The Sanskruti index of trees and unicyclic graphs

Fei Deng; Huiqin Jiang; Jia-Bao Liu; Darja Rupnik Poklukar; Zehui Shao; Pu Wu; Janez Žerovnik

doi:10.1515/chem-2019-0046

Artikel Open Access

The Sanskruti index of trees and unicyclic graphs

Fei Deng , Huiqin Jiang , Jia-Bao Liu , Darja Rupnik Poklukar , Zehui Shao , Pu Wu und Janez Žerovnik

Veröffentlicht/Copyright: 24. August 2019

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Abstract

The Sanskruti index of a graph G is defined as S(G)=∑uv∈E(G)sG(u)sG(v)sG(u)+sG(v)−23,where s_G(u) is the sum of the degrees of the neighbors of a vertex u in G. Let P_n, C_n, S_n and S_n + e be the path, cycle, star and star plus an edge of n vertices, respectively. The Sanskruti index of a molecular graph of a compounds can model the bioactivity of compounds, has a strong correlation with entropy of octane isomers and its prediction power is higher than many existing topological descriptors.

In this paper, we investigate the extremal trees and unicyclic graphs with respect to Sanskruti index. More precisely, we show that

(1) 51227n−1726883375≤S(T)≤(n−1)78(n−2)3for an n-vertex tree T with n ≤ 3, with equalities if and only if T ≌P_n (left) and T ≌ S_n (right);

(2) 51227n≤S(G)≤(n−3)(n+1)38+3(n+1)68n3for an n-vertex unicyclic graph with n ≥ 4, with equalities if and only if G ≌C_n (left) and G ≌ S_n + e (right).

Keywords: Topological index; Molecular descriptor; Sanskruti index; Tree; Unicyclic graph

JEL Classification: 05C90; 05C05; 92E10

1 Introduction

In theoretical chemistry, topological indices (or molecular structure descriptors) are utilized as a standard tool to study structure-property relations, especially in quantitative structure-property relationship (QSPR) and quantitative structure-activity relationship (QSAR) applications [9, 20]. These topological indices are studied on chemical graphs, whose vertices correspond to the atoms of molecules and edges correspond to chemical bonds [15, 17, 18, 19]. In past decades, many topological indices have found important relations between the graph structures and physico-chemical properties [2]. Because of their significant applications, they have been widely studied and applied in many contexts, for example with nanostructures [5, 10], nanomaterials [14], molecular sciences [13], chemistry networks [11], molecular design [1], drug structure analysis [7], fractal graphs [12], and mathematical chemistry [3]. The literature is exhaustive; for example, one of the indices, the Wiener index, along with its applications, is considered in thousands of papers: as of this writing, the seminal paper of Harold Wiener [20] is cited 3535 times according to Google Scholar.

Application of topological indices in biology and chemistry began in 1947 with the work of Harold Wiener [20], who introduced the Wiener index to show correlations between physico-chemical properties of organic compounds and the index of their molecular graphs. This index reveals the correlations of physico-chemical properties of alkanes, alcohols, amines and their analogous compounds [13]. Estrada et al. [4] proposed what is now a well-known atom-bond connectivity (ABC) index, which provides a good model for the stability of linear and branched alkanes as well as the strain energy of cycloalkanes. Inspired by applications of the ABC index, Furtula et al. [6] introduced the augmented Zagreb index, whose prediction power was found to be better than that of the ABC index in the study of heat of formation for heptanes and octanes. More recently, Hosamani [10] proposed the Sanskruti index of a molecular graph and showed that it can model the bioactivity of chemical compounds and showed a correlation with entropy of octane isomers that is comparable to or better than some other well-used descriptors. More precisely, according to [10], the model entropy = 1.7857S±81.4286 models the data from dataset found at http://www.moleculardescriptors.eu/dataset.htm with correlation coefficient 0.829 and with standard error 17.837. Soon after, the Sanskruti indices of some graph families of interest in chemical graph theory were established [8, 16].

