proved in [3].

Theorem A ([3]). If G is a claw-free graph, then

(1) a graph obtained from G by local completion is also claw-free, and(2) cl(G) is uniquely determined.

Theorem B ([3]). Let G be a claw-free graph. Then G, is Hamiltonian if andonly if cl(G) is Hamiltonian.

Recently, several other properties on paths and cycles of a claw-free graph andthose of its closure were studied in [1]. In particular, the following theorem wasproved.

Theorem C ([1]).

(1) A claw-free graph G is traceable if and only if cl(G) is traceable.(2) There exist infinitely many claw-free graphsG such that cl(G) is Hamiltonian-

connected, while G is not Hamiltonian-connected.(3) For any positive integer k, there exists a k-connected claw-free graphG such

that cl(G) is pancyclic, while G is not pancyclic.

LetH1, . . . , Hk be subgraphs ofG. ThenG is said to be covered byH1, . . . , Hk

if V (G) = V (H1) ∪ · · · ∪ V (Hk).We consider two interpretations of a Hamiltonian cycle. First, a Hamiltonian

cycle of a graph G is a cycle that covers G. Second, it is considered as a 2-factorwith one component. These interpretations may lead us to possible extensions ofTheorem B to cycle coverings and 2-factors. This is the motivation of this article.

2. MAIN RESULTS

We prove the following theorems as generalizations of Theorem B.

Theorem 1. Let G be a claw-free graph. Then G is covered by k cycles if andonly if cl(G) is covered by k cycles.

CLOSURE, 2-FACTORS, AND CYCLE COVERINGS 111

Theorem 2. Let G be a claw-free graph. If cl(G) has a 2-factor with k compo-nents, then G has a 2-factor with at most k components.

Note that the conclusion of Theorem 2 says G has a 2-factor with ‘‘at most’’ kcomponents. Under the assumption of Theorem 2,G does not always have a 2-factorwith exactly k components, if k ≥ 2. Let G be a graph with k − 1 componentsH1, . . . , Hk−1, where H1 is the graph shown in Fig. 1, and H2, . . . , Hk−1 arecycles of arbitrary lengths. Then G is claw-free and cl(G) = cl(H1) ∪ H2 ∪· · · ∪Hk−1, where cl(H1) is isomorphic to K9. Since K9 has a 2-factor with twocomponents, cl(G) has a 2-factor with k components. However, G has no 2-factorwith k components, since H1 does not have a 2-factor with two components.

Before proving the above theorems, we introduce some notation that is used inthe subsequent arguments. For a graphG and ∅ /= S ⊂ V (G), the subgraph inducedby S is denoted by G[S]. When we consider a path or a cycle, we always assign anorientation. Let P = x0x1 · · ·xm. We call x0 and xm the starting vertex and theterminal vertex of P , respectively. The set of internal vertices of P is denoted byint(P ): int(P ) = {x1, x2, . . . , xm−1}. The length ofP is the number of edges inP ,

and is denoted by l(P ). We define x+(P )i = xi+1 and x−(P )

i = xi−1. Furthermore,

we define x++(P )i = xi+2. When it is obvious which path is considered in the

context, we sometimes write x+i and x−i instead of x+(P )

i and x−(P )i , respectively.

For xi, xj ∈ V (P ) with i ≤ j, we denote the subpath xixi+1 · · ·xj by xi→Pxj . The

same path traversed in the opposite direction is denoted by xj←Pxi. We use similar

notations with respect to cycles with a given orientation.We present several lemmas before proving the main theorems.

Lemma 1. Let G be a claw-free graph, and let x be a locally connected vertex.Let T1, T2 ⊂ V (G) with T1 ∩ T2 = {x}. Suppose that both G[T1] and G[T2]are Hamiltonian, but G[T1 ∪ T2] is not Hamiltonian. Choose cycles C1 and C2with V (C1) ∪ V (C2) = T1 ∪ T2 and V (C1) ∩ V (C2) = {x} and a path P inG[NG(x)] with starting vertex in {x+(C1), x−(C1)} and terminal vertex in {x+(C2),x−(C2)}, so that P is as short as possible. Then 2 ≤ l(P ) ≤ 3 and int(P ) ∩(T1 ∪ T2) = ∅.

FIGURE 1.


