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BY 4.0 license Open Access Published by De Gruyter Open Access August 24, 2019

The Sanskruti index of trees and unicyclic graphs

  • Fei Deng , Huiqin Jiang , Jia-Bao Liu EMAIL logo , Darja Rupnik Poklukar EMAIL logo , Zehui Shao , Pu Wu and Janez Žerovnik
From the journal Open Chemistry

Abstract

The Sanskruti index of a graph G is defined as S(G)=uvE(G)sG(u)sG(v)sG(u)+sG(v)23,where sG(u) is the sum of the degrees of the neighbors of a vertex u in G. Let Pn, Cn, Sn and Sn + 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) 51227n1726883375S(T)(n1)78(n2)3for an n-vertex tree T with n ≤ 3, with equalities if and only if T ≌Pn (left) and TSn (right);

(2) 51227nS(G)(n3)(n+1)38+3(n+1)68n3for an n-vertex unicyclic graph with n ≥ 4, with equalities if and only if G ≌Cn (left) and GSn + e (right).

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, Pn, Cn, Sn and Sn + e stand for the path, cycle, star and star plus an edge of n vertices, respectively (see Figure 1). We denote by dG(v) the degree of a vertex v of a graph G and by NG(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 sG(u) the sum of the degrees of the neighbors of u, i.e., sG(u)=vNG(u)dG(v).

Figure 1 The graphs Pn, Cn, Sn and Sn + e
Figure 1

The graphs Pn, Cn, Sn and Sn + 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+y23.

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)=uvE(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)=(n1)78(n2)3and

S(Pn)=16,n=3,175364,n=4,51227n1726883375,n5.

Proposition 2Let n ≥ 4, S(Sn+e)=(n3)(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(t2)3.Moreover, f(x,y)=t664(t2)3if and only ifx=y=t2.

Lemma 4Let t ≠2 andg(t)=f(t2,t2)=t664(t2)3,theng(t)=3(t4)t564(t2)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(y2)(x+y2)40andf(x,y)y=3y2x3(x2)(x+y2)40.

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

  1. N(u) ∩ N(v) = ∅ and N2(u) ∩ N2(v) = .

  2. Σv∈N(u)|N(v) {u}| = |N2(u)|.

  3. sT(u) = |N2(u)| + |N(u)| = |N2(u)| + dT(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 wN2(u) ∩ N2(v). Denote with usw and vtw the shortest uw path and vw path. Then sv. Otherwise wN(v), a contradiction. Analogously, tu. Now if st,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, v2N(u). Therefore, Σ vN(u)|N(v) {u}| = |N2(u)|.

(c) It can be seen that sT(u) = ΣvN(u)dT(v) = Σ vN(u) |N(v)| = ΣvN(u)(|N(v) {u}| + 1) = Σ vN(u) |N(v) {u}| + dT(u). From the result of part (b), we have ΣvN(u)|N(v) {u}| = |N2(u)|, and thus sT(u) = |N2(u)| + dT(u) for any uV(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 sG(u) = |N2(u)| + |N(u)| = |N2(u)| + dG(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,uV(C4),N2u+Nu,uV(C4).

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

sGu=vNuNv=vN(u)|N(v){u}|+dG(u)=vN(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)(n1)78(n2)3,

with equality if and only if T ≌Sn.

Proof. By Lemma 6 (c), we have sT(u) + sT(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, sT(u) + sT(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)|2n2.

Moreover,

(5)sT(u)+sT(v)=2n2|N2(u)N2(v)|=n2,|N(u)N(v)|=n.

Recall that f(x,y)=(xyx+y2)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 sT(u) + sT(v) ≥ 4, then by applying Lemma 3 with t = 2n − 2, we have

(T)=uvE(T)sT(u)sT(v)sT(u)+sT(v)23=uvE(T)f(sT(u),sT(v))uvE(T)f(x,y)|x=y=n1=|E(T)|f(x,y)|x=y=n1=(n1)78(n2)3.

Conversely, if S(T)=(n1)78(n2)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

51227n1726883375S(T),

with equality if and only if T ≌Pn.

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 Tn be the family of trees on n vertices and let Fn = {T|S(T) ≤ S(T′′ )foranyT′′ ∈ Tn , TPn}. 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 ∈ Fn and P = x1x2 · · · xp 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, · · · , yq}, where q ≥ 1. Now construct a new tree T′′ with V(T′′ ) = V(T) and E(T′′ ) = E(T)∪{x1yi|i = 1, 2, · · · , q}∖ {x2yi|i = 1, 2, · · · , q} (see Figure 2).

Figure 2 The trees T′ and T′′
Figure 2

The trees T and T′′

Then we have sT (x3) > sT′′ (x3) and sT (v) ≥ sT′′ (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 ∈ Fn. So, d(x2) = 2.

Claim 12. d(x3) = 2.

Proof. Assume to the contrary that N(x3) = {x2, x4, y1, y2, · · · , yq}, 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

(6)sT(x3)>sT(x3),sT(x2)=sT(x2),h(x3y1|T)=dT(x3)sT(x3)dT(x3)+sT(x3)23,h(x1x2|T)=8,h(x1y1|T)=8,h(x1x2|T)=3(1+dT(x3))dT(x3)+23,h(x2x3|T)<h(x2x3|T)

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)+23(1+dT(x3))dT(x3)+2, thus we have

(8)sT(x3)4dT(x3)sT(x3)dT(x3)+sT(x3)233(1+dT(x3))dT(x3)+23h(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)+s313+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))(s32)dT(x3)+s333+1253+8.

