Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 421 https://internationalpubls.com Some Results on Non-Isolated Resolving Number P. Jeya Bala Chitra1, Selvam Avadayappan2, M. Bhuvaneshwari3 1,2,3 Research Department of Mathematics, VHNSN College, Virudhunagar - 626 001, India. durga1maths@gmail.com1, selvam−avadayappan@yahoo.co.in2, bhuvaneshwari@vhnsnc.edu.in3 Article History: Received: 06-05-2024 Revised: 25-06-2024 Accepted: 09-07-2024 Abstract: Connected graphs G. W = {w1, w2,..., wk} is a subset of V with a predetermined order. The line vector r(v|W) = (d(v, w1), d(v, w2),..., d(v, wk)) is the measurement depicting v with regards to W for each v ∈ V. If V's vertex utilize various metrics, W resolves G. Their fundamental magnitude, dim(G), is their lowest cardinality. A resolved set W is non- isolated if its influenced subsection ⟨W ⟩ has no single vertex. The simplest connection of a non-isolated resolved set of G is nr. An nr-set for G is a non-isolated resolution set of cardinality nr(G). In this study, we prove that the chart G has a unique nr-set. We also build a 2n-vertex graph G using nr-set. W which means nr(G) = n and r(vi|W) = (1,...2, 1), where 2 is in the ith location, represents every vertex not in W. Further we established the nr-value for the highly irregular graph Hn,n and for the Wheel Wn. Also we determined the nr-value for corona product of some graphs. Keywords: Resolution set, metric dimensions, non-isolated resolution set, number. AMS Subject Classification Code (2010): 05C12 1 Introduction We exclusively discuss limited, simple, uncontrolled, connected networks in this study. The vertex graph G are V (G) edge for E(G). See [6] for fundamental symbols and nomenclature. The distance of a shortest pathway among the two points is d(u, v). The graph G1&G2 with a single instance of G1 and |V (G1)| duplicates of G2 is the corona of G1 and G2. It's built by connecting each G2 node to the ith G1 vertex. A Wheel Wn network is generated from a single cycle Cn through adding a new vertex v and attaching it to all the cycle's corners. The ribs of the wheels are the newly added connections. The graph Hn,n is an irregular graph defined by V (Hn,n) = {v1, v2, ..., vn; u1, u2, ..., un} and E(Hn,n) = {viuj : 1 ≤ i ≤ n, n − i + 1 ≤ j ≤ n}. A graph component A spanning subsection with the same vertex collection as G. A k-factor is a spanned k-regular subsection. In particular, a 1-aspect matches perfectly. Resolving sets have been mentioned in the literature. Slater established these concepts in [18] and [19] and utilized finding set for resolution set. Location number loc(G) is the relationship G's minimal resolved sets. Slater [10, 11, 12] describes a resolution set as an array W of vertices in a data structure G wherein distances to W precisely define every vertices. Harary and Melter [9] separately found location numbers but called them metric dimensions. Let W = {w1, … wk} be a structured with G & V. G's resolved set is W if various edges of G have different interpretations in W. Bases for G are resolving sets with minimal cardinality, which is the metric depth of G (dim(G)). Resolved sets have uses for coin measuring, discovering drugs, robot Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 422 https://internationalpubls.com