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

In this paper we study cycle coverings and 2-factors of a claw-free graph and those of its closure, which has been de ned by the rst author (On a closure concept in claw-free graphs, J. Combinatorial 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 by k cycles in G if and only if V (cl(G)) is covered by k cycles of cl(G), and (2) G has a 2-factor with at most k components if and only if cl(G) has a 2-factor with at most k components. 2 For graph theoretic notation not de ned in this paper, we refer the reader to [2]. A vertex 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 be eligible if NG(x) induces a noncomplete graph. Let x be an eligible vertex of a graph G. Consider the operation of joining every pair of nonadjacent vertices in NG(x) by an edge so that NG(x) induces a complete graph in the resulting graph. This operation is called local completion of G at x. For a graph G, let G0 = G. For i 0, if Gi is de ned and it has an eligible vertex xi, then apply local completion of Gi at xi to obtain a new graph Gi+1. If Gi has no eligible vertex, let cl(G) = Gi and call it the closure of G. The above operation was introduced and the following theorems 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 and only if cl(G) is hamiltonian. Recently, several other properties on paths and cycles of a claw-free graph and those of its closure were studied in [1]. In particular, the following theorem was proved. Theorem C ([1]). (1) A claw-free graph G is traceable if and only if cl(G) is traceable. (2) There exist in nitely many claw-free graphs G 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 graph G such that cl(G) is pancyclic while G is not pancyclic. Let H1; : : : ; Hk be subgraphs of G. Then G is said to be covered by H1; : : : ; 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 which covers G. Second, it is considered as a 2-factor with one component. These interpretations may lead us to possible extensions of Theorem B to cycle coverings and 2-factors. This is the motivation of this paper. 3 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 and only if cl(G) is covered by k cycles. Theorem 2. Let G be a claw-free graph. If cl(G) has a 2-factor with k components, 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" k components. Under the assumption of Theorem 2, G does not always have a 2-factor with exactly k components if k 2. Let G be a graph with k 1 components H1; : : : ; Hk 1, where H1 is the graph shown in Figure 1 and H2; : : : ; Hk 1 are cycles 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 two components, cl(G) has a 2-factor with k components. However, G has no 2-factor with k components since H1 does not have a 2-factor with two components. insert gure 1 Before proving the above theorems we introduce some notation which is used in the subsequent arguments. For a graph G and ; 6= S V (G), the subgraph induced by S is denoted by G[S]. When we consider a path or a cycle, we always assign an orientation. Let P = x0x1 xm. We call x0 and xm the starting vertex and the terminal vertex of P , respectively. The set of internal vertices of P is denoted by int(P ): int(P ) = fx1; x2; : : : ; xm 1g. The length of P is the number of edges in P , and is denoted by l(P ). We de ne x+(P ) i = xi+1 and x (P ) i = xi 1. Furthermore, we de ne 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 2 