Tadpole Domination in Graphs

: A new type of the connected domination parameters called tadpole domination number of a graph is introduced. Tadpole domination number for some standard graphs is determined, and some bounds for this number are obtained. Additionally, a new graph, finite, simple, undirected and connected, is introduced named weaver graph. Tadpole domination is calculated for this graph with other families of graphs.


Introduction:
Let = ( , ) be finite, simple, connected and undirected graph where denotes its vertices set and its edges set. A degree of a vertex of any graph is defined as the number of edges incident on . It is denoted by ( ). The minimum and maximum degrees of vertices in are denoted by ( ) and ∆( ), respectively. A graph is called connected if there is a path from any vertex to any other vertex in the graph. A graph which is not a connected one is called a disconnected graph, the disconnected graph can contain at least two of the connected graphs, each of these is called component. A cut-vertex of a graph is a vertex where removing it from the graph would increase the number of components.
The girth of a graph is the length of the shortest cycle contained in a graph. A Hamiltonian graph is a graph possessing a cycle that goes through all the vertices of . A set ⊆ of vertices in a graph = ( , ) is called a dominating set if every vertex ∈ is either an element of or is adjacent to an element of . The domination number of ( ) is the minimum cardinality taken over all dominating sets in G (1,2). Many authors have introduced different types of domination parameters through adding conditions on the dominating set. Sampathkumar  In network you need the communication between its members. Such as the internet or electricity or any social network in daily life, facilitating the business task and reducing the cost. Refer to (4,5,6,7,8,9,10,11,12) for more types of domination in graphs.
In this work, a new type of connected graph domination is introduced. If the vertices of a dominating set in a graph form a sub graph as a tadpole graph then these vertices represent a tadpole dominating set in this graph. Some results for this new domination are determined. Also, a new model of graph called weaver is initiated and its tadpole domination, is calculated.  if ∈ is a cut vertex then, ∈ . Proof. Let has tadpole dominating set. Then has , as a sub graph such that each vertex of set − is adjacent to at least one vertex in . Let be a cut-vertex in . Since is a cut vertex of , − is disconnected and has at least two components. From a partition of − { } by letting consist of the vertices of one of these components and the vertices of the others. Then any two vertices ∈ and ∈ lie in different components of − . Therefore, every − path in contains . If ∈ − , and is dominated by some vertices say in then the vertices of are not dominated by any vertex in . Therefore, ∈ . Proposition 2.8. For any complete graph of order ≥ 4, ( )= 4.

Main results
Proof. The girth of the complete graph is a cycle of order three. Since is a complete graph then each vertex from these three vertices is adjacent to all other vertices of we need exactly to one other vertex to be the vertex of the path that adjacent to this cycle to obtain a tadpole dominating set. Therefore, ( )= 4.

Theorem 2.11. A connected graph has a tadpole domination if and only if:
i) There exist a maximal path such that ( ) dominates .
ii) The maximal path dominates a cycle in such that there exists at most one path of order greater than 2, which is common with one vertex with this cycle.
Proof. If a graph is connected and has a tadpole domination number, then there is a minimum tadpole dominating set say such that it contains a cycle of order and a path of order such that all vertices of are dominated by the vertices these cycle and the path. [ ] contains a path forms from the vertices of the cycle and the path, where they have one vertex in common. Then contains a path of order + , say 1 . This path is included in a maximal path say in . Therefore, (i) is holds. It's clear that (ii) is holds from the above proof. Conversely, it's clear that if (i) and (ii) are satisfied, then is connected and has a tadpole dominating set, so it has a tadpole domination number Corollary 2.12. Let be a graph containing at least three simple cycles joined with a common vertex with at least four vertices in every cycle then there is no tadpole dominating set for this graph. Proposition 2.16. Let and be two graphs such that + is not isomorphic to a cycle nor to a tree then ( + ) = 4,5. Proof. There are three cases as follows: Case 1. If is 1 and is a null graph of order , then + ≅ +1 ( +1 , is a star). Thus, in this case + has no tadpole dominating set. Now, if and are two null graphs of order two, then + ≅ 4 , again + has no tadpole dominating set.

Case 2.
If at least one of the two graphs or is not a null graph then without loss of generality say . So, there is an edge say in , which forms a cycle in + with any vertex in . At least one of the vertices of this cycle becomes of degree greater than or equal to three. Thus, there is a path that is not contained in this cycle and joins with it by one edge. One can conclude that, this path is of order one such that a sub graph 3,1 that contains these cycle and path which are mentioned above dominates all other vertices in + . Thus, ( + ) = 4.

Proof.
Suppose that has a tadpole domination number ( )= 4. Since is connected and is not isomorphic to 4 nor to 4 , then one can conclude that ≅ 4 , 4 − { }, ∈ 4 and 3,1 . Conversely, If ≅ 4 , 4 − { }, ∈ 4 , and 3,1 then ( )=4, Since each of these graphs contains spanning sub graph isomorphic to 3,1 . (Characterizations 2.15.(4)).    Proof. Every vertex in is adjacent to two vertices in the outer ring. Let the vertices of are V 1 ={ 1 , 2 , 3 , … , } and the outer ring 2 = { 1 , 2 , 3 , … , }. Since is complete, so one vertex dominates all its vertices. The girth in is 3, therefore, a cycle of length 3 is chosen to belong to a dominating set. Every vertex in 1 with odd label is adjacent to different and independent vertices from 2 . These vertices are contained in a path, choosing any three of these vertices forms a cycle, such that −1 belongs to the dominating set. So, let 1 , 3 , and −1 are the vertices belonging to a set and the remaining vertices of set 1 with odd labels representing a path that belongs to as follows:

Tadpole domination in some graphs
= { 2 −1 , = 1,2, … , ⌈ 2 ⁄ ⌉ }, vertices of that dominate all the vertices of thus, To prove the reverse inequality, the induction method on the number of vertices is used. The results are clear if = 7. Suppose that the result is true for all sun graphs of order less than . Then = 1 ∪ 2 where 1 is the minimum tadpole dominating set that dominates the sun graph with order less than , when is odd. By induction | 1 | ≥ ⌈ −1 2 ⌉, and hence when one vertex is added to the sun graph ( − 1 order of the complete graph) one vertex is needed to dominate it. Therefore, | | ≥ ⌈     For the same reason in Lemma (3.6), the graph is dominated by vertices of 2 .

Case 3. If
Hence, the vertices in represent the minimum tadpole dominating set. (As an example see Fig. 6 c). Example 3.8. For the graph Y m,n , ( 9,6 ) = 23, ( 7,7 ) = 22 , ( 5,4 ) = 9. In the following definition, a new graph is introduced named "weaver graph". Definition 3.9. The "weaver graph" is a simple graph denoted by ,ℎ . It is formed by warping the path of the tadpole graph ,ℎ around itself ( is the order of cycle in ,ℎ and ℎ is the order of path in each copies of the path).

Proof.
The proof is in the same manner that is used in Theorem 3.7.
Conflicts of Interest: None.