Information Processing Letters 109 (2009) 811–815 Contents lists available at ScienceDirect Information Processing Letters www.elsevier.com/locate/ipl On fractional ( f , n)-critical graphs ✩ Sizhong Zhou ∗ , Qiqing Shen School of Mathematics and Physics, Jiangsu University of Science and Technology, Mengxi Road 2, Zhenjiang, Jiangsu 212003, People’s Republic of China a r t i c l e i n f o Article history: Received 19 October 2008 Accepted 30 March 2009 Available online 1 April 2009 Communicated by M. Yamashita Keywords: Graph Binding number Fractional f -factor Fractional ( f , n)-critical graph Combinatorial problems a b s t r a c t Let G be a graph of order p, and let a, b and n be nonnegative integers with b  a  2, and let f be an integer-valued function defined on V (G ) such that a  f (x)  b for all x ∈ V (G ). A fractional f -factor is a function h that assigns to each edge of a graph G a  number in [0, 1], so that for each vertex x we have dhG (x) = f (x), where dhG (x) = e∋x h(e ) (the sum is taken over all edges incident to x) is a fractional degree of x in G. Then a graph G is called a fractional ( f , n)-critical graph if after deleting any n vertices of G the remaining graph of G has a fractional f -factor. The binding number bind(G ) is defined as follows,  bind(G ) = min  | N G ( X )| : ∅ = X ⊆ V (G ), N G ( X ) = V (G ) . |X| In this paper, it is proved that G is a fractional ( f , n)-critical graph if bind(G ) > (a + b − 1)( p − 1)/(ap − (a + b) − bn + 2) and p  (a + b)(a + b − 3)/a + bn/(a − 1). Furthermore, it is showed that the result in this paper is best possible in some sense.  2009 Elsevier B.V. All rights reserved. 1. Introduction We investigate the fractional factor problem in graphs, which can be considered as a relaxation of the well-known cardinality matching problem. The fractional factor problem has wide-ranging applications in areas such as network design, scheduling and the combinatorial polyhedron. For example, in a communication network if we allow several large data packets to be sent to various destinations through several channels, the efficiency of the network will be improved if we allow the large data packets to be partitioned into small parcels. The feasible assignment of data packets can be seen as a fractional flow problem and it becomes a fractional matching problem when the destinations and sources of a network are disjoint (i.e., the underlying graph is bipartite). ✩ This research was supported by Jiangsu Provincial Educational Department and was sponsored by Qing Lan Project of Jiangsu Province. Corresponding author. E-mail address: zsz_cumt@163.com (S. Zhou). * 0020-0190/$ – see front matter doi:10.1016/j.ipl.2009.03.026  2009 Elsevier B.V. All rights reserved. In this paper, we consider only finite undirected simple graph G with vertex set V (G ) and edge set E (G ). For x ∈ V (G ), the degree of x in G is denoted by d G (x). The minimum vertex degree of G is denoted by δ(G ). For any S ⊆ V (G ), we denote by N G ( S ) the neighborhood set of S in G, by G [ S ] the subgraph of G induced by S, by G − S the subgraph obtained from G by deleting vertices in S together with the edges incident to vertices in S. A vertex set S ⊆ V (G ) is called independent if G [ S ] has no edges. The binding number of G is defined by Woodall [10] as  bind(G ) = min  | N G ( X )| : ∅ = X ⊆ V (G ), N G ( X ) = V (G ) . |X| Let f be a nonnegative integer-valued function defined on V (G ). Then a spanning subgraph F of G is called an f -factor if d F (x) = f (x) for all x ∈ V (G ). If f (x) = k for each x ∈ V (G ), then an f -factor is simply called a k-factor. A fractional f -factor is a function h that assigns to each edge of a graph G a number in [0, 1], so that for each vertex x we have dhG (x) = f (x), where dhG (x) = e∋x h(e ) (the sum is taken over all edges incident to x) is a fractional degree of x in G. If f (x) = k for each x ∈ V (G ), then a frac- 812 S. Zhou, Q. Shen / Information Processing Letters 109 (2009) 811–815 tional f -factor is a fractional k-factor. A graph G is called a fractional ( f , n)-critical graph if after deleting any n vertices of G the remaining graph of G has a fractional f -factor. If G is a fractional ( f , n)-critical graph, then we also say that G is fractional ( f , n)-critical. If f (x) = k for each x ∈ V (G ), then a fractional ( f , n)-critical graph is a fractional (k, n)-critical graph. A fractional (k, n)-critical graph is also called a fractional n-critical graph if k = 1. Other terminology and notation not given in this paper can be found in [2,8]. Many authors have investigated k-factors [4,5,7,9], and fractional factors [3,6,11]. Recently, Sizhong Zhou obtained some sufficient conditions for graphs to have factors or fractional factors [12–16]. The following results on fractional f -factors and fractional k-factors are known. In [3], J. Cai and G. Liu gave a sufficient condition for the existence of a fractional f -factor in a connected graph in terms of its stability number and minimum degree. Theorem 1. (See [3].) Let G be a connected graph and let f be a nonnegative integer-valued function defined on V (G ) such that 0  a  f (x)  b for each x ∈ V (G ). If minimum degree δ(G )  b and stability number α (G )  4a(δ(G ) − b)/(b + 1)2 , then G has a fractional f -factor. In [11], J. Yu and G. Liu gave a degree condition for graphs to have fractional k-factors. Theorem 2. (See [11].) Let k be an integer with k  1, and let G be a connected graph of order p with p  4k − 3, δ(G )  k. p If max{d G (x), d G ( y )}  2 for each pair of nonadjacent vertices x, y of G, then G has a fractional k-factor. In [14], S. Zhou and Z. Duan gave a binding number condition for graphs to have fractional k-factors. Theorem 3. (See [14].) Let k  2 be an integer, and let G be a graph of order p such that p  4k − 6. Then (a + b) − bn + 2) and p  (a + b)(a + b − 3)/a + bn/(a − 1), then G is fractional ( f , n)-critical. In Theorem 5, if n = 0, then we get the following corollary. Corollary 1. Let G be a graph of order p, and let a, b be nonnegative integers such that 2  a  b, and let f be an integer-valued function defined on V (G ) such that a  f (x)  b for each x ∈ V (G ). If bind(G ) > (a + b − 1)( p − 1)/(ap − (a + b) + 2) and p  (a + b)(a + b − 3)/a, then G has a fractional f -factor. In Theorem 5, if a = b = k, then we obtain the following corollary. Corollary 2. Let G be a graph of order p, and let k and n be nonnegative integers such that k  2. If bind(G ) > (2k − 1)( p − 1)/(k( p − 2) − kn + 2) and p  4k − 6 + kn/ (k − 1), then G is fractional (k, n)-critical. In Corollary 2, if n = 0, then we have the following corollary. Corollary 3. Let G be a graph of order p, and let k be an integers such that k  2. If bind(G ) > (2k − 1)( p − 1)/(k( p − 2) + 2) and p  4k − 6, then G has a fractional k-factor. 