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BRIEF RESEARCH REPORT article

Front. Phys., 28 July 2020
Sec. Statistical and Computational Physics
This article is part of the Research Topic Mathematical Treatment of Nanomaterials and Neural Networks View all 28 articles

Computing the Mixed Metric Dimension of a Generalized Petersen Graph P(n, 2)

\nHassan RazaHassan RazaYing Ji
Ying Ji*
  • Business School, University of Shanghai for Science and Technology, Shanghai, China

Let Γ = (V, E) be a connected graph. A vertex iV recognizes two elements (vertices or edges) j, kEV, if dΓ(i, j) ≠ dΓ(i, k). A set S of vertices in a connected graph Γ is a mixed metric generator for Γ if every two distinct elements (vertices or edges) of Γ are recognized by some vertex of S. The smallest cardinality of a mixed metric generator for Γ is called the mixed metric dimension and is denoted by βm. In this paper, the mixed metric dimension of a generalized Petersen graph P(n, 2) is calculated. We established that a generalized Petersen graph P(n, 2) has a mixed metric dimension equivalent to 4 for n ≡ 0, 2(mod4), and, for n ≡ 1, 3(mod4), the mixed metric dimension is 5. We thus determine that each graph of the family of a generalized Petersen graph P(n, 2) has a constant mixed metric dimension.

2010 Mathematics Subject Classification: 05C12, 05C90

1. Introduction

The aim of robot navigation functionality is to attain the coveted position promptly whenever it is desired. Let us imagine that robot navigation in a sensor network that can obtain the distances to a collection of landmarks. A robot's position is solely resolved by determining the subset of nodes in the sensor network. It can be achieved by the concept of landmarks in the graphs introduced in Khuller et al. [1]; this idea was later named the metric dimension. All the graphs considered here have no loops and are simple, measurable, and undirected.

Let Γ = (V, E) be the graph of the distance dΓ(a, b) (or d(a, b)) among the vertices a, bV(Γ) the minimum length of paths between them. For a vertex aV, distinguish two vertices in a graph, say b and c, if the condition dΓ(a, b) ≠ dΓ(a, c) holds. A set RV(Γ) is the metric generator if some chosen vertices of the set R recognizes a pair of distinguished vertices. The metric basis with the least number of elements is called the metric generator, and the cardinality of its metric basis is termed the metric dimension. The notation employed here is β(Γ). The fundamental concept of the metric dimension was instated by Slater [2], and the notation of the metric dimension was initiated by Haray and Melter [3]. This concept was later studied by many researchers with unique modifications; for reference, see [48]. Some of the recent results on metric dimension and its further variations are studied in Shao et al. [9] and Raza et al. [1013].

Lemma 1. Suppose R is the distinguishing set of Γ and the vertices a, bV(Γ). If dΓ(a, c) ≠ dΓ(a, c), for all vertices cV(Γ)\{a, b}, then {a, b}⋂R ≠ ∅.

Analogous to this definition, Kelenc et al. [14] introduced the concept of edge metric dimension, and this was further studied in Zubrilina [15], Peterin and Yero [16], and Zhu et al. [17]. This distance between an edge e = ab and a vertex c is given as follows

d(e,c)=min{d(a,c),d(b,c)}

A vertex cV(Γ) distinguishes two edges of a graph e1, e2E(Γ) if dΓ(e1, u) ≠ dΓ(e2, u). The set ReV is termed as the edge metric generator if some distinct edges of Γ are distinguished by the vertex set Re. The cardinality of an edge metric generator is called an edge metric dimension, and it is depicted as βe(Γ). Having defined the notion of an edge metric generator, which distinctly recognizes every edge in a graph, a common assumption would be that any edge metric generator Re would be a metric dimension as well. This assumption is far from reality, as indicated in Kelenc et al. [14], but there are several families of graphs where this occurs, that is, β(Γ) = βe(Γ). Some other distance related parameters are studied in Liu et al. [1822].