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 3. Let G be a claw-free graph and let x be a locally connected vertex. Let T1, 4 T2 V (G) with T1 \ T2 = fxg. Suppose both G[T1] and G[T2] are hamiltonian but G[T1[T2] is not hamiltonian. Choose cycles C1 and C2 with V (C1)[V (C2) = T1[T2 and V (C1) \ V (C2) = fxg and a path P in G[NG(x)] with starting vertex in fx+(C1); x (C1)g and terminal vertex in fx+(C2); x (C2)g so that P is as short as possible. Then 2 l(P ) 3 and int(P ) \ (T1 [ T2) = ;. Proof. First, note that each hamiltonian cycle Di in G[Ti] (i = 1, 2) satis es V (D1) [ V (D2) = T1 [ T2 and V (D1) \ V (D2) = fxg. Furthermore, since x is a locally connected vertex of G, there exists a path in G[NG(x)] with starting vertex in fx+(D1); x (D1)g and terminal vertex in fx+(D2); x (D2)g. Therefore, we can make a choice for (C1; C2; P ). Let u1 = x+(C1), v1 = x (C1), u2 = x+(C2) and v2 = x (C2). We may assume the starting and terminal vertices of P are u1 and u2, respectively. If u1u2 2 E(G), then C 0 = xv1()C1u1u2!C2v2x is a cycle in G with V (C 0) = V (C1) [ V (C2) = T1[T2. This contradicts the assumption. Hence we have u1u2 = 2 E(G). Similarly we have u1v2, v1u2, v1v2 = 2 E(G). Since fu1; v1; u2g NG(x) and G is claw-free, we have u1v1 2 E(G). Similarly u2v2 2 E(G). Let w = u+(P ) 1 . We claim w = 2 V (C1) [ V (C2). Assume w 2 V (C1) [ V (C2). Since w 2 V (P ) NG(x), w 6= x. Thus, w 2 u1!C1v1 [ u2!C2v2. First, suppose w 2 u1!C1v1. Then by the choice of P , w 2 u+1 !C1v 1 . Since fx;w+; w g NG(w) and G is claw-free, we have fxw+; xw ; w+w g \ E(G) 6= ;. If w+w 2 E(G), let C 0 1 = xwu1!C1w w+!C1v1x, C 0 2 = C2 and P 0 = w!Pu2. If w x 2 E(G), then let C 0 1 = xw!C1v1u1!C1w x, C 0 2 = C2 and P 0 = w!Pu2. If w+x 2 E(G), then let C 0 1 = xw()C1u1v1()C1w+x, C 0 2 = C2 and P 0 = w!Pu2. Then in each case, since V (C 0 1) = V (C1), we have V (C 0 1)[ V (C 0 2) = V (C1)[ V (C2) = T1 [ T2 and V (C 0 1)\V (C 0 2) = fxg. Furthermore, w = x+(C0 1) and l(P 0) < l(P ). This contradicts the choice of (C1; C2; P ). Now, suppose w 2 u2!C2v2. Since fu2; v2g \NG(u1) = ;, we have w 2 u+2!C2v 2 . Since fx;w ; w+g NG(w) and G is claw-free, fxw ; xw+; w w+g\E(G) 6= ;. If xw 2 E(G), let C = xv1()C1u1w!C2v2u2!C2w x. If xw+ 2 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+ 2 E(G), then let C 0 1 = xwu1!C1v1x, C 0 2 = xu2!C2w w+!C2v2x and P 0 = w!Pu2. Then V (C 0 1)[V (C 0 2) = V (C1)[V (C2) = T1 [T2, V (C 0 1)\V (C 0 2) = fxg, 5 w = x+(C0 1) and l(P 0) < l(P ). This contradicts the choice of (C1; C2; P ). Therefore, w = 2 V (C1) [ V (C2). Let w0 = u (P ) 2 . (Possibly w0 = w.) Then by the same arguments we have w0 = 2 V (C1) [ V (C2). By the choice of (C1; C2; P ), P is an induced path. Hence if l(P ) 4, then fu1; u++(P ) 1 ; u2g is an independent set. Since V (P ) NG(x) and G is claw-free, this is a contradiction. Thus, l(P ) 3. Since u1u2 = 2 E(G), l(P ) 2. These imply int(P ) \ (T1 [ T2) = ;. By similar arguments, we have the following lemma. Lemma 4. Let G be a claw-free graph and let x be a locally connected vertex of G. Let T V (G) with x 2 T , and let u 2 NG(x) T . Suppose G[T ] is hamiltonian but G[T [fug] is not hamiltonian. Choose a hamiltonian cycle C in G[T ] and a path P in G[NG(x)] with starting vertex in fx+(C); x (C)g and terminal vertex u so that P is as short as possible. Then 2 l(P ) 3 and int(P ) \ (T [ fug) = ;. We prove one more lemma. Lemma 5. Let G be a claw-free graph and let x be an eligible vertex