Motivated by the new proposed Sanskruti index, we investigate the extremal trees and extremal unicyclic graphs with respect to this topological index. Here, we consider only simple graphs, i.e., undirected graphs without loops and multiple edges. Let G be a graph. We denote by V(G) and E(G) the vertex set and edge set of G, respectively. As usual, P_n, C_n, S_n and S_n + e stand for the path, cycle, star and star plus an edge of n vertices, respectively (see Figure 1). We denote by d_G(v) the degree of a vertex v of a graph G and by N_G(v) (or simply N(v)) the set of neighbors of v. For two vertices u, v ∈ V(G), the distance between u and v is the length of a shortest path between u and v. We denote by N₂(v) the set of vertices of distance two from v and by s_G(u) the sum of the degrees of the neighbors of u, i.e., sG(u)=∑v∈NG(u)dG(v).

Figure 1

The graphs P_n, C_n, S_n and S_n + e

Trees are connected graphs without cycles. A vertex in a tree is called a leaf if it has degree one, and a vertex is called a support vertex if it has a leaf neighbor.

In Section 2, we give some definitions and some preliminary observations. The main results are proved in Section 3: first, we give lower and upper bounds for the Sanskruti index on trees and provide the extremal graphs (Theorems 9 and 10), then we give lower and upper bounds for unicyclic graphs and provide the extremal graphs (Theorems 14 and 15).

2 Preliminaries

The following functions and definitions will be used throughout the paper:

(1)f(x,y)=xyx+y−23.

For a graph G and an edge uv ∈ E(G), we define

(2)h(uv|G)=f(sG(u),sG(v))=sG(u)sG(v)sG(u)+sG(v)−23,

and the Sanskruti index of a graph G is defined as

(3)(G)=∑uv∈E(G)h(uv|G).

Based on the above definitions, the following results are immediate, and the proofs are omitted.

Proposition 1Let n ≥ 3, S(Sn)=(n−1)78(n−2)3and

S(Pn)=16,n=3,175364,n=4,51227n−1726883375,n≥5.

Proposition 2Let n ≥ 4, S(Sn+e)=(n−3)(n+1)38+3(n+1)68n3andS(Cn)=51227n.

Lemma 3Let t ≥ 3, x, y > 0 and x + y = t, then f (x, y) ≤ f(x,y)≤t664(t−2)3.Moreover, f(x,y)=t664(t−2)3if and only ifx=y=t2.

Lemma 4Let t ≠2 andg(t)=f(t2,t2)=t664(t−2)3,theng′(t)=3(t−4)t564(t−2)4.

Lemma 5For x ≥ 3 and y ≥ 3, f is an increasing function (as a function of one variable, either x or y). In particular, ∂f(x,y)∂x=3x2y3(y−2)(x+y−2)4≥0and∂f(x,y)∂y=3y2x3(x−2)(x+y−2)4≥0.

The following properties that hold on trees will be useful later.

Lemma 6Let T be an n-vertex tree. Then for any edge uv ∈ E(T), we have

N(u) ∩ N(v) = ∅ and N₂(u) ∩ N₂(v) = ∅.
Σ_v∈N_(u)|N(v)∖ {u}| = |N₂(u)|.
s_T(u) = |N₂(u)| + |N(u)| = |N₂(u)| + d_T(u) for any u ∈ V(T).

Proof. (a) Since T contains no C₃,we have N(u)∩N(v) = ∅. Suppose to the contrary that N₂(u)∩N₂(v) =∅ and let w ∈ N₂(u) ∩ N₂(v). Denote with usw and vtw the shortest u−w path and v−w path. Then s ≠v. Otherwise w ∈ N(v), a contradiction. Analogously, t ≠u. Now if s ≠t,we obtain that uvtws is a cycle of length five in T, a contradiction. If s = t, it follows that uvs is a cycle of length three in T, a contradiction.