In this paper, our focus is on mixed metric dimension, which is a mixed version of metric and edge metric dimension. A set Rm of vertices of a graph Γ is known as a mixed metric generator if any two distinct elements (vertices or edges) of a graph are recognized by some the vertex set of Rm. The least cardinality of a mixed metric generator for a graph is termed as a mixed metric dimension, denoted as βm(Γ). The idea is recently brought forward by Kelenc et al. [23].

Lemma 2. Let for some vertex aV(Γ), and let Rm = V(Γ)\a, and if there is an element bN(a), also for some cRm, dΓ(ab, c) ≠ dΓ(b, c), then Rm is the mixed metric generator for the graph Γ.

The notion of a mixed metric dimension clearly indicates that a mixed metric generator is also a standard metric generator and an edge metric generator, The following relationship is given in [23],

βm(Γ)max{β(Γ),βe(Γ)}

The following remark shows the structure of mixed metric dimension:

Remark 1: [23] Suppose for some graph Γ we have 2 ≤ βmn. Recently, this concept has attracted some attention, and it has been studied by Raza et al. [24]. The authors discussed the mixed metric dimension among the prism graphs, which are commonly known as generalizes Petersen graphs P(n, 1), and two families of convex polytopes An, Rn, further presenting the problem of finding βm(P(n, 2)).

Some of the results regarding metric and edge metric dimension are given:

Remark 2: [14] For n ≥ 2, the metric and edge metric dimension are, β(Pn)=βe(Pn)=1; for cycle graph, β(Cn)=βe(Cn)=2; for complete graph, β(Kn)=βe(Kn)=n-1; and for any complete bipartite graph (Kr,t) different from (K1,1), β(Kr,t)=βe(Kr,t)=r+t-2.

1.1. Known Results

Next, we present some already known results for βm,

Proposition 1: [23] For a path graph (Pn) order n ≥ 4, we have βm(Pn)=2.

Proposition 2: [23] Let us consider any two positive integers: e, f

βm(Ke,f)={e+f-1,if e=2 or f=2;e+f-2,otherwise.

Proposition 3: [23] For a grid graphs, PmPn, with mn ≥ 2, βm = 3.

Proposition 4: [23] Let us assume cycle graph (Cn) of order n ≥ 4, then βm(Cn) = 3.

Lemma 3. [24]The mixed metric generator Rm must contain vertices from both the outer and inner cycle for the prism graph Dn.

Proof: For P(n, 1), this holds, and, by the same intuition, this must be true for P(n, 2). The mixed metric resolving set comprises of vertices from both the cycles, which contain vertices of outer and inner cycle, respectively. 

2. Main Result

The generalized Petersen graphs is introduced by Watkins [25]. The P(n, ℓ), where n ≥ 3 and 1n-12 (see Figure 2), which is the cubic graph consists of vertices and edges, is shown below.

V(P(n,))={q0,q1,,qn-1,p0,p1,,pn-1}
ε(P(n,))={qiqi+1,pipi+,qiqi|i=0,1,,n-1}

Example: We used the graph of P(n, 8), as can be seen in Figure 1. The mixed metric generator for P(n, 8) is βm = {q0, q1, p4, p5}, and it can been seen from Table 1 that all the representation of vertices and edges are distinct.

FIGURE 1
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Figure 1. The generalized Petersen graph P(8, 2).

FIGURE 2
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Figure 2. (A) The generalized Petersen graph P(5, 2), (B) The generalized Petersen graph P(6, 2).

TABLE 1
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Table 1. Codes for P(n, 8).

The graph of the generalized Petersen graph comprises of three types of edges, external edges, internal edges, and spokes between qi and qi+1, pi and pi+m, and qi and pi, respectively. The vertices qi and pi (0 ≤ in − 1) are termed as external and internal vertices, respectively.

The prism graph Dn is known as P(n, 1) for m = 1. Some of the already known results are given as

Theorem 1. [26] The metric dimension of Dn, for n ≥ 4:

β(Dn)={2,n is odd;3,n is even.

