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BY 4.0 license Open Access Published by De Gruyter Open Access April 3, 2019

Minimal Energy Tree with 4 Branched Vertices

  • Saira Hameed and Uzma Ahmad EMAIL logo
From the journal Open Chemistry

Abstract

The energy of a graph is defined as the sum of absolute values of the eigenvalues of its adjacency matrix. Let Ω(n,4) be the class of all trees of order n and having 4 branched vertices. In this paper, the minimal energy tree in the class Ω(n,4) is found.

1 Introduction

The neighboring atoms in a molecule play a significant role in studying the properties of chemical compounds, e.g. the boiling point of alkanes can be related to the branching pattern of their molecular skeleton. Therefore, based on the number of substituents attached to any carbon atom other than hydrogen in a chemical compound, carbon atoms are characterized as primary, secondary, tertiary or quaternary. The stability of the carbon compound depends much on the primary, secondary, tertiary or quaternary atoms. Moreover, the stability of a chemical compound is inversely related to the energy of the compound, i.e. the molecules with lesser energy are more stable. In a molecular graph of a carbon compound, the tertiary and quaternary atoms are refereed as branched vertices. Thus the analysis of minimal energy trees with certain branched vertices is worth studying. This leads to the motivation behind the study of minimal energy trees with certain branched vertices.

Let G be a simple connected undirected graph of order n having adjacency matrix A(G). Let λr, r=1, 2, 3, … , n be the eigenvalues of the adjacency matrix of G. The energy of G is denoted by E(G) and is defined as

E(G)=r=1n|λr|.

The characteristic polynomial of G, denoted by Ψ(G,x), is the same as the characteristic polynomial of A(G) and is defined as

Ψ(G,x)=r=0narxr.

If a graph G is bipartite, then Ψ(G,x) is of the form

Ψ(G,x)=r=0|n2|(1)rm(G,r)xn2r,

where m(G,r)=1,2,...,|n2|are the numbers of r-matchings of G. Using Coulson´s integral formula [1], the energy of a bipartite graph G of order n is defined as

(1)E(G)=2π0+1x2ln(r=0|n2|m(G,r)x2r)dx.

The energy is a graph-spectral-invariant which is a focus of great attention nowadays. It has a chemical motivation and is concerned with the total π-electron energy of graphs which represent a carbon-atom skeleton of conjugated hydrocarbons [2]. From a mathematical point of view, it was first introduced by Gutman in 1978 [3] and he proved that in the class of all trees on n vertices, the star tree has the smallest energy and the path tree has the largest energy [2].

After that, the tree with the minimal energy having a diameter of at least d for a given positive integer d [5], the tree with minimal energy among the trees with k pendent vertices [6] , the tree with the minimal energy in T(n,p) where T(n,p) is the set of trees of order n with at most p pendent vertices [7], were characterized. Moreover, the trees with a given domination number [11] and the trees with a given parameter [15], were also characterized. In 1999 Zhang and Li [21] studied the class of trees with n vertices and perfect matching and proved that the minimal energy tree is a tree which is obtained by joining a pendent vertex to each vertex of K(1,n21). Let Ω(n,m) be the class of all trees with n vertices and exactly m branched vertices.

In 2008 [8], 2010 [9] and 2012 [14], the authors determined up to the twelfth smallest energy trees of order n which belong to either Ω(n,1) or Ω(n,2). Moreover, the trees with larger energies were also characterized in [12], [16], [17] and [18].

In 2015, MArín, Monsalve and Rada [19] showed that the trees T(2,n - 4), given in Figure 1 and T(2,1, n - 6), given in Figure 2 have minimal energy in the classes Ω(n,2) and Ω(n,3), respectively. They also found the trees with maximal energies in these classes which are given in Figures 3 & 4. Recently, in 2018, Zhu and Yang [20] characterized the first four minimum energy trees in the class of all trees of order n ≥ 27 and three branched vertices, that is,

Figure 1 T(2,n - 4).
Figure 1

T(2,n - 4).

Figure 2 T(2,1, n - 6).
Figure 2

T(2,1, n - 6).

Figure 3 Maximal energy tree in Ω(n,2).
Figure 3

Maximal energy tree in Ω(n,2).

Figure 4 Maximal energy tree in Ω(n,3).
Figure 4

Maximal energy tree in Ω(n,3).

