IJCOM International Journal of Combinatorics 1687-9171 1687-9163 Hindawi Publishing Corporation 10.1155/2016/2508156 2508156 Research Article On Self-Centeredness of Product of Graphs http://orcid.org/0000-0003-0293-6649 Singh Priyanka 1 Panigrahi Pratima 1 Szekely Laszlo A. Department of Mathematics Indian Institute of Technology Kharagpur Kharagpur 721302 India iitkgp.ac.in 2016 782016 2016 04 04 2016 28 06 2016 12 07 2016 2016 Copyright © 2016 Priyanka Singh and Pratima Panigrahi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

A graph G is said to be a self-centered graph if the eccentricity of every vertex of the graph is the same. In other words, a graph is a self-centered graph if radius and diameter of the graph are equal. In this paper, self-centeredness of strong product, co-normal product, and lexicographic product of graphs is studied in detail. The necessary and sufficient conditions for these products of graphs to be a self-centered graph are also discussed. The distance between any two vertices in the co-normal product of a finite number of graphs is also computed analytically.

1. Introduction

The concept of self-centered graphs is widely used in applications, for example, the facility location problem. The facility location problem is to locate facilities in a locality (network) so that these facilities can be used efficiently. All graphs in this paper are simple and connected graphs. The distance between two vertices u and v in a graph G, denoted by dG(u,v) (or simply d(u,v)), is the minimum length of u-v path in the graph. The eccentricity of a vertex v in G, denoted by eccG(v), is defined as the distance between v and a vertex farthest from v; that is, eccG(v)=max{dG(v,u):uV(G)}. The radius rad(G) and diameter diam(G) of the graph G are, respectively, the minimum and maximum eccentricity of the vertices of graph G; that is, rad(G)=min{ecc(v):vV(G)} and diam(G)=max{ecc(v):vV(G)}. The center C(G) of graph G is the induced subgraph of G on the set of all vertices with minimum eccentricity. A graph G is said to be a self-centered graph if the eccentricity of every vertex is the same; that is, C(G)=G or rad(G)=diam(G). If the eccentricity of every vertex is equal to d, then G is called d-self-centered graph.

For any kind of graph product G of the graphs G1,G2,,Gn, the vertex set is taken as V(G)={(x1,x2,,xn):xiV(Gi)}. Because of their adjacency rules, product names are different. Let x=(x1,,xn) and y=(y1,,yn) be two vertices in V(G). Then the product is called

Cartesian  product, denoted by G=G1G2Gn, where x~y if and only if xiyiE(Gi) for exactly one index i, 1in, and xj=yj for each index ji,

strong  product, denoted by G=G1Gn, where x~y if and only if xiyiE(Gi) or xi=yi, for every i, 1in,

lexicographic  product, denoted by G=G1Gn, where x~y if and only if, for some j{1,2,,n},xjyjE(Gj) and xi=yi for each 1i<j,

co-normal  product, denoted by G=G1G2Gn, where x~y if and only if xi~yi for some i{1,2,,n}.

Self-centered graphs have been broadly studied and surveyed in . In , the authors described several algorithms to construct self-centered graphs. Stanic  proved that the Cartesian product of two self-centered graphs is a self-centered graph. Inductively, one can prove that Cartesian product of n-self-centered graphs is also a self-centered graph.

In this paper, we find conditions for self-centeredness of strong product, co-normal product, and lexicographic product of graphs.

2. Main Results

In this section, we will discuss the self-centeredness of different types of product graphs. As mentioned before, all graphs considered here are simple and connected. The following result is given by Stanic .

Theorem 1.

If G1 and G2 are m- and n-self-centered graphs, respectively, then G1G2 is (m+n)-self-centered graph. Reciprocally, if G1G2 is self-centered, then both graphs G1 and G2 are self-centered.

By method of induction, one can extend the above theorem and get the result given below.

Theorem 2.

Let G=G1G2Gn be the Cartesian product of graphs G1,G2,,Gn. If every Gi is di-self-centered graph, then G is m-self-centered graph, where m=i=1ndi, 1in. Conversely, if G is a self-centered graph, then every Gi is a self-centered graph.

Next we will discuss self-centeredness of strong product of graphs.

Theorem 3.

Let G=G1Gn be the strong product of graphs G1,G2,,Gn. Then G is d-self-centered graph if and only if, for some k{1,,n}, Gk is d-self-centered graph and diam(Gi)d for every i, 1in.

Proof.

For any two vertices x=(x1,,xn) and y=(y1,,yn), the distance between them is given in : (1)dx,y=max1indGixi,yi. Now, the eccentricity of any vertex x of G is given by (2)eccx=maxdx,y:yVG=maxmax1indGixi,yi:yVG,where x=(x1,,xn) and y=(y1,,yn).