Theorem 2. [27] When, n ≥ 4, βe(Dn)=3.

Theorem 3. [24] For n ≥ 5,

βm(P(n,1))={3,n is even;4,n is odd.

The known results for P(n, 2) concerning metric and an edge metric dimension are

Theorem 4. [28] For n ≥ 5, the metric dimension is β(P(n, 2)) = 3.

Theorem 5. [27] For n ≥ 5, βe(P(n, 2)) = 3.

It is quite natural to investigate the mixed metric dimension of P(n, 2). Now, we will find mixed the metric dimension of (P(n, 2)), and for this the following lemmas are presented.

Lemma 4. Case 1: If n ≡ 0(mod)4, then βm(P(n, 2)) ≤ 4.

Proof: The proof is n = 4r, r ≥ 4, where r ∈ ℤ+. The distinguishing vertices that will distinguish the whole vertices and edges of the graph are Rm = {q0, q1, p2r, p2r+1}. The following representations are presented with respect to Rm.

Representation of external vertices:

CRm(q2s)={(2s,1,rs+1,r+1),0s1;(2s,s+1,rs+1,r),s=2;(s+2,s+2,rs+1,rs+2),3sr;(2rs+2,2rs+3,sr+1,r+1s2r2;          sr+1), (2,3,sr+1,sr+1),s=2r1.

and,

CRm(q2s+1)={(2s+1,2s,sr+1,rs+1),0s2;(s+3,s+2,2,rs+1),3sr1;(r+2,r+2,2,1),s=r;(2rs+2,2rs+2,r+1s2r3;          sr+2,sr+1), (3,4,sr+2,rs+1),s=2r2;(1,2,sr+2,sr+1),s=2r1.

Representation of internal vertices:

CRm(p2s)={(s+1,2,rs,r+2),0s1;(s+1,s+1,rs,rs+3),2sr;(2rs+1,2rs+2,r+1s2r1.sr,sr+2), 

and,

CRm(p2s+1)={(s+2,s+1,rs+2,rs),0sr1;(2rs+1,r,sr+3,sr),rsr+1;(2rs+1,2rs+1,r+2s2r1.          sr+3,sr), 

Representation of external edges:

CRm(q2sq2s+1)={(2s,s,rs+1,rs+1),0s1;(2s,s+1,rs+1,s=2;          rs+1), (s+2,s+2,rs+1,3sr;          rs+1), (2rs+2,2rs+2,r+1s2r3;          sr+1,sr+1), (3,4,sr+1,sr+1),s=2r2;(1,2,sr+1,sr+1),s=2r1.

and,

CRm(q2s+1q2s+2)={(2s+1,2s,rs,rs+1),0s2;(s+3,s+2,rs,rs+1),3sr1;(2rs+1,2rs+2,rs2r3;          sr+2,sr+1), (2,3,sr+2,sr+1),s=2r2;(0,1,sr+2,sr+1),s=2r1.

Representation of external and internal edges:

CRm(q2sp2s)={(2s,1,rs,r+1),0s1;(s+1,s+1,rs,rs+2),2sr;(2rs+1,2rs+2,sr,r+1s2r1.          sr+1), 

and,

CRm(q2s+1q2s+1)={(2s+1,2s,rs+1,rs),0s1;(s+2,s+1,rs+1,rs),2sr1;(r+1,r+1,2,0),s=r;(2rs+1,2rs+1,r+1s2r2;          sr+2,sr), (1,2,sr+2,sr),s=2r1.

Representation of internal edges:

CRm(p2sp2s+2)={(1,2,rs1,rs+2),s=0;(s+1,s+1,rs1,1sr1;          rs+2), (2rs,r,sr,3),rsr+1;(2rs,2rs+1,sr,r+2s2r1.          sr+2), 

and,

CRm(p2s+1p2s+3)={(2,1,rs+1,rs1),s=0;(s+2,s+1,rs+1,1sr3;          rs1), (s+2,s+1,3,rs1),r2sr1;(2rs,2rs,sr+3,rs2r2;          sr), (2,1,sr+3,sr),s=2r1.

