AM  Vol.10 No.10 , October 2019
On the First and Second Locating Zagreb Indices of Graphs
ABSTRACT
By the distance or degree of vertices of the molecular graph, we can define graph invariant called topological indices. Which are used in chemical graph to describe the structures and predicting some physicochemical properties of chemical compound? In this paper, by introducing two new topological indices under the name first and second Zagreb locating indices of a graph G, we establish the exact values of those indices for some standard families of graphs included the firefly graph.

1. Introduction

Topological indices play a significant role mainly in chemistry, pharmacology, etc. (see [1] - [7] ). Many of the topological indices of current interest in mathematical chemistry are defined in terms of vertex degrees of the molecular graph. Two of the most famous topological indices of graphs are the first and second Zagreb indices which have been introduced by Gutman and Trinajstic in [8] , and defined as M 1 ( G ) = u V ( G ) ( d ( u ) ) 2 and M 2 ( G ) = u v E ( G ) d ( u ) d ( v ) , respectively. The Zagreb indices have been studied extensively due to their numerous applications in the place of existing chemical methods which need more time and increase the costs. Many new reformulated and extended versions of the Zagreb indices have been introduced for several similar reasons (cf. [9] - [17] ).

One of the present authors Saleh [18] has recently introduced a new matrix representation for a graph G by defining the locating matrix L o ( G ) over G. We will redefine this representation as in the following.

Definition 1 ( [18] ) Let G = ( V , E ) be a connected graph with vertex set V = { v 1 , v 2 , , v n } . A locating function of G denoted by L ( G ) is a function L ( G ) : V ( G ) ( + { 0 } ) n such that L ( v i ) = v i = ( d ( v 1 , v i ) , d ( v 2 , v i ) , , d ( v n , v i ) ) , where d ( v i , v j ) is the distance between the vertices v i and v j in G. The vector v i is called the locating vector corresponding to the vertex v i , where v i v j is actually the dot product of the vectors v i and v j in the integers space ( + { 0 } ) n such that v i is adjacent to v j .

The above locating function and huge applications of Zagreb indices motivated us to introduce two new topological indices, namely first and second locating indices, based on the locating vectors.

Definition 2. Let G = ( V , E ) be a connected graph with a vertex set V = { v 1 , v 2 , , v n } and an edge set E ( G ) . Then we define the first and second locating indices as

M 1 L ( G ) = v i V ( G ) ( v i ) 2 and M 2 L ( G ) = v i v j E ( G ) v i v j ,

respectively.

All graphs in this paper will be assumed simple, undirected and connected unless stated otherwise. For graph theoretical terminologies, we refer [19] to the readers.

2. Some Exact Values in Terms of Locating Indices

In this section, by considering Definition 2, we determine the first and second locating indices for the standard graphs K n , C n , K n , m , W n , P n , and also for the join graph G G 1 + G 2 such that G 1 and G 2 are both connected graphs with diameter 2 and G will be assumed as C 3 , C 5 -free graphs.

Theorem 3. Let G K n be the complete graph with a vertex set V ( G ) = { v 1 , v 2 , , v n } , where n 2 . Then M 1 L ( K n ) = n ( n 1 ) and M 2 L ( K n ) = n ( n 1 ) ( n 2 ) 2 .

Proof. Let v i be a locating vector corresponding to the vertex v i V ( G ) . Then v i = ( a 1 , a 2 , , a n ) such that a i = 0 and a i + 1 = 1 . Thus ( v i ) 2 = n 1 . But we have total n vertices in V ( G ) , and so M 1 L ( K n ) = n ( n 1 ) , as required. On the other hand, for any two locating vectors v i and v j , where i j , we

definitely have v i v j = n 2 . Hence M 2 L ( K n ) = n ( n 1 ) ( n 2 ) 2 .+

In the next two Theorems, we investigate the cycle C n depends on the status of n.

Theorem 4. For an even integer n 2 , let G C n . Then M 1 L ( C n ) = n ( n 2 + 2 ) 12 and M 2 L ( C n ) = n 2 ( n 2 ) 2 12 .

