Chemical Trees with Extreme Values of Zagreb Indices and Coindices

Chemical Trees with Extreme Values of Zagreb Indices and Coindices

Ž. K. VUKIĆEVIĆ ANDG. POPIVODA^

(COMMUNICATED BYA^LIR^EZAA^SHRAFI)

Department of Mathematics, University of Montenegro 81000 Podgorica, P. O. Box 211, Montenegro

ABSTRACT. We give sharp upper bounds on the Zagreb indices and lower bounds on the Zagreb coindices of chemical trees and characterize the case of equality for each of these topological invariants.

Keywords: Zagreb index, Zagreb coindex, chemical tree.

1. I

NTRODUCTION

Let G be chemical graph with vertex and edge sets V(G) and E(G), respectively. For each u,vV(G) the edge connecting u and v is denoted by uv and d_G(u) denotes the degree of u in G. We will omit subscript G when the graph is clear from the context. The Zagreb indices are defined as follows:

2 ) 1(G)= ( d(u) M

G V

u





).

( ) (

= ) (

)

2 G ( d u d v

M _{G G}

G E uv





Here, M₁(G) and M₂(G) denote the first and the second Zagreb index, respectively.

The first Zagreb index can be also expressed as a sum of vertex degrees over edges of G,



( ) ( )



= ) (

)

1 G ( d u d v

M _{G G}

G E uv







.

The proof of this fact can be found in [8]. Many authors find this definition more useful than the original one. Nevertheless, in this paper we will use its original form.

Corresponding author (Email:goc@t-com.me).

Received : October 18, 2012; Accepted: May 28, 2013.

Zagreb indices belong to better researched topological invariants. Due to their chemical relevance, they have been studied for more than 30 years in numerous paper in chemical literature [3, 5, 6, 7, 8, 9]. Recently, Došlić ([4]) gave a generalisation of these numbers. He introduced two new graph invariants, the first and the second Zagreb coindices, defined as follows:



( ) ( )



= ) (

) (

1 G d u d v

M _{G G}

G E uv







and

) ( ) (

= ) (

) (

2 G d u d v

M _{G G}

G E uv





.

Ashrafi, Došlić and Hamzeh in [1] determined the extremal graphs with respect to the Zagreb coindices. They wrote that it would be interesting to extend results presented in [1] to some other classes of graphs of chemical interest and emphasized that their "result leave open the question of the minimum values of Zagreb coindices over chemical trees".

In this paper we give an answer to this question.

For n=3k6 let T_3k be the family of chemical trees with n vertices such that:

1

k vertices have degree 4, 1 vertex has degree 2 and remaining vertices are pendant.

Denote by T~_3k T_3k family of trees G from T_3k such that for the unique vertex wV(G) of degeree 2 exactly one of its neighbours is pendant. For n=3k17, denote by T_3k1 the family of chemical trees with n vertices such that: k1 vertices have degree 4, 1 veretex has degree 3 and all other vertices are pendant, while ~_{3 1}

k

T denotes the family of trees G from T₃k₁ such that for the unique vertex wV(G) of degeree 3 exactly one of its neighbours is pendant. For n=3k25, T₃k₂ denotes the family of chemical trees with

n vertices such that: k vertices have degree 4 and remaining are pendant.

Our main results are the next two theorems:

Theorem 1.Let G be chemical tree with n5 vertices. Then











 

).

3 mod ( 2 ,

10 6

3) mod ( 0,1 ,

12 ) 6

1( n n

n G n

M

The equality is attained if and only if GT_n.

Theorem 2.Let G be chemical tree with n5 vertices. Then









 

otherwise.

, 24 8

3) mod ( 0,1 ,

26 ) 8

2( n

n G n

M

The equality is attained if and only if n0,1(mod3) and ~ ,

GTn or n2(mod3) and GTn.

2. O

N THE

F

IRST

Z

AGREB

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C

HEMICAL

T

REES

In this section our goal is to obtain sharp upper bound on the first Zagreb index of chemical trees and, accordingly, to characterize the chemical trees with maximal values of the first Zagreb index.

Let G be a chemical tree with n vertices. For each i{1,2,3,4} let n_i denote the number of its vertices of degeree i. Then,

n n n n

n₁ ₂  ₃ ₄ = (1)

and from handshaking lemma

1).

2(

= 4 3

2 _{2 3 4}

1 n  n  n n

n (2)

From (1) and (2) we conclude that

2.

