EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 14, No. 3, 2021, 773-782 ISSN 1307-5543 – ejpam.com Published by New York Business Global On 2-Resolving Sets in the Join and Corona of Graphs† Jean Cabaro1, Helen Rara2 1 Mathematics Department, College of Natural Sciences and Mathematics, Mindanao State University-Main Campus, 9700 Marawi City, Philippines 2 Department of Mathematics and Statistics, College of Science and Mathematics, Center of Graph Theory, Algebra, and Analysis-Premier Research Institute of Science and Mathematics, Mindanao State University-Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. Let G be a connected graph. An ordered set of vertices {v1, ..., vl} is a 2-resolving set in G if, for any distinct vertices u,w ∈ V (G), the lists of distances (dG(u, v1), ..., dG(u, vl)) and (dG(w, v1), ..., dG(w, vl)) differ in at least 2 positions. If G has a 2-resolving set, we denote the least size of a 2-resolving set by dim2(G), the 2-metric dimension of G. A 2-resolving set of size dim2(G) is called a 2-metric basis for G. This study deals with the concept of 2-resolving set of a graph. It characterizes the 2-resolving set in the join and corona of graphs and determine the exact values of the 2-metric dimension of these graphs. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: 2-resolving set, 2-metric dimension, 2-metric basis, corona 1. Introduction The problem of uniquely determining the location of an intruder in a network was the principal motivation of introducing the concept of metric dimension in graphs by Slater [10], where the metric generators were called locating sets. The concept of metric dimension of a graph was also introduced independently by Harary and Melter in [4] where metric generators were called resolving sets. In [6], Monsanto, Acal and Rara discussed the strong resolving dominating sets in the join and corona of graphs while in [5], Monsanto and Rara discussed the resolving restrained domination in graphs. Bailey and Yero in [1] demonstrated a construction of error-correcting codes from graphs by means of k-resolving sets, and present a decoding algorithm which makes use of covering designs. The distance between two vertices u and v of a graph is the length of a shortest path †The authors would like to thank the Commission on Higher Education (CHED) and Mindanao State University-Marawi and MSU-Iligan Institute of Technology, Philippines. DOI: https://doi.org/10.29020/nybg.ejpam.v14i3.3977 Email addresses: amerjean1228@gmail.com (J. Cabaro), helenrara@gmail.com (H. Rara) http://www.ejpam.com 773 © 2021 EJPAM All rights reserved. J. Cabaro, H. Rara / Eur. J. Pure Appl. Math, 14 (3) (2021), 773-782 774 between u and v, and we denote this by dG(u, v). In recent years, much attention has been paid to the metric dimension of graphs: this is the smallest size of a subset of vertices (called a resolving set) with the property that the list of distances from any vertex to those in the set uniquely identifies that vertex and is denoted by dim(G). According to the paper of Saenpholphat et al. [9], for an ordered set of vertices W = {w1, w2, ..., wk} ⊆ V (G) and a vertex v in G, the k-vector (ordered k-tuple) r(v/W ) = (dG(v, w1), dG(v, w2), ..., dG(v, wk)) is referred to as the (metric) representation of v with respect to W . The set W is called a resolving set for G if distinct vertices have distinct representation with respect to W . Hence, if W is a resolving set of cardinality k for a graph G of order n, then the set {r(v/W ) : v ∈ V (G)} consists of n distinct k-vectors. A resolving set of minimum cardi- nality is called a minimum resolving set or a basis, and the cardinality of a basis for G is the dimension dim(G) of G. In the paper of Bailey et al.