Volume 2022, Issue 1 6906316
Research Article
Open Access

[Retracted] On Adjacency Metric Dimension of Some Families of Graph

Ali N. A. Koam

Ali N. A. Koam

Department of Mathematics, College of Science, Jazan University, New Campus, Jazan 2097, Saudi Arabia jazanu.edu.sa

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Ali Ahmad

Ali Ahmad

College of Computer Science & Information Technology, Jazan University, Jazan, Saudi Arabia jazanu.edu.sa

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Muhammad Azeem

Corresponding Author

Muhammad Azeem

Department of Mathematics, Riphah Institute of Computing and Applied Sciences, Riphah International University, Lahore, Pakistan riphah.edu.pk

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Adnan Khalil

Adnan Khalil

Department of Computer Science, Al-Razi Institute Saeed Park, Near New Ravi Bridge, Shahdarah, Lahore, Pakistan

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Muhammad Faisal Nadeem

Muhammad Faisal Nadeem

Department of Mathematics, COMSATS University Islamabad, Lahore Campus, Lahore, Pakistan comsats.edu.pk

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First published: 25 June 2022
Citations: 5
Academic Editor: Muhammad Gulzar

Abstract

Metric dimension of a graph is a well-studied concept. Recently, adjacency metric dimension of graph has been introduced. A set QaV(G) is considered to be an adjacency metric generator for G if u1, u2V\Qa (supposing each pair); there must exist a vertex qQa along with the condition that q is indeed adjacent to one of u1, u2. The minimum number of elements in adjacency metric generator is the adjacency metric dimension of G, denoted by dima(G). In this work, we compute exact values of the adjacency metric dimension of circulant graph Cn(1, 2), Möbius ladder, hexagonal Möbius ladder, and the ladder graph.

1. Introduction

Let G be a simple connected graph with vertex set V and edge set E. Let N be a set of nonnegative integers; we assume a function dG : V × VN defined as
(1)

Then, (V, dG) is a metric space. A subset AV is called a metric generator for G if it is the generator of the metric space (V, dG); that is, every point of the space is uniquely determined by its distances from the elements of A. A minimum metric generator is the metric basis, and its cardinality is the metric dimension of G, denoted by dim(G); for further detail of metric and their parameters, see [16]. The concept of metric dimension was first introduced by [7] in the problem of uniquely determining the location of an intruder in a network and was named as a locating set instead of metric generators. The same concept of metric generators or locating set was introduced by [8], and this time, they named it as resolving sets. Application of metric dimension to the navigation of robots in networks is studied in [9] and to chemistry in [10, 11]. This concept was further studied by many researchers (for instance, see [12, 13]); different studies of metric and its related parameters are studied in literature; for example, metric dimension of bilinear form graphs is found in [14], distance-regular graph is available in [15], in terms of dominating set [16], study of the corona product is found in [17], and some computer fields are attached to this topic found in [18, 19]. Several metric generators have since been developed and researched, including independent resolving sets [20], resolving dominant sets [21], strong resolving sets [22], and local metric sets [23].

A subset QaV of the set of vertices is an adjacency metric generator of G if for every two vertices u1, u2V\Qa, there exists a vertex qQa such that q is indeed adjacent to one of u1, u2. The minimum number of elements in the adjacency metric generator is the adjacency metric dimension of G and symbolized by dima(G). The idea of adjacency metric dimension was put forward by [24], and it is closed related to 1-locating dominating set [25]. The goal of this definition is to look at the metric dimension of the lexicographic product of graphs in terms of graph adjacency. For two graphs with n, K-orders, the researchers in [26, 27] had taken the corona product of both graphs; the resulted graph has (local) metric dimension n times the (local) adjacency metric dimension. They demonstrated that calculating the adjacency metric dimension is NP-hard as a result of this tight relationship.

