(3,k)-Factor-Critical Graphs and Toughness

Graphs and Combinatorics (1999) 15 : 463±471

Graphs and Combinatorics ( Springer-Verlag 1999

…3; k†-Factor-Critical Graphs and Toughness* MinyÇong Shi1, Xudong Yuan 1y , Mao-cheng Cai1, and Odile Favaron2 1 Institute of Systems Science, Academia Sinica, Beijing 100080, China 2 LRI, URA 410 CNRS, BaÃt. 410, Universite de Paris-Sud, 91405 Orsay cedex, France

Abstract. A graph is …r; k†-factor-critical if the removal of any set of k vertices results in a graph with an r-factor (i.e. an r-regular spanning subgraph). Let t…G† denote the toughness of graph G. In this paper, we show that if t…G† V 4, then G is …3; k†-factor-critical for every non-negative integer k such that n ‡ k even, k < 2t…G† ÿ 2 and k U n ÿ 7.

1. Introduction All graphs G ˆ …V ; E† under consideration are simple and ®nite. The following notation and de®nitions will be used. If A J V , we denote by hAi the subgraph of G induced by A and write G ÿ A for hV nAi. Sometimes we use T to denote the subgraph hTi of G. For a vertex x of G, NA …x† is the set of neighbors of x in A and dA …x† ˆ jNA …x†j. We denote by d…G† the minimum degree of G. For any two subsets A and B of V, E…A; B† is the set of edges between A and B and e…A; B† ˆ jE…A; B†j. In particular, for B ˆ A, we simply write E…A† for E…A; A† and e…A† for e…A; A†. For an S J V , let o…G ÿ S† denote the number of connected components of G its toughness is de®ned by t…G† :ˆ  ÿ S. When G is not complete,  jSj j S is a cutset of G . Obviously by the de®nition of t…G†, if G is min o…G ÿ S† not complete, then d…G† V 2t…G†. An r-factor of G, where r is a positive integer, is an r-regular spanning subgraph of G. A graph G is said to be (r; k)-factor-critical if G ÿ X admits an r-factor for every subset X of k elements of V. Tutte gave necessary and su½cient conditions for a graph to have an r-factor ([7] for r ˆ 1 and [8] for r V 2). Let us recall these conditions for r V 2. For a given positive integer r and a pair S; T of disjoint subsets of V, we call a component C of G ÿ …S U T† odd if rjCj ‡ e…T; C† is odd. Let o1 …S; T† and l ˆ o…U† denote the numbers of odd components P and components of U :ˆ G ÿ …S U T†, respectively. Let qG …S; T† :ˆ rjSj ÿ rjTj ‡ v A T dGÿS …v† ÿ o1 …S; T†. * Research partially supported by National Natural Science Foundation of China y Present address: Department of Mathematics, Guanxi Normal University, Guilin 541004, China

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Theorem A. (Tutte [8]): (i) qG …S; T† has always the same parity as nr. (ii) G has an r-factor if and only if qG …S; T† V 0 for every pair S; T of disjoint subsets of V. Using Tutte's Theorem, Liu and Yu found the following characterization of …r; k†-factor-critical graphs. Theorem B. (Liu and Yu [6]): Let r; k be integers with r V 2 and k V 0, and G a graph of order n V r ‡ k ‡ 1. Then G is …r; k†-factor-critical if and only if qG …S; T† V rk for any pair S; T of disjoint subsets of V with jSj V k. The following result was conjectured by Liu and Yu [6] and was proved by Cai, Favaron and Li [1] and by Enomoto [2] independently. Theorem C. Let G be a graph of order n with t…G† V 2. Then G is …2; k†factor-critical for every non-negative integer k such that k U 2t…G† ÿ 2 and k U n ÿ 3. The following open problem was proposed in [1]. Open problem: For larger r, determine functions t0 …r† and k0 …t…G†; r† such that any graph with t…G† > t0 …r† is …r; k†-factor-critical for every non-negative integer k with …n ‡ k†r even, k U k0 …t…G†; r† and k U n ÿ …r ‡ 1†. The purpose of this paper is to present the following main result concerning this problem for r ˆ 3. Theorem 1. Suppose t…G† V 4. Then G is …3; k†-factor-critical for every non-negative integer k with n ‡ k even, k < 2t…G† ÿ 2 and k U n ÿ 7. In the next section, we give the proof of Theorem 1. In the last section, we indicate directions for further research. 2. Proof of Theorem 1 The proof is by contradiction. Suppose Theorem 1 were not true, let G be a counterexample, that is, G satis®es the conditions of Theorem 1 but is not a …3; k†factor-critical for some k, where k is an integer with n ‡ k even, k < 2t…G† ÿ 2 and n V k ‡ 7. Obviously, G is not complete. Let us choose a pair of disjoint subsets …S; T† of V with jSj V k such that: (i) qG …S; T† is minimum; and (ii) subject to (i), T is minimal under inclusion. Let jSj ˆ s; jTj ˆ t; s 0 ˆ s ÿ k: P Then, by Theorem B, qG …S; T† ˆ 3s ÿ 3t ‡ v A T dGÿS …v† ÿ o1 …S; T† < 3k and, by Theorem A (i), qG …S; T† U 3k ÿ 2. U :ˆ G ÿ …S U T†;