[1], an ordered set of vertices W = {w1, ..., wl} is a k- resolving set for G if, for any distinct vertices u, v ∈ V (G), the (metric) representations r(u/W ) and r(v/W ) of u and v, respectively differ in at least k positions. If k = 1, then the k-resolving set is called a resolving set for G. If G has a k-resolving set, the minimum cardinality dimk(G) is called the k-metric dimension of G. In this paper, the concept of 2-resolving set in the join and corona of graphs is dis- cussed. 2. Preliminary Results In this study, we consider finite, simple and connected undirected graphs. For basic graph-theoretic concepts, we refer readers to [3]. Remark 1. Let G be any connected graph of order n ≥ 2. Then the vertex set of G is a 2-resolving set in G. Hence, 2 ≤ dim2(G) ≤ n. Proposition 1.[7] dim2(G) = 2 if and only if G ∼= Pn, n ≥ 2. Proposition 2. For any complete graph Kn of order n ≥ 2, dim2(Kn) = n. Theorem 1. Every 2-resolving set in a connected graph G is a resolving set in G. Hence, dim(G) ≤ dim2(G). Remark 2. A superset of a 2-resolving set is a 2-resolving set. Remark 3. Let S ⊆ V (G). For any pair of vertices x, y ∈ S, r(x/S) and r(y/S) differ in at least 2 positions. Hence, to prove that S is a 2-resolving set in G, we only need to show that for every pair of vertices x, y ∈ V (G) where x ∈ S and y ∈ V (G)\S or both x, y ∈ V (G)\S, r(x/S) and r(y/S) differ in at least 2 positions. J. Cabaro, H. Rara / Eur. J. Pure Appl. Math, 14 (3) (2021), 773-782 775 3. 2-Resolving Sets in the Join of Graphs Definition 1.[2] The join G + H of two graphs G and H is the graph with vertex set V (G + H) = V (G) ∪ V (H) and edge set E(G + H) = E(G) ∪ E(H) ∪ {uv : u ∈ V (G), v ∈ V (H)} . Note that the star K1,n can be expressed as the join of the trivial graph K1 and the empty graph Kn of order n, that is, K1,n = K1 + Kn. The graphs Fn = K1 + Pn and Wn = K1 + Cn of orders n + 1 are called fan and wheel, respectively. Definition 2. Let G = ((V (G), E(G)) be a connected graph. The open neighborhood NG(v) = {u ∈ V (G) : uv ∈ E(G)}. Any element u of NG(v) is called a neighbor of v. The notation x ∈ V (G)\S means that x ∈ V (G) but not in S. Definition 3. Let G be any nontrivial connected graph and S ⊆ V (G). Then S is a 2-locating set of G if ∀x, y ∈ V (G), x 6= y, the following are satisfied: (i) If x, y ∈ V (G)\S, then ∃w, z ∈ S, w 6= z such that either: (a) w, z ∈ (NG(x))\NG(y), or (b) w, z ∈ (NG(y))\NG(x), or (c) w ∈ (NG(x))\NG(y) and z ∈ (NG(y))\NG(x). (ii) If x ∈ S, y ∈ V (G)\S, then ∃p ∈ (NG(x) ∩ S)\NG(y) or p ∈ (NG(y) ∩ S)\NG(x). The 2-locating number of G, denoted by ln2(G), is the smallest cardinality of a 2-locating set of G. A 2-locating set of G of cardinality ln2(G) is referred to as ln2-set of G. Example 1. The sets S1 = {c, d, e, f} and S2 = {a, b, c, f} are 2-locating sets in G in Figure1. Moreover, S1 and S2 are ln2-set in G. Thus, ln2(G) = |S1| = |S2| = 4. .................................... .................................... .................................... .................................... .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ .... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ .... ................... .................. .................. .................. .................. .................. .................. .................. .................. .................. .................. .................. ...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ .... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ....G: a b c d e f Figure 1: A graph G with ln2 = 4 Remark 4. Every 2-locating set in G is a 2-resolving set in G. However, a 2-resolving set in G need not be a 2-locating set in G. Thus, dim2(G) ≤ ln2(G). J. Cabaro, H. Rara / Eur. J. Pure Appl. Math, 14 (3) (2021), 773-782 776 Example 2. Let P6 = [v1, v2, ..., v6] be a path of order 6 and S1 = {v1, v3, v5, v6}. Then S1 is both 2-locating and 2-resolving set in P6. On the other hand, S2 = {v2, v4, v6} is a 2-resolving set but not 2-locating. Example 3. For all n ≥ 2, ln2(Pn) = ⌈ n + 1 2 ⌉ . Example 4. For all n ≥ 5, ln2(Cn) = ⌈n 2 ⌉ and ln2(C3) = 3, ln2(C4) = 4. Definition 4. Let G be any nontrivial connected graph and S ⊆ V (G). S is a strictly 2-locating (strictly 