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# Open Physics

### formerly Central European Journal of Physics

Editor-in-Chief: Seidel, Sally

Managing Editor: Lesna-Szreter, Paulina

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Volume 15, Issue 1

# Topics on data transmission problem in software definition network

Wei Gao
/ Li Liang
/ Tianwei Xu
/ Jianhou Gan
Published Online: 2017-08-03 | DOI: https://doi.org/10.1515/phys-2017-0057

## Abstract

In normal computer networks, the data transmission between two sites go through the shortest path between two corresponding vertices. However, in the setting of software definition network (SDN), it should monitor the network traffic flow in each site and channel timely, and the data transmission path between two sites in SDN should consider the congestion in current networks. Hence, the difference of available data transmission theory between normal computer network and software definition network is that we should consider the prohibit graph structures in SDN, and these forbidden subgraphs represent the sites and channels in which data can’t be passed by the serious congestion. Inspired by theoretical analysis of an available data transmission in SDN, we consider some computational problems from the perspective of the graph theory. Several results determined in the paper imply the sufficient conditions of data transmission in SDN in the various graph settings.

PACS: 02.10.Ox; 07.05.Mh

## 1 Introduction

We equate the fractional factor problem with a relaxation of the famous cardinality matching problem which is one of the core problems in operation research. It has widely applications in various fields such as network design, combinatorial polyhedron and scheduling. For example, several large data packets are sent to different destinations through some channels in a data transmission network. This work helps to partition the large data packets into small ones and then make it efficiency improved. The available allocations of data packets is equated with the problem of fractional flow, which converts to fractional factor problem in a graph generated by a network.

Specifically, there is a graph that can model the complete network and the graph needs to meet the requirements that each site corresponds to a vertex and each channel corresponds to an edge in it. In normal network, the path of data transmission is selected by the shortest path between vertices. Several contributions on data transmission in networks are presented recently. Rolim et al. [1] studied urban sensing problem by means of opportunistic networks to support the data transmission. Vahidi et al. [2] provided the high-mobility airborne hyperspectral data transmission algorithm in view of unmanned aerial vehicles approach. Miridakis et al. [3] determined a rather cost-effective solution for the dual-hop cognitive secondary relaying system. Lee et al. [4] considered streaming data transmission on a discrete memoryless channel. However, in the setting of software definition network, the data transmission depends on the network flow computation. The transmission path is selected with minimum transmission congestion in the current moment. From this perspective, the model of data transmission problem in SDN is becoming what makes the fractional factor avoid certain subgraphs.

It is only the simple graphs that are chosen in our article. Let G = (V(G), E(G)) be a graph, in which V(G) and E(G) are the vertex set and the edge set, respectively. Let n = |V(G)| be the order of graph G. The standard notations and terminologies used but undefined clearly in this article can be found in Bondy and Mutry [5], Basavanagoud et al. [6], Wang [7], and Gao and Wang [8]. The toughness t(G) of a graph G can be stated in the below: $t(G)=min{|S|ω(G−S)|S⊆V(G),ω(G−S)≥2},$ if G is not complete; otherwise, t(G) = +∞. It is an important parameter, which is used to measure the vulnerability of networks.

Let g and f be two positive integer-valued functions on V(G) such that 0 ≤ g(x) ≤ f(x) for any xV(G). A fractional (g, f)-factor is a function h which assigns a number in [0,1] to each edge so that $g\left(x\right)\le {d}_{G}^{h}\left(x\right)\le f\left(x\right)$ for each xV(G), where ${d}_{G}^{h}\left(x\right)=\sum _{e\in E\left(x\right)}h\left(e\right)$ is called the fractional degree of x in G. A fractional (a, b)-factor (here a and b are both positive integers and ab) is a function h that assigned a number [0,1] to each edge of a graph G so that for each vertex x we have $a\le {d}_{G}^{h}\left(x\right)\le b$. Clearly, fractional (a, b)-factor is a particular case of fractional (g, f)-factor when g(x) = a and f(x) = b for any xV(G). If a = b = k(k ≥ 1 is an integer) for all xV(G), then a fractional (a, b)-factor is just a fractional k-factor.

