An Upper Bound on the First Zagreb Index in Trees

(1)

An Upper Bound on the First Zagreb Index in Trees

R.RASI1,S.M.SHEIKHOLESLAMI1, AND A.BEHMARAM2

1

Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, I. R. Iran 2

Faculty of Mathematical Sciences, University of Tabriz, Tabriz, I. R. Iran

ARTICLE INFO ABSTRACT

Article History:

Received: 1 March, 2016 Accepted: 10 June 2016 Published online 6 February 2017 Academic Editor: Tomislav Došlić

The first Zagreb index M1(G) is equal to the sum of squares of the degrees of the vertices and the first Zagreb coindex ( ) is equal to the sum of sums of vertex degrees of the pairs of non-adjacent vertices. Kovijanić Vukićević and G. Popivoda (Iran. J. Math. Chem. 5 (2014) 19–29) proved that for any chemical tree of order n, n 5,

  

   

. o

10 6

3) m ( 0,1 12

6 ) ( 1

therwise n

od n

n T M

IIn this paper, we generalize the aforementioned bound for all trees in terms of their order and maximum degree. Moreover, we give a lower bound on the first Zagreb coindex of trees.

first Zagreb index first Zagreb coindex tree

Chemical tree

1. I

NTRODUCTION

In this paper, G is a simple connected graph with vertex set V = V(G) and edge set E = E(G). The order |V| of G is denoted by n = n(G). For every vertex vV , the open neighborhood N(v) is the set {uV(G)|uvE(G)} and the closed neighborhood of v is the set N[v]= N(v){v}. The degree of a vertex vV is d_v =

|

N(v)

|

. The minimum and maximum degree of a graph G are denoted by  =(G) and =(G), respectively. Trees with the property 4 are called chemical trees.

The Zagreb indices have been introduced more than thirty years ago by Gutman and Trinajestić in [6]. They are important molecular descriptors and have been closely correlated with many chemical properties [6, 7]. Thus, it attracted more and more attention from chemists and mathematicians [2, 3, 4, 8, 10, 11].

The first Zagreb index M₁(G) is defined as follows:



Corresponding Author: (Email address: ([email protected]) DOI: 10.22052/ijmc.2017.42995

(2)

. = )

( 2

1 v

V v

d G

M

_



The first Zagreb index can be also expressed as the sum of vertex degree over edges of G, that is, M₁(G)=_{ }_uv _E₍_G₎(d_u d_v). Došlić in [5] introduced a new graph invariant called the first Zagreb coindex, as M1(G)= uv E₍G₎(dudv). Next we introduce a family of trees. For n=(1)k p (k 2), let T_n be the family of trees of order n with maximum degree  such that:

 If p=0, k1 vertices have degree , 1 vertex has degree 2 and remaining vertices are pendant.

 If p=1, k1 vertices have degree , 1 vertex has degree 1 and remaining vertices are pendant.

 If p=2, k vertices have degree  and remaining vertices are pendant.

 If p3, k vertices have degree , 1 vertex has degree p1, and nk1 remaining vertices are pendant.

Kovijanić Vukićević and Popivoda [9] proved the following upper bound on the first Zagreb index of chemical trees and characterized all extreme chemical trees.

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

  



 



, o

10 6

3) m ( 0,1 12

6 ) ( 1

therwise n

od n

n T

M

with equality if and only if GT_n.

In this paper, we establish an upper bound on the first Zagreb index of trees in terms of the order and maximum degree, as a generalization of aforementioned bound. As a consequence, we obtain a lower bound on the first Zagreb coindex for trees.

2. M

AIN

R

ESULTS

In this section, we prove the following result:

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      

 

     

    

   

    



3, 2)

( 3 2 2) (

2 = 2

2 2) (

1 = 3

2) (

0 = 4

4 2) ( ) ( 1

p p

p n

T M

with equality if and only if GT_n .

