Hindawi Publishing Corporation e Scientific World Journal Volume 2014, Article ID 369798, 7 pages http://dx.doi.org/10.1155/2014/369798

Research Article Toughness Condition for a Graph to Be a Fractional (𝑔, 𝑓, 𝑛)-Critical Deleted Graph Wei Gao1 and Yun Gao2 1 2

School of Information Science and Technology, Yunnan Normal University, Kunming 650500, China Editorial Department of Yunnan Normal University, Kunming 650092, China

Correspondence should be addressed to Yun Gao; [email protected] Received 16 January 2014; Revised 21 June 2014; Accepted 21 June 2014; Published 9 July 2014 Academic Editor: Abdelalim A. Elsadany Copyright © 2014 W. Gao and Y. Gao. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. A graph 𝐺 is called a fractional (𝑔, 𝑓)-deleted graph if 𝐺 − {𝑒} admits a fractional (𝑔, 𝑓)-factor for any 𝑒 ∈ 𝐸(𝐺). A graph 𝐺 is called a fractional (𝑔, 𝑓, 𝑛)-critical deleted graph if, after deleting any 𝑛 vertices from 𝐺, the resulting graph is still a fractional (𝑔, 𝑓)-deleted graph. The toughness, as the parameter for measuring the vulnerability of communication networks, has received significant attention in computer science. In this paper, we present the relationship between toughness and fractional (𝑔, 𝑓, 𝑛)critical deleted graphs. It is determined that 𝐺 is fractional (𝑔, 𝑓, 𝑛)-critical deleted if 𝑡(𝐺) ≥ ((𝑏2 − 1 + 𝑏𝑛)/𝑎).

1. Introduction All graphs considered in this paper are finite, are loopless, and are without multiple edges. The notation and terminology used but undefined in this paper can be found in [1]. Let 𝐺 be a graph with the vertex set 𝑉(𝐺) and the edge set 𝐸(𝐺). For a vertex 𝑥 ∈ 𝑉(𝐺), we use 𝑑𝐺(𝑥) and 𝑁𝐺(𝑥) to denote the degree and the neighborhood of 𝑥 in 𝐺, respectively. Let 𝛿(𝐺) denote the minimum degree of 𝐺. For any 𝑆 ⊆ 𝑉(𝐺), the subgraph of 𝐺 induced by 𝑆 is denoted by 𝐺[𝑆]. The problem of fractional factor can be considered as a relaxation of the well-known cardinality matching problem. It has wide-ranging applications in areas such as scheduling, network design, and the combinatorial polyhedron. For instance, several large data packets are to be sent to various destinations through several channels in a communication network. The efficiency of this work can be improved if large data packets are 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 factor problem when the destinations and sources of a network are disjoint. Suppose that 𝑔 and 𝑓 are two integer-valued functions on 𝑉(𝐺) such that 0 ≤ 𝑔(𝑥) ≤ 𝑓(𝑥) for all 𝑥 ∈ 𝑉(𝐺). A fractional (𝑔, 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 𝑔(𝑥) ≤ ∑𝑒∈𝐸(𝑥) ℎ(𝑒) ≤ 𝑓(𝑥). If 𝑔(𝑥) = 𝑎 and 𝑓(𝑥) = 𝑏 for all 𝑥 ∈ 𝑉(𝐺), then a fractional (𝑔, 𝑓)-factor is a fractional [𝑎, 𝑏]-factor. Moreover, if 𝑔(𝑥) = 𝑓(𝑥) = 𝑘 for all 𝑥 ∈ 𝑉(𝐺), then a fractional (𝑔, 𝑓)-factor is just a fractional 𝑘-factor. Throughout this paper, 𝑘 ≥ 1 is an integer, and we will not reiterate it again. A graph 𝐺 is called a fractional (𝑔, 𝑓, 𝑛)-critical graph if, after deleting any 𝑛 vertices from 𝐺, the resulting graph still has a fractional (𝑔, 𝑓)-factor. A graph 𝐺 is called a fractional (𝑔, 𝑓, 𝑚)-deleted graph if, after deleting any 𝑚 edges, the resulting graph still has a (𝑔, 𝑓)-factor. Fractional deleted graph and fractional critical graph, as extensions of the concept of fractional factor, describe the existence of fractional factor in communication networks when certain channels or certain sites are damaged. Gao [2] proposed a new concept to deal with the combination situation when some channels and some sites are unavailable in networks. A graph 𝐺 is called a fractional (𝑔, 𝑓, 𝑛, 𝑚)-critical deleted graph if, after deleting any 𝑛 vertices from 𝐺, the resulting graph is still a fractional (𝑔, 𝑓, 𝑚)-deleted graph. In particular, the fractional (𝑔, 𝑓, 𝑛, 𝑚)-critical deleted graph is just fractional (𝑔, 𝑓, 𝑛)critical deleted graph if 𝑚 = 1.

2

The Scientific World Journal Let 2, { { { {1, 𝜀 (𝑆, 𝑇) = { { { { {0,

𝑇 is not an independent set 𝑇 is an independent set, and 𝑒𝐺 (𝑇, 𝑉 (𝐺) \ (𝑆 ∪ 𝑇)) ≥ 1 otherwise.

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We heavily depend on the following lemma to prove our main result, which determined a necessary and sufficient condition for a graph to be a fractional (𝑔, 𝑓, 𝑛)-critical deleted graph. Lemma 1 (Gao [2]). Let 𝐺 be a graph and let 𝑔, 𝑓 be two nonnegative integer-valued functions defined on 𝑉(𝐺) satisfying 𝑔(𝑥) ≤ 𝑓(𝑥) for all 𝑥 ∈ 𝑉(𝐺). Let 𝑛 be a nonnegative integer. Then 𝐺 is a fractional (𝑔, 𝑓, 𝑛)-critical deleted graph if and only if 𝑓 (𝑆) − 𝑔 (𝑇) + 𝑑𝐺−𝑆 (𝑇) ≥ max {𝑓 (𝑈) : 𝑈 ⊆ 𝑆, |𝑈| = 𝑛} + 𝜀 (𝑆, 𝑇) ,

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for any disjoint subsets 𝑆 and 𝑇 of 𝑉(𝐺) with |𝑆| ≥ 𝑛. The notion of toughness was first introduced by Chv´atal in [3] to measure the vulnerable of networks: if 𝐺 is complete graph, 𝑡(𝐺) = ∞; if 𝐺 is not complete, |𝑆| 𝜔 (𝐺 − 𝑆) ≥ 2} , 𝑡 (𝐺) = min { 𝜔 (𝐺 − 𝑆)

