Domination in transformation graph G – +–

Abstract Let G = (V, E) be a simple undirected graph of order n and size m. The transformation graph of G is a simple graph with vertex set V(G) ∪ E (G) in which adjacency is defined as follows: (a) two elements in V(G) are adjacent if and only if they are non-adjacent in G (b) two elements in E (G) are adjacent if and only if they are adjacent in G and (c) one element in V(G) and one element in E (G) are adjacent if and only if they are non-incident in G. It is denoted by G – + –. In this paper, we investigate the domination number of transformation graph. We prove that γ (G – + –) ≤ 3 and characterise the graphs for which this number is 1, 2 or 3.


Let
( , ) G V E = be a simple undirected graph of order n and size m.
, then the neighbourhood of v is the set ( ) N v consisting of all ver tices which are adjacent to v. The closed neighbourhood is N v and is denoted by The diameter of G if whenever u and v are vertices of F and uv is an edge of G, then uv is an edge of F as well. A set ( ) S V G 3 is a dominating set if every vertex in V S -is adjacent to at least one vertex in S. The minimum cardinality taken over all dominating sets of G is called the domination number of G and is denoted by l is an edge dominating set of G if every edge in E S -l is adjacent to at least one edge in .
Sl The minimum cardinality taken over all edge dominating sets of G is called the edge domination number of G and is denoted by '( ).
G c An edge dominating set Sl is said to be an independent edge dominating set if no two edges are adjacent in . Sl The minimum cardinality taken over all independent edge dominating sets of G is called the independent edge domination number of G and is denoted by ( ). In [7], Xu and Wu studied about connectivity and independence number of G -+-. They also obtained a necessary and suff icient condition for G -+to be hamiltonian. Several authors [1,2,4,6], have studied domination parameters for diff erent classes of graphs. In this paper we study about domination number in the transformation graph . G -+-Terms not defi ned are used in the sense of [3]. For the graph G in Figure 1.1, u 1 is an isolated vertex in G but it is full vertex in G -+and hence ( ) .
For the graph C e 5 + in Figure 1 Now v will not be adjacent to u and e in G -+and e will not be adjacent to u and v in G -+-. Hence which is a contradition. Therefore G has an isolated vertex. Conversely, assume that G has an isolated vertex say, v. Then by defi nition of G -+-, it is adjacent to all the elements in ( ) is a dominating set. Hence ( ) . Proof. Let D be a minimum dominating set of G -+-. Since each element of ( ) V G is incident with at least one element of ( ) E G and each element of ( ) E G is incident with two elements of ( ) V G , neither a single vertex nor a single edge of G can be a dominating set in G -+-. Hence ( ) . be a cycle on n vertices and D be a dominating set of C -+n . Since Cn is a connected graph, by Theorem 2.3, then any two non-adjacent edges of G dominates all the vertices of .
$ then in C n -+-, v 1 is adjacent to all the elements of ( ) and e n . Then Theorem 2.7. Let G be an isolate free graph of order 4 or 5. If G is not isomorphic to a star, then ( ).
If G is disconnected, then it has two components only say, C 1 and .
! dominates all the vertices and edges of C 2 and any one vertex dominates all the vertices and edges of C 1 . Thus is a dominating set in . G -+-If G is connected, by hypothesis, there are two independent edges 1 e and e 2 of G. Then every edge of G is adjacent to any one of these two edges and every vertex of G is not incident with at least one of these edges. Therefore { , } e e 1 2 is a dominating set in . (ii) two adjacent edges of G.
(iii) one edge and one of its end vertices in G.
x and y are adjacent vertices of G, then the edge ( ) xy E G ! is adjacent to neither x nor y in . G -+-(ii) If x and y are adjacent edges of G, then they are incident with a common vertex, say then v is adjacent to neither x nor y in . G -+-Thus in all cases, S is not a dominating set in . G -+-Lemma 2.9. If ( ) , diam G 2 = then the following 2-element sets are not dominating sets in . G -+-(i) two non-adjacent vertices of G.
(ii) one edge and a vertex other than the end vertices of the edge.
d x y 2 = and hence x and y are adjacent to a common vertex z in G. Then z is adjacent to neither x nor y in .
(a) y is adjacent to at least one of the end vertices of x.
(b) y is adjacent to a vertex yl which is adjacent to xl and xm.
If (a) is true, then both y and x are not adjacent to xl in . G -+-If (b) is true, then yyl is adjacent to neither x nor y in . G -+-If (c) is true, then the edges incident with y in G are adjacent to neither y nor x in . G -+-Thus in all the cases, S is not a dominating set in . G -+-Theorem 2.10. Let G be a connected graph with diam(G) = 2 and .
be a dominating set of . G -+-Since ( ) , diam G 2 = by Lemmas 2.8 and 2.9, x and y are two independent edges of G and D dominates all the edges of G. Thus D is a independent edge dominating set of G and hence ( ) $ there is no edge which dominates all the edges. Hence ( ) . Proof. If , Since v is adjacent to all the vertices and incident with all the edges of , K 1,r v is an isolated vertex in . K -+-1,r Therefore D contains v and hence . Proof. If , n 6 = then any two non-adjacent edges in the rim form a dominating set for W -+- 6 and hence ( ) .
be the vertices of degree 3 and v be the vertex of degree n 1 -in .
, diam W 2 n = by Lemmas 2.8 and 2.9, it is enough to verify that no two independent edges of W n is a dominating set. If the two independent edges are from the rim of the wheel, then n 7 $ implies that there exists a spoke which is not adjacent to any of them. If the two independent edges consist of one spoke and one edge of the rim, then there exists an edge of the rim which is not adjacent to any of them. Hence It is an easy exercise to see that ( ) .
Since any two independent edges in K 4 and K5 form a dominating set in the corresponding transformation graph, we get ( Theorem 2.14. For , n 5 2 .
Proof. Since Kn has no isolated vertex, by Theorem 2.2, can be a dominating set in .
By Lemma 2.8, no two vertices, no two adjacent edges and no set containing one edge and one of it's end vertices form a dominating set in .
is not adjacent to any two independent edges of . Kn Therefore no two independent edges of G is a dominating set of .
such that y is not an end vertex of x, then the end vertices of x are adjacent to neither x nor y in .
Then no set containing one vertex and one edge of G is a dominating set of . G -+-Hence the claim and ( ) .