It’s stated that G includes a Hamiltonian fractional (g, f)-factor if G has a fractional (g, f)-factor containing a Hamiltonian cycle. A graph G is said to be an ID- Hamiltonian graph if the remaining graph of G admits a Hamiltonian cycle, when we delete any independent set of G. We say that G has an ID-Hamiltonian fractional (g, f)-factor if the remaining graph of G includes a Hamiltonian fractional (g, f)-factor when we delete any independent set of G.

In SDN, the independent set I is corresponding to the website with high transmission congestion and a Hamiltonian cycle corresponding to several channels with high transmission congestion. Thus, we study the data transmission problem by considering the Hamiltonian fractional (g, f)-factor and ID-Hamiltonian fractional (g, f)- factor in the networks model. Several related results can refer to [9-14].

The sufficient conditions of Hamiltonian fractional (g, f)-factors and ID-Hamiltonian fractional (g, f)-factors in graphs are studied and at last we obtain three results that are stated in the below.

#### Theorem 1

Let a, b be two integers with 3 ≤ ab, and G be a Hamiltonian graph of order $n>\frac{\left(a+b-5\right)\left(a+b-3\right)}{a-2}.\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}g,f$ are taken to be two integer-valued functions that are defined on V(G) such that ag(x) ≤ f(x) ≤ b for each xV(G). If $δ(G)≥(b−1)n+a+b−3a+b−3$ and $δ(G)>(b−2)n+2α(G)−1a+b−4,$ then G has a Hamiltonian fractional (g, f)-factor.

#### Theorem 2

Let a, b be two integers with 3 ≤ ab, and G be a Hamiltonian graph of order nb + 2. g, f are taken to be two integer-valued functions that are defined on V(G) such that ag(x) ≤ f(x) ≤ b for each xV(G). If t(G) ≥ b$1+\frac{b-1}{a-2}$, then G has a Hamiltonian fractional (g, f )-factor.

#### Theorem 3

a, b are taken to be two integers with 3 ≤ ab, and G be an ID-Hamiltonian graph. And g, f are taken to be two integer-valued functions that are defined on V(G) such that ag(x) ≤ f(x) ≤ b for each xV(G). If $κ(G)≥max(b+1)2+4(a−2)2,(a+b+1)(a+b−3)4(a−2)+(a+b+1)(a+b−3)(b+1)2$(1) and $κ(G)≥(b+1)2+4(a−2)4(a−2)α(G),$ then G has an ID-Hamiltonian fractional (g, f)-factor.

The ways to prove our major results are on the basis of the listed lemmas below:

#### Lemma 1

(Liu and Zhang [15]) Since G is a graph and we have H = G[T], so δ(H) ≥ 1 and 1 ≤ dG(x) ≤ k − 1 for every xV(H) where TV(G) and k ≥ 2. Let T1,..., Tk-1 be a partition of the vertices of H satisfying dG(x) = j for each xTj where some Tj are allowed to be empty on purpose. 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 $∑j=1k−1(k−j)cj≤∑j=1k−1(k−2)(k−j)ij,$ where cj = |CTj | and ij = |ITj| for every j = 1,..., k − 1.

#### Lemma 2

(Liu and Zhang [15]) Since G is a graph and H = G[T], so dG(x) = k − 1 for every xV(H) and no component of H is isomorphic to Kk where TV(G) and k ≥ 2. Then a maximal independent set I is obtained and the covering set C = V(H) − I of H satisfy $|V(H)|≤(k−1k+1)|I|,$ and $|C|≤(k−1−1k+1)|I|.$

#### Lemma 3

(Anstee [16]) Assume f and g be two integervalued functions that are defined on the vertex set of a graph G such that 0 ≤ g(x) ≤ f(x) for each xV(G). Then G has a fractional (g, f)-factor if and only if for every subset S of V(G), g(T) − dG-S(T) ≤ f(S),where T = {xV(G)\S | dG-S(x) ≤ g(x) − 1}.

Clearly, Lemma 3 is equal to the following version.

#### Lemma 4

Suppose that f and g are two integer-valued functions that are defined on the vertex set of a graph G such that 0 ≤ g(x) ≤ f(x) for each xV(G). Then G has a fractional (g, f )-factor if and only if $f(S)+dG−S(T)−g(T)≥0$ for all disjoint subsets S, T of V(G).

#### Lemma 5

(Katerinis [17]) If the graph G isn’t complete, then $t\left(G\right)\le \frac{\delta \left(G\right)}{2}$.

