Minimum Partition of an Independence System

Abstract

Graph partitioning has been studied in the discipline between computer science and applied mathematics. It is a technique to distribute the whole graph data as a disjoint subset to a different device. The minimum graph partition problem with respect to an independence system of a graph has been studied in this paper. The considered independence system consists of one of the independent sets defined by Boutin. We solve the minimum partition problem in path graphs, cycle graphs, and wheel graphs. We supply a relation of twin vertices of a graph with its independence system. We see that a maximal independent set is not always a minimal set in some situations. We also provide realizations about the maximum cardinality of a minimum partition of the independence system. Furthermore, we study the comparison of the metric dimension problem of a graph with the minimum partition problem of that graph.

1. Introduction

An abstract idea of representing any objects which are connected to each other in a form of relation is a graph. In this representation, the object is called as a vertex and their relation denotes as an edge. Partition of a graph is the distribution of the whole graph data into disjoint subsets to different devices. The need of distributing huge graph data set is to process data efficiently and faster process of any graph related applications. Where graph partitioning is essential and applicable are given as follows:(1)Complex networks which include biological networks (in solving biological interaction problem in a huge a biological network), social networks (Facebook, Twitter, and LinkedIn etc., and graph partitioning technology is used to process user query efficiently, as replying a query in a distributed manner is very handy and effective) [1], and transportation networks (graph partitioning can speed up and could be effective in planning a route by using a GPS (global positioning system) tool in the digital era).(2)PageRank, which is an application used to compute the rank of web rank from web network.(3)VLSI design: Very large-scale integration (VLSI) system is one of the graph partitioning problems in order to reduce the connection between circuits in designing VLSI. The main objective of this partitioning is to reduce the VLSI design complexity by splitting them into a smaller component.(4)Image processing: Graph partitioning is one of the most attractive tools to split into several components of a picture, where pixels are denoted by vertices and if there are similarities between pixels are represented as edges [2].

Inspired by these interesting applications of graph partition, we consider a graph partition in the context of resolving set of a graph, which is a well-known parameter in graph theory and having remarkable application in network discovery and verification.

A set system is a finite set together with a family of subsets of and is denoted by the pair . A set system is said to be an independence system if for every subset of possessing property , each proper subset of also possesses the property , ., for each such that , for all . Actually, in an independence system , identified with the family of subsets of possessing the property . A subset of which possess the property is said to be an independent set and dependent set otherwise. The chromatic number of is the smallest natural number such that can be partitioned into independent sets and is denoted by . Clearly, a partition of into independent sets of can be identified by a coloring of such that for each color , the color class has the property , and vice versa. The coloring of is called a coloring of . Thus, is the least number of colors required by a coloring of and is also called the chromatic number of [3].

The chromatic number has been extensively studied by various graph theorists. Remarkable work has been done when is or for a graph having vertex set and edge set , and is a hereditary graphical property. For example, if is the property of being a vertex independent set, then is the ordinary chromatic number of ; if is the property of being an edge independent set, then is the edge chromatic number of ; if is the property of being a forest, then is the arboricity of . In the next section, we consider as the property of being a resolving set for and define the chromatic number of associated with an independence system .

2. Independence System

Hereafter, we consider nontrivial, simple, and connected graph with vertex set and edge set . We denote two adjacent vertices and in by and nonadjacent vertices by . The distance is the length of a shortest path between two vertices in the pair and is denoted by . The maximum distance between the vertices of is called the diameter of , denoted by . Two vertices and in are antipodal or diametral if ; otherwise, they are nonantipodal.

Let be a graph. For any vertex of , the metric code or code of with respect to an ordered subset of is defined as

An ordered subset of is a resolving set for if for every pair of vertices . The cardinality of a minimum resolving set for is called the metric dimension of , denoted by or . A resolving set for of cardinality is called a metric basis or a basis of [4–8]. In [9], it was found and, in [6], an explicit construction was given that finding the metric dimension of a graph is NP-hard. The concept of a resolving set, other than graph theory, is applied in many other areas such as coin-weighing problems [10], network discovery and verification [2], strategies for mastermind games [11], pharmaceutical chemistry [12], robot navigation [6], connected joins in graphs and combinatorial optimization [13], and sonar and coast guard Loran [8].