To prove Theorem 2, we proceed with some definitions and lemmas. If n is a positive integer, then an integer partition of n is a non-increasing sequence of positive integers (a₁,a₂,,a_t) whose sum is n. If 1a₁a₂a_t a, then (a₁,a₂,,a_t) is called an integer partition of n on N_a ={1,2,,a}. An integer partition (a₁,a₂,,a_t) of n on N_a is called an integer a-partition if the number of a in this partition is as large as possible. In other words, if n=ka, then (a,,a) is the integer a–partition and if n=kab where 0

<

b

<

a then (b,a,,a) is the integer a–partition. The proof of the next result is straightforward and therefore omitted.

Lemma 3. For positive integers n,t and a (_i 1it), we have a) If n=a₁a₂a_t and t 1, then 2 2

2 2 1 2

t

a a

a

n    .

b) If a_i a_j, then (a_i1)2(a_j1)2a_i2a2_j 2.

Lemma 4. If (a₁,a₂,,a_t) is an integer partition of n=kab(0ba) on N , then _a

. 2 2 2

1 =

b ka a_i t

i

 



Proof. Let (a₁,a₂,,a_t) be an partition of n on N_a. If ai aj a for some 1i jt, then by switching (ai,aj) to (ai 1,aj 1), we get a new integer partition of n on Na. Note that if a_i 1=0, then we will remove a_i1 from the new partition. Applying Lemma 3 (a), we obtain

. 1)

( 1)

( 2 2 2

2 1 2

1 =

t j

i i

t

i

a a

a         



  

By repeating this process, we arrive at an integer a–partition of n on N_a. It follows from Lemma 2 that 2 2 2

1

=ai ka b t

i  

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Lemma 5. Let n=kab where 0ba and let (a₁,a₂,,a_t) be an integer partition of n on N_a which is not a–partition. Then the following statements holds:

a. If b0, then 2 2 2

1

=( 1)  ( 1) ( 1)



ai k a b

t

i .

b. If b=0, then 2 2 1

=( 1)  ( 1)



t a_i k a

i .

Proof. (a) Since n=a  a_t =ba a=kab k

1     , we have tk1. First let

1 = k

t . Then we have

, 1) ( 1) ( =

1) ( 1)

( 1) ( =

Lemma3) (by

) 2( )

(

) 2( )

( = 1) ( 1)

(

2 2

1 2

2 1

  

     

    

     

  

b a

k

k t b

a k

b ka t b ka

b ka t

a a

a

a  t  t

as desired. Now let tk1. Repeating the switching process described in the proof of Lemma 4, i.e. for any pair (ai,aj) where 1ai aj a and using the fact that

2 1) ( 1)

( 2 2

2 2

    

 _j _i _j

i a a a

a , we get a_i =0 or a_j =a. To achieving an integer a– partition, we need to apply the switching process at least t(k1) times. This implies that

1)). ( 2( 2 2 2 2

1  a ka b  t k

a  _t (1) Thus

. 1) ( 1) (

1)) ( ( 1) ( 1) ( =

(1)) b

( ) 2( 1))

( 2(

) 2( )

( = 1) ( 1)

(

2 2

1 2

2 1

   

     

       

     

  

b a

k

k t b

a k

inequality y

b ka t

k t b ka

b ka t

a a

a

a  t  t

(b) If b=0 ,then = = = .

k

1 a a a ka

a

n  _t __{ }__{ }_ Since (a₁,,a_t) is not a–partition, we have tk. Applying (1), we obtain

. 1) (

1) ( =

2 ) 2(

2 ) (

= 1) ( 1)

(

2 2 2

2 2

1 2

2 1

 

  

    

    

  

a k

t k a

k

ka t k t ka

ka t a a

a

a  _t  _t

(5)

Remark 6. Let T be a tree of order n and maximum degree . For each i{1,2,,}, let n_i denote the number of vertices of degree i. Then

n n n

n₁ ₂ _ = (2) and

2. 2 = 2 ₂

1 n  n n

n  (3) Subtracting (2) from (3), yields

2. = 1) ( 2 ₃

2 n    n n

n  (4)

By (4), we obtain the following integer partition

), 1 , 1, , , ,2 2, , ,1 1, (

3 2

   

  



   

   



  



n n

n

(5)

of n2 on N__₁={1,2,,1}. It follows from Lemma 4 that 22n₂ 32n₃2n_ is maximum if and only if the partition (5) obtained from (4), is an (1)−partition of

2 

n on N__₁. In that case, n₁ (the number of leaves) will be maximum. Next result is an immediate consequence of above discussion.