(b) From the result of part (a), we have (N(v₁)∖ {u}) ∩ (N(v₂)∖ {u}) = ∅ for any v₁, v₂ ∈ N(u). Therefore, Σ _v_∈N(u)|N(v)∖ {u}| = |N₂(u)|.

(c) It can be seen that s_T(u) = Σ_v_∈N(u)d_T(v) = Σ _v_∈N(u) |N(v)| = Σ_v_∈N(u)(|N(v)∖ {u}| + 1) = Σ _v_∈N(u) |N(v)∖ {u}| + d_T(u). From the result of part (b), we have Σ_v_∈N(u)|N(v)∖ {u}| = |N₂(u)|, and thus s_T(u) = |N₂(u)| + d_T(u) for any u ∈ V(T).

It is obvious that the last property also holds for general graphs without triangles and C₄. We write it as a lemma for later reference.

Lemma 7Let G be a graph with no triangles and no C₄. Then s_G(u) = |N₂(u)| + |N(u)| = |N₂(u)| + d_G(u) for any u ∈ V(G).

In a special case used explicitly in a proof later, we have

Lemma 8Let G be an n-vertex unicyclic graph. If G contains a C₄, then for any u ∈ V(G)

sG(u)=N2u+Nu+1,u∈V(C4),N2u+Nu,u∉V(C4).

Proof. If u ∈ V(C₄), we can write C₄ = uv₁v₂v₃ and

sGu=∑v∈NuNv=∑v∈N(u)|N(v)∖{u}|+dG(u)=∑v∈N(u)∖{v1,v3}|N(v)∖{u}|+|N(v1)∖{u}|+|N(v3)∖{u}|+|N(u)|=|N2(u)|+1+|N(u)|,

because (N(v₁)∖ {u}) ∩ (N(v₃)∖ {u}) = {v₂}.

For u ∉V(C₄), the assertion is obvious.

3 Main results

3.1 Extremal trees with respect to Sanskruti index

Theorem 9Let T be an n-vertex tree with n ≥ 3. Then, we have

S(T)≤(n−1)78(n−2)3,

with equality if and only if T ≌S_n.

Proof. By Lemma 6 (c), we have s_T(u) + s_T(v) = |N₂(u)| + |N₂(v)| + |N(u)| + |N(v)| for any edge uv ∈ E(T). By Lemma 6 (a),we have N(u)∩N(v) = ∅ and N₂(u)∩N₂(v) = ∅. Therefore, s_T(u) + s_T(v) = |N₂(u) ∪ N₂(v)| + |N(u) ∪ N(v)|. Since u, v ∉N₂(u) ∪ N₂(v), we have |N₂(u) ∪ N₂(v)| ≤ n − 2. It is clear that |N(u) ∪ N(v)| ≤ n, then we have

(4)sT(u)+sT(v)=|N2(u)∪N2(v)|+|N(u)∪N(v)|≤2n−2.

Moreover,

(5)sT(u)+sT(v)=2n−2⟺|N2(u)∪N2(v)|=n−2,|N(u)∪N(v)|=n.

Recall that f(x,y)=(xyx+y−2)3.From Lemmas 3 and 4 it follows that g(t) is an increasing function on the variable t if t ≥ 4. Note that for any uv ∈ E(T) with at least three vertices, we have s_T(u) + s_T(v) ≥ 4, then by applying Lemma 3 with t = 2n − 2, we have

(T)=∑uv∈E(T)sT(u)sT(v)sT(u)+sT(v)−23=∑uv∈E(T)f(sT(u),sT(v))≤∑uv∈E(T)f(x,y)|x=y=n−1=|E(T)|⋅f(x,y)|x=y=n−1=(n−1)78(n−2)3.

Conversely, if S(T)=(n−1)78(n−2)3,then formula (5) holds for any uv ∈ E(T). It is easy to see that this implies that T must be a star.

