A MEASURE OF GRAPH VULNERABILITY: SCATTERING NUMBER

The scattering number of a graph G, denoted sc(G), is defined by sc(G)=max{c(G−S)− |S| : S ⊆ V(G) and c(G−S)≠ 1} where c(G−S) denotes the number of components in G− S. It is one measure of graph vulnerability. In this paper, general results on the scattering number of a graph are considered. Firstly, some bounds on the scattering number are given. Further, scattering number of a binomial tree is calculated. Also several results are given about binomial trees and graph operations.


Introduction.
In a communication network, vulnerability measures the resistance of the network to disruption of operation after the failure of certain stations or communication links.To measure vulnerability we have some parameters that are toughness, binding number, vertex integrity, and scattering number [5].In this paper, we discuss the scattering number of a graph.
The scattering number of a graph G, denoted sc(G), was introduced in [4].Formally the scattering number is defined by where c(G − S) denotes the number of components in G − S. A cutset S of a graph G fulfilling sc(G) = c(G − S) − |S| is said to be a scattering set.The problem "given a graph G, decide whether the scattering number is larger than zero" is NP-complete.
The scattering number of a graph is closely related to the toughness of a graph and to the existence of Hamilton cycles and paths.The toughness of a graph G, denoted t(G), was defined by Chvátal [1]: for the complete graph K n we have t( for any cutset S. It follows from the definitions that t(G) ≥ 1 if and only if sc(G) ≤ 0 for any graph G [3].Moreover, Jung [4] calls the scattering number the "additive dual" of the toughness.
A Hamilton cycle in a graph G is a cycle containing every vertex of G. Similarly, a Hamilton path in a graph G is a path that contains every vertex of G.If a graph G has a Hamilton cycle, then sc(G) ≤ 0; and if a graph G has a Hamilton path, then sc(G) ≤ 1 [3].
The following theorem is given by Deogun et al. [3].
The path cover number of a graph G is the smallest number of disjoint paths covering the vertex set of G and is denoted by π(G).For the next theorem a short proof is given in [3] and this theorem was also proven by Lehel without using order-theoretic tools [6].
Now we give some definitions.
Definition 1.3.The connectivity κ = κ(G) of a graph G is the minimum number of vertices whose removal results in a disconnected or trivial graph.
The independence number of G, β(G), is the number of vertices in a maximum independent set of G.
In Section 2, some bounds on the scattering number are given.Section 3 gives several results about the scattering number and graph operations.

2.
Bounds.Firstly, we give two theorems showing the relation between the toughness and the scattering number.
Proof.For any cutset S, we have and so Hence, On the other hand, for every cutset S. Since S is a cutset, we have and so The proof is completed. (2.9) The proof is completed.
Next, we give two theorems containing the relation between some graph parameters and the scattering number.

Theorem 2.3. If a graph G does not contain graph 2K 2 as an induced subgraph, then
(2.10) Proof.If a graph G does not contain graph 2K 2 as an induced subgraph, then we have α (2.11) By (2.11) the proof is completed. (2.12) The proof is completed.

Binomial trees and scattering number.
In this section, we consider the binomial tree B n (Figure 3.1) (see [2]).The binomial tree B n is an ordered tree defined recursively.The binomial tree B 0 consists of a single vertex.The binomial tree B n consists of two binomial trees B n−1 that are linked together: the root of one is the leftmost child of the root of the other.Now we give the scattering number of a binomial tree.
But we can show that max (3.9) But we can show that max The proof is completed.
Proof.To prove this theorem we have two cases.Case 1.If we remove the vertices of graphs B 1 ,B 3 ,...,B n−1 , then the remaining components are B 0 ,B 2 ,...,B n and the number of removing vertices is Moreover, we must delete 2 2(i−1) more vertices from each B 2i where 0 < i ≤ n/2 (except B 0 ).Hence 2 * 2 2(i−1) components are obtained from each B 2i where 0 < i ≤ n/2 (except B 0 ).Then the number of removing vertices is exactly and the number of components is exactly and the number of components is exactly

.22)
Since a t = a t /(a − 1) + c(t) where a ≠ 1, we have The proof is completed.

