Graphs and Combinatorics (1999) 15 : 137±142
Graphs and Combinatorics ( Springer-Verlag 1999
2; k-Factor-Critical Gr...
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Graphs and Combinatorics (1999) 15 : 137±142
Graphs and Combinatorics ( Springer-Verlag 1999
2; k-Factor-Critical Graphs and Toughness Mao-Cheng Cai1*, Odile Favaron2, and Hao Li2 1 Institute 2 LRI,
of Systems Science, Academia Sinica, Beijing 100080, China Bat. 490, Universite 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. with an r-regular spanning subgraph). We show that every ttough graph of order n with t V 2 is
2; k-factor-critical for every non-negative integer k U minf2t ÿ 2; n ÿ 3g, thus proving a conjecture as well as generalizing the main result of Liu and Yu in [4].
1. Introduction All graphs G
V ; E under consideration are simple and ®nite of order jV j n. If A J V , we denote by G ÿ A the subgraph obtained from G by deleting the vertices in A together with the edges incident with vertices in A. For a vertex x of G; NA
x is the set of neighbors of x in A and dA
x jNA
xj. If the two subsets A and B of V are disjoint, E
A; B is the set of edges between A and B and e
A; B jE
A; Bj. We denote by d
G and k
G the minimum degree and the vertex connectivity of G. If S is a cutset of G, o
G ÿ S is the number of connected components of G ÿ S. When G is not complete, its toughness is de®ned by jSj S is a cutset of G . It is clear from the de®nition that t
G : min o
G ÿ S
d
G V k
G V 2t
G. The graph G is said to be t-tough for every positive t U t
G. For a clique Kn , we usually put k
Kn n ÿ 1 and t
Kn y. An rfactor of G, where r is a positive integer, is an r-regular spanning subgraph of G. In particular, a 1-factor is a perfect matching. When r is odd, only graphs of even order can admit an r-factor. Tutte gave necessary and su½cient conditions for a graph to have an r-factor ([6] for r 1 and [7] 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, a component C of G ÿ
S U T is said to be odd if rjV
Cj e
T; V
C is odd. We denote by o1
S; T the number of odd components of G ÿ
S U T and let * Research partially supported by National Natural Science Foundation of China and by the CoopeÂration Franco-Chinoise PRA 93-M10
138
M.-C. Cai et al.
qG
S; T : rjSj ÿ rjTj
X
dGÿS
v ÿ o1
S; T:
vAT
Note that when r is even, the odd components C are de®ned by e
T; V
C odd and, inP particular, e
T; V
C V 1 for an odd component C. This implies that for r dGÿS
v ÿ o1
S; T V 0 and thus qG
S; T V rjSj ÿ rjTj. even, vAT
Theorem A. (Tutte [7]): (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. The graph G is said
r; k-factor-critical if G ÿ X admits an r-factor for every subset X of k elements of V (when r 1, we usually simply say k-factor-critical). For r V 2, the study of these graphs has been initialized by Liu and Yu [4] under the name of
r; k-extendable graphs. We prefer to keep here the term factorcritical because usually, in the term extendable, X is not any subset of k vertices but must satisfy some given properties. Using Tutte's Theorem, Liu and Yu found the following characterization of
r; k-factor-critical graphs. Theorem B. (Liu and Yu [4]): 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. They also gave a su½cient condition for a graph to be
2; k-factor-critical in terms of its toughness. Theorem C. (Liu and Yu [4]): Let G be a graph of order n and toughness t
G V 3. Then G is
2; k-factor-critical for every integer k such that 3 U k U t
G and k U n ÿ 3. However they think Theorem C is not best possible and propose Conjecture D. (Liu and Yu [4]): Let G be a graph of order n and toughness t
G. If t
G V q and n V 2q 1 for some integer q V 1, then G is
2; 2q ÿ 2factor-critical. Our purpose is to prove this conjecture with the necessary restriction t
G V 2 since it is proved in [1] that for any positive real number e there exists an