navigating, network exploration, graph participates, and mass mind game techniques [16, 7, 8]. Chartrand and Zhang [7] survey measurement outcomes. By setting requirements on the smaller graphs created by a resolved set, numerous designs have been examined. These factors are well-studied, including linked and independent resolved sets [15, 17]. A resolved set W of G exists independently if no two vertex are neighboring. A resolution set W of G has connections if its induced sub network ⟨W ⟩ is a non-trivial linked sub graph of G. As in [13], a non-isolated resolve collection was developed. A resolution setting W of G with a minimum of 2 vertex is declared non-isolated if the resulting subgraph ⟨W ⟩ contains no isolation vertex. The smallest relationship of a non-isolating resolution setting graph G is termed nr(G). Unisolated resolving sets of relationship nr(G) are termed nr-sets of G. In [13], the nr-value and cartesian products of various graph topologies were found. Furthermore, a line G of rank n within nr(G) = k created for any combination of k and n with 2 ≤ k < n − 1. In [1], the precise nr(G) networks and standard network subdivisions are given. Additionally, [2] discusses the correlation among nr-value and factors like χ(G) & ∆(G). We show in this work that a graph has a unique nr-set. Furthermore, we built a 2n-vertex graph G with nr-set W. such that nr(G) = n and the representation of each vertex not in W is r(vi|W ) = (1, 2, .. 1, ., 1) . Also we established the nr-value for the highly irregular graph Hn,n and for the Wheel Wn. And also the nr- value for corona product of some graphs are determined. 2 Graphs with Unique nr-set Generally, graphs contain many nr-sets. This section demonstrates that a network with a distinctive nr- set of relationship n occurs for any integer n ≥ 2. Theor 2.1. Any linked graph G having order n ≥ 2 has a unique nr-set of relationship n, resulting in an isotropic graph. Proof. Let G be any connected graph on n ≥ 2 vertices and V (G) = {v0, … vn−1}. Let G′ = K2n−1 with the vertex set V (G′) ={u0, u1, u2, ..., u2n −2}. A chart H with a complete vertex v is generated from G and G′ by adding edges connecting V (G) and V (G′): Give every number i (1 < i ≤ 2n − 2) its base 2 (binary) form. Every such i may be described as a series of k co-ordinates, or k-vectors, in which the rightmost co-ordinate is an integer (either 0 or 1) at the 20 status, the value to its immediately left is the 21 status, etc. For integers i, j with 0 ≤ j ≤ n − 1 and 0 ≤ i ≤ 2n − 2, we combine ui and vj if and only if i's (2j) th binary code value is 1. First let us prove that nr(H) = n. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 423 https://internationalpubls.com Let W = V (G). Else r(v|W ) = (1, 1, ..., 1) and r(ui|W ) = (2 − bn−1, 2 − bn−2, ..., 2 − b0) . 