T(2,1,n6)T(2,n7,2)T(3,1,n7)T(2,2,n7).

The motivation is to find the minimal energy tree in the class Ω(n,4). In this paper, it is found that T(2,2, n - 8:0), given in Figure 5, is the minimal energy tree in the class Ω(n,4).

Figure 5 T(2,2, n - 8:0).
Figure 5

T(2,2, n - 8:0).

One of the methods used for the characterization of trees with respect to their energies is the quasi-order method which is defined as follows:

Definition 1.1

Let T1 and T2 be two trees of order n having characteristic polynomials

Ψ(T1,x)=r=0|n2|(1)rm(T1,r)xn2r

and

Ψ(T2,x)=r=0|n2|(1)rm(T2,r)xn2r.
1.Ifm(T1,r)m(T2,r),1r|n2|thenT1T2.2.Ifm(T1,r)<m(T2,r)for some r,T2T2.3.Ifm(T1,r)=m(T2,r),for all r,thenT2~T2.

Now, using equation (1),

1.ifT1T2then E(T1)E(T2)2.ifT1T2then E(T1)<E(T2).

Lemma 1.1

([4], [13]) Let A and X be trees different from P1. Then

1.AXn,iAXn,3forall2in2andi=3;2.AXn,2AXn,n1foralln2,n=3;3.Xn,2Xn,4...Xn,5Xn,3Xn,1.

where Xn,i=Pn(i,x)X, is obtained by identifying the ith vertex of the path Pn to a vertex x of X and AXn,i=A(a,n)Xn,i, is obtained by identifying the ith and nth vertex of the path Pn with a vertex x of X and a vertex a of A, respectively.

Theorem 1.1

[19]Xn,2Xn,1for all n ≥ 2.

The class Ω (n,4) contains two subclasses, namely Ωʹ (n,4) consisting of all trees of the form (2) given in Figure 6 and Ω* (n,4)consisting of all trees T* of the form (3) given in Figure 8. In section 2 and 3, the mathematical computations are carried out to find the minimal energy tree with in the general family of trees with four branched vertices. In section 4, the obtained results are implemented on branched alkanes with 10 and 11 carbon atoms.

Figure 6 T′α1,α2,..,αa;x;β1,β2,…,βb;y;γ1,γ2,…,γc;z;δ1,δ2,…,δd${T}'\left( {{\alpha }_{1}},{{\alpha }_{2}},..,{{\alpha }_{a}};x;{{\beta }_{1}},{{\beta }_{2}},\ldots ,{{\beta }_{b}};y;{{\gamma}_{1}},{{\gamma}_{2}},\ldots, {{\gamma}_{c}};z;{{\delta }_{1}},{{\delta }_{2}},\ldots ,{{\delta }_{d}} \right)$.
Figure 6

Tα1,α2,..,αa;x;β1,β2,,βb;y;γ1,γ2,,γc;z;δ1,δ2,,δd.

Figure 7 T ‘ (a,b,c,d).
Figure 7

T ‘ (a,b,c,d).

Figure 8 T*(α1,α2,…,αa;x; β1,β2,…,βb;y;δ1,δ2,…,δd;z;,γ1,γ2,…,γc)${{T}^{*}}\left( {{\alpha }_{1}},{{\alpha }_{2}},\ldots ,{{\alpha }_{a}};x;\,{{\beta }_{1}},{{\beta }_{2}},\ldots ,{{\beta }_{b}};y;{{\delta }_{1}},{{\delta }_{2}},\ldots ,{{\delta }_{d}};z;,{{\gamma }_{1}},{{\gamma }_{2}},\ldots ,{{\gamma }_{c}} \right)$.
Figure 8

T*(α1,α2,,αa;x;β1,β2,,βb;y;δ1,δ2,,δd;z;,γ1,γ2,,γc).

2 Minimum Energy Tree in Ω′ (n, 4)

In this section, we will find the minimal energy tree in the class Ωʹ (n,4) ⊂ Ω (n,4) consisting of all trees of the form

(2)Tα1,α2,..,αa;x;β1,β2,,βb;y;γ1,γ2,,γc;z;δ1,δ2,,δd

given in Figure 6, where α1, α2, ... ,αa, β1, β2, … , βb, γ1, γ2, …, γc, δ1, δ2, … ,δd are non-negative integers and x, y, z are positive integers.