1-locating) set in G if S is 2-locating and |NG(y) ∩ S| ≤ |S| − 2 (|NG(y) ∩ S| ≤ |S| − 1), ∀y ∈ V (G). The strictly 2-locating (strictly 1-locating) num- ber of G, denoted by sln2(G) (sln1(G)), is the smallest cardinality of a strictly 2-locating (strictly 1-locating) set in G. A strictly 2-locating (strictly 1-locating) set in G of cardi- nality sln2(G) (sln1(G)) is referred to as sln2-set (sln1-set) in G. Example 5. The set S2 = {a, b, c, f} is a strictly 1-locating set in G in Figure 1. Moreover, S2 is a sln1-set in G. Thus, sln1(G) = 4. Example 6. The set S = {u1, u3, u5, u7} is a strictly 2-locating set in P7 in Figure 2. Moreover, S is a sln2-set in P7. Thus, sln2(P7) = 4. .................................... .................................... .................................... .................................... .................................... .................................... ................................................................................................................................. ............................................................................................. ............................................................................................. ............................................................................................. ............................................................................................. .............................................................................................G: u1 u2 u3 u4 u5 u6 u7 Figure 2: A graph P7 with sln2 = 4 Example 7. For all n ≥ 4, sln1(Pn) = { n 2 + 1, n is even⌈ n 2 ⌉ , n is odd Example 8. For all n ≥ 5, sln1(Cn) = { n 2 , n is even⌈ n 2 ⌉ , n is odd Example 9. For all n ≥ 6, sln2(Pn) = { n 2 + 1, n is even⌈ n 2 ⌉ , n is odd Example 10. For all n ≥ 7, sln2(Cn) = { n 2 , n is even⌈ n 2 ⌉ , n is odd Remark 5. Every strictly 2-locating set in G is strictly 1-locating. However, strictly 1-locating set in G need not be a strictly 2-locating set in G. Theorem 2. A proper subset S of V (K1 + Kn) is a 2-resolving set in K1 + Kn if and only if S = V (Kn), ∀n ≥ 2. J. Cabaro, H. Rara / Eur. J. Pure Appl. Math, 14 (3) (2021), 773-782 777 Proof. Let S be a proper subset of V (K1 + Kn). Suppose S is a 2-resolving set in K1 + Kn and suppose ∃x ∈ V (Kn)\S. Then r(x/S) and r(y/S) differ in at most one position for each y ∈ V (Kn). Thus, S = V (Kn). Conversely, let S = V (Kn) and x ∈ V (K1). Then, r(x/S) = (1, ..., 1) and r(y/S) = (..., 2, 2, 2, 0, 2, ...) for each y ∈ V (Kn). Thus, r(x/S) and r(y/S) differ in at least two positions. Therefore S is a 2-resolving set of K1 + Kn. Corollary 1. dim2(K1 + Kn) = |V (Kn)|. Theorem 3. Let G be a connected non-trivial graph and let K1 = {v}. Then S ⊆ V (K1 + G) is a 2-resolving set of K1 + G if and only if either v /∈ S and S is strictly 2-locating set of G or S = {v} ∪ T , where T is a strictly 1-locating set in G. Proof. Let S ⊆ V (K1 +G) be a 2-resolving set of K1 +G. If v /∈ S, then S ⊆ V (G) is 2-locating set in G. Suppose there exists y ∈ V (G) such that |NG(y)∩S| > |S| − 2. Then r(v/S) and r(y/S) differ in at most one position, contrary to our assumption that S is a 2-resolving set in K1 + G. Hence, S is a strictly 2-locating set of K1 + G. Next, suppose that S = T ∪ {v}, where T = V (G) ∩ S. Then ∅ 6= T ⊆ V (G). Thus, T is a 2-locating set in G. Since S is a 2-resolving set and v ∈ S, T is strictly 1-locating set in G. For the converse, let x, y ∈ V (K1 + G). First, assume that v /∈ S and S is a strictly 2-locating set in G. Consider the following cases. Case 1. x, y ∈ S By Remark 3, rK1+G(x/S) and rK1+G(y/S) differ in at least 2 positions, the xth and yth positions. Case 2. x, y ∈ V (G)\S By Definition 3(i), rK1+G(x/S) and rK1+G(y/S) differ in the zth and wth positions, for some distinct vertices z, w ∈ S. Case 3. x ∈ S, y ∈ V (G)\S By Definition 3(ii), there exists z ∈ (NG(x) ∩ S)\NG(y) or z ∈ (NG(y) ∩ S)\NG(x). Hence, rK1+G(x/S) and rK1+G(y/S) differ in