The adjacency metric generator of any graph G can be a metric generator in a correctly selected metric space, as pointed out in [26, 27]. Assume q is a positive number. Assume the dG,q : V × VN is a distance function, which is defined as
(2)

Then, the metric dimension of (V, dG,2) is equal to the adjacency metric dimension of G. A subset QaV is the k adjacency metric generator for a graph G, if for each pair of vertices u1, u2V, there are at least k vertices v1, v2, ⋯, vkQa as a result dG,2(u1, vi) ≠ dG,2(u2, vi), for each 1 ≤ ik. For the minimum number (k) of members in the adjacency metric generator, the definition will be called as k adjacency metric dimension of a graph G, and here, it is symbolized by dima,j(G).

The adjacency metric dimension is an NP-hard problem [25, 28], and it is very important to determine its exact values for well-known families of graphs. The primary goal of this work is to determine the exact values of the adjacency metric dimension of particular graph families, notably circulant graphs Cn(1, 2), ladder, Möbius ladder, and hexagonal ladder graphs. To compute the adjacency metric dimension of these classes, we need the following proposition by Moreno et al. [29].

Proposition 1 (see [29].)If G is a graph with |G| = n ≥ 2, then dima,j(G) = j if and only if j ∈ {1, 2} and G ∈ {P1, P2, }.

Remark 2 (see [29].)If G is a graph with |G| = n ≥ 7, then dima,1(G) ≥ 3.

2. Construction of Graphs

Circulant graphs are a type of graph that may be utilised in the construction of local area networks. Let v1, v2, ⋯, vzz and n be the nonnegative with given conditions, vμvν∀ 1 ≤ μ < νz, where 1 ≤ vμ ≤ ⌊n/2⌋. With the collection of vertices in an undirected graph V = {v1, v2, ⋯, vn} and the set of edges E = {vμvμ+vν : 1 ≤ μn, 1 ≤ νz}, the indices are considered to be taken in modulo n condition, called a circulant graph, and it is symbolized by Cn(v1, v2, ⋯, vz). The generators are v1, v2, ⋯, vz-numbers, and the edge vμvμ+vν is of type vν. Actually, we can observe that the Cn(v1, v2, ⋯, vz) circulant graph is an r-regular graph, and the r given is
(3)

Möbius ladder MLm is built by a grid of m × 1, and twist this grid at 180°; now, paste the extreme most left and right path of vertices as seen in Figure 1. It contains m-horizontal cycles of order four. For more study on ladder-type networks, see [30, 31]. The metric dimension of MLm is three [32].

Details are in the caption following the image
Möbius ladder graph MLm.

Hexagonal Möbius ladder HMLm is built in [33], it can construct by dividing each horizontal edge of a square grid by inserting a new vertex, and it becomes a grid of m × 1 with each cycle having order six; now, twist this grid at 180° and paste the extreme most left and right path of vertices as shown in Figure 2. This graph contains m-horizontal cycles of order six. The metric dimension of the hexagonal Möbius ladder network is three [33].

Details are in the caption following the image
Hexagonal Möbius ladder graph HMLm.

Let Lm be a ladder graph [34], with n ≥ 3. We label the ladder graph as shown in Figure 3. The order and size of the ladder graph are 2n and 3n − 2, respectively.

Details are in the caption following the image
Ladder graph Ln.

3. Results on Adjacency Metric Dimension of Graphs

We compute the adjacency metric dimension of certain graph families in this section. Let Cn(1, 2) be a circulant graph obtained from Cn by joining all the vertices at distance 2. Let n ≥ 6; then, |V(Cn(1, 2))| = n and |E(Cn(1, 2))| = 2n. In the next theorem, we compute the exact value of the adjacency metric dimension of the circulant graph Cn(1, 2).

Theorem 3. Let G = Cn(1, 2) be a circulant graph. Then,

(4)

Proof. We divide the proof of the theorem into three parts.

Part A. For the inequality dima(G) ≤ 3, when n = 6, 7, 8, let Qa = {v1, v2, v3} as an adjacency metric generator; then, the representation of v1, v2, v3 are as follows v1 = (0, 1, 1), v2 = (1, 0, 1), and v3 = (1, 1, 0). For any vertex vkv1, v2, v3, we have the following representation:

(5)
where a = 2, when k = 4, ⋯, n − 2; otherwise, a = 1. Also, b = 1 when k = 4, n, and c = 1, when k = 4, 5; otherwise, both b, c are 0.