If α12 = ⵈ = αa = β1 = β2 = ⵈ = βb = γ1 = γ2 =ⵈ = γc = δ1 = δ2 = ⵈ = δd = x = y = z = 1, then is of the form (a,b,c,d), given in Figure 7.

Let

An,4=Ta,b,c,d:da2,b1,c1,a+b++c+d=n4Ωn,4

and

Bn,4=Ωn,4An,4.

Then

Ω(n,4)=A(n,4)B(n,4)

Lemma 2.1

LetT=Tα1,α2,..,aa;x;β1,β2,...,βb;y;γ1,γ2,...,γc;z;δ1,δ2,...,δd∈∈ (n,4), then there exists a treeT1=T(p1,p2,p3,p4)A(n,4)such thatT1T.

Proof. The proof follows by the successive application of Lemma 1.1 and Theorem 1.1.

Lemma 2.2

Let Tʹ (n,4), then

T2=T(p11,p2,p3,p4+1)T1=T(p1,p2,p3,p4).

Proof. First we compute the characteristic polynomials of T1ʹ and T2ʹ as follows,

ΨT1=xnn1xn2+p1p2+p1p3+p1p4+p2p3+p2p4+p3p4+2p1+p2+p3+2p4+1xn4p1p2p3+p1p2p4+p1p3p4+p2p3p4+p1p4+p1p2+p3p4xn6+p1p2p3p4xn8

and

ΨT2=xnn1xn2+p1p2+p1p3+p1p4+p2p3+p2p4+p3p4+2p1+p2+p3+2p4+1+p1p41xn4p1p2p3+p1p2p4+p1p3p4+p2p3p4+p1p4+p1p2+p3p4+p2p1p42+p3p1p4+p1p41xn6+p1p2p3p4+p2p3p1p41xn8.

We can easily see that p1-p40,p1-p4-1<0andp1-p4-2< 0. Therefore,T2T1.

Lemma 2.3

LetT3=T(2,p2,p3,p4)A(n,4).

1.Ifp4p22,thenT4=T2,p21,p3,p4+1T3.2.Ifp2>p42,thenT5=T2,p2+1,p3,p41T3.3.Ifp4p32,thenT6=T2,p2,p31,p4+1T3.4.Ifp3>p42,thenT7=T2,p2,p3+1,p41T3.5.Ifp2>p32,thenT8=T2,p2+1,p31,p4T3.6.Ifp3p22,thenT9=T2,p21,p3+1,p4T3.

Proof. 1. The characteristic polynomials of T3ʹ and T4ʹ are,

ΨT3=xnn1xn2+3p2+3p3+4p4+p2p3+p2p4+p3p4+5xn42p2+2p4+2p2p3+2p2p4+3p3p4+p2p3p4xn6+2p2p3p4xn8

and

ΨT4=xnn1xn2+{3p2+3p3+4p4+p2p3+p2p4+p3p4+5+p2p4}xn4{2p2+2p4+2p2p3+2p2p4+3p3p4+p2p3p4+p3p2p4+2p2p41}xn6+{2p2p3p4+2p3p2p41}xn8.

Since p2 - p4 ≤ 0 and p2 - p4 - 1 < 0, therefore T4'T3'.

2. The characteristic polynomial of T5ʹ is

ΨT5=xnn1xn2+{3p2+3p3+4p4+p2p3+p2p4+p3p4+5+p4p22xn42p2+2p4+2p2p3+2p2p4+3p3p4+p2p3p4+p3p4p22+2p4p21xn6+2p2p3p4+2p3p4p21xn8.

Since p4 - p2 - 1 < 0 and p4 - p2 - 2 < 0, therefore T5'T3'.

3. The characteristic polynomial of T6ʹ is

ΨT6=xnn1xn2+3p2+3p3+4p4+p2p3+p2p4+p3p4+5+(p3p4xn42p2+2p4+2p2p3+2p2p4+3p3p4+p2p3p4+p2(p3p41)+3p3p41xn6+2p2p3p4+2p2p3p41xn8.

Since p3 - p4 ≤ 0 and p3 - p4 - 1 < 0, therefore T6'T3'.