Theorem 10Let T be an n-vertex tree with n ≥ 3, then we have

51227n−1726883375≤S(T),

with equality if and only if T ≌P_n.

Proof. First observe that the lower bound holds for trees with 3 ≤ n ≤ 6 vertices (for example, by explicitly computing the values for all cases). Therefore, we only need to consider the case n ≥ 7. Let T_n be the family of trees on n vertices and let F_n = {T|S(T) ≤ S(T^′′ )foranyT^′′ ∈ T_n , TP_n}. Now we need to prove that F _n = ∅ for any n ≥ 7.

Suppose to the contrary that there exists an n such that F _n =∅, and let n be the minimal number with this property. Let T^′ ∈ F_n and P = x₁x₂ · · · x_p be a longest path in T^′. Then we claim that

Claim 11. d(x₂) = 2.

Proof. If d(x₂) ≥ 3 write N(x₂) = {x₁, x₃, y₁, y₂, · · · , y_q}, where q ≥ 1. Now construct a new tree T^′′ with V(T^′′ ) = V(T^′) and E(T^′′ ) = E(T^′)∪{x₁y_i|i = 1, 2, · · · , q}∖ {x₂y_i|i = 1, 2, · · · , q} (see Figure 2).

Figure 2

The trees T^′ and T^′′

Then we have s_T^′ (x₃) > s_T^′′ (x₃) and s_T^′ (v) ≥ s_T^′′ (v) for any v ∈ V(T^′)∖ {x₃}. By Lemma 5, and by comparing the contributions of pairs of corresponding edges, we have S(T^′) > S(T^′′ ). Tree T^′′ thus has a smaller value of Sanskruti index than T^′, which contradicts the fact that T^′ ∈ F_n. So, d(x₂) = 2.

Claim 12. d(x₃) = 2.

Proof. Assume to the contrary that N(x₃) = {x₂, x₄, y₁, y₂, · · · , y_q}, where q ≥ 1. We should consider two cases:

• Case 1: y₁is a leaf neighbor of x₃.

Now let V(T^′′ ) = V(T^′) and E(T^′′ ) = E(T^′) ∪ {x₁y₁}∖ {x₃y₁}. Then we have

and

(7)h(uv|T″)≤h(uv|T′)foranyuv∉{x1x2,x1y1,x3y1,x2x3}.

Note that dT′(x3)≥3if and only if 4dT′(x3)dT′(x3)+2≥3(1+dT′(x3))dT′(x3)+2, thus we have

(8)sT′(x3)≥4⇒dT′(x3)sT′(x3)dT′(x3)+sT′(x3)−23≥3(1+dT′(x3))dT′(x3)+23⇔h(x3y1|T′)+h(x1x2|T′)≥h(x1y1|T″)+h(x1x2|T″).

From Eqs. (6), (7) and (8) we have that S(T^′) > S(T^′′ ), which is a contradiction. Thus, we have already proved that d(x₃) = 2 in case y₁ is a leaf neighbor of x₃.

• Case 2: x₃ has a pendent P₂ = y₁z₁.

Now let V(T^′′ ) = V(T^′) and E(T^′′ ) = E(T^′) ∪ {x₁y₁}∖ {x₃y₁}. Then we have

sT″(x3)=sT′(x3)−2,dT″(x3)=dT′(x3)−1,sT″(x2)=sT′(x2),sT″(y1)=3,

and h(uv|T^′′ ) ≤ h(uv|T^′) for any uv ∉ {x₁x₂, x₂x₃, x₃y₁, y₁z₁, x₁y₁}. Denote s3=sT′(x3),then s₃ ≥ 5 and

(9)h1=h(x1x2|T′)+h(x2x3|T′)+h(x3y1|T′)+h(y1z1|T′)=8+2(1+dT′(x3))s3dT′(x3)+s3−13+8,

(10)h2=h(x1x2|T″)+h(x2x3|T″)+h(x1y1|T″)+h(y1z1|T″)=4(1+dT′(x3))3+dT′(x3)3+(1+dT′(x3))(s3−2)dT′(x3)+s3−33+1253+8.