Corollary 7. For any tree T of order n with maximum degree , the first Zagreb index

   

 n n

n T

M₁( )= ₁ 22 ₂  2 is maximum if and only if the integer partition (5) is an 1)

( –partition of n2 on N__₁. In that case, the integer partition (n₁,n₂,,n_) is called an optimal solution of (4).

Theorem 8. Let T be a tree of order n and maximum degree  with n 0 (mod 1). Then M₁(T)(2)n44, with equality if and only if TT_n.

Proof. Assume that n=(1)k. By (4),

, = ) 1

2 2)

( 2

(

=k n2 n3 n 1 k r

n 

 

 

   

 





where

1

2 1 2) ( 3

2 2 =

 

      

 n n

n

r  . Then 1rk1 and 1n_ k1. We consider three cases as follows:

Case 1.r=1. Then clearly n_ =k1. It follows that

2 1) ( = 1) 1)( ( 2)

(

2 ₃ ₁

2 n    n   k  k

n 

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3. = 2). (

2 ₃ ₁

2 n    n 

n 

Thus n__₁=0 and so

3. = 3). (

2 ₃ ₂

2 n    n 

n  (6) According to Corollary 6, the optimal solution of (6) is n₂ =n₃==n__₃=0 and

1 = 2

 

n . Since n₁n₂n_ =n, we conclude that n₁=(2)k. By Corollary 7, 1)

,0,1,0, ,0,

2) (( = ) , , , , , ,

(n₁ n₂ n__₃ n__₂ n__₁ n_  k  k

is the optimal solution and so M₁(T) is maximum. Therefore,

4. 4 2) ( =

4 4 1) 2)( ( =

1) ( 2 2 2) ( 2) ( =

. 2 1 . 2 1) ( 2 . 2 2) ( 2

2 2 1 ) ( 1

    

      

       

     

    

   



n k

k k

n n

n T

M 

Case 2. 2r. Then n₂2n₃(2)n__₁=(2)r(r2). Since 2

2 

r , it follows from Corollary 7 that

) , ,0, ,0,1,0, 1,0,

2) (( = ) , , , , , , , , ,

(n₁ n₂  n_r_₂ n_r_₁ n_r  n__₂ n__₁ n_  k   r kr is an optimal solution in this case. Since 2r< and 4, we have

4 4 < 1) 2

(r   

r and so

4. 4 2) (

1) 2 ( 1) 2)( ( =

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

     

       

           

n

r r k

r k r

r k

T M

Case 3. rk1. Then n₂2n₃(2).n__₁=(2)r(r2). There are non-negative integers t,s such that (r2)=t(2)s and 0s2. Hence

. ) 2)( ( = 2). (

2 ₃ ₁

2 n n r t s

n    __    If 0s2, then ) , ,0, ,0,1,0, 1),0,

( 2) (( = ) , , , , , , , , ,

(n₁ n₂  n_s n_s_₁ n_s_₂ n__₂ n__₁ n_  k t   rt kr is the optimal solution and since (s)<0 and 4r, we obtain

. 4 4 2) (

2) )( ( 2) (

2) )( ( 2) ( =

2) ( ) 2 (1 2) ( 1) 2)( ( =

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

     

         

        

            

    

        

n

r r s

s n

r r s

s n

t r

s s k

r k t

r s

t k T

M

If s=0, then the optimal solution is

). , ,0, ,0, 2)

(( = ) , , , , ,

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Since t(2)=r2s, (s2)0 and 4r, we conclude that

