A topological index of a graph is a numerical quantity which is structural invariant, i.e. it is fixed under graph automorphism. The simplest topological indices are the number of vertices and edges of a graph. In this paper, we define and study a new topological index called multiplicative degree distance. All graphs considered are simple and connected graphs.

We denote the vertex and the edge set of a graph G by V ( G ) and E ( G ) , respectively. d G ( v ) denotes the degree of a vertex v in G. The number of elements in the vertex set of a graph G is called the order of G and is denoted by v ( G ) . The number of elements in the edge set of a graph G is called the size of G and is denoted by e ( G ) . A graph with order n and size m edges is called a ( n , m ) -graph. For any u , v ∈ V ( G ) , the distance between u and v in G, denoted by d G ( u , v ) , is the length of a shortest ( u , v ) -path in G. The edge connective the vertices u and v will be denoted by uv. The complement G ¯ of the graph G is the graph with vertex set V ( G ) , in which two vertices in G ¯ are adjacent if and only if they are not adjacent in G.

The join of graphs G 1 and G 2 is denoted by G 1 + G 2 , and it is the graph with vertex set V ( G 1 ) ∪ V ( G 2 ) and the edge set E ( G 1 + G 2 ) = E ( G 1 ) ∪ E ( G 2 ) ∪ { u 1 u 2 | u 1 ∈ V ( G 1 ) , u 2 ∈ V ( G 2 ) } . The composition of graphs G 1 and G 2 is denoted by G 1 [ G 2 ] , and it is the graph with vertex set V ( G 1 ) × V ( G 2 ) , and two vertices u = ( u 1 , u 2 ) and v = ( v 1 , v 2 ) are adjacent if ( u 1 is adjacent to v 1 ) or ( u 1 = v 1 and u 2 and v 2 are adjacent). The disjunction of graphs G 1 and G 2 is denoted by G 1 ∨ G 2 , and it is the graph with vertex set V ( G 1 ) × V ( G 2 ) and E ( G 1 ∨ G 2 ) = { ( u 1 , u 2 ) ( v 1 , v 2 ) | u 1 v 1 ∈ E ( G 1 )   or   u 2 v 2 ∈ E ( G 2 ) } . The symmetric di- fference of graphs G 1 and G 2 is denoted by G 1 ⊕ G 2 , and it is the graph with vertex set V ( G 1 ) × V ( G 2 ) and edge set E ( G 1 ⊕ G 2 ) = { ( u 1 , u 2 ) ( v 1 , v 2 ) ) | u 1 v 1 ∈ E ( G 1 )   or   u 2 v 2 ∈ E ( G 2 )   butnotboth } .

Let G be a connected graph. The Wiener index W ( G ) of a graph G is defined as

W ( G ) = ∑ { u , v } ⊆ V ( G ) d G ( u , v ) = 1 2 ∑ u , v ∈ V ( G ) d G ( u , v ) .

Dobrynin and Kochetova [1] and Gutman [2] independently proposed a vertex-degree-Weighted version of Wiener index called degree distance or Schultz molecular topological index, which is defined for a connected graph G as

D D ( G ) = ∑ { u , v } ⊆ V ( G ) d G ( u , v ) [ d G ( u ) + d G ( v ) ] = 1 2 ∑ u , v ∈ V ( G ) d G ( u , v ) [ d G ( u ) + d G ( v ) ] .

The Zagreb indices have been introduced more than thirth years ago by Gutman and Trianjestic [3] . The first Zagreb index M 1 ( G ) of a graph G is defined as

M 1 ( G ) = ∑ u v ∈ E ( G ) [ d G ( u ) + d G ( v ) ] = ∑ v ∈ V ( G ) d G 2 ( v ) .

The second Zagreb index M 2 ( G ) of a graph G is defined as

M 2 ( G ) = ∑ u v ∈ E ( G ) d G ( u ) d G ( v ) .

The Zagreb indices are found to have applications in QSPR and QSAR studies as well, see [4] .

Note that contribution of nonadjacent vertex pair should be taken into account when computing the Weighted Wiener Polynomials of certain Composite graphs, see [5] . A.R. Ashrafi, T. Doslic, A. Hamzeha, [6] [7] defined the first Zagreb coindex of a graph G is

M ¯ 1 ( G ) = ∑ u v ∉ E ( G ) [ d G ( u ) + d G ( v ) ]

The second Zagreb coindex of a graph G is

M ¯ 2 ( G ) = ∑ u v ∉ E ( G ) d G ( u ) d G ( v ) ,

respectively.

In [8] , Hamzeh, Iranmanesh Hossein-Zadeh and M.V. Diudea recently introduced the generalized degree distance of graphs. Asma Hamzeh, Ali Iranmanesh and Samaneh Hossein-Zadeh, Cartesian product, composition, join, disjunction and symmetric difference of graphs and introduce generalized and modified generalized degree distance Polynomials of graphs, such that their first derivatives at x = 1 , see [9] .

In this paper, we defne a new graph invariant named multiplicative version of degree distance of a graph denoted by D D * ( G ) and defined by

[ D D * ( G ) ] 2 = ∏ u , v ∈ V ( G ) , u ≠ v d G ( u , v ) [ d G ( u ) + d G ( v ) ] .

Therefore the study of this new topological index is important and we have obtained Sharp upper bounds for the graph operations such as join, disjunction, composition, symmetric difference of graphs.

Lemma 2.1. [10] [11] , Let G 1 and G 2 be two simple connected graphs. The number of vertices and edges of graph G i is denoted by n i and e i respectively for i = 1 , 2 . Then we have

1. d G 1 + G 2 ( u , v ) = ( 1, u v ∈ E ( G 1 )   or   u v ∈ E ( G 2 )   or   ( u ∈ V ( G 1 )   and   v ∈ V ( G 2 ) ) 2, otherwise

For a vertex u of G 1 , d G 1 + G 2 ( u ) = d G 1 ( u ) + n 2 , and for a vertex v of G 2 , d G 1 + G 2 ( v ) = d G 2 ( v ) + n 1 .

2. d G 1 [ G 2 ] ( ( u 1 , v 1 ) , ( u 2 , v 2 ) ) = ( d G 1 ( u 1 , u 2 ) ,               u 1 ≠ u 2 1, u 1 = u 2 , v 1 v 2 ∈ E ( G 2 ) 2,           otherwise

d G 1 [ G 2 ] ( u , v ) = n 2 d G 1 ( u ) + d G 2 ( v ) .

3. d G 1 ∨ G 2 ( ( u 1 , v 1 ) , ( u 2 , v 2 ) ) = ( 1 , u 1 u 2 ∈ E ( G 1 )   or   v 1 v 2 ∈ E ( G 2 ) 2 , otherwise

d G 1 ∨ G 2 ( ( u , v ) ) = n 2 d G 1 ( u ) + n 1 d G 2 ( v ) − d G 1 ( u ) d G 2 ( v ) .

4. d G 1 ⊕ G 2 ( ( u 1 , v 1 ) , ( u 2 , v 2 ) ) = ( 1 , u 1 u 2 ∈ E ( G 1 )   or   v 1 v 2 ∈ E ( G 2 )   butnotboth 2 , otherwise

d G 1 ⊕ G 2 ( ( u , v ) ) = n 2 d G 1 ( u ) + n 1 d G 2 ( v ) − 2 d G 1 ( u ) d G 2 ( v ) .

