EDGE VERSION OF HARMONIC INDEX AND HARMONIC POLYNOMIAL OF SOME CLASSES OF GRAPHS

Abstract

In this paper we define the edge version of harmonic index and harmonic polynomial of a graph G. We computed explicit formulas for the edge version of harmonic index and harmonic polynomial of many well known classes of graphs.

1. Introduction and Preliminaries

Let G be a simple graph, with vertex set V (G) and edge set E(G). The degree dv of a vertex v is the number of vertices joining to v and the degree of an edge e ∈ E(G), de is the number of its adjacent vertices in V (L(G)), where L(G) is the line graph of a graph G which is defined as the graph whose vertices are the edges of G, with two vertices are adjacent if the corresponding edges have one vertex common in G.

Line graphs are very useful in structural chemistry, but in recent years they were considered very little in chemical graph theory. In 1981, Bertz introduced the first topological index on the basis of the line graph in [1], when he was working on molecular branching. After that many topological indices based on line graphs were introduced (see [4,5,9]). For more details about the applications of line graphs in chemistry, we refer the articles (see [6,7,8]).

In chemistry, molecular structure descriptors are used to model information of molecules, which are known as topological indices. They are invariant under graph isomorphisms. There are many topological indices defined on the basis of the vertex-degrees of graphs (see [11,12,14,15,16,19]). One of the vertex-degree based index namely harmonic index H(G) is first time introduced in [2]:

For more results on harmonic index we refer to the articles [3,13,17,18,20,21,22,23]. The harmonic polynomial is defined in [10] as follows

Note that

In a natural way, we introduce the edge version of harmonic index on the basis of the end-vertex degrees of edges in a line graph of G which is defined as:

Similarly the edge version of harmonic polynomial is defined as

Clearly

The following lemma is helpful for computing the degree of a vertex of line graph.

Lemma 1.1. Let G be a graph with u, v ∈ V (G) and e = uv ∈ E(G). Then:

In order to calculate the number of edges of an arbitrary graph, the following lemma is significant for us.

Lemma 1.2. Let G be a graph. Then

This is also known as handshaking Lemma.

In this paper we computed edge version of harmonic index and harmonic polynomial for the case of regular, complete bipartite, wheel, helm, ladder and [n]-pentacene graphs.

 

2. Main results

Proposition 2.1. Let G be a k-regular graph of n vertices, then

Proof. Since G is a k- regular graph, then each vertex of G has degree k and by Lemma 1.2 we have edges. Therefore in L(G) we have vertices and by using Lemma 1.1, all the vertices have degree 2k−2. Lemma 1.2 implies that we have edges in L(G). Consequently we get He(G,x) = kn(k -1)x4k-5. □

Let Kn, Cn, Πn and An denotes the complete graph on n vertices, the cycle on n vertices, the n-sided prism and the n-sided antiprism as shown in Fig. 1.

FIGURE 1.The graphs Kn, Cn, πn and An for n = 4 from left to right

Proposition 2.2. We have

Proof. This proof can be obtained by using Proposition 2.1. □

Proposition 2.3. Let G be a complete bipartite graph Km,n of m + n vertices and mn edges, then

Proof. In G there are m vertices of degree n and remaining n vertices of degree m. In L(G) there are mn vertices and each vertex have degree m + n − 2 by Lemma 1.1. By Lemma 1.2, L(G) has edges. Consequently we get He(km,n, x) = mn(m + n − 2)x2m+2n−5 and □

Proposition 2.4. Let Wn be a graph of wheel, then

Proof. In the wheel graph Wn, the total number of vertices and edges are n +1 and 2n respectively (see Fig. 2). Therefore in L(Wn), the total number of vertices are 2n, out of which n vertices of degree 4 and remaining n vertices of degree n+1 (see Fig. 3). It is easily seen from Lemma 1.2 that the total number of edges in L(Wn) are The edge partition of E(L(Wn)) based on the degree of the vertices is shown in Table 1.

FIGURE 2.The wheel graph W7

FIGURE 3.The line graph of wheel graph W7

TABLE 1.The edge partition of L(Wn)

Hence we get He(Wn, x) = 2nx7 + n(n − 1)x2n+1 + 4nxn+4 and □

Proposition 2.5. Let Hn be a helm graph, then

Proof. In the helm graph Hn, the total number of vertices and edges are 2n+1 and 3n respectively (see Fig. 4). Therefore in L(Hn), the total number of vertices are 3n, out of which n vertices of degree 3, n vertices of degree 6 and n vertices of degree n + 2 (see Fig. 5). It is easily seen from Lemma 1.2 that the total number of edges in L(Hn) are The edge partition of E(L(Hn)) based on the degree of the vertices is shown in Table 2.

FIGURE 4The helm graph H7

FIGURE 5The line graph of helm graph H7

TABLE 2.The edge partition of L(Hn)

Therefore we get He(Hn, x) = 4nx8 + 4nxn+7 + n(n − 1)x2n+3 + 2nxn+4 + 2nx11 and □

Proposition 2.6. Let Ln be a graph of ladder, then

Proof. The ladder graph Ln for n = 1 is a cycle C4 which is known from Proposition 2.1. For L2 we have the edge partition of E(L(L2)) based on the degree of the vertices is shown in Table 3.

TABLE 3.The edge partition of L(L2)

In the ladder graph Ln for n > 2, the total number of vertices and edges are 2n+2 and 3n+1 respectively (see Fig. 6). Therefore in L(Ln), the total number of vertices are 3n + 1, out of which 2 vertices of degree 2, 4 vertices of degree 3 and 3n − 5 vertices of degree 4 (see Fig. 7). It is easily seen from Lemma 1.2 that the total number of edges in L(Ln) are 6n − 2. The edge partition of E(L(Ln)) based on the degree of the vertices is shown in Table 4.

FIGURE 6The ladder graph L5

FIGURE 7The line graph of ladder graph L5

TABLE 4.The edge partition of L(Ln)

Consequently, we get He(Ln, x) = 8x4 + 16x6 + 2x7(6n − 14) and □

In chemistry, the pentacene compound is a hydrocarbon consists of five benzene rings. It is purple powder organic semiconductor and gradually degrades when exposed to light and air. The linear [n]-Pentacene Pn for n = 2 is shown in Fig. 8.

FIGURE 8The pentacene graph P2

Proposition 2.7. Let Pn be a graph of linear [n]-pentacene, then

Proof. In the pentacene graph Pn the total number of vertices and edges are 22n and 28n − 2 respectively (see Fig. 8). Therefore in line graph L(Pn), the total number of vertices are 28n − 2 out of which 6 vertices are of degree 2, 20n − 4 vertices are of degree 3 and 8n − 4 vertices are of degree 4 (see Fig. 9). It is easily seen from Lemma 1.2 that the total number of edges in L(Pn) are 46n−8. The edge partition of E(L(Pn)) based on the degree of the vertices in shown in Table 5.

FIGURE 9The line graph of pentacene P2

TABLE 5.The edge partition of L(Pn)

Hence we get He(Pn, x) = 8x3 + 8x4 + (36n − 8)x5 + (48n − 16)x6 + (8n − 8)x7 and □