On Lanzhou Index of Graphs

Somnath Bera1, Kinkar Chandra Das2
1School of Automation, Huazhong University of Science and Technology, Wuhan 430074, China.
2Department of Mathematics, Sungkyunkwan University, Suwon 16419, Republic of Korea.

Abstract

Let G=(V,E) be a simple graph with vertex set V(G) and edge set E(G). The Lanzhou index of a graph G is defined by Lz(G)=uV(G)du2d¯u, where du (d¯u resp.) denotes the degree of the vertex u in G (G¯, the complement graph of G resp.). It has predictive powers to provide insights of chemical relevant properties of chemical graph structures. In this paper we discuss some properties of Lanzhou index. Several inequalities having lower and upper bound for the Lanzhou index in terms of first, second and third Zagreb indices, radius of graph, eccentric connectivity index, Schultz index, inverse sum indeg index and symmetric division deg index, are discussed. At the end the Lanzhou index of corona and join of graphs have been derived.

Keywords: Lanzhou index, Zagreb indices, forgotten topological index, irregularity index, eccentric connectivity index, Schultz index

1. Introduction and Motivation

Throughout the paper we consider simple, undirected, unweighted graphs only unless it is specified. Let G=(V,E) be a graph with vertex set V(G) and edge set E(G). The number of vertices and edges are denoted by |V(G)| and |E(G)|, respectively. The degree of a vertex uV(G), denoted by dG(u) (simply du whenever understood) is the number of adjacent vertices to u in G. The distance between any two vertices u and v, denoted by d(u,v) is defined as the length of shortest path between u and v in G. The complement graph G¯ of a graph G is the graph with the same vertex set V(G) and the vertices are adjacent in G¯ if and only if they are not adjacent in G.

The Zagreb indices were first introduced by Gutman and Trinajstić [1], they are important molecular descriptors and have been closely correlated with many chemical properties[2]. The first Zagreb index M1(G) of a graph G is defined as M1(G)=uV(G)du2=uvE(G)(du+dv). while the second Zagreb index M2(G) is defined as M2(G)=uvE(G)dudv. Furtula and Gutman[3] introduced forgotten topological index (also called F-index) which is defined as F(G)=uV(G)du3=uvE(G)(du2+dv2).

In [4], Vukičević et al. considered a linear combination of M1(G) and F(G) of the form M1(G)+λF(G), where λ was a free parameter ranging from 20 to 20. From the above linear combination, Vukičević et al. introduced in the same paper a new topological index named as Lanzhou index. It is denoted by Lz(G) and defined by Lz(G)=uV(G)du2d¯u, where d¯u is the degree of the vertex u in G¯. For its mathematical properties see the paper[4]. For self complimentary graphs, du=d¯u, implying that the Lanzhou index is same as forgotten topological index. In chemical graph theory, many vertex degree based topological indices and their properties have been investigated in[5,6,7,8,9,10,11,12,13,14,15,16,17,18].

In this paper we first discuss some properties of Lanzhou index in Section 2. An upper bound of Lanzhou index for unicyclic graphs has been obtained. The relationships between Lanzhou index and other topological indices such as graph radius, eccentric connectivity index, Schultz index, inverse sum indeg index and symmetric division deg index are derived in Section 3. At the end in Section 4 the Lanzhou index of the join and corona of graphs are provided.

2. Some properties of Lanzhou index

In this section we discuss the properties of the Lanzhou index.

Proposition 1. For a graph G, Lz(G)=uvE(G)(dud¯u+dvd¯v).

Proof. We have Lz(G)=auV(G)du2(n1du)=uvE(G)(du(n1du)+dv(n1dv))=uvE(G)(dud¯u+dvd¯v). 

From the definition of Lanzhou index it is clear that the value of Lz(G) is a positive integer. The next result shows that the Lanzhou index of a graph is a positive even integer.

Theorem 1. For any graph G, the Lanzhou index Lz(G) is even.

Proof. Let G be a graph with n vertices. By definition of Lanzhou index we have (1)Lz(G)=uV(G)du2d¯u=uV(G),du is even du2d¯u+uV(G),du is odd du2d¯u.