4. The characteristic polynomial of T7ʹ is

ΨT7=xnn1xn2+3p2+3p3+4p4+p2p3+p2p4+p3p4+5+p4p32xn42p2+2p4+2p2p3+2p2p4+3p2+p4+p2p3p4+p2p4p31+3p4p35xn6+2p2p3p4+2p2p4p31xn8.

Since p4 - p3 ≤ 0 and p4 - p3 - 1 < 0, therefore T7'T3'.

5. The characteristic polynomial of T8ʹ is

ΨT8=xnn1xn2+3p2+3p3+4p4+p2p3+p2p4+p3p4+5+p3p21xn42p2+2p4+2p2p3+2p2p4+3p3p4+p2p3p4+p4p3p22+2p3p2xn6+2p2p3p4+2p4p3p21xn8.

Since (p3-p2-2)<0,(p3-p2-1)<0and p3-p20,therefore, T8'T3'.

6. The characteristic polynomial of T9ʹ is

ΨT9=xnn1xn2+3p2+3p3+4p4+p2p3+p2p4+p3p4+5+p2p31xn42p2+2p4+2p2p3+2p2p4+3p3p4+p2p3p4+p4p2p3+2p2p32xn6+2p2p3p4+2p4p2p31xn8.

Since (p2-p3-2)<0,(p2-p3-1)<0and(p2-p3)0therefore, T9'T3'.

Lemma 2.4

LetT10=T(2,p2,p3,2)A(n,4)..

  1. Ifp3p22,thenT11=T(2,p21,p3+1,2)T10.

  2. Ifp2p32,thenT12=T(2,p2+1,p31,2)T10.

Proof. 1. The characteristic polynomials of T10ʹ and T11ʹ are

ΨT10=xnn1xn2+5p2+5p3+p2p3+13xn46p2+6p3+4p2p3+4xn6+4p2p3xn8

and

ψ(T11)=xn(n1)xn2+{5p2+5p3+p2p3+13+(p2p31)}xn4{6p2+6p3+4p2p3+4+4(p2p31)}xn6+{4p2p3+4(p2p31)}xn8.

Since p2 - p3 - 1 < 0, therefore T11T10.

  1. The characteristic polynomial of T12ʹ is

ΨT12=xnn1xn2+5p2+5p3+p2p3+13+p3p21xn46p2+6p3+4p2p3+4+4p3p21xn6+4p2p3+4p3p21xn8.

Since p3 - p2 - 1 < 0, therefore T12T10.

Lemma 2.5

LetT13=T(2,1,p3,p4)A(n,4).

  1. Ifp4p32,thenT14=T(2,1,p31,p4+1)T13.

  2. Ifp3p42,thenT15=T(2,1,p3+1,p41)T13.

Proof. 1. We compute the characteristic polynomials of T13ʹand T14ʹ as follows,

Ψ(T13)=xn(n1)xn2+(4p3+5p4+p3p4+8)xn4(2+2p3+4p4+4p3p4)xn6+(2p3p4)xn8

and

Ψ(T14)=xn(n1)xn2+{4p3+5p4+p3p4+8+(p3p4)}xn4{2+2p3+4p4+4p3p4+4(p3p4)2}xn6+{2p3p4+2(p3p41)xn8.

Since p3 - p4 ≤ 0 and p3 - p4 - 1 < 0, therefore, T14T13.

1. The characteristic polynomial of T15ʹ is

Ψ(T15)=xn(n1)xn2+{4p3+5p4+p3p4+8+(p4p32)}xn4{2+2p3+4p4+4p3p4+4(p4p31)2}xn6+{2p3p4+2(p4p31)xn8.

Since p4 - p3 - 1 < 0 and p4 - p3 - 2 < 0, therefore, T15T13.

Lemma 2.6

LetT16=T(2,p2,1,p4)A(n,4).

  1. Ifp4p22,thenT17=T(2,p21,1,p4+1)T16.

  2. Ifp2>p42,thenT18=T(2,p2+1,1,p41)T16.

Proof. 1. The characteristic polynomials of T16ʹ and T17ʹ are

Ψ(T16)=xn(n1)xn2+(4p2+5p4+p2p4+8)xn4(4p2+5p4+3p2p4)xn6+(2p2p4)xn8

and

Ψ(T17)=xn(n1)xn2+{4p2+5p4+p2p4+8+(p2p4)}xn4{4p2+5p4+3p2p4+3(p2p4)2}xn6+{2p2p4+2(p2p41)}xn8.