Call for Papers
Space dynamics is a very general title that can accommodate a long list of activities.This kind of research started with the study of the motion of the stars and the planets back to the origin of astronomy, and nowadays it has a large list of topics.It is possible to make a division in two main categories: astronomy and astrodynamics.By astronomy, we can relate topics that deal with the motion of the planets, natural satellites, comets, and so forth.Many important topics of research nowadays are related to those subjects.By astrodynamics, we mean topics related to spaceflight dynamics.
It means topics where a satellite, a rocket, or any kind of man-made object is travelling in space governed by the gravitational forces of celestial bodies and/or forces generated by propulsion systems that are available in those objects.Many topics are related to orbit determination, propagation, and orbital maneuvers related to those spacecrafts.Several other topics that are related to this subject are numerical methods, nonlinear dynamics, chaos, and control.
The main objective of this Special Issue is to publish topics that are under study in one of those lines.The idea is to get the most recent researches and published them in a very short time, so we can give a step in order to help scientists and engineers that work in this field to be aware of actual research.All the published papers have to be peer reviewed, but in a fast and accurate way so that the topics are not outdated by the large speed that the information flows nowadays.
Before submission authors should carefully read over the journal's Author Guidelines, which are located at http://www .hindawi.com/journals/mpe/guidelines.html.Prospective authors should submit an electronic copy of their complete manuscript through the journal Manuscript Tracking System at http://mts.hindawi.com/according to the following timetable:

Theorem 3 . 1 . 1 .
Let n ≥ 3 be a positive integer.Then sc(B n ) = 2 n−2 .Proof.In Figure 3.1, we call the vertex u top vertex of B n .Let r be the number of removing vertices of B n .If we remove top vertex u of B n , then B n−1 ,B n−2 ,...,B 1 ,B 0 are components.Hence the number of components is n.Now if we remove top vertex of B n−1 , then we obtain the components B n−2 ,B n−3 ,...,B 1 ,B 0 .Then we have 2(n − 1) components.If we continue to remove the top vertex of each component, then we have two cases.Case If r = 2 i where 0 ≤ i ≤ n−1, then the number of remaining components is exactly

. 4 )Definition 3 . 2 .Theorem 3 . 3 . 1 .
The function (n−i)2 i −2 i takes its maximum value at i = n−1/ ln 2−1 .It is obvious that n − 1/ ln 2 − 1 = n − 2 for every n ≥ 3. Hence if we substitute this value in the function (n − i)2 i − 2 i , then the proof is completed.The tensor product of two graphs G = (V (G), E(G)) and H = (V (H), E(H)), denoted by G⊗H, has the vertex set V (G)×V (H), the Cartesian product of V (G) and V (H), and an edge between vertices (x, y) and (u, v), if and only if {x, u} ∈ E(G) and {y,v} ∈ E(H).Let m ≥ 4 and n ≥ 4 be positive integers.Thensc B m ⊗ B n = max 2 n−1 , 2 m−1 .(3.5)Proof.The graph B m ⊗ B n has the graphs B m and B n as subgraphs.We consider these graphs, respectively.Let r be the number of removing vertices from B m ⊗ B n .Then we have two cases, depending on B m or B n .Case Let u 1 ,u 2 ,...,u 2 m be the vertices of B m and let v be the top vertex of B n .If we remove the vertices u i v (i = 1, 2,...,2 m ), then the remaining components areB m ⊗ B n−1 , B m ⊗ B n−2 ,...,B m ⊗ B 1 ,and 2 m B 0 .Now let the top vertex of B n−1 be v .If we remove the vertices u i v (i = 1, 2,...,2 m ), then we obtain the components B m ⊗ B n−2 ,...,B m ⊗B 1 and 2 m B 0 .If we continue to remove the vertices as mentioned above, then we obtain the following cases for r .(a) If r = 2 m 2 i where 0 ≤ i ≤ n−2, then the number of components is

Case 2 .
The function 2 i (n − i) takes its maximum value at i = n − 1/ ln 2 .It is obvious that n − 1/ ln 2 = n − 1 for every n ≥ 4 and so sc B m ⊗ B n = 2 n−1 .(3.10) Let v 1 ,v 2 ,...,v 2 n be the vertices of B n and let u be the top vertex of B m .If we remove the vertices uv i (i = 1, 2,...,2 n ), then the remaining components are B m−1 ⊗ B n , B m−2 ⊗ B n ,...,B 1 ⊗ B n , and 2 n B 0 .Now let the top vertex of B m−1 be u .If we remove the vertices u v i (i = 1, 2,...,2 n ), then we obtain the components B m−2 ⊗ B n ,...,B 1 ⊗ B n and 2 n B 0 .If we continue to remove the vertices as mentioned above, then we obtain the following cases for r .(a) If r = 2 n 2 i where 0 .15) By (3.10) and (3.15) we have sc(B m ⊗