It is clear to see that the first term of the sum in (1) is even. For the second term of the sum in (1) we have the following two cases:

Case1: n is even. Since du is odd, we have d¯u=n1du is even. And in this case du2d¯u is even. Therefore the second term of the sum in (1) is even.

Case2: n is odd. Since du is odd, we have d¯u=n1du is odd. And in this case du2d¯u is odd. It is well known that the number of odd degree vertices in a graph is even. Using the fact that the sum of even number of odd numbers is even, it implies that the second term of the sum in (1) is even.

Hence, for any graph, the Lanzhou index Lz(G) is even. 

The union of two graphs G and H denoted by GH is the graph with vertex set V(G)V(H) and edge set E(G)E(H). We know that complete graphs and null graphs (graphs with isolated vertices) are the only example of graphs with minimum Lanzhou index 0. Likewise, the path of length 2 and K2K1 are the graphs with second minimum Lanzhou index 2.

Proposition 2. For any graph G, Lz(G)=2 if and only if GP3 or GK2K1.

Proof. For GP3 or GK2K1, we have Lz(G)=2. Moreover, Lz(Kn)=0=Lz(K¯n). Otherwise, we have n4 and there exist two vertices v and w in G such that 1dvn2 and 1dwn2. Then Lz(G)=uV(G)du2(n1du)dv2(n1dv)+dw2(n1dw)>2  as n4. This completes the proof. 

The bound for Lanzhou index of any graph is provided and the graph with extremal value have been characterized in [4].

Proposition 3. [4] Let G be any graph with n vertices. Then 0Lz(G)427n(n1)3.

In the following we give some lower and upper bounds for any graph with n vertices and m edges having minimum degree δ and maximum degree Δ.

Theorem 2. Let G be a connected graph with n vertices and m edges having minimum degree δ and maximum degree Δ. Then (n1)δΔ2Lz(G)2m(n1)Δδ2 with both equalities hold if and only if G is a regular graph.

Proof. Since δduΔ, by definition of Lanzhou index, we have Lz(G)=uvE(G)(dud¯u+dvd¯v)=uvE(G)((n1)(du+dv)(du2+dv2))uvE(G)((n1)2δ2Δ2)=2m((n1)δΔ2). From the above result, we get the lower bound. Moreover, the equality holds in the lower bound if and only if du=dv=δ=Δ for any edge uvE(G), that is, if and only if G is a regular graph. Again since δduΔ, similarly, we can get the upper bound on the Lanzhou index of graph G. Moreover, the right equality holds if and only if G is a regular graph. 

Theorem 3. Let G be a graph of order n with m edges. Then Lz(G)+Lz(G¯)=(n1)uV(G)dud¯u=(n1)[2(n1)mM1(G)], where M1(G) is the first Zagreb index of graph G.

Proof. We have Lz(G)+Lz(G¯)=uV(G)du2(n1du)+uV(G)d¯u2(n1d¯u)=uV(G)du2(n1du)+uV(G)(n1du)2du=uV(G)(n1du)du((n1du)+du)=(n1)uV(G)[(n1)dudu2]=(n1)[2(n1)mM1(G)]. 

Corollary 1. For self complementary graph G, Lz(G)+Lz(G¯)=(n1)M1(G).

Theorem 4. Let G be a graph with n vertices and e=uv be an edge in G. If H=Ge, then Lz(G)Lz(H)=(2n+1)(du+dv)3(du2+dv2)2n.

Proof. By the definition of Lanzhou index, we have Lz(G)=xV(G){u,v}dx2(n1dx)+du2(n1du)+dv2(n1dv),Lz(H)=xV(G){u,v}dx2(n1dx)+(du1)2(ndu)+(dv1)2(ndv).

Therefore we have Lz(G)Lz(H)=(du2+dv2)+(2du1)(ndu)+(2dv1)(ndv)=(2n+1)(du+dv)3(du2+dv2)2n. This completes the proof of the theorem. 

Let Tn be a tree with n vertices and TnΔ denotes the set of all trees on n vertices with maximum degree at most Δ. Vukičević et al. [4] obtained the following result:

Proposition 4. [4] Let n8 be an integer and TnTn4. Then 4n218n+20Lz(Tn)6n2+O(n).

Theorem 5. Let G be a unicyclic graph with n(8) vertices and Δ(G)=4. Then Lz(G)6n2+O(n).