A subset of the vertex set of a graph is an independent set if no proper subset of is a resolving set for . We denote as the property of being an independent set. That is, a subset of possesses the property if and only if is an independent set. This concept was firstly introduced by Boutin and used the term -independent set [14]. For simplicity, we use the term independent set rather than -independent set. A family of subsets of possessing the property is defined as

Thus, we have a set system consisting of those subsets of which are possessing the property and is called the independence system. All the subsets possessing the property may or may not be resolving. This was an error made by Boutin in [14], and we rectified it in [15].

Remark 1. For a set , the following assertions are equivalent:(1)(2) possesses the property (3) is an independent set

Remark 2. Let be a graph with vertex set of order . Then,(1) for each obviously(2), because every subset of is a resolving set for (3)a minimal resolving set for is a maximal independent set, but converse is not always true [15]

2.1. Minimum Partition Problem

For a connected graph and the r-independence system , the minimum partition problem is to make a partition of V into the minimum number of subsets possessing the property .

The least natural number , such that can be partitioned into subsets possessing the property , is called the resolving chromatic number of associated with , denoted by . A coloring of such that for each color , the color class possesses the property is called an coloring of . Thus, is the least number of colors required by an coloring of and is also called the chromatic number of .

Example 1. Let be a graph with and . Then only two colors are needed to properly color , and it follows that the ordinary chromatic number with color classes and . The metric dimension of is 2 and two nonantipodal vertices of form a basis of [4]. Accordingly, no element subset of possess the property and so the independence system isThe minimum partition of according to coloring of consists of two element subsets of from , and hence, .

In the above example, we obtained that the chromatic number and chromatic number of a graph are same. But, it is not necessary that these numbers are always same.

Example 2. Let be a graph with and . Then only two colors are needed to properly color , and it follows that the ordinary chromatic number with color classes and . The metric dimension of is 1, and are the only two bases of [4, 6]. Accordingly, each element subset of is a resolving set for , and so no element subset of possess the property . Thus, the independence system isThe minimum partition of according to coloring of consists of two bases sets and the set from . Hence, .

It is observed, from Examples 1 and 2, that . But, it is not true generally as, in the next example, we have a have a graph such that .

Example 3. Let be a graph with and . Then, three colors are required to properly color , and it follows that the ordinary chromatic number with color classes and . The metric dimension of is 2 and any two vertices of can form a basis of [4]. Accordingly, the element set does not possess the property and so the independence system isThe minimum partition of according to coloring of consists of one singleton set and one element set from , and hence .

Example 4. Let be a graph with and . Then two colors are required to properly color , and it follows that the ordinary chromatic number with color classes and . The metric dimension of is 2 and is a set of basis of [4]. But the set of three elements which is not resolving set of possess the property and so the independence system isThe minimum partition of according to coloring of consists of one element set and one element set from , and hence .

3. Three Well-Known Families

In this section, we consider families of path graphs, cycle graphs, and wheel graphs and solve the minimum partition problem for each family.

3.1. Path Graphs

A path graph , for , has vertex set and edge set . The following result describes which subset of the vertex set of a path graph possesses the property .

Lemma 1. No element subset of the vertex set of a path graph possess the property .

Proof. As and only each end vertex of forms a basis of [6], every element subset of is a resolving set for . Consequently, no element subset of possess the property .

The next result investigates the number of subsets of the vertex set of a path graph possessing the property .

Lemma 2. For all , if is a path graph with vertex set , then .

Proof. According to Lemma 1, each singleton subset of as well as each element subset of possesses the property . It follows that for , , and for all ,where denotes the collection of all the and element subsets of . Hence,

The following result solves the minimum partition problem for a path graph.

Theorem 1. For all , the vertex set of a path graph can be partitioned into minimum number of subsets possessing the property .

Proof. Let the color classes, due to a coloring of , are When is even, then and ;  When is odd, then and ; In both cases, all these color classes are lying in , by Lemma 2. It follows that is an coloring of , and these color classes define a partition of into the sets possessing the property . Further, any partition of of cardinality less than contains at least one element or element subset of such that . Thus, a minimum partition of has subsets of possessing the property .