= 3

2 _{3 4}

2 n  n n

n (3)

Using definition of the first Zagreb index we get that for any chemical tree G: .

16 9 4

= )

( _{1 2 3 4}

1 G n n n n

M    (4)

Let G_n be nvertex chemical tree such that n=3k, k 2. Then (3) reduces to 2

3

= 3

2 _{3 4}

2  n  n k

n

from which follows that n₄ k1 and for n₄ =k1 hold n₂ =1, n₃ =0 and so, due to (1), n₁ =2k. Hence, using Eq. (4)

12.

6

= )

1(G n

M _n Otherwise, n₄ k2, that is 2

4 n3

n and using equations (3) and (1) we get

14.

6

3 2 3 8

5

3 2)

4(

=

3 ) (

) 3 2 4(

=

16 9 4

= ) (

2

4 2

4 4 3 1 4 3 2

4 3 2 1 1







 

 



































n n n n

n n n n

n n n n n n n

n n n n G

M _n

Therefore, for n 0(mod3), n6, if Gn is nvertex chemical tree 12

6 ) 1(Gn  n

M and equality is attained if and only if in V(G_n) there are 1 3

n vertices of degree 4, 1 vertex of degree 2 and remaining vertices are pendant, that is G_nT_n.

Now, let G_n be nvertex chemical tree such that n=3k1, k2. Then (3) reduces to n₂ 2n₃3n₄ =3k1 and, as in the previous case, follows that n₄ k1.

For n₄ =k1, we have the next two possibilities: n₂ =0, n₃ =1 or n₂ =2, n₃ =0. In the first case, M₁(G_n)=6n12 and in the second one M₁(G_n)=6n14. So, the second case can no give the maximal M₁ index among chemical trees.

Similarly with the above discussion for n 0(mod3), when n₄ k2 we obtain that

15.

6

3 2 3 1 8

5

3 2)

4(

= ) (

2

4 2 1







 



  















 n

n n n

n n n n

G

M _n

Hence, for n1(mod3), n7, if G_n is nvertex chemical tree 12

6 )

1(G  n

M _n and equality is attained if and only if in V(G_n) there are 1 3

1 n vertices of degree 4, 1 vertex of degree 3 and remaining vertices are pendant, i.e. G_nT_n.

Finally, let G_n be chemical tree with n=3k2, k 1, vertices. Equality (3) reduces to n₂ 2n₃3n₄ =3k and so n₄ k.

Similarly with the above discussion, for n₄ =k, we have that n₂ =n₃ =0 and the first Zagreb index on this class of trees takes value M₁(G_n)=6n10.

Otherwise, n₄ k1 and we get:

13.

6

3 1 3 2 8

5

3 2)

4(

= ) (

2

4 2 1







 



  















 n

n n n

n n n n

G

M _n

It follows that for n2(mod3), n5, if G_n is nvertex chemical tree, then 10

6 )

1(G  n

M _n and equality is attained if and only if in V(G_n) there are 3

2

n vertices of degree 4 and remaining vertices are pendant.

3. O

N THE

S

ECOND

Z

AGREB

I

NDEX AMONG

C

HEMICAL

T

REES

Let G₁ and G₂ be vertex-disjoint graphs. Suppose that a₁V(G₁) and a₂V(G₂). We denote by G₁a₁a₂ G₂ the graph with the vertex set V(G₁)V(G₂) and the edge set

} { ) ( )

(G₁ E G₂ a₁a₂

E   , that is the graph obtained from the union G₁G₂ connecting its two components G₁ and G₂ by an edge a₁a₂. G₁ a₁a₂ G₂ is edge-joinof the graphs

G1 and G₂ across the vertices a₁ and a₂.

Lemma 1. If G is chemical tree with at least two vertices of degeree 3, then its second Zagreb index cannot be maximal.