Since p2 - p4 - 1 < 0 and p2 - p4 ≤ 0, therefore, T17T16.

2. The characteristic polynomial of T18ʹ is

Ψ(T18)=xn(n1)xn2+{4p2+5p4+p2p4+8+(p4p22)}xn4{4p2+5p4+3p2p4+3(p4p21)1}xn6+{2p2p4+2(p4p21)xn8.

Since p4- p2 - 1 < 0 and p4- p2 - 2 < 0, therefore, T18T16.

Lemma 2.7

Let Mʹ = Tʹ (2,1,1, n - 8) and Lʹ = Tʹ (2,1,n - 9,2) belong to Aʹ (n,4), then MʹLʹ for n ≥ 10.

Proof. The characteristic polynomials of L and M are

Ψ(L)=xn(n1)xn2+(6n36)xn4+(10n80)xn6+(4n36)xn8Ψ(M)=xn(n1)xn2+(6n36)xn4(8n60)xn6+(2n16)xn8

For n ≥ 10,

(10n-80)-(8n-60)=2n-200

and

(4n-36)-(2n-16)=2n-200.

So, ML.

Theorem 2.1

Let Tʹ ∈ Ωʹ (n,4), Tʹ ≠ Mʹ and n ≥ 10. ThenMT.

Proof. Since Tʹ ∈ Ωʹ (n,4) and Tʹ ≠ Mʹ, so by Lemma 2.1, there exists a tree

T1=Tp1,p2,p3,p4An,4Ωn,4,

such that TT1.By Lemma 2.2, there also exists a tree

T2=T(2,p2,p3,p5)A(n,4)

such that T1T2.

  1. If p2p5, then by Lemma 2.3 (1), we can find a tree

T3=T2,1,p3,p6An,4

such that T2ʹT3.

  1. If p3≤ p6, then by Lemma 2.5, T3M

  2. If p3> p6, then by Lemma 2.5, T3ʹ and by Lemma 2.7 LM.integers.

2. If p2 > p5, then by Lemma 2.3 (2), we can find a tree

T4=T2,p7,p3,2An,4

such that T2T4..Then either p7p3 or p3 > p. In both cases, by Lemma 2.4, we can find such that T4L.

Then using Lemma 2.7, .

3. If p3p5, then by Lemma 2.3 (3), we can find a tree

T5=T(2,p2,1,p8)A(n,4)

such that T2T5.

  1. If p2p8, then by Lemma 2.6, T5M.

  2. If p2 > p8, then by Lemma 2.6, T5Land by Lemma 2.7 LM.

4. If p3 > p5, then by Lemma 2.3 (4), we can find a tree

T6=T(2,p2,p9,2)A(n,4)

such that T2T6Then either p9p2 or p2 > p9. In both cases, by Lemma 2.4, we can find such that T6L.

Then using Lemma 2.7, LM.

5. If p2p3. Then by Lemma 2.3 (6), we can find a tree

T7=T(2,1,p10,p4)A(n,4)

such that T2T7.

  1. If p10p4, then by Lemma 2.5, T7M

  2. If p10p4,then by Lemma 2.5, T7Land by Lemma 2.7 .

6. If p2 > p3 then by Lemma 2.3 (5), we can find a tree

T8=T(2,p11,1,p4)A(n,4)

such that T2T8.

  1. If p11p4, then by Lemma 2.6, T8M.

  2. If p11p4,then by Lemma 2.6, T8ʹ and by Lemma

2.7 LM.

Hence TM.

3 Minimum Energy Tree in Ω* (n,4)

In this section, we will find the minimal energy tree in the class Ω* (n,4) consisting of all trees T* of the form

(3)Tα1,α2,αa;x;β1,β2βb;y;δ1,δ2,δd;z;,γ1,γ2,γc,

given in Figure 8, where α1,α2,,αa,β1,β2,,βb,δ1,δ2,,δd,γ1,γ2,γcare non-negative integers and x, y, z are positive

If α1 = α2 = … = αa = β1 = β2= ⵈ =δ1 = δ2 = ⵈ = δd = γ1 = γ2= ⵈ = γc = x =y = z = 1, then T* is of the form T* (a,b,d:c), given in Figure 9.