l ˆ o…U†;

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P Note that T 6ˆ q, for otherwise v A T dGÿS …v† ˆ 0. Then o1 …S; T† V 3s ÿ …3k ÿ 2† V 2. So S is a cutset of G, and s ˆ jSj V t…G† o1 …S; q† > k‡2 o1 …S; q†. Since o1 …S; q† V 2, s > k ‡ o1 …S; q† and thus 3k ÿ 2 V 2 qG …S; T† ˆ 3s ÿ o1 …S; q† > 3k ‡ 2o1 …S; q†, a contradiction. Claim 1. For any v A T, dGÿS …v† U o1 …S; T† ÿ o1 …S; T ÿ v† ‡ 1, and dT …v† U 1. In particular, dGÿS …v† U l ‡ 1 for any v A T. Proof. For any v A T, let T 0 ˆ T ÿ fvg. Then qG …S; T 0 † V qG …S; T† ‡ 2 by the choice of …S; T† and by parity. So 2 U qG …S; T 0 † ÿ qG …S; T† ˆ 3 ÿ dGÿS …v† ‡ o1 …S; T† ÿ o1 …S; T 0 †, implying dGÿS …v† U o1 …S; T† ÿ o1 …S; T 0 † ‡ 1 U o1 …S; T† ‡ 1 U l ‡ 1. Obviously e…v; U† V o1 …S; T† ÿ o1 …S; T 0 †, yielding dT …v† ˆ dGÿS …v† ÿ e…v; U† U dGÿS …v† ÿ …o1 …S; T† ÿ o1 …S; T 0 †† U 1. By Claim 1, hTi consists of some isolated vertices and some independent edges. Let T1 be a maximal independent vertex set in hTi and T2 ˆ T ÿ T1 . Then T2 is also independent if T2 0 q, and jT2 j U jT1 j. Claim 2. U 0 q. Proof. Suppose U ˆ q. If jT1 j U 1, then jTj U 2; jSj ˆ n ÿ jTj V n ÿ 2 V k ‡ 5. So qG …S; T† V 3jSj ÿ 3jTj V 3…k ‡ 5† ÿ 6 > 3k, a contradiction. Hence jT1 j V 2, S is a cutset of G and G ÿ S contains P jT1 j components. We have jSj V t…G†jT1 j and qG …S; T† ˆ 3jSj ÿ 3jTj ‡ u A T dGÿS …u† ÿ o1 …S; T† V 3t…G†jT1 j ÿ 3…jT1 j ‡ jT2 j† ‡ 2jT2 j V 3t…G†jT1 j ÿ 4jT1 j V 2…3t…G† ÿ 4† > 3…k ‡ 2† ÿ8 ˆ 3k ÿ 2, a contradiction. By Claim 2, we have l V 1. Claim 3. (i) If there exists A H …G ÿ S† such that S U A is a cutset of G, then s 0 V 3 ÿ jAj. In particular, s 0 V 1 if jAj U 2. If G ÿ …S U A† contains at least three components, then s 0 V 7 ÿ jAj. In particular, s 0 V 1 if jAj U 6. (ii) s 0 ‡ dGÿS …x† V 3 for each x A U U T. Proof. (i) If there exists A H …G ÿ S† such that S U A is a cutset of G, then jS U Aj V 2t…G† > k ‡ 2 and thus s 0 ˆ s ÿ k > …k ‡ 2† ÿ jAj ÿ k ˆ 2 ÿ jAj. Since s 0 is an integer, s 0 V 3 ÿ jAj. If jAj U 2, then s 0 V 1. If G ÿ …S U A† contains at least three components, then, since t…G† V 4 and k‡2 ; jS U Aj V 3t…G† V 4 ‡ 2t…G† > 4 ‡ …k ‡ 2† ˆ k ‡ 6. Hence, s 0 ‡ t…G† > 2 k ‡ jAj > k ‡ 6 and since s 0 is an integer, s 0 V 7 ÿ jAj. And if jAj U 6, then s 0 V 1. (ii) For each x A U U T, s ‡ dGÿS …x† V d…G† V 2t…G† > k ‡ 2, yielding (ii) since s ‡ dGÿS …x† is an integer. Claim 4. jTj ˆ t V 4. Proof. Suppose ®rst t U 2, then,Pby Claim 3 (ii), 3k ÿ 2 V qG …S; T† V 3 s ‡ P 0 0 0 x A T …dGÿS …x† ÿ 3† ÿ l V 3k ‡ x A T …s ‡ dGÿS …x† ÿ 3† ‡ s ÿ l V 3k ‡ s ÿ l, k‡2 implying s 0 U l ÿ 2. Hence l V 2 and thus s ‡ t V t…G† l > l V k ‡ l. There2 0 fore s ˆ s ÿ k > l ÿ t V l ÿ 2, a contradiction.