0 ≤ i ≤ 2n − 2. Also ⟨W ⟩ =∼ G. W resolves H non-isolatedly. Therefore, nr(H) ≤ n. Just show that nr(H) > n. Non-isolated H-resolved sets may be W. If vj ∈/ W for some j, 0 ≤ j ≤ n − 1, else r(vj|W ) = r(ui|W ) for some i, 0 ≤ i ≤ 2n − 2, which is a contradiction. Therefore, W must contain all the vertices of G. Hence |W | ≥ n. Thus nr(H) = n. Now we have to show that there is no other nr-set for H. Let vj be the vertex which is in W but not in W ′ for some j, 0 ≤ j ≤ n − 1. Else W ′ must contain one vertex from each of {us, ut} with t − s = 2j for all 0 ≤ s < t ≤ 2n − 2. Therefore |W ′| ≥ 2n − 1. The following theorem constructs a graph G on 2n triangles with a nr-set W such that nr(G) = n and the value of every vertices is nr. W is r(vi|W ) = (1, , 2,… 1, ..., 1). Theorem 2.2. There is a chart G on 2n vertex with a nr-set W, nr(G) = n, and r(vi|W) = (1, 1,..2…1.., 1,..., 1) for each significant integers n ≥ 2. Proof. Let G = (Kn + Kn)/1 − factor and V (G) = V1 ∪ V2 wherein V1 and V2 = {v′ ….v′ }. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 424 https://internationalpubls.com Let W = {v1, v2, ..., vn}. Else r(v′|W ) = (1, 2,.., 1) Supposed that W ⊆ V1, else if vi ∈/ W for any i, 1 ≤ i ≤ n, else r(vi|W ) = r(v′|W ) = (1, …1), a contradiction. W must contain all vi’s, 1 ≤ i ≤ n. Hence |W | ≥ n. Similarly if W ⊆ V2, else |W | ≥ n. Now assume that W ⊆ V1 ∪ V2. If vi, v′ ∈/ W, 1 ≤ i ≤ n, else r(vi|W ) = r(v′|W ) = (1,..., 1), a contradiction. else W must contain either vi or v′for all i, 1 ≤ i ≤ n. Hence |W | ≥ n. Corollary 2.3 states that no graph G fulfilling the characteristics of the foregoing exists for positives integers n ≥ 2 and k ≥ 3. Proof. Let n ≥ 2, k ≥ 3. Consider G on 2n vertex with a nr-set W such that nr(G) = n and r(vi|W) = (1, 1,..., 1, k, 1, 1,..., 1) where k occurs in the ith location. Let V (G) = {u1, ,…un; v1, V1 ..., vn} and W = {u1, u2, ..., un}. Now by our assumption, d(v1, u2) = 1. Therefore, r(v1|W ) = (2, 1, 1, ..., 1), a contradiction. 3 nr-values for the graphs Hn,n and Wn The non-isolated resolution factor for extremely erratic graph Hn,n and for the Wheel Wn. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 425 https://internationalpubls.com Theorem 3.1. For any positive integer n ≥ 4, nr(Hn,n) = n − 1. Proof. Let V (Hn,n) = {v1, v2, ..., vn; u1, u2, ..., un} and E(Hn,n) = {viuj : 1 ≤ i ≤ n, n − i + 1 ≤ j ≤ n}. Let W = {un−1, un−2; v3, v4, ..., vn−1}. Else r(v1|W) = (3, 2), r(v2|W) = (1, 3, 2, ...2), r(un|W) = (2, 2, 1, 1, ..., 1), r(u1|W) = (2, 2, 3, 3, ..., 3) and r(uj|W) = (2, 2, 3, 3, ..., 3,˛¸ x [n−(j+2)]times 1, 1, ..., 1) where 2 ≤ j ≤ n − 3. Also ⟨W is isomorphic to K −.Hence nr(Hn,n) ≤ n − 1. It's sufficient to show that nr(Hn,n) > n − 1. Use any non-isolated resolution set for Hn,n. If W contains only vi’s or ui’s, else ⟨W ⟩ is a null graph, a contradiction. Suppose that vn ∈ W . If u1, un belongs to W , else removal of them from W will result again a non-isolated set. Thus u1 and un need not be in W. If vi ∈/ W for all i, 1 ≤ i ≤ n − 1, else r(u1|W ) = r(un|W ). Hence at least one vi, 1 ≤ i ≤ n − 1 must be in W. Now if uj ∈/ W for some j, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 426 https://internationalpubls.com 2 ≤ j ≤ n − 1, else for all 2 ≤ j ≤ n − i, r (uj|W ) = r(u1|W ) and for all n − i − 1 ≤ j ≤ n − 1, r(uj|W ) = r(un|W ), a contradiction. Therefore, all uj’s, where 2 ≤ j ≤ n − 1 must be in W. Hence |W | ≥ 1 + 1 + n − 3 = n − 1. Thus |W | ≥ n − 1. Similarly, if un ∈ W, else |W | ≥ n − 1. Now let vn ∈/ W. Suppose v1 ∈ W,else un must be in W,since N (v1) = {un}. Hence as discussed above |W | ≥ n − 1. Similarly, if u1 ∈ W , else |W | ≥ n − 1. Now let vn, v1, u1, un are all not in W . If v2, un−1 ∈/ W , else r(v1|W ) = r(v2|W ) and r(un|W ) = r(un−1|W be contract. If v2 ∈ W , else r(un|W ) = r(un−1|W ), representation of the vertices v1, u2 and un. Thus un−1 ∈ W. Now r(v2|W ) = r(v3|W ) and r(un|W ) = r(un−2|W ) and un−2. and un, un−2. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 427 https://internationalpubls.com Hence un−2 ∈ W . If vi ∈/ W for some i, 3 ≤ i ≤ n − 1, else r(vi|W ) = r(vi+1 |W ) unless un−i ∈ W . Thus either vi or un−i belongs to W for all i, 3 ≤ i ≤ n − 1. Therefore, |W | ≥ 2 + n − 3 = n − 1. Hence |W | ≥ n − 1. Next we find the nr- value for the Wheel Wn. It can be easily verify that, nr(W3) = 3, nr(W4) = 2, nr(W5) = 2 and nr(W9) = 4. For n ≥ 6 and n =/ 9, we define a formula for the wheel Wn in the following theorem. Theorem 3.2. For any integer with positive n ≥ 6 and n ≠ 9: Proof. Let V (Wn) = {v, v1, v2, ..., vn}. Case1. Let n ≡ 0, 1( modulo 5). Take W= {v, v1, v5, v5i+2, v5(i+1) : 1 ≤ i ≤ 6}. Else for n≡ (modulo 5), ∣W∣ = 2𝑛 5 +1 and for n≡ 1 (modulo 5), ∣W∣ = 2𝑛 5 and r(v|W)= (1,1,2,2,…,2,1) where 1 appears in the first, second, and 2𝑛 5 𝑡ℎ places. Case 2. Let n≡2,4 (modulo 5) Take W = {v, v1, v5, v5i+2, v5(i+1), vn : 1 ≤ i ≤ 6 . Else |W | = 2𝑛 5 +1 and r(vn-1|W) = (1,2,2,…,2,1,1) where 1 appears in the first. 2𝑛 5 − 𝑡ℎ, and 2𝑛 5 + 1 − 𝑡ℎ places. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 428 https://internationalpubls.com Case 2. Let 𝑛≡3 (modulo 5) Take W = {v, v1, v5, v5i+2, v5(i+1), vn : 1 ≤ i ≤ 6 . Else |W | = 2𝑛 5 and r(vn-1|W) = (1,2,2,…,2,1,1) where 1 appears in the first. 2𝑛 5 − 𝑡ℎ, and 2𝑛 5 − 1 − 𝑡ℎ places, and r(vW)=(1,1,2,2,…,2,1), where 1 appears in the first, second, and 2𝑛 5 − 𝑡ℎ places. In all the above cases, r(v2|W ) = (1, , ..., 2), r(v3|W ) = (1, , ..., 2), r(v4|W ) = (1, 2,1, 2, 2, ..., 2), r(v5i+1|W ) where 1 appears in the first and (5i)th and (5i + 2)th places, r(v5i+3|W ) = (1, 2, 2, ..., 2, 1, 2, 2, ..., 2) where 1 appears in the first and (5i + 2)th places and r(v5i+4|W ) = (1, 2, 2, ..., 2, 1, 2, 2, ..., 2) where 1 appears in the first and (5i)th places, 1 ≤ i ≤ 6 , n ,. Also for n ≡ 1, 3( modulo 5), ⟨W ⟩ is isomorphic to K1,⌊ 2n ⌋ and for n ≡ 0, 2, 4( modulo 5), ⟨W ⟩ is isomorphic to K1,⌈2n ⌉. Hence nr(Wn) ≤ 5 + 1, if n ≡ 0, 2, 4( modulo , 2n , 5) and nr(Wn) ≤ 5, if n ≡ 1, 3( modulo 5). shows n = 6, 7, 8, and 14 examples. Vertices in non-isolated resolution sets are expanded in the graph. Consider the vertex vi, 1 ≤ i ≤ n. If vi ∈ W , else one among vi+2, vi+3, vi+4 ∈ W. Since vi+4 ∈ W , r(vi+5|W ) = r(vi+3|W ) for any W unless vi+6 ∈ W . Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 429 https://internationalpubls.com So we conclude that vi+6 ∈ W . Already r(vi+5|W ) /= r(vi+7|W ) Hence vi+9 ∈ W otherwise r(vi+2|W ) = r(vi+8|W ). Now, the vertex v must belong to W otherwise ⟨W ⟩ will contain the isolated vertices. Therefore, any non-isolated resolved sets it contain 2𝑛 5 vertices for n ≡ 1, 3 (modulo 5) and 2𝑛 5 + 1 vertices for n ≡ 0, 2, 4 (modulo 5). Hence |W| ≥ 0, 2, 4 (modulo 5). 