Graphs of diameter two
In this section, we characterize such graphs. We need the following.

Notation: Let
, be three disjoint stars with centers , x y and z respectively. Then the graph obtained from T T x y , by joining x to one or more vertices of Ty or y to one or more vertices of Tx is denoted by . Txy The graph of the form Txy in which x and y are adjacent is denoted by . Tx y + The graph obtained from T T x y , + z by joining edges from x or y to one or more vertices of Tz or z to one or more vertices of Tx y Let D be a minimum dominating set of . G -+-By Lemmas 2.8 and 2.9, D contains only two independent edges of G.
Then any edge in

Case (i):
Only one vertex of S is of degree greater than or equal to 3. If

Case (ii):
Only two vertices of S are of degree greater than or equal to 3.
Let v i and j v be the vertices of degree greater than or equal  If v i and j v are not adjacent to any vertex of [ ] No vertex of S is of degree greater than or equal to 3.
By Lemmas 2.8 and 2.9, any minimum dominating set D in G -+contains only two independent edges of G.
, then also we get contradiction. Hence the claim. Therefore ,.
Then v i and j v may be adjacent to vertices of ( ) ,. Then Proof. Assume that and v 4 be the only neighbours of v and let v v 1 2 and v v 3 4 be the independent edges in ( ) N v . v v 1 2 dominates all the vertices except v 1 and v 2 and all the edges incident with v 1 and v 2 . But v 3 v 4 dominates v 1 and v 2 and all the edges incident with v 3 and v 4 . Hence ( ) Conversely, assume that ( ) By hypothesis, n 6 $ .
Claim: n 7 = and ( ) N v contains two independent edges.

Suppose
( ) N v does not contain any edge. By Lemmas 2.8 and 2.9, any minimum dominating set D in G -+contains only two independent edges of G. Since ! and since there exists , respectively. Hence ( ) which is a contradiction. Therefore ( ) N v has two or more edges. Let Hence the claim and proof. , . Since G has no isolated vertex, by Theorem 2. , is a dominting set of G -+-. Hence .
By hypothesis, . n 5 $ By Lemmas 2.8 and 2.9, any dominating set D of G -+contains only two independent edges of G. Since u u C ! 1 2 is not adjacent to any element of B, no two edges of B can form a dominating set in G -+-. For the same reason, one edge of A and one edge of B can not serve the purpose. Then we consider the following three cases.