Lemma 2.2. (Arithmetic Geometric inequality)

Let x 1 , x 2 , ⋯ , x n be non-negative numbers. Then x 1 + x 2 + ⋯ + x n n ≥ x 1 x 2 ⋯ x n n

Remark 2.3. For a graph G, let A ( G ) = { ( x , y ) ∈ V ( G ) × V ( G ) | x and y are adjacent in G } and let B ( G ) = { ( x , y ) ∈ V ( G ) × V ( G ) | x and y are not adjacent in G }. For each x ∈ V ( G ) , ( x , x ) ∈ B ( G ) . Clearly, A ( G ) ∪ B ( G ) = V ( G ) × V ( G ) . Let C ( G ) = { ( x , x ) | x ∈ V ( G ) } and D ( G ) = B ( G ) − C ( G ) . Clearly B ( G ) = C ( G ) ∪ D ( G ) , C ( G ) ∩ D ( G ) = ∅ .

The summation ∑ ( x , y ) ∈ A ( G ) runs over the ordered pairs of A ( G ) . For simplicity, we write the summation ∑ ( x , y ) ∈ A ( G ) as ∑ x y ∈ G . Similarly, we write the summation ∑ ( x , y ) ∈ B ( G ) as ∑ x y ∉ G . Also the summation ∑ x y ∈ E ( G ) runs over the edges of G. We denote the summation ∑ x , y ∈ V ( G ) by ∑ x , y ∈ G and similarly ∑ x , y ∈ V ( G ) by ∏ x , y ∈ G . The summation ∑ ( x , y ) ∈ D ( G ) eqivalent to ∑ x y ∉ G , x ≠ y and similarly the summation ∑ ( x , y ) ∈ C ( G ) eqivalent to ∑ x y ∉ G , x = y .

Lemma 2.4. Let G be a graph. Then

∑ x y ∈ G 1 = 2 e ( G )

Proof:

∑ x y ∈ G 1 = 2 ∑ x y ∈ E ( G ) 1 = 2 e ( G )

Lemma 2.5.

∑ x y ∈ G d G ( x ) = M 1 ( G )

Proof: Let x ∈ V ( G ) and t = d G ( x ) . Let y 1 , y 2 , ⋯ , y t be the neighbours of x. Each orderd pair ( x , y i ) ,   1 ≤ i ≤ t , contributes d G ( x ) to the sum. Thus these orderd pairs contribute d G 2 ( x ) to the sum. Hence

∑ x y ∈ G d G ( x ) = ∑ x ∈ V ( G ) d G 2 ( x ) = M 1 ( G )

Lemma 2.6.

∑ x y ∈ G d G ( x ) d G ( y ) = 2 M 2 ( G )

Proof: Clearly,

∑ x y ∈ G d G ( x ) d G ( y ) = 2 ∑ x y ∈ E ( G ) d G ( x ) d G ( y ) = 2 M 2 ( G ) .

Lemma 2.7.

∑ x y ∉ G 1 = 2 e ( G ¯ ) + v ( G )

Proof:

∑ x y ∉ G 1 = ∑ ( x , y ) ∈ D ( G ) 1 + ∑ ( x , x ) ∈ C ( G ) 1 = 2 e ( G ¯ ) + v ( G )

Lemma 2.8.

∑ x y ∉ G d G ( x ) = 2 e ( G ¯ ) ( v ( G ) − 1 ) + 2 e ( G ) − M 1 ( G ¯ )

Proof.

∑ x y ∉ G d G ( x ) = ∑ ( x , y ) ∈ D ( G ) d G ( x ) + ∑ ( x , x ) ∈ C ( G ) d G ( x ) = ∑ ( x , y ) ∈ D ( G ) { v ( G ) − 1 − d G ¯ ( x ) } + ∑ ( x , x ) ∈ C ( G ) d G ( x ) = ( v ( G ) − 1 ) ∑ ( x , y ) ∈ D ( G ) 1 − ∑ ( x , y ) ∈ D ( G ) d G ¯ ( x ) + 2 e ( G ) = ( v ( G ) − 1 ) 2 e ( G ¯ ) − ∑ ( x , y ) ∈ A ( G ¯ ) d G ¯ 2 ( x ) + 2 e ( G ) = ( v ( G ) − 1 ) 2 e ( G ¯ ) − ∑ x y ∈ G ¯ d G ¯ 2 ( x ) + 2 e ( G ) = 2 e ( G ¯ ) ( v ( G ) − 1 ) + 2 e ( G ) − M 1 ( G ¯ )         by   Lemma       2.5

Lemma 2.9.

∑ x y ∉ G d G ( x ) d G ( y ) = 2 M ¯ 2 ( G ) + M 1 ( G )

Proof:

∑ x y ∉ G d G ( x ) d G ( y ) = ∑ ( x , y ) ∈ D ( G ) d G ( x ) d G ( y ) + ∑ ( x , x ) ∈ C ( G ) d G ( x ) d G ( x ) = 2 ∑ x y ∉ E ( G ) d G ( x ) d G ( y ) + ∑ x ∈ V ( G ) d G 2 ( x ) = 2 M ¯ 2 ( G ) + M 1 ( G )

Lemma 2.10.

∑ x y ∉ G [ d G ( x ) + d G ( y ) ] = 2 M ¯ 1 ( G ) + 4 e ( G )

Proof:

∑ x y ∉ G [ d G ( x ) + d G ( y ) ] = ∑ ( x , y ) ∈ C ( G ) [ d G ( x ) + d G ( y ) ] + ∑ ( x , y ) ∈ D ( G ) [ d G ( x ) + d G ( y ) ] = ∑ x ∈ V ( G ) 2 d G ( x ) + 2 ∑ x y ∉ E ( G ) [ d G ( x ) + d G ( y ) ] = 4 e ( G ) + 2 M ¯ 1 ( G )

Theorem 3.1. Let G i , i = 1 , 2 , be a ( n i , m i ) -graph. Then

[ D D * ( G 1 [ G 2 ] ) ] 2 ≤ { 1 n 1 n 2 ( n 1 n 2 − 1 ) [ 4 M 1 ( G 2 ) W ( G 1 ) + 4 n 2 m 2 D D ( G 1 ) + 4 n 1 M ¯ 1 ( G 2 )         + 8 n 2 2 m 1 ( n 2 − 1 ) + 2 n 1 M 1 ( G 2 ) + 8 m 1 n 2 m 2 + 4 W ( G 1 ) M ¯ 1 ( G 2 )   + 2 n 2 D D ( G 1 ) ( 2 m ¯ 2 + n 2 ) ] } n 1 n 2 ( n 1 n 2 − 1 )

Proof:

[ D D * G 1 [ G 2 ] ] 2 ≤ { 1 n 1 n 2 ( n 1 n 2 − 1 ) ( S 3 + S 1 + S 2 + S 4 ) } n 1 n 2 ( n 1 n 2 − 1 )

where S 3 , S 1 , S 2 , S 4 are terms of the above sums taken in order.

Next we calculate S 1 , S 2 , S 3 and S 4 separately.