2. Preliminary lemmas Let f be an integer-valued function defined on the vertex-set V (G ) of a graph G. If S ⊆ V (G ) then we define f ( S ) = x∈ S f (x). If S and T are disjoint subsets of V (G ) define δG ( S , T ) = f ( S ) + d G − S ( T ) − f ( T ), and if | S |  n define   f n ( S ) = max f (U ): U ⊆ S and |U | = n . (1) if kp is even, and bind(G ) > (2k − 1)( p − 1)/(k( p − 2) + 3), then G has a fractional k-factor; and (2) if kp is odd, and bind(G ) > (2k − 1)( p − 1)/(k( p − 2) + 2), then G has a fractional k-factor. In [13], S. Zhou gave a binding number condition for graphs to be fractional n-critical graphs. Theorem 4. (See [13].) Let n  1 be an integer, and let G be a graph. If vertex-connectivity of G κ (G )  n − 1, and bind(G )  n, then G is fractional n-critical. In this paper, we prove the following result, which is a extension of Theorems 3 and 4. We extend Theorems 3 and 4 to fractional ( f , n)-critical graphs. Theorem 5. Let G be a graph of order p, and let a, b and n be nonnegative integers such that 2  a  b, and let f be an integer-valued function defined on V (G ) such that a  f (x)  b for each x ∈ V (G ). If bind(G ) > (a + b − 1)( p − 1)/(ap − (1) Anstee [1] used an algorithm to obtain a necessary and sufficient condition for a graph to have a fractional f -factor. Liu [6] proved the same condition by graphical methods. Lemma 2.1. (See [1,6].) Let G be a graph, and let f be a nonnegative integer-valued function defined on V (G ). Then G has a fractional f -factor if and only if δG ( S , T )  0 for every S ⊆ V (G ), where   T = x ∈ V (G ) − S: d G − S (x)  f (x) . (2) Lemma 2.2. Let G be a graph, and let n be a nonnegative integer, and let f be a nonnegative integer-valued function defined on V (G ). Then G is fractional ( f , n)-critical if and only if δG ( S , T )  f n ( S ) for all subsets S of V (G ) with | S |  n, where T is defined by (2). Proof. Suppose U ⊆ S ⊆ V (G ) where |U | = n, and S ′ = S \ U and G ′ = G − U . Then G ′ − S ′ = G − S, and if 813 S. Zhou, Q. Shen / Information Processing Letters 109 (2009) 811–815 T ′ = x ∈ V (G ′ ) − S ′ : d G ′ − S ′ (x)  f (x) ,  then T ′ = T , as defined by (2). Also  (3) Case 1. 1  h  b − 1 Let Y = ( V (G ) \ S ) \ N G − S (x1 ). Then x1 ∈ Y \ N G (Y ), so Y = ∅ and N G (Y ) = V (G ), and | N G (Y )|  bind(G )|Y |. Thus, we obtain δG ′ ( S ′ , T ′ ) = f ( S ′ ) + d G ′ − S ′ ( T ′ ) − f ( T ′ ) = f ( S ) − f (U ) + d G − S ( T ) − f ( T ) = δG ( S , T ) − f (U ),    p − 1   N G (Y )  bind(G )|Y | = bind(G ) p − h − | S | , so that ′ Choose x1 ∈ T such that d G − S (x1 ) = h. The proof splits into two cases. i.e. ′ δG ′ ( S , T )  0 ⇐⇒ δG ( S , T )  f (U ). (4) Suppose first that G is fractional ( f , n)-critical, and let S , S ′ , U , T ′ , G ′ be as above. Then G ′ = G − U has a fractional f -factor, and so δG ′ ( S ′ , T ′ )  0 by Lemma 2.1. Thus δG ( S , T )  f (U ) by (4). This holds for every subset U ⊆ S with |U | = n, and so δG ( S , T )  f n ( S ); and this holds for every subset S ⊆ V (G ) with | S |  n. Suppose conversely that δG ( S , T )  f n ( S ) for every subset S ⊆ V (G ) with | S |  n, and let U ⊆ V (G ) with |U | = n, so that δG ( S , T )  f (U ). Let G ′ = G − U . For any S ′ ⊆ V (G ′ ), define T ′ by (3) and S = S ′ ∪ U , so that S ′ = S \ U as before. Then δG ′ ( S ′ , T ′ )  0 by (4). This holds for every subset S ′ ⊆ V (G ′ ), and so G ′ = G − U has a fractional f factor by Lemma 2.1. This holds for every subset U ⊆ V (G ) with |U | = n, and so G is fractional ( f , n)-critical. This completes the proof of Lemma 2.2. ✷ p−1 |S|  p − h − bind(G ) ap − (a + b) − bn + 2 > p −h− a+b−1 . (7) ✷ Subcase 1.1. 