The connected domination polynomial of some graph constructions

The connected domination polynomial of a connected graph G of order n is the polynomial Dc[G,x]=∑i=γcndc(G,i)xi , where d c (G, i) is the number of connected dominating sets of G of cardinality i and γc (G) is the connected domination number of G [5]. In this paper we find the polynomial D c (G, x) for some constructive graphs.


Introduction
The domination polynomial of a graph is introduced by Saeid Alikhani and Yee-hock Peng in [3]. While extending the concept of domination polynomial in view of connected dominating set(cd−set), we came across with many interesting relations among the connected domination polynomials of different graphs [4].
Let Let G be a simple connected graph of order n. A connected dominating set (cd − set) of G is a set S of vertices of G such that every vertex in V \ S is adjacent to some vertex in S and the induced subgraph < S > is connected. The connected domination number γ c (G) is the minimum cardinality of a connected dominating set in G [4]. Considering the polynomial idea of Alikhani et.al., we studied the connected domination(cd− polynomial) polynomial of a connected graph and the information about the graph that we can obtain from the polynomial. For the basic concepts in graph theory we refer mainly Bondy and Murthy [1]. The graphs considered here are all connected and simple of order n.

Conneted domination polynomial
The minimum cardinality of a cd− set is γ c (G) and the maximum cardinality is n. Definition 1. [4]. Let G be a connected graph of order n. The connected domination polynomial