S 1 = ∑ x , y ∈ G 1 , x ≠ y         ∑ u v ∈ G 2 d G 1 [ G 2 ] ( ( x , u ) , ( y , v ) ) [ d G 1 [ G 2 ] ( x , u ) + d G 1 [ G 2 ] ( y , v ) ] = ∑ x , y ∈ G 1 , x ≠ y         ∑ u v ∈ G 2 d G 1 ( x , y ) [ d G 2 ( u ) + n 2 d G 1 ( x ) + d G 2 ( v ) + n 2 d G 1 ( y ) ]         by   Lemma   2 .1 = n 2 ∑ x , y ∈ G 1 , x ≠ y d G 1 ( x , y ) [ d G 1 ( x ) + d G 1 ( y ) ] ∑ u v ∈ G 2 1 + ∑ x , y ∈ G 1 , x ≠ y d G 1 ( x , y ) ∑ u v ∈ G 2 [ d G 2 ( u ) + d G 2 ( v ) ] = 4 n 2 m 2 D D ( G 1 ) + 4 M 1 ( G 2 ) W ( G 1 )

S 2 = ∑ x , y ∈ G 1 , x = y         ∑ u v ∉ G 2 d G 1 [ G 2 ] ( ( x , u ) , ( y , v ) ) [ d G 1 [ G 2 ] ( x , u ) + d G 1 [ G 2 ] ( y , v ) ] = ∑ x , y ∈ G 1 , x = y { ∑ u v ∉ G 2 , u = v d G 1 [ G 2 ] ( ( x , u ) , ( y , v ) ) [ d G 1 [ G 2 ] ( x , u ) + d G 1 [ G 2 ] ( y , v ) ]     + ∑ u v ∉ G 2 , u ≠ v d G 1 [ G 2 ] ( ( x , u ) , ( y , v ) ) [ d G 1 [ G 2 ] ( x , u ) + d G 1 [ G 2 ] ( y , v ) ] } = 0 + ∑ x , y ∈ G 1 , x = y         ∑ u v ∉ G 2 , u ≠ v d G 1 [ G 2 ] ( ( x , u ) , ( y , v ) ) [ d G 1 [ G 2 ] ( x , u ) + d G 1 [ G 2 ] ( y , v ) ] = ∑ x , y ∈ G 1 , x = y         ∑ u v ∉ G 2 , u ≠ v d G 1 ( x , y ) [ d G 2 ( u ) + n 2 d G 1 ( x ) + d G 2 ( v ) + n 2 d G 1 ( y ) ]         by   Lemma   2 .1   = 2 ∑ u v ∉ G 2 , u ≠ v [ d G 2 ( u ) + d G 2 ( v ) ] ∑ x , y ∈ G 1 , x = y 1 + 2 n 2 ( ∑ u v ∉ G 2 , u ≠ v 1 ) ∑ x , y ∈ G 1 , x = y [ d G 1 ( x ) + d G 1 ( y ) ] = 4 n 1 M ¯ 1 ( G 2 ) + 8 n 2 m 1 n 2 ( n 2 − 1 )

S 3 = ∑ x , y ∈ G 1 , x = y         ∑ u v ∈ G 2 d G 1 [ G 2 ] ( ( x , u ) , ( y , v ) ) [ d G 1 [ G 2 ] ( x , u ) + d G 1 [ G 2 ] ( y , v ) ] = 1 ⋅ ∑ x , y ∈ G 1 , x = y         ∑ u v ∈ G 2 [ d G 1 [ G 2 ] ( x , u ) + d G 1 [ G 2 ] ( y , v ) ] = ∑ x , y ∈ G 1 , x = y         ∑ u v ∈ G 2 [ d G 2 ( u ) + d G 1 ( x ) n 2 + d G 2 ( v ) + n 2 d G 1 ( y ) ]         by   Lemma   2 .1   = ( ∑ x , y ∈ G 1 , x = y 1 ) ∑ u v ∈ G 2 [ d G 2 ( u ) + d G 2 ( v ) ] + n 2 ∑ x , y ∈ G 1 , x = y [ d G 1 ( x ) + d G 1 ( y ) ] ( ∑ u v ∈ G 2 1 ) = 2 n 1 M 1 ( G 2 ) + 8 n 2 m 1 m 2

S 4 = ∑ x , y ∈ G 1 , x ≠ y         ∑ u v ∉ G 2 d G 1 [ G 2 ] ( ( x , u ) , ( y , v ) ) [ d G 1 [ G 2 ] ( x , u ) + d G 1 [ G 2 ] ( y , v ) ] = ∑ x , y ∈ G 1 , x ≠ y         ∑ u v ∉ G 2 d G 1 ( x , y ) [ d G 1 [ G 2 ] ( x , u ) + d G 1 [ G 2 ] ( y , v ) ] = ∑ x , y ∈ G 1 , x ≠ y         ∑ u v ∉ G 2 d G 1 ( x , y ) [ d G 2 ( u ) + d G 1 ( x ) n 2 + d G 2 ( v ) + n 2 d G 1 ( y ) ]         by   Lemma   2 .1   = ( ∑ x , y ∈ G 1 , x ≠ y d G 1 ( x , y ) ) ∑ u v ∉ G 2 [ d G 2 ( u ) + d G 2 ( v ) ]     + n 2 ∑ x , y ∈ G 1 , x ≠ y d G 1 ( x , y ) [ d G 1 ( x ) + d G 1 ( y ) ] ( ∑ u v ∉ G 2 1 ) = 4 W ( G 1 ) M ¯ 1 ( G 2 ) + 2 n 2 D D ( G 1 ) ( 2 m ¯ 2 + n 2 )

[ D D * ( G 1 [ G 2 ] ) ] 2 ≤ { 1 n 1 n 2 ( n 1 n 2 − 1 ) [ 4 M 1 ( G 2 ) W ( G 1 ) + 4 n 2 m 2 D D ( G 1 ) + 4 n 1 M ¯ 1 ( G 2 )         + 8 n 2 2 m 1 ( n 2 − 1 ) + 2 n 1 M 1 ( G 2 ) + 8 m 1 n 2 m 2 + 4 W ( G 1 ) M ¯ 1 ( G 2 )   + 2 n 2 D D ( G 1 ) ( 2 m ¯ 2 + n 2 ) ] } n 1 n 2 ( n 1 n 2 − 1 )

Lemma 3.2.

D D * K n [ K r ] = [ 2 ( n r − 1 ) ] n r ( n r − 1 ) 2

Proof: Clearly the graph K n [ K r ] is the complete graph K n r .

∴ D D * ( K n [ K r ] ) = D D * ( K n r ) = [ 2 ( n r − 1 ) ] n r ( n r − 1 ) 2 (1)

Remark 3.3. Let G 1 = K n and G 2 = K r . We get,

D D ( G 1 ) = 2 ( n − 1 ) n ( n − 1 ) 2 = n ( n − 1 ) 2 ,     m 1 = n ( n − 1 ) 2 ,         W ( G 1 ) = n ( n − 1 ) 2

M 1 ( G 2 ) = r ( r − 1 ) 2 ,     M ¯ 1 ( G 2 ) = 0 ,   m ¯ 2 = 0 ,     n 1 = n ,   n 2 = r ,     m 2 = r ( r − 1 ) 2

∴ In Theorem 3.1, put G 1 = K n and G 2 = K r , we get

D D * K n [ K r ] ≤ [ 2 ( n r − 1 ) ] n r ( n r − 1 ) 2 (2)

From (1) and (2) our bound is tight

Theorem 4.1. Let G i , i = 1 , 2 , be a ( n i , m i ) -graph and let m ¯ i = e ( G ¯ i ) . Then

[ D D * ( G 1 + G 2 ) ] 2 ≤ { 1 ( n 1 + n 2 ) ( n 1 + n 2 − 1 ) [ 2 M 1 ( G 1 ) + 4 n 2 m 1 + 4 M ¯ 1 ( G 1 ) + 8 n 2 m ¯ 1     + 2 m 1 n 2 + 2 m 2 n 1 + n 1 n 2 ( n 1 + n 2 ) + 2 M 1 ( G 2 ) + 4 n 1 m 2 + 4 M ¯ 1 ( G 2 ) + 8 n 1 m ¯ 2 ] } ( n 1 + n 2 ) ( n 1 + n 2 − 1 )

Proof:

[ D D * ( G 1 + G 2 ) ] 2 ≤ [ 1 ( n 1 + n 2 ) ( n 1 + n 2 − 1 ) ( J 1 + 2 J 2 + J 3 ) ] ( n 1 + n 2 ) ( n 1 + n 2 − 1 )

where J 1 , J 2 , J 3 are terms of the above sums taken in order.

