The problem of finding one optimal location for schools, drug stores, police stations, and hospitals requires facilities to be placed near the users in order to minimize, for example, the distance traveled to reach them. Location theory deals with this type of optimization problem. Location functions such as the median, the center, and the mean have been used to solve these type of problems. On the other hand, there are circumstances where placing one or more facilities as far as possible from the users is the best solution. For instance, it is necessary to locate nuclear power plants far from cities or towns to minimize the risk of radiation problems. Similar problems include the determination of suitable locations for observatories, radio stations, airports, and chemical plants. The solution to the problem of finding an optimal location for these types of obnoxious facilities on networks has been studied by Church and Garfinkel [1] , Minieka [2] , Ting [3] , and Zelinka [4] . In these investigations two solutions to the problem are given from an algorithmic perspective. The most appealing solution is called the antimedian, the points that maximizes the total distance from the facility to the users. Another solution is the anticenter, the points that maximizes the total distance from facilities to users. For more information about obnoxious facilities the reader is remitted to [5] -[7] . In the case of tree graphs, Ting [3] published a linear algorithm to find the antimedian of a tree, and Zelinka [4] proved that the set of leaves of a tree contains an antimedian. This problem can be approached through the axiomatization of location functions. The input of a location function consists of some information with respect to the users of the facilities, and the output is related to the consensus reached based on the given information. The rationality of this process is supported by the fact that location functions must satisfy a number of consensus axioms. The mean function on tree graphs was the first location function studied from the axiomatic point of view by Holzman [8] in the continuous case (in the continuous case a tree contains an infinity number of elements, the edges of the tree are considered to be rectifiable curves, and a profile and its members are allowed to be located anywhere on edges). After that Vohra [9] also characterized the median function in the continuous case; in addition the reader can see [10] . In the discrete case, the center, the median, and the mean function have been characterized axiomatically on trees (see for example [11] -[20] ). Not much research has been done with respect to the axiomatic characterization of obnoxious location functions, but recently Balakrishnan et al. [21] published a characterization of the antimedian function on paths. In what follows, we present a different axiomatic characterization, also on paths, of the antimedian function. Ortega and Wang have recently sent for publication an axiomatic characterization of the antimean function on paths. For more information about location theory and axiomatization we refer the reader to the following references [22] -[27] .

Let be a finite metric space and set

where is the cartesian product of. The elements of are called and usually denoted by. Location theory and consensus theory are related to solve the following problem: Given a collection of users (voters, customers, clients, etc.) with each user having a preferred location point in, one attempts to find a set of elements of that satisfy the preferences of the users with respect to some well-defined criteria. Modeling this situation requires the use of a location function on, which is a function, where denotes the set of all subsets of. Three well known examples of location functions are:

a) the center function, denoted by Cen, and defined as

where.

b) the median function, denoted by Med, and defined as

where.

c) the mean function, denoted by Mean, and defined as

where.

We are interested in finite metric spaces defined in terms of connected graphs. Let be a finite connected graph, and let be the usual distance on, where is the length of a shortest path between and. It is well known that is a metric space, and observe that a profile in a graph is simply a sequence of vertices where repetitions are allowed. We will investigate some properties of the antimedian function on finite metric spaces defined in terms of a very special type of connected graphs, namely paths.

In this section or will denote a path of length. We will label the vertices of as and assume that the order that the vertices have in the path is given by the order of the numbers. Hence, will be represented as

Notice that the set of vertices is and also that vertex is adjacent to vertex, vertex is adjacent to vertex and so on. In the case has an even number of vertices, we will write. In the case has an odd number of vertices, we will write. Let be a profile on; for any we define the status of with respect to to be the number

A vertex is called an antimedian of if

The antimedian of is the set

In order to study the antimedian function on, we will divide the paths in two classes. The set of paths that have an odd number of vertices will be called odd paths, and the set of paths with an even number of vertices will be called even paths. Let be a profile on, the notation

will indicate that there is such that. We also use to denote the set of all the different vertices included in, and the number of vertices in the profile counting repetitions is denoted by.

For example consider the profile on with. In this case, , and.

If and are profiles on, denote by the profile

The profile is called the concatenation of and. The following result related to the antimedian function has been proved in [21] .

Lemma 1 Let and be profiles on. If, then

The definition of the antimedian function implies the following characteristic of this function.

Lemma 2 Let be a profile on, and let be any permutation of. Then

where

The median function on finite tree graphs satisfies the following property that was proved in [13] , and will be important in the proof of several results.

Lemma 3 Let be a profile on a finite tree. If and if is a path contained in such that, then

The property of the median function described by Lemma 3 will be called the increasing status property.