Proof.Let G be chemical tree and x,yV(G) such that d(x)=d(y)=3. Consider first in detail the case when the vertices x and y are not neighbours. Denote by x_i and y_i, i=1,3 its neighbours, respectively. Let e_i = x_ix and g_i = y_iy be appropriate edges for each

1,3

=

i . Without loss of generality, suppose that

) ( ) ( ) ( ) ( ) ( )

(x₁ d x₂ d x₃ d y₁ d y₂ d y₃

d      (5)

and the unique path between x and y goes toward the vertices x₁ and y₁. Denote by e3

G the subgraph of G obtained by deleting the edge e₃. The graph Ge₃ is forest with two components H_x3 and H_x such that x₃V(H_x3) and xV(H_x). Let

x

x x y H

H

G  ₃ 

= 3 be edge-join of the trees H_x3 and H_x across the edge x₃y. We are going to prove that M₂(G)<M₂(G). Let S ={e₁,e₂,e₃,g₁,g₂,g₃}. It holds



( ) ( ) ( )

 

3 ( ) ( ) ( )



3 ) ( ) (

= )

( _{1 2 3 1 2 3}

2 G d u d v d x d x d x d y d y d y

M

S uv

















and M₂(G)= d(u)d(v) 2



d(x₁) d(x₂)

 

4d(y₁) d(y₂) d(y₃) d(x₃)



S uv



















. Therefore

)]

( ) ( ) ( [ ) ( ) ( ) (

= ) ( )

( _{2 1 2 3 1 2 3}

2 G M G d x d x d x d y d y d y

M       

and due to inequality (5)

0,

<

3) ( 2

= 3)]

( 2) ( 1) ( [ 3) ( 2) ( 1) ( ) 2( )

2(G M G d x d x d x d x d x d x d x

M         

since d(x₃)1.

Now, suppose that the vertices x and y are neighbours. Then, the vertices x₁ and y1 from the above construction are the vertices y and x, respectively, and the edges e₁ and g₁ are one and the same edge xy. Now, inequality (5) reduces to:

), ( ) ( ) ( )

(x₂ d x₃ d y₂ d y₃

d    (6)

and the next hold:

)]

( ) ( 3[

9 )]

( ) ( 3[

) ( ) (

= )

( _{2 3 2 3}

2 G d u d v d x d x d y d y

M

S uv















and

)].

( ) ( ) ( 4[

8 ) ( 2 ) ( ) (

= )

( _{2 2 3 3}

2 G d u d v d x d y d y d x

M

S uv

















Due to (6)

0,

<

) ( 2 1

=

)]

( ) ( [ 1 ) ( ) (

)]

( ) ( [ 1 ) ( ) (

= ) ( ) (

3

3 2

3 2

3 2

3 2

2 2

x d

x d x d x

d x d

y d y d x

d x d G

M G M



















 



since d(x₃)1.

 Lemma 2. If G is chemical tree with at least two vertices of degeree 2, then its second Zagreb index cannot be maximal.

Proof.Let G be chemical tree and x,yV(G) such that d(x)=d(y)=2. Suppose that x and y are not neighbours and denote by x₁, x₂ and y₁, y₂ its neighbours, respectively.

Let e_i =x_ix, g_i = y_iy be appropriate edges for each i=1,2. Without loose of generality suppose that

) ( ) ( ) ( )

(x₁ d x₂ d y₁ d y₂

d    (7)

and the unique x-y path goes toward the vertices x₁ and y₁. The graph Ge₂ is forest with two components H_x2 and H_x such that x₂V(H_x2) and xV(H_x). Let

x

x x y H

H

G  ₂  

= 2 be edgejoin of the trees H_x2 and H_x across the edge x₂y. We are going to prove that M₂(G)<M₂(G). Let S ={e₁,e₂,g₁,g₂}. Then,

)]

( ) ( 2[

)]

( ) ( 2[

) ( ) (

= )

( _{1 2 1 2}

2 G d u d v d x d x d y d y

M

S uv













and

)]

( ) ( ) ( 3[

) ( ) ( ) (

= )

( _{1 1 2 2}

2 G d u d v d x d y d y d x

M

S uv















Using the inequality (7),

0,

<

) ( 2

=

)]

( ) ( [ ) ( ) (

)]

( ) ( [ ) ( ) (

= ) ( ) (

2

2 1

2 1

2 1

2 1

2 2

x d

x d x d x d x d

y d y d x d x d G

M G M















 



since d(x₂)1.

When the vertices x and y are neighbours, the vertices x₁ and y₁ from the upper construction are the vertices y and x, respectively, and the edges e₁ and g₁ are the one and the same edge xy. Now, inequality (7) is equivalent with

) ( ) (x₂ d y₂

d  (8)

and we have that:

) ( 2 4 ) ( 2 ) ( ) (

= )

( _{2 2}

2 G d u d v d x d y

M

S uv











and

)].