Define F(x)=f(x,6)−f(x,4)=8x3(x−2)(19x2+104x+148)(x+2)3(x+4)3.It is obvious that F(4)=21727and that F(x) is an increasing function for x ≥ 4.

If s₃ ≥ 6, since f(1+dT′(x3),s3)−f(1+dT′(x3),s3−2)≥0, we have

S(T′)−S(T″)=h1−h2≥8+f(1+dT′(x3),6)−f(1+dT′(x3),4)+f(1+dT′(x3),s3)−f(1+dT′(x3),s3−2)−1253=8−1253+F(1+dT′(x3))+f(1+dT′(x3),s3)−f(1+dT′(x3),s3−2)≥8−1253+F(4)=74693375>0,

a contradiction. It follows s₃ < 6. Since s₃ ≥ 5, it is obvious that s₃ = 5. But in this case dT′(x₄) = 1. Thus T is isomorphic to a tree with six vertices, which contradicts with n ≥ 7. So, we have proved that d(x₃) = 2 in case x₃ has a pendent P₂ = y₁z₁. Together with Case 1 this means that d(x₃) = 2.

Claim 13. d(x₄) = 2.

Proof. Suppose to the contrary that d(x₄) ≥ 3. Let T^′′ = T^′−{x₁}. By the minimum assumption of n and T^′ ∈ F _n,we have S(T^′′ )≥ S(P_n₋₁). Denote s1=sT′(x3)ands2=sT′(x4)then we have s₁ ≥ 5 and s₂ ≥ 4. It can be verified that

(11)3s1s1+13+s1s2s1+s2−23+s2(s1−1)s1+s2−33>833.

On the other hand, by Proposition 1, we have

(12)S(Pn)−S(Pn−1)=833.

Therefore, we have

S(T′)≥S(T″)+8+3s1s1+13+s1s2s1+s2−23−8−s2(s1−1)s1+s2−33>S(Pn−1)+3s1s1+13+s1s2s1+s2−23−s2(s1−1)s1+s2−33>S(Pn),

a contradiction. Thus, d(x₄) = 2.

Now we have d(x4)=2for any i = 2, 3, 4, and let T^′′′ = T^′ − {x₁}. By the minimality assumption on n and T^′ ∈ F_n, we have S(T^′′′ ) > S(P_n₋₁). Moreover, it can be seen that S(T‴)=S(T′)+(125)3.Together with Eq. (12), we have S(T^′) > S(P_n), a contradiction with minimality of n. Hence F _n = ∅ for any n ≥ 7, which completes the proof of Theorem 10.

3.2 Extremal unicyclic graphs with respect to Sanskruti index

Theorem 14Let G be an n-vertex unicyclic graph with n ≥ 4, then we have

S(G)≤(n−3)(n+1)38+3(n+1)68n3,

with equality if and only if G ≌S_n + e.

Proof. By Proposition 2, in case G ≌S_n + e we have S(G)=(n−3)(n+1)38+3(n+1)68n3.Now we will show that S(G) ≤ S(G)≤(n−3)(n+1)38+3(n+1)68n3and if S(G)=(n−3)(n+1)38+3(n+1)68n3,then G ≌ S_n + e.

Suppose to the contrary that Gis a graph with maximum Sanskruti index, but GS_n + e,We consider the following four cases.

• Case 1: Gcontains a C₃.

Let C₃ = v₁v₂v₃. Then for any u ∈ V(G)∖ {v₁, v₂, v₃} we have

(13)sG(u)=|N2(u)|+|N(u)|≤n−1,

and for any i ∈ {1, 2, 3}

(14)sG(vi)=|N2(vi)|+|N(vi)|+2.