Lemma 4 Let be a profile on a path of length. Then

a) implies,

b) implies.

Proof. Notice first that a path is also a tree; consequently, we can apply to the increasing status property. We first obtain the set; if for some, then we define the paths

and

By the increasing status property we have

and

Observe that

if, then, and

if, then.

On the other hand, assume. Define the paths

and

By the increasing status property we have

and

If, then, and

if, then.

We say that a profile on is of the form for some integer, if contains exactly times the vertices and. For example the profile is of the form. Profiles of the form are special for the antimedian functions because.

Lemma 5 Let be a profile the form for some integer on a path of length. Then.

Proof. It is well known that if is a profile on a finite tree, the median of consists of all the vertices in the path

from to. This implies that

Since a path is also a tree, and if, then we have

and the definition of the antimedian function implies that. Because is of the form, then, and we can reorder the vertices of to define the profile

By Lemmas 1 and 2 we obtain

The next result characterizes profiles on a paths of length that satisfy the condition, and that are not of the form.

Lemma 6 Let be a path of length, and let be a profile that is not of the form

for some integer. If, then.

Proof. Since is a tree we can apply the increasing status property. We start by obtaining the set. If for some, then we define the paths

and

By the increasing status property we obtain

and

Observe that:

if, then. On the other hand, assume. Define the paths

and

By the increasing status property we have

and

Note that implies.

From Lemmas 4, 5, and 6 we obtain the following important result that characterizes the output of the antimedian function on paths of length.

Lemma 7 If is a profile on a path of length, then

Assume

is a path of length. If and, then. Similarly, if and, then. Denote by the set of vertices such that; similarly we define the set. Using the sets

we define a partition of the profile as follows: denote by the profile such that and each vertex in is included in the profile as many times it appears in. In a similar way we define the profile using the set. Notice that can be seen as the concatenation of profiles and, in other words. The following number

will play an important role in the following sections.

In this section

represents a path such that, and note that in this case. Let be a profile on; we will use to define a new profile that will be denoted. This profile contains the vertex repeated times. In other words we are assuming that for all. So, is the profile

We want to establish a relationship between and. From the definition of profiles and we derive the identities

and

Using (3), (4), and the definitions of and, we get

In terms of, defined by (2), and we deduce the following relation for

The next result is corollary to the definition of the number.

Corollary 1 If is a profile on, then if and only if.

The definition of and implies

These relations imply

The definition of and the fact that indicate

The following three lemmas establish an important relationship between the numbers, , and. These results will be used to characterize the antimedian of profiles on.

Lemma 8 If is a profile on, then if and only if.

Proof. Assume first, and notice

This implies.

Because of (5) and the fact that, we obtain

Replacing the equal sign with and in the proof of Lemma 8, we obtain the next two results.

Lemma 9 If is a profile on, then if and only if.

Lemma 10 If is a profile on, then if and only if.

We end this section with an important result that characterizes the antimedian of a profile on odd paths that is not of the form for some integer.

Lemma 11 Let be a profile on an odd path. If is not of the form for some integer, then

Proof. Assuming and because is not of the form, then Lemma 8 implies, and Lemma 6 proves.

If, then Lemma 10 shows, and Lemma 4 demonstrates.

If, then Lemma 9 shows, and Lemma 4 proves.

In this section

represents a path where; so, we have, and in this case. Let be a

profile on. Using similar ideas as in the last section, we can obtain a relationship between the numbers, , , and. Since the profile contains all the vertices of whose index is less or equal to, then

Using the profile we obtain

From (6) and (7) and the definition of and, we deduce

In terms of, we have the relation

Observe that

Using a similar argument as above, we obtain

The definition of implies the identity

This identity provides the following relation between and.

The next three results show some properties of the numbers, , , and.

Lemma 12 If is a profile on, then if and only if.

Proof. Assume first, and notice that (8) and (9) indicate

Conversely, if, then

By replacing the equal sign with and in the proof of Lemma 12, we obtain the following two results.

Lemma 13 If is a profile on, then if and only if.

Lemma 14 If is a profile on, then if and only if.

The next lemma is an important result because it characterizes the antimedian of profiles, on even paths, that are not of the form for some integer.

Lemma 15 Let be a profile on an even path. If and is not a profile of the form for some integer, then

Proof. Assuming and since is not of the form, Lemma 12 shows,

and Lemma 6 indicates.

If, then Lemma 14 implies. Now Lemma 4 proves.

If, then Lemma 13 shows, and Lemma 4 implies.

The next result is a corollary to Lemma 15.

Corollary 2 Let be a profile on. If is of the form for some integer, then .