Case (i): D contains two edges of C
Since two edges are independent, u u D g Since there exists some a k which is not dominated by , Without loss of generality, let Since D is an edge dominating set of G, every edge B must be adjacent with a i and/or c j 2 and hence the end vertices of such edges must be adjacent to v i and/or j v . If B has no such vertices, then ( ) N v K , D . If such vertices are adjacent to v i or j v alone, then    Proof. Assume that , , , , .
. Then every G i has two independent edges v v 2 1 and v v 3 4 which are adjacent to all the edges of G.
. By Lemmas 2.8 and 2.9, any minimum dominating set must contain two independent edges.

Claim 1:
( ) N v contains two independent edges.

Suppose not. Then
Therefore v 1 can not be adjacent to both u 1 and u 2 . Since ( ) , diam G 2 = both u 1 and u 2 are adjacent to all the vertices of ( ) N v Since diam(G) 2 = , v 3 and v 4 must be adjacent to u 1 and u 2 . Therefore ( ) N u 3 $ i . For the same reason, at least one of v 1 and v 2 is adjacent to u 1 and u 2 . Since , n 7 = no two edges of C can form a dominating set of G -+-. Since is not a dominating set of G -+-. Therefore any 2-element set containing one element of A and one element of B can not serve as a dominating set. Since ( ) N u 3 $ i no 2-element set containing one edge of A and one edge of C can be a dominating set of G -+-. If Since diam(G) 2 = , v 4 must be adjacent to u 1 and u 2 . Since deg(v 2 ) 3 = , in ( ) , N v v 2 is adjacent to at most one of u 1 and u 2 . If v 2 is not adjacent to u 1 and u 2 , then u 1 and u 2 are adjacent to at least one of v 1 and v 3 . Let u 1 be adjacent to v 1 and u 2 be adjacent to v 3 . If v 2 is adjacent to u 1 , then v 1 and v 3 must be adjacent to u 2 . In both cases, no two independent edges form a dominating set of G -+which is a contradiction. Hence the claim.

Claim 2:
( ) , N v K P 2 , 4 2 or . K e + , 1 3 Suppose not. Then ( ) N v K K e -= 4 4 , or C 4 and all the vertices of ( ) N v are of degree at least 3 in [ ] . N v Hence u 1 and u 2 can not be adjacent to a common vertex in ( ) Hence Claim 2.
, 2 = each u i is adjacent to at least two non-adjacent vertices of ( )

Subcase (a):
( ) . Then v 2 must be adjacent to v 1 and v 4 . Since diam(G) , 2 = u 1 must be adjacent to at least one of v 1 and v 4 . If u 1 is adjacent to v 1 , then

Subcase (c): At least one of v 2 and v 3 is adjacent to a vertex of
Let v 2 be adjacent to u 1 . Then u 2 must be adjacent to v 1 and v 4 . Since diam(G) If v 3 is adjacent to u 1 , then also we get G 8 or . Let v 1 be of degree 3 and v 4 be of degree 2 in [ ] N v . Then v 1 can not adjacent to u 1 and u 2 . Also v 2 and v 3 are adjacent to at most one of u 2 and u 1 . Let v 2 be adjacent to u 1 and v 3 be adjacent to u 2 . Since diam(G) , 2 = u 1 and u 2 are adjacent to v 4 . In this case, . G G , 11 Theorem 3.8. Let G be a graph of order n 8 $ with diam(G)=2 and v be a vertex of degree

H T T T T T T x y z N u
Since G has no isolated vertex, by Theorem 2.2, ( ) .  , is a dominating set of G -+-. Thus ( ) Let D be a minimum dominating set of G -+-. By Lemmas 2.8 and 2.9, D contains only two independent edges of G.
$ there is an edge a i which is not dominated by any 2-element set of B, or any 2-element set of C or any 2-element set containing one element of B and one element of C. Also there is an edge c ij which is not dominated by any 2-element set containing one element of A and one element of B or any set containing one element of A and one element of C. Thus ( ) G 2 ! c -+which is a contradiction. Hence the claim.

Since
, n 8 $ no two edges of B form a dominating set of G -+-. For the same reason no two edges of C and no set containing one edge of B and one edge of C can form a dominating set of G -+-. Then we consider the following two cases.

