A 2-factor in which each cycle has long length in claw-free graphs

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1 A -factor in which each cycle has long length in claw-free graphs Roman Čada Shuya Chiba Kiyoshi Yoshimoto 3 Department of Mathematics University of West Bohemia and Institute of Theoretical Computer Science (ITI) Charles University Univerzitní Plzeň Czech Republic. Department of Mathematical Information Science, Tokyo University of Science, -3 Kagurazaka, Shinjuku-ku, Tokyo 6-860, Japan. 3 Department of Mathematics, College of Science and Technology, Nihon University, Tokyo , Japan. Abstract A graph G is said to be claw-free if G has no induced subgraph isomorphic to K,3. In this article, we prove that every claw-free graph G with minimum degree δ(g) 4 has a -factor in which each cycle contains at least δ(g) vertices and every -connected claw-free graph G with δ(g) 3 has a -factor in which each cycle contains at least δ(g) vertices. For the case where G is -connected, the lower bound on the length of a cycle is best possible. Keywords: -factor, Claw-free graph, Minimum degree AMS Subject Classification: 05C70 Introduction In this paper, we consider finite graphs. A simple graph means an undirected graph without loops or multiple edges. A multigraph may contain multiple edges but no loops. Let G be a graph. For a vertex v of G, the degree of v in G is the number of edges incident with v. Let V (G), E(G) and δ(g) be the vertex set, the edge set and the minimum degree of G, respectively. We refer to the number of vertices of G as the order of G and denote it by G. A graph G is said to be claw-free if G has no induced subgraph isomorphic to K,3 (here K,3 denotes the An extended abstract published in Eurocomb0, Electronic Notes in Discrete Mathematics Vol. 38 (0) address: cadar@kma.zcu.cz This work was supported by No. M of the Czech Ministry of Education address: chiba@rs.kagu.tus.ac.jp This work was supported by Grant-in-Aid for Young Scientists (B) ( ) address: yosimoto@math.cst.nihon-u.ac.jp This work was supported by KAKENHI (5405)

2 complete bipartite graph with partite sets of cardinalities and 3, respectively). A -factor of G is a spanning subgraph of G in which every component is a cycle. It is a well-known conjecture that every 4-connected claw-free graph is Hamiltonian due to Matthews and Sumner []. Since a Hamilton cycle is a connected -factor, there are many results on -factors of claw-free graphs. For instance, results of both Choudum and Paulraj [4] and Egawa and Ota [5] imply that a moderate minimum degree condition already guarantees that a claw-free graph has a -factor. Theorem A ([4, 5]) Every claw-free graph G with δ(g) 4 has a -factor. For -connected claw-free graphs, the third author [5] improved the condition on minimum degree as follows. Theorem B ([5]) Every -connected claw-free graph G with δ(g) 3 has a -factor. In particular, in [], it is shown that Matthews and Sumner s conjecture and the statement that every 4-connected claw-free graph of order n has a -factor with at most f(n) components (here f(n) is a function of n with the property that lim n f(n)/n = 0) are equivalent. This indicates that it is important to study about the number of components of -factors in claw-free graphs to solve Matthews and Sumner s conjecture. Concerning the upper bound on the number of components, Broersma, Paulusma and the third author [3] and Jackson and the third author [8] proved the following, respectively (other related results can be found in [6, 9, 0,, 5]). Theorem C ([3]) Every claw-free graph G with δ(g) 4 has a -factor with at most max { G 3 δ(g), } components. Theorem D ([8]) Every -connected claw-free graph G with δ(g) 4 has a -factor with at most G + 4 components. In this paper, we approach this problem in terms of the lengths of cycles in a -factor of a clawfree graph. More precisely, we consider the functions f : {d Z d 4} {m Z m 3} and f : {d Z d 3} {m Z m 3} defined as follows: for an integer d with d 4, let f (d) := max{m every connected claw-free graph with minimum degree d has a -factor in which each cycle contains at least m vertices}, and for an integer d 3, let f (d) := max{m every -connected claw-free graph with minimum degree d has a -factor in which each cycle contains at least m vertices}. Concerning the determinations of f (d) and f (d), we will show the following two results in Sections 3 and 4. Theorem Every claw-free graph G with δ(g) 4 has a -factor in which each cycle contains at least δ(g) vertices.