Proof. By labeling the vertices of the cycle C n as { v 1 , v 2 , , v n } in the anticlockwise direction, we obtain

v 1 = ( 0 , 1 , 2 , 3 , , n 2 , n 2 1 , n 2 2 , , 1 ) ,

v 2 = ( 1 , 0 , 1 , 2 , , n 2 1 , n 2 , n 2 1 , , 2 ) ,

v 3 = ( 2 , 1 , 0 , 1 , , n 2 2 , n 2 1 , n 2 , , 3 ) ,

v n = ( 1 , 2 , 3 , , n 2 , n 2 1 , n 2 2 , n 2 2 , , 0 ) ,

and hence v i 2 = 2 i = 1 n 2 i 2 n 2 4 . It is not difficult to see that each v i has the same components within different location, and so each v i 2 has the same sum as the form of v i 2 = n ( n + 1 ) ( n + 2 ) 3 n 2 12 . Therefore M 1 L ( C n ) = n ( n 2 + 2 ) 12 . In addition, by the symmetry,

v i v i + 1 = 2 i = 2 n 2 i ( i 1 ) = 2 ( n 2 ( n 2 + 1 ) ( n + 1 ) 6 1 ) 2 ( n 2 ( n 2 + 1 ) 2 1 ) = n ( n 2 ) 2 12

which gives M 2 L ( C n ) = n 2 ( n 2 ) 2 12 .+

Theorem 5. For an odd integer n 3 , let G C n . Then M 1 L ( C n ) = n 2 ( n 2 1 ) 12 and M 2 L ( C n ) = n ( n 1 ) ( n 2 ) ( n + 3 ) 12 .

Proof. With a similar procedure as in the proof of Theorem 4, we get

v 1 = ( 0 , 1 , 2 , 3 , , n 1 2 , n 1 2 1 , n 1 2 2 , , 1 ) ,

v 2 = ( 1 , 0 , 1 , 2 , , n 1 2 1 , n 1 2 , n 1 2 1 , , 2 ) ,

v 3 = ( 2 , 1 , 0 , 1 , , n 1 2 2 , n 1 2 1 , n 1 2 , , 3 ) ,

v n = ( 1 , 2 , 3 , , n 1 2 , n 1 2 1 , n 1 2 2 , , 0 )

which implies

v i 2 = 2 i = 1 n 1 2 i 2 = n ( n 2 1 ) 12 ,

and so M 1 L ( C n ) = n 2 ( n 2 1 ) 12 . Also, by the symmetry,

v i v i + 1 = 2 i = 2 n 1 2 i ( i 1 ) + ( n 1 ) 2 4 = ( 2 n 1 2 ( n 1 2 + 1 ) ( 2 n 1 2 + 1 ) 6 ) 1 ( 2 n 1 2 ( n 1 2 + 1 ) 2 1 ) + ( n 1 ) 2 4 = ( n 1 ) ( n 2 ) ( n + 3 ) 12

which gives the exact value of M 2 L ( C n ) as depicted in the statement of theorem.+

Now we will take into account the complete bipartite graphs to determine the locating indices.

Theorem 6. Let G K n , m , where 1 n m . Then M 1 L ( K n , m ) = 4 ( n 2 + m 2 ) 4 ( n + m ) + 2 n m and M 2 L ( K n , m ) = 2 n m ( n + m 2 ) .

Proof. For all 1 i n and 1 j m , by labeling the adjacent vertices v i and v n + j of K n , m , the locating vectors v i of v i are given by:

v 1 = ( 0 , 2 , , 2 n 1 , 1 , 1 , , 1 m ) , v 2 = ( 2 , 0 , 2 , , 2 n 2 , 1 , 1 , , 1 m ) , v 3 = ( 2 , 2 , 0 , 2 , , 2 n 3 , 1 , 1 , , 1 m ) , , v n = ( 2 , , 2 n 1 , 0 , 1 , 1 , , 1 m ) , v n + 1 = ( 1 , , 1 n , 0 , 2 , 2 , , 2 m 1 ) , v n + 2 = ( 1 , , 1 n , 2 , 0 , 2 , , 2 m 2 ) , , v n + m = ( 1 , , 1 n , 2 , , 2 m 1 , 0 ) .

In here, for any i = 1 , 2 , , n , we have v i 2 = 4 ( n 1 ) + m and for any i = n + 1 , n + 2 , , n + m , we get v i 2 = 4 ( m 1 ) + n . Therefore

M 1 L ( K n , m ) = n ( 4 ( n 1 ) + m ) + m ( 4 ( m 1 ) + n ) = 4 ( n 2 + m 2 ) 4 ( n + m ) + 2 n m .

