Abstract
A graph 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 and in a graph , denoted by (or simply ), is the minimum length of path in the graph. The eccentricity of a vertex in , denoted by , is defined as the distance between and a vertex farthest from ; that is, . The radius and diameter of the graph are, respectively, the minimum and maximum eccentricity of the vertices of graph ; that is, and . The center of graph is the induced subgraph of on the set of all vertices with minimum eccentricity. A graph is said to be a self-centered graph if the eccentricity of every vertex is the same; that is, or . If the eccentricity of every vertex is equal to , then is called -self-centered graph.
For any kind of graph product of the graphs , the vertex set is taken as . Because of their adjacency rules, product names are different. Let and be two vertices in . Then the product is called(i)Cartesian  product, denoted by , where if and only if for exactly one index , , and for each index ,(ii)strong  product, denoted by , where if and only if or , for every , ,(iii)lexicographic  product, denoted by , where if and only if, for some and for each ,(iv)co-normal  product, denoted by , where if and only if for some . Self-centered graphs have been broadly studied and surveyed in [1–3]. In [4], the authors described several algorithms to construct self-centered graphs. Stanic [5] proved that the Cartesian product of two self-centered graphs is a self-centered graph. Inductively, one can prove that Cartesian product of -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 [5].
Theorem 1. If and are - and -self-centered graphs, respectively, then is ()-self-centered graph. Reciprocally, if is self-centered, then both graphs and are self-centered.
By method of induction, one can extend the above theorem and get the result given below.
Theorem 2. Let be the Cartesian product of graphs . If every is -self-centered graph, then is -self-centered graph, where , . Conversely, if is a self-centered graph, then every is a self-centered graph.
Next we will discuss self-centeredness of strong product of graphs.
Theorem 3. Let be the strong product of graphs . Then is -self-centered graph if and only if, for some , is -self-centered graph and for every , .
Proof. For any two vertices and , the distance between them is given in [6]: Now, the eccentricity of any vertex of is given by where and .
First, let be -self-centered graph for some and for all , . Since is -self-centered, and there exists some in such that . As for all , , the distance between any two vertices in any cannot exceed . Hence, for all and thus is -self-centered graph.
Conversely, let be a -self-centered graph. If, for some , , then there exist vertices and in such that . Now for and in , and so . This contradicts the fact that is -self-centered graph and thus it is proven that for all . Now, our claim is that there exists such that is -self-centered graph. On the contrary, suppose that none of is -self-centered graph. Then there exist vertices for all such that . Let . Then , which contradicts the fact that is -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 be the co-normal product of graphs . The distance between and in is
Proof. Consider two vertices and of . If, for some , , then by the definition of co-normal product and thus .
Next, let for all . In this case, for any path between and , every adjacent pair of vertices in differ only in the coordinate. So . For the third option of the distance formula, we have vertices and as and such that and for some . Since is connected graph, there exists a vertex such that and thus we get a vertex such that and (because ) and is a path of length two and hence .
Finally, consider the case, where, for at least two indices and , and ; that is, for at least two indices and , and . Since , , and , then from the connectivity of graphs and there exist vertices and such that in and in . Then we have a vertex such that and . Thus will be an - path of length two and this proves that .
The following theorem gives necessary and sufficient conditions for a co-normal product of graphs to be a self-centered graph.
Theorem 5. Let be the co-normal product of graphs with . Then the following hold:(i)Let and for all . Then is -self-centered graph if and only if is -self-centered graph.(ii)Let there be at least two values of such that . Then is 2-self-centered graph if and only if there exists an index such that , where is the maximum degree of a vertex in .
Proof. (i) The result is true because is isomorphic to in this case through the isomorphism with .
(ii) Let be a 2-self-centered graph. If, for all the indices , , then there are vertices , , such that . Now, the vertex , , which contradicts the fact that is 2-self-centered graph. Hence there exists an index such that .
Conversely, let there be an index such that . Then for any vertex in there exists another vertex , where and . Since , from the third option of the distance formula given in Lemma 4, . Since is an arbitrary vertex, is 2-self-centered graph.
In the following two theorems, we discuss self-centeredness of lexicographic product of graphs.
Theorem 6. Let be the lexicographic product of graphs and let be the smallest index for which . If is -self-centered graph, where , then is -self-centered graph. The converse is true for .
Proof. For vertices and of , the following distance formula is due to Hammack et al. [6]:where is the smallest index for which .
Let for and let be -self-centered graph, where . First let . Since ,   is connected and degree of no vertex in is zero; then the second option in the distance formula will not arise. Then the above formula to calculate the distance reduces towhere is the smallest index for which . For , let . Then . Since , we get . Now, for ,because and there exists such that . This proves that for all and hence is a -self-centered graph.
Next, let . Since , there is no such that . So, first option in the distance formula will not arise. Since the degree of the vertex in for is zero, if in the above distance formula then . Since and is connected . So if in the above formula, and thus the above formula to calculate the distance reduces towhere is the smallest index for which . For let . Then . Since , we get . Thus, for any vertex , we haveThis proves that for all and hence is a -self-centered graph.
Conversely, let be a -self-centered graph, where . Then for all . Notice that, for any vertex in ,where and are as defined above. Since (which is the maximum of and or ) is equal to , and , we get for all . So is -self-centered graph.
If we take , then may not imply that (there may be and or is equal to 2; see example below).
Example 7. Here we consider the lexicographic product of three graphs, , , and , where , , and . Let , , and . The lexicographic product of graphs , , and is shown in Figure 1. One can check that the eccentricity of every vertex of is two and hence is a 2-self-centered graph. However, 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 be the lexicographic product of graphs with , let be 1-self-centered graph for some , and let (if it exists) be for all . Then is a 2-self-centered graph if and only if for some .
Proof. First let be a 2-self-centered graph. It is given that, for some , is 1-self-centered graph and let be for all . Our claim is that for some . On the contrary, let for all . Then there are vertices such that for every , . Now, by using above distance formula, for every in , one gets . This contradicts the fact that is a 2-self-centered graph.
Conversely, let for some . Then for any vertex there exists such that . For any vertex there exists a vertex such that . So, . Since for all (if any), the distance formula will bewhere is the smallest index for which . Since is a 1-self-centered graph, if . Also, for , . Thus eccentricity of no vertex is more than two and we get for every . Hence is a 2-self-centered graph.
Competing Interests
The authors declare that there are no competing interests regarding the publication of this paper.