Subcase (a): ( )
Since diam(G) = 2, u 1 and u 2 must be adjacent to j v . Hence Case (ii): D contains one edge of A and one edge of C.
Since D is an edge dominating set of G, every edge of B must be adjacent to a i and/or j 1 c and hence every vertex of B must be adjacent to v i and/or j v .
Since G has no isolated vertex, by , there exists at least one vertex in ( ) ( ) N u N u + 2 1 and let it be j v . If H T T x , , y or Txy , then every edge of A and every edge of B which is adjacent to v i and c 1i or 2 c i or both are adjacent to a i . Further all other edges are incident with j v and u 2 . Therefore then every edge of A and every edge of B and C which is incident with v k are adjacent to a k . By assumption, every edge of B and C which is adjacent Let D be a minimum dominating set of . G -+-By Lemmas 2.8 and 2.9, D contains only two independent edges of G.

Claim:
( ) Since u 1 and u 2 are non-adjacent and diam(G) 2 = , $ there is an edge a i which is not dominated by any 2-element subset of B or 2-elements subset of C. Also there exists an edge c i 1 which is not dominated by j { , } a b rs and there exists an edge 2 c i which is not dominated by k { , }. a b rs Therefore any 2-element set containing one element of A and one element of B can not form a dominating set. Similarly, no 2-element set containing one element of B and one element of C can form a dominating set. Since there exists an edge c 1j which is adjacent to neither a i nor c 2k and there exists an edge c 2j which is adjacent to neither a i nor 1 c k , any 2-element set containing one edge of A and one edge of C can not be a dominating set. Thus which is a contradiction. Hence the claim. Since , n 8 $ no two edges of B or no two edges of C is a dominating set of G -+and also any set containing one edge in B and one edge in C is not a dominating set. We consider the following two cases. Then Since u 1 and u 2 are non-adjacent and diam(G) = 2, v i must be adjacent to v k or j .
v If j v is adjacent to all the elements in S 1 , then v i and v k may be adjacent to some element of S 1 . If v i and v k are not adjacent to any element of , If j v is not adjacent to all elements of S 1 , then v i and v k must be adjacent to some elements of S 1 . Then Let v i be adjacent to . v k Also every element in S 1 must be adjacent to at least one of Since G has no isolated vertex, by Theorem 2.2, is adjacent to v i and j v only. Then all the edges of A and all the edges of B and C which are incident with v i are adjacent to a i . Further all the edges of B and C which are incident with j v are adjacent to c j 1 . Therefore { , } a c 1j i is a dominating set of G -+-.
If H T( Then all the edges of A and all the edges of B and C which are incident with v l are adjacent to l a . Further all the other edges of B and C are adjacent to b ij . Therefore Let D be a minimum dominating set of G -+-. By Lemmas 2.8 and 2.9, D contains only two independent edges of G. Since n 8 $ , there is an edge a i which is not adjacent to any two edges of C. Therefore no two edges of C can form a dominating set of G -+-. For the same reason one edge of B and one edge of C can not form a dominating set of G -+-. Then we consider the following cases.

Case (i): D contains one edge of A and one edge of B
. Since diam(G) = 2 and must be adjacent to 2-element subset or 3-element subset of S. Let S 1 be the set of vertices in ( ) N v S -which are adjacent to one or more vertices of S and . S n k 3 = -- 1 Since diam(G) = 2, ( ) .
We claim that ( ) .
which is a contradition and hence the claim. Since D is a dominating set, ( ) N S 1 must be a 2-element subset or 3-element subset of S. If ( ) , Case (ii): D contains one edge of A and one edge of C.
Case (iii): D contains two edges of B.
Therefore B contains only two edges which are independent. Hence Theorem 3.11. Let G be a graph of order n 8 $ with ( ) diam G 2 = and let v be a vertex of degree , N v is adjacent to u i or/and j u . Then is a dominating set of . G -+-Then v l may be adjacent to vertices of ( ).
N v If v l is adjacent to one or more vertices of ( ), Theorem 3.12. Let G be a graph of order n 8 $ with ( ) diam G 2 = and let v be a vertex of degree , .
Since G has no isolated vertex, by Theorem 2. $ there exists a i which is not adjacent to any two edges of C. For the same reason, { , } b u u ij r s is not a dominating set. Then the following cases are possible.
u j which is a contradiction. If j , u u D ! l then every vertex of ( ) [ ] V G N v -must be adjacent to j u and/or l u . Therefore