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If t ˆ 3, then 3k ÿ 2 V qG …S; T† V 3s ‡

X

…dGÿS …x† ÿ 3† ÿ l

xAT

V 3k ‡

X

…s 0 ‡ dGÿS …x† ÿ 3† ÿ l:

…†

xAT

So l V 2 and if l ˆ 2, then s 0 ‡ dGÿS …x† ˆ 3 for each x A T. If l V 3 then, by Claim 3 (i) applied with A ˆ T, s 0 V 7 ÿ jTj ˆ 4 and thus 3k ÿ 2 V qG …S; T† V 3…k ‡ 4† ÿ 9 ÿ l. Hence l V 5. On the other hand, since s ‡ t V t…G†l > t…G†…l ÿ 2† ‡ k ‡ 2, we have 3k ÿ 2 V qG …S; T† > 3t…G† …l ÿ 2† ‡ 3…k ‡ 2† ÿ 9 ÿ 9 ÿ l V 11l ‡ 3k ÿ 36 since t…G† V 4. This yields 11l < 34, a contradiction. Hence l ˆ 2. Under this condition, we have s 0 ‡ dGÿS …x† ˆ 3 for each x A T and l ˆ o1 …S; T† by (*), thus two components are odd. Let C1 , C2 be the two components of hUi, We check three cases. If s 0 V 2, then dGÿS …x† U 1 for any x A T. Let l 0 ˆ o…hU U Ti†, then l 0 V 2. Hence s V l 0 t…G†, s 0 ˆ s ÿ k > l 0 t…G† ÿ …2t…G† ÿ 2† ˆ 2 ‡ …l 0 ÿ 2†t…G†. Considering s 0 ‡ dGÿS …x† ˆ 3 for each x A T, we get 3 > 2 ‡ …l 0 ÿ 2†t…G† ‡ dGÿS …x† and thus dGÿS …x† ˆ 0 for each x A T. This implies l 0 V 3 and contradicts with 1 > …l 0 ÿ 2†t…G† since t…G† V 4. When s 0 U 1, notice t ˆ 3 and dT …x† U 1 for any x A T, then there exist two non-adjacent vertices in T. If s 0 ˆ 1, then dGÿS …x† ˆ 2 for any x A T, and jU U Tj V k ‡ 7 ÿ …k ‡ 1† ˆ 6. If jU U Tj V 7, let x; y be two non-adjacent vertices in T, and A ˆ NGÿS …fx; yg†, then T U U ÿ …A U fx; yg† 0 q, G ÿ …S U A† has at least three components. As jAj U 4, s 0 V 3 by Claim 3 (i), a contradiction. If jU U Tj ˆ 6, then n ‡ k ˆ s ‡ 6 ‡ k ˆ 2k ‡ 7 is odd, a contradiction. If s 0 ˆ 0, then dGÿS …x† ˆ 3 for any x A T, and jU U Tj V 7. If jU U Tj V 9, similarly we arrive at a contradiction by Claim 3 (i). So what remain to be considered are the cases jUj ˆ 4 or 5. If jUj ˆ 4, then n ‡ k ˆ s ‡ 7 ‡ k ˆ 2k ‡ 7 is odd, a contradiction. If jUj ˆ 5, we may assume jC1 j < jC2 j, then jC1 j U 2. If jC1 j ˆ 1, say C1 ˆ fxg, then jC2 j ˆ 4 and e…x; T† is even as C1 is odd, thus e…C1 ; T† U 2 for jTj ˆ 3. But, by Claim 3 (ii), e…x; T† ˆ dGÿS …x† V 3, a contradiction. Thus jC1 j ˆ 2 and jC2 j ˆ 3. As C1 is odd, e…C1 ; T† is odd too. By Claim 3 (ii), e…C1 ; T† V 4, implying e…C1 ; T† V 5. Similarly, e…C2 ; T† is even and e…C2 ; T† V 3, yielding e…C2 ; T† V 4. Therefore e…T† ˆ 0. So G ÿ …S U U† consists of three components, by Claim 3 (i) s 0 V 2, a contradiction. Claim 5. (i) jSj ‡ jT2 j ‡ e…T1 ; U† ÿ l V t…G†jT1 j. (ii) If 1 U jT2 j < jT1 j, then jSj ‡ jT2 j ‡ e…T2 ; U† V t…G†…jT2 j ‡ 1†. Proof. Let C1 ; C2 ; . . . ; Cl be the components of U, L1 ˆ fCi j e…Ci ; T1 † > 0g and L2 ˆ fCi j e…ui ; T1 † ˆ 1 for some ui A Ci g. Clearly L2 J L1 . We may assume L1 ˆ fC1 ; C2 ; . . . ; Cl1 g and L2 ˆ fC1 ; C2 ; . . . ; Cl2 g (l2 U l1 ). For each Ci A L2 ,