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where 𝜔(𝐺 − 𝑆) is the number of connected components of 𝐺 − 𝑆. Liu and Zhang [4] determined a necessary and sufficient condition for a graph to have a fractional (𝑔, 𝑓)-factor. For several characters on fractional (𝑔, 𝑓)-factor one can refer to Liu and Zhang [4, 5] for more details. Liu [6] investigated the necessary and sufficient condition for a graph 𝐺 to be a fractional (𝑔, 𝑓, 𝑛)-critical graph. For more recent results for fractional deleted graph and fractional critical graph one can refer to [7–12]. Some toughness conditions for a graph to have a fraction factor were given in [13, 14]. Liu et al. [6] studied the relationship between toughness and fractional (𝑔, 𝑓, 𝑛)-critical graphs and proved that 𝐺 is a fractional (𝑔, 𝑓, 𝑛)-critical graph if 𝑡(𝐺) ≥ ((𝑏2 − 1)(𝑛 + 1)/𝑎) for 𝑎 ≤ 𝑔(𝑥) ≤ 𝑓(𝑥) ≤ 𝑏 with 1 ≤ 𝑎 ≤ 𝑏 and 𝑏 ≥ 2. Zhou et al. [15] studied the toughness condition for fractional (𝑘, 𝑚)-deleted graph. It is determined that 𝐺 is a fractional (𝑘, 𝑚)-deleted graph if 𝑡(𝐺) ≥ 𝑘 + ((2𝑚 − 1)/𝑘). Recently, in [16], Gao et al. derived a new bound for graphs to be fractional (𝑔, 𝑓, 𝑛)-critical. It is verified that 𝐺 is a fractional (𝑔, 𝑓, 𝑛)-critical graph if 𝑡(𝐺) ≥ ((𝑏2 − 1 + 𝑏𝑛)/𝑎). This inspires us to think about the 𝑡(𝐺) for fractional (𝑔, 𝑓, 𝑛)-critical deleted graphs. In this paper, we determine that such bound of toughness as above is sufficient for a graph 𝐺 to be a fractional (𝑔, 𝑓, 𝑛)-critical deleted graph. Our main result to be proved in next section can be stated as follows. Theorem 2. Let 𝐺 be a graph and let 𝑔, 𝑓 be two integer-valued functions defined on 𝑉(𝐺) satisfying 𝑎 ≤ 𝑔(𝑥) ≤ 𝑓(𝑥) ≤ 𝑏 with

1 ≤ 𝑎 ≤ 𝑏 and 𝑏 ≥ 2 for all 𝑥 ∈ 𝑉(𝐺), where 𝑎, 𝑏 are positive integers. Let 𝑛 be a nonnegative integer. |𝑉(𝐺)| ≥ 𝑛 + 𝑏 + 2 if 𝐺 is complete. If 𝑡(𝐺) ≥ ((𝑏2 − 1 + 𝑏𝑛)/𝑎), then 𝐺 is a fractional (𝑔, 𝑓, 𝑛)-critical deleted graph. Clearly, our result strengthened the previous conclusions, and it is sharp if 𝑎 = 𝑏 and 𝑛 = 0 according to the sharpness example in Liu and Zhang [17]. The proof strategy is similar to the one in Liu and Zhang [17], but we need to cope with the more detailed case now and hence new methods are necessary. Before proving Theorem 2, we would like to show some useful lemmas. Lemma 3 (Chv´atal [3]). If a graph 𝐺 is not complete, then 𝑡(𝐺) ≤ (1/2)𝛿(𝐺). Lemma 4 (Liu and Zhang [17]). Let 𝐺 be a graph and let 𝐻 = 𝐺[𝑇] such that 𝛿(𝐻) ≥ 1 and 1 ≤ 𝑑𝐺(𝑥) ≤ 𝑘 − 1 for every 𝑥 ∈ 𝑉(𝐻), where 𝑇 ⊆ 𝑉(𝐺) and 𝑘 ≥ 2. Let 𝑇1 , . . . , 𝑇𝑘−1 be a partition of the vertices of 𝐻 satisfying 𝑑𝐺(𝑥) = 𝑗 for each 𝑥 ∈ 𝑇𝑗 , where one allows some 𝑇𝑗 to be empty. If each component of 𝐻 has a vertex of degree at most 𝑘 − 2 in 𝐺, then 𝐻 has a maximal independent set 𝐼 and a covering set 𝐶 = 𝑉(𝐻) − 𝐼 such that 𝑘−1

𝑘−1

𝑗=1

𝑗=1

∑ (𝑘 − 𝑗) 𝑐𝑗 ≤ ∑ (𝑘 − 2) (𝑘 − 𝑗) 𝑖𝑗 ,

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where 𝑐𝑗 = |𝐶 ∩ 𝑇𝑗 | and 𝑖𝑗 = |𝐼 ∩ 𝑇𝑗 | for every 𝑗 = 1, . . . , 𝑘 − 1. The lemma below can be deduced from Lemma 2.2 in [17]. Lemma 5 (Liu and Zhang [17]). Let 𝐺 be a graph and let 𝐻 = 𝐺[𝑇] such that 𝑑𝐺(𝑥) = 𝑘 − 1 for every 𝑥 ∈ 𝑉(𝐻) and no component of 𝐻 is isomorphic to 𝐾𝑘 , where 𝑇 ⊆ 𝑉(𝐺) and 𝑘 ≥ 2. Then there exist an independent set 𝐼 and the covering set 𝐶 = 𝑉(𝐻) − 𝐼 of 𝐻 satisfying 󵄨󵄨 (1) 󵄨󵄨 𝑘 󵄨 󵄨 󵄨󵄨𝐼 󵄨󵄨 |𝑉 (𝐻)| ≤ ∑ (𝑘 − 𝑖 + 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 − 󵄨 󵄨 , 2 𝑖=1 (5) 󵄨󵄨 (1) 󵄨󵄨 𝑘 󵄨󵄨 (𝑖) 󵄨󵄨 󵄨󵄨󵄨𝐼 󵄨󵄨󵄨 , |𝐶| ≤ ∑ (𝑘 − 𝑖) 󵄨󵄨󵄨𝐼 󵄨󵄨󵄨 − 2 𝑖=1 where 𝐼(𝑖) = {𝑥 ∈ 𝐼, 𝑑𝐻(𝑥) = 𝑘 − 𝑖}, 1 ≤ 𝑖 ≤ 𝑘, and ∑𝑘𝑖=1 |𝐼(𝑖) | = |𝐼|.

