Minimal Energy Tree with 4 Branched Vertices

Saira Hameed; Uzma Ahmad

doi:10.1515/chem-2019-0013

Article Open Access

Minimal Energy Tree with 4 Branched Vertices

Saira Hameed and Uzma Ahmad

Published/Copyright: April 3, 2019

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From the journal Open Chemistry Volume 17 Issue 1

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.

Keywords: Energy; Tree; Branch; Characteristic Polynomial; r- matching

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 λ_r, 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)xn−2r,

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)=2π∫0+∞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,n2−1). 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 2

T(2,1, n - 6).

Figure 3

Maximal energy tree in Ω(n,2).

Figure 4

Maximal energy tree in Ω(n,3).

T(2,1,n−6)≺T(2,n−7,2)≺T(3,1,n−7)≺T(2,2,n−7).

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).

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)xn−2r

and

Ψ(T2,x)=∑r=0|n2|(−1)rm(T2,r)xn−2r.

1. If m(T1,r)≤m(T2,r),1≤r≤|n2|then T1≼T2.2. If m(T1,r)<m(T2,r) for some r,T2≺T2.3. If m(T1,r)=m(T2,r), for all r,then T2~T2.

Now, using equation (1),

1. if T1≼T2 then E(T1)≤E(T2)2. if T1≺T2 then E(T1)<E(T2).

Lemma 1.1

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

1. AXn,i≺AXn,3 for all 2≤i≤n−2 and i =3 ;2. AXn,2≺AXn,n−1 for all n ≥2 ,n=3 ;3. Xn,2≺Xn,4≺... ≺Xn,5≺Xn,3≺Xn,1 .

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

Theorem 1.1

[19]Xn,2≺Xn,1for all n ≥ 2.

The class Ω (n,4) contains two subclasses, namely Ωʹ (n,4) consisting of all trees Tʹ 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

T′α1,α2,..,αa;x;β1,β2,…,βb;y;γ1,γ2,…,γc;z;δ1,δ2,…,δd.

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 Tʹ of the form

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

given in Figure 6, where α₁, α₂, ... ,α_a, β₁, β₂, … , β_b, γ₁, γ₂, …, γ_c, δ₁, δ₂, … ,δ_d are non-negative integers and x, y, z are positive integers.

If α₁=α₂ = ⵈ = α_a = β₁ = β₂ = ⵈ = β_b = γ₁ = γ₂ =ⵈ = γ_c = δ₁ = δ₂ = ⵈ = δ_d = x = y = z = 1, then Tʹ is of the form Tʹ (a,b,c,d), given in Figure 7.

Let

A′n,4=T′a,b,c,d:d≥a≥2,b≥1,c≥1,a+b++c+d=n−4⊂Ω′n,4

and

B′n,4=Ω′n,4∖A′n,4.

Then

Ω′(n,4)=A′(n,4)∪B′(n,4)

Lemma 2.1

LetT′=T′α1,α2,..,aa;x;β1,β2,...,βb;y;γ1,γ2,...,γc;z;δ1,δ2,...,δd∈∈ Bʹ (n,4), then there exists a treeT′1=T′(p1,p2,p3,p4)∈A′(n,4)such thatT1′≺T′.

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

Lemma 2.2

Let Tʹ ∈ Aʹ (n,4), then

T2′=T′(p1−1,p2,p3,p4+1)≺T1′=T′(p1,p2,p3,p4).

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

ΨT′1=xn−n−1xn−2+p1p2+p1p3+p1p4+p2p3+p2p4+p3p4+2p1+p2+p3+2p4+1xn−4−p1p2p3+p1p2p4+p1p3p4+p2p3p4+p1p4+p1p2+p3p4xn−6+p1p2p3p4xn−8

and

ΨT′2=xn−n−1xn−2+p1p2+p1p3+p1p4+p2p3+p2p4+p3p4+2p1+p2+p3+2p4+1+p1−p4−1xn−4−p1p2p3+p1p2p4+p1p3p4+p2p3p4+p1p4+p1p2+p3p4+p2p1−p4−2+p3p1−p4+p1−p4−1xn−6+p1p2p3p4+p2p3p1−p4−1xn−8.