Next we calculate J 1 , J 2 and J 3 separately one by one. Now,

J 1 = ∑ x ∈ V ( G 1 )         ∑ y ∈ V ( G 1 ) d ( G 1 + G 2 ) ( x , y ) [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ] = ∑ x , y ∈ V ( G 1 ) d ( G 1 + G 2 ) ( x , y ) [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ] = ∑ x y ∈ G 1 d ( G 1 + G 2 ) ( x , y ) [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ]     + ∑ x y ∉ G 1 , x ≠ y d ( G 1 + G 2 ) ( x , y ) [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ]     + ∑ x y ∉ G 1 , x = y d ( G 1 + G 2 ) ( x , y ) [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ] = 1 ⋅ ∑ x y ∈ G 1 [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ]     + 2 ⋅ ∑ x y ∉ G 1 , x ≠ y [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ] + 0 = ∑ x y ∈ G 1 [ d G 1 ( x ) + n 2 + d G 2 ( y ) + n 2 ]     + 2 ∑ x y ∉ G 1 , x ≠ y [ d G 1 ( x ) + n 2 + d G 2 ( y ) + n 2 ]         by   Lemma   2 .1 = ∑ x y ∈ G 1 [ d G 1 ( x ) + d G 2 ( y ) ] + 2 n 2 ∑ x y ∈ G 1 1     + 2 { ∑ x y ∉ G 1 , x ≠ y [ d G 1 ( x ) + d G 2 ( y ) ] + 2 n 2 ∑ x y ∉ G 1 , x ≠ y 1 } = 2 M 1 ( G 1 ) + 4 n 2 m 1 + 4 M ¯ 1 ( G 1 ) + 8 n 2 m ¯ 1

J 2 = ∑ x ∈ V ( G 1 )         ∑ y ∈ V ( G 2 ) d ( G 1 + G 2 ) ( x , y ) [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ] = 1 ∑ x ∈ V ( G 1 )         ∑ y ∈ V ( G 2 ) [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ] = ∑ x ∈ V ( G 1 )         ∑ y ∈ V ( G 2 ) [ d G 1 ( x ) + n 2 + d G 2 ( y ) + n 1 ]           by   Lemma   2 .1   = ∑ x ∈ V ( G 1 ) d G 1 ( x ) ∑ y ∈ V ( G 2 ) 1 + ∑ x ∈ V ( G 1 ) 1 ∑ y ∈ V ( G 2 ) d G 2 ( y ) + ( n 1 + n 2 ) ∑ x ∈ V ( G 1 ) 1 ∑ y ∈ V ( G 2 ) 1 = 2 m 1 n 2 + 2 m 2 n 1 + ( n 1 + n 2 ) n 1 n 2

J 3 = ∑ x ∈ V ( G 2 )         ∑ y ∈ V ( G 2 ) d ( G 1 + G 2 ) ( x , y ) [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ] = ∑ x , y ∈ V ( G 2 ) d ( G 1 + G 2 ) ( x , y ) [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ] = ∑ x y ∈ G 2 d ( G 1 + G 2 ) ( x , y ) [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ]     + ∑ x y ∉ G 2 , x ≠ y d ( G 1 + G 2 ) ( x , y ) [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ]     + ∑ x y ∉ G 2 , x = y d ( G 1 + G 2 ) ( x , y ) [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ]

= 1 ∑ x y ∈ G 2 [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ]     + 2 ∑ x y ∉ G 2 , x ≠ y [ d ( G 1 + G 2 ) ( x ) + d ( G 1 + G 2 ) ( y ) ] + 0 = ∑ x y ∈ G 2 [ d G 2 ( x ) + n 1 + d G 2 ( y ) + n 1 ]     + 2 ∑ x y ∉ G 2 , x ≠ y [ d G 2 ( x ) + n 1 + d G 2 ( y ) + n 1 ]         by   Lemma   2 .1 = ∑ x y ∈ G 2 [ d G 2 ( x ) + d G 2 ( y ) ] + 2 n 1 ∑ x y ∈ G 2 1     + 2 { ∑ x y ∉ G 2 , x ≠ y [ d G 2 ( x ) + d G 2 ( y ) ] + 2 n 1 ∑ x y ∉ G 2 , x ≠ y 1 } = 2 M 1 ( G 2 ) + 4 n 1 m 2 + 4 M ¯ 1 ( G 2 ) + 8 n 1 m ¯ 2

[ D D * ( G 1 + G 2 ) ] 2 ≤ { 1 ( n 1 + n 2 ) ( n 1 + n 2 − 1 ) [ 2 M 1 ( G 1 ) + 4 n 2 m 1 + 4 M ¯ 1 ( G 1 ) + 8 n 2 m ¯ 1     + 2 m 1 n 2 + 2 m 2 n 1 + n 1 n 2 ( n 1 + n 2 ) + 2 M 1 ( G 2 ) + 4 n 1 m 2 + 4 M ¯ 1 ( G 2 ) + 8 n 1 m ¯ 2 ] } ( n 1 + n 2 ) ( n 1 + n 2 − 1 )

Lemma 4.2.

D D * [ K n + K r ] = [ 2 ( n + r − 1 ) ] ( n + r ) ( n + r − 1 ) 2

Proof: Clearly the graph K n + K r is the complete graph K n + r

D D * [ K n + K r ] = D D * [ K n + r ] = [ 2 ( n + r − 1 ) ] ( n + r ) ( n + r − 1 ) 2 (3)

Remark 4.3. Let G 1 = K n and G 2 = K r . We get,

M 1 ( G 1 ) = n ( n − 1 ) 2 ,     m 1 = n ( n − 1 ) 2 ,     M 1 ( G 2 ) = r ( r − 1 ) 2 ,     M ¯ 1 ( G 2 ) = 0 ,

m 2 = r ( r − 1 ) 2 ,     M ¯ 1 ( G 1 ) = 0 ,     n 1 = n ,     n 2 = r ,     m ¯ 1 = 0 ,     m ¯ 2 = 0.

∴ In Theorem 4.1, put G 1 = K n ,     G 2 = K r , we get

D D * [ K n + K r ] ≤ [ 2 ( n + r − 1 ) ] ( n + r ) ( n + r − 1 ) 2 (4)

From (3) and (4) our bound is tight.

Theorem 5.1. Let G i , i = 1 , 2 , be a ( n i , m i ) -graph and let m ¯ i = e ( G ¯ i ) . Then

[ D D * ( G 1 ∨ G 2 ) ] 2 ≤ [ 1 n 1 n 2 ( n 1 n 2 − 1 ) { 2 m 2 n 2 ( 2 M ¯ 1 ( G 1 ) + 4 m 1 ) + 2 n 1 ( 2 m ¯ 1 + n 1 ) M 1 ( G 2 )       − 2 M 1 ( G 2 ) ( 2 m ¯ 1 ( n 1 − 1 ) + 2 m 1 − M 1 ( G ¯ 1 ) ) + 2 n 2 M 1 ( G 1 ) ( 2 m ¯ 2 + n 2 )       + 2 m 1 n 1 ( 2 M ¯ 1 ( G 2 ) + 4 m 2 ) − 2 M 1 ( G 1 ) ( 2 ( n 2 − 1 ) m ¯ 2 + 2 m 2 − M 1 ( G ¯ 2 ) )       + 4 n 2 M 1 ( G 1 ) m 2 + 4 n 1 m 1 M 1 ( G 2 ) − 2 M 1 ( G 1 ) M 1 ( G 2 )       + 2 n 2 ( 2 M ¯ 1 ( G 1 ) + 4 m 1 ) ( 2 m ¯ 2 + 2 n 2 ) + 2 n 1 ( 2 M ¯ 1 ( G 2 ) + 4 m 2 ) ( 2 m ¯ 1 + n 1 )       − 4 ( 2 m ¯ 1 ( n 1 − 1 ) + 2 m 1 − M 1 ( G ¯ 1 ) ) ( 2 m ¯ 2 ( n 2 − 1 ) + 2 m 2 − M 1 ( G ¯ 2 ) )   − 8 m 1 n 2 2 − 4 n 1 2 m 2 + 16 m 1 m 2 } ] n 1 n 2 ( n 1 n 2 − 1 )

Proof:

[ D D * ( G 1 ∨ G 2 ) ] 2 ≤ [ 1 n 1 n 2 ( n 1 n 2 − 1 ) ( A 3 + A 1 + A 2 + A 4 ) ] n 1 n 2 ( n 1 n 2 − 1 )

where A 3 , A 1 , A 2 , A 4 are terms of the above sums taken in order.