the xth and zth positions. Case 4. x = v, y ∈ V (G). By Definition 4, ∃u,w ∈ S\NG(y), u 6= w. Thus, rK1+G(x/S) and rK1+G(y/S) differ in the uth and wth positions. Next, suppose S = {v} ∪ T where T is strictly 1-locating set in G. Consider the following cases. Case 1. x, y ∈ S By Remark 3, rK1+G(x/S) and rK1+G(y/S) differ in at least 2 positions, the xth and yth positions. Case 2. x, y ∈ V (K1 + G)\S. Then x, y ∈ V (G)\T . By Definition 3(i), rK1+G(x/S) and rK1+G(y/S) differ in at least 2 positions. Case 3. x = v, y ∈ V (G). By Definition 4, ∃z ∈ T\NG(y). Thus, rK1+G(x/S) and rK1+G(y/S) differ in the xth J. Cabaro, H. Rara / Eur. J. Pure Appl. Math, 14 (3) (2021), 773-782 778 and zth positions. Case 4. x ∈ T , y ∈ V (G)\T . Since T is 2-locating set in G, rG(x/T ) and rG(y/T ) differ in at least 2 positions. Hence, rK1+G(x/S) and rK1+G(y/S) differ also in at least 2 positions. Therefore, S is a 2-resolving set in K1 + G. The sets {u, u1, u3, u4} and {v, v1, v3, v5} are 2-resolving sets in the join 〈u〉+ P5 and 〈v〉+ C6, respectively, in Figure 3. .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. .. ................................. ................................ ................................ ................................ ................................ ................................ ........... ............................................................................................................................................................................................................ .......................................................................................................................................................................................................................................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ .... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ .... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ .... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ... .............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. .. .......................... ......................... ......................... ......................... ......................... ......................... ......................... ......................... ......................... ......................... ......................... ......................... ................ ............................................................................................................................................................................................................................................................................................................................. .......................................................................................................................................................................................................................................................... ........................................................................................................................................ ................... .................. .................. .................. .................. ............... .......................................................................................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ..... ................... .................. .................. .................. .................. ......................................................................................................................... {u}+ P5: u1 u2 u3 u4 u {v}+ C6: v1 v2 v3 v4 v5v6 v Figure 3: The join {u}+ P5 with dim2({u}+ P5) = 4 and the join {v}+ C6 with dim2({v}+ C6) = 4 The next result follows immediately from Theorem 3. Corollary 2. dim2(K1 + G) = min {sln2(G), sln1(G) + 1}. Example 11.[8] For any integer n ≥ 6, dim2(F1,n) = ⌈ (n + 1) 2 ⌉ =sln2(Pn). Example 12.