Proof. Let e=uv be an edge in G such that H=Ge, where H is a tree of order n. Then by Theorem 1, we see that Lz(G)Lz(H)+(2n+1)(du+dv), where du and dv are the degrees of u and v, respectively.

Since du,dv4, from the above result, we have Lz(G)Lz(H)+8(2n+1). Since H is a tree, by Proposition 1, we obtain Lz(G)6n2+O(n). This completes the proof of the theorem. 

3. Relationships between Lanzhou index and other topological indices

The eccentricity of a vertex v in a graph G is defined as eccG(v)=max{dG(v,u)|uV(G)}. The radius of a graph G, denoted by r(G) is r(G)=min{eccG(v)|vV(G)}.

Lemma 1. [19] Let G be a nontrivial connected graph of order n. For each vertex v in G, it holds eccG(v)ndv. Moreover, the above equality holds together for all vertices in G if and only if GP4 or GKniK2 (0in2), where KniK2 denotes the graph obtained by removing i independent edges from Kn.

We now give a relation between Lz and M1.

Theorem 6. For any graph G, (2)Lz(G)(r(G)1)M1(G). Moreover, the equality holds in (2) if and only if GKn or GKnn2K2 (n is even).

Proof. For any vertex v in G we have ndveccG(v)r(G). Now by the definition of Lanzhou index, we have Lz(G)=uvE(G)(du(n1du)+dv(n1dv))uvE(G)(r(G)1)(du+dv)=(r(G)1)uvE(G)(du+dv)=(r(G)1)M1(G).

Suppose that equality holds in (2). Then we have ndv=eccG(v)=r(G) for all vV(G). This implies that G is self-centered graph. By Lemma 1, we have GKniK2 (0in2). Hence GKn or GKnn2K2 (n is even).

Conversely, one can easily see that the equality holds in (2) for Kn or for Knn2K2 (n is even). 

Here we give a relation between Lz, M1 and M2.

Theorem 7. Let G be a graph of order n. If du+dvn for each edge uvE(G), then Lz(G)2M2(G)M1(G).

Proof. From the definition of Lanzhou index with the given condition, we have Lz(G)=uvE(G)(du(n1du)+dv(n1dv))uvE(G)(du(dv1)+dv(du1))=2uvE(G)dudvuvE(G)(du+dv)=2M2(G)M1(G). 

We now mention two more relations between Lz, M1 and M2.

Theorem 8. Let G be a graph of order n. Then Lz(G)(n1)M1(G)2M2(G) with equality holding if and only if each connected component of G is regular. Moreover, (n1Δ)M1(G)Lz(G)(n1δ)M1(G) with both equalities hold if and only if G is a regular graph.

Proof. One can easily see that F(G)2M2(G) with equality holding if and only if each connected component of G is regular, and δM1(G)F(G)ΔM1(G) with both equalities hold if and only if G is a regular graph. Since Lz(G)=(n1)M1(G)F(G), using the above results, we get the required results. This completes the proof of the theorem. 

The eccentric connectivity index [18] of a graph G, denoted by ξc(G) is defined as ξc(G)=uV(G)dueccG(u).

Here we give a relation between Lz, M1 and ξc.

Theorem 9. Let G be a graph of order n and minimum degree δ. Then Lz(G)δξc(G)M1(G) with equality holding if and only if GKn or GKnn2K2 (n is even).

Proof. From the definition of Lanzhou index, we have Lz(G)=uvE(G)(du(n1du)+dv(n1dv))uvE(G)(du(eccG(u)1)+dv(eccG(v)1))=uvE(G)(dueccG(u)+dveccG(v))uvE(G)(du+dv)δuV(G)dueccG(u)M1(G)=δξc(G)M1(G). By the proof of the Theorem 6, one can easily see that the equality holds if and only if GKn or GKnn2K2 (n is even)

We now give a relation between Lz and M2.

Theorem 10. Let G be a connected graph with n vertices and minimum degree δ. Then 2(n2)ΔM2(G)Lz(G)2(n2)δM2(G) with both equalities hold if and only if G is a regular graph.

Proof. We construct an auxiliary real valued function of two variables x and y as g(x,y)=(n1)(x+y)(x2+y2)xy=(n1)(1y+1x)(xy+yx),  where δxyΔn1.