Proof. First, note that each Hamiltonian cycle Di in G[Ti] (i = 1, 2) satisfiesV (D1)∪V (D2) = T1 ∪T2 and V (D1)∩V (D2) = {x}. Furthermore, since x is alocally connected vertex of G, there exists a path in G[NG(x)] with starting vertexin {x+(D1), x−(D1)} and terminal vertex in {x+(D2), x−(D2)}. Therefore, we canmake a choice for (C1, C2, P ). Let u1 = x+(C1), v1 = x−(C1), u2 = x+(C2), andv2 = x−(C2). We may assume that the starting and terminal vertices of P are u1and u2, respectively.

If u1u2 ∈ E(G), then C ′ = xv1←C1u1u2

→C2v2x is a cycle in G with V (C ′) =

V (C1) ∪ V (C2) = T1 ∪ T2. This contradicts the assumption. Hence, we haveu1u2 /∈ E(G). Similarly, we have u1v2, v1u2, v1v2 /∈ E(G). Since {u1, v1, u2} ⊂NG(x) and G is claw-free, we have u1v1 ∈ E(G). Similarly, u2v2 ∈ E(G).

Let w = u+(P )1 . We claim w /∈ V (C1) ∪ V (C2). Assume that w ∈ V (C1) ∪

V (C2). Since w ∈ V (P ) ⊂ NG(x), w /= x. Thus, w ∈ u1→C1v1 ∪ u2

→C2v2.

First, suppose that w ∈ u1→C1v1. Then by the choice of P,w ∈ u+

1→C1v

−1 .

Since {x,w+, w−} ⊂ NG(w) andG is claw-free, we have {xw+, xw−, w+w−}∩E(G) /= ∅. If w+w− ∈ E(G), let C ′1 = xwu1

→C1w

−w+→C1v1x,C′2 = C2, and

P ′ = w→Pu2. If w−x ∈ E(G), then let C ′1 = xw

→C1v1u1

→C1w

−x,C ′2 = C2, andP ′ = w

→Pu2. If w+x ∈ E(G), then let C ′1 = xw

←C1u1v1

←C1w

+x,C ′2 = C2, andP ′ = w

→Pu2. Then in each case, since V (C ′1) = V (C1), we have V (C ′1)∪V (C ′2) =

V (C1)∪V (C2) = T1∪T2, and V (C ′1)∩V (C ′2) = {x}. Furthermore,w = x+(C′1)

and l(P ′) < l(P ). This contradicts the choice of (C1, C2, P ).Now, suppose that w ∈ u2

→C2v2. Since {u2, v2} ∩ NG(u1) = ∅, we have w ∈

u+2→C2v

−2 . Since {x,w−, w+} ⊂ NG(w) and G is claw-free, {xw−, xw+, w−w+}

∩ E(G) /= ∅. If xw− ∈ E(G), let C = xv1←C1u1w

→C2v2u2

→C2w

−x. If xw+ ∈E(G), let C = xw+→C2v2u2

→C2wu1

→C1v1x. Then, in either case, C is a cycle in

G with V (C) = V (C1) ∪ V (C2) = T1 ∪ T2. This contradicts the assumption.If w−w+ ∈ E(G), then let C ′1 = xwu1

→C1v1x,C

′2 = xu2

→C2w

−w+→C2v2x, andP ′ = w

→Pu2. ThenV (C ′1)∪V (C ′2) = V (C1)∪V (C2) = T1∪T2, V (C ′1)∩V (C ′2) =

{x}, w = x+(C′1) and l(P ′) < l(P ). This contradicts the choice of (C1, C2, P ).Therefore, w /∈ V (C1) ∪ V (C2).

Let w′ = u−(P )2 . (Possibly w′ = w.) Then, by the same arguments, we have

w′ /∈ V (C1) ∪ V (C2).By the choice of (C1, C2, P ), P is an induced path. Hence, if l(P ) ≥ 4, then

{u1, u++(P )1 , u2} is an independent set. Since V (P ) ⊂ NG(x) and G is claw-free,

this is a contradiction. Thus, l(P ) ≤ 3. Since u1u2 /∈ E(G), l(P ) ≥ 2. Theseimply int(P ) ∩ (T1 ∪ T2) = ∅.

By similar arguments, we have the following lemma.