r ÿ etough graph which has no r-factor. 2. The main result Theorem 1. Let G be a t-tough graph of order n with t V 2. Then G is
2; k-factorcritical for every non-negative integer k U minf2t ÿ 2; n ÿ 3g and the bound 2t ÿ 2 on k is sharp. Proof: First we may assume G to be not complete for otherwise the result is obvious. Now suppose, to the contrary, that G is not
2; k-factor-critical. Then, by
2; k-Factor-Critical Graphs and Toughness
139
Theorem B, there exist a pair S; T of disjoint subsets of V with jSj V k such that qG
S; T < 2k, implying by Theorem A(i) qG
S; T U 2k ÿ 2: Among all such pairs, we choose a pair S; T with additional properties: (i) qG
S; T is minimum and (ii) subject to (i), T is minimal under inclusion. P dGÿS
v 0; o1
S; q 0, and thus qG
S; T Then T 0 q. For otherwise vAT
2jSj V 2k, a contradiction. We write U : V
G ÿ
S U T;
jSj s;
jTj t:
Thus s V k and t V 1. In order to prove the theorem we need the following claims, whose ideas already appeared in [4]. Claim 1: (a) T is independent. (b) For all y A T, each edge of E
y; U joins y to an odd component of G ÿ
S U T, and, moreover, di¨erent edges of E
y; U join y to di¨erent odd components of G ÿ
S U T. Proof of Claim 1: For any y A T, let L
y be the component of G ÿ
S U
T n fyg containing y, and put h
y 1 if L
y is odd for the pair S; T n fyg, 0 otherwise. Let also u
y be the number of odd components of GP ÿ
S U T joined to y by some edges. Then qG
S; T n fyg 2s ÿ 2
t ÿ 1 dGÿS
v ÿ
o1
S; T ÿ u
y h
y. On the other hand, qG
S; T n fyg V v A Tnfyg
qG
S; T 2 by the choice of
S; T and by parity. Hence 2 UqG
S; T n fyg ÿ qG
S; T 2 ÿ dGÿS
y u
y ÿ h
y 2 ÿ dT
y ÿ e
y; U u
y ÿ h
y. But dT
y V 0, e
y; U ÿ u
y V 0 and h
y V 0. Therefore, dT
y 0, implying (a), h
y 0 and e
y; U u
y, implying (b). Note that by (b), there is no edge between T and the even components of G ÿ
S U T. Claim 2: For t V 2, s e
T; U ÿ o1
S; T V tt. Proof of Claim 2: Let us label the odd components Ci of G ÿ
S U T in such a way that for i U o2 , each component Ci contains at least one vertex with exactly one neighbor in T, and for i > o2 , no vertex of Ci has exactly one neighbor in T. For short we put o1
S; T o1 and note that 0 U o2 U o1 . For each i U o2 , we choose in Ci one vertex ui such that dT
ui 1 and put L fu1 ; u2 ; ; uo2 g. If o2 0 then L q. We denote by W the set NU
T (by Claim 1(b), W J 6 V
Ci ) and put Z W nL. Each vertex of W has at least one 1UiUo1
neighbor in T. Moreover by the de®nition of o2 and the fact that e
T; Ci is odd for all i, each component Ci with i > o2 contains at least one vertex having at least three neighbors in T. Therefore e
T; U V jW j 2
o1 ÿ o2 jZj o1
140
M.-C. Cai et al.
o1 ÿ o2 V jZj o1 . The equality e
T; U jZj o1 occurs if and only if o2 o1 and each vertex of W has exactly one neighbor in T. On the other hand, by the construction of Z, the t vertices of the set T, which is independent by Claim 1, belong to t di¨erent components of G ÿ
S U Z. Since t V 2, S U Z is a cutset of G and thus jS U Zj V tt. Hence s e
T; U ÿ o1
S; T V tt, as required. Claim 3: For t V 2 and s 2; s e
T; U ÿ o1
S; T > tt. Proof of Claim 3: Suppose s e
T; U ÿ o1
S; T tt. Then by the proof of Claim 2, o2 o1 , every vertex of W NU
T has exactly one neighbor in T, and tt jS U Zj, that is, the number of components of G ÿ
S U Z is exactly t (in particular G ÿ
S U T has no even component). Now, since d
G V 2t
G V 4 and s 2, and by Claim 1(b), some vertex y of T has one neighbor in at least two di¨erent components Ci , say y1 in C1 and y2 in C2 . In the choice of L, let us take u1 y1 and u2 y2 . For the set R S U Z U fyg of cardinality jZj 3, the number of components of G ÿ R is at least
t ÿ 1 2 t 1, and thus t
t 1 U jZj 3. This contradicts tt jS U Zj jZj 2 since t V 2. We distinguish three cases according to the values of t. Case t 1. Then s V 2t ÿ 1. Indeed, say T f yg. If e
y; U U 1, clearly s V 2t ÿ 1. And if e
y; U V 2, then, by Claim 1(b), o1
S; T e