3.2. Cycle Graphs

A cycle graph , for , has vertex set and edge set . In the next result, we investigate which subset of the vertex set of a cycle graph possesses the property is.

Lemma 3. A subset of the vertex set of a cycle graph which possess the property is a singleton set or a element set.

Proof. In [4, 5], it was shown that and any two nonantipodal vertices of form a basis of . Further note that, a element subset of , whether containing two antipodal vertices or not, is not an independent set. It completes the proof.

The number of subsets of the vertex set of a cycle graph possessing the property is counted in the following result.

Lemma 4. For all , if is a cycle graph with vertex set , then .

Proof. Lemma 3 yields that each singleton subset of as well as each element subset of possesses the property . It follows that for all ,where denotes the collection of all the , element subsets of . Hence,

The minimum partition problem for a cycle graph is solved in the following result.

Theorem 2. For all , the vertex set of a cycle graph can be partitioned into minimum number of subsets possessing the property .

Proof. Let the color classes, due to a coloring of , are as follows: When is even, then ; . When is odd, then and ; .In both the cases, all these color classes are lying in , by Lemma 4. It follows that is an coloring of , and these color classes define a partition of into the sets possessing the property . Further, any partition of of cardinality less than contains at least one element subset of such that . Thus, a minimum partition of has subsets of possessing the property .

3.3. Wheel Graphs

For , let be a cycle and be the trivial graph with vertex . Then a wheel graph is the sum with vertex set and edge set . For fixed ; , let a path of order on the cycle of , where the indices greater than or less than zero will be take modulo . The following result describes the sets in possessing the property for .

Remark 3. For , let be a wheel graph with vertex set . Let be set of any consecutive vertices of wheel graph, then is a maximal independent set which is not a minimal resolving set.

Lemma 5. For , let be a wheel graph with vertex set . Then,(1)every element subset of the set belongs to for (2)every element subset of the set belongs to for

Proof. (1)For fixed ; , . Since and for each , for any . It follows the required result.(2)Note that . For any and for each . This implies that for any . Hence, the required result followed.

For wheel graphs, the following result solves the minimum partition problem.

Theorem 3. For , let be a wheel graph with vertex set . Then,

Proof. It can be easily seen that a partition(i) is a minimum partition of having sets possessing the property in ,(ii) is a minimum partition of having sets possessing the property in ,(iii) is a minimum partition of having sets possessing the property in ,(iv) is a minimum partition of having sets possessing the property in , and(v) is a minimum partition of having sets possessing the property in .It follows that is 2 when and is 3 when .
For all , let be a coloring of and let the corresponding color classes are , where for any fixed , and . Then and define a partition of . Also, Lemma 5 yields that both and possess the property . Therefore, is an coloring of , and hence .

4. Twins and Independence

Let be a vertex of a graph having vertex set . Then the open neighborhood of is and the closed neighborhood of is . Two distinct vertices and of are adjacent twins if and nonadjacent twins if . Observe that if are adjacent twins, then in and if are nonadjacent twins, then in . Adjacent twins are called true twins and nonadjacent twins are called false twins. Either are adjacent or nonadjacent twins, they are called twins. A vertex is called self twin if neither nor , for all . Each self twin in a graph makes a set of singleton twins. A set is called a twin set in if are twins in for every pair of distinct vertices . The next lemma follows from the above definitions [16].

Lemma 6 (see [16]). If and are twins in a graph , then for every vertex .

Due to Lemma 6, we have the following remark.

Remark 4. (1)If and are twins in a graph and is a resolving set for . Then either or .(2)If is a twin set in a graph of order , then every resolving set for contains at least elements of .

Removal of two twins from the vertex set makes it independent as given in the next result.

Lemma 7. Let be a twin set of order in a graph having the vertex set . Then, for any two elements , the set possesses the property .

Proof. Since for all , by Lemma 6, so no subset of is a resolving set for . It follows the result.

Remarks 2 and 4 yield the following two results.

Lemma 8. Let be a twin set of order in a graph . Then, for each , any element subset of and the set both possess the property .