On the other hand, for any two consecutive locating vertices v i , v i + 1 in K n , m , since v i v i + 1 = 2 ( n + m 2 ) , we obtain M 2 L ( K n , m ) = 2 n m ( n + m 2 ) .+

Since the following consequences of Theorem 6 are very special cases and clear, we will omit their proofs.

Corollary 7. Let G K n , n , where n 1 . Then M 1 L ( K n , n ) = 2 n ( 5 n 4 ) and M 2 L ( K n , n ) = 2 n 2 ( 2 n 2 ) .

Corollary 8. Let G K 1, m . Then M 1 L ( K 1 , m ) = 2 m ( 2 m 1 ) and M 2 L ( K 1 , m ) = 2 m ( m 1 ) .

The case for wheel graphs will be investigated in the following result.

Theorem 9. Let us consider G as the wheel graph W n ( n 4 ) with n + 1 vertices. Then we have M 1 L ( W n ) = 4 n ( n 2 ) and M 2 L ( W n ) = n ( 6 n 15 ) .

Proof. With a similar approximation as in the previous results, by labeling the vertices of V ( G ) in the anticlockwise direction as v 1 , v 2 , , v n , v n + 1 such that v n + 1 is the center of the wheel, we obtain

v 1 = ( 0 , 1 , 2 , 2 , , 2 n 3 , 1 , 1 ) , v 2 = ( 1 , 0 , 1 , 2 , 2 , , 2 n 3 , 1 ) , v 3 = ( 2 , 1 , 0 , 1 , 2 , , 2 n 2 , 1 ) , , v n = ( 1 , 2 , 2 , , 2 n 3 , 1 , 0 , 1 ) , v n + 1 = ( 1 , 1 , , 1 n , 0 ) .

Now for any locating vector v i corresponding to a vertex v i ( i { 1,2, , n } ) , we have v i 2 = 4 n + 9 and v n + 1 2 = n . Hence M 1 L ( W n ) = 4 n ( n 2 ) .

For M 2 L ( W n ) , by labeling the vertices as above, we have

v 1 = ( 0 , 1 , 2,2, ,2 n 3 ,1, 1 ) , v 2 = ( 1 , 0 , 1 , 2,2, ,2 n 3 ,1 ) , v 3 = ( 2, 1 , 0 , 1 , 2, ,2 n 2 ,1 ) , v 4 = ( 2,2, 1 , 0 , 1 , 2, ,2 n 1 ,1 ) , v 5 = ( 2,2,2, 1 , 0 , 1 , 2, ,2 n ,1 ) , , v n = ( 1, 2,2, ,2 n 3 , 1 , 0 , 1 ) , v n + 1 = ( 1,1, ,1 n ,0 ) .

Bearing in mind the permutation of components 1 , 0 , 1 in each vector v i , where i = 1 , 2 , , n , it is easy to see that any two adjacent vertices v i and v j ( i , j { 1,2, , n } ) satisfy v i v j = 4 n 11 and v i v i + 1 = 2 n 4 for i = 1 , 2 , , n . Hence M 2 L ( W n ) = n ( 6 n 15 ) .+

The result for determining of locating indices on path graphs can be given as in the following.

Theorem 10. Let G P n ( n 3 ) . Then

M 1 L ( P n ) = j = 1 n 1 ( n j ) ( n j + 1 ) ( 2 n 2 j + 1 ) 3 ,

and

M 2 L ( P n ) = 2 j = 1 n 1 ( n j ) ( n j + 1 ) ( n j 1 ) 3 .

Proof. Assume that G is the graph P n ( n 3 ). By labeling the vertices from left to right as v 1 , v 2 , , v n according to the locating function, the corresponding vector for each vertex v i V ( G ) ( i = 1 , , n ) will be the form of

v 1 = ( 0 , 1 , 2 , 3 , , n 1 ) , v 2 = ( 1 , 0 , 1 , 2 , , n 2 ) , , v n 1 = ( n 2 , n 1 , , 0 , 1 ) , v n = ( n 1 , n 2 , n 3 , , 0 ) .

By applying the symmetry on components between the vector pairs v 1 , v n and v 2 , v n 1 , and so on, we can see that

M 1 L ( P n ) = 2 j = 1 n 1 i = 1 n j i 2 = j = 1 n 1 ( n j ) ( n j + 1 ) ( 2 n 2 j + 1 ) 3 .