Figure 9 T* (a,b,d:c).
Figure 9

T* (a,b,d:c).

Let

A*(n,4)={T*(a,b,d:c):dba2,c0,a+b+d++c=n4}Ω*(n,4)

and

B*(n,4)=Ω*(n,4)\A*(n,4)

Then

Ω*(n,4)=A(n,4)B*(n,4)

Lemma 3.1

LetT*=T*(α1,α2,,αa;x;β1,β2,βb;y;δ1,δ2,δd;z;γ1,γ2,,γc)B*(n,4)then there exists a treeT1*=T*(q1,q2,q3:q4)A*(n,4)such thatT1T.

Proof. The proof follows by the successive application

of Lemma 1.1 and Theorem 1.1.

Lemma 3.2

. Let T*A* (n,4), then

T2=Tq11,q2,q3+1:q4T1=Tq1,q2,q3:q4.

Proof. First we compute the characteristic polynomials of T1* and T2* as follows,

ΨT1=xnn1xn2+q1q2+q1q3+q1q4+q2q3+q2q4+q3q4+2q1+2q2+2q3xn4q1q2q3+q1q2q4+q1q3q4+q2q3q4+q1q2+q1q3+q2q3xn6+q1q2q3q4xn8

and

Ψ(T2*)=xn(n1)xn2+{(q1q2+q1q3+q1q4+q2q3+q2q4+q3q4+2q1+2q2+2q3)+(q1q31)}xn4{(q1q2q3+q1q2q4+q1q3q4+q2q3q4+q1q2+q1q3+q2q3)+(q1q31)(q2+q4+1)}xn6+{(q1q2q3q4)+q2q4(q1q31)}xn8

Since q1 - q3 - 1 < 0, therefore T2T1.

Lemma 3.3

Let T*A* (n,4), then

T4=T2,q21,q3+1:q4T3=T2,q2,q3:q4

Proof. First we compute the characteristic polynomials of T3* and T4* as follows,

ΨT3=xnn1xn2+4q2+4q3+2q4+q2q3+q2q4+q3q4+4xn4q2q3q4+3q2q3+2q2q4+2q3q4+2q2+2q3xn6+2q2q3q4xn8

and

ΨT4=xnn1xn2+4q2+4q3+2q4+q2q3+q2q4+q3q4+4+q2q31xn4q2q3q4+3q2q3+2q2q4+2q3q4+2q2+2q3+q4+3q2q31xn6+2q2q3q4+2q4q2q31xn8.

Since q2-q3-10,therefore T4T3.

Lemma 3.4

Let T*=T* (2,2, q: q4) ∈A* (n,4).

  1. Ifq3>q4,thenT6=T2,2,q3+1:q4-1T5.

  2. Ifq4q3,thenT7=T(2,2,q3-1:q4+1)T5.

Proof. 1. First we compute the characteristic polynomials of T5* and T6* as follows,

ΨT5=xnn1xn2+6q3+4q4+q3q4+12xn44q3q4+8q3+4q4+4xn6+4q3q4xn8

and

Ψ(T6*)=xn(n1)xn2+{(6q3+4q4+q3q4+12)+(q4q3+1)}xn4{(4q3q4+8q3+4q4+4)+4(q4q3)}xn6+{(4q3q4)+4(q4q31)}xn8

Since q4-q3+10,q4-q3<0andq4-q31<0,therefore T6T5.

2 . The characteristic polynomial of T7* is

ΨT7=xnn1xn2+6q3+4q4+q3q4+12+q3q43xn44q3q4+8q3+4q4+4+4q3q42xn6+4q3q4+4q3q41xn8

Since q3-q4-3<0,q3-q42<0andq3-q4-1<0,therefore T7T5.

Lemma 3.5

LetM*=T*(2,2,n8:0)andT8*=T*(2,2,2:n10)belong to A* (n,4) ,thenMT8for n > 10. In particular, M*∼T8* for n=10.

Proof. The characteristic polynomials of T8* and M* are

ΨT8=xnn1xn2+6n36xn412n100xn6+8n80xn8

and

ΨM=xnn1xn2+6n36xn48n80xn6

respectively. Hence MT8.

Theorem 3.1

Let T* ∈ Ω* (n,4), T*≠ M* and n ≥ 10. ThenMT.