3 G u H K d+ K d+ K d+ v u H K d+ K d+ K d+ v u 3 H 3 K d+ K d+ K d+ v 3 Figure : The graph G Theorem Every -connected claw-free graph G with δ(g) 3 has a -factor in which each cycle contains at least δ(g) vertices. As a corollary of Theorem, we can get the following which is slightly stronger than Theorems C and D for -connected claw-free graphs. Corollary 3 Every -connected claw-free graph G with δ(g) 3 has a -factor with at most G δ(g) components. Moreover, in Theorem, the lower bound on the length of a cycle is best possible because we can construct a -connected claw-free graph G such that for each -factor of G, the minimum length of cycles in it is at most δ(g) as follows: let d 3 be an integer, and let H, H and H 3 be graphs with the connectivity one such that each block of H i ( i 3) is a complete graph of order d +. For each i 3, let u i and v i be vertices contained in distinct end blocks of H i, respectively, and u i and v i are not cut vertices of H i. Let G be a graph obtained from H H H 3 by joining u i and u i+ for i 3 and joining v i and v i+ for i 3, where let u 4 := u and v 4 := v, see Figure (here K m denotes the complete graph of order m). Then G is a -connected claw-free graph with δ(g ) = d, and for each -factor of G, the minimum length of cycles in it is at most d. So this together with Theorem implies that f (d) = d holds. On the other hand, we can obtain slightly stronger statement than Theorem if we consider only line graphs of trees as follows, which will be shown in Section 5 (here a clique-factor is a spanning subgraph in which every component is a clique). 3

4 Theorem 4 The line graph G of a tree has a clique-factor in which each clique contains at least vertices. δ(g)+ In view of Theorem 4, the following statement might hold. Conjecture 5 Every claw-free graph G with δ(g) 4 has a -factor in which each cycle contains at least δ(g)+ vertices. In Conjecture 5, the lower bound on the length of a cycle is best possible if it is true. So this together with Theorem implies that d f (d) d+ holds. In section 5, we give a graph showing the sharpness of the lower bound in Conjecture 5. Now we have that f (d) is completely determined and f (d) is almost determined, in particular, f (d) = d and d f (d) d+ hold even if we consider only graphs with sufficiently large orders compared to d (see Figure and see the second paragraph and Figure 4 in Section 5). Therefore it would be natural to consider the case where the connectivity is at least 3 as the next step. If Matthews and Sumner s conjecture is true, we can always take Hamilton cycles in 4-connected claw-free graphs. So what will happen when the connectivity is just 3? For instance, concerning the number of components of -factors in 3-connected claw-free graphs, Kužel, Ozeki and the third author [0] proved that every 3-connected claw-free graph G has a -factor with at most max { G } δ(g)+, components. Furthermore, most recently, Ozeki, Ryjáček and the third author [] proved that every 3-connected claw-free graph G has a -factor with at most 4 G 5(δ(G)+) + 5 components. In view of this result, one might expect that the coefficient of d in the lower bound on the length of a cycle would be greater than for 3-connected claw-free graphs. Problem Determine f 3 (d) := max{m every sufficiently large 3-connected claw-free graph with minimum degree d has a -factor in which each cycle contains at least m vertices}. In particular, is there an constant c > such that f 3 (d) cd holds? It is of course important to consider about the case of 4-connected claw-free graphs. As mentioned above, if there exists a function f(n) of n such that lim n f(n)/n = 0 and every 4-connected claw-free graph of order n has a -factor with at most f(n) components, then every 4-connected claw-free graph is Hamiltonian, i.e., to solve Matthews and Sumner s conjecture, it suffices to show the existence of a -factor with small number of components (not necessarily component). Therefore it could be also another possible approach to study about the lengths of cycles in -factors of 4-connected claw-free graphs. In addition, for a result similar to Theorems, and 4 concerning path-factors (here a path-factor is a spanning subgraph in which every component is a path), Ando, Egawa, Kaneko, Kawarabayashi and Matsuda [] proved that every claw-free graph G has a path-factor in which 4

