ISSN 2148-838Xhttps://doi.org/10.13069/jacodesmath.1000842 J. Algebra Comb. Discrete Appl. 8(3) • 197–212 Received: 30 September 2020 Accepted: 20 May 2021 Journal of Algebra Combinatorics Discrete Structures and Applications On metric dimension of plane graphs Jn, Kn and Ln Research Article Sunny Kumar Sharma, Vijay Kumar Bhat Abstract: Let Γ = Γ(V,E) be a simple (i.e., multiple edges and loops and are not allowed), connected (i.e., there exists a path between every pair of vertices), and an undirected (i.e., all the edges are bidirectional) graph. Let dΓ(%i,%j ) denotes the geodesic distance between two nodes %i,%j ∈ V. The problem of characterizing the classes of plane graphs with constant metric dimensions is of great interest nowadays. In this article, we characterize three classes of plane graphs (viz., Jn, Kn, and Ln) which are generated by taking n-copies of the complete bipartite graph (or a star) K1,5, and all of these plane graphs are radially symmetrical with the constant metric dimension. We show that three vertices is a minimal requirement for the unique identification of all vertices of these three classes of plane graphs. 2010 MSC: 05C10, 05C12 Keywords: Resolving set, Metric basis, Independent set, Metric dimension, Planar graph 1. Introduction Let Γ be a simple connected graph with the vertex set V and the edge set E, and let dΓ(%i,%j) denotes the geodesic distance between two vertices %i,%j ∈ V. A subset of vertices R ⊆ V is said to be a resolving set (metric generator or locating set) if for every pair of distinct vertices ς,% ∈ V there exists at least one α ∈ R such that dΓ(α,ς) 6= dΓ(α,%). In other words, for an ordered subset of vertices R = {%1,%2,%3, ...,%k} of Γ, any vertex α ∈ V may be represented uniquely in the form of the vector γ(α|R) = (dΓ(α,%1),dΓ(α,%2), ...,dΓ(α,%k)). Then R is the metric generator of Γ if γ(δ|R) = γ(β|R) implies that δ = β, ∀ β,δ ∈ V. The metric generator R with the minimum possible cardinality is the metric basis for Γ, and this minimum cardinality is known as the metric dimension of Γ, denoted by β(Γ). A set S consisting of vertices of the graph Γ is said to be an independent metric generator for Γ, if S is both metric generator and independent. For the given ordered subset of vertices, R = {ρ1,ρ2,ρ3, ...,ρk} of Γ, the fth component (or distance coordinate) of the code γ(ρ|R) is zero iff ρ = ρf . Subsequently, to see that the set R is the metric generator, it is sufficient to prove that ζ(ρ|R) 6= ζ(%|R) for any pair of distinguishable nodes %,ρ ∈ V(Γ) \R. Sunny Kumar Sharma, Vijay Kumar Bhat (Corresponding Author); School of Mathematics, Shri Mata Vaishno Devi University, Katra-182320, India (email: sunnysrrm94@gmail.com, vijaykumarbhat2000@yahoo.com). 197 https://orcid.org/0000-0001-8423-9067 S. K. Sharma, V. K. Bhat / J. Algebra Comb. Discrete Appl. 8(3) (2021) 197–212 The notions of locating or resolving set and that of the metric dimension go back to 1950s. These were characterized by Blumenthal [2] with regard to metric space, and were acquainted with graph networks independently by Melter and Harary in 1976 [6], and Slater in 1975 [11]. Graph theory has applications in numerous zones of figuring, social, and normal sciences and is likewise an affable play area for the investigation of the verification procedure in discrete science. Utilizations of this invariant to the problem of picture preparing (or image processing) and design acknowledgment (or pattern recognition) are given in [9], to the route of exploring specialist (navigating agent or robots) in systems (or networks) are examined in [8], applications to science are given in [5], application to combinatorial enhancement (or optimization) is yielded in [10], and to problems of network discovery and verification in [1]. Let ℵ represent a family of connected graphs. We say that the family ℵ has constant metric dimension if dim(Γ) does not depend upon the choice of the graph Γ in ℵ and is finite. In other words, if all the graphs in ℵ have an indistinguishable metric dimension, at that point ℵ is known as a family with a constant (steady) metric dimension [12]. Chartrand et al. in [5], demonstrated that graphs on n vertices have metric dimension one iff it is a path ℘n. Additionally, cycle Cn has metric dimension two for each positive integer n; n ≥ 3. With this, Cn (n ≥ 3) and ℘n (n ≥ 2) establish a family of graphs with a steady metric dimension. Additionally, Harary graphs H4,n and generalized Petersen graphs P(n, 2), are also the families of graphs with constant metric dimension [7]. By joining of two graphs Γ1 = Γ1(V1,E1) and Γ2 = Γ2(V2,E2), denoted by Γ = Γ1 + Γ2, we mean a graph Γ = Γ(V,E) such that V = V1 ∪V2 and E = E1 ∪E2 ∪{%ς : % ∈ V1 and ς ∈ V2}. Then a Fan graph Fm is characterized as Fm = K1 + ℘m for m ≥ 1, a Wheel graph Wm is characterized as Wm = K1 +Cm, for m ≥ 3, and the Jahangir graph J2m (m ≥ 2) is obtained from the Wheel graph W2m by alternately deleting m spokes of the Wheel graph (which is