Case 2: For n ≡ 2(mod)4, we have βm(P(n, 2) ≤ 4

Proof: Now we can see n = 4r + 2, r ≥ 4, where r ∈ ℤ+. The set of vertices that will distinguish the whole vertices and edges of the graph are Rm = {q0, q1, p2r+1, p2r+2}. The following representations are presented with respect to Rm.

Representation of external vertices:

CRm(q2s)={(2s,2s,rs+2,r+1),0s1;(2s,3,rs+2,rs+2),s=2;(s+2,s+2,rs+2,rs+2),3sr;(2rs+3,2rs+4,r+1s2r1;          sr+1,sr), (2,3,sr+1,sr),s=2r.

and,

CRm(q2s+1)={(2s+1,2s,sr+1,rs+2),0s2;(s+3,s+2,rs+1,rs+1),3sr;(2rs+3,2rs+4,r+1s2r2;          sr+1,sr+1), (3,5,sr+1,sr+1),s=2r1;(1,3,sr+1,sr+1),s=2r.

Representation of internal vertices:

CRm(p2s)={(s+1,2,rs+3,r),0s1;(s+1,r+1,rs+3,rs+1),2sr;(2rs+2,2rs+3,r+1s2r.          sr+2,sr1), 

and,

CRm(p2s+1)={(s+2,s+1,rs,rs+3),0sr;(2rs+2,2rs+3,r+1s2r.          sr,sr+2), 

Representation of external edges:

CRm(q2sq2s+1)={(2s,s,rs+1,r+1),0s1;(2s,s+1,rs+1,rs+2),s=2;(s+2,s+2,rs+1,3sr;          rs+2), (2rs+3,2rs+3,r+1s2r2;          sr+1,sr), (3,4,sr+1,sr),s=2r1;(1,2,sr+1,sr),s=2r.

and,

CRm(q2s+1q2s+2)={(2s+1,2s,rs+1,rs+1),0s2;(s+3,s+2,rs+1,rs+1),3sr1;(2rs+2,r+2,sr+1,rsr+1;          sr+1), (2rs+2,2rs+3,sr+1,r+2s2r2;          sr+1), (2,3,sr+1,sr+1),s=2r1;(0,1,sr+1,sr+1),s=2r.

Representation of external and internal edges:

CRm(q2sp2s)={(2s,1,rs+2,r),0s1;(s+1,s+1,rs+2,rs+1),2sr;(2rs+2,2rs+3,r+1s2r.          sr+1,sr1), 

and,

CRm(q2s+1q2s+1)={(2s+1,2s,rs,rs+2),0s1;(s+2,s+1,rs,rs+2),2sr;(2rs+2,2rs+2,r+1s2r1;     sr,sr+1), (1,2,sr,sr+1),s=2r.

Representation of internal edges:

CRm(p2sp2s+2)={(s+1,2,rs+2,rs),0s1;(s+1,s+1,rs+2,rs),2sr1;(2rs+1,r+1,3,0),rsr+1;(2rs+1,2rs+2,r+2s2r.          sr+2,sr1), 

and,

CRm(p2s+1p2s+3)={(s+2,s+1,rs1,0sr1;          rs+2), (2rs+1,2rs+1,rsr+1;          sr,3), (2rs+1,2rs+1,r+2s2r1;          sr,sr+2), (2,1,sr,sr+2),s=2r.

Now from lemma3, the resolving set Rm contains vertices from external and internal cycles; that is, the resolving set cannot comprise either external or internal vertices.

Lemma 5. When n ≡ 0, 2(mod4), then βm(P(n, 2)) ≥ 4.

Proof: Suppose that βm(P(n, 2)) = 3, the following contradictions arises:

Case 1: This is when the two fixed vertices are in the external cycle, {q0, q1}, and the other vertex lie in internal cycle p, that is, Rm = {q0, q1, p}.