of G. Let G0 be the graph obtained from G by local completion at x. Let C 0 be a cycle in G0 with x 2 V (C 0). Then either (1) or (2) follows. (1) There exists a cycle C in G with V (C) = V (C 0). (2) There exist T1, T2 V (G) such that (2.1) T1 [ T2 = V (C 0) and T1 \ T2 = fxg, and (2.2) G[Ti] is hamiltonian or isomorphic to K2 (i = 1, 2). Proof. Let B = E(G0) E(G). Note that for each uv 2 B, fu; vg NG(x). Choose a cycle C inG0 with V (C) = V (C 0) so that jE(C)\Bj is as small as possible. If E(C)\B = ;, then C is a cycle satisfying (1). Therefore, we may assume E(C) \ B 6= ;. We claim jE(C)\Bj = 1. Assume, to the contrary, jE(C)\Bj 2, say e1, e2 2 E(C)\B, e1 6= 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 in NG(x). Since G is claw-free, fx1x2; x1x ; x2x g \ E(G) 6= ;. If x1x2 2 E(G), let C0 = y2!Cx1x2()Cy1y2. 6 Then V (C0) = V (C) and E(C0) = (E(C) fx1y1; x2y2g) [ fx1x2; y1y2g. This implies jE(C0)\Bj < jE(C)\Bj, which contradicts the minimality of jE(C)\Bj. If x1x 2 E(G), let C0 = x!Cx1x ()Cy1x. Then V (C0) = V (C) and E(C0) = (E(C) fx1y1; xx g) [ fxy1; x1x g. Since xy1 2 E(G), we have jE(C0)\Bj < jE(C)\Bj, again a contradiction. We have a similar contradiction if x x2 2 E(G). Therefore, the claim is proved. Let E(C) \ B = fx1y1g. We may assume x, x1 and y1 appear in this order along C. Let T1 = x!Cx1 and T2 = y1!Cx. Then T1 [T2 = V (C) and T1 \T2 = fxg. Since x1y1 2 B, xx1, xy1 2 E(G). If x1 6= x+, then x!Cx1x is a hamiltonian cycle in G[T1]. If x1 = x+, then G[T1] ' K2. Similarly, G[T2] is either hamiltonian or isomorphic to K2. Let G0 be a graph obtained from a claw-free graph by local completion at a vertex. Using Lemmas 3, 4, 5 we prove that for each cycle in G0 there exists a cycle in G which contains it. We can also impose some restriction on its length. Theorem 6. Let G be a claw-free graph and let x be a locally connected vertex of G. Let G0 be the graph obtained from G by local completion at x. Then for each cycle C 0 in G0 there exists a cycle C in G with V (C 0) V (C) and l(C 0) l(C) l(C 0) + 3. Proof. If E(C 0)\ (E(G0) E(G)) = ;, then C 0 is a required cycle. Hence we may assume E(C 0) \ (E(G0) E(G)) 6= ;. If x 2 V (C 0), let C 0 1 = C 0. Suppose x = 2 V (C 0). Let e = uu+(C0) 2 E(C 0) \ (E(G0) E(G)). Then fu; u+(C0)g NG(x). Let C 0 1 = u+(C0)!C 0uxu+(C0). In either case, we have a cycle C 0 1 with V (C 0) [ fxg V (C 0 1) and l(C 0) l(C 0 1) l(C 0) + 1. If there exists a cycle C in G with V (C 0 1) = V (C), then C is a required cycle. Therefore, we may assume G has no such cycle. Then by Lemma 5, there exist T1, T2 V (G) with T1 \ T2 = fxg and T1 [ T2 = V (C 0 1) such that G[Ti] is hamiltonian or G[Ti] ' K2. Suppose both G[T1] and G[T2] are hamiltonian. Then by Lemma 3 there exist 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 0 1), V (C1) \ V (C2) = fxg, and (2) P joins fx+(C1); x (C1)g and fx+(C2); x (C2)g, 2 l(P ) 3 and int(P )\ (T1 [T2) = ;. Let u = x+(C1) and v = x+(C2). We may assume P joins u and v. Let C = x()C1u!Pv!C2x. Then C is a cycle in G, V (C 0 1) V (C) and l(C 0 1) l(C) l(C 0 1) + 2. Therefore, 7 V (C 0) V (C 0 1) V (C) and l(C 0) l(C 0 1) l(C) l(C 0 1) + 2 l(C 0) + 3. Using Lemma 4 instead of Lemma 3, we can, by similar arguments, deal with the case in which G[T1] or G[T2] is isomorphic to K2. Now Theorem 1 is a consequence of the following corollary of Theorem 6. Corollary 7. Let G be a claw-free graph and let x be an eligible verex of G. Let G0 be