otherwise known as the Gear graph). In [4], Caceres et al. decided the location number of the Fan graph Fm (m ≥ 1) which is b2m+25 c for m /∈ {1, 2, 3, 6}. Tomescu and Javaid [13] acquired the location number of the Jahangir graph J2m (m ≥ 4) which is b2m 3 c, and in [3] Chartrand et al. decided the location number of the Wheel graph Wm (m ≥ 3) which is b2m+2 5 c for m /∈{3, 6}. It is important to note that the metric dimension of these three graphs depend upon the number of vertices in the graph and thus these three families of graphs do not constitute the families with constant metric dimensions. For the simple connected graphs with the metric dimension 2, Khuller et al. [8] proved the following important result: Theorem 1.1. [8] Let A ⊆ V(Γ) be the basis set of the connected graph Γ = Γ(V,E) of cardinality two i.e., |A| = dim(Γ) = β(Γ) = 2, and say A = {$,ξ}. Then, the following are true: 1. Between the vertices $ and ξ, there exists a unique shortest path ℘. 2. The valencies (or degrees) of the nodes $ and ξ can never exceed 3. 3. The valency of any other node on ℘ can never exceed 5. The main motivation in characterizing the families of plane graphs with constant metric dimension (or with non-constant metric dimension) is towards making metric dimension of possibly all plane graphs known. In this article, we determine the metric dimension of three classes of plane graphs (viz., Jn, Kn, and Ln) which are generated by taking n-copies of the complete bipartite graph (or a star) K1,5 (see Figure 1). These classes of plane graphs are radially symmetric and possess an independent minimum resolving set with cardinality three i.e., three vertices is a minimal requirement for the unique identification of all vertices of these three classes of plane graphs. Throughout this article, all vertex indices are taken to be modulo n. In the accompanying section, we acquire the exact metric dimension of the radially symmetrical plane graph Jn (see Figure 2), and for each positive integer n: n ≥ 6 we prove that β(Jn) = 3. 198 S. K. Sharma, V. K. Bhat / J. Algebra Comb. Discrete Appl. 8(3) (2021) 197–212 Figure 1. n-copies of the star K1,5 2. Metric dimension of the planar graph Jn The construction of the plane graph Jn can be done in the following four steps: 1. Construct n-copies of the complete bipartite graph (or the star) K1,5. Denote the central node of each star by rl and the outer nodes of the star K1,5 by pl, ql, sl, al, and bl (1 ≤ l ≤ n). This results in a disconnected graph on 6n nodes with 5n edges (rlpl, rlql, rlsl, rlal, and rlbl for 1 ≤ l ≤ n). 2. Placing new edges between these stars as blal+1 and slql+1 for 1 ≤ l ≤ n. This adds 2n new edges. 3. Adding n new edges in each star as slql for 1 ≤ l ≤ n. 4. Finally, adding n new nodes {cl : 1 6 l 6 n}, and 2n new edges as plcl and clpl+1 for 1 ≤ l ≤ n. Thus, the radially symmetrical plane graph Jn comprises of 7n nodes and 10n edges. It has n 7-sided cycles, n 6-sided cycles, n 3-sided cycles, and a pair of 2n-sided faces (see Figure 2). Figure 2. The radially symmetrical graph Jn For our purpose, we name the cycle generated by the set of vertices {pl : 1 6 l 6 n}∪{cl : 1 6 l 6 n} in the graph, Jn as the inner cycle, the cycle generated by the set of vertices {rl : 1 6 l 6 n}∪{al : 1 6 l 6 n}∪{bl : 1 6 l 6 n} in the graph, Jn as the middle cycle, and the cycle generated by the set of vertices {ql : 1 6 l 6 n}∪{sl : 1 6 l 6 n} in the graph, Jn as the outer cycle. In the following theorem, we obtain that the minimum cardinality of resolving set for the plane graph, Jn is 3 i.e., three vertices is a minimal requirement for the unique identification of all vertices in the graph Jn. Theorem 2.1. Let Jn be the planar graph on 7n vertices as defined above. Then for each n ≥ 6, we have β(Jn) = 3 i.e., it has metric dimension 3. 199 S. K. Sharma, V. K. Bhat / J. Algebra Comb. Discrete Appl. 8(3) (2021) 197–212 Proof. To establish this, we study the following two cases relying upon the positive integer n i.e., when n is even and when it is odd. Case(1) When the integer n is even. In this case, the positive integer n can be written as n = 2w, where w ∈ N and w ≥ 3. Let R = {q1,qw+1,p1}⊂ V(Jn). Now, to unveil that R is the resolving set for the graph Jn, we consign the metric codes for each vertex of the graph Jn r R regarding the set R. Now, the metric codes for the vertices of inner cycle {pl : 1 6 l 6 n}∪{cl : 1 6 l 6 n} are γ(pl|R) = { (2l, 2w − 2l + 3, 2l− 2), 2 ≤ l ≤ w (4w − 2l + 3, 2l− 2w, 4w − 2l + 2), w + 1 ≤ l ≤ 2w. and γ(cl|R) =   (2l + 1, 2w − 2l + 2, 2l− 1), 1 ≤ l ≤ w − 1 (2w + 1, 3, 2w − 1), l = w; (4w − 2l + 2, 2l− 2w + 1, 4w − 2l + 1), w + 1 ≤ l ≤ 2w − 1; (3, 2l− 2w + 1, 4w − 2l + 1), l = 2w. The metric codes for the vertices of the middle cycle {rl : 1 6 l 6 n}∪{al : 1 6 l 6 n}∪{bl : 1 6 l 6 n} are γ(rl|R) =   (1, 2w, 1), l = 1; (2l− 1, 2w − 2l + 2, 2l− 1), 2 ≤ l ≤ w; (2w, 1, 2w + 1), l = w + 1; (4w − 2l + 2, 2l− 2w − 1, 4w − 2l + 3), w + 2 ≤ l ≤ 2w. γ(al|R) =   (2, 2w + 1, 