For M 2 L ( P n ) , we see that

v 1 v 2 = ( 0 1 ) + ( 1 0 ) + + ( n 1 ) ( n 2 ) = i = 1 n 1 i ( i 1 ) ,

v 2 v 3 = ( 1 2 ) + ( 0 1 ) + + ( n 2 ) ( n 3 ) = i = 1 n 1 i ( i 1 ) ,

v 3 v 4 = i = 1 n 1 i ( i 1 ) ,

However, by the symmetry between the components of the vectors as mentioned above, we get

M 2 L ( P n ) = 2 j = 1 n 1 i = 1 n j i ( i 1 ) = 2 j = 1 n 1 ( i = 1 n j i 2 i = 1 n j i )

which can be rewritten as in the form

M 2 L ( P n ) = 2 j = 1 n 1 ( ( n j ) ( n j + 1 ) ( 2 n 2 j + 1 ) 6 ( n j ) ( n j + 1 ) 2 ) = 2 j = 1 n 1 ( n j ) ( n j + 1 ) ( n j 1 ) 3 .

This complete the proof.+

It is known that from the elementary textbooks the join G = G 1 + G 2 of graphs G 1 and G 2 with disjoint vertex sets V 1 and V 2 and edge sets E 1 and E 2 is the graph union G 1 G 2 together with all the edges joining V 1 and V 2 . In the following theorem we find first and second locating indices for the join graph G.

Theorem 11. Let G G 1 + G 2 such that G 1 and G 2 are both connected graphs with diameter 2 and G is a C 3 or C 5 -free graph. Assume that G 1 has n 1 vertices and m 1 edges while G 2 has n 2 vertices and m 2 edges. Then

M 1 L ( G ) = 2 n 1 n 2 + 4 ( n 1 2 + n 2 2 n 1 n 2 ) 6 ( m 1 + m 2 ) ,

and

M 2 L ( G ) = 2 ( m 1 n 1 + m 2 n 2 ) 4 ( m 1 + m 2 ) + 2 n 1 n 2 ( n 1 + n 2 2 ) .

Proof. Assume that G satisfies the conditions in the statement of theorem. Let us label the vertices of the graph G as

v 1 , v 2 , , v n 1 , v n 1 + 1 , v n 1 + 2 , , v n 1 + n 2 ,

where v 1 , v 2 , , v n 1 V ( G 1 ) and v n 1 + 1 , v n 1 + 2 , , v n 1 + n 2 V ( G 2 ) . Also let v be the locating vector corresponding to the vertex v such that v V ( G 1 ) :

v = ( 0, 1, ,1 deg ( v ) , 2, ,2 n 1 deg ( v ) 1 , 1, ,1 n 2 ) .

Then v 2 = n 2 + 4 n 1 4 3 deg ( v ) .

Similarly, for any vertex w V ( G 2 ) , the locating vector w corresponding to w:

w = ( 1, ,1 n 1 ,0, 1, ,1 deg ( w ) , 2, ,2 n 2 deg ( w ) 1 ) .

So w 2 = n 1 + 4 n 2 4 3 deg ( w ) . Therefore, by the above equalities on v 2 and w 2 , we obtain

M 1 L ( G ) = n 1 ( n 2 + 4 n 1 4 ) 6 m 1 + n 2 ( n 1 + 4 n 2 4 ) 6 m 2 = 2 n 1 n 2 + 4 ( n 1 2 + n 2 2 n 1 n 2 ) 6 ( m 1 + m 2 ) .

Now, let us make partition to the set of vertices of G as

A = { u v : u , v V ( G 1 ) } ,

B = { u v : u , v V ( G 2 ) } ,

C = { u v : u V ( G 1 ) , v V ( G 2 ) } .

Hence M 2 L ( G ) can be written as u v A u v + u v B u v + u v C u v . To get u v A u v for any two adjacent vertices u , v V ( G 1 ) , let us consider

u = ( 0 , 1 , , 1 deg ( u ) , 2 , , 2 n 1 deg ( u ) 1 , 1 , , 1 n 2 )

v = ( 1,0, 2, ,2 deg ( u ) 1 , 1, ,1 n 1 deg ( u ) 1 , 1, ,1 n 2 ) .

We then have

u v = 2 ( deg ( u ) 1 ) + 2 ( n 1 deg ( u ) 1 ) + n 2 = n 2 + 2 n 1 4

which implies u v A u v = m 1 ( n 2 + 2 n 1 4 ) . With a similar calculation, we get u v B u v = m 2 ( n 1 + 2 n 2 4 ) .