(i) If 0 ≤ ℓ ≤ 1 then, rm{q0|q0, q1, p} = rm{q0qn−1|q0, q1, p} = (0, 1, ℓ + 1).

(ii) If ℓ = 2, 4, …, 2r, then rm{q0|q0, q1, p} = rm{q0qn−1|q0, q1, p}.

(iii) If ℓ = 3, 5, …, 2r−1, then rm{q0|q0, q1, p} = rm{q0qn−1|q0, q1, p}.

Case 2: When internal cycle contains two fixed vertices that is {p0, p1}, and the other vertex lie in external cycle q. That is Rm = {p0, p1, q}.

(i) If 0 ≤ ℓ ≤ 3, then rm{q0|p0, p1, q} = rm{q0qn−1|p0, p1, q} = (1, 2, ℓ).

(ii) If ℓ = 4, 6, …, 2r, then rm{q0|p0, p1, q} = rm{q0qn−1|p0, p1, q}.

(iii)If ℓ = 5, then rm{q0|p0, p1, q} = rm{q0qn−1|p0, p1, q} = (1, 2, ℓ).

(iv) If ℓ = 7, 9, …, 4r − 1, then rm{q1|p0, p1, q} = rm{q1q2|p0, p1, q}.

Similarly, other contradictions can be assumed; all the cases mentioned above suggest that βm(P(n, 2)) ≥ 4, which clearly indicates that βm(P(n, 2)) = 4 for n ≡ 0(mod4). Similar kind of contradictions can be proved for n ≡ 2(mod4). 

Remark 3: From the above cases, it can be deduced that if the mixed metric generator Rm for P(n, 2) contains two vertices of one cycle, then Rm contain at least two vertices of another cycle.

Lemma 6. βm(P(n, 2) ≤ 5, for n ≡ 1(mod)4

Proof: Case 1: Now we can write, if n = 4r + 1, r ≥ 4, where r ∈ ℤ+. The set of vertices that will distinguish the whole vertices and the edges of the graph are Rm = {q0, q1, p1, p2r+1, p2r+2}. The following representations are presented with respect to Rm.

Representation of external vertices:

CRm(q2s)={(2s,22s,2,r+1,r+1),0s1;(2s,s,s+1,rs+2,rs+2),s=2;(r+2,r+2,r+1,2,1),s=r+1;(2rs+3,2rs+4,r+2s2r2;          2rs+2,sr+1,sr), (3,5,3,sr+1,sr),s=2r1;(1,3,2,sr+1,sr),s=2r.

and,

CRm(q2s+1)={(2s+1,1,s+1,rs+1,r+1),0s1;(2s+1,s+1,s+1,rs+1,s=2;          rs+2), (s+3,s+2,s+1,rs+1,3sr1;          rs+2), (2rs+2,r+2,r+1,sr+1,2),rsr+1;(2rs+2,2rs+3,2rs+2,r+2s2r2;          sr+1,sr+1), (2,4,3,sr+1,sr+1),s=2r1.

Representation of internal vertices:

CRm(p2s)={(s+1,2s,3,r+s,rs+1),0s1;(s+1,s+1,s+2,rs+3,2sr1;          rs+1), (r+1,s,2rs+1,3,rs+1),rsr+1;(2rs+2,2rs+3,2rs+1,r+2s2r1;          sr+2,sr1), (2,3,1,sr+1,sr1),s=2r.

and,

CRm(p2s+1)={(s+2,2s,s,rs,r+s),0s1;(s+2,s+1,s,rs,rs+3),2sr1;(2rs+1,r+2,s,sr,3),rsr+1;(2rs+1,2rs+2,r+2s2r1.          2rs+3,sr,sr+2), 

Representation of external edges:

CRm(q2sq2s+1)={(2s,1s,s+1,rs+1,r+1),0s1;(2s,s,s+1,rs+1,rs+2),s=2;(s+2,s+1,s+1,rs+1,3sr;          rs+2), (2rs+2,2rs+3,r+1s2r2;          2rs+2,sr+1,sr), (2,4,3,sr+1,sr),s=2r1;(0,2,2,sr+1,sr),s=2r.

and,

CRm(q2s+1q2s+2)={(2s+1,s,s+1,rs+1,0s1;          ars+1), (2s+1,s+1,s+1,rs+1,s=2;          rs+1), (2rs+1,s+2,s+1,3sr1;          rs+1,rs+1), (2rs+2,r+1,2rs+1,rs2r4;          sr+1,sr+1), (5,6,4,sr+1,sr+1),s=2r3;(3,5,3,sr+1,sr+1),s=2r2;(1,2,3,sr+1,sr+1),s=2r1.

Representation of external and internal edges:

CRm(q2sp2s)={(2s,2s,2,r+s,rs+1),0s1;(s+1,s,s+1,rs+2,2sr;          rs+1), (2rs+2,r+1,2rs+1,r+3s2r1;          sr+1,sr1), (1,3,1,sr+1,sr1),s=2r.

and,

CRm(q2s+1p2s+1)={(2s+1,1,s,rs,r+s),0s1;(s+2,s+1,s,rs,rs+2),2sr1;(r+1,r+1,2rs,0,2),s=r;(2rs+1,2rs+2,r+1s2r1.     2rs+2,sr,sr+1), 

Representation of internal edges:

CRm(p2sp2s+2)={(s+1,1,3,r+s,rs),0s1;(s+1,s,s+2,rs+2,rs),2sr1;(2rs+1,s,2rs,3,0),rsr+1;(2rs+1,2rs+2,2rs,r+2s2r1;          sr+2,sr1), (2,2,0,sr,sr1),s=2r.

and,

CRm(p2s+1p2s+3)={(s+2,2,s,rs1,r+s),0s1;(s+2,s+1,s,rs1,2sr1;          rs+2), (2rs,2rs+1,s,rs,3),rsr+1;(2rs,2rs+1,2rs+2,r+2s2r1.          sr,sr+2), 

Proof: Case 2: Now we can write, if n = 4r + 3, r ≥ 4, where r ∈ ℤ+. The set of vertices which will distinguish the whole vertices, and edges of graph are Rm = {q0, q1, p1, p2r+3, p2r+4}. The following representations are presented with respect to Rm.

Representation of external vertices:

CRm(q2s)={(2s,22s,2,r+s+1,r+s+1),0s1;(s+2,2s2,s+1,rs+3,2s3;          rs+3), (s+2,s+1,s+1,rs+3,rs+3),4sr+1;(2rs+4,2rs+5,2rs+3,r+2s2r1;          sr,sr1), (3,5,3,sr,sr1),s=2r;(1,3,2,sr,sr1),s=2r+1.

and,

CRm(q2s+1)={(2s+1,1,s+1,rs+2,0s1;          r+s+1), (s+3,2s1,s+1,rs+2,2s3;          rs+3), (s+3,s+2,s+1,rs+2,4sr;          rs+3), (2rs+3,2rs+4,r+1sr+2;          2rs+3,sr,2), (2rs+3,2rs+4,r+3s2r1;          2rs+3,sr,sr), (2,4,3,sr,sr),s=2r.

Representation of internal vertices:

CRm(p2s)={(s+1,2s,3,r+s,rs+2),0s1;(s+1,s,s+2,rs+4,rs+2),2sr;(2rs+3,s,2rs+1,3,rs+1),r+1sr+2;(2rs+3,2rs+4,2rs+2,r+3s2r+1.          sr+1,sr2), 

and,

CRm(p2s+1)={(s+2,2,s,rs+1,r+s),0s1;(s+2,s+1,s,rs+1,rs+4),2sr;(2rs+1,2rs+3,2rs+5,r+1sr+2;          sr1,3), (2rs+2,2rs+3,2rs+4,r+3s2r.          sr1,sr+1), 