Furthermore, because u₁ and u₂ are not on a cycle, we have for any u₁u₂ ∉{v₁v₂, v₁v₃, v₂v₃}

(15)sG(u1)+sG(u2)≤2n.

By Eq. (13), it is impossible that s_G(u₁) = s_G(u₂) = n. Therefore,

h(u1u2|G)≤f(n+1,n−1).

Further, we have

(16)sG(v1)+sG(v2)≤2n+2

and h(v₁v₂|G) ≤ f (n + 1, n + 1). It follows

S(G)=∑uv∈E(G)h(uv|G)≤3f(n+1,n+1)+(n−3)f(n+1,n−1)=S(Sn+e).

If the equality S(G) = S(S_n + e) holds, then the equalities in (15) and (16) hold and s_G(v₁) = s_G(v₂) = n + 1 and h(u₁u₂|G) = f (n − 1, n + 1) for any u₁u₂ ∉ {v₁v₂, v₁v₃, v₂v₃}. From these results it follows that G is a graph obtained by adding some pendent vertices to a C₃ = v₁v₂v₃ and, in addition, all vertices must be attached to the same vertex in {v₁, v₂, v₃}. Such a graph is isomorphic to S _n + e, which is in contradiction with our assumption.

• Case 2: G contains a C₄.

For any edge uv on the cycle, by Lemma 8, we have

(17)sG(u)≤|N2(u)|+1+|N(u)|,

and

(18)sG(v)≤|N2(v)|+1+|N(v)|.

Then we have N₂(u) ∩ N₂(v) = ∅, otherwise G contains a C₅. Together with the assumption that G contains C₄ this is a contradiction with the fact that G is unicyclic. Furthermore, we have N (u) ∩ N (v) = ∅. Otherwise G contains a C₃ and this is again a contradiction. Since u and v are not in N₂(u) ∩ N₂(v), we have |N₂(u) ∩ N₂(v)| ≤ n − 2. As |N(u) ∩ N(v)| ≤ n, from Eqs. (17-18) it follows

(19)sG(u)+sG(v)≤|N2(u)|+1+|N(u)|+|N2(v)|+1+|N(v)|≤2n.

Now we have to consider two separate subcases.

– Case 2.1: There exists no edge u^′v^′ ∈ E (G) such that s_G(u^′) = s_G(v^′) = n.

In this case, we have h(uv|G) ≤ f (n + 1, n − 1) for any edge uv ∈ E(G). Then

S(G)=∑uv∈E(G)h(uv|G)≤|E(G)|⋅f(n+1,n−1)=nf(n+1,n−1)<3f(n+1,n+1)+(n−3)f(n+1,n−1)=S(Sn+e),

a contradiction.

– Case 2.2: There exists an edge uv ∈ E(G) such that s_G(u) = s_G(v) = n.

Since u ∉N₂(u) ∪ N(u) for any u ∈ V(G), we have |N₂(u) ∪ N(u)| ≤ n − 1. Therefore, the equalities hold in Eqs. (17-18) and uv must be in C₄. Then, for any w ∈ V(G) − {u, v}, d(w, v) = 1 and d(w, u) = 2 or d(w, v) = 2 and d(w, u) = 1. Let C₄ = uvst, G is a graph isomorphic to a graph obtained by adding some pendent vertices to u or v of C₄. Then there are at most two edges xy with s_G(x) = s_G(y) = n, and hence

S(G)=∑uv∈E(G)h(uv|G)≤2f(n,n)+(n−2)f(n+1,n−1)<3f(n+1,n+1)+(n−3)f(n+1,n−1)=S(Sn+e),

a contradiction.

From Cases 2.1 and 2.2, it follows that G does not contain C₄.

• Case 3: G contains a C₅.

By Lemma 7, for any uv ∈ E (G), we have

(20)sG(u)≤|N2(u)|+|N(u)|,

and

(21)sG(v)≤|N2(v)|+|N(v)|.