Next, we need to calculate u v C u v . To do that let us take u V ( G 1 ) and v V ( G 2 ) , and then labeling as

u = ( 0 , 1 , , 1 deg ( u ) , 2 , , 2 n 1 deg ( u ) 1 , 1 , , 1 n 2 )

v = ( 1, ,1 deg ( u ) + 1 , 1, ,1 n 1 deg ( u ) 1 ,0, 1, ,1 deg ( v ) , 2, ,2 n 2 deg ( v ) 1 ) .

Hence we get

u v = deg ( u ) + 2 ( n 1 deg ( u ) 1 ) + deg ( v ) + 2 ( n 2 deg ( v ) 1 ) = 2 ( n 1 + n 2 ) 4 ( deg ( u ) deg (v))

and so u v C u v = n 1 n 2 ( 2 ( n 1 + n 2 ) 4 ) 2 n 2 m 1 2 n 1 m 2 .

After all above calculations, we finally obtain

M 2 L ( G ) = m 1 ( n 2 + 2 n 1 4 ) + m 2 ( n 1 + 2 n 2 4 ) + n 1 n 2 ( 2 ( n 1 + n 2 ) 4 ) 2 n 2 m 1 2 n 1 m 2 = 2 ( m 1 n 1 + m 2 n 2 ) 4 ( m 1 + m 2 ) + 2 n 1 n 2 ( n 1 + n 2 2 ) .

Hence the result.+

3. Locating Indices of Firefly Graphs

We recall that a firefly graph F s , t , n 2 s 2 t 1 ( s 0, t 0 and n 2 s 2 t 1 0 ) is a graph of order n that consists of s triangles, t pendant paths of length 2 and n 2 s 2 t 1 pendent edges that are sharing a common vertex (cf. [20] ). Let F n be the set of all firefly graphs F s , t , n 2 s 2 t 1 . Note that F n contains the stars S n ( F 0,0, n 1 ) , stretched stars ( F 0, t , n 2 t 1 ) , friendship graphs ( F n 1 2 ,0,0 ) and butterfly graphs ( F s ,0, n 2 s 1 ) .

In the next theorem we present the first and second locating indices for the firefly graph. In our calculations, for simplicity, we denote n 2 s 2 t 1 by a single letter l.

Theorem 12. Let G F s , t , l ( s 0, t 0 and l 0 ) be a firefly graph of order n. Then

M 1 L ( G ) = 4 l 2 + 16 l s + 26 l t 2 l + 16 s 2 + 52 s t 10 s + 38 t 2 28 t ,

and

M 2 L ( G ) = 2 l 2 + 16 l s + 13 l t 2 l + 24 s 2 + 52 s t 20 s + 22 t 2 17 t .

Proof. Let G F s , t , l ( s 0, t 0 and l 0 ) is a firefly graph of order n. Let us label the vertices with clockwise direction as

v 1 , v 2 , , v 2 s + 1 , v 2 s + 2 , v 2 s + 3 , , v 2 s + l + 1 , v 2 s + l + 2 , v 2 s + l + 3 , , v 2 s + l + t + 1 , v 2 s + l + t + 2 , v 2 s + l + t + 3 , , v 2 s + 2 t + l + 1 ,

where v 1 is the center of the firefly graph and

v 2 , v 3 , , v 2 s + 1 2 s : verticesoftriangles,

v 2 s + 2 , v 2 s + 3 , , v 2 s + l + 1 l : verticesofpendentedges,

v 2 s + l + 2 , v 2 s + l + 3 , , v 2 s + l + t + 1 t : verticesofpendentpathoflength1,

v 2 s + l + t + 2 , v 2 s + l + t + 3 , , v 2 s + 2 t + l + 1 t : verticesofpendentpathoflength2 .