Lemma 2. Let G be a claw-free graph, and let x be a locally connected vertexof G. Let T ⊂ V (G) with x ∈ T, and let u ∈ NG(x) − T. Suppose that G[T ]is Hamiltonian, but G[T ∪ {u}] is not Hamiltonian. Choose a Hamiltonian cycleC in G[T ], and a path P in G[NG(x)] with starting vertex in {x+(C), x−(C)} andterminal vertex u, so that P is as short as possible. Then 2 ≤ l(P ) ≤ 3 andint(P ) ∩ (T ∪ {u}) = ∅.


We prove one more lemma.

Lemma 3. Let G be a claw-free graph, and let x be an eligible vertex of G. LetG′ be the graph obtained from G by local completion at x. Let C ′ be a cycle in G′with x ∈ V (C ′). Then either (1) or (2) follows:

(1) There exists a cycle C in G with V (C) = V (C ′).(2) There exist T1, T2 ⊂ V (G) such that

(2.1) T1 ∪ T2 = V (C ′) and T1 ∩ T2 = {x}, and(2.2) G[Ti] is Hamiltonian or isomorphic to K2 (i = 1, 2).

Proof. Let B = E(G′)− E(G). Note that, for each uv ∈ B, {u, v} ⊂ NG(x).Choose a cycle C in G′ with V (C) = V (C ′), so that |E(C) ∩ B| is as small aspossible. If E(C) ∩ B = ∅, then C is a cycle satisfying (1). Therefore, we mayassume E(C) ∩B /= ∅.

We claim |E(C) ∩ B| = 1. Assume, to the contrary, that |E(C) ∩ B| ≥ 2, saye1, e2 ∈ E(C) ∩ B, e1 /= e2. Let ei = xiyi (i = 1, 2). We may assume x1, y1,x2, y2 and x appear in this order along C. (Possibly, y1 = x2.) Then x1, x2,and x− are distinct vertices inNG(x). SinceG is claw-free, {x1x2, x1x

−, x2x−}∩

E(G) /= ∅. If x1x2 ∈ E(G), let C0 = y2→Cx1x2

←Cy1y2. Then V (C0) = V (C)

and E(C0) = (E(C) − {x1y1, x2y2}) ∪ {x1x2, y1y2}. This implies |E(C0) ∩B| < |E(C) ∩ B|, which contradicts the minimality of |E(C) ∩ B|. If x1x

− ∈E(G), let C0 = x

→Cx1x

−←Cy1x. Then V (C0) = V (C) and E(C0) = (E(C) −{x1y1, xx

−}) ∪ {xy1, x1x−}. Since xy1 ∈ E(G), we have |E(C0) ∩ B| <

|E(C) ∩ B|, again a contradiction. We have a similar contradiction, if x−x2 ∈E(G). Therefore, the claim is proved.

Let E(C) ∩ B = {x1y1}. We may assume that x, x1, and y1 appear in thisorder along C. Let T1 = x

→Cx1 and T2 = y1

→Cx. Then T1 ∪ T2 = V (C) and

T1 ∩ T2 = {x}. Since x1y1 ∈ B, xx1, xy1 ∈ E(G). If x1 /= x+, then x→Cx1x is

a Hamiltonian cycle in G[T1]. If x1 = x+, then G[T1] ' K2. Similarly, G[T2] iseither Hamiltonian or isomorphic to K2.

Let G′ be a graph obtained from a claw-free graph by local completion at avertex. Using Lemmas 1, 2, 3 we prove that for each cycle inG′ there exists a cyclein G that contains it. We can also impose some restriction on its length.

Theorem 3. Let G be a claw-free graph, and let x be a locally connected vertexof G. Let G′ be the graph obtained from G by local completion at x. Then for eachcycleC ′ inG′ there exists a cycleC inG with V (C ′) ⊂ V (C) and l(C ′) ≤ l(C) ≤l(C ′) + 3.

Proof. If E(C ′) ∩ (E(G′) − E(G)) = ∅, then C ′ is a required cycle. Hence,we may assume E(C ′) ∩ (E(G′)− E(G)) /= ∅.