Proof. If , then each subset of of order less that as well as the set both are singleton, and so Remark 2(1) yields the result. If , then as vertices of must belong to any resolving set for , by Remark 4(2), so no subset of of order is not a resolving set, because there are at least two twins are remained in form that one of them must belong to a resolving set for , by Remark 4(1). It follows that any element subset of possesses the property for each . Further, since the set is either singleton or contains at least more than one twins from , no subset of is a resolving set for . Thus, it must possess the property .

Lemma 8 can be generalized with the similar proof when a graph has more than one twin sets of order at least two, and this generalization is stated in the following result.

Theorem 4. For , let are twin sets in a graph of orders , respectively, where each . If is a element subset of for , then(1)each possesses the property ,(2) possesses the property , and(3) possesses the property .

Remark 5. Let be a family of graph and . For , let are twin sets in a graph of orders , respectively, where each . Let a nonempty set is union of singleton twin sets in , and let belong to any one of , then is a maximal independent set which is not minimal resolving set.

The following result states the relationship between twins and independence.

Theorem 5. The chromatic number of a graph (except ) of order having a nonsingleton twin set is two.

Proof. Let be a graph of order with vertex set , and let be a twin set in . Let be a coloring of , and let the corresponding color classes are for any and . Then and define a partition of . Also, Lemma 7 implies that . Further, since no path graph of order more than three has a twin set, so is not a path graph. It follows that no singleton subset of is resolving, because a path graph only has a singleton resolving set [4]. Thus, . Hence, is an coloring of , and so .

Remark 6. (1)The converse of Theorem 5 is not true generally. Theorem 3 describes that the chromatic number of a wheel graph is two, but has no twin class for any .(2)In Theorem 5, if is , then has one twin set containing two end vertices. But, the chromatic number of is three, by Theorem 1.

Next, we provide two well-known families of graphs as in the favor of Theorem 5.

4.1. Complete Multipartite Graphs

Let be a complete multipartite graph with partite sets of cardinality , respectively, where each .(i)If for all , then is a complete graph having vertex as the twin set.(ii)If some of is not equal to one. Let us suppose, without loss of generality, that for and . Then, has twin sets and .(iii)If and , then is .

As, the chromatic number of is 3, by Theorem 1, so we receive the following consequence from Theorem 5.

Corollary 1. The chromatic number of a complete multipartite graph (which is not ) is two.

Example 5. (Circulant networks).
The family of circulant networks is an important family of graphs, which is useful in the design of local area networks [17]. These networks are the special case of Cayley graphs when the group is (an additive group of integers modulo ) and [18]. These graphs are defined as follows: let and be positive integers, and for all . An undirected graph with the set of vertices , and the set of edges is called a circulant graph, denoted by . The numbers are called the generators, and we say that the edge is of type . The indices after will be taken modulo . The cycle in is called the principal cycle. Consider a class of circulant networks , for . Then there are twin sets for . Thus, as a consequence of Theorem 5, the chromatic number of is two.

5. Some Realizations

Remark 2(3) describes that there is no element independent set in a connected graph of order . Lemma 7 illustrates that a connected graph of order having twins (other than self twins) can have an element set as an independent set. In the result to follow, we characterize all the connected graphs of order in which every element subset of the vertex set is independent.

Theorem 6. Let be a connected graph of order with vertex set . Then any element subset of possesses the property if and only if is a complete graph.

Proof. Suppose that is a complete graph. Then every two vertices of are twins and itself is the twin set in . So Lemma 8 yields the required result.
Conversely, suppose that any element subset, say , of possesses the property . Then is a resolving set for , because for any , 0 lies at the th position in the code , whereas the code of the element has all nonzero coordinates. Further, is a minimum resolving set for , because no element subset of is a resolving set for any , by our supposition. Thus, . In [4, 6], it was shown that a graph of order has if and only if is a complete graph. It completes the proof.

Since any singleton set is an independent, so we have the following consequences for a complete graphs.

Corollary 2. The chromatic number of a complete graph is two.

If is a connected graph of order with vertex set , then , by Remark 2(3). The next result characterizes all the connected graphs of order having chromatic number .

Theorem 7. Let be a connected graph of order with vertex set . Then, if and only if is either or .