Now we calculate the corresponding vectors v i for each vertex v i V ( G ) , where i = 1 , 2 , , 2 s + 2 t + l + 1 , as in the following:

v 1 = ( 0 , 1 , 1 , , 1 2 s , 1 , 1 , , 1 l , 1 , 1 , , 1 t , 2 , 2 , , 2 t ) ,

v 2 = ( 1 , 0 , 1 , 2 , 2 , , 2 2 s 2 , 2 , 2 , , 2 l , 2 , 2 , , 2 t , 3 , 3 , , 3 t ) ,

v 3 = ( 1 , 1 , 0 , 2 , 2 , , 2 2 s 2 , 2 , 2 , , 2 l , 2 , 2 , , 2 t , 3 , 3 , , 3 t ) ,

v 4 = ( 1 , 2 , 2 , 0 , 1 , 2 , 2 , , 2 2 s 4 , 2 , 2 , , 2 l , 2 , 2 , , 2 t , 3 , 3 , , 3 t ) ,

v 5 = ( 1,2,2,1,0, 2,2, ,2 2 s 4 , 2,2, ,2 l , 2,2, ,2 t , 3,3, ,3 t ) ,

v 2 s + 1 = ( 1, 2,2, ,2 2 s 2 ,1,0, 2,2, ,2 l , 2,2, ,2 t , 3,3, ,3 t ) ,

v 2 s + 2 = ( 1, 2,2, ,2 2 s ,0, 2,2, ,2 l 1 , 2,2, ,2 t , 3,3, ,3 t ) ,

v 2 s + 3 = ( 1, 2,2, ,2 2 s ,2,0, 2,2, ,2 l 2 , 2,2, ,2 t , 3,3, ,3 t ) ,

v 2 s + l + 1 = ( 1, 2,2, ,2 2 s , 2,2, ,2 l 1 ,0, 2,2, ,2 t , 3,3, ,3 t )

v 2 s + l + 2 = ( 1, 2,2, ,2 2 s , 2,2, ,2 l ,0, 2,2, ,2 t 1 ,1, 3,3, ,3 t 1 )

v 2 s + l + t + 1 = ( 1, 2,2, ,2 2 s , 2,2, ,2 l , 2,2, ,2 t 1 ,0, 3,3, ,3 t 1 ,1 ) ,

v 2 s + l + t + 2 = ( 2, 3,3, ,3 2 s , 3,3, ,3 l ,1, 3,3, ,3 t 1 ,0, 4,4, ,4 t 1 ) ,

v 2 s + l + t + 3 = ( 2, 3,3, ,3 2 s , 3,3, ,3 l ,3,1, 3,3, ,3 t 2 ,0, 4,4, ,4 t 2 ) ,

v 2 s + l + 2 t + 1 = ( 2, 3,3, ,3 2 s , 3,3, ,3 l , 3,3, ,3 t 1 ,1, 4,4, ,4 t 1 ,0 ) .

Suppose that A , B , C , D V ( G ) such that

A = { v 2 , v 3 , , v 2 s + 1 } , C = { v 2 s + l + 2 , v 2 s + l + 3 , , v 2 s + l + t + 1 } , B = { v 2 s + 2 , v 2 s + 3 , , v 2 s + l + 1 } , D = { v 2 s + l + t + 2 , v 2 s + l + t + 3 , , v 2 s + 2 t + l + 1 } .

Therefore we can write

M 1 L ( G ) = v A v 2 + v B v 2 + v C v 2 + v D v 2 .

For the calculation of v A v 2 , we have the cases v 1 2 = 2 s + l + t + 4 t = 2 s + l + 5 t and

v i 2 = 2 + 4 ( 2 s 2 ) + 4 l + 4 t + 9 t = 4 l + 8 s + 13 t 6 ,

where i = 2 , 3 , , 2 s + 1 . Hence

v A v 2 = 2 s + l + 5 t + 2 s ( 4 l + 8 s + 13 t 6 ) = 2 s + l + 5 t + 8 s l + 16 s 2 + 26 s t 12 s = l 10 s + 5 t + 8 l s + 26 s t + 16 s 2 .

On the other hand, for the calculation of v B v 2 , we have

v i 2 = 1 + 4 ( 2 s ) + 4 ( l 1 ) + 4 t + 9 t = 4 l + 8 s + 13 t 3,

where for i = 2 s + 1 , 2 s + 2 , , 2 s + l + 1 . Thus

v B v 2 = l ( 4 l + 8 s + 13 t 3 ) = 4 l 2 + 8 l s + 13 l t 3 l .

Thirdly to calculate v C v 2 , we have

v i 2 = 2 + 4 ( 2 s ) + 4 l + 4 ( t 1 ) + 9 ( t 1 ) = 4 l + 8 s + 13 t 11,

where i = 2 s + l + 2 , 1 , 2 s + l + 3 , , 2 s + l + t + 1 , and so

v C v 2 = t ( 4 l + 8 s + 13 t 11 ) = 4 t l + 8 t s + 13 t 2 11 t .