If x ∈ V (C ′), let C ′1 = C ′. Suppose that x /∈ V (C ′). Let e = u+(C1) ∈E(C ′) ∩ (E(G′) − E(G)). Then {u, u+(C′)} ⊂ NG(x). Let C ′1 = u+(C′)→C′uxu+(C′). In either case, we have a cycle C ′1 with V (C ′) ∪ {x} ⊂ V (C ′1)and l(C ′) ≤ l(C ′1) ≤ l(C ′) + 1.


If there exists a cycle C in G with V (C ′1) = V (C), then C is a required cycle.Therefore, we may assume that G has no such cycle. Then, by Lemma 3, thereexist T1, T2 ⊂ V (G) with T1 ∩ T2 = {x}, and T1 ∪ T2 = V (C ′1) such that G[T] isHamiltonian or G[Ti] ' K2.

Suppose that both G[T1] and G[T2] are Hamiltonian. Then by Lemma 1 thereexist cycles C1 and C2 in G and a path P in G[NG(x)] such that

(1) V (C1) ∪ V (C2) = T1 ∪ T2 = V (C ′1), V (C1) ∩ V (C2) = {x}, and

(2) P joins {x+(C1), x−(C1)} and {x+(C2), x−(C2)}, 2 ≤ l(P ) ≤ 3 and int(P ) ∩(T1 ∪ T2) = ∅.

Let u = x+(C1) and v = x+(C2). We may assume that P joins u and v. LetC = x

←C1u

→Pv→C2x. Then C is a cycle in G,V (C ′1) ⊂ V (C), and l(C ′1) ≤ l(C) ≤

l(C ′1) + 2. Therefore, V (C ′) ⊂ V (C ′1) ⊂ V (C), and l(C ′) ≤ l(C ′1) ≤ l(C) ≤l(C ′1) + 2 ≤ l(C ′) + 3.

Using Lemma 2 instead of Lemma 1, we can, by similar arguments, deal withthe case in which G[T1] or G[T2] is isomorphic to K2.

Now Theorem 1 is a consequence of the following corollary of Theorem 3.

Corollary 1. Let G be a claw-free graph, and let x be an eligible vertex of G.Let G′ be the graph obtained from G by local completion at x. Then G is coveredby k cycles if and only if G′ is covered by k cycles.

Proof. Since the ‘‘only if’’ part is trivial, we have only to prove the ‘‘if’’ partof the corollary. Suppose G′ is covered by k cycles, say V (G′) = V (C ′1) ∪ · · · ∪V (C ′k) for cycles C ′1, . . . , C ′k in G′. By Theorem 3 for each C ′i there existsa cycle Ci in G with V (C ′i) ⊂ V (Ci) (1 ≤ i ≤ k). Then V (G) = V (C1) ∪· · · ∪ V (Ck).

Now we prove Theorem 2. Actually, we prove a stronger statement.

Theorem 4. Let G be a claw-free graph, and let x be an eligible vertex of G.LetG′ be the graph obtained fromG by local completion at x. Then for each set ofk disjoint cycles {D1, . . . , Dk} in G′ there exists a set of at most k disjoint cycles{C1, . . . , Cl} (l ≤ k) in G with ∪ki=1V (Di) ⊂ ∪li=1V (Ci).

Proof. Let S0 = ∪ki=1V (Di). Assume, to the contrary, that G[S] has no 2-factor with at most k components for any S ⊂ V (G) with S0 ⊂ S. Let B =E(G′)− E(G). Note {a, b} ⊂ NG(x) for each ab ∈ B. Let

F = {(S, F ):S0 ⊂ S ⊂ V (G) and F is a 2-factor of G′[S]}.Since (S0,∪ki=1E(Di)) ∈ F ,F /= ∅. LetF0 be the set of pairs (S, F ) ∈ F , chosenso that


(a) the number of components of F is as small as possible, and

(b) |F ∩B| is as small as possible, subject to (a).

Let (S, F ) ∈ F0. Suppose that F consists of l components (cycles) C1, . . . , Cl:F = E(C1) ∪ · · · ∪ E(Cl) (disjoint). Since (S0,∪ki=1E(Di)) ∈ F , l ≤ k. By theassumption F ∩B /= ∅.

