In this paper, we determine unique metro domination number of circulant graphs

B. Sooryanarayana

John Sherra

English

ZENODO

International J.Math. Combin. Vol.1(2019), 53-61Unique Metro Domination Number of Circulant GraphsB. Sooryanarayana(Department of Mathematics, Dr.Ambedkar Institute of Technology, Bengaluru, Karnataka 560 056, India)John Sherra(Department of Mathematics, St. Aloysius College, Mangaluru, Karnataka 575 003, India)E-mail: dr bsnrao@yahoo.co.in, johnsherra@gmail.comAbstract: A dominating set D of G which is also a resolving set of G is called a metrodominating set. A metro dominating set D of a graph G(V,E) as a unique metro dominatingset (in short an UMD-set) if |N(v) ∩D| = 1 for each vertex v ∈ V −D and the minimumcardinality of an UMD-set of G is the unique metro domination number of G. In this paper,we determine unique metro domination number of circulant graphs.KeyWords: Domination, Smarandachely k-dominating set, metric dimension, metro dom-ination, uni-metro domination.AMS(2010): 05C20, 05C26.§1. IntroductionAll the graphs considered in this paper are simple, connected and undirected. The length ofa shortest path between two vertices u and v in a graph G is called the distance between uand v and is denoted by d(u, v). For a vertex v of a graph, N(v) denotes the set of all verticesadjacent to v and is called open neighborhood of v. Similarly, the closed neighborhood of v isdefined as N [v] = N(v) ∪ {v}.Let G(V,E) be a graph. For each ordered subset S = {v1, v2, · · · , vk} of V , each vertexv ∈ V can be associated by a vector of distances denoted by Γ(v/S) = (d(s1, v), d(s2, v), · · · ,d(sk, v)). The set S is said to be a resolving set of G, if Γ(v/S) 6= Γ(u/S), for every u, v ∈V − S. A resolving set of minimum cardinality is the metric basis and cardinality of a metricbasis is the metric dimension of G. The k-tuple, Γ(v/S) associated to the vertex v ∈ Vwith respect to a Metric basis S, is referred as a code generated by S for that vertex v. IfΓ(v/S) = {v1, v2, · · · , vk}, then v1, v2, · · · , vk are called components of the code of v generatedby S and in particular vi, 1 ≤ i ≤ k, is called ith-component of the code of v generated by S.A dominating set D of a graph G(V,E) is the subset of V having the property that foreach vertex v ∈ V −D there exists a vertex u in D such that uv ∈ E. Generally, a set D ⊆ Vof G is said to be a Smarandachely k-dominating set if each vertex of G is dominated by atleast k vertices of S with k ≥ 1. Clearly, a dominating set is a Smarandachely 1-dominating1Received April 22, 2018, Accepted February 24, 2019.54 B. Sooryanarayana and John Sherraset. A dominating set D of G which is also a resolving set of G is called a metro dominatingset or in short an MD− set. A metro dominating set D of a graph G(V,E) as a unique metrodominating set (in short an UMD-set) if |N(v) ∩ D| = 1 for each vertex v ∈ V − D andthe minimum cardinality of an UMD-set of G is the unique metro domination number of G,denoted by γuβ(G).Metric dimensions and locating dominating sets of certain classes of graphs were studiedin [1-14].