We can easily see that p1-p4≤0,p1-p4-1<0andp1-p4-2< 0. Therefore,T2′≺T1′.

Lemma 2.3

LetT3′=T′(2,p2,p3,p4)∈A′(n,4).

1.Ifp4≥p2≥2,thenT′4=T′2,p2−1,p3,p4+1≺T′3.2.Ifp2>p4≥2,thenT′5=T′2,p2+1,p3,p4−1≺T′3.3.Ifp4≥p3≥2,thenT′6=T′2,p2,p3−1,p4+1≺T′3.4.Ifp3>p4≥2,thenT′7=T′2,p2,p3+1,p4−1≺T′3.5.Ifp2>p3≥2,thenT′8=T′2,p2+1,p3−1,p4≺T′3.6.Ifp3≥p2≥2,thenT′9=T′2,p2−1,p3+1,p4≺T′3.

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

Ψ T3′=xn−n−1xn−2+3p2+3p3+4p4+p2p3+p2p4+p3p4+5xn−4−2p2+2p4+2p2p3+2p2p4+3p3p4+p2p3p4xn−6+2p2p3p4xn−8

and

Ψ T4′=xn−n−1xn−2+{3p2+3p3+4p4+p2p3+p2p4+p3p4+5+p2−p4}xn−4−{2p2+2p4+2p2p3+2p2p4+3p3p4+p2p3p4+p3p2−p4+2p2−p4−1}xn−6+{2p2p3p4+2p3p2−p4−1}xn−8.

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

2. The characteristic polynomial of T₅ʹ is

Ψ T5′=xn−n−1xn−2+{3p2+3p3+4p4+p2p3+p2p4+p3p4+5+p4−p2−2xn−4−2p2+2p4+2p2p3+2p2p4+3p3p4+p2p3p4+p3p4−p2−2+2p4−p2−1xn−6+2p2p3p4+2p3p4−p2−1xn−8.

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

3. The characteristic polynomial of T₆ʹ is

Ψ T6′=xn−n−1xn−2+3p2+3p3+4p4+p2p3+p2p4+p3p4+5+(p3−p4xn−4−2p2+2p4+2p2p3+2p2p4+3p3p4+p2p3p4+p2(p3−p4−1)+3p3−p4−1xn−6+2p2p3p4+2p2p3−p4−1xn−8.

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

4. The characteristic polynomial of T₇ʹ is

Ψ T7′=xn−n−1xn−2+3p2+3p3+4p4+p2p3+p2p4+p3p4+5+p4−p3−2xn−4−2p2+2p4+2p2p3+2p2p4+3p2+p4+p2p3p4+p2p4−p3−1+3p4−p3−5xn−6+2p2p3p4+2p2p4−p3−1xn−8.

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

5. The characteristic polynomial of T₈ʹ is

Ψ T8′=xnn−1xn−2+3p2+3p3+4p4+p2p3+p2p4+p3p4+5+p3−p2−1xn−4−2p2+2p4+2p2p3+2p2p4+3p3p4+p2p3p4+p4p3−p2−2+2p3−p2xn−6+2p2p3p4+2p4p3−p2−1xn−8.

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

6. The characteristic polynomial of T₉ʹ is

Ψ T9′=xn−n−1xn−2+3p2+3p3+4p4+p2p3+p2p4+p3p4+5+p2−p3−1xn−4−2p2+2p4+2p2p3+2p2p4+3p3p4+p2p3p4+p4p2−p3+2p2−p3−2xn−6+2p2p3p4+2p4p2−p3−1xn−8.

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

Lemma 2.4

LetT′10=T′(2,p2,p3,2)∈A′(n,4)..