Now, g(x,y)x=(n1)(1x2)(1yyx2)=1x2(n1y)1y<0,  for δxyΔn1.

Therefore, g(x,y) is monotonically decreasing in the variable x. Since the function g(x,y) is symmetric in both x and y, it is also monotonically decreasing in the variable y. Thus we have g(x,y) attains its maximum value at (δ,δ) and the minimum value at (Δ,Δ). Hence g(δ,δ)2(n2)δ  and  g(Δ,Δ)2(n2)Δ.

This implies that (n1)(du+dv)(du2+dv2)2(n2)Δdudv and (n1)(du+dv)(du2+dv2)2(n2)δdudv.

Using the above results with the definition of the Lanzhou index, we have Lz(G)=uvE(G)((n1)(du+dv)(du2+dv2))uvE(G)2(n2)Δdudv=2(n2)ΔM2(G) and Lz(G)=uvE(G)((n1)(du+dv)(du2+dv2))2(n2)δM2(G)

Moreover, both equalities hold in (2) if and only if du=dv=δ or du=dv=Δ for any edge uvE(G). Since G is connected, both equalities hold in (2) if and only if G is a regular graph. 

The irregularity of a graph G, denoted by irr(G) is defined as irr(G)=uvE(G)|dudv|. It is also called third Zagreb index of graph. More results on irregularity, one can find in[20,21,22]. Here we give a relation between Lz with M1(G),M2(G),irr(G) and F(G) of graph G.

Theorem 11. Let G be a connected graph with n vertices. Then (n1)M1(G)irr(G)2+Lz(G)+2M2(G)(n1)M1(G)+2(m2)(Δδ)2, with left (right) equality holding if and only if G is a regular graph (G is a regular graph or a bipartite semiregular graph).

Proof. From the definition of irregularity, we have irr(G)2=(uvE(G)|dudv|)2=uvE(G)(dudv)2+2uv,xyE(G)uvxy|dudv||dxdy|uvE(G)(du2+dv2)2uvE(G)dudv+2(m2)(Δδ)2=(n1)M1(G)Lz(G)2M2(G)+2(m2)(Δδ)2 as Lz(G)=(n1)M1(G)F(G). Hence we get the right inequality. Moreover, the equality holds if and only if |dudv|=Δδ for all uvE(G), that is, if and only if G is a regular graph or a bipartite semiregular graph as G is connected.

Now, irr(G)2=(uvE(G)|dudv|)2uvE(G)(dudv)2=uvE(G)(du2+dv2)2uvE(G)du.dv=F(G)2M2(G)=(n1)M1(G)Lz(G)2M2(G) as Lz(G)=(n1)M1(G)F(G). Hence we get the left inequality. Moreover, the left equality holds if and only if du=dv for all edges uvE(G), that is, if and only if G is a regular graph as G is connected. 

The Schultz index of a molecular graph G, introduced by Schultz [23], is defined as SI(G)=12{u,v}V(G)(du+dv)d(u,v), where d(u,v) denotes the distance between the vertices u and v. The Schultz indices have been shown as useful descriptor for molecular design and characterization with desired properties in[2, 24].

The join GH of two simple graphs G and H is the graph with the vertex set V(GH)=V(G)V(H) and the edge set E(GH)=E(G)E(H){uv:uV(G),vV(H)}. The following theorem gives a relation between Schultz index and Lanzhou index.

Theorem 12. Let G be a connected graph of order n. Then SI(G)12(n1)(Lz(G)+F(G))+(n(n1)2m)δ with equality holding if and only if GKn or GKδ(nδ)K1 (δ<n1) or G is a regular graph with diameter 2.

Proof. By definition of Schultz index we have SI(G)=12{u,v}V(G)(du+dv)d(u,v)=12uvE(G)(du+dv)+12{u,v}V(G),d(u,v)2(du+dv)d(u,v)12M1(G)+{u,v}V(G),d(u,v)2(du+dv)12M1(G)+2((n2)m)δ=12(n1)(Lz(G)+F(G))+(n(n1)2m)δ. The first part of the proof is done.