Next we calculate A 1 , A 2 , A 3 and A 4 separately one by one. Now,

A 1 = ∑ x y ∉ G 1     ∑ u v ∈ G 2 d ( G 1 ∨ G 2 ) ( ( x , u ) , ( y , v ) ) [ d ( G 1 ∨ G 2 ) ( x , u ) + d ( G 1 ∨ G 2 ) ( y , v ) ] = ∑ x y ∉ G 1     ∑ u v ∈ G 2 1 ⋅ [ n 2 d G 1 ( x ) + n 1 d G 2 ( u ) − d G 1 ( x ) d G 2 ( u ) + n 2 d G 1 ( y )     + n 1 d G 2 ( v ) − d G 1 ( y ) d G 2 ( v ) ]         by   Lemma   2 .1   = n 2 ∑ x y ∉ G 1     ∑ u v ∈ G 2 [ d G 1 ( x ) + d G 1 ( y ) ] + n 1 ∑ x y ∉ G 1     ∑ u v ∈ G 2 [ d G 2 ( u ) + d G 2 ( v ) ]     − ∑ x y ∉ G 1     ∑ u v ∈ G 2 d G 1 ( x ) d G 2 ( u ) − ∑ x y ∉ G 1     ∑ u v ∈ G 2 d G 1 ( y ) d G 2 ( v ) = n 2 ( ∑ x y ∉ G 1 [ d G 1 ( x ) + d G 1 ( y ) ] ) ( ∑ u v ∈ G 2 1 ) + n 1 ( ∑ x y ∉ G 1 1 ) ( ∑ u v ∈ G 2 [ d G 2 ( u ) + d G 2 ( v ) ] )     − ( ∑ x y ∉ G 1 d G 1 ( x ) ) ( ∑ u v ∈ G 2 d G 2 ( u ) ) − ( ∑ x y ∉ G 1 d G 1 ( y ) ) ( ∑ u v ∈ G 2 d G 2 ( v ) ) = 2 m 2 n 2 ( 2 M ¯ 1 ( G 1 ) + 4 m 1 ) + 2 n 1 ( 2 m ¯ 1 + n 1 ) M 1 ( G 2 )     − 2 M 1 ( G 2 ) ( 2 m ¯ 1 ( n 1 − 1 ) + 2 m 1 − M 1 ( G ¯ 1 ) )

A 2 = ∑ x y ∈ G 1     ∑ u v ∉ G 2 d ( G 1 ∨ G 2 ) ( ( x , u ) , ( y , v ) ) [ d ( G 1 ∨ G 2 ) ( x , u ) + d ( G 1 ∨ G 2 ) ( y , v ) ] = ∑ x y ∈ G 1     ∑ u v ∉ G 2 [ n 2 d G 1 ( x ) + n 1 d G 2 ( u ) − d G 1 ( x ) d G 2 ( u ) + n 2 d G 1 ( y )     + n 1 d G 2 ( v ) − d G 1 ( y ) d G 2 ( v ) ]         by   Lemma   2 .1 = n 2 ∑ x y ∈ G 1     ∑ u v ∉ G 2 [ d G 1 ( x ) + d G 2 ( y ) ] + n 1 ∑ x y ∈ G 1     ∑ u v ∉ G 2 [ d G 2 ( u ) + d G 2 ( v ) ]     − ∑ x y ∈ G 1     ∑ u v ∉ G 2 d G 1 ( x ) d G 2 ( u ) − ∑ x y ∈ G 1     ∑ u v ∉ G 2 d G 1 ( y ) d G 2 ( v ) = n 2 ( ∑ x y ∈ G 1 [ d G 1 ( x ) + d G 2 ( y ) ] ) ( ∑ u v ∉ G 2 1 ) + n 1 ( ∑ x y ∈ G 1 1 ) ( ∑ u v ∉ G 2 [ d G 2 ( u ) + d G 2 ( v ) ] )     − ( ∑ x y ∈ G 1 d G 1 ( x ) ) ( ∑ u v ∉ G 2 d G 2 ( u ) ) − ( ∑ x y ∈ G 1 d G 1 ( y ) ) ( ∑ u v ∉ G 2 d G 2 ( v ) ) = 2 n 2 M 1 ( G 1 ) ( 2 m ¯ 2 + n 2 ) + 2 n 1 m 1 ( 2 M ¯ 1 ( G 2 ) + 4 m 2 )     − 2 M 1 ( G 1 ) [ 2 ( n 2 − 1 ) m ¯ 2 + 2 m 2 − M 1 ( G ¯ 2 ) ]

A 3 = ∑ x y ∈ G 1     ∑ u v ∈ G 2 d ( G 1 ∨ G 2 ) ( ( x , u ) , ( y , v ) ) [ d ( G 1 ∨ G 2 ) ( x , u ) + d ( G 1 ∨ G 2 ) ( y , v ) ] = ∑ x y ∈ G 1     ∑ u v ∈ G 2 [ n 2 d G 1 ( x ) + n 1 d G 2 ( u ) − d G 1 ( x ) d G 2 ( u ) + n 2 d G 1 ( y )     + n 1 d G 2 ( v ) − d G 1 ( y ) d G 2 ( v ) ]         by   Lemma   2 .1   = n 2 ∑ x y ∈ G 1     ∑ u v ∈ G 2 [ d G 1 ( x ) + d G 2 ( y ) ] + n 1 ∑ x y ∈ G 1     ∑ u v ∈ G 2 [ d G 2 ( u ) + d G 2 ( v ) ]     − ∑ x y ∈ G 1     ∑ u v ∈ G 2 d G 1 ( x ) d G 2 ( u ) − ∑ x y ∈ G 1     ∑ u v ∈ G 2 d G 1 ( y ) d G 2 ( v ) = n 2 ( ∑ x y ∈ G 1 [ d G 1 ( x ) + d G 2 ( y ) ] ) ( ∑ u v ∈ G 2 1 ) + n 1 ( ∑ x y ∈ G 1 1 ) ( ∑ u v ∈ G 2 [ d G 2 ( u ) + d G 2 ( v ) ] )     − ( ∑ x y ∈ G 1 d G 1 ( x ) ) ( ∑ u v ∈ G 2 d G 2 ( u ) ) − ( ∑ x y ∈ G 1 d G 1 ( y ) ) ( ∑ u v ∈ G 2 d G 2 ( v ) ) = 4 n 2 m 2 M 1 ( G 1 ) + 4 n 1 m 1 M 1 ( G 2 ) − 2 M 1 ( G 1 ) M 1 ( G 2 )

A 4 = ∑ x y ∉ G 1     ∑ u v ∉ G 2 d ( G 1 ∨ G 2 ) ( ( x , u ) , ( y , v ) ) [ d ( G 1 ∨ G 2 ) ( x , u ) + d ( G 1 ∨ G 2 ) ( y , v ) ] = 2 ∑ x y ∉ G 1     ∑ u v ∉ G 2 [ d ( G 1 ∨ G 2 ) ( x , u ) + d ( G 1 ∨ G 2 ) ( y , v ) ]     − 2 ∑ x y ∉ G 1 , x = y         ∑ u v ∉ G 2 , u = v [ d ( G 1 ∨ G 2 ) ( x , u ) + d ( G 1 ∨ G 2 ) ( y , v ) ]

A 4 = 2 A 5 − 2 A 6 , where A 5 and A 6 are terms of the above sums taken in order.