Ifp3≥p2≥2, then T′11=T′(2,p2−1, p3+1,2)≺T′10.
Ifp2≥p3≥2, then T′12=T′(2,p2+1,p3−1,2)≺T′10.

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

Ψ T′10=xn−n−1xn−2+5p2+5p3+p2p3+13xn−4−6p2+6p3+4p2p3+4xn−6+4p2p3xn−8

and

ψ(T′11)=xn−(n−1)xn−2+{5p2+5p3+p2p3+13+(p2−p3−1)}xn−4−{6p2+6p3 +4p2p3+4+4(p2−p3−1)}xn−6+{4p2p3+4(p2−p3−1)}xn−8.

Since p₂ - p₃ - 1 < 0, therefore T′11≺T′10.

The characteristic polynomial of T₁₂ʹ is

Ψ T′12=xn−n−1xn−2+5p2+5p3+p2p3+13+p3−p2−1xn−4−6p2+6p3+4p2p3+4+4p3−p2−1xn−6+4p2p3+4p3−p2−1xn−8.

Since p₃ - p₂ - 1 < 0, therefore T′12≺T′10.

Lemma 2.5

LetT′13=T′(2,1,p3,p4)∈A′(n,4).

Ifp4≥p3≥2, then T′14=T′(2,1,p3−1,p4+1)≺T′13.
^Ifp3≻p4≥2, then T′15=T′(2,1,p3+1,p4−1)≺T′13.

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

Ψ(T′13)=xn−(n−1)xn−2+(4p3+5p4+p3p4+8)xn−4−(2+2p3+4p4+4p3p4)xn−6 +(2p3p4)xn−8

and

Ψ(T′14)=xn−(n−1)xn−2+{4p3+5p4+p3p4+8+(p3−p4)}xn−4−{2+2p3+4p4+ 4p3p4+4(p3−p4)−2}xn−6+{2p3p4+2(p3−p4−1)xn−8.

Since p₃ - p₄ ≤ 0 and p₃ - p₄ - 1 < 0, therefore, T′14≺T′13.

1. The characteristic polynomial of T₁₅ʹ is

Ψ(T′15)=xn−(n−1)xn−2+{4p3+5p4+p3p4+8+(p4−p3−2)}xn−4 −{2+2p3+4p4+4p3p4+4(p4−p3−1)−2}xn−6+{2p3p4 +2(p4−p3−1)xn−8.

Since p₄ - p₃ - 1 < 0 and p₄ - p₃ - 2 < 0, therefore, T′15≺T′13.

Lemma 2.6

LetT′16=T′(2,p2,1,p4)∈A′(n, 4).

If p4≥p2≥2, then T′17=T′(2, p2−1,1, p4+1)≺T′16.
If p2>p4≥2, then T′18=T′(2, p2+1,1, p4−1)≺T′16.

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

Ψ(T′16)=xn−(n−1)xn−2+(4p2+5p4+p2p4+8)xn−4−(4p2+5p4+3p2p4)xn−6 +(2p2p4)xn−8

and

Ψ(T′17)=xn−(n−1)xn−2+{4p2+5p4+p2p4+8+(p2−p4)}xn−4 −{4p2+5p4+3p2p4+3(p2−p4)−2}xn−6+{2p2p4+2(p2−p4−1)}xn−8.

Since p₂ - p₄ - 1 < 0 and p₂ - p₄ ≤ 0, therefore, T17′≺T16′.

2. The characteristic polynomial of T₁₈ʹ is

Ψ(T′18)=xn−(n−1)xn−2+{4p2+5p4+p2p4+8+(p4−p2−2)}xn−4 −{4p2+5p4+3p2p4+3(p4−p2−1)−1}xn−6+{2p2p4 +2(p4−p2−1)xn−8.

Since p₄- p₂ - 1 < 0 and p₄- p₂ - 2 < 0, therefore, T18′≺T16′.

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 Mʹ ≺ Lʹ for n ≥ 10.