If x /∈ S, choose i with E(Ci) ∩ B /= ∅, say e = uv ∈ E(Ci) ∩ B andv = u+(Ci). Let C ′i = xv

→Ciux and F ′ = (F − E(Ci)) ∪ E(C ′i). Then F ′ is a

2-factor of G′[S ∪ {x}] with l components and |F ′ ∩ B| = |F ∩ B| − 1. Thiscontradicts the choice of (S, F ) given in (b). Therefore, we have x ∈ S. We mayassume x ∈ V (C1).

We claim B ∩ (∪li=2E(Ci)) = ∅. Assume that B ∩ (∪li=2E(Ci)) /= ∅, sayf = u′v′ ∈ B ∩ E(Cj) (j ≥ 2). Then {u′, v′, x+(C1)} ⊂ NG(x) and, hence,u′x+(C1) ∈ E(G′). We may assume that j = 2 and v′ = u′+(C2). Let C ′ =xv′

→C2u

′x+(C1)→C1x and F ′ = (F − (E(C1) ∪ E(C2))) ∪ E(C ′). Then F ′ is a2-factor of G′[S] with l − 1 components. This contradicts the choice of (S, F ).

Since F ∩ B /= ∅, B ∩ E(C1) /= ∅. If there exists a cycle C ′1 in G withV (C ′1) = V (C1), then (F − E(C1)) ∪ E(C ′1) is a 2-factor of G[S] with l com-ponents. This contradicts the assumption. Since x ∈ V (C1), by Lemma 3, thereexist T0, T1 ⊂ V (G) such that T0 ∪ T1 = V (C1), T0 ∩ T1 = {x}, and G[T] isHamiltonian or isomorphic to K2 (i = 0, 1).

First, consider the case in which both G[T0] and G[T1] are Hamiltonian. LetC ′0 and C ′1 be cycles in G[T0 ∪ T1] with V (C ′0) ∪ V (C ′1) = T0 ∪ T1 = V (C1)and V (C ′0) ∩ V (C ′1) = {x}. Let ui = x+(C′i) and vi = x−(C′i) (i = 0, 1). Sincex is a locally connected vertex of G,G[NG(x)] has a path P with starting vertexin {u0, v0} and terminal vertex in {u1, v1}. Since G[S] has no 2-factors with lcomponents, u0u1, u0v1, v0u1, v0v1 ∈ B. By the choice of (S, F ) given in (b),|E(C1) ∩B| = 1.

Now choose (S, F ) ∈ F0, C′0, C

′1, and P so that

(c) P is as short as possible.

Then by Lemma 1, 2 ≤ l(P ) ≤ 3 and int(P ) ∩ V (C1) = ∅. We may assume thatthe starting vertex and the terminal vertex of P are v0 and u1, respectively.

Let a = v+(P )0 . Then a /∈ V (C1). Assume that a /∈ S. Since V (P ) ⊂ NG(x),

ax ∈ E(G) and, hence, au1 ∈ E(G′). Let C ′ = xu0→C′0v0au1

→C′1v1x and F ′ =

(F − E(C1)) ∪ E(C ′). Then F ′ is a 2-factor of G′[S ∪ {a}] with l components,andF ′∩B ⊂ {au1}. Since |B∩E(C1)| = 1, |F ′∩B| = |F∩B| = 1. Furthermore,C ′′0 = xu0

→C′0v0ax andC ′′1 = C ′1 are two cycles inGwithV (C ′′0 )∪V (C ′′1 ) = V (C ′)

and V (C ′′0 ) ∩ V (C ′′1 ) = {x}. Since a→Pu1 is shorter than P , this contradicts the

choice of (S, F ) given in (c). Therefore, we have a ∈ S. We may assume thata ∈ V (C2). Let a′ = a+(C2) and a′′ = a−(C2).

If a′x ∈ E(G), then {a′, u1} ⊂ NG(x) and, hence, a′u1 ∈ E(G′). Let

C ′ = xu0→C′0v0a

←C2a

′u1→C′1v1x


and F ′ = (F − (E(C1) ∪ E(C2))) ∪ E(C ′). Then F ′ is a 2-factor of G′[S] withl − 1 components. This contradicts the choice of (S, F ). Hence, we have a′x /∈E(G). By the same argument, we have a′′x /∈ E(G). Since a and {x, a′, a′′} donot form a claw in G, a′a′′ ∈ E(G). If l(C2) ≥ 4, let C ′ = xu0