2), l = 1; (2l− 1, 2w − 2l + 3, 2l− 1), 2 ≤ l ≤ w; (2w + 1, 2, 2w + 1), l = w + 1; (4w − 2l + 3, 2l− 2w − 1, 4w − 2l + 4), w + 2 ≤ l ≤ 2w. and γ(bl|R) =   (2, 2w, 2), l = 1; (2l, 2w − 2l + 2, 2l), 2 ≤ l ≤ w − 1; (2w, 3, 2w), l = w; (2w, 2, 2w + 1), l = w + 1; (4w − 2l + 2, 2l− 2w, 4w − 2l + 3), w + 2 ≤ l ≤ 2w − 1; (3, 2l− 2w, 4w − 2l + 3), l = 2w. At last, the metric codes for the vertices of outer cycle {sl : 1 6 l 6 n}∪{ql : 1 6 l 6 n} are γ(sl|R) = { (2l− 1, 2w − 2l + 1, 2l), 1 ≤ l ≤ w; (4w − 2l + 1, 2l− 2w − 1, 4w − 2l + 3), w + 1 ≤ l ≤ 2w. and γ(ql|R) = { (2l− 2, 2w − 2l + 2, 2l− 1), 2 ≤ l ≤ w; (4w − 2l + 2, 2l− 2w − 2, 4w − 2l + 4), w + 2 ≤ l ≤ 2w. 200 S. K. Sharma, V. K. Bhat / J. Algebra Comb. Discrete Appl. 8(3) (2021) 197–212 From above, we find that there do not exist two vertices with the same metric codes, which suggest that β(Jn) ≤ 3 i.e., the location number of the plane graph Jn is less than or equal to 3. Now, to finish the evidence for this case, we show that β(Jn) ≥ 3 by working out that there does not exist a resolving set R such that |R| = 2. On contrary, suppose that β(Jn) = 2. Now, from Theorem 1.1, we find that the degree of basis vertices can be at most 3. But except the vertices pl, ql, sl, al, bl, and cl (1 ≤ l ≤ n), all other vertices of the radially symmetrical plane graph Jn have a degree 5. Then, without loss of generality, we suppose that the first resolving vertex is one of the vertices p1, q1, s1, a1, bl, or c1 and other nodes lie in the inner cycle, the middle cycle, and in the outer cycle. Therefore, we have the following possibilities to be discussed. Resolving sets Contradictions {p1,ph}, ph (2 ≤ h ≤ n) For 2 ≤ h ≤ w, we have γ(r1|{p1,ph}) = γ(cn|{p1,ph}); and for h = w + 1, we have γ(c1|{p1,pw+1}) = γ(cn|{p1,pw+1}), a contradiction. {c1,ch}, ch (2 ≤ h ≤ n) For 2 ≤ h ≤ w, we have γ(r1|{c1,ch}) = γ(cn|{c1,ch}); and for h = w + 1, we have γ(p1|{c1,cw+1}) = γ(p2|{c1,cw+1}), a contradiction. {a1,ah}, ah (2 ≤ h ≤ n) For 2 ≤ h ≤ w + 1, we have γ(p1|{a1,ah}) = γ(q1|{a1,ah}), a contradiction. {b1,bh}, bh (2 ≤ h ≤ n) For 2 ≤ h ≤ w, we have γ(p1|{b1,bh}) = γ(q1|{b1,bh}); and for h = w + 1, we have γ(q1|{b1,bw+1}) = γ(s1|{b1,bw+1}), a contradiction. {q1,qh}, qh (2 ≤ h ≤ n) For 2 ≤ h ≤ w, we have γ(bn|{q1,qh}) = γ(cn|{q1,qh}); and for h = w + 1, we have γ(sn|{q1,qw+1}) = γ(s1|{q1,qw+1}), a contradiction. {s1,sh}, sh (2 ≤ h ≤ n) For 2 ≤ h ≤ w, we have γ(bn|{s1,sh}) = γ(cn|{s1,sh}); and for h = w + 1, we have γ(q2|{s1,sw+1}) = γ(q1|{s1,sw+1}), a contradiction. {p1,ch}, ch (1 ≤ h ≤ n) For 1 ≤ h ≤ w, we have γ(cn|{p1,ch}) = γ(r1|{p1,ch}); and for h = w + 1, we have γ(q2|{p1,cw+1}) = γ(a2|{p1,cw+1}), a contradiction. {p1,ah}, ah (1 ≤ h ≤ n) For h = 1, we have γ(q1|{p1,a1}) = γ(b1|{p1,a1}); when h = 2, we have γ(s2|{p1,a2}) = γ(b2|{p1,a2}); when h = 3, we have γ(s3|{p1,a3}) = γ(b3|{p1,a3}); and for 4 ≤ h ≤ w + 1, we have γ(q1|{p1,ah}) = γ(b1|{p1,ah}), a contradiction. {p1,bh}, bh (1 ≤ h ≤ n) For h = 1, we have γ(q1|{p1,b1}) = γ(s1|{p1,b1}); when h = 2, we have γ(a2|{p1,b2}) = γ(q2|{p1,b2}); when 3 ≤ h ≤ w, we have γ(q1|{p1,bh}) = γ(b1|{p1,bh}); and for h = w + 1, we have γ(r2|{p1,bw+1}) = γ(bn|{p1,bw+1}), a contradiction. {p1,qh}, qh (1 ≤ h ≤ n) For h = 1, we have γ(a1|{p1,q1}) = γ(b1|{p1,q1}); when h = 2, we have γ(p2|{p1,q2}) = γ(q1|{p1,q2}); and for 3 ≤ h ≤ w + 1, we have γ(q1|{p1,qh}) = γ(b1|{p1,qh}), a contradiction. {p1,sh}, sh (1 ≤ h ≤ n) For h = 1, we have γ(a1|{p1,s1}) = γ(b1|{p1,s1}); when 2 ≤ h ≤ w, we have γ(q1|{p1,sh}) = γ(b1|{p1,sh}); and for h = w + 1, we have γ(r2|{p1,sw+1}) = γ(bn|{p1,sw+1}), a contradiction. {c1,ah}, ah (1 ≤ h ≤ n) For 1 ≤ h ≤ 2, we have γ(s1|{c1,ah}) = γ(q1|{c1,ah}); when h = 3, we have γ(a2|{c1,a3}) = γ(q2|{c1,a3}); and for 4 ≤ h ≤ w + 1, we have γ(a2|{c1,ah}) = γ(s1|{c1,ah}), a contradiction. {c1,bh}, bh (1 ≤ h ≤ n) For h = 1, we have γ(s1|{c1,b1}) = γ(q1|{c1,b1}); when h = 2, we have γ(a2|{c1,a2}) = γ(q2|{c1,a2}); and for 3 ≤ h ≤ w + 1, we have γ(a2|{c1,bh}) = γ(s1|{c1,bh}), a contradiction. {c1,sh}, sh (1 ≤ h ≤ n) For h = 1, we have γ(a1|{c1,s1}) = γ(b1|{c1,s1}); and for 2 ≤ h ≤ w + 1, we have γ(a2|{c1,sh}) = γ(s1|{c1,sh}), a contradiction. {c1, th}, th (1 ≤ h ≤ n) For h = 1, we have γ(a1|{c1, t1}) = γ(b1|{c1, t1}); when h = 2, we have γ(a2|{c1, t2}) = γ(b2|{c1, t2}); and for 3 ≤ h ≤ w + 1, we have γ(a2|{c1, th}) = γ(s1|{c1, th}), a contradiction. {a1,bh}, bh (1 ≤ h ≤ n) For h = 1, we have γ(s1|{a1,b1}) = γ(q1|{a1,b1}); when h = 2, we have γ(a2|{a1,b2}) = γ(q2|{a1,b2}); when h = 3, we have γ(a3|{a1,b3}) = γ(q2|{a1,b3}); when 4 ≤ h ≤ w, we have γ(q1|{a1,bh}) = γ(b1|{a1,bh}); and for h = w + 1, we have γ(an|{a1,bw+1}) = γ(sn|{a1,bw+1}), a contradiction. {a1,sh}, sh (1 ≤ h ≤ n) For h = 1, we have γ(b1|{a1,s1}) = γ(p1|{a1,s1}); when 2 ≤ h ≤ w, we have γ(q1|{a1,sh}) = γ(b1|{a1,sh}); and for h = w + 1, we have γ(an|{a1,sw+1}) = γ(sn|{a1,sw+1}), a contradiction. 