2. Proof of Theorem 2 If 𝐺 is complete, then 𝐺 is a fractional (𝑔, 𝑓, 𝑛)-critical deleted graph due to |𝑉(𝐺)| ≥ 𝑛 + 𝑏 + 2. In what follows, we assume that 𝐺 is not complete. Suppose that 𝐺 satisfies the conditions of Theorem 2 but is not a fractional (𝑔, 𝑓, 𝑛)-critical graph. By Lemma 1 and 𝜀(𝑆, 𝑇) ≤ 2, there exist subsets 𝑆 and 𝑇 of 𝑉(𝐺) such that 𝑎 |𝑆| + ∑ 𝑑𝐺−𝑆 (𝑥) − 𝑏 |𝑇| 𝑥∈𝑇

≤ 𝑓 (𝑆) − 𝑔 (𝑇) + 𝑑𝐺−𝑆 (𝑇) ≤ 𝑏𝑛 + 1.

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The Scientific World Journal

3

We choose subsets 𝑆 and 𝑇 such that |𝑇| is minimum. Obviously, we deduce 𝑇 ≠ 0 and 𝑑𝐺−𝑆 (𝑥) ≤ 𝑔(𝑥) − 1 ≤ 𝑏 − 1 for any 𝑥 ∈ 𝑇. Let 𝑙 be the number of the components of 𝐻󸀠 = 𝐺[𝑇] which are isomorphic to 𝐾𝑏 and let 𝑇0 = {𝑥 ∈ 𝑉(𝐻󸀠 ) | 𝑑𝐺−𝑆 (𝑥) = 0}. Let 𝐻 be the subgraph obtained from 𝐻󸀠 − 𝑇0 by deleting those 𝑙 components isomorphic to 𝐾𝑏 . If |𝑉(𝐻)| = 0, then by virtue of (6) we infer 󵄨 󵄨 𝑎 |𝑆| ≤ 𝑏 󵄨󵄨󵄨𝑇0 󵄨󵄨󵄨 + 𝑏𝑙 + 𝑏𝑛 + 1,

(7)

or 󵄨 󵄨 𝑏 (󵄨󵄨󵄨𝑇0 󵄨󵄨󵄨 + 𝑙) + 𝑏𝑛 + 1 ≤ . |𝑆| 𝑎

𝑏−1 𝑏 󵄨 󵄨 (13) 󵄨 󵄨 ≥ ∑ (𝑡 − 𝑗) 𝑖𝑗 + 𝑡 (𝑡0 + 𝑙) + 𝑡 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 − ∑ (𝑖 − 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 . 𝑗=1

𝑖=1

In view of 𝑏|𝑇| − 𝑑𝐺−𝑆 (𝑇) ≥ 𝑎|𝑆| − 𝑏𝑛 − 1, we obtain

󵄨󵄨 (1) 󵄨󵄨 󵄨󵄨 (𝑖) 󵄨󵄨 󵄨󵄨󵄨𝐼 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 𝐼 − ≤ 𝐶 , − 𝑖) ∑ (𝑏 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 𝑖=1

(9)

𝑗=1

Combining with (13), we deduce 𝑏−1

󵄨 󵄨 𝑏𝑡0 + 𝑏𝑙 + 󵄨󵄨󵄨𝑉 (𝐻1 )󵄨󵄨󵄨 + ∑ (𝑏 − 𝑗) 𝑖𝑗 𝑗=1

𝑏−1

󵄨 󵄨 + ∑ (𝑏 − 𝑗) 𝑐𝑗 + 𝑎 󵄨󵄨󵄨𝐶1 󵄨󵄨󵄨 + 𝑏𝑛 + 1 𝑗=1

𝑗=1

𝑖=1

(10)

󵄨 󵄨 󵄨 󵄨 |𝑈| ≤ |𝑆| + 󵄨󵄨󵄨𝐶1 󵄨󵄨󵄨 + ∑ 𝑗𝑖𝑗 + ∑ (𝑖 − 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 , 𝑏−1

≥ ∑ (𝑎𝑡 − 𝑎𝑗) 𝑖𝑗 + 𝑎𝑡 (𝑡0 + 𝑙) 𝑗=1

𝑏 󵄨 󵄨 󵄨 󵄨 + 𝑎𝑡 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 − 𝑎∑ (𝑖 − 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 .

Therefore,

𝑏−1

󵄨 󵄨 ≥ ∑ (𝑎𝑡 − 𝑎𝑗 − 𝑏 + 𝑗) 𝑖𝑗 + (𝑎𝑡 − 𝑏) (𝑡0 + 𝑙) + 𝑎𝑡 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 (16) 𝑗=1

𝑏

󵄨 󵄨 − 𝑎∑ (𝑖 − 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 − 𝑏𝑛 − 1. 𝑖=1

By virtue of (9), we have 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨𝑉 (𝐻1 )󵄨󵄨󵄨 + 𝑎 󵄨󵄨󵄨𝐶1 󵄨󵄨󵄨

󵄨󵄨 (1) 󵄨󵄨 󵄨󵄨 (𝑖) 󵄨󵄨 (𝑎 + 1) 󵄨󵄨󵄨𝐼 󵄨󵄨󵄨 󵄨 󵄨 ≤ ∑ (𝑎𝑏 − 𝑎𝑖 + 𝑏 − 𝑖 + 1) 󵄨󵄨𝐼 󵄨󵄨 − . 2 𝑏

(17)

𝑖=1

Using (10), (16), and (17), we get 𝑏−1

𝑏 󵄨 󵄨 ∑ (𝑏 − 2) (𝑏 − 𝑗) 𝑖𝑗 + ∑ (𝑎𝑏 − 𝑎𝑖 + 𝑏 − 𝑖 + 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨

(11)

𝑗=1

𝑖=1

󵄨󵄨 (1) 󵄨󵄨 󵄨󵄨 󵄨󵄨 (𝑎 + 1) 󵄨󵄨󵄨𝐼 󵄨󵄨󵄨 ≥ ∑ (𝑎𝑡 − 𝑎𝑗 − 𝑏 + 𝑗) 𝑖𝑗 + 𝑎𝑡 󵄨󵄨𝐼1 󵄨󵄨 + 2 𝑏−1

󵄨 󵄨 𝜔 (𝐺 − 𝑈) ≥ 𝑡0 + 𝑙 + 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 + ∑ 𝑖𝑗 , 𝑗=1

𝑗=1

where 𝑡0 = |𝑇0 |. Let 𝑡(𝐺) = 𝑡. Then when 𝜔(𝐺 − 𝑈) > 1, we have |𝑈| ≥ 𝑡𝜔 (𝐺 − 𝑈) ,

(15)

𝑏−1

𝑗=1

where 𝑐𝑗 = |𝐶2 ∩𝑇𝑗 | and 𝑖𝑗 = |𝐼2 ∩𝑇𝑗 | for every 𝑗 = 1, . . . , 𝑏−1. Set 𝑊 = 𝑉(𝐺) − 𝑆 − 𝑇 and 𝑈 = 𝑆 ∪ 𝐶1 ∪ (𝑁𝐺(𝐼1 ) ∩ 𝑊) ∪ 𝐶2 ∪ (𝑁𝐺(𝐼2 ) ∩ 𝑊). We infer 𝑏