Observations
For any connected graph with n vertices, we have

Main Theorems
The cd− polynomial of the star graph K 1,n is The cartesian product of two graphs G and H is the graph G H with vertex set V (G)×V (H) and the vertices (u, v) and (x, y) are adjacent ⇔ either u = x and vy ∈ E(H) or ux ∈ E(G) and v = y.
An n − book graph is obtained as the cartesian product of the star graph K 1,n and the path graph P 2 .
Theorem 2. For n > 1, the cd− polynomial of the n − book graph is given by Proof. We know that B n is isomorphic to K 1,n P 2 . Let u, v be the two vertices of the common edge. Let {u 1 , u 2 , ..., u n } be the vertices adjacent to u and {v 1 , v 2 , ..., v n } be the vertices adjacent to v. Note thatγ c (B n ) = 2 and since n > 1 every connected dominating set of cardinality i, i ≤ n must include both the vertices u and v and n ≥ 2 the set {u, v} is the only connected dominating set of cardinality 2.
Let i ≤ n. Then every cd− set must contain both u and v. Since u and v dominate all vertices, we can choose the remaining i − 2 vertices arbitrarily from the 2n vertices {u 1 , u 2 , ..., u n , v 1 , v 2 , ..., v n }. Therefore for i ≤ n, there are 2n i−2 cd− sets of cardinality i.
Consider i ≥ n + 1. There are cd− sets containing both u and v and cd− sets containing exactly one of u and v. For cd− sets containing both u and v, we have 2n i−2 choices as above. If a cd− set S contain u, but not v every u i must be in S, in order to dominate v i and to be connected. The remaining vertices can be selected arbitrarily from {v 1 , ..., v n }. So that there are For a vertex w of a graph G, the vertex switching G w of G is obtained by taking a vertex w of G, removing the entire edges incident with w and adding edges joining w to every vertex which are not adjacent to w in G [6].
Theorem 3. The cd− polynomial of the vertex switching of the cycle C n , for n > 6 is Proof. Let {u 1 , u 2 , ..., u n } be the vertices of the cycle C n . Without loss of generality we take the vertex u 1 as the switching vertex of C n , then u 1 is adjacent to all vertices u i except u 2 and u n and denote the resulting graph by G. The cd− number of the vertex switching graph cycle G is 3. For every cycle G, n > 6 there exist only one cd− set of cardinality 3, namely {u 1 , u n−1 , u 3 }. Any cd− set must include the two vertices u n−1 and u 3 to dominate u 2 and u n respectively and for connection.
A cd− set that does not contain u 1 must include all the n − 3 vertices {u 3 , u 4 , ..., u n−1 }. On the other hand any cd− set of cardinality less than n − 3, must include the vertex u 1 . So that a cd− set of cardinality i < n − 3, must include the three vertices u 1 , u 3 and u n−1 . The remaining i − 3 vertices can be chosen arbitrarily from the n − 3 vertices. Hence the number of such sets is n−3 i−3 .
Consider n − 3 ≤ i < n. There are cd− sets containing u 1 and not containing u 1 . For a cd− set containing u 1 , we have n−3 i−3 choices as above. If a cd− set does not contain u 1 , the cd− set For i = n we have one choice, for convenient of the proof we take it as n−3 n−3 choices. Therefore the polynomial is Theorem 4. For an end vertex switching graph P n , n ≥ 5 the cd− polynomial is Proof. Let v n be the switching end vertex of P n . If G denote the switching graph, we have γ c (G) = 2, and {v n , v n−2 } is the only cd− set of cardinality 2. Note that a connected dominating set of G that does not contain v n , must include all the n − 3 vertices {v 2 , v 3 , ..., v n−2 }.
So that any cd− set of less than n − 3 elements must include v n and v n−2 ; and since these two elements are always to dominate the entire graph, the remaining vertices in the cd− set can be chosen arbitrarily from the n − 2 vertices. Thus there are n−2 i−2 cd− sets of cardinality i, for 1 < i < n − 3. Similarly there are two cd− sets of cardinality n − 2, that does not contain v n and n−2 n−4 cd− sets containing v n . Hence the polynomial is Theorem 5. For the vertex switching of the path P n , n ≥ 5, with respect to a vertex which is neither an end vertex, nor a support vertex, the cd− polynomial is . Proof. Let {v 1 , v 2 , ..., v n } be the vertices of the path P n and assume that v i (3 ≤ i ≤ n − 2) be the switching vertex. Note that {v i , v i+2 , v i−2 } is a cd− set and every cd− set must contain the vertices {v i , v i+2 , v i−2 } . So the number of cd− sets of cardinality i have n−3 i−3 choices. Therefore the polynomial is Now the vertex switching graph of K 1,n is disconnected if n = 1 and K 1,2 itself if n = 2. For n > 2, we have the following result. Theorem 6. The cd− polynomial of the vertex switching of the star graph K 1,n , n ≥ 2, with respect to a vertex of degree 1 is Proof. The switching vertex of the star graph should be any vertex other than the centre vertex to retain the graph connected, and the graph is denoted by G. The cd− number of this vertex switching graph of K 1,n is 2. Let {v, v 1 , v 2 , ..., v n } be the vertices of K 1,n with centre v. Without loss of generality we assume that the vertex v 1 is the switching vertex. There are two types of cd− sets of cardinality two, one which contains v but not v 1 and the other which contains v 1 but not v. Both of these together have 2 n−1 1 choices.