5 each path contains at least δ(g) + vertices, and they conjectured that every -connected claw-free graph G has a path-factor in which each path contains at least 3δ(G) + 3 vertices. In the rest of this section, we prepare notation which we use in subsequent sections. Let G be a graph. We denote the number of edges of G by e(g) and the line graph of G by L(G). For X V (G), we let G[X] denote the subgraph induced by X in G, and let G \ X := G[V (G) \ X]. If H is a subgraph of G, then let G \ H := G \ V (H). A subset X of V (G) is called a independent set of G if G[X] is edgeless. Let H and H be subgraphs of G or subsets of V (G), respectively. If H and H have no any common vertex in G, we define E G (H, H ) to be the set of edges of G between H and H, and let e G (H, H ) := E G (H, H ). For a vertex v of G, we denote by N G (v) and d G (v) the neighborhood and the degree of v in G, respectively. Let o(g) the set of vertices with odd degree in G. For a positive integer l, we define V l (G) := {v V (G) d G (v) = l}, and let V l (G) := m l V m(g) and V l (G) := m l V m(g). Let e = uv E(G). We denote by V (e) the set of end vertices of e, i.e., V (e) = {u, v}. The edge degree of e in G is defined by the number of edges incident with e, and is denoted by ξ G (e), i.e., ξ G (e) = {f E(G) f e, V (f) V (e) }. Note that if G is a simple graph, then ξ G (e) = e G (V (e), G \ e) = d G (u) + d G (v). Let ξ(g) be the minimum edge degree of G. Outline of the proofs of Theorems, and 4 In this section, we give an outline of the proofs of Theorems, and 4. To prove Theorems and, we use Ryjáček closure which is defined as follows. Let G be a claw-free graph. For each vertex v of G, G[N G (v)] has at most two components; otherwise G contains a K,3 as an induced subgraph. If G[N G (v)] has two components, both of them must be cliques. In the case that G[N G (v)] is connected, we add edges joining all pairs of nonadjacent vertices in N G (v). The closure cl(g) of G is a graph obtained by recursively repeating this operation, as long as this is possible. In [3], it is shown that cl(g) is uniquely defined and cl(g) is the line graph of some triangle-free simple graph. On the other hand, in [4, Theoerm 4], Ryjáček, Saito and Schelp proved that for any vertex-disjoint cycles D,..., D q in cl(g), G has vertex-disjoint cycles C,..., C p with p q such that q i= V (D i) p i= V (C i). By modifying the proof, we can easily improve this result as follows (see [0, Lemma 3]): for any vertex-disjoint cycles D,..., D q in cl(g), G has vertex-disjoint cycles C,..., C p with p q such that for each j with j q, there exists i with i p such that V (D j ) V (C i ). These imply that for a claw-free graph G, G has a -factor in which each cycle contains at least m vertices if and only if cl(g) has a -factor in which each cycle contains at least m vertices. By this fact, to prove Theorems and, it suffices to show that the following hold (here a graph H is called essentially k-edge-connected if e(h) k + and H \ F has at most one component which contains an edge 5

6 for every F E(H) with F < k. It is easy to see that for a graph H such that L(H) is not complete, H is essentially k-edge-connected if and only if L(H) is k-connected). Theorem 6 Let G be the line graph of some triangle-free simple graph and δ(g) 4. Then G has a -factor in which each cycle contains at least δ(g) vertices. Theorem 7 Let G be the line graph of some essentially -edge-connected triangle-free simple graph and δ(g) 3. Then G has a -factor in which each cycle contains at least δ(g) vertices. Moreover, to prove Theorems 7 and 4, we consider about some particular set of subgraphs. Now let H be a multigraph. We call a subgraph S of H a star if S consists of a vertex (called a center) and edges incident with the center. So a star in this paper is not necessary a tree. A closed trail T in H is called a dominating closed trail if H \ T is edgeless. It is known that for a connected multigraph H with e(h) 3, H has a dominating closed trail if and only if L(H) is Hamiltonian (see [7]). A set D is called a cover set of E(H) if (i) D is a set of edge-disjoint connected subgraphs in H such that D D E(D) = E(H), and (ii) for each D D, D is a star or D has a dominating closed trail. Let D be a cover set of E(H). If D is a star for all D D, then D is called a star cover set of E(H). Note that for each D D, if D is a star, then L(D) is a clique. Therefore, to prove Theorems 7 and 4, it suffices to show that the following hold, respectively. Theorem 8 Let G be an essentially -edge-connected triangle-free simple graph with ξ(g) 3. Then G has a cover set D of E(G) such that e(d) ξ(g) for all D D. Theorem 9 A tree G with ξ(g) has a star cover set S of E(G) such that e(s) ξ(g)+ for all S S. The following lemma is used to find dominating closed trails in the proof of Theorem 8. Lemma ([5]) If G is an essentially -edge-connected graph with ξ(g) 3, then there exists a set T of mutually vertex-disjoint closed trails in G such that V 3 (G \ V (G)) T T V (T ). 3 Proof of Theorem 8 If G is a star, then the assertion clearly holds. Thus we may assume that G is not a star. Then since G is essentially -edge-connected, we have δ(g \ V (G)). By Lemma, there exists a set T of mutually vertex-disjoint closed trails in G such that V 3 (G \ V (G)) T T V (T ). 6