Suppose that equality holds. Then d(u,v)=1 or 2 for any pair of vertices (u,v). Moreover, du=dv=δ when d(u,v)=2 for any pair of vertices (u,v), that is, all the vertices in G have degree either n1 or δ. If Δ=n1, then GKn or GKδ(nδ)K1 (δ<n1). Otherwise, Δ<n1. Then all the vertices in G are of degree δ. Hence G is a regular graph with diameter 2.

Conversely, let GKn. Then Lz(G)=0, F(G)=n(n1)3 and SI(G)=n(n1)2/2. Hence the equality holds.

Let GKδ(nδ)K1 (δ<n1). Then d1=d2==dδ=n1 and dδ+1=dδ+2==dn=δ. Then Lz(G)+F(G)=(n1)3δ+(n1)(nδ)δ2,2m=δ(n1)+(nδ)δ and SI(G)=δ(δ1)(n1)2+(nδ)(nδ1)δ+δ(nδ)(n1+δ)2. Now, 12(n1)(Lz(G)+F(G))+(n(n1)2m)δ=δ(n1)2+(nδ)δ22+(n(n1)δ(n1)(nδ)δ)δ=δ(δ1)(n1)2+(nδ)(nδ1)δ+δ(nδ)(n1+δ)2=SI(G).

Let G be a regular graph with diameter 2. Then d1=d2==dn=δ. Then Lz(G)+F(G)=(n1)nδ2,2m=nδ  and  SI(G)=δ[m+(n(n1)2m)2]. Now, 12(n1)(Lz(G)+F(G))+(n(n1)2m)δ=nδ22+(n(n1)nδ)δ=SI(G). This completes the proof of the theorem. 

The inverse sum indeg (ISI) index [16] is used as a significant predictor of total surface area for octane isomers. The ISI index is defined as ISI(G)=uvE(G)11du+1dv.

Theorem 13. Let G be a graph with n vertices and maximum degree Δ. Then Lz(G)4(n1Δ)ISI(G) with equality holding if and only if G is a regular graph.

Proof. Since Δ is the maximum degree in G, we have du(Δdu)+dv(Δdv)0,  that is,  du2+dv2du+dvΔ with equality holding if and only if du=dv=Δ.

One can easily see that (dudv)20,  that is, (du+dv)24dudv,  that is, (du+dv)4dudvdu+dv=41du+1dv with equality holding if and only if du=dv.

Using the above results with the definition of the Lanzhou index, we have Lz(G)=uvE(G)((n1)(du+dv)(du2+dv2))=uvE(G)(du+dv)(n1du2+dv2du+dv)uvE(G)(du+dv)(n1Δ)4(n1Δ)uvE(G)11du+1dv=4(n1Δ)ISI(G). Moreover, the equality holds if and only if du=dv=Δ for any edge uvE(G), that is, if and only if G is a regular graph. 

Corollary 1. Let G be a graph with n vertices, m edges and maximum degree Δ. Then Lz(G)4m2(n1Δ)n with equality holding if and only if G is a regular graph.

Proof. Since ISI(G)m2n, from Theorem 13, we obtain the required result. Moreover, the equality holds if and only if G is a regular graph. 

The symmetric division deg index, SDD, was defined in [17] as SDD(G)=uvE(G)du2+dv2dudv. For recent results on SDD(G) see the papers [25,26,27,28,29] and the references cited therein. Here we give a relation between Lz and SDD.

Theorem 14. Let G be a graph with n vertices and maximum degree Δ. Then Lz(G)Δ2(n(n1)SDD(G)) with equality holding if and only if G is a regular graph.

Proof. Let GHpK1 (p0), where H is a graph of order np. Since n1du0 for all uV(G), by the definition of Lanzhou index, we obtain Lz(G)=uvE(G)[(n1)(du+dv)(du2+dv2)]=uvE(G)dudv(n1)(du+dv)(du2+dv2)dudvΔ2uvE(G)((n1)(du+dv)dudvdu2+dv2dudv)=Δ2[(n1)uvE(G)(1du+1dv)uvE(G)du2+dv2dudv]=Δ2((np)(n1)SDD(G))Δ2(n(n1)SDD(G)). The first part of the proof is done.

The equality holds if and only if du=dv=Δ for any edge uvE(H) and p=0, that is, if and only if G is a regular graph. 

4. The join and corona of graphs

The following theorem provides the formula to find the Lanzhou index of the join of two graphs.