First, let Gk be d-self-centered graph for some k{1,2,,n} and diam(Gi)d for all i, 1in. Since Gk is d-self-centered, ecc(xk)=d and there exists some yk in Gk such that d(xk,yk)=d. As diam(Gi)d for all i, 1in, the distance between any two vertices in any Gi cannot exceed d. Hence, ecc(x)=d for all xV(G) and thus G is d-self-centered graph.

Conversely, let G be a d-self-centered graph. If, for some l{1,,n}, diam(Gl)=dl>d, then there exist vertices xl and yl in Gl such that d(xl,yl)=dl. Now for x=(x1,,xl,,xn) and y=(y1,,yl,,yn) in V(G), d(x,y)d(xl,yl)=dl>d and so ecc(x)dl>d. This contradicts the fact that G is d-self-centered graph and thus it is proven that diam(Gi)d for all i. Now, our claim is that there exists k{1,,n} such that Gk is d-self-centered graph. On the contrary, suppose that none of Gi is d-self-centered graph. Then there exist vertices xiV(Gi) for all i such that ecc(xi)=di<d. Let x=(x1,,xn). Then ecc(x)=max1in{di}<d, which contradicts the fact that G is d-self-centered graph.

In the following lemma, we determine the formula for the distance between two vertices in the co-normal product of a finite number of graphs.

Lemma 4.

Let G=G1G2Gn be the co-normal product of graphs G1,G2,,Gn. The distance between x=(x1,,xn) and y=(y1,,yn) in G is(3)dx,y=1if  xi~yi  for  some  i1,2,,ndxi,yiif  Gj=K1,  ji2if  xy,  xlyl  for  exactly  one  index  l  and  GjK1  for  some  jl2if  xy  and    at  least  two  indicesk,l  s.t.  xkyk  and  xlyl.

Proof.

Consider two vertices x=(x1,,xn) and y=(y1,,yn) of G. If, for some i{1,2,,n}, xi~yi, then by the definition of co-normal product x~y and thus d(x,y)=1.

Next, let Gj=K1 for all ji. In this case, for any path P between x and y, every adjacent pair of vertices in P differ only in the ith coordinate. So d(x,y)=d(xi,yi). For the third option of the distance formula, we have vertices x and y as x=(x1,x2,,xl,,xj,,xn) and y=(x1,x2,,yl,,xj,,xn) such that xlyl and GjK1 for some jl. Since Gj is connected graph, there exists a vertex zjV(Gj) such that xj~zj and thus we get a vertex z=(x1,x2,,xl,,zj,,xn)G such that x~z and z~y (because xj=yj) and xzy is a path of length two and hence d(x,y)=2.

Finally, consider the case, where, for at least two indices k and l, xkyk and xlyl; that is, for at least two indices k and l, GkK1 and GlK1. Since xy, xkyk, and xlyl, then from the connectivity of graphs Gk and Gl there exist vertices zkV(Gk) and zlV(Gl) such that zk~xk in Gk and zl~yl in Gl. Then we have a vertex z=(x1,,zk,,zl,,xn)V(G) such that x~z and z~y. Thus xzy will be an x-y path of length two and this proves that d(x,y)=2.

The following theorem gives necessary and sufficient conditions for a co-normal product of graphs to be a self-centered graph.

Theorem 5.

Let G=G1G2Gn be the co-normal product of graphs G1,G2,,Gn with VGi=ni. Then the following hold:

Let GiK1 and Gj=K1 for all ji. Then G is d-self-centered graph if and only if Gi is d-self-centered graph.

Let there be at least two values of i such that GiK1. Then G is 2-self-centered graph if and only if there exists an index l such that Δ(Gl)nl-1, where Δ(G) is the maximum degree of a vertex in G.

Proof.

(i) The result is true because G is isomorphic to Gi in this case through the isomorphism (4)f:VGVGi with f(x1,,xi,,xn)=xi.

(ii) Let G be a 2-self-centered graph. If, for all the indices i, Δ(Gi)=ni-1, then there are vertices xiV(Gi), 1in, such that deg(xi)=ni-1. Now, the vertex x=(x1,x2,,xn), ecc(x)=1, which contradicts the fact that G is 2-self-centered graph. Hence there exists an index l such that Δ(Gl)nl-1.

Conversely, let there be an index l such that Δ(Gl)nl-1. Then for any vertex x=(x1,x2,,xl,xl+1,,xn) in G there exists another vertex y=(x1,x2,,yl,xl+1,,xn), where ylV(Gl) and xlyl. Since xy, from the third option of the distance formula given in Lemma 4, ecc(x)=2. Since x is an arbitrary vertex, G is 2-self-centered graph.

In the following two theorems, we discuss self-centeredness of lexicographic product of graphs.

Theorem 6.

Let G=G1G2Gn be the lexicographic product of graphs G1,G2,,Gn and let k1 be the smallest index for which GkK1. If Gk is d-self-centered graph, where d2, then G is d-self-centered graph. The converse is true for d3.

Proof.