7 G T G v T T v T T v T Figure : The graph G Since δ(g \ V (G)) and V 3 (G \ V (G)) T T V (T ), v V (G \ V (G)) for all v V 3 (G) \ T T V (T ). This implies that for each v V 3(G) \ T T V (T ), there exist exactly two vertices u, u N G (v) with u u such that u i V (G) for i =, and N G (v) \ {u, u } V (G); (3.) in particular, N G (v) \ {u, u } = since v V 3 (G), that is, N G (v) V (G), and hence N G (v) \ {u, u } d G (v) ξ(g). Let G be a graph obtained from G by contracting an induced subgraph G[V (T )] of G to a vertex v T for each T T (note that G may be a multigraph, see Figure ). Let X := {v T T T }. Note that X = T since T is a set of mutually vertex-disjoint closed trails in G. By the definition of G, d G (v) = d G (v) for all v V (G ) \ X. (3.) By (3.) and since V (G) ( T T V (T )) =, V (G) = V (G ) \ X. (3.3) By the definition of G and (3.) and since ξ(g) 3, we also have that V (G ) \ X is an independent set of G. To show the existence of a cover set D of E(G), we find a mapping ϕ : E(G ) V (G ) so that (M) ϕ(e) = u or ϕ(e) = v for all e = uv E(G ), (M) ϕ (v) = 0 for all v V (G ) \ X, (M3) ϕ (v) ξ(g) for all v V 3 (G ) \ X, (M4) ϕ (v T ) + e(g[v (T )]) ξ(g) for all v T X. 7

8 Suppose that there exists such a mapping ϕ. Let S := {S v v V 3 (G ) X} be a set of stars such that for each S v S, S v is a star consisting of a vertex v (as the center) and the edges in ϕ (v). Then by the conditions (M) and (M), S is a star cover set of E(G ). Furthermore by the conditions (M3) and (M4), e(s v ) = ϕ (v) ξ(g) for each v V 3 (G ) \ X and e(s vt ) = ϕ (v T ) ξ(g) e(g[v (T )]) for each v T X. Hence by the definition of G, and by considering a subset of E(G) corresponding to E(S v ) for each v V 3 (G ) X and replacing v T with G[V (T )] for each v T X, we can find a cover set D of E(G) such that e(d) ξ(g) for all D D. Thus it suffices to show the existence of such a mapping ϕ. Let G := G \ (V (G ) \ X). To show the existence of a mapping ϕ satisfying the conditions (M) (M4), we define a mapping ϕ : E(G ) V (G ) as follows. Since the number of vertices of odd degree is even in G, there exists a collection of paths P,..., P l in G such that each vertex in o(g ) appears in the set of end vertices of them exactly one (note that l = o(g ) /). By considering the symmetric difference of them, we may assume that P,..., P l are pairwise edge-disjoint. For each i l, let x i xi xi 3... xi P i xi P i := P i, and let e i j := xi j xi j+ for each j P i, and we define ϕ (e i j ) := xi j+ for each j P i. Let G := G \ l i= E(P i). By the definitions of P,..., P l, we have o(g ) =. Hence the edges of G can be covered by pairwise edge-disjoint cycles. For each cycle, written by y y y 3... y m y m (= y ), we define ϕ (e j ) := y j+, where e j := y j y j+ for each j m. Then by the definition of ϕ, for each v V (G ), ( (dg ϕ (v) (d G (v) )/ = (v) (N G (v) V (G )) \ X ) ) /. (3.4) Now we define a mapping ϕ : E(G ) V (G ) as follows: for each e = uv E(G ), we let { u if v V (G ) \ X ϕ(e) := ϕ (e) otherwise. Since V (G ) \ X is an independent set of G, ϕ is well defined. By the definitions of ϕ and ϕ, we can easily see that ϕ satisfies the conditions (M) and (M). So we show that ϕ satisfies the conditions (M3) and (M4). Since (V (G ) \ X) V (G ) = by the definition of G, it follows from the definition of ϕ that ϕ (v) ϕ (v) + (N G (v) V (G )) \ X for all v X. (3.5) We first show that ϕ satisfies the condition (M3). Let v V 3 (G ) \ X. Then by (3.) and the definitions of G and X, v V 3 (G) \ T T V (T ). Hence by (3.), there exist exactly two vertices u, u N G (v) with u u such that u i V (G) for i =,, N G (v)\{u, u } V (G) and N G (v) \ {u, u } ξ(g). Therefore by (3.), (3.3) and the definition of G, there exist two distinct edges vu, vu E(G ) such that u i V (G ) X for i =,, N G (v) \ {u, u } 8