§2. Resolvability of Circulant GraphsA graph whose vertex set is {vi|i ∈ Z+} and two vertices vi and vj are adjacent if and only ifi − j (mod n) ∈ C, for a given C ⊆ Zn with 0 6∈ C, is called a circulant diagraph. If the setC has the property that C = −C, then the underlying graph is called circulant graph, and wedenote it by Xn,∆, where |C| = ∆. The set C is referred to as a connected set. The circulantgraph Xn,∆ is a ∆-regular graph. In this paper, we consider a family of circulant graph Xn,3with connection set C = {1, n2 , n− 1}, where n is even.We state the following lemma whose proof follows directly by the definition of domination,and is most helpful to find UMD-sets.Lemma 2.1 In the circulant graph Xn,3, n is even, with connection set C = {1, n2 , n − 1}, avertex vi dominates vi−1, vi+1 and vi+n2 , where i+n2 is under modulo n.Now we consider G = Xn,3, n is even, where the connection set C = {1, n2 , n− 1}. Let Sbe a dominating set of G. Then by Lemma 2.1, a vertex vi ∈ S can dominate at most 3 verticesin V − S. Hence |V − S| 6 3|S|. Therefore,|V | − |S| 6 3|S| ⇒ 4|S| > |V | ⇒ |S| > n4.Thus we have the following lemma.Lemma 2.2 For any positive even integer n,γ(Xn,3) >⌈n4⌉.Let R be a set of two or more vertices of the principal cycle. Consider two distinct verticesu and v of R. Let P , P ′ be two distinct uv-path on the principal cycle. The vertices u and vare said to be neighboring vertices if u and v are the only vertices of S contained in one of thepaths P , P ′. If P (or P ′) is the path containing only u, v from S, then the set of all verticesof P − {u, v} is called a gap of S determined by u and v and is denoted by γ. The number ofvertices in the gap is called order of the gap and is denoted by o(γ).Notice that it is shown that X26,3 with a unique metro dominating set S = {vi, vi+4, vi+8,vi+15, vi+19, vi+23, vi+24, vi+25} in Figure 1. We observe that vi, vi+4 are neighboring verticesUnique Metro Domination Number of Circulant Graphs 55of S and the gap determined by vi and vi+4 is of order 3. Similarly, the gap determined by vi+8and vi+15 is of order 6. The gap between vi+24 and vi+25 is of order 0 or empty gap.8?iv4?iv15?iv19?iv23?iv24?iv25?ivivFigure 1 Circulant graph X26,3 with C = {1, 13, 25}.Consider G = Xn,3, n is even. Let S be a UMD-set of G. Suppose there is a gap of S oforder 1. Let u, v be two neighboring vertices and w be the only vertex of the gap. Then w isdominated by both u and v and hence w is not uniquely dominated. Thus, we haveLemma 2.1 If γ is a gap of a UMD-set S of the graph G = Xn,3, then 0(γ) 6= 1.In the discussion to follow, we want to find suitable gaps of a dominating set S, so thatS becomes a UMD-set. Gaps of order 4 or more will introduce gaps of order 0 and therebyincrease |S|; for, consider a gap γ of order 4 between the neighboring vertices vi and vi+5.Vertices in the gaps are vi+1, vi+2, vi+3 and vi+4. Vertices vi+1 and vi+4 are dominated by viand vi+5 respectively. It is therefore obvious that vi+2 and vi+3 should be dominated by vi+2+ n2and vi+3+ n2 . Thus, vi+2+n2and vi+3+ n2 belongs to S. The gap between them is empty. Hence|S| is increased.If all the gaps of S are of order 3, then |S| is the least; for, if vi, vi+4 ∈ S and are neighboringvertices, then the gap determined by them is of order 3. As vi+1 and vi+3 are dominated by viand vi+4 respectively, the vertex vi+2 has to be dominated by vi+2+ n2 . Thus, vi+2+n2∈ S.Observe that vi+ n2+4 is dominated by vi+4. Therefore, we take vi+n2+6∈ S, so that vi+n2 +6and vi+ n2+2 are neighboring vertices of a gap of order 3. As vi+ n2+6 dominates vi+6, we includevi+8 in S so that vi+4 and vi+8 are neighboring vertices of gap of order 3. Thus, S = {vi, vi+4,vi+2+ n2, vi+8, · · · }.If the above set of vertices has vi +n2 − 2 as the last vertex, then the