201 S. K. Sharma, V. K. Bhat / J. Algebra Comb. Discrete Appl. 8(3) (2021) 197–212 Resolving sets Contradictions {a1,qh}, qh (1 ≤ h ≤ n) For h = 1, we have γ(b1|{a1,q1}) = γ(p1|{a1,q1}); when h = 2, we have γ(sn|{a1,q2}) = γ(c1|{a1,q2}); and for 3 ≤ h ≤ w + 1, we have γ(q1|{a1,qh}) = γ(b1|{a1,qh}), a contradiction. {b1,sh}, sh (1 ≤ h ≤ n) For h = 1, we have γ(a1|{b1,s1}) = γ(p1|{b1,s1}); when h = 2, we have γ(b2|{b1,s2}) = γ(p2|{b1,s2}); and for 3 ≤ h ≤ w + 1, we have γ(q2|{b1,sh}) = γ(b2|{b1,sh}), a contradiction. {b1,qh}, qh (1 ≤ h ≤ n) For h = 1, we have γ(a1|{b1,q1}) = γ(p1|{b1,q1}); when 2 ≤ h ≤ 3, we have γ(b2|{b1,q2}) = γ(p2|{b1,q2}); and for 4 ≤ h ≤ w + 1, we have γ(q2|{b1,qh}) = γ(b2|{b1,qh}), a contradiction. {s1,qh}, qh (1 ≤ h ≤ n) For h = 1, we have γ(a1|{s1,q1}) = γ(b1|{s1,q1}); and for 2 ≤ h ≤ w + 1, we have γ(r1|{s1,qh}) = γ(q1|{s1,qh}), a contradiction. Thus, from the above table, we obtain that there does not exist a resolving set consisting of two vertices for V(Jn), suggesting that β(Jn) = 3 in this case. Case(2) When the integer n is odd. In this case, the positive integer n can be written as n = 2w + 1, where w ∈ N and w ≥ 3. Let R = {q1,qw+1,p1}⊂ V(Jn). Now, to unveil that R is the resolving set for the graph Jn, we consign the metric codes for each vertex of the graph Jn r R regarding the set R. Now, the metric codes for the vertices of inner cycle {pl : 1 6 l 6 n}∪{cl : 1 6 l 6 n} are γ(pl|R) =   (2l, 2w − 2l + 3, 2l− 2), 2 ≤ l ≤ w (2w + 2, 2, 2w), l = w + 1; (4w − 2l + 5, 2l− 2w, 4w − 2l + 4), w + 2 ≤ l ≤ 2w + 1. and γ(cl|R) =   (2l + 1, 2w − 2l + 2, 2l− 1), 1 ≤ l ≤ w − 1 (2w + 1, 3, 2w − 1), l = w; (4w − 2l + 4, 2l− 2w + 1, 4w − 2l + 3), w + 1 ≤ l ≤ 2w; (3, 2l− 2w, 1), l = 2w + 1. The metric codes for the vertices of the middle cycle {rl : 1 6 l 6 n}∪{al : 1 6 l 6 n}∪{bl : 1 6 l 6 n} are γ(rl|R) =   (1, 2w, 1), l = 1; (2l− 1, 2w − 2l + 2, 2l− 1), 2 ≤ l ≤ w; (2w + 1, 1, 2w + 1), l = w + 1; (4w − 2l + 4, 2l− 2w − 1, 4w − 2l + 5), w + 2 ≤ l ≤ 2w + 1. γ(al|R) =   (2, 2w + 1, 2), l = 1; (2l− 1, 2w − 2l + 3, 2l− 1), 2 ≤ l ≤ w; (2w + 1, 2, 2w + 1), l = w + 1; (4w − 2l + 5, 2l− 2w − 1, 4w − 2l + 6), w + 2 ≤ l ≤ 2w + 1. and γ(bl|R) =   (2, 2w, 2), l = 1; (2l, 2w − 2l + 2, 2l), 2 ≤ l ≤ w − 1; (2w, 3, 2w), l = w; (2w + 2, 2, 2w + 2), l = w + 1; (4w − 2l + 4, 2l− 2w, 4w − 2l + 5), w + 2 ≤ l ≤ 2w; (3, 2l− 2w, 4w − 2l + 5), l = 2w + 1. 202 S. K. Sharma, V. K. Bhat / J. Algebra Comb. Discrete Appl. 8(3) (2021) 197–212 At last, the metric codes for the vertices of outer cycle {sl : 1 6 l 6 n}∪{ql : 1 6 l 6 n} are γ(sl|R) =   (2l− 1, 2w − 2l + 1, 2l), 1 ≤ l ≤ w; (2w + 1, 1, 2w + 2), l = w + 1; (4w − 2l + 3, 2l− 2w − 1, 4w − 2l + 5), w + 2 ≤ l ≤ 2w + 1. and γ(ql|R) = { (2l− 2, 2w − 2l + 2, 2l− 1), 2 ≤ l ≤ w; (4w − 2l + 4, 2l− 2w − 2, 4w − 2l + 6), w + 2 ≤ l ≤ 2w + 1. Again, we find that there do not exist two vertices with the same metric codes, which suggest that β(Jn) ≤ 3 i.e., the location number of the plane graph Jn is less than or equal to 3. Now, on assuming that β(Jn) = 2, we get the same eventualities as in Case(1), and similarly, the contradiction can be obtained. So, in this case, we have β(Jn) = 3 as well and hence the theorem. Now, in terms of independent resolving set, we have the following result: Theorem 2.2. Let Jn be the planar graph on 7n vertices as defined above. Then for every positive integer n; n ≥ 6, its independent resolving number is 3. Proof. For proof, refer to Theorem 2.1. In the accompanying section, we acquire the exact metric dimension of the radially symmetrical plane graph Kn (see Figure 3), and for each positive integer n: n ≥ 6 we prove that β(Kn) = 3. 3. Metric dimension of the planar graph Kn The construction of the plane graph Kn can be done in the following four steps: 1. Construct n-copies of the complete bipartite graph (or the star) K1,5. Denote the central node of each star by rl and the outer nodes of the star K1,5 by pl, ql, sl, al, and bl (1 ≤ l ≤ n). This results in a disconnected graph on 6n nodes with 5n edges (rlpl, rlql, rlsl, rlal, and rlbl for 1 ≤ l ≤ n). 2. Placing new edges between these stars as qlpl+1 and blal+1 for 1 ≤ l ≤ n. This adds 2n new edges. 3. Adding n new edges in each of these stars as plql for 1 ≤ l ≤ n. 4. Finally, adding n new nodes {tl : 1 6 l 6 n}, and 2n new edges as sltl and tlsl+1 for 1 ≤ l ≤ n. Thus, the radially symmetrical plane graph Kn comprises of 7n nodes and 10n edges. It has n 7-sided cycles, n 6-sided cycles, n 3-sided cycles, and a pair of 2n-sided faces (see Figure 3). For our purpose, we name the cycle generated by the set of vertices {pl : 1 6 l 6 n}∪{ql : 1 6 l 6 n} in the graph, Kn as the inner cycle, the cycle generated by the set of vertices {rl : 1 6 l 6 n}∪{al : 1 6 l 6 n}∪{bl : 1 6 l 6 n} in the graph, Kn as the middle cycle, and the cycle generated by the set of vertices {tl : 1 6 l 6 n}∪{sl : 1 6 l 6 n} in the graph, Kn as the outer cycle. In the following theorem, we obtain that the minimum cardinality of resolving set for the plane graph, Kn is 3 i.e., three vertices is a minimal requirement for the unique identification of all vertices in the graph Kn. Theorem 3.1. Let Kn be the planar graph on 7n vertices as defined above. Then for each n ≥ 6, we have β(Kn) = 3 i.e., it has metric dimension 3. 