Abstract

A profile is a finite sequence of vertices of a graph. The set of all vertices of the graph which minimises the sum of the distances to the vertices of the profile is the median of the profile. Any subset of the vertex set such that it is the median of some profile is called a median set. The number of median sets of a graph is defined to be the median number of the graph. In this paper, we identify the median sets of various classes of graphs such as , for , and wheel graph and so forth. The median numbers of these graphs and hypercubes are found out, and an upper bound for the median number of even cycles is established. We also express the median number of a product graph in terms of the median number of their factors.

1. Introduction

We consider only nonempty finite simple undirected connected graphs. For the graph , , and denote its vertex set and edge set, respectively. When the underlying graph is obvious, we will use and for and , respectively. A finite sequence of vertices is called a profile. For the profile and , the remoteness is [1]. The set of all vertices for which is minimum is the median of in and is denoted by . A set such that for some profile is called a Median set of . The Interval between vertices and of consists of all vertices which lie in some shortest path between and . The number of intervals of a graph is denoted by in . The Hypercube is the graph with vertex set , two vertices being adjacent if they differ exactly in one coordinate. A Subcube of the hypercube is an induced subgraph of , isomorphic to for some . A graph is a Median graph if, for every , contains a unique vertex, see Avann [2]. Trees, hypercubes, and grid graphs are all median graphs. For further references concerning median graphs, see Mulder and Schrijver [3], Mulder [4], and Bandelt and Hedlíková [5]. The graph on vertices formed by joining all the vertices of a -cycle to a vertex is a wheel graph and is denoted by . The vertex which is joined to all the vertices of the -cycle is the universal vertex. The Cartesian product of two graphs and has vertex set , two vertices and being adjacent if either and or and . The eccentricity of a vertex is . A vertex is an eccentric vertex of if . Graph products form a well-studied area in graph theory, see [6, 7]. The problem of finding the median of a profile is very much significant in location theory, particularly in efficiency-oriented model [8], as it corresponds to finding the most desirable service point for a group of customers when the minimisation of the total cost is a primary objective. Here we use profiles of vertices instead of a set of vertices which gives us the liberty to repeat a vertex any number of times. This makes the problem more relevant in location theory as it enables us to assign different weights to different customers. The problem of finding the median of a profile has been thoroughly investigated by many authors; see [914]. The objective of this paper is to identify median sets of some classes of graphs and to enumerate them.

2. Median Number

On the set of all profiles on , , we define a relation by if . This is an equivalence relation. This equivalence relation gives a partition of the whole set of profiles. The number of equivalence classes in this partition is defined as the Median number of graph and is denoted by . That is, it is the number of distinct median sets in .

Proposition 2.1. For any graph on vertices, .

Proof. The upper bound is obvious as it is the number of nonempty subsets of the vertex set. For every , is a median set of the profile . For every , the set is the median set of the profile . Therefore .

Proposition 2.2. , where is the complete graph on vertices.

Proof. In , each nonempty subset of the vertex set is a median set, namely, of the profile formed by taking all the elements of the set exactly once. Therefore the number of distinct median sets is the number of nonempty subsets of which is .

Proposition 2.3. If is an edge of , , .

Proof. Let . For every vertex set such that , there exists a profile which has as its median set, namely, the profile formed by taking the vertices of exactly once. Let be a profile which does not simultaneously contain and . Then is a subset of the set of vertices corresponding to the profile and hence does not contain and . Now, let be profile which contains both and . Then if or is repeated more than the other in the profile, then and so they cannot appear together in the . Assume that contains both and where both are repeated the same number of times. Let the profile be , . For , , . Also, and . Therefore does not contains both and . Now the profile has as its median set. Hence is the only median set which contains both and . Therefore the class of all median sets of the graph consists of and all subsets of which do not simultaneously contain and . Hence,

Proposition 2.4 (Bandelt and Barthélémy [10]). Let be a median graph. For any profile in the median set is an interval in .

Proposition 2.5. The median number of a tree on vertices is .

Proof. Since is a median graph, by the above proposition all median sets are intervals. As observed in the proof of Proposition 2.1, all intervals are median sets. Therefore, class of median sets of is precisely the class of intervals of which is the class of all paths in . Hence the median number is the number of distinct paths in which is .

Proposition 2.6 (Imrich and Klavžar [6]). Let be a hypercube. Then, for any pair of vertices the subgraph induced by the interval is a hypercube of dimension .

Theorem 2.7. For the hypercube ,

Proof. Since is a median graph, by Propositions 2.4 and 2.6, every median set of is a subcube. Also in any graph , is the median set of the profile , where . Thus in a hypercube every subcube is a median set. Therefore, the median sets of are precisely the induced subcubes. So the Median number of is the number of subcubes of . Every vertex of contains coordinates where each coordinate is either or . Keeping co-ordinates fixed and varying and over the other positions, we get a subcube of dimension . By varying 's and 's over these positions we get such subcubes. The positions, to be fixed can be chosen in ways. So the total number of subcubes of dimension is . Therefore the total number of subcubes of is .