. 4 4 2) (

2) (

2) ( 2) ( =

2 2 2

2 2

=

) ( 2 ) ( 2 1) ( ) 2) (( =

. 2 3

9 2 4 1 ) ( 1

     

      

                

    

    

      

n

r r n

r r s

n

r k t t t r r r t k k

r k t

r t

k

n n

n n T

M 

Therefore, in all cases M₁(T)(2)n44. If TT_n, then clearly 4

4 2) ( = ) (

1T  n 

M . Conversely, let T be a tree of order n with n 0 (mod 1) and M₁(T)=(2)n44. This occurs only in Case 1, that is, T has

1 1 =

1

 

  

 n

k

vertices of degree , one vertex of degree 2 and (2)k leaves. Hence TT_n and the proof is complete.

Theorem 9. Let T be a tree of order n with maximum degree  and n1(mod 1). Then M₁(T)(2)n3, with equality if and only if TT_n.

Proof. Let n=(1)k1. Set

1

1 2)

( 2

= 2 3 1

 

 

  

 n n__

n

r  . By (4),

. = ) 1

1 2)

( 2

(

=k n2 n3 n 1 k r

n 

 

 

   

 





Then clearly 1rk1 and 1n_ k1. We consider three cases.

Case 1. r=1. Since n_ =k1, it follows from (4) that 2)

( = 2).

( ₁

2   n 

n  and by Corollary 7

1) ,0,1, 1,0, 2)

(( = ) , , , , ,

(n₁ n₂  n__₂ n__₁ n_  k  k

is the optimal solution. Thus

. 3 2) ( =

1) ( 2 (1) 2 1) ( 1) 2) (( =

. 2 1 . 2 1) ( 2 . 2 2) ( 2

2 2 1 ) ( 1

   

   

    

     

    

   



n

k k

n n

n T

M 

Case 2. 2r1. As above, n₂(2).n__₁=(2)r(r1). Since 2

1 

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) , ,0, ,0,1,0, ,0,

2) (( = ) , , , , , , , , ,

(n₁ n₂  n_r_₁ n_r n_r_₁ n__₂ n__₁ n_  k   r kr

is the optimal soloution. Since 2r1, it is easy to see that

0 2) (

) (1

2 r  r2 r  and we have

. 3 2) (

2) 2

( ) (1 2 3 2) ( =

2 2

1) 2)( ( =

) ( 2 2 1) ( (1) 2 2) ( =

. 2 1 . 2 1) ( 2 . 2 2) ( 2

4 1 = ) ( 1

    

         

       

      

 

     

    

   

n

r r r n

r r r k

r k r

r k

n n

n T

M 

Case 3. 1rk1. There are non-negative integers t,s such that s

t

r1= (2) , t1 and s1. By substituting in (4), we have s

t r n

n

n₂2 ₃(2) __₁=(2)(  ) . First let 0 s. Since s2, it follows from Corollary 7 that

) , ,0,0, ,0,1,0, ,0,

2) (( = ) , , , , , , , , ,

(n₁ n₂ n_s n_s_₁ n_s_₂  n__₂ n__₁ n_  kt   rt kr is the optimal solution. Thus

. 3 2) (

1) 1)( ( 2) (

3 2) ( =

2) ( ) 2 (1 2 1) ( 1) 2)( ( =

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

( ) ( 1

    

           

            

    

       

n

r s

s n

t r

s k

r k t

r s

t k T

M

Now let s=0. Then the optimal solution is

) , ,0, 1,0, 2)

(( = ) , , , , ,

(n₁ n₂  n__₂ n__₁ n_  kt  rt kr and we have

. 3 2) (

1) 1)( ( 3 2) ( =

2) ( 1 1) (2 1) 2)( ( =

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

( ) ( 1

    

       

           

    

      

n

r n

t r

k

r k t

r t

k T

M

As in the proof of Theorem 8 we can see that M₁(T)=(2)n3 if and only if

n

TT .