Finally, for the case of v D v 2 , we get

v i 2 = 3 + 9 ( 2 s ) + 9 l + 9 ( t 1 ) + 16 ( t 1 ) = 9 l + 18 s + 25 t 22 ,

where i = v 2 s + l + t + 2 , v 2 s + l + t + 3 , , v 2 s + 2 t + l + 1 . This gives

v D v 2 = t ( 9 l + 18 s + 25 t 22 ) = 9 t l + 18 t s + 25 t 2 22 t .

By collecting all above calculations, we obtain

M 1 L ( G ) = v A v 2 + v B v 2 + v C v 2 + v D v 2 = l 10 s + 5 t + 8 l s + 26 s t + 16 s 2 + 4 l 2 + 8 l s + 13 l t 3 l + 4 t l + 8 t s + 13 t 2 11 t + 9 t l + 18 t s + 25 t 2 22 t = 4 l 2 + 16 l s + 26 l t 2 l + 16 s 2 + 52 s t 10 s + 38 t 2 28 t ,

as required.

Before starting to calculate the index M 2 L ( G ) = v i u j E ( G ) v i u j , we should remind that for any two adjacent vertices u and v will be denoted by u v . Now, let us again consider the same subsets A, B, C and D of V ( G ) . Therefore we firstly have

v i A v 1 v i = 2 s ( 1 + 2 ( 2 s 2 + l + t ) + 3 t ) = 2 s ( 2 l + 4 s + 5 t 3 ) = 4 s l + 8 s 2 + 10 s t 6 s .

v i B v 1 v i = l ( 2 ( 2 s + l 1 + t ) + 3 t ) = l ( 2 l + 4 s + 5 t 2 ) = 2 l 2 + 4 s l + 5 l t 2 l .

v i C v 1 v i = t ( 1 + 2 ( 2 s + l + t 1 ) + 2 t ) = t ( 2 l + 4 s + 4 t 1 ) = 2 t l + 4 t s + 4 t 2 t .

Secondly,

v i A { v 1 } v i v i + 1 v i v i + 1 = 2 s ( 1 + 4 ( 2 s 2 + l + t ) + 9 t ) = 2 s ( 4 l + 8 s + 13 t 7 ) = 8 s l + 16 s 2 + 26 s t 14 s .

Thirdly,

v i C , v i + t D v i v i + t v i v i + t = t ( 2 + 6 ( 2 s + l + t 1 ) + 12 ( t 1 ) ) = t ( 6 l + 12 s + 18 t 16 ) = 6 t l + 12 t s + 18 t 2 16 t .

Again, by collecting all above calculations, we obtain

M 2 L ( G ) = v i A v 1 v i + v i B v 1 v i + v i C v 1 v i + v i A { v 1 } v i v i + 1 v i v i + 1 + u i C , v i + t D v i v i + t v i v i + t = 4 s l + 8 s 2 + 10 s t 6 s + 2 l 2 + 4 s l + 5 l t 2 l + 2 t l + 4 t s + 4 t 2 t + 8 s l + 16 s 2 + 26 s t 14 s + 6 t l + 12 t s + 18 t 2 16 t = 2 l 2 + 16 l s + 13 l t 2 l + 24 s 2 + 52 s t 20 s + 22 t 2 17 t .

These all above processes complete the proof.+

Corollary 13. 1) For any friendship graph of order n,

M 1 L ( G ) = 4 n 2 13 n + 9 and M 2 L ( G ) = 6 n 2 22 n + 16.

2) For any butterfly graph of order n,

M 1 L ( G ) = 4 n 2 10 n 6 s + 6 and M 2 L ( G ) = 8 n s 24 s 6 n + 2 n 2 + 4.

4. Conclusion

In this paper, two new topological indices based on Zagreb indices are proposed. The exact values of these new topological indices are calculated for some standard graphs and for the firefly graphs. These new indices can be used to investigate the chemical properties for some chemical compound such as drugs, bridge molecular graph etc. For the future work, instead of defining these new topological indices based on the degrees of the vertices, we can redefine them based on the degrees of the edges by defining them on the line graph of any graph. Similar calculations can be computed to indicate different properties of the graph.

Cite this paper
Wazzan, S. and Saleh, A. (2019) On the First and Second Locating Zagreb Indices of Graphs. Applied Mathematics, 10, 805-816. doi: 10.4236/am.2019.1010057.
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