Research Article | Open Access
On Self-Centeredness of Product of Graphs
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.
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 , 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 -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 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 : 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. :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.
The authors declare that there are no competing interests regarding the publication of this paper.
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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.