Now,

A 5 = ∑ x y ∉ G 1     ∑ u v ∉ G 2 [ d ( G 1 ∨ G 2 ) ( x , u ) + d ( G 1 ∨ G 2 ) ( y , v ) ] = ∑ x y ∉ G 1     ∑ u v ∉ G 2 [ n 2 d G 1 ( x ) + n 1 d G 2 ( u ) − d G 1 ( x ) d G 2 ( u ) + n 2 d G 1 ( y )     + n 1 d G 2 ( v ) − d G 1 ( y ) d G 2 ( v ) ]         by   Lemma   2 .1   = n 2 ∑ x y ∉ G 1     ∑ u v ∉ G 2 [ d G 1 ( x ) + d G 1 ( y ) ] + n 1 ∑ x y ∉ G 1 ∑ u v ∉ G 2 [ d G 2 ( u ) + d G 2 ( v ) ]     − ∑ x y ∉ G 1     ∑ u v ∉ G 2 d G 1 ( x ) d G 2 ( u ) − ∑ x y ∉ G 1     ∑ u v ∉ G 2 d G 1 ( y ) d G 2 ( v ) = n 2 ( ∑ x y ∉ G 1 [ d G 1 ( x ) + d G 2 ( y ) ] ) ( ∑ u v ∉ G 2 1 ) + n 1 ( ∑ x y ∉ G 1 1 ) ( ∑ u v ∉ G 2 [ d G 2 ( u ) + d G 2 ( v ) ] )     − ( ∑ x y ∉ G 1 d G 1 ( x ) ) ( ∑ u v ∉ G 2 d G 2 ( u ) ) − ( ∑ x y ∉ G 1 d G 1 ( y ) ) ( ∑ u v ∉ G 2 d G 2 ( v ) ) = n 2 ( 2 M ¯ 1 ( G 1 ) + 4 m 1 ) ( 2 m ¯ 2 + n 2 ) + n 1 ( 2 M ¯ 1 ( G 2 ) + 4 m 2 ) ( 2 m ¯ 1 + n 1 )     − 2 ( 2 m ¯ 1 ( n 1 − 1 ) + 2 m 1 − M 1 ( G ¯ 1 ) ) ( 2 m ¯ 2 ( n 2 − 1 ) + 2 m 2 − M 1 ( G ¯ 2 ) )

A 6 = ∑ x y ∉ G 1 , x = y     ∑ u v ∉ G 2 , u = v [ d ( G 1 ∨ G 2 ) ( x , u ) + d ( G 1 ∨ G 2 ) ( y , v ) ] = ∑ x y ∉ G 1 , u = v     ∑ u v ∉ G 2 , u = v [ n 2 d G 1 ( x ) + n 1 d G 2 ( u ) − d G 1 ( x ) d G 2 ( u ) + n 2 d G 1 ( y )     + n 1 d G 2 ( v ) − d G 1 ( y ) d G 2 ( v ) ]         by   Lemma   2 .1   = n 2 ∑ x y ∉ G 1 , x = y     ∑ u v ∉ G 2 , u = v [ d G 1 ( x ) + d G 1 ( y ) ] + n 1 ∑ x y ∉ G 1 , x = y     ∑ u v ∉ G 2 , u = v [ d G 2 ( u ) + d G 2 ( v ) ]     − ∑ x y ∉ G 1 , x = y     ∑ u v ∉ G 2 , u = v d G 1 ( x ) d G 2 ( u ) − ∑ x y ∉ G 1 , x = y     ∑ u v ∉ G 2 , u = v d G 1 ( y ) d G 2 ( v ) = n 2 ( ∑ x y ∉ G 1 , x = y [ d G 1 ( x ) + d G 2 ( y ) ] ) ( ∑ u v ∉ G 2 , u = v 1 ) + n 1 ( ∑ x y ∉ G 1 , x = y 1 ) ( ∑ u v ∉ G 2 , u = v [ d G 2 ( u ) + d G 2 ( v ) ] )     − ( ∑ x y ∉ G 1 , x = y d G 1 ( x ) ) ( ∑ u v ∉ G 2 , u = v d G 2 ( u ) ) − ( ∑ x y ∉ G 1 , x = y d G 1 ( y ) ) ( ∑ u v ∉ G 2 , u = v d G 2 ( v ) ) = 4 m 1 n 2 2 + 4 m 2 n 1 2 − 8 m 1 m 2

[ D D * ( G 1 ∨ G 2 ) ] 2 ≤ [ 1 n 1 n 2 ( n 1 n 2 − 1 ) { 2 m 2 n 2 ( 2 M ¯ 1 ( G 1 ) + 4 m 1 ) + 2 n 1 ( 2 m ¯ 1 + n 1 ) M 1 ( G 2 )       − 2 M 1 ( G 2 ) ( 2 m ¯ 1 ( n 1 − 1 ) + 2 m 1 − M 1 ( G ¯ 1 ) ) + 2 n 2 M 1 ( G 1 ) ( 2 m ¯ 2 + n 2 )       + 2 m 1 n 1 ( 2 M ¯ 1 ( G 2 ) + 4 m 2 ) − 2 M 1 ( G 1 ) ( 2 ( n 2 − 1 ) m ¯ 2 + 2 m 2 − M 1 ( G ¯ 2 ) )       + 4 n 2 M 1 ( G 1 ) m 2 + 4 n 1 m 1 M 1 ( G 2 ) − 2 M 1 ( G 1 ) M 1 ( G 2 )       + 2 n 2 ( 2 M ¯ 1 ( G 1 ) + 4 m 1 ) ( 2 m ¯ 2 + 2 n 2 ) + 2 n 1 ( 2 M ¯ 1 ( G 2 ) + 4 m 2 ) ( 2 m ¯ 1 + n 1 )       − 4 ( 2 m ¯ 1 ( n 1 − 1 ) + 2 m 1 − M 1 ( G ¯ 1 ) ) ( 2 m ¯ 2 ( n 2 − 1 ) + 2 m 2 − M 1 ( G ¯ 2 ) )   − 8 m 1 n 2 2 − 4 n 1 2 m 2 + 16 m 1 m 2 } ] n 1 n 2 ( n 1 n 2 − 1 )

Lemma 5.2.

D D * [ K m ∨ K n ] = ( 2 m n − 2 ) m n ( m n − 1 ) 2

Proof: Clearly the graph K m ∨ K n is the complete graph K m n .

D D * ( K m ∨ K n ) = D D * ( K m n ) = ( 2 m n − 2 ) m n ( m n − 1 ) 2 (5)

Remark 5.3. Let G 1 = K m and G 2 = K n . We get

n 1 = m ,     n 2 = n ,     m 1 = m ( m − 1 ) 2 ,     m 2 = n ( n − 1 ) 2 ,     m ¯ 1 = 0 ,     m ¯ 2 = 0

M 1 ( G 1 ) = M 1 ( K m ) = m ( m − 1 ) 2 ,     M 1 ( G 2 ) = M 1 ( K n ) = n ( n − 1 ) 2

M 1 ( G ¯ 1 ) = M 1 ( K ¯ m ) = 0 ,     M 1 ( G ¯ 2 ) = M 1 ( K ¯ n ) = 0 ,     M ¯ 1 ( G 1 ) = M ¯ 1 ( K m ) = 0

∴ In Theorem 5.1, put G 1 = K m and G 2 = K n , we get

D D * [ K m ∨ K n ] ≤ ( 2 m n − 2 ) m n ( m n − 1 ) 2 (6)

From (5) and (6) our bound is tight.

Theorem 6.1.