From the above discussion, we can say that dima(G) ≤ 3.

Converse: on the contrary, assume that dima(G) = 2, for n = 7, 8. Let be a resolving set for adjacency metric dimension. Then, for any , the distance (d(v1, v), d(v2, v)) ∈ {(1, 1), (1, 2), (2, 1), (2, 2)}. Given , by the principle of Dirichlet’s box, two or more components of resulted in identical vectors of distance, and further, this implied as a contradiction. Now, for n = 6, let 2 ≤ i ≤ 6; we have the same representations, that is, either or Let the adjacency metric generator 3 ≤ i ≤ 6; then, we have the same representations, that is, either or Let 3 ≤ i ≤ 5, 4 ≤ j ≤ 6; then, we have the same representations, that is, , , or Therefore, dima(G) ≥ 3, concluding that dima(G) = 3.

Part B. When n ≡ 0, 3, 4, 5(mod6), n ≥ 9, we intend to use the induction method on the order of cycle; set the base step as n = 9, which implies that dima(C9) = 4. Let Qa = {v1, v2, v3, v4} be a resolving set for adjacency metric dimension. All vertices have the following representations:

(6)
where a, b, c, d = 0, when k = 1, 2, 3, 4, respectively, and a = 1 when k = 2, 3, 8, 9, b = 1 when k = 1, 3, 4, 9, c = 1 when k = 1, 2, 4, 5, and d = 1 when k = 2, 3, 5, 6; otherwise, all values of a, b, c, d are 2. All the representations are different so
(7)

On the contrary, assume that dima(C9) = 3. On this contradiction, the following are some cases to defend it.

Case 1. Let the adjacency metric generator 1 ≤ i, j, k ≤ 9, taking vertices in with 0-size gap (consecutive vetices). Then, we have the same representations in ; it means that have the same representations with the vertices which are not adjacent to any of the vertex of

Case 2. Let 1 ≤ i ≤ 6, taking vertices in with 1- and 0-size gap, respectively. Then, we have the same representations in ; it means that have the same representations with the vertices which are adjacent to vi only.

Case 3. Let , 1 ≤ i ≤ 6, taking vertices again in with 0- and 1-size gap, respectively. Then, we have the same representations in vi+3 ~ {vi, vj}; it means that have the same representations with the vertices which are adjacent to vi+3 only.

Case 4. Let , taking vertices in with 1-size gap. Then, we have the same representations in vi+4 ~ {vi, vj}; it means that have the same representations with the vertices which are adjacent to vi+4 only.

Case 5. Let , with 0- and 2-size gap, respectively. Then, we have the same representations in , which simplifies that have the same representations with the vertices which are adjacent to vi+4 only.

Case 6. Let , taking vertices in with 1- and 2-size gap, respectively. Then, we have the same representations in , which means that have the same representations with the vertices which are adjacent to vi+5 only.

Due to the nature of the adjacency metric generator with three cardinalities, have the following gap size possibilities:

  • (i)

    The first gap is even, and second is odd

  • (ii)

    The first gap is odd and the next is even

  • (iii)

    Both gaps are even

  • (iv)

    Both gaps are odd

Case 7. Let , taking vertices in above any sizes of gap possibilities. Then, we have the same representations in either or only single vertex of adjacent to both vertices, which have the same representations. From all above cases, dima(C9) ≠ 3. Hence, our base step is true:

(8)

Now, assume that it is also true for n = m, and we have to show that it is also true for n = m + 1 which implies that

(9)

Assume the assertion that

(10)

Putting equations (8) and (9) in equation (10), we have

(11)
where ⌊(n + 3)/6⌋ = ⌊(n + 4)/6⌋.

Part C. Remaining cases when n ≡ 1, 2(mod6), n ≥ 13.