3  h  b − 1 According to (6) and the fact that | T |  p − | S |,  bn > a| S | − (b − h) p − | S | (8) = (a + b − h)| S | − (b − h) p . Multiplying (8) by (a + b − 1) and rearranging, and then using (7), 0 > (a + b − 1) (a + b − h)| S | − (b − h) p − bn > (a + b − h) (a + b − 1)( p − h) − ap + (a + b) + bn − 2 − (a + b − 1) (b − h) p + bn 3. The proof of Theorem 5 = (h − 1) ap − bn − (a + b − h)(a + b − 1) Proof. Suppose that G is not fractional ( f , n)-critical. Then, by Lemma 2.2, there exists some subset S ⊆ V (G ) with | S |  n such that δG ( S , T )  f n ( S ) − 1, (5) where T = {x: x ∈ V (G ) − S , d G − S (x)  f (x)}. We choose such subsets S and T so that | T | is as small as possible. Claim 1. d G − S (x)  f (x) − 1  b − 1 for all x ∈ T . > (h − 2)(a + b − h) (10) since, by the hypotheses of the theorem, ap  (a + b)(a + b − 3) + abn a−1 > (a + b)(a + b − h) + bn. (In verifying the coefficient of p in (9), it may help to note that (x − h)(b − 1) − (x − 1)(b − h) = (h − 1)(x − b), Proof. If d G − S (x)  f (x) for some x ∈ T , then the subsets S and T \ {x} satisfy (5). This contradicts the choice of S and T . This completes the proof of Claim 1. ✷ If T = ∅, then by (1) and (5), f ( S ) − 1  f n ( S ) − 1  δG ( S , T ) = f ( S ), a contradiction. Hence, T = ∅. Let   h = min d G − S (x): x ∈ T . and to substitute x = a + b.) But (10) is impossible, since 3  h  b − 1 < a + b. Subcase 1.2. h = 2 Obviously, b  3. We prove the following claim. Claim 2. (b − 2)(a + b − 1)| T |  a[( p − 2)(a + b − 1) − ap + a + b + bn − 2] − (a + b − 1)bn + (a + b − 1), that is, |T |  According to Claim 1, we have a b−2 bn − 0  h  b − 1. Since f (x)  b for every vertex x, it follows from (1) and (5) that bn − 1  f n ( S ) − 1  δG ( S , T ), and |T |  so that (6) p−2− b−2 + ap − (a + b) − bn + 2 a+b−1 1 b−2 . Proof. If (b − 2)(a + b − 1)| T |  a[( p − 2)(a + b − 1) − ap + a + b + bn − 2] − (a + b − 1)bn + (a + b − 1) + 1, we deduce δG ( S , T ) = f ( S ) + d G − S ( T ) − f ( T )  a| S | + h| T | − b| T |, bn − 1  a| S | − (b − h)| T |. (9) − (a + b − h) a b−2 bn − p−2− b−2 + ap − (a + b) − bn + 2 1 b−2 + a+b−1 1 (b − 2)(a + b − 1) . 814 S. Zhou, Q. Shen / Information Processing Letters 109 (2009) 811–815 Then, by (7) and p  (a + b)(a + b − 3)/a + bn/(a − 1), we obtain | S | + |T | > p − 2 − ap − (a + b) − bn + 2 a+b−1 ap − (a + b) − bn + 2 a p−2− + b−2 a+b−1 − bn 1 + 1 + b−2 + 1 − b−2 bn 1 + 1 + b−2 b−2 (b − 2)(a + b − 1) (a + b − 1)bn − (a + b) p+ (b − 2)(a + b − 1) bn b−2 + 1 b−2 + This contradicts | S | + | T |  p. This completes the proof of Claim 2. ✷ p−1− ap − (a + b) − bn + 2 a+b−1 bn − 1 b−1 p−1− ap − (a + b) − bn + 2 a+b−1 bn − 1 This contradicts | S | + | T |  p. Case 2. h = 0 At first, we prove the following claim. Claim 3. = p. b−1 a+b−1 b−1 (b − 1) p + bn − 1 bn − 1 − = p. = b−1 b−1 1 (b − 2)(a + b − 1) a b−1 − (b − 2)(a + b − 1) (a + b)(a + b − 3) + a−a 1 bn ap −(a+b)−bn+2 −1 ) − bn a+b−1 b−1 ap − (a + b) − bn + 2 a+b−1 = 1 + (b − 2)(a + b − 1) (a + b)(a + b − 2) − (a + b − 2)bn − (b − 2)(a + b − 1) − + (b − 2)(a + b − 1) bn p+ − | S | + |T | > p − 1 − b−2 b−2 (b − 2)(a + b − 1) ap − (a + b)(a + b − 2) + (a + b − 2)bn =p+ − a Subcase 1.3.2. | T | > b− (p − 1 − 1 In view of (7), we obtain ap −(a+b)−bn+2 p −1 > 1. Proof. Since p  (a + b)(a + b − 3)/a + bn/(a − 1), then we have ap − (a + b) − bn + 2 − ( p − 1) = (a − 1) p − (a + b) − bn + 3 By combining Claim 2 with (6) and (7), we obtain  (a − 1) (a + b)(a + b − 3) a bn − 1  a| S | − (b − 2)| T | >a p−2− − (b − 2) − bn b−2 + ap − (a + b) − bn + 2 a+b−1 a b−2 p−2− = ap − (a + b) − bn + 2 a+b−1 b−2 Thus, we have ap − (a + b) − bn + 2 p−1 ap −(a+b)−bn+2 a −1 Subcase 1.3.1. | T |  b− (p − 1 − ) − bn 1 a+b−1 b−1 By combining this with (6) and (7), we have that bn − 1  a| S | − (b − 1)| T | a+b−1 b−1 p−1−  = bn − 1, a contradiction.   p − m   N G (Y )  bind(G )|Y | = bind(G ) p − | S | . ap − (a + b) − bn + 2 a ✷ Let m be the number of vertices x in T such that d G − S (x) = 0, and let Y = V (G ) \ S. Then N G (Y ) = V (G ) since h = 0, and Y = ∅ by T = ∅, and so | N G (Y )|  bind(G )|Y |. Thus So ap − (a + b) − bn + 2 a+b−1 |S|  p − bn − 1 b−1 > 1. This completes the proof of Claim 3. Subcase 1.3. h = 1 − − (a + b) + 3  2(a + b − 3) − (a + b) + 3 a contradiction. − (b − 1) a bn a−1 = a + b − 3 > 0. 1 = bn − 1, >a p−1− − (a + b) − bn + 3 (a − 1)(a + b)(a + b − 3) + >p− p −m bind(G ) ( p − m)(ap − (a + b) − bn + 2) (a + b − 1)( p − 1) . (11) In view of (5) and (11), and | T |  p − | S |, and Claim 3, we get that 815 S. Zhou, Q. Shen / Information Processing Letters 109 (2009) 811–815 bn − 1  f n ( S ) − 1  δG ( S , T ) = f ( S ) + d G − S ( T ) − f ( T )  a| S | + | T | − m − b| T | = a| S | − (b − 1)| T | − m   a| S | − (b − 1) p − | S | − m δ H ( S , T ) = f ( S ) + d H − S ( T ) − f ( T ) = a| S | + | T | − b | T | = a| S | − (b − 1)| T | =a ( p − m)(ap − (a + b) − bn + 2) (a + b − 1)( p − 1) − (b − 1) p − m = ap − ( p − m)(ap − (a + b) − bn + 2) −m p−1  ap − ( p − 1)(ap − (a + b) − bn + 2) −1 p−1 = bn − 1 < bn = f n ( S ). By Lemma 2.2, H is not a fractional ( f , n)-critical graph. In the above sense, the result of Theorem 5 is best possible. Acknowledgements = bn + (a + b) − 3 The author would like to express his gratitude to the anonymous referees for their very helpful comments and suggestions in improving this paper. > bn. References This is a contradiction. ✷ From all the cases above, we deduced the contradiction. Hence, G is a fractional ( f , n)-critical graph. This completes the proof of Theorem 5. Remark. Let us show that the condition bind(G ) > (a + b − 1)( p − 1)/(ap − (a + b) − bn + 2) in Theorem 5 cannot be replaced by bind(G )  (a + b − 1)( p − 1)/(ap − (a + b) − bn + 2). Let 2  a = b, n  0 be three integers such that n is odd, p= a − (b − 1)(a + b + n − 1) = (a + b − 1)| S | − (b − 1) p − m > (a + b − 1) p − (a + b − 1)(b − 2) + (a + b − 2) + (2b − 1)n (a + b − 1)(a + b − 2) + (a + b − 2) + (a + 2b − 1)n a be an integer, and let l = (a + b + n − 1)/2 and m = p − 2l = p − (a + b + n − 1) = ((a + b − 1)(b − 2) + (a + b − 2) + (2b − 1)n)/a. Clearly, m is an integer. Let H = K m ∨ lK 2 . 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