The tricks used to prove our main results are borrowed from Zhou et al. [18], [19] and [20], and more new technologies are introduced here to deal with complex problems.

## 2 Proof of Theorem 1

In terms of the condition of Theorem 1 and the definition of Hamiltonian graph, G admits a Hamiltonian cycle C. Let G′ = GE(C). Clearly, G includes the desired fractional factor if H contains a fractional (g − 2, f − 2)-factor. In view of contradiction, we assume that G′ has no fractional (g − 2, f − 2)-factor. Hence, concerning Lemma 4, there are some disjoint subset S and T of V(H) satisfying $(a−2)|S|+dG′−S(T)−(b−2)|T|≤(f−2)(S)+dG′−S(T)−(g−2)(T)≤−1.$(2) We select suitable subsets S and T with the smallest |T|.

If T = ∅, then using (2) we obtain −1 ≥ (a − 2)|S| ≥ 0, a contradiction. Thus, T ≠ ∅. Define h = min{dR-S(x) : xT}. We infer, $δ(G)≤dG(x1)≤dG−S(x1)+|S|=h+|S|.$(3) Now, the following facts are to be concerned.

#### Claim 1.1

dG′-S(x) ≤ b − 3 for any xT.

#### Proof

If dG′-S(x) ≥ b − 2 for certain xT, then the subset S and T − {x} satisfy (2) as well. This is contradictive with the selection of T. □

#### Claim 1.2

dG-S(x) ≤ dG′-S(x) + 2 ≤ b − 1 for all xT.

#### Proof

From Claim 1.1 and G′ = GE(C), we deduce $dG−S(x)≤dG′−S(x)+2≤b−1$ for any xT. □

Taking the definition of h and Claim 1.2 into consideration, we yield 0 ≤ hb − 1. The following proof is composed of two cases in terms of the value of h.

#### Case 1

h = 0.

Set X = {xT : dG-S(x) = 0}, Y = {xT : dG-S(x) = 1}, Y1 = {xY : NG-S(x) ⊆ T} and Y2 = YY1· G[Y1] has maximum degree at most 1 in GS. Let Z be a maximum independent set of G[Y1]. We get $|Z|\ge \frac{1}{2}|{Y}_{1}|$. On the basis of definitions, XZY2 is an independent set of G. Therefore, we get $α(G)≥|X|+|Z|+|Y2|≥|X|+12|Y1|+12|Y2|=|X|+12|Y|.$(4) In terms of (2), (3), (4) and Claim 1.2, we infer $α(G)−1≥|X|+12|Y|+(a−2)|S|+dG′−S(T)−(b−2)|T|≥|X|+12|Y|+(a−2)|S|+dG−S(T)−2|T|−(b−2)|T|=|X|+12|Y|+(a−2)|S|+dG−S(T)−b|T|=|X|+12|Y|+(a−2)|S|+dG−S(T−X∪Y)+|Y|−b|T|=|X|+32|Y|+(a−2)|S|+dG−S(T−X∪Y)−b|T|≥|X|+32|Y|+(a−2)|S|+2|T−X∪Y|−b|T|=(a−2)|S|−(b−2)|T|−(|X|+12|Y|)≥(a−2)δ(G)−(b−2)|T|−α(G),$ i.e. $(b−2)|T|≥(a−2)δ(G)−2α(G)+1.$(5)

For b ≥ 3, (3), (5) and |S| + |T| ≤ n, we get $0≤n−|S|−|T|≤n−δ(G)−(a−2)δ(G)−2α(G)+1b−2=(b−2)n−(a+b−4)δ(G)+2α(G)−1b−2,$ that is $δ(G)≤(b−2)n+2α(G)−1a+b−4.$ This contradicts $\delta \left(G\right)>\frac{\left(b-2\right)n+2\alpha \left(G\right)-1}{a+b-4}$.

#### Case 2

1 ≤ hb − 1.

In light of (2) and |S| + |T| ≤ n, we derive $−1≥(a−2)|S|+dG′−S(T)−(b−2)|T|≥(a−2)|S|+dG−S(T)−2|T|−(b−2)|T|=(a−2)|S|+dG−S(T)−b|T|≥(a−2)|S|+h|T|−b|T|=(a−2)|S|−(b−h)|T|≥(a−2)|S|−(b−h)(n−|S|)=(a+b−h−2)|S|−(b−h)n,$ which implies $|S|≤(b−h)n−1a+b−h−2.$(6) Using (3) and (6), we deduce $δ(G)≤|S|+h≤(b−h)n−1a+b−h−2+h.$(7)

#### Subcase 2.1

h = 1.