9 V (G ) \ X and N G (v) \ {u, u } ξ(g) (possibly u = u, i.e., vu and vu may be parallel edges in G ). If u i V (G ) \ X for some i = or, then by the definition of ϕ, we have ϕ (v) (N G (v) V (G )) \ X + (N G (v) V (G )) \ X N G (v) \ {u, u } + (ξ(g) )+ = ξ(g). Thus we may assume that u i V 3(G ) X for i =,. Note that by the definition of G, u i V (G ) for i =,. Since N G (v)\{u, u } V (G )\X, we have v V (G ) and vu i E(G ) for i =,. Hence by the definition of ϕ, ϕ (vu i ) = v and ϕ (vu 3 i ) = u 3 i for some i = or. Then by the definition of ϕ and since {v, u, u } V 3(G ) X, ϕ(vu i ) = v for some i = or. Therefore ϕ (v) (N G (v) V (G ))\X + {vu i } N G (v)\{u, u } + (ξ(g) ) + = ξ(g) holds for some i = or. Thus ϕ satisfies the condition (M3). We next show that ϕ satisfies the condition (M4). Let T T. Since G is triangle-free, all cycle which is contained in T has order at least four. Hence there exist two independent edges e and e in T. For each i =,, let ɛ i := {e E(G[V (T )]) e ei, V (e) V (e i ) }. Since G is triangle-free, {e E(G[V (T )]) V (e) V (ei ) for i =, }. Hence by the definition of ɛ i, e(g[v (T )]) (ɛ + ɛ ) + {e, e } = ɛ + ɛ. (3.6) We also have e G (V (e i ), G \ T ) ξ(g) ɛ i for each i =,. Hence since d G (v T ) = e G (T, G \ T ) by the definition of G and since E G (V (e ), G \ T ) E G (V (e ), G \ T ) =, we obtain d G (v T ) = e G (T, G \ T ) e G (V (e ), G \ T ) + e G (V (e ), G \ T ) ξ(g) (ɛ + ɛ ). (3.7) Then by (3.4), (3.5), (3.6) and (3.7), ϕ (v T ) ϕ (v T ) + (N G (v T ) V (G )) \ X ( d G (v T ) (N G (v T ) V (G )) \ X ) / + (N G (v T ) V (G )) \ X ( ξ(g) (ɛ + ɛ ) ) / = ξ(g) ( (ɛ + ɛ ) + ) / ξ(g) ( e(g[v (T )]) + ) /. Since e(g[v (T )]) 4, we get ϕ (v T ) +e(g[v (T )]) ξ(g) ( e(g[v (T )])+ ) /+e(g[v (T )]) ξ(g) + ( e(g[v (T )]) ) / > ξ(g). Thus ϕ satisfies the condition (M4). This completes the proof of Theorem 8. 9

10 T B 0 = {B 0 } C 0 B C B C C m B m Figure 3: Subsets B 0,..., B m and subsets C 0,..., C m 4 Proof of Theorem 6 To prove Theorem 6, we use Theorem and the following proposition. Proposition 0 Let G be a line graph with δ(g) 4. Then there exists a set H of mutually vertex-disjoint subgraphs in G such that H H V (H) = V (G), and H is a -connected claw-free graph and δ(h) δ(g) for all H H. Proof of Proposition 0. It suffices to consider the case where G is connected. By the assumption, we may assume that the connectivity of G is one. Let B be the set of blocks of G and C be the set of cut vertices of G. We consider a block cut tree T of G, i.e., V (T ) = B C and E(T ) = {Bv B B, v V (B) C}. Since G is a line graph, for each v V (G), G[N G (v)] has a clique-factor with at most components. (4.) Hence d T (v) = for all v C. Let B 0 B be a root of T. We consider T as an oriented tree from B 0 to leaves, and denote it T. For each v C, we let B v + (resp. B v ) denote a successor (resp. a predecessor) of v along T (note that B v + and B v are uniquely defined because d T (v) =, and note also that B v +, B v B). For each B B \ {B 0 }, we let v B denote a predecessor of B along T (note that v B is uniquely defined). We define subsets B 0,..., B m of B and subsets C 0,..., C m of C inductively by the following procedure. First let B 0 := {B 0 }, and let C 0 := {v C B v = B 0 }. Now let i, and assume that we have defined B 0,..., B i and C 0,..., C i. If C i, then let B i := {B v + v C i }, and let C i := {v C B v B i }; if C i =, we let m := i and terminate the procedure (see Figure 3). Then it follows from 0