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choose ui A Ci with e…ui ; T1 † ˆ 1. Let U2 ˆ fui j i ˆ 1; 2; . . . ; l2 g, U1 ˆ fu A U j e…u; T1 † > 0g and U3 ˆ U1 ÿ U2 . Then jU3 j U e…T1 ; U† ÿ l1 . Let G1 ˆ G ÿ …S U T2 U U3 †. Then vertices in T1 belong to di¨erent components of G1 since T1 is independent. Hence o…G1 † V jT1 j ‡ …l ÿ l1 †. First let us show (i). Seeing t V 4, jT1 j V 2, implying o…G1 † V 2. Thus jS U T2 U U3 j V t…G† o…G1 † V t…G†…jT1 j ‡ l ÿ l1 †. On the other hand, jS U T2 U U3 j U jSj ‡ jT2 j ‡ e…T1 ; U† ÿ l1 . We have jSj ‡ jT2 j ‡ e…T1 ; U† ÿ l1 V t…G†o…G1 † V jSj ‡ jT2 j ‡ e…T1 ; U† ÿ l V t…G†jT1 j ‡ …t…G† ÿ 1†…l ÿ l1 † V t…G†…jT1 j ‡ l ÿ l1 †, t…G†jT1 j, as required. Now we show (ii). Let T0 be the set of vertices in T1 which are not adjacent with vertices in T2 . Since jT2 j < jT1 j, T0 0 q by Claim 1. Let G2 ˆ G ÿ …S U NG …T2 ††. Then G2 contains at least jT2 j ‡ 1 V 2 components. Therefore, jSj ‡ e…T2 ; U† ‡ jT2 j V jSj ‡ e…T2 ; U† ‡ e…T2 ; T1 † V jS U NG …T2 †j V t…G†…jT2 j ‡ 1†. Claim 6. T is not independent in G. Proof. Assuming T is independent, then T ˆ T1 and T2 ˆ q. By Claim 5P (i), jSj ‡ e…T; U† ÿ l V t…G†jTj. Therefore 3k ÿ 2 V qG …S; T† ˆ 3s ÿ 3t ‡ 0 0 x A T dGÿS …x† ÿ o1 …S; T† V 2…k ‡ s † ÿ 3t ‡ jSj ‡ e…T; U† ÿ l V 2k ‡ 2s ‡ k‡2 , this …t…G† ÿ 3†t V 2k ‡ 4…t…G† ÿ 3† as t V 4. On one hand, since t…G† > 2 yields 3k ÿ 2 > 2k ‡ 2…k ‡ 2† ÿ 12 ˆ 4k ÿ 8 and thus k < 6; on the other hand, since t…G† V 4; 3k ÿ 2 V 2k ‡ 4 and thus k V 6, a contradiction. In order to complete the proof, we distinguish two cases according to