Then s_G(u) + s_G(v) ≤ 2n − 1 and h(uv|G) ≤ f (n, n − 1) for any edge uv ∈ E(G). It follows

S(G)=∑uv∈E(G)h(uv|G)≤|E(G)|⋅f(n,n−1)=nf(n,n−1)<3f(n+1,n+1)+(n−3)f(n+1,n−1)=S(Sn+e),

a contradiction.

• Case 4: G contains C _k for some k ≥ 6.

In this case, for any edge uv ∈ E(G) we have s_G(u) + s_G(v) ≤ 2n − 2 and h(uv|G) ≤ f (n − 1, n − 1). Then

S(G)=∑uv∈E(G)h(uv|G)≤|E(G)|⋅f(n−1,n−1)=n⋅f(n−1,n−1)<3⋅f(n+1,n+1)+(n−3)⋅f(n+1,n−1)=S(Sn+e),

a contradiction.

Summing up, we have proved that G does not contain any C _k for all k ≥ 3, which contradicts the fact that G is an unicyclic graph.

Theorem 15Let Gbe an n-vertex unicyclic graph with n ≥ 4, then we have

51227n≤S(G),

with equality if and only if G ≌C_n.

Proof. Suppose G is a graph with minimum Sanskruti index. Let C = v₁v₂ · · · v_k, 3 ≤ k ≤ n, be the unique cycle of G. We consider the following cases.

• Case 1: for any edge v_iv_i₊₁ in C we have s_G(v_i) = 4 or s_G(v_i₊₁) = 4.

In this case, we have d(v_j) = 2 for any v_j in C, and hence G ≌C.

Observe that if v_i ∈ C and s_G(v_i) > 4, then there must be a neighbor of v_i, say u ∈ C with s_G(u) > 4.

• Case 2: there exists an edge v_iv_i₊₁ in C such that s_G(v_i)≥ 5 and s_G(v_i₊₁) ≥ 5.

Now we claim that

Claim 16. s_G(v_i) = 5 and s_G(v_i₊₁) = 5.

Proof. Otherwise, we assume without loss of generality that s_G(v_i₊₁) ≥ 6. Let T = G − {v_iv_i₊₁}, then T is a tree. By Theorem 9 we have S(T) ≥ S(P_n).

Denote s₃ = s_G(v_i), s₁ = s_G(v_i₊₁), s₂ = s_G(v_i₊₂).

* First,we have s₃ = 5. Otherwise, s₃ 6. Let q(x,y)=f(x,y)−f(x−2,y)=(xyx+y−2)3−((x−2)yx+y−4)3.Then

(22)∂q(x,y)∂y=3y2(x−2)x3(x+y−2)4−(x−4)(x−2)3)(x+y−4)4>0,

which means that function q(x, y) is an increasing function on variable y for fixed x ≥ 4. Note that s₂ ≥ 4, so from s₁ ≥ 6 it follows q(s₁, s₂)≥ q(s₁, 4) and

S(G)≥S(T)+h(vivi+1|G)+h(vi+1vi+2|G)−h(vi+1vi+2|T)≥S(Pn)+s3s1s3+s1−23+q(s1,s2)≥S(Pn)+6s16+s1−23+q(s1,4)>S(Cn),

contradicting with the assumption that G is a graph with minimum Sanskruti index. Thus, s₃ = 5.

* Now we have s₂ ≠4. Otherwise, we have d_G(v_i₊₁) = d_G(v_i₊₂) = 2, d_G(v_i)≥ 4 and thus s₃ ≥ 6, contradicting with the above result s₃ = 5.

* Also, we have s₂ ≠ 5. Otherwise, s₂ = 5. Then we have d_G(v_i₊₁) ≤ 3.