Proof. The characteristic polynomials of L_′ and M_′ are

Ψ(L′)=xn−(n−1)xn−2+(6n−36)xn−4+(10n−80)xn−6+(4n−36)xn−8Ψ(M′)=xn−(n−1)xn−2+(6n−36)xn−4−(8n−60)xn−6+(2n−16)xn−8

For n ≥ 10,

(10n - 80)-(8n - 60)=2n - 20≥0

and

(4n - 36) - (2n - 16)=2n - 20≥ 0.

So, M′≺L′.

Theorem 2.1

Let Tʹ ∈ Ωʹ (n,4), Tʹ ≠ Mʹ and n ≥ 10. ThenM′≺T′.

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

T1′=T′p1,p2,p3,p4∈A′n,4⊂Ω′n,4,

such that T′≻T1′.By Lemma 2.2, there also exists a tree

T2′=T′(2,p2,p3,p5)∈A′(n,4)

such that T1′≻T2′.

If p₂ ≤ p₅, then by Lemma 2.3 (1), we can find a tree

T′3=T′2,1,p3,p6∈A′n,4

such that T₂ʹ ≺ T₃.

If p₃≤ p₆, then by Lemma 2.5, T′3≻M′
If p₃> p₆, then by Lemma 2.5, T₃ʹ ≺ Lʹ and by Lemma 2.7 L′≻M′.integers.

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

T′4=T′2,p7,p3,2∈A′n,4

such that T2′≻T4′..Then either p₇ ≥ p₃ or p₃ > p. In both cases, by Lemma 2.4, we can find Lʹ such that T4′≻L′.

Then using Lemma 2.7, Lʹ ≺ Mʹ.

3. If p₃ ≤ p₅, then by Lemma 2.3 (3), we can find a tree

T5′=T′(2,p2,1,p8)∈A′(n,4)

such that T2′≻T5′.

If p₂ ≤ p₈, then by Lemma 2.6, T5′≻M′.
If p₂ > p₈, then by Lemma 2.6, T5′≻L′and by Lemma 2.7 L′≻M′.

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 T2′≻T6Then either p₉ ≥ p₂ or p₂ > p₉. In both cases, by Lemma 2.4, we can find Lʹ such that T6′≻L′.

Then using Lemma 2.7, L′≻M′.

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

T7′=T′(2,1,p10,p4)∈A′(n,4)

such that T2′≻T7′.

If p₁₀ ≤ p₄, then by Lemma 2.5, T7′≻M′
If p10≻p4,then by Lemma 2.5, T7′≻L′and by Lemma 2.7 Lʹ ≺ Mʹ.

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 T2′≻T8′.

If p₁₁ ≤ p₄, then by Lemma 2.6, T8′≻M′.
If p11≻p4,then by Lemma 2.6, T₈ʹ ≺ Lʹ and by Lemma

2.7 L′≻M′.

Hence T′≻M′.

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)T∗α1,α2…,αa;x;β1,β2…βb;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 = … = α_a = β₁ = β₂= ⵈ =δ₁ = δ₂ = ⵈ = δ_d = γ₁ = γ₂= ⵈ = γ_c = 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).

Let

A*(n,4)={T*(a,b,d:c):d≥b≥a≥2, c≥0, a+b+d+ +c=n−4}⊂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 thatT1⋆≺T⋆.

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∗=T∗q1−1,q2,q3+1:q4≺T1∗=T∗q1,q2,q3:q4.

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

ΨT1∗=xn−n−1xn−2+q1q2+q1q3+q1q4+q2q3+q2q4+q3q4+2q1+2q2+2q3xn−4−q1q2q3+q1q2q4+q1q3q4+q2q3q4+q1q2+q1q3+q2q3xn−6+q1q2q3q4xn−8

and

Ψ(T2*)=xn−(n−1)xn−2+{(q1q2+q1q3+q1q4+q2q3+q2q4+q3q4+2q1+2q2+ 2q3)+(q1−q3−1)}xn−4−{(q1q2q3+q1q2q4+q1q3q4+q2q3q4+q1q2+q1q3+ q2q3)+(q1−q3−1)(q2+q4+1)}xn−6+{(q1q2q3q4)+q2q4(q1−q3−1)}xn−8

Since q₁ - q₃ - 1 < 0, therefore T2⋆≺T1⋆.