Define F(x)=f(x,6)f(x,4)=8x3(x2)(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),s32)0, we have

S(T)S(T)=h1h28+f(1+dT(x3),6)f(1+dT(x3),4)+f(1+dT(x3),s3)f(1+dT(x3),s32)1253=81253+F(1+dT(x3))+f(1+dT(x3),s3)f(1+dT(x3),s32)81253+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(Pn−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+s223+s2(s11)s1+s233>833.

On the other hand, by Proposition 1, we have

(12)S(Pn)S(Pn1)=833.

Therefore, we have

S(T)S(T)+8+3s1s1+13+s1s2s1+s2238s2(s11)s1+s233>S(Pn1)+3s1s1+13+s1s2s1+s223s2(s11)s1+s233>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 TFn, we have S(T′′′ ) > S(Pn−1). Moreover, it can be seen that S(T)=S(T)+(125)3.Together with Eq. (12), we have S(T) > S(Pn), 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)(n3)(n+1)38+3(n+1)68n3,

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

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

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

• Case 1: Gcontains a C3.

Let C3 = v1v2v3. Then for any uV(G) {v1, v2, v3} we have

(13)sG(u)=|N2(u)|+|N(u)|n1,

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 sG(u1) = sG(u2) = n. Therefore,

h(u1u2|G)f(n+1,n1).

Further, we have

(16)sG(v1)+sG(v2)2n+2

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

S(G)=uvE(G)h(uv|G)3f(n+1,n+1)+(n3)f(n+1,n1)=S(Sn+e).

If the equality S(G) = S(Sn + e) holds, then the equalities in (15) and (16) hold and sG(v1) = sG(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 uvE (G) such that sG(u) = sG(v) = n.

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

S(G)=uvE(G)h(uv|G)|E(G)|f(n+1,n1)=nf(n+1,n1)<3f(n+1,n+1)+(n3)f(n+1,n1)=S(Sn+e),

a contradiction.

– Case 2.2: There exists an edge uv ∈ E(G) such that sG(u) = sG(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 sG(x) = sG(y) = n, and hence

S(G)=uvE(G)h(uv|G)2f(n,n)+(n2)f(n+1,n1)<3f(n+1,n+1)+(n3)f(n+1,n1)=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 sG(u) + sG(v) ≤ 2n − 1 and h(uv|G) ≤ f (n, n − 1) for any edge uv ∈ E(G). It follows

S(G)=uvE(G)h(uv|G)|E(G)|f(n,n1)=nf(n,n1)<3f(n+1,n+1)+(n3)f(n+1,n1)=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 sG(u) + sG(v) ≤ 2n − 2 and h(uv|G) ≤ f (n − 1, n − 1). Then

S(G)=uvE(G)h(uv|G)|E(G)|f(n1,n1)=nf(n1,n1)<3f(n+1,n+1)+(n3)f(n+1,n1)=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

51227nS(G),

with equality if and only if G ≌Cn.

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

• Case 1: for any edge vivi+1 in C we have sG(vi) = 4 or sG(vi+1) = 4.

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

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

• Case 2: there exists an edge vivi+1 in C such that sG(vi)≥ 5 and sG(vi+1) ≥ 5.

Now we claim that

Claim 16. sG(vi) = 5 and sG(vi+1) = 5.

Proof. Otherwise, we assume without loss of generality that sG(vi+1) ≥ 6. Let T = G{vivi+1}, then T is a tree. By Theorem 9 we have S(T) ≥ S(Pn).

Denote s3 = sG(vi), s1 = sG(vi+1), s2 = sG(vi+2).

* First,we have s3 = 5. Otherwise, s3 6. Let q(x,y)=f(x,y)f(x2,y)=(xyx+y2)3((x2)yx+y4)3.Then

(22)q(x,y)y=3y2(x2)x3(x+y2)4(x4)(x2)3)(x+y4)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+s123+q(s1,s2)S(Pn)+6s16+s123+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 dG(vi+1) = dG(vi+2) = 2, dG(vi)≥ 4 and thus s3 ≥ 6, contradicting with the above result s3 = 5.

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

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

S(G)S(T)+h(vivi+1|G)+h(vi+1vi+2|G)h(vi+1vi+2|T)+h(viv|G)h(viv|T)S(Pn)+s3s1s3+s123+q(s1,s2)+15638S(Pn)+6s16+s123+q(s1,5)+15638>S(Cn),

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

– If dG(vi+1) = 3, then dG(vi) = dG(vi+2) = 2. Let N(vi+1) = {vi , vi+2, u} and s2 = sG(u). Then s2 ≥ 4 and

S(G)S(T)+h(vivi+1|G)+h(vi+1vi+2|G)h(vi+1vi+2|T)+h(vi+1u|G)h(vi+1u|T)S(Pn)+s3s1s3+s123+q(s1,s2)+q(s1,s2)S(Pn)+s3s1s3+s123+q(s1,5)+q(s1,4)>S(Cn),

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 vi and vi+2. Since s3 = sG(vi) = 5, we also have s2 = sG(vi+2) = 5, a contradiction.

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

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

– Subcase 2.1: dG(vi) = 2 and dG(vi+1) = 2.

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

Figure 3 The case dG(vi) = 2 and dG(vi+1) = 2 (left); the case dG(vi) = 3 and dG(vi+1) = 2 (right).
Figure 3

The case dG(vi) = 2 and dG(vi+1) = 2 (left); the case dG(vi) = 3 and dG(vi+1) = 2 (right).

S(G)S(T)+h(vi1w|G)h(vi1w|T)+h(vi1vi|G)>S(Cn),

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

– Subcase 2.2: dG(vi) = 3 and dG(vi+1) = 2.

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

Let T = G{vivi+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.

  1. Ethical approval

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

  2. 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

© 2019 F. Deng et al., published by De Gruyter

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

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