– If d_G(v_i₊₁) = 2, then we have d_G(v_i) = d_G(v_i₊₂) = 3 and both v_i and v_i₊₂ have a leaf neighbor. Let the leaf neighbor of v_i be v^′, then we have

contradicting with the assumption that G is a graph with minimum Sanskruti index.

– If d_G(v_i₊₁) = 3, then d_G(v_i) = d_G(v_i₊₂) = 2. Let N(v_i₊₁) = {v_i , v_i₊₂, u^′} and s^′₂ = s_G(u^′). Then s^′₂ ≥ 4 and

contradicting with the assumption that G is a graph with minimum Sanskruti index. Thus, we have proved that s₂ ≠5.

* If s₂ ≥ 6, we exchange the role of v_i and v_i₊₂. Since s₃ = s_G(v_i) = 5, we also have s₂ = s_G(v_i₊₂) = 5, a contradiction.

Concluding: Claim 16 was proved, we have s_G(v_i) = 5 and s_G(v_i₊₁) = 5.

Continuing with the proof of Theorem 15 in Case 2 it is sufficient to consider the following two cases.

– Subcase 2.1: d_G(v_i) = 2 and d_G(v_i₊₁) = 2.

In this case, obviously d_G(v_i₋₁) = 3 and by Claim 16, we have s_G(v_i₋₁) = 5, thus v_i₋₁ has a leaf neighbor w (see Figure 3 (left)). Let T = G − {v_i₋₁v_i}, then

Figure 3

The case d_G(v_i) = 2 and d_G(v_i₊₁) = 2 (left); the case d_G(v_i) = 3 and d_G(v_i₊₁) = 2 (right).

S(G)≥S(T)+h(vi−1w|G)−h(vi−1w|T)+h(vi−1vi|G)>S(Cn),

contradicting with the assumption that G is a graph with minimum Sanskruti index.

– Subcase 2.2: d_G(v_i) = 3 and d_G(v_i₊₁) = 2.

In this case, v_i has a leaf neighbor w. (see Figure 3 (right)).

Let T = G − {v_iv_i₊₁}, then we have

S(G)≥S(T)+h(viw|G)−h(viw|T)+h(vivi+1|G)>S(Cn),

contradicting with the assumption that Gis a graph with minimum Sanskruti index.

This concludes the proof of Theorem 15.

4 Conclusions and future work

This paper reveals the idea that the structure of a molecular tree or unicyclic graph with minimal Sanskruti index has a path as long as possible. Similarly, a tree or a unicyclic graph with maximal value of Sanskruti index has a path as short as possible, These results may also hold in other families of molecular graphs. Moreover, there are several research avenues that may naturally extend the results of this paper. A natural generalization of trees and unicyclic graphs are cactus graphs, and it may be possible to find the extremal graphs among cacti applying the methods used here. Another idea that may be worth investigation is the following: the exponent 3 in the definition of Sanskruti index seems to be a rather arbitrary and lucky choice. One could replace 3 with arbitrary exponent α > 0 and perhaps obtain similar mathematical properties, and for some other α maybe even better correlation with some chemical properties of the corresponding chemical graphs.

Ethical approval
The conducted research is not related to either human or animal use.
Conflict of interest
Authors state no conflict of interest.

Acknowledgement

This work is supported by the Natural Science Foundation of Guangdong Province under grant 2018A0303130115, and the China Postdoctoral Science Foundation under Grant 2017M621579; the Postdoctoral Science Foundation of Jiangsu Province under Grant 1701081B; Project of Anhui Jianzhu University under Grant no. 2016QD116 and 2017dc03. Research of J. Žerovnik and D. Rupnik Poklukarwas supported in part by Slovenian Research Agency under grants P2-0248, N1-0071 and J1-8155.

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Received: 2018-10-13

Accepted: 2018-11-19

Published Online: 2019-08-24

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

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https://doi.org/10.1515/chem-2019-0046

Schlagwörter für diesen Artikel

Topological index; Molecular descriptor; Sanskruti index; Tree; Unicyclic graph

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