Theorem 10. Let T be a tree of order n with maximum degree  and n p(mod 1)

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  

 

     

     

3, 2)

( 3 2 2) (

2 = 2

2 2) ( ) ( 1

p p

p n

T M

with equality if and only if TT_n.

Proof. Let n=(1)k p. Suppose that

1

) (2 2)

( 2

= 2 3 1

 

  

  

 n n__ p

n

r  . By (4),

we have

. = ) 1

) (2 2)

( 2

(

=k n2 n3 n 1 p k r

n 

 

  

   

 





Then clearly 0rk1 and 1 n_  k. We consider four cases. Case 1. r=0. Then n_ =k and by (4) we have

2. = 1) ( 2) 1)

(( = ) 1). (( 2) ( = 2). (

2 ₃ ₁

2 n    n n   n  k p   k p

n 

If p= 2, then n₂2n₃(2).n__₁=0. This implies that n₂=n₃==n__₁=0 and n₁=nk by (2). Thus

2. 2 2) ( =

2) 1)( ( =

1) 1)( ( =

2 ) ( =

. 2 1 . 2 1) ( 2

2 2 1 ) ( 1

    

   

    

  

     

   



n n n

k n

k k

n

n n

T

M 

Now let 2 p2. Since 1 p24 and 2

= 2). (

2 ₃ ₁

2 n   n p

n  , it follows from Corollary 7 that

) ,0, ,0,1,0, 1,0,

( = ) , , , ,

, , ,

(n₁ n₂  n_p_₂ n_p_₁n_p  n__₁ n_ nk   k

is the optimal solution and so

. 3 2 2)

( =

2 2 )

1)( ( =

2 2 1)

1)( ( =

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

=

. 2 1 . 2 1) ( 2

4 1 ) ( 1

p p p n

p p n p n

p p n k

k p

k n

n n

n n T max M

     

     

      

  

  

     

   

 

Case 2. r =1. Then n_ =k1 and

1) ,0,1, ,0,1,0,

1,0, 2)

(( = ) , , , , , , , , ,

(n₁ n₂  n_p_₁ n_p n_p_₁  n__₂ n__₁ n_  k p   k

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. 3 2 2) ( 2 2 ) 2)( ( = 2 2 1) 2)( ( = 2 2 1 2 2 2 1 2 = 1) ( 2 2 1) ( 2 1 2) ( = . 2 1 . 2 1) ( 2 4 1 = ) ( 1 p p p n p p p n p p k k p p k k k p p k n n n n T M                                                          

Case 3. 2r p. By (4), we have n₂2n₃(2)n__₁ 2)

( 2) (

=  r pr . Since r22, it follows from Corollary 7 that

) , ,0, ,0,1,0, 1,0, 2) (( = ) , , , , , , , , ,

(n₁ n₂  n_p__r_₂ n_p__r_₁ n_p__r  n__₂ n__₁ n_  kp   r kr

is the optimal solution. On the other hand, we deduce from p2 and r< p that

0 1 = ) 2( 1 ) 2(

1        

 p p p p

r and so r(r12(p))0. Thus

. ) ( 1 = 3 2 2) ( )) 2( 1 ( 3 2 2) ( = 1) 2 ( 2 2 ) 2)( ( = 1) 2 ( 2 2 1) 2)( ( = 2 2 2 2 2 2 2 1 2 2 1 2 = ) ( 2 ) ( 2 1) ( (1) 2 1) ( 1) 2) (( = . 2 1 . 2 1) ( 2 4 1 ) ( 1 T max M p p p n p r r p p p n p r r r p p p n p r r r p p k r k r r r r p rp r p p k k r k r r p p k n n n n T M                                                                                   

Case 4.  prk1. Let pr2=t(2)s._{By substituting in (4), we have}

s t r n

n

n₂2 ₃(2) __₁=(2)(  ) . If s=0 then by Corollary 7, ) , ,0, ,0, 2) (( = ) , , , , ,