whether jT2 j < jT1 j or jT2 j ˆ jT1 j. Case 1. jT2 j < jT1 j. ‡ e…T1 ; U† ÿ l V t…G†jT1 j and jSj ‡ jT2 j ‡ By Claim 5 (i) and (ii), jSj ‡ jT2 j P e…T2 ; U† V t…G†…jT2 j ‡ 1†. So 2jSj ‡ u A T dGÿS …u† ÿ o1 …S; T† V 2jSj ‡ 2jT2 j ‡ e…T1 ; U† ‡ e…T2 ; U† ÿ l V t…G†…jT1 j ‡ jT2 j ‡ 1† ˆ t…G†…t ‡ 1†. We have 3k ÿ 2 V qG …S; T† V k ‡ s 0 ÿ 3t ‡ t…G†…t ‡ 1† ˆ k ‡ s 0 ‡ …t…G† ÿ 3†…t ÿ 4† ‡ 4…t…G† ÿ 3† ‡ t…G† and thus 2k V s 0 ‡ …t…G† ÿ 3†…t ÿ 4† ‡ 5t…G† ÿ 10. If k U 5, then, since t…G† V 4; 10 V 2k V s 0 ‡ …t…G† ÿ 3†…t ÿ 4† ‡ 10 and thus, since t V 4, we get s 0 ˆ 0 and t ˆ 4. By Claim 6, hTi consists of one edge and two isolated vertices. Let x1 and x2 be two non-adjacent vertices. Clearly o…G ÿ …S U NGÿS …x1 † U NGÿS …x2 ††† V 3, thus s ‡ dGÿS …x1 † ‡ dGÿS …x2 † V 3t…G† > k ‡ 6, yielding dGÿS …x1 † ‡ dGÿS …x2 † V 7. We may assume dGÿS …x1 † V 4, then l V dGÿS …x1 † ÿ 1 V 3 by Claim 1. But, by Claim 3 (i), s 0 V 3 since G ÿ …S U T† contains at least three components, a contradiction. k‡2 , 2k > s 0 ‡ …t…G† ÿ 3†…t ÿ 4† ‡ t…G† ‡ 2k ÿ 6, If k V 6 then, since t…G† > 2 and thus 2 > s 0 ‡ …t…G† ÿ 3†…t ÿ 4† as t…G† V 4. If t V 5, then s 0 ˆ 0. As in the case k U 5, we can obtain s 0 V 2, a contradiction. Then t ˆ 4 by Claim 4, and thus s 0 U 1. If s 0 ˆ 0 or l V 3, similarly, we can obtain a contradiction. Hence, s 0 ˆ 1 and l U 2. Now let T1 ˆ fx1 ; x2 ; x3 g J T, with no two vertices adjacent in T. Let A ˆ NGÿS …T1 † and for 1 U i 0 j U 3; Aij ˆ NGÿS …fxi ; xj g†. By Claim 1, jAij j U dGÿS …xi † ‡ dGÿS …xj † U 6. Notice G ÿ …S U Aij † has at least three components, by