203 S. K. Sharma, V. K. Bhat / J. Algebra Comb. Discrete Appl. 8(3) (2021) 197–212 Figure 3. The radially symmetrical graph Kn Proof. To establish this, we study the following two cases relying upon the positive integer n i.e., when n is even and when it is odd. Case(1) When the integer n is even. In this case, the positive integer n can be written as n = 2w, where w ∈ N and w ≥ 3. Let R = {p1,pw+1,s2}⊂ V(Kn). Now, to unveil that R is the resolving set for the graph Kn, we consign the metric codes for each vertex of the graph Kn r R regarding R. Now, the metric codes for the vertices of inner cycle {pl : 1 6 l 6 n}∪{ql : 1 6 l 6 n} are γ(pl|R) =   (2, 2w − 2, 2), l = 2; (2l− 2, 2w − 2l + 2, 2l− 3), 3 ≤ l ≤ w (4w − 2l + 2, 2, 2w + 1), l = w + 2; (4w − 2l + 2, 2l− 2w − 2, 4w − 2l + 6), w + 3 ≤ l ≤ 2w. and γ(ql|R) =   (1, 2w − 1, 3), l = 1; (2l− 1, 2w − 2l + 1, 2l− 2), 2 ≤ l ≤ w (4w − 2l + 1, 1, 2w), l = w + 1; (4w − 2l + 1, 2l− 2w − 1, 4w − 2l + 5), w + 2 ≤ l ≤ 2w. The metric codes for the vertices of the middle cycle {rl : 1 6 l 6 n}∪{al : 1 6 l 6 n}∪{bl : 1 6 l 6 n} are γ(rl|R) =   (1, 2w, 3), l = 1; (2l− 1, 2w − 2l + 2, 2l− 3), 2 ≤ l ≤ w; (2w, 1, 2w − 1), l = w + 1; (4w − 2l + 2, 2l− 2w − 1, 4w − 2l + 5), w + 2 ≤ l ≤ 2w. γ(al|R) =   (2, 2w + 1, 4), l = 1; (3, 2w − 1, 2), l = 2; (2l− 1, 2w − 2l + 3, 2l− 3), 3 ≤ l ≤ w; (2w + 1, 2, 2w − 1), l = w + 1; (4w − 2l + 3, 2l− 2w − 1, 2w + 1), l = w + 2; (4w − 2l + 3, 2l− 2w − 1, 4w − 2l + 6), w + 3 ≤ l ≤ 2w. 204 S. K. Sharma, V. K. Bhat / J. Algebra Comb. Discrete Appl. 8(3) (2021) 197–212 and γ(bl|R) =   (2, 2w, 3), l = 1; (2l, 2w − 2l + 2, 2l− 2), 2 ≤ l ≤ w − 1; (2w, 3, 2w − 2), l = w; (2w, 2, 2w), l = w + 1; (4w − 2l + 2, 2l− 2w, 4w − 2l + 5), w + 2 ≤ l ≤ 2w − 1; (3, 2l− 2w, 4w − 2l + 5), l = 2w. At last, metric codes for the vertices of outer cycle {sl : 1 6 l 6 n}∪{tl : 1 6 l 6 n} are γ(sl|R) =   (2l, 2w − 2l + 3, 2), l = 1; (2l, 2w − 2l + 3, 2l− 4), 3 ≤ l ≤ w; (2w + 1, 2, 2w − 2), l = w + 1; (4w − 2l + 3, 2l− 2w, 4w − 2l + 4), w + 2 ≤ l ≤ 2w. and γ(tl|R) =   (3, 2w, 1), l = 1; (2l + 1, 2w − 2l + 2, 2l− 3), 2 ≤ l ≤ w − 1; (2w + 1, 3, 2w − 3), l = w; (4w − 2l + 2, 3, 2w − 1), l = w + 1; (4w − 2l + 2, 2l− 2w + 1, 4w − 2l + 3), w + 2 ≤ l ≤ 2w − 1; (3, 2l− 2w + 1, 4w − 2l + 3), l = 2w. From above, we find that there do not exist two vertices with the same metric codes, which suggest that β(Kn) ≤ 3 i.e., the location number of the plane graph Kn is less than or equal to 3. Now, so as to finish the evidence for this case, we show that β(Kn) ≥ 3 by working out that there does not exist a resolving set R such that |R| = 2. On contrary, suppose that β(Kn) = 2. Now, from Theorem 1.1, we find that the degree of basis vertices can be at most 3. But except the vertices pl, ql, sl, al, bl, and tl (1 ≤ l ≤ n), all other vertices of the radially symmetrical plane graph Kn have a degree 5. Then, without loss of generality, we suppose that the first resolving vertex is one of the vertices p1, q1, s1, a1, bl or t1 and other nodes lie in the inner cycle, the middle cycle, and in the outer cycle. Therefore, we have the following possibilities to be discussed. Resolving sets Contradictions {p1,ph}, ph (2 ≤ h ≤ n) For 2 ≤ h ≤ w + 1, we have γ(a1|{p1,ph}) = γ(s1|{p1,ph}), a contradiction. {q1,qh}, qh (2 ≤ h ≤ n) For 2 ≤ h ≤ w, we have γ(a1|{q1,qh}) = γ(s1|{q1,qh}); and for h = w + 1, we have γ(p2|{q1,qw+1}) = γ(p1|{q1,qw+1}), a contradiction. {a1,ah}, ah (2 ≤ h ≤ n) For 2 ≤ h ≤ w + 1, we have γ(p2|{a1,ah}) = γ(s2|{a1,ah}), a contradiction. {b1,bh}, bh (2 ≤ h ≤ n) For 2 ≤ h ≤ w + 1, we have γ(p2|{b1,bh}) = γ(s2|{b1,bh}), a contradiction. {s1,sh}, sh (2 ≤ h ≤ n) For 2 ≤ h ≤ w, we have γ(r1|{s1,sh}) = γ(tn|{s1,sh}); and for h = w + 1, we have γ(t1|{s1,sw+1}) = γ(tn|{s1,sw+1}), a contradiction. {t1, th}, th (2 ≤ h ≤ n) For 2 ≤ h ≤ w, we have γ(r1|{t1, th}) = γ(tn|{t1, th}); and for h = w + 1, we have γ(s1|{t1, tw+1}) = γ(s2|{t1, tw+1}), a contradiction. {p1,qh}, qh (1 ≤ h ≤ n) For 1 ≤ h ≤ w, we have γ(s1|{p1,qh}) = γ(a1|{p1,qh}); and for h = w + 1, we have γ(bn|{p1,qw+1}) = γ(r2|{p1,qw+1}), a contradiction. {p1,ah}, ah (1 ≤ h ≤ n) For h = 1, we have γ(s1|{p1,a1}) = γ(b1|{p1,a1}); when h = 2, we have γ(s2|{p1,a2}) = γ(b2|{p1,a2}); when h = 3, we have γ(s3|{p1,a3}) = γ(b3|{p1,a3}); and for 4 ≤ h ≤ w + 1, we have γ(b1|{p1,ah}) = γ(s1|{p1,ah}), a contradiction. {p1,bh}, bh (1 ≤ h ≤ n) For h = 1, we have γ(s1|{p1,b1}) = γ(a1|{p1,b1}); when h = 2, we have γ(sn|{p1,b2}) = γ(bn|{p1,b2}); when 3 ≤ h ≤ w, we have γ(s1|{p1,bh}) = γ(b1|{p1,bh}); and for h = w + 1, we have γ(b1|{p1,bh}) = γ(s1|{p1,bh}), a contradiction. 