Again, the method of induction can be used, and this implies that the base step becomes true for n = 13 and as well dima(C13) = 5; for this purpose, let Qa = {v1, v2, v5, v7, v11} be a resolving set for adjacency metric dimension; all the vertices have the following representations:

(12)
where a, b, c, d = 0 when k = 1, 2, 5, 7, 11, respectively, and a = 1 when k = 2, 3, 12, 13, b = 1 when k = 1, 3, 4, 13, c = 1 when k = 3, 4, 6, 7, d = 1 when k = 5, 6, 8, 9, and e = 1 when k = 9,10,12,13; otherwise, all values of a, b, c, d are 2.
(13)

On the contrary, assume that dima(C13) = 4. Let be the adjacency metric generator with cardinality 4 having the following gap size possibilities.

  • (i)

    All the gap sizes are even

  • (ii)

    All the gap sizes are odd

  • (iii)

    One gap size is odd, and others are even

  • (iv)

    One gap size is even, and others are odd

Case 1. Let , 1 ≤ i, j, k, l ≤ 13, with any sizes of gap possibilities above defined. Then, we have the same representations in either {vi, vj, vk, vl}­{vi, vj} or only single vertex of adjacent to both vertices, which have the same representations. From all above cases, dima(C13) ≠ 4. Hence,

(14)

Now, assume that it is also true for n = m, and we have to show that it is also true for n = m + 1, which leads to the assertion that

(15)

Assuming that

(16)

Putting equations (14) and (15) in equation (16), we have

(17)
where ⌊(n + 3)/6⌋ = ⌊(n + 4)/6⌋. This completes the proof of the theorem with all possible cases on order of the Cn graph.

Theorem 4. Let MLm be a Möbius ladder graph. Then,

(18)

Proof. We divide the proof of the theorem into three parts.

Part A. Let Qa = {v1, v2, v5} be an adjacency metric generator, and those shown in Table 1 are the representations of all vertices for dima(ML4) ≤ 3.

All the vertices have different representations, and it proves that dima(ML4) ≤ 3. On the contrary, suppose that dima(ML4) = 2. By using Proposition 1, it is not possible, concluding that

(19)

Part B. When m = 3, 5, this adjacency metric generator is Qa = {v1, v2, v4, v4+⌊m/2⌋} which claims that dima(MLm) ≤ 4. All vertices have the following representations:

(20)
where a, b, c, d = 0 when k = 1, 2, 4, 4 + ⌊m/2⌋, respectively, and a = 1 when k = 2, m + 1, 2m, b = 1 when k = 1, 3, m + 2, c = 1 when k = m − 2, n, 2m − 1, and d = 1 when m − 1, m + 1, 2m; otherwise, all a, b, c, d are 2. It proves the claim that dima(MLm) ≤ 4; on the other hand, suppose that dima(MLm) = 3, when m = 3, 5.

Case 1. Let , 1 ≤ i, j, k ≤ 2m, taking vertices in with 0-size gap. Then, we have the same representations in and m + 2 ≤ l ≤ 2m.

Case 2. Let , with the vertices in with any size of gap discussed in Theorem 3, part 6 Then, we have the same representations in and m + 2 ≤ l ≤ 2m. It means that when choosing the resolving set for adjacency metric dimension with cardinality three and vertices with any gap size, there must exist two vertices with the same representations which have one belonging to l-domain and the second from l-domain. From all above cases, dima(MLm) ≠ 3 for the values of m = 3, 5. Hence,

(21)

Part C. Here, we will use the induction method. For the base step, we choose m = 6; the dima(ML6) = 4 having the adjacency metric generator Qa = {v1, v3, v5, v7}, and the following are the representations of all vertices:

(22)
where a, b, c, d = 0 when k = 1, 3, 5, 7, respectively, and a = 1 when k = 2, 7, 12, b = 1 when k = 2, 4, 9, c = 1 when k = 4, 6, 11, and d = 1 when k = 1, 6, 8; otherwise, all values of a, b, c, d are 2. It concludes that
(23)

The contradiction side gives us dima(ML6) = 3. For this contradiction, possible cases are discussed in the converse of part b of the theorem. One can evaluate from the discussion that Qa with cardinality three is not possible which implies that

(24)

Now, assume that it is also true for m = l, and we have to show that m = l + 1 which leads to the inductive step:

(25)

Assume that

(26)

Using equations (24) and (25) in equation (26), we have

(27)

It completes the proof of the theorem with all possible cases on m-cycles of the Möbius ladder graph.