By means of (7), we yield $δ(G)≤(b−1)n−1a+b−3+1=(b−1)n+a+b−4a+b−3,$ which contradicts $\delta \left(G\right)\ge \frac{\left(b-1\right)n+a+b-3}{a+b-3}$.

#### Subcase 2.2

2 ≤ h = b − 1.

Set $f\left(h\right)=\frac{\left(b-h\right)n-1}{a+b-h-2}+h$. From 2 ≤ hb − 1 and n > $\frac{\left(a+b-5\right)\left(a+b-3\right)}{a-2}$, we get $f′(h)=−n(a+b−h−2)+(b−h)n−1(a+b−h−2)2+1=−n(a−2)−1(a+b−h−2)2+1≤−n(a−2)−1(a+b−4)2+1<−(a+b−5)(a+b−3)−1(a+b−4)2+1=0.$ Hence, max{f(h)} = f(2). According to (7) and n > $\frac{\left(a+b-5\right)\left(a+b-3\right)}{a-2}$, we have $δ(G)≤f(h)≤f(2)=(b−2)n−1a+b−4+2<(b−1)n+a+b−3a+b−3,$ which contradicts $\delta \left(G\right)\ge \frac{\left(b-1\right)n+a+b-3}{a+b-3}$.

From what we discussed above, we can draw the conclusion that G′ has a fractional (g − 2, f − 2)-factor. Hence, the proof of Theorem 1 is finished successfully. □

## 3 Proof of Theorem 2

If G is complete, clearly G has a Hamiltonian fractional (g, f)-factor by the order of graph. We always assume G is not a complete graph below.

Considering the situation of Theorem 2 and the definition of Hamiltonian graph, G admits a Hamiltonian cycle C. Let G′ = GE(C). Clearly, G includes the desired fractional factor if H includes a fractional (g − 2, f − 2)-factor. Using contradiction, we assume that G′ doesn’t have fractional (g − 2, f − 2)-factor.

Then under the help of Lemma 3, there exists some subset S of V(G′) satisfying $(a−2)|S|+dG′−S(T)−(b−2)|T|≤(f−2)(S)+dG′−S(T)−(g−2)(T)≤−1,$(8) where T = {xV(G′) − S, dG′-S(x) ≤ b − 3}. Moreover, for each xT, we get $dG−S(x)≤dG′−S(x)+2≤b−1.$(9) In light of (8) and (9), we infer $b|T|−dG−S(T)>(a−2)|S|.$(10)

Since G is not complete, by Lemma 5 we ensure δ(G) ≥ $2t\left(G\right)\ge 2\left(b-1\right)+\frac{2\left(b-1\right)}{a-2}>b$. Hence, from (10), we get S ≠ ∅ and |S| ≠ 1. Let λ be the amount of the components of H′ = G[T] which are isomorphic to Kb and let T0 = {x : xV(H′), dG-S(x) = 0}. Let H be the subgraph gotten from H′ − T0 by deleting these λ components isomorphic to Kb.

If |V(H)| = 0, in view of (10), we obtain $b|T0|+bλ>(a−2)|S|,$ and $2≤|S| Hence, we infer $|T0|+λ>2(a−2)b.$ If ω(GS) ≥ |T0| + λ > 1, we deduce $t(G)≤|S|ω(G−S)≤|S||T0|+λ which contradicts to $t\left(G\right)\ge b-1+\frac{b-1}{a-2}$.

If |T0| + λ = 1, then dG-S(x) + |S| ≥ dG(x) ≥ δ(G) ≥ 2t(G) ≥ $2\left(b-2\right)+\frac{2\left(b-1\right)}{a-2}$. We have ${d}_{G-S}\left(x\right)\ge 2\left(b-2\right)+\frac{2\left(b-1\right)}{a-2}-|S|>2\left(b-2\right)+\frac{2\left(b-1\right)}{a-2}-\frac{b}{a-2}=2\left(b-2\right)+\frac{b-2}{a-2}$, which contradicts to (9).