For vertices x=(x1,,xn) and y=(y1,,yn) of G, the following distance formula is due to Hammack et al. :(5)dx,y=dG1x1,y1if  x1y1dGixi,yiif  dGlxl=0  1l<imindGixi,yi,2if  dGlxl0  for  some  1l<i,where i is the smallest index for which xiyi.

Let VGi=1 for i=1,2,,k-1 and let Gk be d-self-centered graph, where d2. First let k=1. Since VG1>1,  G1 is connected and degree of no vertex in G1 is zero; then the second option in the distance formula will not arise. Then the above formula to calculate the distance reduces to(6)dx,y=dG1x1,y1if  i=1mindGixi,yi,2ifi2,where i is the smallest index for which xiyi. For i2, let r=min{dGi(xi,yi),2}. Then r2. Since d2, we get rd. Now, for xV(G),(7)eccx=maxdx,y:yVG=maxdG1x1,y1,r:y1VG1=d,because ecc(x1)=d and there exists y1G1 such that d(x1,y1)=d. This proves that ecc(x)=d for all xV(G) and hence G is a d-self-centered graph.

Next, let k>1. Since VG1=1, there is no y1G1 such that x1y1. So, first option in the distance formula will not arise. Since the degree of the vertex in Gj for j=1,2,,k-1 is zero, if i=k in the above distance formula then d(x,y)=dGk(xk,yk). Since GkK1 and is connected deg(xk)0. So if ik+1 in the above formula, d(x,y)=min{dGi(xi,yi),2} and thus the above formula to calculate the distance reduces to(8)dx,y=dGkxk,ykif  i=kmindGixi,yi,2if  ik+1,where i is the smallest index for which xiyi. For ik+1 let r1=min{dGi(xi,yi),2}. Then r12. Since d2, we get r1d. Thus, for any vertex xV(G), we have(9)eccx=maxdx,y:yVG=maxdGkxk,yk,r1:ykVGk=d.This proves that ecc(x)=d for all xV(G) and hence G is a d-self-centered graph.

Conversely, let G be a d-self-centered graph, where d3. Then ecc(x)=d for all xV(G). Notice that, for any vertex x=(x1,x2,,xn) in G,(10)eccx=maxdx,y:yVG=maxeccx1,r:x1VG1if  k=1maxeccxk,r1:xkVGkif  k>1,where r and r1 are as defined above. Since ecc(x) (which is the maximum of ecc(xk) and r or r1) is equal to d, d3 and r,r12, we get ecc(xk)=d for all xkV(Gk). So Gk is d-self-centered graph.

If we take d=2, then ecc(x)=2 may not imply that ecc(xk)=2 (there may be ecc(xk)<2 and r or r1 is equal to 2; see example below).

Example 7.

Here we consider the lexicographic product of three graphs, G1, G2, and G3, where G1=K2, G2=P4, and G3=K2. Let V(G1)={x,y}, V(G2)={a,b,c,d}, and V(G3)={1,2}. The lexicographic product G=K2P4K2 of graphs K2, P4, and K2 is shown in Figure 1. One can check that the eccentricity of every vertex of G is two and hence G is a 2-self-centered graph. However, G1 is not a 2-self-centered graph.

In the theorem below, we present the general version of the 2-self-centered product graphs included in the previous example.

Theorem 8.

Let G=G1G2Gn be the lexicographic product of graphs G1,G2,,Gn with VGi=ni, let Gk be 1-self-centered graph for some k{1,,n-1}, and let Gi (if it exists) be K1 for all i<k. Then G is a 2-self-centered graph if and only if Δ(Gj)nj-1 for some jk+1.

Proof.

First let G be a 2-self-centered graph. It is given that, for some k{1,,n-1}, Gk is 1-self-centered graph and let Gi be K1 for all i<k. Our claim is that Δ(Gj)nj-1 for some jk+1. On the contrary, let Δ(Gj)=nj-1 for all jk+1. Then there are vertices giGi such that ecc(gi)=1 for every i, kin. Now, by using above distance formula, for every x=(x1,,xk-1,gk,,gn) in G, one gets ecc(x)=1. This contradicts the fact that G is a 2-self-centered graph.

Conversely, let Δ(Gl)nl-1 for some lk+1. Then for any vertex xlGl there exists ylGl such that xlyl. For any vertex x=(x1,,xk,,xl,,xn) there exists a vertex y=(x1,,xk,,yl,,xn) such that xy. So, ecc(x)2. Since Gi=K1 for all i<k (if any), the distance formula will be(11)dx,y=dGixi,yiif  i=kmindGixi,yi,2if  ik+1,where i is the smallest index for which xiyi. Since Gk is a 1-self-centered graph, dGi(xi,yi)=1 if i=k. Also, for ik+1, min{dGi(xi,yi),2}2. Thus eccentricity of no vertex is more than two and we get ecc(x)=2 for every xG. Hence G is a 2-self-centered graph.

Competing Interests

The authors declare that there are no competing interests regarding the publication of this paper.

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