Proof. If , then , by Corollary 2.
If , then , by Theorem 1.
Conversely, suppose that for a connected graph of order with vertex set , we have . Then,(i)for , the only connected graph is such that , by Corollary 2,(ii)for , is either or . Since the chromatic number of is , by Theorem 2, and the chromatic number of is , by Theorem 1, so in this case.(ii)for , either has a twin set or no twin set exists in . In the former case, , by Theorem 5. In the latter case, let . Then, implies that the minimum partition of the independence system is . It follows that no element subset of belongs to for . Otherwise, . But, in every connected graph of order , at least one element subset of must possess the property , because singleton resolving sets exist in a path graph only (and in the case of path graph , we have element subsets of in , by Lemma 2). Therefore, no connected graph of order exists such that .From the above three cases, we conclude that is either or .

From Theorem 7, it concludes that if is a connected graph of order with vertex set and , then . All the connected graphs of order having chromatic number are characterized in the following result.

Theorem 8. Let be a connected graph of order with vertex set . Then, if and only if is either or or .

Proof. If , then Theorem 2 yields that .
If or , then , by Theorem 1.
Conversely, suppose that for a connected graph of order with vertex set , we have . Then,(i)for , is either or . is not true, by Theorem 7. So , by Theorem 2,(ii)for , , where and can be obtained by deleting one and two edges from , respectively. In this case, except , all the graphs has a twin set and so , by Theorem 5. Thus , by Theorem 1.(iii)for , either has no twin set or has a twin set. in the latter case, by Theorem 5. In the former case, except , a minimum resolving set for is of at least two order, which implies that every element subset of belongs to . It follows that a minimum partition of according to the independence system contains at least two element subsets of , which implies that , a contradiction. However, when , then only for and for , by Theorem 1.From the above three case, we conclude that is either or or .

Theorem 8 concludes that if is a connected graph of order with vertex set and , then .

6. Metric Dimension and Independence

In this section, we develop a relationship between the metric dimension and chromatic number of a connected graph by providing three existing type results.

There exists a connected graph whose metric dimension is different from its chromatic number by one.

Theorem 9. For even , there exists a connected graph with vertex set such that .

Proof. Let be a cycle graph on even vertices and a path . Then is a graph obtained by taking the product of and . Let the vertex set of be , and the edge set is . The resultant graph consists of two cycles: one is outer cycle , and the other one is inner cycle . It is shown, in [19], that . Next, we investigate the chromatic number of with the help of the following five claims:Claim 1.Every singleton, element and element subset of possesses the property . Based on Remark 2(1) and due to [19], this claim is true, because a minimum resolving set for is of cardinality 3, and so no singleton and element subset of is a resolving set for .Claim 2.For fixed , the sets and are minimum resolving sets for only when . Otherwise, and .Claim 3.For fixed , the set possesses the property . By Claim 2, no subset of is a resolving set for . It follows the required claim.Claim 4.No independent set (other than ) in of cardinality greater than 3 contains any of the pairs and . Let be an independent set in of cardinality greater than 3. Suppose, without loss of generality, contains the pair . Then a subset of for is a resolving set for , by Claim 2, which contradicts the independence of . Thus, cannot contain the pair . Similarly, the pair will not contained in .Claim 5.For each , there is a element subset of possessing the property . Claim 1 yields the result of . For , we have a set in , by Claim 3. Next, keeping Claim 4 in mind, let us consider two subset of of cardinality as follows: for fixed , where the indices greater than or less than or equal to zero will be taken modulo . Then for all and for all . It follows that for any and for any . Hence, both the and each subset of any cardinality all are the subsets of possessing the property , and of course, they are element subsets of for . Now, let be a coloring of , and let the corresponding color classes are, for fixed , where the indices greater than or less than or equal to zero will be taken modulo . Then and define a partition of . Also, Claim 5 yields that . Hence, is an coloring of , and so . Therefore, for every even value of .

Remark 7. It is not necessary that the difference is constant (fixed) always. It can arbitrarily large depending upon the order of the graph. For instance, let be a wheel graph of order , then [20], and , by Theorem 3. So, it can be seen that the difference , which is depending upon and is not fixed.

The next result shows that there exists a connected graph whose chromatic number is different from its metric dimension by one.

Theorem 10. For odd , there exists a connected graph with vertex set such that .