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Claim 3 (i), then 1 ˆ s 0 V 7 ÿ jAij j. Hence, jAij j ˆ 6 for all 1 U i 0 j U 3. Therefore, jAj ˆ 9 and NU …T1 † ˆ 8. Since n ÿ k is even, s 0 ‡ t ‡ jUj ˆ 5 ‡ jUj is even and thus UnNU …T1 † 0 q. Then the set G ÿ …S U A† has at least four components and 1 ‡ k ‡ 9 V 4t…G† > 2k ‡ 4, yielding k < 6, a contradiction. Case 2. 1 U jT2 j ˆ jT1 j. Claim 7. l ‡ s 0 U 9 ÿ t and t U 6. Proof. Since jT1 j ˆ jT2 j, t is even, T2 is also a maximal independent set of T. By Claim (5) (i), we have jSj ‡ jT2 j ‡ e…T P 1 ; U† ÿ l V t…G†jT1 j and jSj ‡ jT1 j ‡ e…T2 ; U† ÿ l V t…G†jT2 j. So, 2jSj ‡ u A T dGÿS …u† ÿ o1 …S; T† V 2jSj ‡ jT2 j ‡ jT1 j ‡ e…T1 ; U† ‡ e…T2 ; U† ÿ l V t…G†…jT1 j ‡ jT2 j† ‡ l ˆ t…G†t ‡ l. We have 3k ÿ 2 V qG …S; T† V k ‡ s 0 ÿ 3t ‡ t…G†t ‡ l V k ‡ s 0 ‡ …t…G† ÿ 3†…t ÿ 4† ‡ 4…t…G† ÿ 3† ‡ l: k‡2 , then 2k V s 0 ‡ …t ÿ 4† ‡ …2k ‡ 4† ÿ 12 ‡ Since t…G† V 4, t V 4 and t…G† > 2 l ‡ 2, and s 0 ‡ l ‡ t < 10, thus s 0 ‡ l U 9 ÿ t. By Claim 1 and Claim 3 (ii), 3 ÿ s 0 U dGÿS …v† U l ‡ 1 for all v A T. Hence, 0 s ‡ l V 2, yielding 2 ‡ t U 9 and thus, since t is even, t U 6. If l V 3 then, by Claim 3 (i), s 0 V 7 ÿ jTj and thus l ‡ s 0 ‡ t V l ‡ 7 V 10, a contradiction. Therefore l U 2. By Claim 4 and Claim 7, t ˆ 4 or 6 and we consider two cases. Subcase 2.1. t ˆ 6; 2 U l ‡ s 0 U 3 and 1 U l U 2 (which implies by Claim 1, dGÿS …v† U 3 for all v A T). In this case, hTi consists of three independent edges xi yi with 1 U i U 3. If l ˆ 1, then s 0 U 2. By Claim 1, dGÿS …v† U 2 for any v A T. Let A ˆ NGÿS …fx1 ; x2 g†, then jAj U 4 and G ÿ …S U A† contains at least three components. By Claim 3 (i), s 0 V 3, a contradiction. If l ˆ 2 and s 0 ˆ 0, then by Claim 1 and Claim 3 (ii), dGÿS …v† ˆ 3 for any v of T. Let A ˆ NGÿS …fx1 ; x2 g†. Then jAj U 6 and G ÿ …S U A† contains at least three components. By Claim 3 (i), s 0 V 1, a contradiction. If l ˆ 2 and s 0 ˆ 1, then by the proof of Claim 7, we have 2k V s 0 ‡ …t…G† ÿ 3†t ‡ l ‡ 2 V 5 ‡ 2…t…G† ÿ 3† ‡ 4 > 9 ‡ k ‡ 2 ÿ 6 ˆ k ‡ 5 since t…G† V 4, k‡2 and t ˆ 6, thus k V 6. t…G† V 2 We take T1 ˆ fx1 ; x2 ; x3 g; A ˆ NGÿS …T1 † and for 1 U i 0 j U 3, Aij ˆ NGÿS …fxi ; xj g†. Then, as in the proof of Case 1 for k V 6, we can obtain a contradiction. Subcase 2.2. t ˆ 4; 2 U l ‡ s 0 U 5 and 1 U l U 2. In this case, hTi consists of two independent edges xi yi with i ˆ 1; 2. Moreover, if 3s 0 ÿ l V 7, then 3k ÿ 2 V qG …S; T† V 3k ‡ 7 ÿ 3t ‡ …4 ‡ e…T; U†† V 3k ÿ 1, a contradiction. Hence, 3s 0 U 6 ‡ l U 8 and thus s 0 U 2. So we only need to check the following cases. If l ˆ 1, then s 0 V 1.