(14)

𝑏−1

𝑗=1

𝑏−1

𝑗=1

󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨𝑉 (𝐻1 )󵄨󵄨󵄨 + ∑ (𝑏 − 𝑗) 𝑐𝑗 + 𝑎 󵄨󵄨󵄨𝐶1 󵄨󵄨󵄨

𝑏−1

∑ (𝑏 − 𝑗) 𝑐𝑗 ≤ ∑ (𝑏 − 2) (𝑏 − 𝑗) 𝑖𝑗 ,

𝑗=1

𝑖=1

where 𝐼(𝑖) = {𝑥 ∈ 𝐼1 , 𝑑𝐻1 (𝑥) = 𝑏−𝑖}, 1 ≤ 𝑖 ≤ 𝑏, and ∑𝑏𝑖=1 |𝐼(𝑖) | = |𝐼1 |. Let 𝑇𝑗 = {𝑥 ∈ 𝑉(𝐻2 ) | 𝑑𝐺−𝑆 (𝑥) = 𝑗} for 1 ≤ 𝑗 ≤ 𝑏−1. Each component of 𝐻2 has a vertex of degree at most 𝑏 − 2 in 𝐺 − 𝑆 by the definitions of 𝐻 and 𝐻2 . According to Lemma 4, 𝐻2 has a maximal independent set 𝐼2 and the covering set 𝐶2 = 𝑉(𝐻2 ) − 𝐼2 such that 𝑏−1

𝑏−1

≥ 𝑎 |𝑆| − 𝑏𝑛 − 1.

𝑏

𝑏

𝑏−1

󵄨 󵄨 𝑏𝑡0 + 𝑏𝑙 + 󵄨󵄨󵄨𝑉 (𝐻1 )󵄨󵄨󵄨 + ∑ (𝑏 − 𝑗) 𝑖𝑗 + ∑ (𝑏 − 𝑗) 𝑐𝑗

(8)

If 𝜔(𝐺−𝑆) = |𝑇0 |+𝑙 > 1, then 𝑡(𝐺) ≤ (|𝑆|/𝜔(𝐺−𝑆)) ≤ ((𝑏(|𝑇0 |+ 𝑙) + 𝑏𝑛 + 1)/𝑎(|𝑇0 | + 𝑙)) < ((𝑏 + 𝑏𝑛 + 1)/𝑎), which contradicts 𝑡(𝐺) ≥ ((𝑏2 − 1 + 𝑏𝑛)/𝑎) and 𝑏 ≥ 2. If 𝜔(𝐺 − 𝑆) = |𝑇0 | + 𝑙 = 1, then |𝑇0 | + 𝑙 = 1. Since 𝑑𝐺−𝑆 (𝑥) + |𝑆| ≥ 𝑑𝐺(𝑥) ≥ 𝛿(𝐺) ≥ 2𝑡(𝐺), we have 2𝑡(𝐺) ≤ 𝑏 − 1 + |𝑆| ≤ 𝑏 − 1 + ((𝑏(𝑛 + 1) + 1)/𝑎), which contradicts 𝑡(𝐺) ≥ ((𝑏2 − 1 + 𝑏𝑛)/𝑎). Now we consider that |𝑉(𝐻)| > 0. Let 𝐻 = 𝐻1 ∪ 𝐻2 , where 𝐻1 is the union of components of 𝐻 which satisfies that 𝑑𝐺−𝑆 (𝑥) = 𝑏 − 1 for every vertex 𝑥 ∈ 𝑉(𝐻1 ) and 𝐻2 = 𝐻 − 𝐻1 . In terms of Lemma 5, 𝐻1 has a maximum independent set 𝐼1 and the covering set 𝐶1 = 𝑉(𝐻1 ) − 𝐼1 such that 󵄨󵄨 (1) 󵄨󵄨 󵄨󵄨 (𝑖) 󵄨󵄨 󵄨󵄨󵄨𝐼 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 , 󵄨󵄨𝑉 (𝐻1 )󵄨󵄨 ≤ ∑ (𝑏 − 𝑖 + 1) 󵄨󵄨󵄨𝐼 󵄨󵄨󵄨 − 2 𝑖=1

and it also holds when 𝜔(𝐺 − 𝑈) = 1. In terms of (11) and (12), we get 󵄨 󵄨 |𝑆| + 󵄨󵄨󵄨𝐶1 󵄨󵄨󵄨

(12)

𝑏 󵄨 󵄨 − 𝑎∑ (𝑖 − 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 + (𝑎𝑡 − 𝑏) (𝑡0 + 𝑙) − 𝑏𝑛 − 1. 𝑖=1

(18)

4

The Scientific World Journal

The following proof splits into two cases by the value of 𝑡0 + 𝑙.

Let 3 1 ℎ1 (𝑏) = −𝑏2 + (𝑎 + 1) 𝑏 − 𝑎 + . 2 2

2

Case 1 (𝑡0 + 𝑙 ≥ 1). By 𝑎𝑡 ≥ 𝑏 − 1 + 𝑏𝑛, we have (𝑎𝑡 − 𝑏)(𝑡0 + 𝑙) − 𝑏𝑛 − 1 ≥ 𝑏2 − 𝑏 − 2 ≥ 0. Hence, (18) becomes 𝑏−1

𝑏

𝑗=1

𝑖=1

From 𝑏 ≥ 𝑎 and ℎ1 (2) < 0, if 𝑎 = 1, we deduce

󵄨 󵄨 ∑ (𝑏 − 2) (𝑏 − 𝑗) 𝑖𝑗 + ∑ (𝑎𝑏 − 𝑎𝑖 + 𝑏 − 𝑖 + 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨

𝑎 1 max {ℎ1 (𝑏)} = ℎ1 (𝑎) = − + < 0. 2 2

𝑏−1

󵄨 󵄨 ≥ ∑ (𝑎𝑡 − 𝑎𝑗 − 𝑏 + 𝑗) 𝑖𝑗 + 𝑎𝑡 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨

(19)

𝑗=1

+

󵄨 󵄨 (𝑎 + 1) 󵄨󵄨󵄨󵄨𝐼(1) 󵄨󵄨󵄨󵄨 2

ℎ2 (𝑏) = −𝑏2 + (𝑎 + 1) 𝑏 − 𝑎.

And then, at least one of the following two cases must hold. is at least one 𝑗 such that

≥

∑𝑏−1 𝑗=1 (𝑎𝑡 − 𝑎𝑗 − 𝑏 + 𝑗)𝑖𝑗 ).