[ D D * ( G 1 ⊕ G 2 ) ] 2 ≤ { 1 n 1 n 2 ( n 1 n 2 − 1 ) [ 2 n 2 m 2 ( 2 M ¯ 1 ( G 1 ) + 4 m 1 ) + 2 n 1 M 1 ( G 2 ) ( 2 m ¯ 1 + n 1 )       − 4 M 1 ( G 2 ) ( 2 m ¯ 1 ( n 1 − 1 ) + 2 m 1 − M 1 ( G ¯ 1 ) ) + 2 n 2 M 1 ( G 1 ) ( 2 m ¯ 2 + n 2 )       + 2 n 1 m 1 ( 2 M ¯ 1 ( G 2 ) + 4 m 2 ) − 4 M 1 ( G 1 ) ( 2 ( n 2 − 1 ) m ¯ 2 + 2 m 2 − M 1 ( G ¯ 2 ) )       + 4 n 2 M 1 ( G 1 ) m 2 + 4 n 1 m 1 M 1 ( G 2 ) − 4 M 1 ( G 1 ) M 1 ( G 2 )       + 2 [ n 2 ( 2 M ¯ 1 ( G 1 ) + 4 m 1 ) ( 2 m 2 + n 2 ) + n 1 ( 2 M ¯ 1 ( G 2 ) + 4 m 2 ) ( 2 m 1 + n 1 )       − 4 ( 2 m ¯ 1 ( n 1 − 1 ) + 2 m 1 − M 1 ( G ¯ 1 ) ) ( 2 m ¯ 2 ( n 2 − 1 ) + 2 m 2 − M 1 ( G ¯ 2 ) ) ]   − 2 ( 4 n 2 2 m 1 + 4 n 1 2 m 2 − 16 m 1 m 2 ) ] } n 1 n 2 ( n 1 n 2 − 1 )

Proof:

[ D D * ( G 1 ⊕ G 2 ) ] 2 ≤ [ 1 n 1 n 2 ( n 1 n 2 − 1 ) ( C 3 + C 1 + C 2 + C 4 ) ] n 1 n 2 ( n 1 n 2 − 1 )

where C 3 , C 1 , C 2 , C 4 are terms of the above sums taken in order.

Next we calculate C 1 , C 2 , C 3 and C 4 separately.

C 1 = ∑ x y ∉ G 1     ∑ u v ∈ G 2 d ( G 1 ⊕ G 2 ) ( ( x , u ) , ( y , v ) ) [ d ( G 1 ⊕ G 2 ) ( x , u ) + d ( G 1 ⊕ G 2 ) ( y , v ) ] = ∑ x y ∉ G 1     ∑ u v ∈ G 2 1 ⋅ [ n 2 d G 1 ( x ) + n 1 d G 2 ( u ) − 2 d G 1 ( x ) d G 2 ( u ) + n 2 d G 1 ( y )     + n 1 d G 2 ( v ) − 2 d G 1 ( y ) d G 2 ( v ) ]         by   Lemma   2 .1   = n 2 ∑ x y ∉ G 1     ∑ u v ∈ G 2 [ d G 1 ( x ) + d G 1 ( y ) ] + n 1 ∑ x y ∉ G 1 ∑ u v ∈ G 2 [ d G 2 ( u ) + d G 2 ( v ) ]     − 2 ∑ x y ∉ G 1     ∑ u v ∈ G 2 d G 1 ( x ) d G 2 ( u ) − 2 ∑ x y ∉ G 1     ∑ u v ∈ G 2 d G 1 ( y ) d G 2 ( v ) = n 2 ( ∑ x y ∉ G 1 [ d G 1 ( x ) + d G 1 ( y ) ] ) ( ∑ u v ∈ G 2 1 ) + n 1 ( ∑ x y ∉ G 1 1 ) ( ∑ u v ∈ G 2 [ d G 2 ( u ) + d G 2 ( v ) ] )     − 2 ( ∑ x y ∉ G 1 d G 1 ( x ) ) ( ∑ u v ∈ G 2 d G 2 ( u ) ) − 2 ( ∑ x y ∉ G 1 d G 1 ( y ) ) ( ∑ u v ∈ G 2 d G 2 ( v ) ) = 2 n 2 m 2 ( 2 M ¯ 1 ( G 1 ) + 4 m 1 ) + 2 n 1 M 1 ( G 2 ) ( 2 m ¯ 1 + n 1 )     − 4 M 1 ( G 2 ) ( 2 m ¯ 1 ( n 1 − 1 ) + 2 m 1 − M 1 ( G ¯ 1 ) )

C 2 = ∑ x y ∈ G 1     ∑ u v ∉ G 2 d ( G 1 ⊕ G 2 ) ( ( x , u ) , ( y , v ) ) [ d ( G 1 ⊕ G 2 ) ( x , u ) + d ( G 1 ⊕ G 2 ) ( y , v ) ] = ∑ x y ∈ G 1     ∑ u v ∉ G 2 [ n 2 d G 1 ( x ) + n 1 d G 2 ( u ) − 2 d G 1 ( x ) d G 2 ( u )     + n 2 d G 1 ( y ) + n 1 d G 2 ( v ) − 2 d G 1 ( y ) d G 2 ( v ) ]         by   Lemma   2 .1   = n 2 ∑ x y ∈ G 1     ∑ u v ∉ G 2 [ d G 1 ( x ) + d G 2 ( y ) ] + n 1 ∑ x y ∈ G 1 ∑ u v ∉ G 2 [ d G 2 ( u ) + d G 2 ( v ) ]     − 2 ∑ x y ∈ G 1     ∑ u v ∉ G 2 d G 1 ( x ) d G 2 ( u ) − 2 ∑ x y ∈ G 1     ∑ u v ∉ G 2 d G 1 ( y ) d G 2 ( v ) = n 2 ( ∑ x y ∈ G 1 [ d G 1 ( x ) + d G 2 ( y ) ] ) ( ∑ u v ∉ G 2 1 ) + n 1 ( ∑ x y ∈ G 1 1 ) ( ∑ u v ∉ G 2 [ d G 2 ( u ) + d G 2 ( v ) ] )     − 2 ( ∑ x y ∈ G 1 d G 1 ( x ) ) ( ∑ u v ∉ G 2 d G 2 ( u ) ) − 2 ( ∑ x y ∈ G 1 d G 1 ( y ) ) ( ∑ u v ∉ G 2 d G 2 ( v ) ) = 2 n 2 M 1 ( G 1 ) ( m ¯ 2 + n 2 ) + 2 n 1 m 1 ( 2 M ¯ 1 ( G 2 ) + 4 m 2 )     − 4 M 1 ( G 1 ) ( 2 ( n 2 − 1 ) m ¯ 2 + 2 m 2 − M 1 ( G ¯ 2 ) )