11 the definitions of B 0,..., B m and C 0,..., C m that B is disjoint union of B 0,..., B m and C is disjoint union of C 0,..., C m. We define a mapping ϕ : C B inductively as follows. First for each v C 0, let { B ϕ(v) := v if N G (v) V (B v ) δ(g). B v + otherwise Next for each v C i with i m, let ϕ(v) := B v if ϕ(v ) = B Bv v and N G(v) V (B v ) δ(g) B v if ϕ(v ) B Bv v and N G(v) V (B v ) δ(g)+ B v B v + if ϕ(v ) B Bv v, N G(v) V (B v ) = δ(g) and vv Bv otherwise / E(G). Let H := {G[(V (B) \ C) ϕ (B)] B B}. We show that H is a desired set. By the definition of H, it is easy to check that H is a set of mutually vertex-disjoint subgraphs in G such that H H V (H) = V (G) and H is claw-free for all H H. Let H H, and let H = G[(V (B) \ C) ϕ (B)] for some B B i with 0 i m. We first show that δ(h) δ(g) Case. B B 0, i.e., B = B 0.. Let u V (H). Assume for the moment that u ϕ (B 0 ). Then by the definition of ϕ, N G (u) V (B 0 ) δ(g). Since G[N G(u) V (B 0 )] is a complete graph by (4.), N G (v) V (B 0 ) δ(g) for all v N G (u) V (B 0 ). Hence by the definition of ϕ, ϕ(v) = B 0 for all v N G (u) V (B 0 ) C. Since H = G[(V (B 0 ) \ C) ϕ (B 0 )], this implies that (N G (u) V (B 0 )) {u} V (H), and hence d H (u) = N G (u) V (B 0 ) δ(g). Thus we may assume that u V (B 0) \ C. Then d B0 (u) = d G (u) δ(g). Hence by (4.), there exists a complete subgraph F of G[N B0 (u) {u}] such that u V (F ) and F δ(g)+. Then N G (v) V (B 0 ) F δ(g) for all v V (F ). Hence by the definition of ϕ, ϕ(v) = B 0 for all v V (F ) C. Since H = G[(V (B 0 )\C) ϕ (B 0 )], this implies that V (F ) V (H), and hence d H (u) F δ(g). Case. B B i with i m and u ϕ (B). Assume for the moment that u = v B, or u v B and ϕ(v B ) = B. Then by the definition of ϕ, N G (u) V (B) δ(g). Since G[N G(u) V (B)] is a complete graph by (4.), N G (v) V (B) δ(g) for all v N G (u) V (B). Since ϕ(v B ) = B, it follows from the definition of ϕ that ϕ(v) = B for all v N G (u) V (B) C. Since H = G[(V (B) \ C) ϕ (B)], this implies that (N G (u) V (B)) {u} V (H), and hence d H (u) = N G (u) V (B) δ(g). Thus we may

12 assume that u v B and ϕ(v B ) B. Then by the definition of ϕ, (i) N G (u) V (B) δ(g)+, or (ii) N G (u) V (B) = δ(g) and uv B / E(G). Since G[N G (u) V (B)] is a complete graph by (4.), for each v N G (u) V (B), N G (v) V (B) δ(g)+, or N G (v) V (B) = δ(g) and vv B / E(G). Hence by the definition of ϕ, ϕ(v) = B for all v N G (u) V (B) (C \ {v B }). Since H = G[(V (B) \ C) ϕ (B)], this implies that ( (N G (u) V (B)) {u} ) \ {v B } V (H), and hence d H (u) (N G (u) V (B)) \ {v B } δ(g). Case 3. B B i with i m and u / ϕ (B). Then u V (B) \ C, and hence d B (u) = d G (u) δ(g). Therefore by (4.), there exists a complete subgraph F of B[N B (u) {u}] such that (i) u V (F ) and F δ(g)+3, or (ii) u V (F ), F = δ(g)+ and v B / V (F ). Then since F is a complete graph, for each v V (F ), N G (v) V (B) δ(g)+, or N G (v) V (B) = δ(g) and vv B / E(G). Hence by the definition of ϕ, ϕ(v) = B for all v V (F ) (C \ {v B }). Since H = G[(V (B) \ C) ϕ (B)], this implies that V (F ) \ {v B } V (H), and hence d H (u) V (F ) \ {u, v B } δ(g) By Cases, and 3, d H (u) δ(g) We next show that H is -connected. Suppose that H is not -connected. Since δ(h) δ(g).. Since u is an arbitrary vertex in H, δ(h) δ(g), H 3. Hence there exist two distinct subgraphs F and F of H such that V (F j ) \ V (F 3 j ) for j =,, and V (F ) V (F ), V (H) = V (F ) V (F ) and E G (F \ F, F \ F ) =. Since B H 3, we also have that B is -connected. Hence V (B \ H) = (V (B) C) \ ϕ (B). Let w V (F ) V (F ) if V (F ) V (F ) ; otherwise, let w be an arbitrary vertex in B \ H. Let U := {u V (B \ H) N G (u) (V (F ) \ {w}) }. Since B is -connected, U. If there exists u U such that N G (u) (V (F ) \ {w}), then G[V (F F B u +) {u}] contains K,3 as an induced subgraph because E G (B \ {u}, B u +) = since u is a cut vertex of G, a contradiction. Thus N G (u) (V (F )\{w}) = for all u U. Hence since B is -connected, there exist u U and v V (B)\(V (H) U) such that uv E(G). Then G[V (F B u +) {u, v}] contains K,3 as an induced subgraph because E G (B \ {u}, B u +) = since u is a cut vertex of G, a contradiction again. Thus H is -connected. This completes the proof of Proposition 0. Proof of Theorem 6. Let G be as in Theorem 6. By Theorem A, we may assume that δ(g) 6. By Proposition 0, there exists a set H of mutually vertex-disjoint subgraphs in G such that H H V (H) = V (G), and H is a -connected claw-free graph and δ(h) δ(g) 3 for all H H. Then by Theorem, H has a -factor in which each cycle contains at least δ(h) δ(g) vertices for all H H. Since V (G) = H H V (H) (disjoint union), this implies that G has a -factor in which each cycle contains at least δ(g) vertices..