(25)

From 𝑏 ≥ 𝑎, we infer

𝑖=1

Subcase 1 (∑𝑏−1 𝑗=1 (𝑏 − 2)(𝑏 − 𝑗)𝑖𝑗

(24)

Furthermore, 𝑎𝑏 + 𝑏 − 𝑎 − 𝑖 + 2 − 𝑏2 ≤ −𝑏2 + (𝑎 + 1)𝑏 − 𝑎 due to 𝑖 ≥ 2. Let

𝑏

󵄨 󵄨 − 𝑎∑ (𝑖 − 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 .

(23)

There

(𝑏 − 2) (𝑏 − 𝑗) ≥ 𝑎𝑡 − 𝑎𝑗 − 𝑏 + 𝑗,

(20)

which implies

max {ℎ2 (𝑏)} = ℎ2 (𝑎) = 0.

(26)

This is a contradiction. If 𝑛 ≥ 1, we obtain 𝑏 󵄨 󵄨 ∑ (𝑎𝑏 − 𝑎𝑖 + 𝑏 − 𝑖 + 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 𝑖=1

𝑎𝑡 ≤ (𝑏 − 2) (𝑏 − 𝑗) + 𝑎𝑗 + 𝑏 − 𝑗 = 𝑏 (𝑏 − 2) + (𝑎 − 𝑏 + 1) 𝑗 + 𝑏.

󵄨󵄨 (1) 󵄨󵄨 𝑏 󵄨 󵄨 󵄨 󵄨 (𝑎 + 1) 󵄨󵄨󵄨𝐼 󵄨󵄨󵄨 ≥ (𝑏2 − 1 + 𝑏𝑛) 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 + − 𝑎∑ (𝑖 − 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 2 𝑖=1

(21)

If 𝑎 = 𝑏, then 𝑎𝑡 ≤ 𝑎(𝑎 − 2) + 𝑗 + 𝑎 ≤ 𝑎2 − 1. By 𝑡(𝐺) ≥ ((𝑏2 − 1 + 𝑏𝑛)/𝑎), we get 𝑛 = 0 and ∑𝑏−2 𝑗=1 𝑖𝑗 = 0, which contradicts the definition of 𝐻2 and the choice of 𝐼2 (see Lemma 4 proof in [17] such that ∑𝑏−2 𝑗=1 𝑖𝑗 ≠ 0). If 𝑎 < 𝑏, then 𝑎𝑡 ≤ 𝑏(𝑏 − 2) + (𝑎 − 𝑏 + 1) + 𝑏 = (𝑏2 − 1) + (𝑎 − 𝑏)+(2−𝑏) < 𝑏2 −1, which contradicts 𝑡(𝐺) ≥ ((𝑏2 −1+𝑏𝑛)/𝑎).

󵄨󵄨 (1) 󵄨󵄨 𝑏 󵄨 󵄨 󵄨󵄨 󵄨󵄨 (𝑎 + 1) 󵄨󵄨󵄨𝐼 󵄨󵄨󵄨 ≥ (𝑏 − 1) 󵄨󵄨𝐼1 󵄨󵄨 + − 𝑎∑ (𝑖 − 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 + 2. 2 2

𝑖=1

(27) Hence, 𝑏

󵄨 󵄨 ∑ (𝑎𝑏 + 𝑏 − 𝑎 − 𝑖 + 2 − 𝑏2 ) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨

Subcase 2 (∑𝑏𝑖=1 (𝑎𝑏−𝑎𝑖+𝑏−𝑖+1)|𝐼(𝑖) | ≥ 𝑎𝑡|𝐼1 |+((𝑎+1)|𝐼(1) |/2)− 𝑎 ∑𝑏𝑖=1 (𝑖 − 1)|𝐼(𝑖) |). If 𝑡0 + 𝑙 ≥ 2 or 𝑏 ≥ 3, then by (𝑏𝑡 − 𝑎)(𝑡0 + 𝑙) − 𝑏𝑛 − 1 ≥ 1 we have 𝑏 󵄨 󵄨 ∑ (𝑎𝑏 − 𝑎𝑖 + 𝑏 − 𝑖 + 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 𝑖=1

󵄨󵄨 (1) 󵄨󵄨 𝑏 󵄨 󵄨 󵄨󵄨 󵄨󵄨 (𝑎 + 1) 󵄨󵄨󵄨𝐼 󵄨󵄨󵄨 ≥ 𝑎𝑡 󵄨󵄨𝐼1 󵄨󵄨 + − 𝑎∑ (𝑖 − 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 + 1 2 𝑖=1 󵄨󵄨 (1) 󵄨󵄨 𝑏 󵄨 󵄨 󵄨󵄨 󵄨󵄨 (𝑎 + 1) 󵄨󵄨󵄨𝐼 󵄨󵄨󵄨 ≥ (𝑏 − 1 + 𝑏𝑛) 󵄨󵄨𝐼1 󵄨󵄨 + − 𝑎∑ (𝑖 − 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 + 1 2 2

𝑖=1

󵄨󵄨 (1) 󵄨󵄨 𝑏 󵄨 󵄨 󵄨󵄨 󵄨󵄨 (𝑎 + 1) 󵄨󵄨󵄨𝐼 󵄨󵄨󵄨 ≥ (𝑏 − 1) 󵄨󵄨𝐼1 󵄨󵄨 + − 𝑎∑ (𝑖 − 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 + 1, 2

(28)

3 1 󵄨 󵄨 + (𝑎𝑏 + 𝑏 − 𝑎 − 𝑏2 + ) 󵄨󵄨󵄨󵄨𝐼(1) 󵄨󵄨󵄨󵄨 ≥ 2, 2 2 a contradiction. In conclusion, we have 𝑛 = 0, 𝑡0 + 𝑙 = 1, and (𝑎, 𝑏) = (2, 2) (if (𝑎, 𝑏) = (1, 2), then ℎ1 < 0 and ℎ2 < 0, a contradiction). Then the result follows from the main result in [18] which determined that 𝐺 is fractional 2-deleted graph if 𝑡(𝐺) ≥ (3/2). Case 2 (𝑡0 + 𝑙 = 0). In this case, (18) becomes 𝑏−1

𝑏 󵄨 󵄨 ∑ (𝑏 − 2) (𝑏 − 𝑗) 𝑖𝑗 + ∑ (𝑎𝑏 − 𝑎𝑖 + 𝑏 − 𝑖 + 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨

2

𝑗=1

𝑖=1

𝑏

𝑖=1

𝑏−1

󵄨 󵄨 ∑ (𝑎𝑏 + 𝑏 − 𝑎 − 𝑖 + 2 − 𝑏2 ) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨

󵄨 󵄨 ≥ ∑ (𝑎𝑡 − 𝑎𝑗 − 𝑏 + 𝑗) 𝑖𝑗 + 𝑎𝑡 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨

𝑖=2

3 1 󵄨 󵄨 + (𝑎𝑏 + 𝑏 − 𝑎 − 𝑏2 + ) 󵄨󵄨󵄨󵄨𝐼(1) 󵄨󵄨󵄨󵄨 ≥ 1. 2 2

𝑖=2

𝑗=1

(22)

+

󵄨 󵄨 (𝑎 + 1) 󵄨󵄨󵄨󵄨𝐼(1) 󵄨󵄨󵄨󵄨 2

𝑏 󵄨 󵄨 − 𝑎∑ (𝑖 − 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 − 𝑏𝑛 − 1. 𝑖=1

(29)

The Scientific World Journal

5

Subcase 1 (|𝐼1 | = 0). In this subcase, (29) becomes 𝑏−1

∑ ((𝑏 − 2) (𝑏 − 𝑗) − (𝑎𝑡 − 𝑎𝑗 − 𝑏 + 𝑗)) 𝑖𝑗 + 𝑏𝑛 + 1 ≥ 0. (30)

𝑗=1

Let ℎ𝑗 = (𝑏 − 2) (𝑏 − 𝑗) − (𝑎𝑡 − 𝑎𝑗 − 𝑏 + 𝑗) = 𝑏2 + (𝑎 − 𝑏 + 1) 𝑗 − 𝑏 − 𝑎𝑡 ≤ 𝑏2 + (𝑎 − 𝑏 + 1) 𝑗 − 𝑏 − 𝑎 ⋅

𝑏2 − 1 + 𝑏𝑛 𝑎

(31)

(i) If 𝑏 ≥ 𝑎 + 1, then (𝑎 − 𝑏 + 1) 𝑗 − 𝑏 + 1 − 𝑏𝑛

≤ −𝑏 + 1 − 𝑏𝑛

󵄨 󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨𝐶2 󵄨󵄨 ≤ (𝑏 − 2) + (󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨 − 1) (𝑏 − 1 − 1) = 󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨 (𝑏 − 2) , 󵄨 󵄨 |𝑇| ≤ 󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨 (𝑏 − 1) , 󵄨 󵄨 |𝑇| + 1 + 𝑏𝑛 󵄨󵄨 󵄨󵄨 1 − 󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨 ≤ 󵄨󵄨𝐼2 󵄨󵄨 + + 𝑛. |𝑆| ≤ 𝑎 𝑏

𝑏−

1 + 𝑛 ≤ 𝑡 (𝐺) 𝑏 |𝑈| 𝜔 (𝐺 − 𝑈) 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ((1 − 󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨) /𝑏) + 󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨 (𝑏 − 2) + 𝑛 ≤ 󵄨󵄨 󵄨󵄨 󵄨󵄨𝐼2 󵄨󵄨

(32)

≤

= −𝑏𝑛 − 1. This implies (𝑎, 𝑏) = (1, 2), ∑𝑏−1 𝑗=2 𝑖𝑗 = 0, |𝐶2 | ≤ |𝐼2 |, and |𝑇| ≤ 2|𝐼2 |. Thus, |𝑆| ≤

󵄨 󵄨 2 󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨 (𝑏 − 1) + 𝑏𝑛 + 1 . 𝑎

(37)

1 𝑛 1 = (𝑏 − 1 − ) + 󵄨󵄨 󵄨󵄨 + 󵄨󵄨 󵄨󵄨 . 𝑏 𝑏 󵄨󵄨𝐼2 󵄨󵄨 󵄨󵄨𝐼2 󵄨󵄨

(33)

If |𝐼2 | = 1, then |𝑆| ≤ ((2(𝑏 − 1) + 𝑏𝑛 + 1)/𝑎) = 3 + 2𝑛 and 𝛿(𝐺) ≤ |𝑆|+1 ≤ 4+2𝑛. This contradicts 𝛿(𝐺) ≥ 2𝑡(𝐺) ≥ 6+4𝑛. Hence, |𝐼2 | ≥ 2. Let 𝑍 = {𝑥 | 𝑥 ∈ 𝐶2 , 𝑑𝐺−𝑆 (𝑥) = 1} and 𝑧 = |𝑍|. Thus, 0 ≤ 𝑧 ≤ |𝐼2 | and 𝑁𝐺−𝑆 (V) ∈ 𝐼2 if V ∈ 𝑍. We obtain 󵄨 󵄨 󵄨 󵄨 (󵄨󵄨𝐼2 󵄨󵄨 + 𝑧) (𝑏 − 1) + (󵄨󵄨󵄨𝐶2 󵄨󵄨󵄨 − 𝑧) (𝑏 − 2) + 𝑏𝑛 + 1 . (34) |𝑆| ≤ 󵄨 󵄨 𝑎

This reveals 𝑛(1 − (1/|𝐼2 |)) ≤ ((1/𝑏|𝐼2 |) − 1), which contradicts 𝑏 ≥ 2 and |𝐼2 | ≥ 2. Subcase 2 (|𝐼2 | = 0). In this subcase, (29) becomes 󵄨󵄨 (1) 󵄨󵄨 𝑏 󵄨 󵄨 󵄨 󵄨 (𝑎 + 1) 󵄨󵄨󵄨𝐼 󵄨󵄨󵄨 ∑ (𝑎𝑏 − 𝑎𝑖 + 𝑏 − 𝑖 + 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 − 𝑎𝑡 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 − 2 𝑖=1 𝑏

(38)

󵄨 󵄨 + 𝑎∑ (𝑖 − 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 + 𝑏𝑛 + 1 ≥ 0.

Letting 𝑍󸀠 = {𝑥 | 𝑥 ∈ 𝑁𝐺(𝐼2 ) ∩ 𝑊, 𝑑𝐺−𝑆 (𝑥) = 1}, we infer

𝑖=1

𝑏2 − 1 + 𝑏𝑛 𝑎

󵄨󵄨 󵄨 󵄨󵄨𝑈 − 𝑍 ∪ 𝑍󸀠 󵄨󵄨󵄨 󵄨 󵄨 ≤ 𝑡 (𝐺) ≤ 𝜔 (𝐺 − (𝑈 − 𝑍 ∪ 𝑍󸀠 )) 󵄨 󵄨 󵄨 󵄨 (󵄨󵄨𝐼 󵄨󵄨 + 𝑧) (𝑏 − 1) + (󵄨󵄨󵄨𝐶2 󵄨󵄨󵄨 − 𝑧) (𝑏 − 2) + 𝑏𝑛 + 1 ≤ ( 󵄨 2󵄨 𝑎

(36)

If |𝐼2 | = 1, then |𝑆| ≤ 1 + 𝑛 and 𝛿(𝐺) ≤ |𝑆| + (𝑏 − 1) ≤ 𝑏 + 𝑛, which contradicts 𝛿(𝐺) ≥ 2𝑡(𝐺) > 𝑏 + 𝑛. Hence, |𝐼2 | ≥ 2 and

= (𝑎 − 𝑏 + 1) 𝑗 − 𝑏 + 1 − 𝑏𝑛.

