On Resolvability of a Graph Associated to a Finite Vector Space
Abstract
The metric dimension of non-component graph, associated to a finite vector space, is determined. It is proved that the exchange property holds for resolving sets of the graph, except a special case. Some results are also related to an intersection graph.
AMS Subject Classification Number: 05C12, 05C25, 05C62
Keywords: Graph, vector space, metric dimension, resolving set, exchange property
1 Introduction
Let be a vector space over a field with a basis . A vector is expressed, uniquely, as a
linear combination of the form . A graph called Non-Zero
Component graph is associated, by Angsuman Das in [11], to a finite dimensional vector space in the following way:
the vertex set of the graph is the non-zero vectors and two vertices are joined by an edge if they share at least one
with non-zero coefficient in their unique linear combination with respect to . It is proved in [11] that is independent of choice of basis, i.e., isomorphic non-zero component graphs are obtained for two different bases. Other basic properties and results about can be found in [11, 12], for example
Theorem 1.1 ([12], Theoerem 3.6).
If be a -dimensional vector space over a finite field with elements, then the order of is and the size of is
Theorem 1.2 ([11], Theoerem 4.2).
is complete if and only if is a -dimensional vector space.
The study of graphs associated to algebraic structures has, recently, got much attention from researchers to expose the relation between algebra and graph theory: zero divisor graph associated to a commutative ring was discussed in [1, 3]; commuting graphs for groups were studied in [2, 4, 17]; power graphs for groups and semigroups were focused in [5, 8, 21]; the papers [13, 25] are devoted to investigate intersection graphs assigned to a vector space; and so on.
In this paper, we study the metric dimension and the exchange property for resolving sets of of a vector space . The vector space under consideration is of dimension and defined over a finite field having elements.
2 Metric Dimension
Let be a finite simple connected graph. The distance between two vertices is the length of a shortest path between them. Let be an ordered set of vertices of and let be a vertex of . The representation of with respect to is the -tuple . is called a resolving set [10] or locating set [22] if every vertex of is uniquely identified by its distances from the vertices of , or equivalently, if distinct vertices of have distinct representations with respect to . Two vertices and are said to be resolved by a vertex if . A resolving set of minimum cardinality is called a metric basis for and this cardinality is the metric dimension of (see [7, 14, 16, 23, 24]). A resolving set of cardinality is called minimal resolving set if it does not contain a resolving set of cardinality as a proper subset. For other basic terms and concepts, the reader is referred to [26]. An application of the metric dimension and basis of a graph in robot navigation problem is considered in [18] by S. Khuller.
Given a vertex in a graph , the open neighborhood of in , denoted by , is the set
And the closed neighborhood of in , denoted by , is the set
For two vertices and in a graph , define if or . The relation is an equivalence relation (see [15]). If then and are called twins.
Lemma 2.1 ([15], Corollary 2.4. ).
Suppose are twins in a connected graph and resolves . Then or is in . Moreover, if and , then also resolves .
Proof.
For any two vertices a connected graph . We have = for all . Therefore, no vertex can resolve and , which forces either or . Moreover, if two vertices are resolved by then they are also resolved by . Hence, also resolves . ∎
In order avoid confusion, we denote a basis of a vector space by -basis and a minimum resolving set by -basis. We denote the dimension of by and the metric dimension of by . The elements of the set represent non-zero vectors of the vector space and vertices of the graph at the same time.
Definition 2.2 ([12]).
Skeleton of a vector is the set of ’s with non-zero coefficients in the basic representation of with respect to -basis
A vector is not unique for a skeleton , i.e., two distinct vectors may have the same skeleton. The length of a skeleton is the number of vectors of -basis it contains.
For two vertices and in , we define the equivalence relation if . Let denote the equivalence class containing .
The following lemma plays a key role in the proof of our main theorems.
Lemma 2.3.
In , the two equivalence relations and are equivalent, i.e., both relations bring the same partition to the set of vertices of .
Proof.
Let . Then, and have the same neighborhood and we need to show that Suppose on contrary that and have different skeletons. Without loss generality, there exists some and . Consequently, the vertex is adjacent to and not adjacent to in . It contradicts that and have the same neighborhood. Conversely, let then, by definition of the graph , . ∎
Theorem 2.4.
Let be a vector space over a field of elements
with as -basis:
a) if and then
;
b) if and then
; and
c) if then
.
Proof.
a) The two vertices and can be resolved by the vertex , hence -basis
b) For , a class for all is of length , i.e., and, by Lemma 2.3, all vertices are singleton twins.
Any two vertices, being singleton twins, can be resolved by some -basis. Therefore, the -basis is a resolving set. To show that this -basis is a -basis, we need to show that
Let be the vertex whose skeleton is , i.e., and with the property that -basis for .