[8] For any n ≥ 7, dim2(W1,n) = ⌈n 2 ⌉ =sln2(Cn). Theorem 4. Let G and H be nontrivial connected graphs. A proper subset S of V (G+H) is a 2-resolving set in G + H if and only if SG = V (H) ∩ S and SH = V (H) ∩ S are 2- locating sets in G and H respectively where SG or SH is strictly 2-locating set or SG and SH are strictly 1-locating sets. Proof. Suppose S is a proper subset of V (G + H). Let S be a 2-resolving set in G + H. Let SG = V (G) ∩ S and SH = V (H) ∩ S. Then S = SG ∪ SH . Suppose SG = ∅. Then S = SH . Let x, y ∈ V (G), x 6= y. Then rG+H(x/S) = rG+H(y/S) = (1, ..., 1). A contradiction to the assumption of S. Thus, SG 6= ∅. Similarly, SH 6= ∅. Next, suppose SG or SH , say SG is not 2-locating set in G. Then there exist x, y ∈ V (G), x 6= y such that rG(x/SG) and rG(y/SG) differ in at most 1 position. Hence, rG+H(x/S) and rG+H(y/S) differ also in at most one position. Thus, S is not 2-resolving set in G + H, contrary to our assumption. Therefore SG and SH are 2-locating sets in G J. Cabaro, H. Rara / Eur. J. Pure Appl. Math, 14 (3) (2021), 773-782 779 and H, respectively. Now, suppose that both SG and SH are not strictly 2-locating sets. Then |NG(x) ∩ SG| > |SG| − 2, ∀x ∈ V (G) and |NH(y) ∩ SH | > |SH | − 2, ∀y ∈ V (H). Hence either NG(x) ∩ SG = SG or ∃p ∈ SG\NG(x) and either NH(y) ∩ SH = SH or ∃q ∈ SH\NH(y). Since S is a 2-locating set, ∃p ∈ SG\NG(x) and ∃q ∈ SH\NH(y). Thus, SG and SH are both strictly 1-locating sets. For the converse, suppose that SG and SH are 2-locating sets in G and H, respectively where SG or SH is strictly 2-locating set or SG and SH are both strictly 1-locating sets. Let x, y ∈ V (G+H) with x 6= y. If x, y ∈ V (G), then rG(x/SG) and rG(y/SG) differ in at least 2 positions since SG is a 2-locating set in G. Hence, rG+H(x/S) and rG+H(y/S) also differ in at least 2 positions. Similarly, if x, y ∈ V (H), then rG+H(x/S) and rG+H(y/S) differ in at least 2 positions. Suppose that x ∈ V (G) and y ∈ V (H) and SG is strictly 2-locating set. Then, ∃w, z ∈ SG\NG(x). Then rG+H(x/S) and rG+H(y/S) differ in the zth and wth positions. On the other hand, if SG and SH are strictly 1-locating sets, then ∃p ∈ SG\NG(x) and q ∈ SH\NH(y). Hence rG+H(x/S) and rG+H(y/S) differ in pth and qth positions. Therefore, S is a 2-resolving set in G + H. Corollary 3. Let G and H be connected nontrivial graphs. Then, dim2(G + H) = min {sln2(G) + ln2(H), ln2(G) + sln2(H), sln1(G) + sln1(H)} . Proof. Let S be a minimum 2-resolving set of G + H. Let SG = V (G) ∩ S and SH = V (H) ∩ S. By Theorem 4, SG and SH are 2-locating sets in G and H, respec- tively where SG or SH is strictly 2-locating set or SG and SH are strictly 1-locating sets. If SG is strictly 2-locating set in G, then sln2(G) + ln2(H) ≤ |SG| + |SH | = |S| = dim2(G + H). If SH is strictly 2-locating set in H, then sln2(H) + ln2(G) ≤ |SH | + |SG| = |S| = dim2(G + H). If SG and SH are both strictly 1-locating sets, then sln1(G) + sln1(H) ≤ |SG| + |SH | = |S| = dim2(G + H). Thus, dim2(G + H) ≥ min {sln2(G) + ln2(H), ln2(G) + sln2(H), sln1(G) + sln1(H)}. Next suppose that sln1(G) + sln1(H) ≤ sln2(G) + ln2(H) and sln1(G) + sln1(H) ≤ ln2(G) + sln2(H). Let SG be a minimum strictly 1-locating set in G and SH be a min- imum strictly 1-locating set in H. Then S = SG ∪ SH is a 2-resolving set in G + H, by Theorem 4. Hence dim2(G + H) ≤ |S| = |SG| + |SH | = sln1(G) + sln1(H). Therefore, dim2(G + H) ≤ sln1(G) + sln1(H). Similarly, if sln2(G) + ln2(H) ≤ sln1(G) + sln1(H) and sln2(G) + ln2(H) ≤ ln2(G) + sln2(H), then dim2(G + H) ≤ sln2(G) + ln2(H). Also, if ln2(G) + sln2(H) ≤ sln2(G) + ln2(H) and ln2(G) + sln2(H) ≤ sln1(G) + sln1(H), then dim2(G + H) ≤ ln2(G) + sln2(H). Therefore, dim2(G + H) = min {sln2(G) + ln2(H), ln2(G) + sln2(H), sln1(G) + sln1(H)}. Example 13. For any n,m ≥ 4, dim2(Pn + Pm) =  ( n 2 + 1 ) + ( m 2 + 1 ) , if n,m even( n 2 + 1 ) + ⌈ m 2 ⌉ , if n is even,m is odd⌈ n 2 ⌉ + ( m 2 + 1 ) , if n is odd,m is even⌈ n 2 ⌉ + ⌈ m 2 ⌉ , if n,m odd J. Cabaro, H. Rara / Eur. J. Pure Appl. Math, 14 (3) (2021), 773-782 780 In particular, for n = 2, 3 and m = 2, 3, dim2(Pn + Pm) = n + m 4. 