205 S. K. Sharma, V. K. Bhat / J. Algebra Comb. Discrete Appl. 8(3) (2021) 197–212 Resolving sets Contradictions {p1,sh}, sh (1 ≤ h ≤ n) For h = 1, we have γ(b1|{p1,s1}) = γ(a1|{p1,s1}); and for 2 ≤ h ≤ w + 1, we have γ(t1|{p1,sh}) = γ(r2|{p1,sh}), a contradiction. {p1, th}, th (1 ≤ h ≤ n) For h = 1, we have γ(b1|{p1, t1}) = γ(a1|{p1, t1}); and for 2 ≤ h ≤ w + 1, we have γ(t1|{p1, th}) = γ(r2|{p1, th}), a contradiction. {q1,ah}, ah (1 ≤ h ≤ n) For h = 1, we have γ(b1|{q1,a1}) = γ(s1|{q1,a1}); when h = 2, we have γ(a1|{q1,a2}) = γ(s1|{q1,a2}); when h = 3, we have γ(b3|{q1,a3}) = γ(s3|{q1,a3}); and for 4 ≤ h ≤ w + 1, we have γ(s1|{q1,ah}) = γ(b1|{q1,ah}), a contradiction. {q1,bh}, bh (1 ≤ h ≤ n) For h = 1, we have γ(a1|{q1,b1}) = γ(s1|{q1,b1}); when h = 2, we have γ(a2|{q1,b2}) = γ(s2|{q1,b2}); when 3 ≤ h ≤ w, we have γ(b1|{q1,bh}) = γ(s1|{q1,bh}); and for h = w + 1, we have γ(rn|{q1,bw+1}) = γ(a2|{q1,bw+1}), a contradiction. {q1,sh}, sh (1 ≤ h ≤ n) For h = 1, we have γ(b1|{q1,s1}) = γ(a1|{q1,s1}), when h = 2, we have γ(a2|{q1,s2}) = γ(b2|{q1,s2}); and for 3 ≤ h ≤ w + 1, we have γ(r2|{q1,sh}) = γ(q2|{q1,sh}), a contradiction. {q1, th}, th (1 ≤ h ≤ n) For h = 1, we have γ(b1|{q1, t1}) = γ(a1|{q1, t1}); when h = 2, we have γ(a2|{q1, t2}) = γ(b2|{q1, t2}); and for 3 ≤ h ≤ w + 1, we have γ(r2|{q1, th}) = γ(q2|{q1, th}), a contradiction. {a1,bh}, bh (1 ≤ h ≤ n) For h = 1, we have γ(p1|{a1,b1}) = γ(q1|{a1,b1}); when h = 2, we have γ(s2|{a1,b2}) = γ(q2|{a1,b2}); and for 3 ≤ h ≤ w + 1, we have γ(s1|{a1,bh}) = γ(b1|{a1,bh}), a contradiction. {a1,sh}, sh (1 ≤ h ≤ n) For h = 1, we have γ(p1|{a1,s1}) = γ(q1|{a1,s1}), when h = 2, we have γ(a2|{a1,s2}) = γ(p2|{a1,s2}), and for 3 ≤ h ≤ w + 1, we have γ(r2|{a1,sh}) = γ(q2|{a1,sh}), a contradiction. {a1, th}, th (1 ≤ h ≤ n) For h = 1, we have γ(p1|{a1, t1}) = γ(q1|{a1, t1}); when h = 2, we have γ(a2|{a1, t2}) = γ(p2|{a1, t2}); and for 3 ≤ h ≤ w + 1, we have γ(r2|{a1, th}) = γ(q2|{a1, th}), a contradiction. {b1,sh}, sh (1 ≤ h ≤ n) For h = 1, we have γ(p1|{b1,s1}) = γ(q1|{b1,s1}); when h = 2, we have γ(q2|{b1,s2}) = γ(p2|{b1,s2}); and for 3 ≤ h ≤ w + 1, we have γ(b2|{b1,sh}) = γ(t1|{b1,sh}), a contradiction. {b1, th}, th (1 ≤ h ≤ n) For h = 1, we have γ(p1|{b1, t1}) = γ(q1|{b1, t1}); when h = 2, we have γ(q2|{b1, t2}) = γ(p2|{b1, t2}); and for 3 ≤ h ≤ w + 1, we have γ(b2|{b1, th}) = γ(t1|{b1, th}), a contradiction. {s1, th}, th (1 ≤ h ≤ n) For 1 ≤ h ≤ w, we have γ(r1|{s1, th}) = γ(tn|{s1, th}); and for h = w + 1, we have γ(b2|{s1, tw+1}) = γ(r2|{s1, tw+1}), a contradiction. Thus, from the above table, we obtain that there does not exist a resolving set consisting of two vertices for V(Kn), suggesting that β(Kn) = 3 in this case. Case(2) When the integer n is odd. In this case, the positive integer n can be written as n = 2w + 1, where w ∈ N and w ≥ 3. Let R = {p1,pw+1,s2}⊂ V(Kn). Now, to unveil that R is the resolving set for the graph Kn, we consign the metric codes for each vertex of the graph Kn r R regarding the set R. Now, the metric codes for the vertices of inner cycle {pl : 1 6 l 6 n}∪{ql : 1 6 l 6 n} are γ(pl|R) =   (2, 2w − 2, 2), l = 2; (2l− 2, 2w − 2l + 2, 2l− 3), 3 ≤ l ≤ w (4w − 2l + 4, 2, 2w + 1), l = w + 2; (4w − 2l + 4, 2l− 2w − 2, 4w − 2l + 8), w + 3 ≤ l ≤ 2w + 1. 206 S. K. Sharma, V. K. Bhat / J. Algebra Comb. Discrete Appl. 8(3) (2021) 197–212 and γ(ql|R) =   (1, 2w − 1, 3), l = 1; (2l− 1, 2w − 2l + 1, 2l− 2), 2 ≤ l ≤ w (4w − 2l + 3, 1, 2w), l = w + 1; (4w − 2l + 3, 2l− 2w − 1, 2w + 2), l = w + 2; (4w − 2l + 3, 2l− 2w − 1, 4w − 2l + 7), w + 3 ≤ l ≤ 2w + 1. The metric codes for the vertices of the middle cycle {rl : 1 6 l 6 n}∪{al : 1 6 l 6 n}∪{bl : 1 6 l 6 n} are γ(rl|R) =   (1, 2w, 3), l = 1; (2l− 1, 2w − 2l + 2, 2l− 3), 2 ≤ l ≤ w; (2w + 1, 1, 2w − 1), l = w + 1; (4w − 2l + 4, 3, 2w + 1), l = w + 2; (4w − 2l + 4, 2l− 2w − 1, 4w − 2l + 7), w + 3 ≤ l ≤ 2w + 1. γ(al|R) =   (2, 2w + 1, 4), l = 1; (3, 2w − 1, 2), l = 2; (2l− 1, 2w − 2l + 3, 2l− 3), 3 ≤ l ≤ w; (2w + 1, 2, 2w − 1), l = w + 1; (4w − 2l + 5, 2l− 2w − 1, 2w + 1), l = w + 2; (4w − 2l + 5, 2l− 2w − 1, 4w − 2l + 8), w + 3 ≤ l ≤ 2w + 1. and γ(bl|R) =   (2, 2w, 3), l = 1; (2l, 2w − 2l + 2, 2l− 2), 2 ≤ l ≤ w − 1; (2w, 3, 2w − 2), l = w; (2w + 2, 2, 2w), l = w + 1; (4w − 2l + 4, 2l− 2w, 2w + 2), l = w + 2; (4w − 2l + 4, 2l− 2w, 4w − 2l + 7), w + 3 ≤ l ≤ 2w; (3, 2l− 2w, 4w − 2l + 7), l = 2w + 1. At last, the metric codes for the vertices of outer cycle {sl : 1 6 l 6 n}∪{tl : 1 6 l 6 n} are γ(sl|R) =   (2l, 2w − 2l + 3, 2), l = 1; (2l, 2w − 2l + 3, 2l− 4), 3 ≤ l ≤ w; (2w + 2, 2, 2w − 2), l = w + 1; (4w − 2l + 5, 2l− 2w, 2w), l = w + 2; (4w − 2l + 5, 2l− 2w, 4w − 2l + 6), w + 3 ≤ l ≤ 2w + 1. and γ(tl|R) =   (3, 2w, 1), l = 1; (2l + 1, 2w − 2l + 2, 2l− 3), 2 ≤ l ≤ w − 1; (2w + 1, 3, 2w − 3), l = w; (4w − 2l + 4, 3, 2w − 1), l = w + 1; (4w − 2l + 4, 2l− 2w + 1, 4w − 2l + 5), w + 2 ≤ l ≤ 2w; (3, 2w + 2, 4w − 2l + 5), l = 2w + 1. 