Proof. Let a cycle graph on odd vertices and a path . Then is a graph obtained by taking the product of and . Let the vertex set of be and the edge set is . The resultant graph consists of two cycles: one is outer cycle , and the other one is inner cycle . Note that , and it was show, in [19], that , so a minimum resolving set for consists of two vertices of . Next, we investigate the chromatic number of on the base of the following six claims:Claim 1.Every singleton and element subset of possesses the property . Based on Remark 2(1) and due to [19], this claim is true, because a minimum resolving set for is of cardinality 2, and so no singleton subset of is a resolving set for .Claim 2.A minimum resolving set for contains both the vertices either from the outer cycle or from the inner cycle of . Let be a minimum resolving set for . For fixed , let , where . Then,  when ,  when , and  when , where the indices greater than or less than zero will be taken modulo . It follows that is not a resolving set for , a contradiction.Claim 3.For fixed , a minimum resolving set for contains both the vertices: such that . Let be a minimum resolving set for , then must contain both the vertices from the same cycle of , by Claim 2. Let and (in , no two vertices belonging to a same cycle have ). Then, and  if , and  if , where the indices greater than or less than or equal to zero will be taken modulo . A contradiction to the fact that is a resolving set for .Claim 4.No independent set in of cardinality greater than 2 contains two vertices from a same cycle (outer or inner) of such that . Otherwise, the set is such a subset of that independent set which is resolving for , by Claim 3.Claim 5.For each , there is a element subset of possessing the property . Claim 1 follows the claim for . Next, keeping Claim 4 in mind, let us consider two subset of of cardinality as follows: for fixed , where the indices greater than or less than or equal to zero will be taken modulo . Then, for all and for all . It follows that for any and for any . Hence, both the set and each subset of any cardinality all are the subsets of possessing the property , and of course, they are element subsets of for .Claim 6.No independent set of cardinality greater than exists in . Otherwise, Claim 4 will be contradicted. Now, let be a coloring of , and let the corresponding color classes are: for fixed , , where the indices greater than or less than or equal to zero will be taken modulo . Then , and define a partition of . Also, Claims 1 and 5 yield that . Hence, is an coloring of , and so . Therefore, for every odd value of .

Remark 8. It is not necessary that the difference is constant (fixed) always. It can arbitrarily large by depending upon the order of the graph. For instance, let be a cycle graph of order , then [4] and , by Theorem 2. It can be seen that the difference , which is depending upon and is not fixed.

The following result provides the existence of a connected graph whose metric dimension is equal to its chromatic number.

Theorem 11. For odd , there exists a connected graph with vertex set such that .

Proof. Let be a circulant network for odd with vertex set (defined in Example 5), where the indices will be taken modulo . The distance between and on the principal is denoted by (for instance, in a circulant network , where as for any ). The metric dimension, , of is 3 as shown in [21]. On the basis of the following three claims, we investigate the chromatic number of .Claim 1.Every element subset of possesses the property for . As a minimum resolving set for is of cardinality three, so no singleton and element subset of resolves . It follows the claim.Claim 2.No independent set in of cardinality greater than three contains two vertices and such that . If is an independent set in containing two vertices and such that , then the set , for any , is a resolving set for , a contradiction.Claim 3.A maximum independent set in consists of consecutive vertices form the principal cycle. Firstly, for fixed , we show that a set of consecutive vertices form the principal cycle possesses the property . Note that, for all . It follows that for each . Hence, no subset of is a resolving set for . Secondly, if an independent set either contains nonconsecutive vertices or contains more than vertices (consecutive or nonconsecutive) form the principal cycle, then must contradict Claim 2. Now, let be a coloring of , and let the corresponding color classes are: for fixed , , , and , where the indices will be taken modulo . Then and define a partition of . Also, Claims 1 and 3 yield that . Hence, is an coloring of , and so . Therefore, for every odd value of .

Data Availability

All the data and materials used to compute the results are provided in Section 1.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

The first author, as a visiting researcher, is supported by the Punjab Higher Education Commission (PHEC) of Pakistan through International Research Support Fellowship (IRSF) Ref: PHEC/HRD/IRSF/1-20/2020/2674 at the University of Exeter, UK.