2-Resolving Sets in the Corona of Graphs Definition 5. [2] The corona G ◦ H of two graphs G and H is the graph obtained by taking one copy of G of order n and n copies of H, and then joining the ith vertex of G to every vertex in the ith copy of H. For every v ∈ V (G), denote by Hv the copy of H whose vertices are attached one by one to the vertex v. Subsequently, denote by v + Hv the subgraph of the corona G ◦H corresponding to the join 〈{v}〉+ Hv, v ∈ V (G). The sets {u1, u2, v1, v2, w1, w2} and {a1, a3, b1, b3, c1, c3, d1, d3} are 2-resolving sets in the coronas P3 ◦ P2 and C4 ◦ P3, respectively, in Figure 4. .................................... .................................... ........................................................................ .................................... ........................................................................ .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... ............................................................................................................ .......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ................................................................................................................ ............................................................................................. ........................................................................................................................................................... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ................................................................................................................ ............................................................................................. ........................................................................................................................................................... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ................................................................................................................ ............................................................................................. ......... ........ ........ ........ ........ . ......... ........ ........ ........ ........ . ............................................................................................. .......................................................................................................... ................... .................. .................. .................. .................. ............... ........................................................................................................................................ .......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ...... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........................................................................................................................................................ .......................................... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ... ................... .................. .................. .................. .................. ............... ....................................................................................................................................................................................................... ......... ........ ........ ........ ........ . ......... ........ ........ ........ ........ . ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ... ................................................................................................................... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ .... .......