From i = 3 onwards the cd− sets of cardinality i fall in 3 categories. One which contains v but not v 1 , second which contains v 1 but not v and the third which contains both v and v 1 . The first 2 types together would contains n−1 i−1 cd− sets in each as above and the third type has n−1 i−2 choices. So that for i = 3 onwards we have A spider is a tree with atmost one vertex of degree more than two, called the center of spider.
Proof. We have γ c (K 1,n,n ) = n+1. Let v be the centre vertex of the graph K 1,n,n , {v 1 , v 2 , ..., v n } be the vertices adjacent to v and u i be the vertices adjacent to v i for each i = 1, 2, ..., n. Any cd− set of cardinality i, i ≥ n + 1 must contain all the vertice {v, v 1 , v 2 , ...v n }.First we consider cd− sets of cardinality n + 1, there is only one such set namely {v, v 1 , v 2 , ...v n }. Any cd− sets of cardinality i, i > n + 1 we have n i choices. Therefore the polynomial is The bispider graph is a graph obtained by edge introducing between two star graphs and the introducing is the rooted vertices, which is denoted by S p 1 ,p 2 of order 2p 1 + 2p 2 + 2 [5].
Theorem 8. The cd− polynomial of the bispider graph S p 1 ,p 2 is Proof. Let u, v be the rooted vertices of the spider graph S p 1 ,p 2 and γ cd (S p 1 ,p 2 ) = p 1 + p 2 + 2. All the cd− sets must consist of all the p 1 + p 2 + 2 vertices of this graph. For p 1 + p 2 + 2 ≤ i ≤ 2p 1 + 2p 2 + 2, the cd− set of cardinality i have p 1 +p 2 i−(p 1 +p 2 +2 choices. Therefore the polynomial is Theorem 9. The cd− polynomial of the connected graph G • mK p is Proof. We haveγ c (G • mK p ) = n. All cd− sets must contain the n vertices of thae graph G, say {v 1 , v 2 , ..., v n } otherwise it wont form a cd− set. The total number of vertices of this graph is nmp + n. There are only one set of cardinality n namely the set {v 1 , v 2 , ..., v n }. For a cd− sets The friendship graph F n can be constructed by joining n copies of the cycle graph C 3 with a common vertex.
Theorem 10. The cd− polynomial of the friendship graph F n n ≥ 2 is given by Proof. The cd− number of F n is one, there is only one such set namely the center vertex v.
Since the blocks of F n are connected through the centre vertex only,the center vertex v must be an element of every cd− set. For i = 2 onwards we have 2n i−1 choices exists. Therefore the polynomial is Theorem 11. The cd− polynomial of the wheel graph W n , on n + 1 vertices, where n ≥ 4 is given by .
Proof. Let {v, v 1 , v 2 , ..., v n } be the vertices of the the graph W n . The cd− number of W n is 1, and the only cd− set of one element is {v}, namely the center vertex v. For 1 < i < n − 2, every cd− sets of cardinality i must include the center vertex v. Since all other vertices are adjacent to v,it follows that there are n i−1 such cd− sets exists. For n − 2 ≤ i ≤ n there are two types of cd− sets, one containing v for which there are n i−1 choices as above and the other which does not contain the vertex v, for which there are n choices. Finally i = n + 1 is the order of W n , and we have a unique cd− set of n + 1 elements. Therefore the polynomial is Note that W n = H ∪ G, the edge disjoint union of H and G(factorization) where H = C n and G = K 1,n . From the above theorem we get The graph G (K) is One point union of k copies of G, which is obtained by taking k copies of G and identifying a fixed vertex of G from each each copy [7].
Theorem 12. The cd− polynomial of the W (k) n , k ≥ 2 with respect to the center vertex is given by Proof. Let v be the apex vertex of the graph W A shellgraph S n is obtained from the cycle graph C n by adding the edges corresponding to the n − 3 concurrent chords of the cycle.
Proof. Let {v 1 , v 2 , ..., v n } be the vertices of the shellgraph S n and γ c (S n ) = 1. Without loss of generality we assume that v 1 is adjacent to all other vertices of the graph S n . For 1 ≤ i ≤ n − 1, all the cd− sets must contain the vertex v 1 , and number of cd− sets of cardinality i is n−1 i−1 . There are two types of cd− sets of cardinality n − 1 and n − 2 exists, the one which is which contain v 1 and the other type is which does not contain the vertex v 1 . For i = n − 2 the cd− sets which does not contain v 1 are {v 2 , v 3 , ..., v n−2 } and {v 3 , v 4 , ..., v n−1 } and the cd− sets which contain v 1 has n−1 n−3 choices as above. For i = n − 1, the set {v 2 , v 3 , ..., v n−1 } is the only cd− set which does not contain v 1 , and also there are n−1 n−2 cd− sets which contain v 1 exists. There is one cd− set of cardinality n exists. Therefore the polynomial is Proof. The bow graph B N has 2N − 1 vertices and γ c (B N , x) = 1. Let v be the apex vertex of B N and v must include in all the cd− sets of B N , otherwise that set wont form a cd− set. We can choose any vertices with v because it is adjacent to all other vertices. For 1 ≤ i ≤ 2N − 1, the number of cd− sets of cardinality i is 2N −2 i−1 . Therefore the polynomial is