Now, we discuss |V(H)| ≥ 1. Let H = H1H2, where H1 is the combination of components of H that meets dG-S(x) = b − 1 for each vertex xV(H1) and H2 = HH1. By means of Lemma 2, there exists a maximum independent set I1 and the covering set C1 = V(H1) − I1 of H1 such that $|V(H1)|≤(b−1b+1)|I1|$(11) and $|C1|≤(b−1−1b+1)|I1|.$(12)

Clearly, δ(H2) ≥ 1 and Δ(H2) ≤ b − 1. Let Tj = {xV(H2)|dG-S(x) = j} for 1 ≤ jb − 1. Each component of H2 has a vertex of degree at most b − 2 in GS from the definitions of H and H2. On the ground of Lemma 1, H2 has a maximal independent set I2 and the covering set C2 = V(H2) − I2 such that $∑j=1b−1(b−j)cj≤∑j=1b−1(b−2)(b−j)ij,$(13) where Cj = |C2Tj| and ij = |I2Tj| for each j = 1,..., b−1. Set W = V(G) − ST and U = SC1 ∪ (NG(I1) ∩ W) ∪ C2 ∪ (NG(I2) ∩ W). We infer $|U|≤|S|+|C1|+∑j=1b−1jij$(14) and $ω(G−U)≥|T0|+λ+|I1|+∑j=1b−1ij.$(15) If ω(GU) > 1, we yield $|U|≥t(G)ω(G−U).$(16) If ω(GU) = 1, then (16) also holds.

Using (14), (15) and (16), we obtain $|S|+|C1|+∑j=1b−1jij≥t(G)|T0|+λ+|I1|+∑i=1b−1ij.$(17) In terms of (10), we yield $b|T0|+bλ+|V(H1)|+∑j=1b−1(b−j)ij+∑j=1b−1(b−j)cj>(a−2)|S|.$(18) In light of (17) and (18), we have $|V(H1)|+∑j=1b−1(b−j)cj+(a−2)|C1|>∑j=1b−1((a−2)t(G)−(a−2)j−b+j)ij+((a−2)t(G)−b)(|T0|+λ)+(a−2)t(G)|I1|.$(19)

In view of $t\left(G\right)\ge b-1+\frac{b-1}{a-2}$, we infer (a − 2)t(G) − b ≥ (a − 2)(b − 1) − 1 ≥ 1. Using (19), we deduce $|V(H1)|+∑j=1b−1(b−j)cj+(a−2)|C1|>∑j=1b−1((a−2)t(G)−(a−2)j−b+j)ij+(a−2)t(G)|I1|+1.$(20) According to (11) and (12), we derive $|V(H1)|+(a−2)|C1|≤[(b−1b+1)+(a−2)(b−1−1b+1)]|I1|.$ Combining this with (13) and (20), we have $∑j=1b−1(b−2)(b−j)ij+[(b−1b+1)+(a−2)(b−1−1b+1)]|I1|>∑j=1b−1((a−2)t(G)−(a−2)j−b+j)ij+(a−2)t(G)|I1|+1.$ Therefore, at least one of the two cases below must be proved to be efficient.

#### Case 1

$\left(b-\frac{1}{b+1}\right)+\left(a-2\right)\left(b-1-\frac{1}{b+1}\right)>\left(a-2\right)t\left(G\right)+1$. We get $t(G) which contradicts $t\left(G\right)\ge b-1+\frac{b-1}{a-2}$.

#### Case 2

There exists at least one j such that $(b−2)(b−j)>(a−2)t(G)−(a−2)j−b+j.$ It means $(b−1)(b−j)+(a−2)j>(a−2)t(G).$(21) By jb − 1 and (21), we yield $b−1+(a−2)(b−1)>(a−2)t(G).$ It reveals $t(G) which is conflictive with $t\left(G\right)\ge b-1+\frac{b-1}{a-2}$. □

## 4 Proof of Theorem 3

By means of the hypothesis of Theorem 3, GX admits a Hamiltonian cycle C for any independent set X in G. Set R = GX for any independent set XV(G). Using the hypothesis of Theorem 3 and the definition of ID- Hamiltonian graph, R contains a Hamiltonian cycle C. Set H = RE(C). Hence, G admits the desired fractional factor if H has a fractional (g − 2, f − 2)-factor. We assume that H has no fractional (g − 2, f − 2)-factor. Then, there exists some subset S of V(H) = V(R) with $(a−2)|S|+dH−S(T)−(b−2)|T|≤(f−2)(S)+dH−S(T)−(g−2)(T)≤−1,$(22) where T = {xV(H) − S, dH-S(x) ≤ b − 3}.