y; U and S U fyg is a cutset, implying s V to1
S; T ÿ 1 V 2t ÿ 1. Hence s V k 1 and qG
S; T V 2s ÿ 2t V 2k, a contradiction. Case t 2. By Claim 1, either E
T; U is empty or there exists an odd number of edges, and thus exactly one edge, between T and each odd component of G ÿ
S U T. Therefore the two vertices of T belong to two di¨erent components of G ÿ S and S is a cutset. Hence s V 2t V k 2 and thus qG
S; T V 2s ÿ 2t V 2k, a contradiction, Case t V 3. Then qG
S; T s ÿ 2t
s e
T; U ÿ o1
S; T V s ÿ 2t tt by Claim 2, with a strict inequality when s 2 by Claim 3. As t V 2, then k U s U qG
S; T U 2k ÿ 2
implying k V 2. Let us show k V 3. Indeed, if k 2, the equality occurs everywhere in
and s 2, contradicting Claim 3. k2 , we have As s V k; t V 3 and t V 2 k2 5k ÿ2 ÿ 3 > 2k ÿ 2; qG
S; T V k 3 2 2 a contradiction. Since the assumption qG
S; T < 2k leads to a contradiction in all cases, qG
S; T V 2k and thus G is
2; k-factor-critical by Theorem B. The proof is complete. To show the sharpness of our result, consider the graph G consisting of a clique of vertex set A S U fyg with jSj k V 3, a second clique C of order at least 2,
2; k-Factor-Critical Graphs and Toughness
141
all the edges between C and S, and one edge yz for some vertex z A C. The graph k1 V 2 and is not
2; k-factor-critical since G ÿ S has G has toughness t
G 2 no 2-factor. Hence a t-tough graph with t V 2 and n V 2t 2 is not necessarily
2; 2t ÿ 1-factor-critical. r An obvious consequence of Theorem 1 is the following. Corollary 3. (conjecture D): Let q be an integer V 1 and G a graph of order n V 2q 1 and toughness t
G V maxfq; 2g. Then G is
2; 2q ÿ 2-factor-critical. Note that for the values of t
G belonging to intervals of the form 2q 1 ; q 1 , Theorem 1 shows that G is
2; 2q ÿ 1-factor-critical and thus is 2 slightly stronger than Conjecture D. 3. Open problem For r 1, a theorem similar to Theorem 1 already exists: Theorem E. (Favaron [3]). Let G be a t-tough graph of order n with t > 1. Then G is
1; k-factor-critical for every non-negative integer k such that n k is even, k < 2t and k U n ÿ 2. It would be interesting to generalize Theorems E and 1 to larger values of r and to determine functions t0
f and k0
t; f such that any t-tough graph with t > t0
f is
r; k-factor-critical for every non-negative integer k with
n k f even, k U k0
t; f and k U n ÿ
f 1. Added, as a partial answer to the open problem, the following results have been obtained [5, 8]: (1) Every t-tough graph of order n V 12 with t V 4 is
3; k-factor-critical for every non-negative integer k such that n k even and k U minf2t ÿ 3; n ÿ 7g. (2) Every t-tough graph of order n V 14 with t V 5 is
4; k-factor-critical for every non-negative integer k U minf2t ÿ 4; n ÿ 8g. Acknowledgments. This work was done while the ®rst author was visiting LRI, Universite Paris-Sud, he wishes to thank LRI and Professor Hao Li for their hospitality. The authors are grateful to Professor H. Enomoto for his stimulating discussions, which led to Theorem 1 extended to case t 2.
References 1. Enomoto, H., Jackson, B., Katerinis P., and Saito, A.: Toughness and the existence of k-factors, J. Graph Theory 9, 87±95 (1985) 2. Enomoto, H.: Toughness and the existence of k-factors III, Discrete Math. 189, 277± 282 (1998)
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3. Favaron, O.: On k-factor-critical graphs, Discussiones Mathematicae-Graph Theory 16, 41±51, (1996) 4. Liu G., Yu, Q.: k-factors and extendability with prescribed components, submitted 5. Shi, M., Yuan, X., and Cai, M.:
3; k-Factor-critical graphs and toughness, submitted 6. Tutte, W.T.: The factorization of linear graphs, J. London Math. Soc. 22, 107±111, (1947) 7. Tutte, W.T.: The factors of graphs, Canad. J. Math. 4, 314±328, (1952) 8. Yuan, X., Shi, M., and Cai, M.:
4; k-Factor-critical graphs and toughness, in preparation
Received: December 16, 1996 Revised: September 17, 1997