C 3 = ∑ x y ∈ G 1     ∑ u v ∈ G 2 d ( G 1 ⊕ G 2 ) ( ( x , u ) , ( y , v ) ) [ d ( G 1 ⊕ G 2 ) ( x , u ) + d ( G 1 ⊕ G 2 ) ( y , v ) ] = ∑ x y ∈ G 1     ∑ u v ∈ G 2 [ n 2 d G 1 ( x ) + n 1 d G 2 ( u ) − 2 d G 1 ( x ) d G 2 ( u )     + n 2 d G 1 ( y ) + n 1 d G 2 ( v ) − 2 d G 1 ( y ) d G 2 ( v ) ]         by   Lemma   2 .1   = n 2 ∑ x y ∈ G 1     ∑ u v ∈ G 2 [ d G 1 ( x ) + d G 2 ( y ) ] + n 1 ∑ x y ∈ G 1       ∑ u v ∈ G 2 [ d G 2 ( u ) + d G 2 ( v ) ]     − 2 ∑ x y ∈ G 1     ∑ u v ∈ G 2 d G 1 ( x ) d G 2 ( u ) − 2 ∑ x y ∈ G 1     ∑ u v ∈ G 2 d G 1 ( y ) d G 2 ( v ) = n 2 ( ∑ x y ∈ G 1 [ d G 1 ( x ) + d G 2 ( y ) ] ) ( ∑ u v ∈ G 2 1 ) + n 1 ( ∑ x y ∈ G 1 1 ) ( ∑ u v ∈ G 2 [ d G 2 ( u ) + d G 2 ( v ) ] )     − 2 ( ∑ x y ∈ G 1 d G 1 ( x ) ) ( ∑ u v ∈ G 2 d G 2 ( u ) ) − 2 ( ∑ x y ∈ G 1 d G 1 ( y ) ) ( ∑ u v ∈ G 2 d G 2 ( v ) ) = 4 n 2 M 1 ( G 1 ) m 2 + 4 n 1 m 1 M 1 ( G 2 ) − 4 M 1 ( G 1 ) M 1 ( G 2 )

C 4 = ∑ x y ∉ G 1     ∑ u v ∉ G 2 d ( G 1 ⊕ G 2 ) ( ( x , u ) , ( y , v ) ) [ d ( G 1 ⊕ G 2 ) ( x , u ) + d ( G 1 ⊕ G 2 ) ( y , v ) ] = 2 ∑ x y ∉ G 1     ∑ u v ∉ G 2 [ d ( G 1 ⊕ G 2 ) ( x , u ) + d ( G 1 ⊕ G 2 ) ( y , v ) ]     − 2 ∑ x y ∉ G 1 , x = y       ∑ u v ∉ G 2 , u = v [ d ( G 1 ⊕ G 2 ) ( x , u ) + d ( G 1 ⊕ G 2 ) ( y , v ) ]

C 4 = 2 C 5 − 2 C 6 , where C 5 and C 6 denote the sums of the above terms in order.

Now,

C 5 = ∑ x y ∉ G 1     ∑ u v ∉ G 2 [ d ( G 1 ⊕ G 2 ) ( x , u ) + d ( G 1 ⊕ G 2 ) ( y , v ) ] = ∑ x y ∉ G 1     ∑ u v ∉ G 2 [ n 2 d G 1 ( x ) + n 1 d G 2 ( u ) − 2 d G 1 ( x ) d G 2 ( u )     + n 2 d G 1 ( y ) + n 1 d G 2 ( v ) − 2 d G 1 ( y ) d G 2 ( v ) ]         by   Lemma   2 .1   = n 2 ∑ x y ∉ G 1     ∑ u v ∉ G 2 [ d G 1 ( x ) + d G 2 ( y ) ] + n 1 ∑ x y ∉ G 1     ∑ u v ∉ G 2 [ d G 2 ( u ) + d G 2 ( v ) ]     − 2 ∑ x y ∉ G 1     ∑ u v ∉ G 2 d G 1 ( x ) d G 2 ( u ) − 2 ∑ x y ∉ G 1     ∑ u v ∉ G 2 d G 1 ( y ) d G 2 ( v ) = n 2 ( ∑ x y ∉ G 1 [ d G 1 ( x ) + d G 2 ( y ) ] ) ( ∑ u v ∉ G 2 1 ) + n 1 ( ∑ x y ∉ G 1 1 ) ( ∑ u v ∉ G 2 [ d G 2 ( u ) + d G 2 ( v ) ] )     − 2 ( ∑ x y ∉ G 1 d G 1 ( x ) ) ( ∑ u v ∉ G 2 d G 2 ( u ) ) − 2 ( ∑ x y ∉ G 1 d G 1 ( y ) ) ( ∑ u v ∉ G 2 d G 2 ( v ) ) = n 2 ( 2 M ¯ 1 ( G 1 ) + 4 m 1 ) ( 2 m 2 + n 2 ) + n 1 ( 2 M ¯ 1 ( G 2 ) + 4 m 2 ) ( 2 m 1 + n 1 )     − 4 ( ( 2 m ¯ 1 ( n 1 − 1 ) + 2 m 1 − M 1 ( G ¯ 1 ) ) ( 2 m ¯ 2 ( n 2 − 1 ) + 2 m 2 − M 1 ( G ¯ 2 ) )

[ D D * ( G 1 ⊕ G 2 ) ] 2 ≤ { 1 n 1 n 2 ( n 1 n 2 − 1 ) [ 2 n 2 m 2 ( 2 M ¯ 1 ( G 1 ) + 4 m 1 ) + 2 n 1 M 1 ( G 2 ) ( 2 m ¯ 1 + n 1 )       − 4 M 1 ( G 2 ) ( 2 m ¯ 1 ( n 1 − 1 ) + 2 m 1 − M 1 ( G ¯ 1 ) ) + 2 n 2 M 1 ( G 1 ) ( 2 m ¯ 2 + n 2 )       + 2 n 1 m 1 ( 2 M ¯ 1 ( G 2 ) + 4 m 2 ) − 4 M 1 ( G 1 ) ( 2 ( n 2 − 1 ) m ¯ 2 + 2 m 2 − M 1 ( G ¯ 2 ) )       + 4 n 2 M 1 ( G 1 ) m 2 + 4 n 1 m 1 M 1 ( G 2 ) − 4 M 1 ( G 1 ) M 1 ( G 2 )       + 2 [ n 2 ( 2 M ¯ 1 ( G 1 ) + 4 m 1 ) ( 2 m 2 + n 2 ) + n 1 ( 2 M ¯ 1 ( G 2 ) + 4 m 2 ) ( 2 m 1 + n 1 )       − 4 ( 2 m ¯ 1 ( n 1 − 1 ) + 2 m 1 − M 1 ( G ¯ 1 ) ) ( 2 m ¯ 2 ( n 2 − 1 ) + 2 m 2 − M 1 ( G ¯ 2 ) ) ]   − 2 ( 4 n 2 2 m 1 + 4 n 1 2 m 2 − 16 m 1 m 2 ) ] } n 1 n 2 ( n 1 n 2 − 1 )

Lemma 6.2.

D D * [ K m ⊕ K 1 ] = ( 2 m − 2 ) m ( m − 1 ) 2

Proof: Clearly the graph K m ⊕ K 1 is the complete graph K m

D D * [ K m ⊕ K 1 ] = D D * K m = ( 2 m − 2 ) m ( m − 1 ) 2 (7)

Remark 6.3. Let G 1 = K m and G 2 = K 1 . We get

n 1 = m ,         n 2 = 1 ,     m 1 = m ( m − 1 ) 2 ,         m 2 = 0 ,         m ¯ 1 = 0 ,         m ¯ 2 = 0

M 1 ( G 1 ) = M 1 ( K m ) = m ( m − 1 ) 2 ,     M 1 ( G 2 ) = M 1 ( K 1 ) = 0

M 1 ( G ¯ 1 ) = M 1 ( K ¯ m ) = 0 ,     M 1 ( G ¯ 2 ) = M 1 ( K ¯ 1 ) = 0

M ¯ 1 ( G 1 ) = M ¯ 1 ( K m ) = 0 ,     M ¯ 1 ( G 2 ) = M ¯ 1 ( K 1 ) = 0

∴ In Theorem 6.1, put G 1 = K m and G 2 = K 1 , we get

D D * [ K m ⊕ K 1 ] ≤ ( 2 m − 2 ) m ( m − 1 ) 2 (8)

From (7) and (8) our bound is tight.

Muruganandam, R., Manikandan, R.S. and Aruvi, M. (2017) Sharp Upper Bounds for Multiplicative De- gree Distance of Graph Operations. Open Access Library Journal, 4: e2987. https://doi.org/10.4236/oalib.1102987