If T = ∅, then follow from (22) we obtain $−1≥(a−2)|S|≥0,$ which is a contradiction. Thus, T ≠ ∅.

We select x1T, so x1 is the vertex with the least degree in R[T]. Set N1 = NR[x1] ∩ T and T1 = T. For i ≥ 2, we continue set Ti = T − ∪1≤j<iNj if T − ∪1≤j<iNj ≠ ∅. In the following, we select xiTi such that xi is the vertex with the least degree in R[Ti], and set Ni = NR[xi] ∩ Ti. We continue this procedures until Ti = ∅ for some i, say for i = m + 1. In light of the concept mentioned above, we ensure that {x1, x2,...., xm} is an independent set of R. Obviously, T ≠ ∅ and m ≥ 1.

Let |Ni| = ni, and thus $|T|=\sum _{1\le i\le m}{n}_{i}$. Set U = V(R) − (ST) and κ(RS) = t.

#### Claim 3.1

dR-S(x) ≤ dH-S(x) + 2 ≤ b − 1 for any xT.

#### Proof

Using the definition of T and H = RE(C), we infer $dR−S(x)≤dH−S(x)+2≤(b−3)+2=b−1$ for any xT. □

#### Claim 3.2

$\alpha \left(G\right)\le \kappa \left(G\right)-\frac{\left(a+b+1\right)\left(a+b-3\right)}{4\left(a-2\right)}$.

#### Proof

If $\alpha \left(G\right)<\frac{\left(a+b+1\right)\left(a+b-3\right)}{\left(b+1{\right)}^{2}}$, then by means of (1), we deduce $κ(G)≥(a+b+1)(a+b−3)4(a−2)+(a+b+1)(a+b−3)(b+1)2>(a+b+1)(a+b−3)4(a−2)+α(G).$ If $\alpha \left(G\right)\ge \frac{\left(a+b+1\right)\left(a+b-3\right)}{\left(b+1{\right)}^{2}}$, then in view of κ(G) ≥ $\frac{\left(b+1{\right)}^{2}+4\left(a-2\right)}{4\left(a-2\right)}\alpha \left(G\right)$, we get $κ(G)≥(b+1)2+4(a−2)4(a−2)α(G)=(b+1)24(a−2)α(G)+α(G)≥(b+1)24(a−2)(a+b+1)(a+b−3)(b+1)2+α(G)=(a+b+1)(a+b−3)4(a−2)+α(G).$ Now, Claim 3.2 is completed. □

U ≠ ∅ or m ≠ 1.

#### Proof

Assume that m = 1 and U = ∅. In terms of (22), Claim 3.1 and the selection rule of x1, we yield $−1≥(a−2)|S|+dH−S(T)−(b−2)|T|≥(a−2)|S|+dR−S(T)−2|T|−(b−2)|T|=(a−2)|S|+dR−S(T)−b|T|=(a−2)|S|+n1(n1−1)−bn1,$ which means $|S|≤−n12+(b+1)n1−1a−2.$(23)

In view of (23), Claim 3.2, R = GX and |X| ≤ α(G), we have $|V(G)|=|V(R)|+|X|=|S|+|T|+|X|=|S|+n1+|X|≤−n12+(b+1)n1−1a−2+n1+α=−n12+(a+b−1)n1−1a−2+α≤(a+b−1)2−44(a−2)+α=(a+b−1)(a+b−3)4(a−2)+α≤κ(G),$ which contradicts to |V(G)| > κ(G). Claim 3.3 is hold. □

#### Claim 3.4

${d}_{R-S}\left(T\right)\ge \sum _{1\le i\le m}{n}_{i}\left({n}_{i}-1\right)+\frac{mt}{2}$.