207 S. K. Sharma, V. K. Bhat / J. Algebra Comb. Discrete Appl. 8(3) (2021) 197–212 Again, we find that there do not exist two vertices with the same metric codes, which suggest that β(Kn) ≤ 3 i.e., the location number of the plane graph Kn is less than or equal to 3. Now, on assuming that β(Kn) = 2, we get the same eventualities as in Case(1), and similarly, the contradiction can be obtained. So, in this case, we have β(Kn) = 3 as well and hence the theorem. Now, in terms of independent resolving set, we have the following result: Theorem 3.2. Let Kn be the planar graph on 7n vertices as defined above. Then for every positive integer n; n ≥ 6, its independent resolving number is 3. Proof. For proof, refer to Theorem 3.1. In the accompanying section, we acquire the exact metric dimension of the radially symmetrical plane graph Ln (see Figure 4), and for each positive integer n: n ≥ 6 we prove that β(Ln) = 3. 4. Metric dimension of the planar graph Ln The construction of the plane graph Ln can be done in the following three steps: 1. Construct n-copies of the complete bipartite graph (or the star) K1,5. Denote the central node of each star by rl and the outer nodes of the star K1,5 by pl, ql, sl, al, and bl (1 ≤ l ≤ n). This results in a disconnected graph on 6n nodes with 5n edges (rlpl, rlql, rlsl, rlal, and rlbl for 1 ≤ l ≤ n). 2. Placing new edges between these stars as qlpl+1 and blal+1 for 1 ≤ l ≤ n. This adds 2n new edges. 3. Finally adding 2n new edges in each star as plql and albl for 1 ≤ l ≤ n. Thus, the radially symmetrical plane graph Ln comprises of 6n nodes and 9n edges. It has n 6-sided cycles, 2n 3-sided cycles, and a pair of 2n-sided faces (see Figure 4). Figure 4. The radially symmetrical graph Ln For our purpose, we name the cycle generated by the set of vertices {pl : 1 6 l 6 n}∪{ql : 1 6 l 6 n} in the graph, Ln as the inner cycle, the set of vertices {rl : 1 6 l 6 n}∪{sl : 1 6 l 6 n} in the graph, Ln as the set of middle vertices, and the cycle generated by the set of vertices {al : 1 6 l 6 n}∪{bl : 1 6 l 6 n} in the graph, Ln as the outer cycle. In the following theorem, we obtain that the minimum cardinality of resolving set for the plane graph, Ln is 3 i.e., three vertices is a minimal requirement for the unique identification of all vertices in the graph Ln. 208 S. K. Sharma, V. K. Bhat / J. Algebra Comb. Discrete Appl. 8(3) (2021) 197–212 Theorem 4.1. Let Ln be the planar graph on 6n vertices as defined above. Then for each n > 6, we have β(Ln) = 3 i.e., it has metric dimension 3. Proof. In order to establish this, we study the following two cases relying upon the positive integer n i.e., when n is even natural and when it is odd. Case(1) When the integer n is even. In this case, the positive integer n can be written as n = 2w, where w ∈ N and w ≥ 3. Let R = {p1,pw+1,s1}⊂ V(Ln). Now, to unveil that R is the resolving set for the graph Ln, we consign the metric codes for each vertex of the graph Ln r R regarding the set R. Now, the metric codes for the vertices of inner cycle {pl : 1 6 l 6 n}∪{ql : 1 6 l 6 n} are γ(pl|R) = { (2l− 2, 2w − 2l + 2, 2l− 1), 2 ≤ l ≤ w (4w − 2l + 2, 2l− 2w − 2, 4w − 2l + 4), w + 2 ≤ l ≤ 2w. and γ(ql|R) = { (2l− 1, 2w − 2l + 1, 2l), 1 ≤ l ≤ w (4w − 2l + 1, 2l− 2w − 1, 4w − 2l + 3), w + 1 ≤ l ≤ 2w. The metric codes for the set of middle vertices {rl : 1 6 l 6 n}∪{sl : 1 6 l 6 n} are γ(rl|R) =   (1, 2w, 1), l = 1; (2l− 1, 2w − 2l + 2, 2l), 2 ≤ l ≤ w; (4w − 2l + 2, 2l− 2w − 1, 4w − 2l + 4), w + 1 ≤ l ≤ 2w. , and γ(sl|R) = { (2l, 2w − 2l + 3, 2l + 1), 2 ≤ l ≤ w; (4w − 2l + 3, 2l− 2w, 4w − 2l + 5), w + 1 ≤ l ≤ 2w. At last, the metric codes for the vertices of outer cycle {al : 1 6 l 6 n}∪{bl : 1 6 l 6 n} are γ(al|R) =   (2, 2w + 1, 2), l = 1; (2l− 1, 2w − 2l + 3, 2l− 1), 2 ≤ l ≤ w; (2w + 1, 2, 2w + 1), l = w + 1; (4w − 2l + 3, 2l− 2w − 1, 4w − 2l + 4), w + 2 ≤ l ≤ 2w. and γ(bl|R) =   (2, 2w, 2), l = 1; (2l, 2w − 2l + 2, 2l), 2 ≤ l ≤ w; (4w − 2l + 2, 2l− 2w, 4w − 2l + 3), w + 1 ≤ l ≤ 2w − 1; (3, 2l− 2w, 4w − 2l + 3), l = 2w. From above, we find that there do not exist two vertices with the same metric codes, which suggest that β(Ln) ≤ 3 i.e., the location number of the plane graph Ln is less than or equal to 3. Now, so as to finish the evidence for this case, we show that β(Ln) ≥ 3 by working out that there does not exist a resolving set R such that |R| = 2. On contrary, suppose that β(Ln) = 2. Now, from Theorem 1.1, we find that the degree of basis vertices can be at most 3. But except the vertices pl, ql, sl, al, and bl (1 ≤ l ≤ n), all other vertices of the radially symmetrical plane graph Ln have a degree 5. Then, without loss of generality, we suppose that the first resolving vertex is one of the vertices p1, q1, s1, a1 or bl, and 209 S. K. Sharma, V. K. Bhat / J. Algebra Comb. Discrete Appl. 8(3) (2021) 197–212 other nodes lie in the inner cycle, the set of middle nodes, and in the outer cycle. Therefore, we have the following possibilities to be discussed. Resolving sets Contradictions {p1,ph}, ph (2 ≤ h ≤ n) For 2 ≤ h ≤ w + 1, we have γ(a1|{p1,ph}) = γ(s1|{p1,ph}), a contradiction. {q1,qh}, qh (2 ≤ h ≤ n) For 2 ≤ h ≤ w, we have γ(a1|{q1,qh}) = γ(s1|{q1,qh}); and for h = w + 1, we have γ(p2|{q1,qw+1}) = γ(p1|{q1,qw+1}), a contradiction. {s1,sh}, sh (2 ≤ h ≤ n) For 2 ≤ h ≤ w + 1, we have γ(q1|{s1,sh}) = γ(b1|{s1,sh}), a