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ...... ........................................................................................................................................ .......................................... .......................................... P3 ◦ P2: u1 u2 v1 v2 w1 w2 u v w C4 ◦ P3: a a1 a2 a3 b b1 b2 b3 c c1 c2 c3 d d1 d2 d3 Figure 4: The corona P3 ◦ P2 with dim2(P3 + P2) = 6 and the corona C4 ◦ P3 with dim2(C4 ◦ P3) = 8 Remark 6. Let v ∈ V (G). For every x, y ∈ V (Hv), dG◦H(x,w) = dG◦H(y, w) and dG◦H(v, w) + 1 = dG◦H(x,w) for every w ∈ V (G ◦H)\V (Hv). Remark 7. Let G and H be non-trivial connectd graphs, C ⊆ V (G ◦ H) and Sv = V (Hv) ∩ C where v ∈ V (G). For each x ∈ V (Hv) \ Sv and z ∈ Sv, dG◦H(x, z) = { 1 if z ∈ NHv(x) 2 otherwise Theorem 5. Let G and H be nontrivial connected graphs. A proper subset S of V (G◦H) is a 2-resolving set of G ◦H if and only if S = A ∪B, where A ⊆ V (G) and B = ⋃ {Sv : Sv is a 2-resolving set of Hv, ∀v ∈ V (G)}. J. Cabaro, H. Rara / Eur. J. Pure Appl. Math, 14 (3) (2021), 773-782 781 Proof. Suppose S is a 2-resolving set in G◦H. Let A = V (G)∩C and Sv = S∩V (Hv) for all v ∈ V (G). Then S = A∪ ( ⋃ v∈V (G) Sv ) where A ⊆ V (G) and Sv ⊆ V (Hv). Suppose Sv = ∅ for some v ∈ V (G). Let x, y ∈ V (Hv). Then rG◦H(x/S) = rG◦H(y/S) which is a contradiction to the assumption of S. Thus Sv 6= ∅. Now, we claim that Sv is a 2-resolving set in Hv for each v ∈ V (G). Let p, q ∈ V (Hv) where p 6= q. Since S is a 2-resolving set in G ◦H, rG◦H(p/S) and rG◦H(q/S) differ in at least 2 positions. By Remark 6, rHv(p/Sv) and rHv(q/Sv) must differ in at least 2 positions. Thus Sv is a 2-resolving set in Hv. Conversely, let S = A ∪ ( ⋃ v∈V (G) Sv ) where A ⊆ V (G) and Sv ⊆ V (Hv) satisfying the given conditions. Let x, y ∈ V (G ◦ H) with x 6= y and let u, v ∈ V (G) such that x ∈ V (u + Hu) and y ∈ V (v + Hv). Case 1. u = v Subcase 1.1 x, y ∈ V (Hv) Since Sv is a 2-resolving set, rHv(x/Sv) and rHv(y/Sv) differ in at least 2 positions. By Remark 6, rG◦H(x/S) and rG◦H(y/S) differ in at least 2 positions. Subcase 1.2 x = v and y ∈ V (Hv) Since G is nontrivial and connected, ∃w ∈ NG(v) and |Sw| ≥ 2. By Remark 6, rG◦H(x/S) and rG◦H(y/S) differ in at least 2 positions. Case 2. u 6= v Subcase 2.1 x ∈ V (Hu), y ∈ V (Hv) Note that rG◦H(x/Sv) has components greater than or equal to 3 and rG◦H(y/Sv) has components less than or equal to 2. Since |Sv| ≥ 2, rG◦H(x/S) and rG◦H(y/S) differ in at least 2 positions. Subcase 2.2 x = u, y ∈ V (v + Hv) Since |Su| ≥ 2, rG◦H(x/Su) and rG◦H(y/Su) differ in at least 2 positions. Hence, rG◦H(x/S) and rG◦H(y/S) differ in at least 2 positions. Therefore, in any case, S is a 2-resolving set in G ◦H. Corollary 4. Let G and H be nontrivial connected graphs, where |V (G)| = n. Then dim2(G ◦H) = n · dim2(H). Proof. Let S be a minimum 2-resolving set of G ◦H. Then by Theorem 5, S = A∪B, where A ⊆ V (G) and B = ⋃ Sv, v ∈ V (G) and Sv is a 2-resolving set in H. Hence, dim2(G ◦H) = |S| = |A|+ |B| ≥ |A|+ |V (G)| · dim2(H) = |A|+ n.dim2(H) ≥ n · dim2(H). Now, let C be a minimum 2-resolving set in H. For each v ∈ V (G), choose Cv ⊆ V (Hv) with 〈Cv〉 ∼= 〈C〉. Then D = ⋃ v∈V (G)Cv is a 2-resolving set in G◦H by Theorem 5. Hence, REFERENCES 782 dim2(G ◦H) ≤ |D| = | ⋃ v∈V (G) Cv| = n · |Cv| = n · |C| = n · dim2(H). Therefore, dim2(G ◦H) = n · dim2(H). Example 14. For any integer n ≥ 2 and m ≥ 5, dim2(G ◦ Cm) = { n (⌈ m 2 ⌉) , if m is odd n ( m 2 ) , if m is even Example 15. For any integer n,m ≥ 2, dim2(G ◦ Pm) = { n (⌈ m 2 ⌉) , if m is odd n [( m 2 ) + 1 ] , if m is even References [1] R Bailey and I Yero. Error-correcting codes from k-resolving sets. Discussiones Mathematicae, Graph Theory, 39:341–355, 2019. [2] G Chartrand and P Zhang. Graphs and Digraphs. WMU,Kalamazoo, USA, sixth edition, 2016. [3] F Harary. Graph Theory. Addison-Wesley Publishing Company, USA, 1969. [4] F Harary and R Melter. On the metric dimension of a graph. Ars Combinatoria., 2:191–195, 1976. [5] G Monsanto and H Rara. Resolving restrained domination in graphs. European Journal of Pure and Applied Mathematics., 2021(accepted). [6] P Acal G Monsanto and H Rara. On strong resolving domination in the join and corona of graphs. 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