13 G K d K (d+)/ K (d+)/ K d K d + + Y + + K (d+)/ K (d+)/ y + + x X Y X K d K (d+)/ K (d+)/ K d K d K d K d Figure 4: The graph G 5 Proof of Theorem 9 As is mentioned in Section, in this section, we give a graph showing the sharpness of the lower bound on the length of a cycle in Conjecture 5, and we prove Theorem 9. The graph G illustrated in Figure 4 (here, in Figure 4, + means the join of vertices) is a claw-free graph with minimum degree d ( 4), and for each -factor of G, the minimum length of cycles in it is at most d+, and G has a -factor such that the minimum length of cycles in it is just d+. To see this, let C be a set of mutually vertex-disjoint cycles in G such that C C C is a -factor of G. If x X and y Y are contained in same cycle in C, then by the definition of G, the maximum size of the cycle containing both x and y is d+. Hence by the symmetry, we may assume that for all x X X and all y Y Y, the cycle containing x and the cycle containing y in C are distinct. Then by the definition of G, there exists a subset C of C such that C C V (C) = X or C C V (C) = X. Since X = X = d+, the minimum length of cycles in C is at most d+. Moreover, by these arguments, we can take a set C so that the minimum length of cycles in C is just d+. (Note that by adding X, Y, X and Y iteratively, we can construct a sufficiently large order graph.) To prove Theorem 9, we prove slightly stronger statement than Theorem 9 as follows (note that for each tree G with ξ(g), there exist vertices u V ξ(g) + (G) and u N G (u ) such that N G (u ) \ {u } V (G), and hence Theorem implies Theorem 9). Theorem Let G be a tree with ξ(g), and let u V l+ (G) and u N G (u ) such 3

14 T 0 G G i T 0 X u T 0 u u x x x m x i T T T m T i Figure 5: The tree G i that N G (u ) \ {u } V (G), where let l := ξ(g). Then the following hold. (I) If u V l (G), then G has a star cover set S of E(G) such that e(s) ξ(g)+ for all S S, and there exists S u S such that u u E(S u ) and u is the center of S u. (II) If u V l+ (G), then G has two star cover sets S and S of E(G) such that for each i =,, the following hold: (i) e(s) ξ(g)+ for all S Si. (ii) there exists S i S i such that u u E(S i ) and u i is the center of S i. Proof of Theorem. We prove Theorem by induction on e(g). If N G (u ) = V (G), then S := {G} is a star cover set of E(G) such that S satisfies the statement (I) in Theorem. Thus we may assume that N G (u ) V (G). Write N G (u ) \ {u } = {x, x,..., x m }. Let X be a subset of N G (u ) \ {u } such that N G (u ) \ ({u } X) = ξ(g) (note that d G (u ) ξ(g) + since d G (u ) l + and N G (u ) \ {u } V (G)). Let T 0, T,..., T m be components of G \ {u } such that T 0 contains a vertex u, and T i contains a vertex x i for each i m, and let T 0 := T 0 \ X. We construct m trees G, G,..., G m as follows. For each i m, let V (G i ) := V (T 0 ) V (T i ) and E(G i ) := E(T 0 ) E(T i ) {u x i } (see Figure 5). Then by the definitions of T 0, T,..., T m and G,..., G m, for each i m, E(G i ) < E(G) and d Gi (v) = d G (v) for all v V (G i ) \ {u }. (5.) 4