contradiction. {a1,ah}, ah (2 ≤ h ≤ n) For 2 ≤ h ≤ w, we have γ(an|{a1,ah}) = γ(rn|{a1,ah}); and for h = w + 1, we have γ(b1|{a1,aw+1}) = γ(bn|{a1,aw+1}), a contradiction. {b1,bh}, bh (2 ≤ h ≤ n) For 2 ≤ h ≤ w − 1, we have γ(an|{b1,bh}) = γ(rn|{b1,bh}); when h = w, we have γ(s2|{b1,bw}) = γ(p2|{b1,bw}); and for h = w + 1, we have γ(a1|{b1,bw+1}) = γ(a2|{b1,bw+1}), a contradiction. {p1,qh}, qh (1 ≤ h ≤ n) For 1 ≤ h ≤ w, we have γ(a1|{p1,qh}) = γ(s1|{p1,qh}); and for h = w + 1, we have γ(bn|{p1,qw+1}) = γ(sn|{p1,qw+1}), a contradiction. {p1,sh}, sh (1 ≤ h ≤ n) For 1 ≤ h ≤ w − 1, we have γ(pn|{p1,sh}) = γ(rn|{p1,sh}); and for h = w,w + 1, we have γ(r2|{p1,sh}) = γ(a2|{p1,sh}), a contradiction. {p1,ah}, ah (1 ≤ h ≤ n) For h = 1, we have γ(pn|{p1,a1}) = γ(p2|{p1,a1}); when h = 2, we have γ(p2|{p1,a2}) = γ(a1|{p1,a2}); and for 3 ≤ h ≤ w + 1, we have γ(r1|{p1,ah}) = γ(q1|{p1,ah}), a contradiction. {p1,bh}, bh (1 ≤ h ≤ n) For h = 1, we have γ(rn|{p1,b1}) = γ(p2|{p1,b1}); and for 2 ≤ h ≤ w + 1, we have γ(r1|{p1,bh}) = γ(q1|{p1,bh}), a contradiction. {q1,sh}, sh (1 ≤ h ≤ n) For 1 ≤ h ≤ w − 1, we have γ(pn|{q1,sh}) = γ(rn|{q1,sh}); when h = w, we have γ(p1|{q1,sw}) = γ(r1|{q1,sw}); and for h = w + 1, we have γ(bn|{q1,sw+1}) = γ(rn|{q1,sw+1}), a contradiction. {q1,ah}, ah (1 ≤ h ≤ n) For h = 1, we have γ(qn|{q1,a1}) = γ(r2|{q1,a1}); when h = 2, we have γ(b1|{q1,a2}) = γ(r2|{q1,a2}); when h = 3, we have γ(b1|{q1,a3}) = γ(q2|{q1,a3}); and for 4 ≤ h ≤ w + 1, we have γ(q2|{q1,ah}) = γ(r2|{q1,ah}), a contradiction. {q1,bh}, bh (1 ≤ h ≤ n) For h = 1, we have γ(s1|{q1,b1}) = γ(r2|{q1,b1}); when h = 2, we have γ(b1|{q1,b2}) = γ(q2|{q1,b2}); and for 3 ≤ h ≤ w + 1, we have γ(q2|{q1,bh}) = γ(r2|{q1,bh}), a contradiction. {s1,ah}, ah (1 ≤ h ≤ n) For h = 1, we have γ(p1|{s1,a1}) = γ(q1|{s1,a1}); when h = 2, we have γ(b2|{s1,a2}) = γ(r2|{s1,a2}); and for 3 ≤ h ≤ w + 1, we have γ(q1|{s1,ah}) = γ(a1|{s1,ah}), a contradiction. {s1,bh}, bh (1 ≤ h ≤ n) For h = 1, we have γ(p1|{s1,b1}) = γ(q1|{s1,b1}); and for 2 ≤ h ≤ w + 1, we have γ(a1|{s1,bh}) = γ(q1|{s1,bh}), a contradiction. {a1,bh}, bh (1 ≤ h ≤ n) For 1 ≤ h ≤ w − 1, we have γ(rn|{a1,bh}) = γ(an|{a1,bh}); when h = w, we have γ(qn|{a1,bw}) = γ(sn|{a1,bw}); and for h = w + 1, we have γ(r2|{a1,bw+1}) = γ(qn|{a1,bw+1}), a contradiction. Thus, from the above table, we obtain that there does not exist a resolving set consisting of two vertices for V(Ln), suggesting that β(Ln) = 3 in this case. Case(2) When the integer n is odd. In this case, the positive integer n can be written as n = 2w + 1, where w ∈ N and w ≥ 3. Let R = {p1,qw+1,s1}⊂ V(Ln). Now, to unveil that R is the resolving set for the graph Ln, we consign the metric codes for each vertex of the graph Ln r R regarding the set R. Now, the metric codes for the vertices of inner cycle {pl : 1 6 l 6 n}∪{ql : 1 6 l 6 n} are γ(pl|R) =   (2l− 2, 2w − 2l + 3, 2l− 1), 2 ≤ l ≤ w; (2l− 2, 1, 2l− 1), l = w + 1; (4w − 2l + 4, 2l− 2w − 3, 4w − 2l + 6), w + 2 ≤ l ≤ 2w + 1. 210 S. K. Sharma, V. K. Bhat / J. Algebra Comb. Discrete Appl. 8(3) (2021) 197–212 and γ(ql|R) = { (2l− 1, 2w − 2l + 2, 2l), 1 ≤ l ≤ w; (4w − 2l + 3, 2l− 2w − 2, 4w − 2l + 5), w + 2 ≤ l ≤ 2w + 1. The metric codes for the set of middle vertices {rl : 1 6 l 6 n}∪{sl : 1 6 l 6 n} are γ(rl|R) =   (1, 2w + 1, 1), l = 1; (2l− 1, 2w − 2l + 3, 2l), 2 ≤ l ≤ w; (2w + 1, 1, 2w + 2), l = w + 1; (4w − 2l + 4, 2l− 2w − 2, 4w − 2l + 6), w + 2 ≤ l ≤ 2w + 1. and γ(sl|R) = { (2l, 2w − 2l + 4, 2l + 1), 2 ≤ l ≤ w + 1; (4w − 2l + 5, 2l− 2w − 1, 4w − 2l + 7), w + 2 ≤ l ≤ 2w + 1. At last, the metric codes for the vertices of outer cycle {al : 1 6 l 6 n}∪{bl : 1 6 l 6 n} are γ(al|R) =   (2, 2w + 2, 2), l = 1; (2l− 1, 2w − 2l + 4, 2l− 1), 2 ≤ l ≤ w + 1; (4w − 2l + 5, 3, 4w − 2l + 6), l = w + 2; (4w − 2l + 5, 2l− 2w − 2, 4w − 2l + 6), w + 3 ≤ l ≤ 2w + 1. and γ(bl|R) =   (2, 2w + 1, 2), l = 1; (2l, 2w − 2l + 3, 2l), 2 ≤ l ≤ w; (2w + 2, 2, 2w + 2), l = w + 1; (4w − 2l + 4, 2l− 2w − 1, 4w − 2l + 5), w + 2 ≤ l ≤ 2w; (3, 2l− 2w − 1, 4w − 2l + 5), l = 2w + 1. Again, we find that there do not exist two vertices with the same metric codes, which suggest that β(Ln) ≤ 3 i.e., the location number of the plane graph Ln is less than or equal to 3. Now, on assuming that β(Ln) = 2, we get the same eventualities as in Case(1), and similarly, the contradiction can be obtained. So, in this case, we have β(Ln) = 3 as well and hence the theorem. Now, in terms of independent resolving set, we have the following result: Theorem 4.2. Let Ln be the planar graph on 7n vertices as defined above. Then for every positive integer n; n ≥ 6, its independent resolving number is 3. Proof. For proof, refer to Theorem 4.1. 5. Conclusion In this article, we determined the metric dimension of three classes of plane graphs (viz., Jn, Kn, and Ln), which are generated by taking n-copies of the complete bipartite graph (or a star) K1,5, and are radially symmetric with the constant metric dimension. 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