Representation of external edges:

CRm(q2sq2s+1)={(2s,1s,s+1,r+1,r+s+1),0s1;(2s,s,s+1,rs+2,rs+3),s=2;(s+2,s+1,s+1,rs+2,3sr;          rs+3), (2rs+3,r+2,2rs+3,r+1sr+2;          sr,sr+3), (2rs+3,2rs+4,2rs+3,r+3s2r1;          sr,sr1), (2,4,3,sr,sr1),s=2r;(0,2,2,sr,sr1),s=2r+1.

and,

CRm(q2s+1q2s+2)={(2s+1,s,s+1,rs+2,r+1),0s1;(s+3,2s1,s+1,rs+2,2s3;          rs+2), (s+3,s+2,s+1,rs+2,4sr;          rs+2), (2rs+3,2rs+4,r+1s2r2;          2rs+2,sr,sr), (3,5,3,sr,sr),s=2r1;(1,3,2,sr,sr),s=2r.

Representation of external and internal edges:

CRm(q2sp2s)={(2s,2s,2,r+s,r+1),0s1;(s+1,s,s+1,rs+3,2sr;          rs+2), (2rs+3,s,2rs+2,2,r+1sr+2;          rs+2), (2rs+3,2rs+4,2rs+2,r+3s2r;          sr,sr2), (1,3,1,sr,sr2),s=2r+1.

and,

CRm(q2s+1p2s+1)={(2s+1,1,s,rs+1,r+s),0s1;(s+2,s+1,s,rs+1,2sr;          rs+3), (2rs+2,2rs+3,r+1,s=r+1;          sr1,2), (2rs+2,2rs+3,2rr+2s2r+1.          s+3,           sr2,sr1), 

Representation of internal edges:

CRm(p2sp2s+2)={(s+1,1,3,r+s,rs+1),0s1;(s+1,s,s+2,rs+3,rs+1),2sr1;(r+1,s,2rs+1,3,1),rsr+1;(2rs+1,2rs+3,2rs+1,r+2s2r;          sr+1,sr2), (2,2,0,sr,sr2),s=2r+1.

and,

CRm(p2s+1p2s+3)={(s+2,2,s,rs,r+s),0s1;(s+2,s+1,s,rs,rs+4),2sr1;(2rs+1,r+1,s,0,3),rsr+1;(2rs+1,2rs+2,2rs+3,r+2s2r.          sr1,sr+1), 

Now, from lemma3, the resolving set Rm must contain vertices from both the external and internal cycles of graph.

Lemma 7. Suppose n ≡ 1, 3(mod4), then βm ≥ 5.

Proof: Suppose that βm = 4. If so, the following contradictions are assumed.

Case 1: When the external cycle contain three fixed vertices, {q0, q1, q2}, and other vertex lie in the internal cycle p.

(i) If ℓ = 0, 2, 4, …, 2r, then rm{q0|q0, q1, q2, p} = rm{q0qn−1|q0, q1, q2, p} = (0, 1, 2, ℓ + 1).

(ii) If ℓ = 1, 3, 5, …, 4r − 1, then rm{q0|q0, q1, q2, p} = rm{q0qn−1|q0, q1, q2, p}.

Case 2: When {p0, p1, p2} lie in the internal cycle and the other vertex lie in the external cycle q.

(i) If ℓ = 0, 2, 4, …, 2r, then rm{q0|p0, p1, p2, q} = rm{q0qn−1|p0, p1, p2, q}.

(ii) If ℓ = 1, 3, 5, …, 2r + 3, then rm{q0|p0, p1, p2, q} = rm{q0qn−1|p0, p1, p2, q}.

We already proved that for n ≡ 1, 3(mod4), and the mixed metric dimension is βm ≤ 5. From Remark2, we consider the following cases where the external and internal cycles comprise two vertices each.

Case 3: When two external vertices are fixed {q0, q1}, and the internal vertices are {p0, p}.