( ) ( 3[

3 ) ( ) (

= )

( _{2 2}

2 G d u d v d y d x

M

S uv













Hence, due to d(x₂)1and inequality (8)

0.

<

) ( 2 1 ) ( ) ( 1

= ) ( )

( _{2 2 2 2}

2 G M G d x d y d x

M      

 Lemma 3. If G is chemical tree with at least one vertex of degeree 2 and at least one vertex of degeree 3, then its second Zagreb index cannot be maximal.

Proof.Let G be chemical tree and x,yV(G) such that d(x)=2 and d(y)=3. If x and y are not neighbours, denote by x₁, x₂ and y₁, y₂, y₃ its neighbours, respectively. Let

,

= ₁

1 xx

e e₂ = x₂x, g₁ = y₁y, g₂ = y₂y, and g₃ = y₃y and suppose that x-y path goes toward the vertices x₁ and y₁. The graph Ge₂ is forest with two components H_x2 and

Hx such that x₂V(Hx₂) and xV(Hx). Let G Hx  x₂ y Hx

= 2 be edgejoin of

the trees H_x2 and H_x across the edge x₂y. We are going to prove that M₂(G)<M₂(G). Let S ={e₁,e₂,g₁,g₂,g₃}. Then,

3)]

( 2) ( 1) ( 3[

2)]

( 1) ( 2[

) ( ) (

= )

2( d u d v d x d x d y d y d y

S G uv

M      

 and

2)].

( 3) ( 2) ( 1) ( 4[

1) ( ) ( ) (

= )

2( d u d v d x d y d y d y d x

S G uv

M      



 Therefore,

3)].

( 2) ( 1) ( [ 2) ( 2 1) ( ) 2( )

2(G M G d x d x d y d y d y

M       

Since d(x₂)1, d(y₁)2, d(y₂)1 and d(y₃)1, it follows that M₂(G)M₂(G)<0. In the case when x and y are neighbours, that is x₁ is the same as y and y₁ is the same as x, the next is true:

)], ( ) ( 3[

6 ) ( 2 ) ( ) (

= )

( _{2 2 3}

2 G d u d v d x d y d y

M

S uv













)]

( ) ( ) ( 4[

4 ) ( ) (

= )

( _{2 3 2}

2 G d u d v d y d y d x

M

S uv















, and so

)].

( ) ( [ ) ( 2 2

= ) ( )

( _{2 2 2 3}

2 G M G d x d y d y

M     

Since that degrees d(x₂), d(y₂) and d(y₃) are at least 1, it follows that 0

<

) ( )

( ₂

2 G M G

M   .

 From the upper three lemmas we make the next conclusion:

Corollary 1. If G is chemical tree such that



( ) ischemicaltree



, max

= )

( ₂

2 G M T T

M

then G satisfies one of the next three conditions:

(i)all vertices of the graph G have degrees 1 or 4;

(ii) in V(G) there is exactly one vertex of degeree 2 and remaining have degrees1 or 4;

(iii) in V(G) there is exactly one vertex of degeree 3 and remaining are of degerees 1 or 4 .

Now, we are ready to prove Theorem 2.

Proof of Theorem 2: For edge e with end points u and v denote by w(e) the product )

( ) (u d v

d . Let A denotes the set of pendant edges in the graph G and B=E(G)A. Then,

) ( )

(

= )

2(G w e w e

M

B e A

e





 



First, consider the case n=3k2, k1. From Eq. (3) we have that n₂ 2n₃ 0 3)

mod

( and so, from Corollary 1 we get n₂ =n₃ =0. From (1) and (3) follows that k

n₄ = and n₁ =2k2, that is |A|=2k2 and |B|=k1. Since







 A e

B e e

w 4,

= 16, ) ( we obtain that

24.

8

= 8 24

= ) ( )

(

= )

2(







 





n k

e w e

w G

M

B e A

n e

Now, let n=3k1, k 2. From Eq. (3) we have that n₂ 2n₃ 2(mod3) and so, from Corollary 1 we get n₂ =0, n₃ =1. Hence, n₄ = k1, n₁ =2k1, that is

1 2

|=

|A k and |B|=k1. Let xV(G_n) be unique vertex of degree 3. Then,

















edges pendant with

incident not

is 4,

8

edge pendant 1

with incident is

3, 8

edges pendant 2

with incident is

2, 8

= ) (

x k

x k

x k

e w

A e

and

















edges.