Theorem 15. Let G and H be two graphs. Then Lz(GH)=Lz(G)+Lz(H)+4n2(n11)|E(G)|+4n1(n21)|E(H)|2n2M1(G)2n1M1(H)+n1n2(2n1n2n)2n22|E(G)|2n12|E(H)|.       

Proof. Let G and H be two graphs with n1 and n2 vertices. Then GH has n1+n2 vertices. We have dGH(u)=dG(u)+n2, for uV(G) and dGH(u)=dH(u)+n1, for uV(H). By the definition of Lanzhou index, we obtain Lz(GH)=uV(GH)du2(n1+n21du)=uV(G)dGH(u)2(n1+n21dGH(u))               +uV(H)dGH(u)2(n1+n21dGH(u)). Now, uV(G)dGH(u)2(n1+n21dGH(u))=uV(G)(dG(u)+n2)2(n11dG(u))=uV(G)dG(u)2d¯G(u)+2n2uV(G)dG(u)(n11dG(u))+n22uV(G)(n11dG(u))=Lz(G)+4n2(n11)|E(G)|2n2M1(G)+n22(n1(n11)2|E(G)|). Similarly, we obtain uV(H)dGH(u)2(n1+n21dGH(u))=Lz(H)+4n1(n21)|E(H)|2n1M1(H)+n12(n2(n21)2|E(H)|) Combining two above results, we get the required result. 

The corona product GH of two graphs G and H is defined to be the graph Γ obtained by taking one copy of G (which has n1 vertices) and n1 copies of H (which has n2 vertices), and then joining the i-th vertex of G to every vertex in the i-th copy of H, i=1,2,,n1. The next theorem provides the Lanzhou index of the corona product of two graphs G and H.

Theorem 16. Let G and H be two graphs of order n1 and n2, respectively. Then
Lz(GH)=Lz(G)+n1Lz(H)+A+n1B,
where
A=2n2(2n1n2+2n13n22)|E(G)|3n2M1(G)+n1n22(n11)(n2+1)
and
B=2(2n1n2+2n15)|E(H)|3M1(H)+n2(n1n2+n12).

Proof. For any vertex uV(G), we have dGH(u)=dG(u)+n2 and for any vertex uV(H), we have dGH(u)=dH(u)+1. Now, uV(G)(dGH(u))2(n1n2+n11dGH(u))=uV(G)(dG(u)+n2)2(n1n2+n11dG(u)n2)=uV(G)[dG(u)2d¯G(u)3n2dG(u)2+n2(2n1n2+2n13n22)dG(u)                                             +n22(n11)(n2+1)]=Lz(G)3n2M1(G)+2n2(2n1n2+2n13n22)|E(G)|+n1n22(n11)(n2+1) and uV(H)(dGH(u))2(n1n2+n11dGH(u))=uV(H)(dH(u)+1)2(n1n2+n12dH(u))=uV(H)[dH(u)2d¯H(u)3dH(u)2+(2n1n2+2n15)dH(u)+n1n2+n12]=Lz(H)3M1(H)+2(2n1n2+2n15)|E(H)|+n2(n1n2+n12). Using the above results, we obtain Lz(GH)=uV(GH)(dGH(u))2(n1n2+n11dGH(u))=uV(G)(dG(u)+n2)2(n1n2+n11dG(u)n2)             +uV(G)vV(H)(dH(v)+1)2(n1n2+n12dH(v))=Lz(G)+n1Lz(H)+A+n1B, where
A=2n2(2n1n2+2n13n22)|E(G)|3n2M1(G)+n1n22(n11)(n2+1)
and
B=2(2n1n2+2n15)|E(H)|3M1(H)+n2(n1n2+n12). This completes the proof of the theorem. 

Acknowledgements

The authors would like to thank the referee for his/her valuable comments which led to an improvement of the original manuscript. The first author gratefully acknowledges support by National Natural Science Foundation of China (61320106005 and 61772214) and the Innovation Scientists and Technicians Troop Construction Projects of Henan Province (154200510012). The second author was supported by the Sungkyun research fund, Sungkyunkwan University, 2017, and National Research Foundation of the Korean government with grant No. 2017R1D1A1B03028642.

Conflict of Interest

The authors declare no conflict of interests.

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