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When s 0 ˆ 2, then S is not Pa cutset of G (if S is a cutset of G , then we have s V 3 by Claim 3 (i) ), thus v A T dGÿS …v† V 6. We also have 3k ÿ 2 V qG …S; T† V 3k ÿ 1, a contradiction. When s 0 ˆ 1, then jUj V n ÿ k ÿ 1 ÿ 4 V 2. By Claim 1 and Claim 3 (ii), dGÿS …v† ˆ 2 for each v A T and thus e…U; T† ˆ 4. Then jUj V 3, otherwise U consists of two vertices, then hUi is an even component since e…T; U† ˆ 4, so we have 3k ÿ 2 V qG …S; T† V 3k ‡ 3 ÿ 12 ‡ 8 ˆ 3k ÿ 1, a contradiction. But if jUj V 3, take A ˆ NGÿS …fx1 ; x2 g†, so jAj U dGÿS …x1 † ‡ dGÿS …x2 † U 4 and Gÿ…S U A† contains at least three components. By Claim 3 (i), s 0 V 3, a contradiction. If l ˆ 2, let C1 and C2 be the two components of U with jC1 j U jC2 j. By Claim 1, dGÿS …v† U 3 for each v A T. When s 0 ˆ 2, either e…U; T† V 3 or e…U; T† U 2. In the former case 3k ÿ 2 V qG …S; T† V 3…k ‡ 2† ÿ 12 ‡ 7 ÿ 2 ˆ 3k ÿ 1, and in the latter case S is a cutset of G since l ˆ 2, thus s 0 V 3 by Claim 3 (i), a contradiction. When s 0 ˆ 1, then 3 V dGÿS …v† V 2 for any v A T by Claim 3 (ii), and jUj V n ÿ k ÿ 1 ÿ 4 V 2. Then there are at least three vertices in T with dGÿS …v† ˆ 2 and thus 4 U e…T; U† U 5. For otherwise, there are at most two vertices in T with dGÿS …v† ˆ 2, and then 3k ÿ 2 V qG …S; T† V 3k ‡ 3 ÿ 12 ‡ 10 ÿ 2 ˆ 3k ÿ 1. Similarly if jUj V 3, there exists A H …T U U† with jAj U 4 such that G ÿ …S U A† contains at least three components. By Claim 3, s 0 V 3, a contradiction. If jUj ˆ 2, then each component of U consists of just one vertex. If e…T; U† ˆ 5, then one component of U is even, thus 3k ÿ 2 V qG …S; T† V 3k ‡ 3 ÿ 12 ‡ 5 ‡ 4 ÿ 1 ˆ 3k ÿ 1, a contradiction. If e…T; U† ˆ 4, then dGÿS …v† ˆ 2 for each v A T U U by Claim 3 (ii), and hT U Ui consists of two triangles or a cycle of length 6. The former case implies s 0 V 3, the latter case implies that there exists A H …T U U† with jAj ˆ 3 such that G ÿ …S U A† contains three components. By Claim 3, s 0 V 4, a contradiction. When s 0 ˆ 0, then jUj ˆ n ÿ k ÿ 4 V 3. By Claim 1 and Claim 3 (ii), dGÿS …v† ˆ 3 for each v A T, and e…U; T† ˆ 8. And thus two components C1 and C2 of U are odd, for otherwise, 3k ÿ 2 V qG …S; T† V 3k ÿ 12 ‡ 12 ÿ 1 ˆ 3k ÿ 1. Hence, jUj V 4, otherwise jUj ˆ 3 implies that jC1 j ˆ 1 and jC2 j ˆ 2, and that one of them is an even component since e…C1 ; T† ‡ e…C2 ; T† ˆ 8. If jUj V 5, take A ˆ NGÿS …fx1 ; x2 g†, then A H …T U U† with jAj U 6, and G ÿ …S U A† contains at least three components, by Claim 3 (i), s 0 V 1, a contradiction. If jUj ˆ 4, we distinguish two cases. For jC1 j ˆ 1 and jC2 j ˆ 3, then each vertex in C2 is adjacent to at least two vertices of T. For otherwise, we can choose one vertex v 0 A C2 , and two vertices x; y A T such that any two vertices are non-adjacent in G, let A ˆ …T U U† ÿ fx; y; v 0 g, so jAj ˆ 5 and G ÿ …S U A† contains at least three components. By Claim 3 (i), s 0 V 2, a contradiction. Then e…C2 ; T† V 6, thus dGÿS …v† ˆ e…C1 ; T† U 2, where C1 ˆ fvg, that contradict with Claim 3 (ii). For jC1 j ˆ jC2 j ˆ 2, since both C1 ; C2 are odd components, then both e…C1 ; T†; e…C2 ; T† are odd numbers. Since dGÿS …v† V 3 for any v A T U U by Claim 3 (ii), so for i ˆ 1; 2, e…Ci ; T† V 6 ÿ 2 ˆ 4, and thus e…Ci ; T† V 5, and 8 V e…U; T† V 10, a contradiction. 0