→C′0v0au1

→C′1v1x

(note au1 ∈ E(G′)), C ′′ = a′→C2a

′′a′ and F ′ = (F − (E(C1) ∪ E(C2))) ∪E(C ′) ∪ E(C ′′). Then F ′ is a 2-factor of G′[S] with l components and F ′ ∩B ⊂ {au1}. Since |B ∩ E(C1)| = 1, |F ′ ∩ B| = |F ∩ B|. Furthermore, C ′′0 =xu0

→C′0v0ax and C ′′1 = C ′1 are two cycles in G with V (C ′′0 ) ∪ V (C ′′1 ) = V (C ′)

and V (C ′′0 ) ∩ V (C ′′1 ) = {x}. Since a→Pu1 is shorter than P , this contradicts the

choice of (S, F ) given in (c). Therefore, we have l(C2) = 3, which impliesC2 = aa′a′′a.

If a′ ∈ NG(v0), let C ′ = xu0→C′0v0a

′a′′au1→C′1v1x and F ′ = (F − (E(C1) ∪

E(C2))) ∪ E(C ′). Then F ′ is a 2-factor of G′[S] with l − 1 components. Thiscontradicts the choice of (S, F ). If a′ ∈ NG(u1), letC ′ = xu0

→C′0v0aa

′′a′u1→C′1v1x

and F ′ = F − ((E(C1) ∪ E(C2))) ∪ E(C ′). Then F ′ is a 2-factor of G′[S] withl − 1 components, which contradicts the assumption. Therefore, a′ /∈ NG(v0) ∩NG(u1). Similarly, a′′ /∈ NG(v0) ∪NG(u1).

Let b = u−(P )1 . Let b ∈ V (Ci), 2 ≤ i ≤ l, b′ = b+(Ci), and b′′ = b−(Ci). By

symmetry, we have {b′, b′′} ∩ (NG(x) ∪NG(u1) ∪NG(v0)) = ∅ and l(Ci) = 3.Suppose that l(P ) = 2. Then b = a and, hence, Ci = C2. Since a′ /∈ NG(v0) ∪

NG(u1) and v0u1 /∈ E(G), a and {a′, v0, u1} form a claw in G, a contradiction.Therefore, we have l(P ) = 3. Since u1a

′, u1a′′ /∈ E(G), Ci /= C2. We may assume

that b ∈ V (C3).By the choice of P given in (c), bv0, au1 /∈ E(G). Since v0a

′ /∈ E(G) and aand {a′, b, v0} do not form a claw, a′b ∈ E(G). Similarly, we have a′′b, ab′, ab′′ ∈E(G). Now let C ′ = aa′′a′bb′b′′a and F ′ = (F − (E(C2) ∪ E(C3)) ∪ E(C ′).Then F ′ is a 2-factor of G′[S] with l − 1 components. This contradicts the choiceof (S, F ) given in (a), and the theorem follows in this case.

By replacing Lemma 1 with Lemma 2, we can follow the same arguments toobtain a contradiction ifG[T1] orG[T2] is isomorphic toK2. Therefore, the theoremis proved.

3. CONCLUDING REMARKS

Let S be a set of vertices in a claw-free graph G. Then by Theorem 3, the min-imum number of cycles covering S in G is the same as the minimum num-ber of cycles covering S in cl(G). Furthermore, by Theorem 4, the minimumnumber of disjoint cycles covering S in G is the same as the minimum num-ber of disjoint cycles covering S in cl(G), if there exist such cycles. Therefore,these invariants (and, hence, the minimum number of cycles covering V (G)) arestable in the sense of [1]. Furthermore, the existence of a 2-factor is a stableproperty.


ACKNOWLEDGMENTS

This research was carried out while Z.R. and A.S. visited the Department of Math-ematical Sciences, The University of Memphis. These authors are grateful for thehospitality extended during their stay.

References

[1] S. Brandt, O. Favaron, and Z. Ryjacek, Closure and stable hamiltonian prop-erties in claw-free graphs, preprint.

[2] G. Chartrand and L. Lesniak, Graphs & digraphs, 2nd Ed., Wadsworth &Brooks/Cole, Monterey, CA, 1986.

[3] Z. Ryjacek, On a closure concept in claw-free graphs, J Combin Theory SerB 70 (1997), 217–224.

Documents

Closure, 2-factors, and cycle coverings in claw-free graphs