(n₁ n₂  n__₂ n__₁ n_  k pt  rt kr is the optimal solution. Since  pr and p2, we have

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. 3 2 2) ( ) 2 2 (2 3 2 2) ( = 2 2 2 2) ( = 2 2) ( ) 2)( ( = 2 2) ( ) 2 2 ( = ) ( 2 ) ( 2 1) ( ) 2) (( = . 2 1 . 2 1) ( 2 4 1 = ) ( 1 p p p n r r p p p p p p n r r p r p p p n r r p t p n r r p t k k k r k t r t p k n n n n T M                                                                                  

Now let 0 s. Since s2, it follows from Corollary 7 that

) , ,0,0, ,0,1,0, 1),0, ( 2) (( = ) , , , , , , , , ,

(n₁ n₂ n_s n_s_₁ n_s_₂ n__₂ n__₁ n_  k p t   rt kr is the optimal solution. Since 2 p2 and 0s3, it is straightforward to verify that pp22ps22srr2s0. Thus

. 3 2 2) ( ) 2 2 2 2 2 ( 3 2 2) ( = 2 2 2 2) ( = ) 2 ( 2 2 2 ) 2)( ( = 2) ( 2 2 2 1) 2)( ( = 2 2 2 2 2 ) 2 2 ( = ) ( 2 ) ( 2 1) ( 2 1) ( 1) ( 2) ( = . 2 1 . 2 1) ( 2 4 1 = ) ( 1 p p p n s r r s s p p p p p p n s r r s s p n s r p r r s s p p n t r r s s p k t t r r s s p k k k r k t r s t p k n n n n T M                                                                                                         

Therefore, in all cases M₁(T)2)np p2 3p. As in the proof of Theorem 8, we can see that

                3, 2) ( 3 2 2) ( 2 = 2 2 2) ( = ) ( 1 p p p n p n T M

if and only if TT_n. This completes the proof.

We now present a lower bound on the first Zagreb coindex among all trees. Ashrafi et al. [1] proved that for any conneted graph G of order n and size m

). ( 1) ( 2 = ) ( ₁

1 G m n M G

M  

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Corollary 11. Let T be a tree of order n with maximum degree . If n p (mod 1), then

 

  

 

 

       

      

      



3. 3)

( 2 2

6) (

2. = 4

2 2 6) (

1 = 2

3 2 6) (

0 = 2

4 2 6) (

) (

2 2 2 2

1

p p

n n

p n

n

p n

n

p n

n

T M

R

EFERENCES

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

[2] K. C. Das, Sharp bounds for the sum of the squares of the degrees of a graph, Kragujevac J. Math. 25 (2003), 31–49.

[3] K. C. Das, Maximizing the sum of the squares of the degrees of a graph, Discrete Math. 285 (2004), 57–66.

[4] D. de Caen, An upper bound on the sum of squares in a graph, Discrete Math. 185 (1998), 245–248.

[5] T. Došlić, Vertex-weighted Wiener polynomials for composite graphs, Ars. Math. Contemp. 1 (2008), 66–80.

[6] I. Gutman, N. Trinajstić, Graph theory and molecular orbitals. Total -electron energy of alternant hydrocarbons, Chem. Phys. Lett. 17 (1972), 535–538.

[7] X. L. Li, I. Gutman, Mathematical Aspects of Randić-Type Molecular Structure Descriptors, Mathematical Chemistry Monograph 1, University of Kragujevac, 2006. [8] S. Nikolić, G. Kovačević, A. Miličević, N. Trinajstić, The Zagreb indices 30 years after, Croat. Chem. Acta 76 (2003), 113–124.

[9] Ž. Kovijanić Vukićević, G. Popivoda, Chemical trees with extreme values of Zagreb indices and coindices, Iranian. J. Math. Chem. 5 (2014) 19–29.

[10] S. Zhang, W. Wang, T. C. E. Cheng, Bicyclic graphs with the first three smallest and largest values of the first general Zagreb index, MATCH Commun. Math. Comput. Chem. 56 (2006), 579–592.