above sequence ofvertices in S terminates at vi−4. In this case each vertex in S is uniquely dominating exactly3 vertices in V − S. Thus by Lemma 2.1, |S| is the least. This leads to the lemma.Lemma 2.4 A dominating set S of Xn,3 has a least |S|, when each gap of S is of order 3.Lemma 2.5 If G is a graph of order n having a dominating set S such that every gap of S isof order 3, then n ≡ 4(mod 8).56 B. Sooryanarayana and John SherraProof If every gap is of order 3, then vi+ n2+6, vi+n2+10, · · · , vi−4 are in S. Hence n2+4k+2 ≡0 (mod n)⇒ n ≡ 4 (mod 8). 2Note that when n ≡ 4 (mod 8), there are exactly n4 gaps of order 3 on the principal cycle.Also observe that n4 is an odd integer. From these we conclude the following result.Lemma 2.6 When n ≡ 0(mod 8), n > 8, there is at least one gap of order less than 3.In a circulant graph G = Xn,3, let vi, vi+1, · · · , vi+(n−1), vi+n = vi (subscripts increase inanti-clockwise direction) form the principal cycle. Each vertex vi, on the principal cycle, alsolies on two other cycles:(1) vi, vi+n2 , vi+n2−1, . . . , vi+1, vi (Clockwise) and(2) vi, vi+n2, vi+n2+1, . . . , vi−1, vi (anti-clockwise).Length of these cycles is n2 + 1. Hence maximum distance between any two vertices onthese cycles is 12(n2 + 1), if n2 + 1 is even and is12(n2)if n2 + 1 is odd. Thus we haveLemma 2.7 If x and y are any two vertices of Xn,3 then d(x, y) = k 6⌈n4⌉.Note that the subscripts of the vertex names are in Zn, i.e. congruent modulo n.Lemma 2.8 If n > 8 and n ≡ 0 (mod 4), then for a fixed i, {vi, vi+4, vi+n2 +2} is a resolvingset.Proof For the cases, n = 12, 16 and 20, it is easy to see that the set S is a resolving set.We prove the case when n > 20. Suppose that x and y are vertices on the principal cycle suchthat d(vi, x) = d(vi, y). Then there are four cases, in each case k 6 ⌈n4 ⌉.Case 1. x = vi−k and y = vi+k.In this case we see that d(vi+4, y) = |k−4|. If k 6 n4−4, then d(vi+4, x) = k+4 6= d(vi+4, y);If k > n4 − 4, then d(vi+4, x) = n2 +1− (k+4) = n2 − k− 3 6= d(vi+4, y) for if n2 − k− 3 = k− 4,then 2k = n2 + 1⇒ k = n4 + 12 , which is not possible as n ≡ 0 modulo 4. Thus vi+4 resolves xand y.44??nkkivy??iv44?? nk kivx??44??nkkivy??)1(2????knivy5?k4?iv) 1( 2? ? ? ?kn ivy5 ? k5 ?k) 1(2? ?? ?k ni vxkivx??44nk??5?k12???knivFigure 2. Circulant graph X32,3.Unique Metro Domination Number of Circulant Graphs 57Case 2. x = vi+n2+k−1 and y = vi+ n2−(k−1).If k 6 5, then d(vi+4, x) = 6−k and d(vi+4, y) = 4+k. Now 6−k = 4+k⇒ 2k = 2⇒ k = 1which is not possible. Hence d(vi+4, x) 6= d(vi+4, y); If k > 5, then d(vi+4, x) = k − 4, which isnot equal to d(vi+4, y). Thus vi+4 resolves x and y.Case 3. x = vi−k and y = vi+n2−(k−1).If k 6 ⌈n4 ⌉− 4, then we have d(vi+4, x) = 4+ k and d(vi+4, y) = 4+ k. But d(vi+ n2 +2, x) =3 + k 6= d(vi+ n2+2, y) = k + 1; If n4 − 4 < k 6 n4 then d(vi+4, x) = n2 − k − 3 = d(vi+4, y) andd(vi+ n2+2, x) 6= d(vi+ n2+2, y). Hence in all the cases, vi+ n2+2 resolves x and y.Case 4. If x = vi+n2 +k−1 and y = vi+k then d(vi+4, y) = |k − 4|.If k < 5, then d(vi+4, x) = 6−k 6= d(vi+4, y); If k > 5, then d(vi+4, x) = k−4 = d(vi+4, y).However, d(vi+ n2+2, x) = k − 3 6= d(vi+ n2+2, y) = k − 1. Hence vi+ n2+2 resolves x and y.Note that the Theorem 1 of Muhammad Salman et al in [6], states that for all n > 4 andn ≡ 0 (mod 4), the metric dimension of Xn,3 > 2. Hence {vi, vi+4, vi+ n2 +2} is a resolving set.Hence the lemma. 