Case1. When -basis. The vertex and a vertex can only be resolved by , because for all . Therefore, at least, one vertex from each pair must be part of the -basis. There are such pairs, hence
Case2. When -basis. Any two vertices can only be resolved either by or by , because for . Therefore, at least, one vertex from each pair of pairs, of type , must be part of the -basis along with the vertex and hence the result follows.
c) Since , so for all and, by Lemma 2.3, there is no singleton twin. By Lemma 2.1, a resolving set must contain all vertices, except one, in every class of length for ; there are vertices in a class of length (there are choices and places); and for a fixed , there are distinct classes. Therefore, any resolving set contains, at least, vertices. There are distinct classes of length and, by Lemma 2.1, contains the set , where . Any two vertices having different skeletons can be resolved by some vertex in the set , hence the minimum cardinality of is .
∎
Corollary 2.5.
For , any -basis contains a -basis.
Proof.
The set in the proof part of Theorem 2.4 is a -basis. ∎
For the above corollary is not true. For example, let the set is a -basis, but not a -basis.
3 Exchange Property
Resolving sets are said to have the exchange property in a graph if whenever and are minimal resolving sets for and , then there exists so that is also a minimal resolving set (see [6]). If a graph has the exchange property, then every minimal resolving set for has the same size and algorithmic methods for finding the metric dimension of are more feasible. Similarly exchange property for dominating sets of a graph was also introduced in [6]. In this paper, the exchange property of a graph always means the property for minimal resolving sets.
For a vector space , the exchange property holds for -bases. Resolving sets behave like -bases in a vector space , i.e., each vertex in the graph can be uniquely identified relative to the vertices of a resolving set. But, unlike -bases in vector spaces, resolving sets do not always have the exchange property. To show the exchange property does not hold in a graph, it is sufficient to show that there exist two minimal resolving sets of different size. However, the condition is not necessary, i.e., the exchange property does not hold does not imply that there are minimal resolving sets of different size. Results about the exchange property for different graphs can be found in the literature, for example
Theorem 3.1 ([6], Theorem 3).
The exchange property holds for resolving sets in trees.
Theorem 3.2 ([6], Theorem 7).
For , the exchange property does not hold in wheels .
Another sufficient condition, for a graph not to have the exchange property, can be stated as
Lemma 3.3.
If there is a minimal resolving set of size in a graph . Then, the exchange property does not hold in .
Proof.
Since a minimum resolving set is also minimal so there are two minimal resolving sets of different size (one of size and other is of size ). ∎
Lemma 3.4.
For and , the set is a minimal resolving set in .
Proof.
Any vertex is either or of the form , where Let . Then, there exist some for satisfying , i.e., the representations and are different. Hence, is a resolving set. The set is minimal resolving set because no vertex in can resolve the vertices and ∎
Theorem 3.5.
For and the exchange property does not hold in .
Proof.
Note that, for and , the exchange property holds in as there are only two minimal resolving sets and
Theorem 3.6.
For , the exchange property holds in the graph .
Proof.
For , there is no class of cardinality for all . By Lemma 2.3, there is no singleton twin in the graph . Let be two minimal resolving sets. Let . Now, if then is, obviously, a minimal resolving set. Suppose . Now, since is always connected and has no singleton twin. Therefore, by lemma 2.1, there exists such that . Consequently, again by lemma 2.1, is a resolving set. Since for and for all . Therefore, two vertices are resolved by the vertex if and only if they are resolved by the vertex . Therefore, is minimal. ∎
4 Intersection Graph of for
Let be the collection of sets. The intersection graph of
is the graph whose vertices are
the sets in and two vertices are connected by an edge
if the corresponding sets intersect. Symbolically
and (for details see, [20]). The following theorem reflects the importance of intersection graphs which states that every graph can be represented as an intersection graph.
Theorem 4.1 ([19]).
Let be an arbitrary graph. Then, there is a set and a family of subsets of which can be put into one-one correspondence with the vertices of in such a way that the set of edges is .
The case, where the field has two elements, i.e., . Then, the graph of a vector space can be represented by the intersection graph for , where and the power set of . This representation can be established by identifying the vector with the subset of .
Corollary 4.2.
For , where
a) and
b) holds the exchange property if and only if
5 Conclusion
In this paper, we find the metric dimension of the graph . We study the relation between a metric basis -basis and a vector space basis -basis. We proved that, except the case where the corresponding field has elements and dimension of is , a -basis has the exchange property like a -basis. We also concluded that a -basis contains a -basis if the corresponding field has or more elements.
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