(i) If ℓ = 0, 2, 4, …, 2r, then rm{q0|q0, q1, p0, p} = rm{q0qn−1|q0, q1, p0, p}.

(ii) If ℓ = 1, 3, 5, …, 4r − 1, then rm{q0|q0, q1, p0, p} = rm{q0qn−1|q0, q1, p0, p}.

Because of the symmetry, other possible cases can also be derived. From all the above cases, therefore, it is proven that, for n ≡ 1, 3(mod4), the mixed metric dimension is βm ≥ 5. We can therefore say βm = 5 when n ≡ 1, 3(mod4). 

Theorem 6. For n ≥ 7, we have a mixed metric dimension

βm(P(n,2))={4,n0,2(mod4);5,n1,3(mod4).

Proof:Case 1: When n ≡ 0, 2(mod4).

From lemma 4,5, we have βmP(n, 2) = 4.

Case 1: When n ≡ 1, 3(mod4).

From lemma 6,7, we have βmP(n, 2) = 5. 

For the remainder of the cases, when n ≤ 15, the mixed metric dimension βm(P(n, 2)) is calculated through the total enumeration method, shown in Table 2, along with the mixed metric basis.

TABLE 2
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Table 2. Mixed Metric generator βm for P(n, 2).

3. Conclusion and Further Research

The recently introduced mixed metric dimension is calculated for P(n, 2). It has been shown that P(n, 2) has mixed metric dimension equal to 4 for n ≡ 0, 2(mod4), and, for n ≡ 1, 3(mod4), the mixed metric dimension is 5. This shows that each graph of the family of generalized Petersen P(n, 2) has constant mixed metric dimension.

Theorem 7. [29] For the graph of P(n, 3),

β(P(n,3))={4,when n0(mod6);3,when n1,(mod6).

and,

β(P(n,3)){5,when n2(mod6);4,when n3,4,5(mod6).

Theorem 8. [30] For n ≥ 17, we have,

β(P(n,4)){3,when n0(mod4);4,when n1,2,3(mod4).

Theorem 9. [31] The metric dimension of graph of P(2n, n) is

β(P(2n,n))={3,when n is even;4,otherwise.

The standard metric dimension is examined for these as well as other known classes of generalized Petersen graphs; the mixed metric dimension for these as well as other graphs would therefore be intriguing to investigate. If the other variants of dimension are identified, a comparative study can be carried out; this could evaluate the relationship between β(Γ), βe(Γ), and βm(Γ) in the different families of graphs.

Data Availability Statement

The original contributions presented in the study are included in the article/supplementary materials, further inquiries can be directed to the corresponding author/s.

Author Contributions

The main idea was presented by HR. YJ read and approved the final manuscript.

Funding

The work of HR was supported by Post-Doctoral fund of University of Shanghai for Science and Technology under the grant no. 5B19303001. The work of YJ was supported by the National Science Foundation of China under grant no. 71571055.

Conflict of Interest

The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.

Acknowledgments

We would like to express our sincere gratitude to the referees for a very careful reading of this paper and for all their insightful comments/criticism, which have led to a number of significant improvements to this paper.

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Keywords: mixed metric dimension, metric dimension, edge metric dimension, generalized Petersen graph, exact values

Citation: Raza H and Ji Y (2020) Computing the Mixed Metric Dimension of a Generalized Petersen Graph P(n, 2). Front. Phys. 8:211. doi: 10.3389/fphy.2020.00211

Received: 09 April 2020; Accepted: 18 May 2020;
Published: 28 July 2020.

Edited by:

Shaohui Wang, Louisiana College, United States

Reviewed by:

Muhammad Kamran Siddiqui, COMSATS University Islamabad, Lahore Campus, Pakistan
Azmat Ullah Khan Niazi, University of Lahore, Pakistan
Lidan Pei, Hefei Normal University, China

Copyright © 2020 Raza and Ji. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.

*Correspondence: Ying Ji, jiying@usst.edu.cn

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