the graph obtained from G by local completion at x. Then G is covered by k cycles if and only if G0 is covered by k cycles. Proof. Since the \only if" part is trivial, we have only to prove the \if" part of the corollary. Suppose G0 is covered by k cycles, say V (G0) = V (C 0 1) [ [ V (C 0 k) for cycles C 0 1; : : : ; C 0 k inG0. By Theorem 6 for each C 0 i there exists a cycle Ci inG with V (C 0 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 8. Let G be a claw-free graph and let x be an eligible vertex of G. Let G0 be the graph obtained from G by local completion at x. Then for each set of k disjoint cycles fD1; : : : ; Dkg in G0 there exists a set of at most k disjoint cycles fC1; : : : ; Clg (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(G0) E(G). Note fa; bg NG(x) for each ab 2 B. Let F = f(S; F ) : S0 S V (G) and F is a 2-factor of G0[S]g: Since (S0;[ki=1E(Di)) 2 F, F 6= ;. Let F0 be the set of pairs (S; F ) 2 F chosen so that (a) the number of components of F is as small as possible, and (b) jF \ Bj is as small as possible, subject to (a). Let (S; F ) 2 F0. Suppose F consists of l components (cycles) C1; : : : ; Cl: F = E(C1) [ [ E(Cl) (disjoint). Since (S0;[ki=1E(Di)) 2 F, l k. By the assumption F \ B 6= ;. If x = 2 S, choose i with E(Ci) \ B 6= ;, say e = uv 2 E(Ci) \ B and v = u+(Ci). Let C 0 i = xv!Ciux and F 0 = F E(Ci) [ E(C 0 i). Then F 0 is a 2-factor of G0[S [ fxg] with l 8 components and jF 0 \Bj = jF \Bj 1. This contradicts the choice of (S; F ) given in (b). Therefore, we have x 2 S. We may assume x 2 V (C1). We claim B\([li=2E(Ci)) = ;. Assume B\([li=2E(Ci)) 6= ;, say f = u0v0 2 B\E(Cj) (j 2). Then fu0; v0; x+(C1)g NG(x) and hence u0x+(C1) 2 E(G0). We may assume j = 2 and v0 = u0+(C2). Let C 0 = xv0!C2u0x+(C1)!C1x and F 0 = F (E(C1) [ E(C2)) [ E(C 0). Then F 0 is a 2-factor of G0[S] with l 1 components. This contradicts the choice of (S; F ). Since F \ B 6= ;, B \ E(C1) 6= ;. If there exists a cycle C 0 1 in G with V (C 0 1) = V (C1), then F E(C1) [ E(C 0 1) is a 2-factor of G[S] with l components. This contradicts the assumption. Since x 2 V (C1), by Lemma 5, there exist T0, T1 V (G) such that T0 [ T1 = V (C1), T0 \ T1 = fxg, and G[Ti] is hamiltonian or isomorphic to K2 (i = 0, 1). First, consider the case in which both G[T0] and G[T1] are hamiltonian. Let C 0 0 and C 0 1 be cycles in G[T0 [T1] with V (C 0 0)[V (C 0 1) = T0 [T1 = V (C1) and V (C 0 0)\V (C 0 1) = fxg. Let ui = x+(C0 i) and vi = x (C0 i) (i = 0, 1). Since x is a locally connected vertex of G, G[NG(x)] has a path P with starting vertex in fu0; v0g and terminal vertex in fu1; v1g. Since G[S] has no 2-factors with l components, u0u1, u0v1, v0u1, v0v1 2 B. By the choice of (S; F ) given in (b), jE(C1) \ Bj = 1. Now choose (S; F ) 2 F0, C 0 0, C 0 1 and P so that (c) P is as short as possible. Then by Lemma 3, 2 l(P ) 3 and int(P ) \ V (C1) = ;. We may assume that the starting vertex and the terminal vertex of P are v0 and u1, respectively. Let a = v+(P ) 0 . Then a = 2 V (C1). Assume a = 2 S. Since V (P ) NG(x), ax 2 E(G) and hence au1 2 E(G0). Let C 0 = xu0!C 0 0v0au1!C 0 1v1x and F 0 = F E(C1) [E(C 0). Then F 0 is a 2-factor of G0[S [fag] with l components and F 0 \B fau1g. Since jB \E(C1)j = 1, jF 0 \ Bj = jF \ Bj = 1. Furthermore, C 00 0 = xu0!C 0 0v0ax and C 00 1 = C 0 1 are two cycles in G with V (C 00 0 ) [ V (C 00 1 ) = V (C 0) and V (C 00 0 ) \ V (C 00 1 ) = fxg. Since a!Pu1 is shorter than