Theorem 2.8. For the wheel graph , , .

Proof. Let be the vertex set of with as the universal vertex, and let be the cycle . Each singleton set , , is a median set. The sets , where and are adjacent, are also median sets. The profile , ( means addition modulo ) has as median set. The set is the median set of the profile . Let be a profile of which contains the universal vertex . Then since contains the vertex , . If some , , belongs to , then and this implies at least for some . Also, the number of 's with is less than the number of repetitions of in . Let be such that . Then belongs to that implies number of repetitions of is greater than the number of repetitions of in the profile . But these two statements are contradictory. Thus for a profile which contains the universal vertex the median set cannot contain two vertices which are at distance 2. Hence the only possible median sets for such a profile are (i)sets of type ,(ii)sets of type ,(iii)sets of type .
Now, let be a profile which does not contain . Then . If some , , belongs to , then . Let be such that and . Then . since , the number of zeroes in number of twos in . Similarly, number of zeroes in number of twos in .
Thus, number of repetitions of in = number of repetitions of in .
Now, let . Without loss of generality, we may assume that . If some vertex other than belongs to , then . If , then . Therefore can only be , where and are repeated the same number of times. Since , we have and for all other , . Hence . If then some vertex other than and that belong to will contradict the fact that and belong to . Therefore, in this case also , where and are repeated the same number of times. Here , , and for all other , . In other words, .
Hence the only median sets are(1),(2),(3),(4),(5) (since , such vertices do exist),(6).
Thus, .

Theorem 2.9. , where is the complete bipartite graph with , .

Proof. Let be a bipartition of with and . Let and . Let be a -element subset of with . Without loss of generality, we may assume that .
If , take . For each , , . For each , , . For each , , . Therefore, .
If , then has median set . Therefore, every subset of is a median set.
Now, let with .
If then as in the previous case has median set .
Now, let and let be the profile , where each is repeated the same number of times, (say) .
For each , , , for each , , , and for each , , . Moreover, . That is, if each is repeated times where , then .
Now, let , , .
Take , where each is repeated times and is repeated times.
For each , , , for each , , , for each , , , and for each , , . Any , or , cannot be in ).
Now . Hence for any and such that , the profile , where each is repeated times and is repeated times, has as its median. Therefore, every nonempty subset of is a median set. Hence, .

Lemma 2.10. Let and be profiles in the graphs and , respectively. If and , then is a profile in the graph with .

Proof. Let , , and .
Consider .
If , then , see [9]. For any, = .
Let , that is, and . Then, Therefore, Hence, , for all .
Thus, or .
Now, let If for some , This contradicts .
Therefore Hence, and or . That is, .

Theorem 2.11. .

Proof. By the above lemma the product of median sets of and is again a median set of . Now, let be a median set of , with , where . Let , , , and . Let .
We have In other words, , for all .
or .
Let . Then , for all , for  all .
That is Therefore, which implies or . Thus, the class of all median sets of is the same as the class of all Cartesian products of median sets of and .
Hence, .

The above theorem can be used to find the median number of various classes of graphs. The first of the following corollaries gives another technique to find out the Median number of a hypercube.

Corollary 2.12. If are graphs, then .

Corollary 2.13. For the hypercube , .

Proof. Since , .

Corollary 2.14. If is the Grid graph , .

Corollary 2.15. If is the Hamming graph , .

Lemma 2.16. The only median set of the cycle which contains a vertex and its eccentric vertex is .

Proof. Let and be two eccentric vertices of the cycle which belongs to where . Let . Then Hence . Now, suppose . Then there exists an such that . That is, . Let be the eccentric vertex of . . Therefore, . That is, ( times) or , a contradiction. Therefore, any set distinct from which is a median set cannot contain two eccentric vertices. Also, , since and are diametrical and . Hence the only median set of which contains a vertex and its eccentric vertex is .

Theorem 2.17. For the even cycle , .

Proof. Let be the vertex set of with . Let = and does not contain any pair of eccentric vertices}. By the above lemma, the set of all median sets is a subset of . Hence . Let , . Now consists of all subsets of which does not simultaneously contain both the elements from the same , . The number of ways of choosing a -element subset of so that it belongs to is the product of the number of ways of choosing 's from the 's and the number of ways of choosing one element from each of these chosen 's. That is, . Therefore . Hence = .

3. Conclusion

In this paper, we could identify the median set of certain classes of graphs and evaluate their median numbers. Identifying the median sets of more classes of graphs and finding their median numbers will be interesting and challenging. Another possibility is the realisation problem; given a number of finding a graph whose median number is that number.

Acknowledgment

The authors thank the two unknown referees for their valuable comments that helped them in improving the paper.