Lemma 3.3

Let T* ∈ A* (n,4), then

T4∗=T∗2,q2−1,q3+1:q4≺T3∗=T∗2,q2,q3:q4

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

ΨT3∗=xn−n−1xn−2+4q2+4q3+2q4+q2q3+q2q4+q3q4+4xn−4−q2q3q4+3q2q3+2q2q4+2q3q4+2q2+2q3xn−6+2q2q3q4xn−8

and

ΨT4∗=xn−n−1xn−2+4q2+4q3+2q4+q2q3+q2q4+q3q4+4+q2−q3−1xn−4−q2q3q4+3q2q3+2q2q4+2q3q4+2q2+2q3+q4+3q2−q3−1xn−6+2q2q3q4+2q4q2−q3−1xn−8.

Since q2-q3-1≺0,therefore T⋆4≺T⋆3.

Lemma 3.4

Let T*=T* (2,2, q: q₄) ∈A* (n,4).

Ifq3>q4,thenT6⋆=T⋆2,2,q3+1:q4-1≺T5⋆.
If q4≥q3, then T7⋆=T⋆(2,2,q3 -1:q4 +1)≺T5⋆.

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

ΨT5∗=xn−n−1xn−2+6q3+4q4+q3q4+12xn−4−4q3q4+8q3+4q4+4xn−6+4q3q4xn−8

and

Ψ(T6*)=xn−(n−1)xn−2+{(6q3+4q4+q3q4+12)+(q4−q3+1)}xn−4−{(4q3q4+8q3+4q4+4)+4(q4−q3)}xn−6+{(4q3q4)+4(q4−q3−1)}xn−8

Since q4- q3+1≤0, q4 - q3<0 and q4 - q3−1<0,therefore T6⋆≺T5⋆.

2 . The characteristic polynomial of T₇* is

ΨT7∗=xn−n−1xn−2+6q3+4q4+q3q4+12+q3−q4−3xn−4−4q3q4+8q3+4q4+4+4q3−q4−2xn−6+4q3q4+4q3−q4−1xn−8

Since q3- q4 -3<0, q3- q4 −2<0 and q3 - q4 - 1<0,therefore T7⋆≺T5⋆.

Lemma 3.5

LetM*=T*(2,2,n−8:0)andT8*=T*(2,2,2:n−10)belong to A* (n,4) ,thenM⋆≺T8⋆for n > 10. In particular, M*∼T₈* for n=10.

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

ΨT8∗=xn−n−1xn−2+6n−36xn−4−12n−100xn−6+8n−80xn−8

and

ΨM∗=xn−n−1xn−2+6n−36xn−4−8n−80xn−6

respectively. Hence M⋆≺T8⋆.

Theorem 3.1

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

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 T⋆≻T1⋆.

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 T1⋆≻T2⋆≻T3⋆.

Now, if q₆ > q₄, then by Lemma 3.4, T3⋆≻M⋆.

If q₆ ≤ q₄, then by Lemma 3.4, there exists a tree T4*=T*(2,2,2:n−10) ∈ Α*(n,4),, such that T3⋆≻T4⋆and by Lemma 3.5. T4⋆≻M⋆.

Hence M⋆≺T⋆ .

Theorem 3.2

The minimal energy tree inΩ(n,4) is M*=T*(2,2,n−8:0)∈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 M⋆≺M′immediately 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 C₁₁H₂₄	Energy
	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 C₁₀H₂₂	Energy
	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,n−8: 0)=T*(2,2,n−8,0) ∈ 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.

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

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

Articles in the same Issue

https://doi.org/10.1515/chem-2019-0013

Keywords for this article

Energy; Tree; Branch; Characteristic Polynomial; r- matching

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