P , this contradicts the choice of (S; F ) given in (c). Therefore, we have a 2 S. We may assume a 2 V (C2). Let a0 = a+(C2) and a00 = a (C2). If a0x 2 E(G), then fa0; u1g NG(x) and hence a0u1 2 E(G0). Let C 0 = xu0!C 0 0v0a()C2a0u1!C 0 1v1x 9 and F 0 = F (E(C1) [ E(C2)) [ E(C 0). Then F 0 is a 2-factor of G0[S] with l 1components. This contradicts the choice of (S; F ). Hence we have a0x =2 E(G). By thesame argument we have a00x =2 E(G). Since a and fx; a0; a00g do not form a claw in G,a0a00 2 E(G). If l(C2) 4, let C 0 =xu0!C00v0au1!C01v1x (note au1 2 E(G0)), C 00 =a0!C2a00a0and F 0 = F (E(C1) [ E(C2)) [ E(C 0) [ E(C 00). Then F 0 is a 2-factor of G0[S] with lcomponents and F 0 \B fau1g. Since jB\E(C1)j = 1, jF 0\Bj = jF \Bj. Furthermore,C 000 =xu0!C00v0ax and C 001 = C 01 are two cycles in G with V (C 000 ) [ V (C 001 ) = V (C 0) andV (C 000 )\ V (C 001 ) = fxg. Sincea!Pu1 is shorter than P , this contradicts the choice of (S; F )given in (c). Therefore, we have l(C2) = 3, which implies C2 = aa0a00a.If a0 2 NG(v0), let C 0 =xu0!C00v0a0a00au1!C01v1x and F 0 = F (E(C1)[E(C2)) [E(C 0).Then F 0 is a 2-factor of G0[S] with l 1 components. This contradicts the choice of (S; F ).If a0 2 NG(u1), let C 0 =xu0!C00v0aa00a0u1!C01v1x and F 0 = F (E(C1)[E(C2)) [E(C 0).Then F 0 is a 2-factor of G0[S] with l 1 components, which contradicts the assumption.Therefore, a0 =2 NG(v0) [NG(u1). Similarly, a00 =2 NG(v0) [NG(u1).Let b = u (P )1 . Let b 2 V (Ci), 2 i l, b0 = b+(Ci) and b00 = b (Ci). By symmetry, wehave fb0; b00g \ (NG(x) [NG(u1) [NG(v0)) = ; and l(Ci) = 3.Suppose l(P ) = 2. Then b = a and hence Ci = C2. Since a0 =2 NG(v0) [ NG(u1) andv0u1 =2 E(G), a and fa0; v0; u1g form a claw in G, a contradiction. Therefore, we havel(P ) = 3. Sinceu1a0, u1a00 =2 E(G), Ci 6= C2. We may assume b 2 V (C3).By the choice of P given in (c), bv0, au1 =2 E(G). Sincev0a0 =2 E(G) and a andfa0; b; v0g do not form a claw, a0b 2 E(G). Similarly, we have a00b, ab0, ab00 2 E(G). Nowlet C 0 = aa00a0bb0b00a and F 0 = F (E(C2) [ E(C3) [ E(C 0). Then F 0 is a 2-factor ofG0[S] with l 1 components. This contradicts the choice of (S; F ) given in (a), and thetheorem follows in this case.By replacing Lemma 3 with Lemma 4, we can follow the same arguments to obtain acontradiction if G[T1] or G[T2] is isomorphic to K2. Therefore, the theorem is proved.Concluding Remarks.Let S be a set of vertices in a claw-free graph G. Then by Theorem 6 the minimumnumber of cycles covering S in G is the same as the minimum number of cycles covering10 S in cl(G). Furthermore, by Theorem 8, the minimum number of disjoint cycles coveringS in G is the same as the minimum number of disjoint cycles covering S in cl(G), (if thereexist such cycles). Therefore, these invariants (and hence the minimum number of cyclescovering V (G)) are stable in the sense of [1]. Furthermore, the existence of a 2-factor is astable property.AcknowledgmentsThis research was carried out while Z.R. and A.S. visited the Department of Mathemat-ical Sciences, The University of Memphis. These authors are grateful for the hospitalityextended during their stay.References[1] S. Brandt, O. Favaron and Z. Ryjacek, Closure and stable hamiltonian properties inclaw-free graphs, preprint.[2] G. Chartrand and L. Lesniak, Graphs & Digraphs (2nd ed.), Wadsworth & Brooks/Cole,Monterey, CA, (1986).[3] Z. Ryj acek, On a closure concept in claw-free graphs, J. Combinatorial Theory Ser. B70 (1997) 217{224.11