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The proof is complete.

r

3. Remark Remark 1. The condition k < 2t…G† ÿ 2 is necessary for Theorem 1 to be true. Indeed, if k ˆ 2t…G† ÿ 2, the graph G is not necessarily …3; k†-factor-critical, as shown by the following example. Let graph G consist of a clique Kk‡m‡2 and a vertex fxg which is adjacent to k ‡ 2 vertices of the Kk‡m‡2 , where m is odd. Then k‡2 , i.e. k ˆ 2t…G† ÿ 2, and is not a …3; k†the graph G has toughness t…G† ˆ 2 factor-critical by Theorem B since qG …S; T† ˆ 3k ÿ 2, where S consists of k vertices which are adjacent to x and T ˆ fxg. Remark 2. The condition n V k ‡ 7 is necessary for Theorem 1 to be true. Indeed, if n ˆ k ‡ 6, the graph G is not necessarily …3; k†-factor-critical, as shown by the following example. Let graph G consist of a clique S F Kk‡1 and a cycle with ®ve vertices such that the vertices of the cycle are adjacent to all the vertices of the k‡3 and is not a …3; k†clique. Then the graph G has toughness t…G† ˆ 2 factor-critical by Theorem B since qG …S; T† ˆ 3…k ‡ 1† ÿ 9 ‡ 6 ÿ 2 ˆ 3k ÿ 2, where T consists of 3 vertices of the cycle such that G ÿ …S U T† consists of two isolated vertices. For r ˆ 1, a theorem similar to Theorem C and the main Theorem already exists: Theorem D. (Favaron [5]). Let G be a graph of order n with t…G† > 1. Then G is …1; k†-factor-critical for every non-negative integer k such that n ‡ k is even, k U 2t…G† ÿ 1 and k U n ÿ 2. Conjecture E. Let G be a graph of order n and toughness t…G†. Then there exists an t0 …r† and an integer k0 such that G is …r; k†-factor-critical when t…G† V t0 …r†, n ‡ k even when r is odd, k U 2t…G† ÿ r and k U n ÿ k0 . Note. After this paper was accepted, Professor Enomoto kindly informed us that the problem due to Cai et al. (Conjecture E), had been completely solved [3]. References 1. Cai, M., Favaron O., Li, H.: …2; k†-factor-critical graphs and toughness, Graph. Comb. 15(2), 137±142 2. Enomoto, H.: Toughness and the existence of k-factor (III), Discrete Math. 189, 277± 282 (1998) 3. Enomoto, H., Hagia, M.: Toughness and the existence of k-factor (IV), submitted 4. Enomoto, H., Jackson, B., Katerinis P., Saito, A.: Toughness and the existence of k-factors, J. Graph Theory 9, 87±95 (1985)

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5. Favaron, O.: On k-factor-critical graphs, Discuss. Math. Graph Theory 16, 41±51 (1996) 6. Liu G., Yu, Q.: k-factors and extendability with prescribed components, submitted 7. Tutte, W.T.: The factorization of linear graphs, J. Lond. Math. Soc. 22, 107±111 (1947) 8. Tutte, W.T.: The factors of graphs, Can. J. Math. 4, 314±328 (1952)

Received: July 28, 1997 Revised: September 21, 1998