Theorem 5. Let HMLm be a hexagonal Möbius ladder graph. Then,

(28)

Proof. We break the proof of the theorem into four parts according to the adjacency metric dimension.

Part A. In this part, we claim that dima(HML2) = 3. For the case dima(HML2) ≤ 3, let the adjacency metric generator be Qa = {v1, v2, v6}, and those shown in Table 2 are the representations of all vertices of HML2 with respect to Qa.

In Table 2, the given representations are different, and it proves that dima(HML2) ≤ 3; on the reverse, for inequality which is dima(HML2) = 2, using Proposition 1 is not true, concluding that

(29)

Part B. This part contains the adjacency metric dimension of the hexagonal Möbius ladder graph with m = 3; for this, let the adjacency metric generator be Qa = {v1, v3, v5, v7}, and those shown in Table 3 are the representations of all vertices according to Qa, which are again different, and it proves that dima(HML3) ≤ 4; on the reverse, inequality which is dima(HML3) = 3 is not true, and the following are some discussions to this side.

Case 1. Let , taking vertices in with any size of gap discussed in Theorem 3, part 6 Then, we have the same representations in either or and 2m + 2 ≤ l ≤ 4m. It means that when choosing the resolving set of adjacency metric dimension with cardinality three and vertices with any gap size, there must exist two vertices with the same representations which have one belonging to l-domain and the second from l-domain. It is also possible that the same representations are in two vertices, and v1 is one of them, and the second vertex belongs to l-domain. From all above cases, dima(ML3) ≠ 3. Hence,

(30)

Part C. For dima(HML4) ≤ 6, let the adjacency metric generator be Qa = {v1, v2, v5, v7, v9, v12}, and representations are shown in Table 4.

Table 1. Representations of vertices with respect to Qa = {v1, v2, v5}.
vk r(vk | v1) r(vk | v2) r(vk | v5) vk r(vk | v1) r(vk | v2) r(vk | v5)
1 0 1 2 5 1 2 1
2 1 0 1 6 2 1 0
3 2 1 2 7 2 2 1
4 2 2 2 8 1 2 2
Table 2. Representations of vertices with respect to Qa = {v1, v2, v6}.
vk r(vk | v1) r(vk | v2)  r(vk | v6) vk r(vk | v1) r(vk | v2) r(vk | v6)
1 0 1 2 5 1 2 1
2 1 0 2 6 2 1 0
3 2 1 2 7 2 2 1
4 2 2 2 8 1 2 2
Table 3. Representations of vertices with respect to Qa = {v1, v3, v5, v7}.
vk r(vk | v1) r(vk | v3) r(vk | v5) r(vk | v7) vk r(vk | v1) r(vk | v3) r(vk | v5) r(vk | v7)
1 0 2 2 1 7 1 2 2 0
2 1 1 2 2 8 2 2 2 1
3 2 0 2 2 9 2 1 2 2
4 2 1 1 2 10 2 2 2 2
5 2 2 0 2 11 2 2 1 2
6 2 2 1 1 12 1 2 2 2
Table 4. Representations r(vk | Qa) of vertices with respect to Qa = {v1, v2, v5, v7, v9, v12}.
vk v1 v2 v5 v7 v9 v12 vk v1 v2 v5 v7 v9 v12
1 0 1 2 2 1 2 9 1 2 2 2 0 2
2 1 0 2 2 2 2 10 2 2 2 2 1 2
3 2 1 2 2 2 2 11 2 2 2 2 2 1
4 2 2 1 2 2 2 12 2 2 2 2 2 0
5 2 2 0 2 2 2 13 2 2 1 2 2 1
6 2 2 1 1 2 2 14 2 2 2 2 2 2
7 2 2 2 0 2 2 15 2 2 2 1 2 2
8 2 2 2 1 1 2 16 1 2 2 2 2 2

The representations given in Table 4 prove that dima(HML4) ≤ 6. On contrary, suppose that dima(HML4) = 5; the following are some discussions to the side of this contradiction.