pendant with

incident not

is 28,

16

edge pendant 1

with incident is

24, 16

edges pendant 2

with incident is

20, 16

= ) (

x k

x k

x k

e w

B e

Hence,













edges pendant with

incident not

is 24,

24

edge pendant 1

with incident is

21, 24

edges pendant 2

with incident is

18, 24

= )

2(

x k

x k

x k

G

M _n

i.e.













edges.

pendant with

incident not

is 32,

8

edge pendant 1

with incident is

29, 8

edges pendant 2

with incident is

26, 8

= )

2(

x n

x n

x n

G

M _n

Finally, let n=3k, k 2. Similarly with the discussion from the above two cases we obtain that n₂ =1, n₃ =0, n₁ =2k and n₄ = k1, that is A =2k and B = k1. Let

x be unique vertex of degree 2. It holds





 

 8 , isnotincidentwithpendantedges edge pendant 1

with incident is

2,

= 8 )

( k x

x e k

w

A e

and











 16 32, isnotincidentwithpendantedges.

edge pendant 1

with incident is

24,

= 16 )

( k x

x e k

w

B e

So,









edges pendant with

incident not

is 32,

24

edge pendant 1

with incident is

26,

= 24 )

2( k x

x G k

M _n

that is









edges.

pendant with

incident not

is 32,

8

edge pendant 1

with incident is

26,

= 8 )

2( n x

x G n

M _n

Combining the above results we get that









 

otherwise ,

24 8

3) mod ( 0,1 ,

26 ) 8

2( n

n G n

M

and equality is attained if and only if n0,1(mod3) and ~ ,

GTn or n2 (mod3) and GTn.

4. Z

AGREB

C

OINDICES AMONG

C

HEMICAL

T

REES

In [2] A. R. Ashrafi, T. Došlić and A. Hamzeh proved that for any connected graph G with n verices and m edges hold

) ( 1) ( 2

= )

( ₁

1 G m n M G

M  

and

).

2 ( ) 1 ( 2

= )

( ² _{2 1}

2 G m M G M G

M  

The next two theorems are direct consequence of these equalities and theorems 1 and 2 proved in sections 2 and 3.

Theorem 3.Let G be chemical tree with n5 vertices. Then















 

. 3) mod ( 2 ,

12 10 2

3) mod ( 0,1 ,

14 10 ) 2

( ₂²

1 n n n

n n

G n M

The equality is attained if and only if GT_n.

Theorem 4. Let G be chemical tree with n5 vertices. Then













 

otherwise.

31, 15 2

3) mod ( 0,1 ,

34 15 ) 2

( ₂²

2 n n

n n

G n M

The equality is attained if and only if n0,1(mod3) and ~ ,

GTn or n2 (mod3) and GTn.

R

^EFERENCES

[1] A. R. Ashrafi, T. Došlić, A. Hamzeh, Extremal Graphs with Respect to the Zagreb Coindices, MATCH Commun. Math. Comput. Chem. 65 (2011) 8592.

[2] A. R. Ashrafi, T. Došlić, A. Hamzeh, The Zagreb coindices of graph operations, Discrete Appl. Math. 158 (2010) 15711578.

[3] J. Braun, A. Kerber, M. Meringer, C. Rucker, Similarity of molecular descriptors: the equivalence of Zagreb indices and walk counts, MATCH Commun. Math. Comput. Chem.

54 (2005), 163176.

[4] T. Došlić, Vertex-weighted Wiener polynomials for composite graphs, Ars. Math.

Contemp 1 (2008) 6680.

[5] I. Gutman, K. C. Das,The first Zagreb index 30 years after, MATCH Commun. Math.

Comput. Chem. 50 (2004), 8392.

[6] M. H. Khalifeh, H. YousefiAzari, A. R. Ashrafi, The first and the second Zagreb indices of graph operations, Discrete Appl. Math 157 (2009) 804811.

[7] M. H. Khalifeh, H. YousefiAzari, A. R. Ashrafi, S. Wagner, Some new results on distancebased graph invariantsEur. J. Comb. 30 (2009) 11491163.

[8] S. Nikolić, G. Kovačević, A. Miličević, N. Trinajstić, The Zagreb indices 30 years after, Croat. Chem. Acta 76 (2003) 113124.

[9] B. Zhou, I. Gutman, Relations between Wiener, hyper-Wiener and Zagreb indices, Chem. Phys. Lett. 394 (2004) 9395.