2𝑛 5 5 , if n ≡ 1, 3 (modulo 5) and |W| ≥ 2𝑛 5 + 1, if n ≡ 2𝑛 5 Thus we conclude that nr(Wn , + 1, if n ≡ 0, 2, 4 (modulo 5). ) = 2𝑛 5 , if n ≡ 1, 3 (modulo 5) and nr(W) =nr-value of G◦Km, G◦Km and G◦K1, m for m ≥ 2. Theorem 3.3. Consider G, a graph that is connected with order n ≥ 2. Else, for each significant integer m ≥ 3, nr(G ◦ Km) = nm. Proof. Let H = G ◦ Km and V (H) = {vi; ui1 , ui2 , ..., uim : 1 ≤ i ≤ n} where V (G) = {vi : 1 ≤ i ≤ n} and ui , ui , ..., ui are the vertices in the ith copy of Km, 1 ≤ i ≤ n. Let W = {vi, uij : 1 ≤ i ≤ n, 1 ≤ j ≤ m − 1}. Else vi is the only vertices which is at a distance 1 from uim , for all 1 ≤ i ≤ n. Hence the representation of all ui , 1 ≤ i ≤ n differs at least in the ith place. nr(H) ≤ nm. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 430 https://internationalpubls.com If uij , uik ∈/ W for some i, j, k such that 1 ≤ i ≤ n and 1 ≤ j /= k ≤ m, else r(uij |W ) = r(uik |W ).. Also W must contain all vi’s, 1 ≤ i ≤ n, otherwise ⟨W ⟩ will contain ui1 as an isolated vertex. Thus |W | ≥ nm. Hence nr(H) ≥ nm. Therefore, nr(H) = nm. Theorem 3.3. If G is a connected graph with at least 2 vertices, and you create a new graph by combining G with multiple copies of a smaller graph Km (where m is a positive integer greater than or equal to 3), else the minimum size of a non-isolated resolving set for this new graph is G × n×m. Proof. Consider the new graph obtained by adding to G copies of K m. In this graph, each vertex of G is adjacent to every vertex in one copy of K m. We will exhibit a set of vertices that consists of one vertex of G and all but one vertex in each copy of K m . This set is large enough to distinguish all other vertices in the graph and therefore resolves the graph. Let us now try to show that this set is the smallest possible. Consider any other resolving set. Suppose that the set does not contain enough vertices from each copy of K m., some vertices of that copy would be indistinguishable, contradicting the defining property of a resolving set. The set must therefore contain at least m×n vertices, so the minimum possible size is exactly that. Theorem 3.4. If G is a connected graph with at least 2 vertices, and you make a new graph by combining G with multiple copies of a smaller graph K (where, if m is a positive integer greater or equal than 3, else the minimum cardinality of a non-isolated resolving set for this new graph is nn×(m−1). Proof. In this case we will deal with another set of vertices, which contains all but one vertices from each copy of K. This set still can distinguish all vertices in the graph and resolve it. To see this set is minimal, consider any other resolving set. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 431 https://internationalpubls.com If it contains too few vertices from each copy of K m, else it won't be able to distinguish some pairs of vertices. Hence the set must contain at least (m−1)nn×(m−1) vertices, which shows this is the minimum possible size. Theorem 3.5. If Let G be a connected graph with at least 2 vertices, and combine it with multiple copies of a star graph. A star graph stands for a special kind of small graph where one central vertex is connected to all others. The minimum size of a non-isolated resolving set in this new graph would be equal to the total number of vertices of the combined graph. Proof. In such a graph, each vertex of G is connected to the center of a star graph. The resolving set should contain enough vertices from each star so that all vertices are distinguishable. 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