15 Since e(t0 ) = ξ(g), we also have ξ(g i ) = ξ(g) for each i m. (5.) We divide the proof into three cases according as u V l (G), or u V l+ (G) and N G (u ) V l+ (G), or u V l+ (G) and N G (u ) V l+ (G). Case : u V l (G). Then N G (u ) V l+ (G) (otherwise, ξ G (u x) < ξ(g) for x N G (u ) V l (G), a contradiction). Fix i with i m. Then by (5.) and (5.) and since e(t 0 ) = ξ(g), {u, x i } V l+ (G i ) = V ξ(gi ) + (G i). Therefore by the induction hypothesis and (5.), G i has a star cover set S i of E(G i) such that ξ(gi ) + ξ(g) + e(s) = for all S S i, (5.3) and there exists S xi S i such that u x i E(S xi ) and x i is the center of S xi. Note that T 0 S i because S x i S i and e(t 0 ) = ξ(g i) by (5.). Since i is an arbitrary integer with i m, there exists such a star cover set S i of E(G i) for all i with i m. For each i m, let Sx i be a star consisting of a vertex x i (as the center) and the edges in (E(S xi ) \ {u x i }) {u x i }, and let Su be a star consisting of a vertex u (as the center) and the edges in E(T 0 ) {u u }. Then by the definitions of Sx i and Su, e(sx i ) = e(s xi ) for each i m and e(su ) e(t0 ) = ξ(g). Hence by (5.3) and the definitions of S x i and Su, S := m ( i= S i \ {S xi, T0 }) {Su, Sx, Sx,..., Sx m } is a desired star cover set of E(G) (note that Su is a star in S such that u u E(Su ) and u is the center of Su ). Thus the statement (I) in Theorem holds. Case : u V l+ (G) and N G (u ) V l+ (G). Then we may assume that there exists an integer s with 0 s m such that {x,..., x s } V l+ (G) and {x s+,..., x m } V l (G). Since N G (u ) V l (G), we have ξ(g) + m (otherwise, ξ G (u x i ) < ξ(g) for s + i m, a contradiction). Fix i with i m. Then by (5.) and (5.), x i V l+ (G i ) = V ξ(gi ) + (G i) if i s; otherwise, x i V l (G i ) = V ξ(gi ) (Gi ). Therefore by the induction hypothesis, G i has a star cover set S i of E(G i) such that ξ(gi ) + ξ(g) + e(s) = for all S S i, (5.5) and there exists S i u S i such that u x i E(S i u ) and u is the center of S i u. Since i is an arbitrary integer with i m, there exists such a star cover set S i of E(G i) for all i 5 (5.4)

16 with i m. For each i m, let Su i := {S S i u is the center of S}. Then S Su i E(S) = E(T0 ) {u x i } for each i m because Su i S i. Let Su be a star consisting of a vertex u (as the center) and the edges in E(T 0 ) {u u }, Su be a star consisting of a vertex u (as the center) and the edges in E(T 0 ), Su be a star consisting of a vertex u (as the center) and the edges in {u x i i m}, and S u be a star consisting of a vertex u (as the center) and the edges in {u x i i m} {u u }. Then by the definitions of S u, S u, S u and Su, e(su ) > e(su ) e(t0 ) = ξ(g) and e(s u ) = e(su ) + = m +. Hence by (5.4), (5.5) and the definitions of Su, Su, Su E(T0 ) {u x i } for each i m, we have that S := m S := m i= and S u and since S S i u E(S) = i= ( S i \ S i u ) {S u, S u } and ( S i \ S i u ) {S u, S u } are desired star cover sets of E(G) (note that S u is a star in S such that u u E(Su ) and u is the center of Su, and Su is a star in S such that u u E(Su ) and u is the center of Su ). Thus the statement (II) in Theorem holds. Case 3: u V l+ (G) and N G (u ) V l+ (G). Fix i with i m. Then by (5.), (5.) and the assumption of Case 3, {u, x i } V l+ (G i ) = V ξ(gi ) + (G i). Therefore by the induction hypothesis and (5.), G i has two star cover sets S i and S i of E(G i) such that ξ(gi ) + ξ(g) + e(s) = for all S S i S i, (5.6) and there exists S i u S i such that u x i E(S i u ) and u is the center of S i u, and there exists S xi S i such that u x i E(S xi ) and x i is the center of S xi. Since i is an arbitrary integer with i m, there exist such star cover sets S i and S i of E(G i) for all i with i m. We first find a star cover set S of E(G). Note that T0 S i for each i m because e(t0 ) = ξ(g) and S x i S i. For each i m, let S x i be a star consisting of a vertex x i (as the center) and the edges in (E(S xi ) \ {u x i }) {u x i }, and let Su be a star consisting of a vertex u (as the center) and the edges in E(T 0 ) {u u }. Then by the definitions of Sx i and Su, e(sx i ) = e(s xi ) for each i m and e(su ) e(t0 ) = ξ(g). Hence by (5.6) and the definitions of Sx i and Su, S := m ( i= S i \ {S xi, T0 }) {Su, Sx, Sx,..., Sx m } is a desired star cover set of E(G) (note that Su is a star in S such that u u E(Su ) and u is the center of S u ). We next find a star cover set S of E(G). For each i m, let S i u := {S S i u is the center of S}. Then S S i u E(S) = E(T 0 ) {u x i } for each i m because S i u S i. Let Su be a star consisting of a vertex u (as the center) and the edges in E(T 0 ) and Su be a star consisting of a vertex u (as the center) and the edges in {u x i i m} {u u }. Then by the definitions of Su and Su and the assumption of Case 3, e(su ) e(t0 ) = ξ(g) and e(su ) = m + = d G (u ) l + = ξ(g) +. Hence by (5.6) and the definitions of 6

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