2Lemma 2.9 For any integer n > 8 and n ≡ 2(mod 4), the set {vi, vi+4, vi+n2 +2, vi+ n2+6} is aresolving set.Proof The cases where n 6 22 follows easily. We now suppose n > 22 and, x and y aretwo vertices of G such that d(vi, x) = d(vi, y). Then there are four cases in each case k 6n+24 .Case 1. x = vi−k and y = vi+k.In this case d(vi+4, y) = |k − 4|. If k 6 n+24 − 4, then d(vi+4, x) = k + 4 6= d(vi+4, y); Ifk > n+24 − 4, then d(vi+4, x) = n2 − k − 3. Now d(vi+4, y) = d(vi+4, x)⇒ k − 4 = n2 − k − 3⇒k = n+24 . If k =n+24 , d(vi+ n2 +2, x) = k − 3 6= d(vi+ n2+2, y) = k − 1.Case 2. x = vi+n2+k−1 and y = vi+ n2−(k−1).If k 6 5, d(vi+4, x) = 6 − k 6= d(vi+4, y) = k + 4; except when k = 1. But when k = 1, xand y coincide. Now if 5 < k < n+24 , then d(vi+4, y) = k − 4 6= d(vi+4, x) = k + 4.Case 3. x = vi−k and y = vi+n2−(k−1).When k 6 n+24 − 4, d(vi+4, x) = 4 + k and d(vi+4, y) = 4 + k. But d(vi+ n2 +2, x) =3 + k 6= d(vi+ n2+2, y) = k + 1, and whenn+24 − 4 < k 6 n+24 , d(vi+4, x) = n2 − k − 3 andd(vi+4, y) =n2 − k − 3. However d(vi+ n2 +2, x) = n2 − k − 2 and d(vi+ n2+2, y) = k + 1.Now n2 − k − 2 = k + 1⇒ 2k = n2 − 3⇒ k = n−64 .If k 6= n−64 , then vi+ n2+2 resolves x and y; If k = n−64 , then d(vi+ n2+6, x) = n2 − k − 6 andd(vi+ n2+6, y) =n2 − k − 4, which are not equal. Hence vi+ n2+6 resolves x and y.Case 4. x = vi+n2 +k−1 and y = vi+k.If k 6 5, then d(vi+4, x) = 6 − k 6= d(vi−4, y) = |k − 4|; If 5 < k < n+24 , then d(vi+4, x) =k − 4 = d(vi+4, y). But d(vi+ n2+2, x) = k − 3 6= d(vi+ n2+2, y) = k − 1. Thus vi+ n2+2 resolves xand y.58 B. Sooryanarayana and John SherraNow by the Theorem 2 of Muhammad Salman et al [6], we have for all n > 6 and n ≡ 2(mod 6), the metric dimension of the graph Xn,3 > 3.Hence the set {vi, vi+4, vi+n2+2, vi+n2+6} as per the above four cases becomes a resolvingset. Hence the lemma. 2§4. Algorithm to Extend Circulant Graphs and Resolving SetsWe give an algorithm, which constructs new circulant graph from the old one by increasing itsorder and extending the its resolving set to suit for the newly constructed one (as in Figure 3).112??niv442??? n iv32??? niv22??? niv12???niv2niv ?1?iviv1??iv2??iv3??iv4??ivFigure 3. Algorithmic construction of Xn+8,3 from Xn,3.Input: The graph Xn,3 and a metric basis S with |S| = k.Step 1: Select two neighboring vertices on the principal cycle with a gap of 3. Let vi and vi+4 bethe vertices. Then vi, vi+4 and vi+ n2 +2 ∈ S.Step 2: Delete the edge vivi+1. Add four vertices v′i+1, v′i+2, v′i+3 and v′i+4 between the verticesvi and vi+1. Join the vertices to get the edges viv′i+1, v′i+1v′i+2, v′i+2v′i+3, v′i+3v′i+4 andv′i+4vi+1.Step 3: Delete the edge vi+n2 vi+n2+1. Add four vertices vertices v′i+n2 +1, v′i+ n2 +2, v′i+ n2+3 andv′i+n2 +4 between the vertices vin2and vi+ n2+1. Join these vertices to get the edges vin2v′i+n2 +1,v′i+n2 +1v′i+ n2+2, v′i+ n2+2v′i+ n2 +3, v′i+n2 +3v′i+ n2+4and v′i+ n2 +4vi+n2+1.Step 4: Join these 8 vertices to get the edges, vi+1v′i+ n2+1, vi+2v′i+n2 +2, vi+3v′i+n2 +3and vi+4v′i+ n2 +4.Unique Metro Domination Number of Circulant Graphs 59Step 5: Add v′i+4 and vi+ n2 +2 into S.Output: The graph Xn+8,3 and a metric basis S with |S| = k + 2§4. Unique Metro Domination Number of Xn,3In this section we completely determine unique metro domination number of 3-regular circulantgraphs in the form of following sequence of theorems.Theorem 4.1 If n > 8 is an even integer and n ≡ 6 or 4 (mod 8), then γµβ(Xn,3) =⌈n4⌉.Proof We consider two cases separately following.Case 1. n ≡ 4 (mod 8) and n > 8.We first take