≤ 𝑎 − 2𝑏 + 2 − 𝑏𝑛

of 𝐻2 , choose a vertex with the smallest degree and add it to 𝐼2 . Hence, by the definition of 𝐻2 , we confirm that 𝐻2 is connected; each vertex in 𝐼2 has degree 𝑏 − 1 in 𝐺 − 𝑆 except that one vertex has degree 𝑏 − 2 in 𝐺 − 𝑆. This fact implies

This implies

(35)

󵄨 󵄨 󵄨 󵄨 󵄨 󵄨−1 󵄨 󵄨 + (󵄨󵄨󵄨𝐶2 󵄨󵄨󵄨 − 𝑧) + (󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨 − 󵄨󵄨󵄨𝐶2 󵄨󵄨󵄨)) × 󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨 . By (𝑎, 𝑏) = (1, 2) and |𝐶2 | ≤ |𝐼2 |, we get 2𝑛(1 − (1/|𝐼2 |)) ≤ (1/|𝐼2 |) − 1. By |𝐼2 | ≥ 2, we derive the contradiction. (ii) If 𝑎 = 𝑏, then max{ℎ𝑗 } = ℎ𝑏−1 = −𝑏𝑛 and the second largest value of ℎ𝑗 is ℎ𝑏−2 = −𝑏𝑛 − 1. In terms of the analysis of Lemma 4 in [17]: for each connected component

𝑏

󵄨 󵄨 ∑ (𝑎𝑏 + 𝑏 − 𝑎 − 𝑖 + 2 − 𝑏2 ) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 𝑖=2

(39)

3 1 󵄨 󵄨 + (𝑎𝑏 + 𝑏 − 𝑎 − 𝑏2 + ) 󵄨󵄨󵄨󵄨𝐼(1) 󵄨󵄨󵄨󵄨 + 1 ≥ 0. 2 2 Then, by ℎ1 ≤ −1/2, we get ∑𝑏𝑖=4 |𝐼(𝑖) | = 0, |𝐼(3) | ≤ 1, and |𝐼(1) | ≤ 2. Now, we consider the following three subcases. Subcase 2.1 (|𝐼(1) | = 1). In this subcase, we have ∑𝑏𝑖=3 |𝐼(𝑖) | = 0. By analyzing the proof of Lemma 2.2 in [17]: “for each vertex

6

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𝑥 ∈ 𝐼𝑛 and 𝑑𝐻𝑛 (𝑥) = 𝑘 − 1, there exists a vertex 𝑦 ∈ 𝐼𝑛 such that 𝑁𝐻𝑛 (𝑥) ∩ 𝑁𝐻𝑛 (𝑦) ≠ 0,” we obtain |𝐼1 | ≥ 2, 󵄨 󵄨 󵄨 󵄨 |𝑇| ≤ (𝑏 − 1) + (󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 − 1) (𝑏 − 1) = 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 (𝑏 − 1) , 󵄨 󵄨 |𝑇| + 1 + 𝑏𝑛 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 (𝑏 − 1) + 1 + 𝑏𝑛 ≤ , |𝑆| ≤ 𝑎 𝑎

≤

𝑖=1

󵄨󵄨 󵄨󵄨 󵄨𝐼 󵄨 (𝑏 − 1) + 1 + 𝑏𝑛 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ≤ 󵄨 1󵄨 + 󵄨󵄨𝐼1 󵄨󵄨 (𝑏 − 1) − 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 + (󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 − 1) 𝑎 󵄨󵄨 󵄨󵄨 󵄨𝐼 󵄨 (𝑏 − 1) + 1 + 𝑏𝑛 󵄨󵄨 󵄨󵄨 = 󵄨 1󵄨 + 󵄨󵄨𝐼1 󵄨󵄨 (𝑏 − 1) − 1. 𝑎 (40) Thus, 𝑏2 − 1 + 𝑏𝑛 𝑎

≤

𝑏2 − 1 + 𝑏𝑛 𝑎 ≤ 𝑡 (𝐺) ≤

𝑏 󵄨 󵄨 󵄨 󵄨 |𝑈| ≤ |𝑆| + 󵄨󵄨󵄨𝐶1 󵄨󵄨󵄨 + ∑ (𝑖 − 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨

≤ 𝑡 (𝐺) ≤

By |𝐼1 | ≥ 2, we obtain

|𝑈| 𝜔 (𝐺 − 𝑈)

(45)

󵄨 󵄨 󵄨 󵄨 ((󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 (𝑏 − 1) + 1 + 𝑏𝑛) /𝑎) + 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 (𝑏 − 1) − 1 . 󵄨󵄨 󵄨󵄨 󵄨󵄨𝐼1 󵄨󵄨

This implies 𝑏𝑛(1 − (1/|𝐼1 |)) ≤ (𝑏 − 𝑎)(1 − 𝑏) + ((1 − 𝑎)/|𝐼1 |), a contradiction. If each vertex in 𝑌 is adjacent to at least two vertices in 𝐼1 , we get 󵄨󵄨 󵄨󵄨 󵄨𝐼1 󵄨 󵄨󵄨 󵄨󵄨 |𝑈| ≤ |𝑆| + 󵄨󵄨𝐼1 󵄨󵄨 (𝑏 − 2) + 󵄨 󵄨 2 󵄨󵄨 󵄨󵄨 󵄨𝐼 󵄨 (𝑏 − 1) + 1 + 𝑏𝑛 󵄨󵄨 󵄨󵄨 ≤ 󵄨 1󵄨 + 󵄨󵄨𝐼1 󵄨󵄨 (𝑏 − 2) + 𝑎

󵄨󵄨 󵄨󵄨 󵄨󵄨𝐼1 󵄨󵄨 , 2

(46)

where 𝑈 = 𝑆 ∪ 𝐶1 ∪ (𝑁𝐺(𝐼1 ) ∩ 𝑊). Due to |𝐼1 | ≥ 2, we deduce |𝑈| 𝜔 (𝐺 − 𝑈)

(41)

󵄨 󵄨 󵄨 󵄨 ((󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 (𝑏 − 1) + 1 + 𝑏𝑛) /𝑎) + 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 (𝑏 − 1) − 1 . 󵄨󵄨 󵄨󵄨 󵄨󵄨𝐼1 󵄨󵄨