Proof. Notice that in this case the profile contains times the vertex, and the profile contains times the vertex. Consequently, we have

Finally, Lemma 15 implies.

The axioms listed below are among the desirable properties that a general location function should satisfy, and it is not difficult to verify that the antimedian function satisfies these properties.

Oddness (O): Let be a location function on a path of length with. Let be defined as in (2); if is not a profile of the form for some integer, then

Evenness (E): Let be a location function on a path of length with. Let be defined as in (2); if is not a profile of the form for some integer, then

Consistency (C): Let be a location function on. If and are profiles and with, then.

Extremeness (Ex): Let be a location function, and be a profile on. If, then .

Generalized Extremeness (G-Ex): Let be a location function, and let be a profile on. If is of the form for some integer, then.

Anonymity (A): For any profile on and any permutation of, we

have, where

Some of these axioms are not independent. For example it is clear that (Ex) is a particular case of (G-Ex) when. Next we prove that if a location function satisfies (C) and (Ex), it also satisfies (G-Ex).

Lemma 16 If is a location function on that satisfies axioms (C), (A), and (Ex), then satisfies axiom (G-Ex).

Proof. Let be a profile on of the form for some integer. Corollary 2 implies . Because of axiom (A), we can reorder the vertices of to obtain a profile of the form. We can express as a concatenation of profiles of the form; in other words. Since satisfies axiom (Ex), then, and applying axiom (C) we conclude

With the axioms listed above we will give two axiomatic characterizations for the antimedian function. The next theorem contains the first of these characterizations.

Theorem 1 Let be a location function on. is the antimedian function on if and only if satisfies axioms (O), (E), (Ex), (C), and (A).

Proof. It is clear that if is the antimedian function, then satisfies axioms (O), (E), (Ex), (C), and (A). Assume now is a location function that satisfies axioms (O), (E), (Ex), (C), and (A). To prove that is the antimedian function we consider three cases.

Case 1. Assume first is a profile of the form for some integer, by Lemma 5 we have. Since satisfies axioms (C), (A), and (Ex), then Lemma 16 proves satisfies (G-Ex) which implies. It is clear that in this case.

Case 2. Assume is a path such that. Let be a profile on that is not of the form for some integer. Notice that, and if, then Lemma 8 shows, and Lemma 6 implies. On the other hand, since satisfies axiom (O) and, then. Therefore, means.

If, then Lemma 10 indicates, and Lemma 4 proves. Since satisfies axiom (O) and, we obtain. From this we conclude that if, then. A similar argument can be used to show that if, then.

Case 3. Assume is a path such that which means, and let be a

profile on that is not of the form for some integer. If, then Lemma 12 demon-

strates, and Lemma 6 proves. Since satisfies axiom (E) and,

we get. Therefore, implies.

If, then Lemma 13 indicates, and Lemma 4 shows. Because satisfies axiom (E) and,. Hence, when is a profile that is not of the form and,. A similar argument can be used to show that if, then

. Notice that Cases 1, 2, and 3 demonstrate the theorem.

We leave it to the reader to prove that the axioms used in the proof of Theorem 1 are independent. Notice that in the proof of Theorem 1 we needed to use three axioms to establish Case 1. We want to improve the demonstration of this result using fewer axioms. We achieve this objective using axiom (G-Ex) in the following theorem which is our main result.

Theorem 2 Let be a location function on. is the antimedian function on if and only if satisfies axioms (O), (E), and (G-Ex).

Proof. It is clear that if is the antimedian function, then satisfies axioms (O), (E), and (G-Ex). Assume now that is a location function that satisfies axioms (O), (E), and (G-Ex). To prove that is the antimedian function we consider three cases.

Case 1. Assume first is a profile of the form, then by Lemma 5 we obtain . Because satisfies axiom (G-Ex), we have. So in this case.

Case 2. Assume is a path such that which means. Let be a profile on that is not of the form. If, then Lemma 8 indicates, and Lemma 6 implies. Since satisfies axiom (O) and,. Therefore, indicates. A similar argument can be employed to show that if, then , and if, then.

Case 3. Assume is a path such that, and let be a profile on. Notice that.

If, then Lemmas 12 implies, and Lemma 6 shows. Because satisfies axiom (E), we conclude. Therefore, implies.

A similarly argument can be used to demonstrate that if, then, and if, then. It is clear that Cases 1, 2, and 3 finish the proof of the theorem.

Notice that the definition of axioms (O), (E), and (G-Ex) indicate that they are independent. So it is not necessary to add a proof for the independence of these three axioms. More research is needed to find an axiomatic characterization of the antimedian function on tree graphs.