#### Proof

The selection rule of xi reveals that all vertices in Ni have degree at least ni − 1 in R[Ti]. Therefore, we obtain $∑1≤i≤m(∑x∈NidR[Ti](x))≥∑1≤i≤mni(ni−1).$(24) On the left of (24), an edge joining xNi and yNj (i < j) is counted only once, i.e., it is counted in the term dR[Ti](x) but not in dR[Tj](y). Therefore, we derive $dR−S(T)≥∑1≤i≤mni(ni−1)+∑1≤i(25) In light of κ(RS) = t and Claim 3.3, we yield $eR(Ni,∪j≠iNj)+eR(Ni,U)≥t$(26) for any Ni(i = 1, 2,..., m). By means of (26), we get $∑1≤i≤m(eR(Ni,∪j≠iNj)+eR(Ni,U))=2(∑1≤i which implies $∑1≤i(27) Using (25) and (27), we infer $dR−S(T)≥∑1≤i≤mni(ni−1)+mt2.$ Therefore, Claim 3.4 is proved. □

In terms of $|T|=\sum _{1\le i\le m}{n}_{i}$, (22), Claim 3.1 and Claim 3.4, we have $−1≥(a−2)|S|+dH−S(T)−(b−2)|T|≥(a−2)|S|+dR−S(T)−2|T|−(b−2)|T|=(a−2)|S|+dR−S(T)−b|T|≥(a−2)|S|+∑1≤i≤mni(ni−1)+mt2−b∑1≤i≤mni=(a−2)|S|+∑1≤i≤m(ni2−(b+1)ni)+mt2,$ i.e., $−1≥(a−2)|S|+∑1≤i≤m(ni2−(b+1)ni)+mt2.$(28)

Let $f\left({n}_{i}\right)={n}_{i}^{2}-\left(b+1\right){n}_{i}$. It is easy to say that $min\left\{f\left({n}_{i}\right)\right\}=-\frac{\left(b+1{\right)}^{2}}{4}$. Hence, we infer $ni2−(b+1)ni≥−(b+1)24.$(29) According to (28) and (29), we deduce $−1≥(a−2)|S|+∑1≤i≤m(−(b+1)24)+mt2=(a−2)|S|−(b+1)2m4+mt2.$(30)

#### Claim 3.5

$-\frac{\left(b+1{\right)}^{2}}{4}+\frac{t}{2}<0$.

#### Proof

Suppose that $-\frac{\left(b+1{\right)}^{2}}{4}+\frac{t}{2}\ge 0$. In view of (30), m ≥ 1 and |S|≥ 0, we get $−1≥(a−2)|S|−(b+1)2m4+mt2≥0,$ a contradiction. Then proof of Claim 3.5 is completed. □

Note that {x1, x2,..., xm} is an independent set of R. We obtain $α(R[T])≥m.$(31) In view of R = GX and (32), we yield $α(G)≥α(R)≥α(R[T])≥m.$(32) Note that R = GX and κ(RS) = t. It is obvious that $κ(G)≤κ(G−X)+|X|=κ(R)+|X|≤κ(R−S)+|S|+|X|=t+|S|+|X|.$(33) In light of (30), (32), (33), |X| ≤ α(G), Claim 3.5, κ(G) ≥ $\frac{\left(b+1{\right)}^{2}+4\left(a-2\right)}{2},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\kappa \left(G\right)\ge \frac{\left(b+1{\right)}^{2}+4\left(a-2\right)}{4\left(a-2\right)}\alpha \left(G\right)$, we have $−1≥(a−2)|S|−(b+1)2m4+mt2=(a−2)|S|+(−(b+1)24+t2)m≥(a−2)(κ(G)−|X|−t)+(−(b+1)24+t2)α(G)≥(a−2)(κ(G)−α(G)−t)+(−(b+1)24+t2)α(G)≥(a−2)(κ(G)−4(a−2)κ(G)(b+1)2+4(a−2)−t)+(−(b+1)24+t2)4(a−2)κ(G)(b+1)2+4(a−2)=(a−2)t(2κ(G)(b+1)2+4(a−2)−1)≥0,$ a contradiction. Therefore, we finish proving Theorem 3. □

## Acknowledgement

The constructive comments from the reviewers to improve the paper are sincerely appreciated.

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Accepted: 2017-03-13

Published Online: 2017-08-03

Conflict of interestConflict of Interests: It is declared that there is no conflict of interests regarding the publication of this paper.

Citation Information: Open Physics, Volume 15, Issue 1, Pages 501–508, ISSN (Online) 2391-5471,

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