≤ 𝑡 (𝐺) ≤

This implies 𝑏𝑛(1 − (1/|𝐼1 |)) ≤ (𝑏 − 𝑎)(1 − 𝑏) + ((1 − 𝑎)/|𝐼1 |), a contradiction. Subcase 2.2 (|𝐼(1) | = 2). In this subcase, ∑𝑏𝑖=3 |𝐼(𝑖) | = 0. We can get a contradiction via a similar discussion as in Subcase 2.1. Subcase 2.3 (|𝐼(1) | = 0). In this subcase, we have ∑𝑏𝑖=4 |𝐼(𝑖) | = 0 and |𝐼(3) | ≤ 1. If |𝐼1 | = 1, then |𝑆| ≤ (((𝑏 − 1) + 𝑏𝑛 + 1)/𝑎). Thus, we infer (𝑏 − 1) + 𝑏𝑛 + 1 +𝑏−1 𝑎 ≥ 𝑏 − 1 + |𝑆| ≥ 𝛿 (𝐺) ≥ 2𝑡 (𝐺)

(42)

2

≥

2 (𝑏 − 1 + 𝑏𝑛) 𝑎

a contradiction. Hence, |𝐼1 | ≥ 2. Let 𝑌 = 𝑁𝐺(𝐼1 ) ∩ 𝑊. If there is a vertex 𝑦 ∈ 𝑌 such that 𝑦 only adjacent to one vertex in 𝐼1 . Reset (43)

Then, we have 󵄨 󵄨 |𝑈| ≤ |𝑆| + 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 (𝑏 − 1) − 1 󵄨󵄨󵄨𝐼 󵄨󵄨󵄨 (𝑏 − 1) + 1 + 𝑏𝑛 󵄨 󵄨 ≤ 󵄨 1󵄨 + 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 (𝑏 − 1) − 1. 𝑎

|𝑈| 𝜔 (𝐺 − 𝑈)

󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ((󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 (𝑏 − 1) + 1 + 𝑏𝑛) /𝑎) + 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 (𝑏 − 2) + (󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 /2) ≤ . 󵄨󵄨 󵄨󵄨 󵄨󵄨𝐼1 󵄨󵄨 (47) That is to say, 𝑏𝑛(1−(1/|𝐼1 |)) ≤ (𝑏−𝑎)(1−𝑏)+((1/|𝐼1 |)−(𝑎/2)), which contradicts 𝑎 ≥ 1 and |𝐼1 | ≥ 2. Subcase 3 (|𝐼1 | ≠ 0 and |𝐼2 | ≠ 0). From what we have discussed 𝑏−1 in Subcase 1, we get ∑𝑏−1 𝑗=1 (𝑏 − 2)(𝑏 − 𝑗)𝑖𝑗 ≤ ∑𝑗=1 (𝑎𝑡 − 𝑎𝑗 − 𝑏 + 𝑗)𝑖𝑗 + 𝑏𝑛 + 1. Then, we deduce 𝑏 󵄨 󵄨 ∑ (𝑎𝑏 − 𝑎𝑖 + 𝑏 − 𝑖 + 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 𝑖=1

󵄨 󵄨 ≥ 𝑎𝑡 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 +

,

𝑈 = 𝑆 ∪ 𝐶1 ∪ (𝑁𝐺 (𝐼1 ) ∩ (𝑊 − {𝑦})) .

𝑏2 − 1 + 𝑏𝑛 𝑎

(44)

󵄨 󵄨 (𝑎 + 1) 󵄨󵄨󵄨󵄨𝐼(1) 󵄨󵄨󵄨󵄨 2

𝑏

(48)

󵄨 󵄨 − 𝑎∑ (𝑖 − 1) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 . 𝑖=1

This implies 𝑏

󵄨 󵄨 ∑ (𝑎𝑏 + 𝑏 − 𝑎 − 𝑖 + 2 − 𝑏2 ) 󵄨󵄨󵄨󵄨𝐼(𝑖) 󵄨󵄨󵄨󵄨 𝑖=2

(49)

3 1 󵄨 󵄨 + (𝑎𝑏 + 𝑏 − 𝑎 − 𝑏2 + ) 󵄨󵄨󵄨󵄨𝐼(1) 󵄨󵄨󵄨󵄨 ≥ 0. 2 2 Thus, we have ∑𝑏𝑖=4 |𝐼(𝑖) | = 0, |𝐼(3) | ≤ 1, |𝐼(1) | ≤ 2, and 𝑛 = 0, by what we have discussed in Subcase 2. It is enough to discuss the situation of |𝐼(1) | = 0; other two cases for |𝐼(1) | = 1 and |𝐼(1) | = 2 can be considered in a similar way.

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7

Under the condition of |𝐼(1) | = 0, we get ∑𝑏𝑖=4 |𝐼(𝑖) | = 0, |𝐼 | ≤ 1, (3)

󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 |𝑇| ≤ 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 (𝑏 − 1) + 󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨 (𝑏 − 1) = (𝑏 − 1) (󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨) , 󵄨 󵄨 󵄨 󵄨 (50) |𝑇| + 1 (󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨) (𝑏 − 1) + 1 ≤ . |𝑆| ≤ 𝑎 𝑎 Since |𝐼1 | + |𝐼2 | ≥ 2, we get 𝑏2 − 1 |𝑈| ≤ 𝑡 (𝐺) ≤ 𝑎 𝜔 (𝐺 − 𝑈) 󵄨 󵄨 󵄨 󵄨 |𝑆| + 󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨 (𝑏 − 2) + 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 (𝑏 − 1) . ≤ 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨𝐼1 󵄨󵄨 + 󵄨󵄨𝐼2 󵄨󵄨

(51)

Hence, 󵄨 󵄨 󵄨 󵄨 (𝑏2 − 1) (󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨) 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ≤ (󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨) (𝑏 − 1) + 1 + (𝑎𝑏 − 2𝑎) (󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝐼2 󵄨󵄨󵄨) 󵄨 󵄨 + 󵄨󵄨󵄨𝐼1 󵄨󵄨󵄨 𝑎.

(52)

This implies (𝑏−𝑎)(𝑏−1)(|𝐼1 |+|𝐼2 |) ≤ 1−𝑎|𝐼2 |, a contradiction. We complete the proof of the theorem.

Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments First the authors thank the reviewers for their constructive comments that helped improve the quality of this paper. They also would like to thank the anonymous referees for providing them with constructive comments and suggestions. This work was supported in part by Key Laboratory of Educational Informatization for Nationalities, Ministry of Education, the National Natural Science Foundation of China (60903131), Key Science and Technology Research Project of Education Ministry (210210), and the PHD initial funding of the first author.

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