Case 1. Let , taking vertices in with any size of gap already discussed above in the theorem. Then, we have the same representations in either or and 2m + 2 ≤ l ≤ 4m. It means that when choosing the resolving set for adjacency metric dimension with cardinality five and vertices with any gap size, there must exist two vertices with the same representations which have one belonging to l-domain and the second from l-domain. It is also possible that the same representations are in two vertices and v1 is one of them, and the second vertex belongs to l-domain. From all above cases, dima(ML4) ≠ 5. Hence,

(31)

Part D. In this part for the proof of dima(HMLm) = m + 3 where m ≥ 5, we will apply the induction method and the base case for m = 5, which implies that dima(HML5) = 8, to prove that dima(HML5) ≤ 8; the following are the representations with respect to the adjacency metric generator Qa = {v1, v2, v4, v7, v9, v11, v13, v16}:

(32)
where a, b, c, d, e, f, g, h = 0 when k = 1, 2, 4, 7, 9,11,13,16, and a = 1 when k = 2,11,20, b = 1 when k = 1, 3, c = 1 when k = 3, 5, d = 1 when k = 6, 8, 17, e = 1 when k = 8,10,19, f = 1 when k = 1,10,12, g = 1 when k = 3,12,14, and h = 1 when k = 15, 17; otherwise, all values of a, b, c, d, e, f, g, h are 2. All the representations show that there are no two vertices with the same representation; hence,
(33)

On the contrary, suppose that dima(HML5) = 7; the following are some discussions to the side of contradiction.

Case 1. Let and 1 ≤ j ≤ 7, taking vertices in with any size of gap discussed in Theorem 3, part 6 Then, we have the same representations in either or or and 2m + 2 ≤ l ≤ 4m. It means that when choosing the resolving set of adjacency metric dimension with cardinality seven and vertices with any gap size, there must exist two vertices with the same representations which have one belonging to l-domain and the second from l-domain. It is also possible that the same representations are in two vertices and v1 is one of them, and the second vertex belongs to l-domain. From all above cases, dima(HML5) ≠ 7. Hence,

(34)

Now, assume that the assertion is true for m = l:

(35)

We have to show that it is also true for m = l + 1. Suppose

(36)

Plugging the values of equations (34) and (35) in equation (36), we have

(37)

As a result, the result holds for all positive integers m ≥ 5. It also completes the all possible cases that we assume on the very start of the proof.

Theorem 6. Let Ln be a ladder graph with n ≥ 3. Then,

(38)

Proof. We prove that dima(Ln) = n − ⌊(n + 1)/4⌋ with the induction method and the base step is n = 3 which implies that dima(L3) = 2. If we assume on the contrary it is not possible that dima(L3) = 1, for dima(L3) ≤ 2, all vertices have the following representations with respect to Qa = {v1, v5}:

(39)
where a, b = 0 when k = 1, 5, respectively, and a = 1 when k = 2, 3, b = 1 when k = 3, 6; otherwise, all values of a, b are 2. All the representations show that there are no two vertices with the same representation; hence,
(40)

Now, assume that the assertion is true for n = k:

(41)

We have to show that it is also true for n = k + 1. Suppose

(42)

Using equations (40) and (41) in equation (42), we have

(43)
where ⌊(k + 1)/4⌋ = ⌊(k + 2)/4⌋. As a consequence, the result holds for all positive integers for n ≥ 3, which completes the proof.

4. Conclusion

In this article, we studied the adjacency metric dimension of circulant, ladder, Möbius ladder, and hexagonal Möbius ladder graphs. It is known that adjacency metric dimension is a useful parameter in localization, networking, and some robot navigation ideas. Therefore, it is interesting to find adjacency metric generators for generalized classes of graphs. We demonstrated that in adjacency metric generators, all graphs have inconstant numbers of members and that each graph follows the modification of parameters or order.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

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