Proof. Since T* ∈ Ω* (n,4) and T*≠ M*, so by Lemma 3.1, there exists a tree

T1*=T*(q1,q2,q3:q4)Α*(n,4)Ω*(n,4)

such that TT1.

By Lemmas 3.2 and 3.3, there exist trees

T2*=T*(2,q2,q5:q4)Α*(n,4)

and

T3*=T*(2,2,q6:q4)Α*(n,4),

respectively, such that T1T2T3.

Now, if q6 > q4, then by Lemma 3.4, T3M.

If q6q4, then by Lemma 3.4, there exists a tree T4*=T*(2,2,2:n10)Α*(n,4),, such that T3T4and by Lemma 3.5. T4M.

Hence MT.

Theorem 3.2

The minimal energy tree inΩ(n,4)isM*=T*(2,2,n8:0)Ω*(n,4).

Proof. The minimal energy tree in Ω (n,4) is one of the minimal energy trees of its subclasses Ωʹ (n,4) and Ω* (n,4).

Since Mʹ and M * are the minimal energy trees in Ωʹ (n,4) and Ω* (n,4), respectively, therefore MMimmediately follows from the comparison of the coefficients of Ψ(M) and Ψ(M*).

4 Application in Structural Chemistry

Alkanes are the simplest hydrocarbons but are very useful in the chemical industry. These compounds not only occur naturally in different products like natural gas and petroleum but are also used in the formation of different chemical products like lubricating oil, chloromethane, dichloromethane, trichloroethane (chloroform) and tetrachloromethane. Alkanes basically are of two kinds i.e., straight chain alkanes and branched alkanes. The stability of branched alkanes is better than that of straight chain alkanes. In other words, branched alkanes have less orbital energy as compared to straight ones.

But the question arises, “Among branched alkanes, which one is more stable?”

The proposed result partially answers this question. The branched alkanes contain tertiary or quaternary carbons. Thus, the molecular graphs of these alkanes are trees having branched vertices. According to the proposed

result, the 3 − (2 − methyl ethyl) − 2,2,4 − trimethylpentane and 3−(2−methylethyl) −2,4−diimethylpentane has minimal energy among all isomers of branched alkanes with 10 and 11 carbon atoms, respectively having exactly four branched carbon atoms and hence are more stable. To verify the result, we take all isomers of branched alkanes with 10 and 11 carbon atoms having exactly four branched carbon atoms. Then their energies are computed. The tables 1 & 2 give all possible such branched alkanes with n = 10,11 and their corresponding orbital energies. From Table 1, it can be observed that 3−(2−methylethyl) −2,2,4− trimethylpentane has minimal energy among all branched

Table 1

The values of energies of different isomers of C11H24.

Isomers of C11H24Energy
10.6980
11.4548
11.5756
11.5994
11.8492
11.9086
12.5266
Table 2

The values of energies of different isomers of C10H24.

Isomers of C10H22Energy
10.1290
10.8572

alkanes with 10 carbon atoms and having four branched carbon atoms. Similarly from Table 2, the alkane 3−(2− methylethyl)−2,4−diimethylpentane has minimal energy among all branched alkanes with 11 carbon atoms and having four branched carbon atoms. In both cases, the minimal energy branched alkanes exactly match with the proposed result as given in section 3.

Conclusion 4.1. The minimum energy tree in the class of all trees of order n ≥ 10 with 4 branched vertices isT(2,2,n8:0)=T*(2,2,n8,0)Ω(n,4).. As an application in structural chemistry, energies of 2 classes of alkanes with 10 and 11 carbon atoms respectively and having exactly 4 branched carbon atoms, are computed and it is observed that 3 − (2 − methyl ethyl)−2,2,4−trimethylpentane has minimal energy among all branched alkanes with 10 carbon atoms and having four branched carbon atoms and 3−(2−methylethyl)− 2,4 − diimethylpentane has minimal energy among all branched alkanes with 11 carbon atoms and having four branched carbon atoms. In both cases, the minimal energy branched alkanes exactly match with the proposed result.

  1. Ethical approval: The conducted research is not related to either human or animal use.

  2. Conflict of interest: Authors declare no conflict of interest.

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Received: 2018-07-19
Accepted: 2018-11-27
Published Online: 2019-04-03

© 2019 Saria Hameed, Uzma Ahmad, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 Public License.

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