the smallest possible n, that is n = 12. Define S = {vi, vi+4, vi+n2 +2} for anyi, 1 6 i 6 12. In view of Lemma 2.8, the set S so defined is a resolving set. Further, each gapis of order 3. Hence S is uniquely dominates V − S. So, by Lemma 2.2, γµβ(X12,3) >⌈n4⌉andhere |S| = 3. Thus, S is a UMD-set. Therefore, γµβ(X12,3) =⌈n4⌉, when n = 12.Now we apply the algorithm to construct the next cases for n. At each time algorithmincreases the order by 8 and |S| by 2. Hence γµβ(Xn+8,3) =⌈n4⌉+ 2 =⌈n+84⌉.Case 2. n ≡ 6 (mod 8) and n > 8.For the least possible n = 14, by Lemma 2.8, it follows that the set {vi, vi+4, vi+ n2+2, vi+n2 +6}resolves V −D. Now from Lemma 2.2, γµβ(Xn,3) >⌈n4⌉= 4. There is a gap of order 0 in S.The middle vertex vk of a gap of order 3 is dominated by vk+ n2 ∈ S. Hence S is a UMD-set.We apply the algorithm on X14,3. It increases n = 14 to n = 22. It also increases |S| by 2.Hence γµβ(X22,3) =⌈n4⌉. Repeated application of the algorithm gives the theorem. 2Theorem 4.2 If n is any even integer, n > 8 and n ≡ 0 or 2 (mod 8), then γµβ(Xn,3) =⌈n4⌉+1.Proof We prove the theorem in two different cases as follows:Case 1. n ≡ 0 (mod 8).The graph X16,3 is the graph of least possible order in this case. Invoking Lemma 2.4,consider a gap of order 3, having vi and vi+4 as the neighboring vertices. Then vi, vi+4, andvi+ n2+2 ∈ S.This leads to a gap of order 5 between vi+4 and vi+ n2+2 in which vi+6, vi+7 are notdominated and a gap of order 5 between vi+ n2+2 = vi+10 and vi in which vi−2 and vi−3 arenot dominated. As⌈n4⌉= 4, S contains a minimum of 4 vertices by Lemma 2.2. If vi+6 ∈ S,then vi+5 is not uniquely dominated. If vi+7 ∈ S, then vi−1 is not uniquely dominated andif vi−3 ∈ S, then vi+5 is not uniquely dominated. Hence none of vi+6, vi+7, vi−2, vi−3 can beincluded in S. If vi−1 ∈ S, then vi−2 and vi+7 ∈ V − S are uniquely dominated. Similarly ifvi+5 ∈ S, then vi+6 and vi−3 in V − S are uniquely dominated. However |S| = 5 and is not60 B. Sooryanarayana and John Sherrapossible to reduce it to 4. (Because each gap of order 5 can be converted to two gaps of order2, But it will have |S| = 5).Hence γµβ(Xn,3) = 5. Application of algorithm now will increase the order by 8 and |S|increases by 2. Therefore γµβ(Xn,3) =⌈n4⌉+ 1.Case 2. n ≡ 2 (mod 8).As in Case 1, we take vi, vi+4 and vi+n2+2 ∈ S. If vi+ n2+6 ∈ S, then it leaves a gap of order2 between vi and vi−3 and a gap of order 6 between vi+4 and vi+ n2+2. In this gap of order 6,the vertices vi+5, vi+6, vi+9 and vi+10 are uniquely dominated. If vi+7 ∈ S, then vi−2 and vi+6are not uniquely dominated. If vi+8 ∈ S, then vi−1 and vi+9 are not uniquely dominated. Ifwe include vi−2 and vi−1 in S, then the domination of all vertices in V − S is unique. However|S| = 6. It can not be reduced to 5 = ⌈n4 ⌉. If the gap of order 6 between vi+4 and vi+11 isconverted into 2 gaps by including vi+8 ∈ S , then vi−1 is not uniquely dominated. Similarlyincluding vi+4 fails. Hence γµβ(Xn,3) = 6. As before we apply algorithm to conclude the restof the theorem. 2Acknowledgment The authors are thankful to the learned referees for their valuable sugges-tions for the improvement of the paper.References[1] Jose´ Ca´ceres, Mary´a L. 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Slater, Leaves of trees, In Proc. 6th Southeastern Conf. on Combinatorics, GraphTheory and Computing, Congressus Numerantium, Vol. 14 (1975), 549-559.[8] P. J. Slater, Domination and location in acyclic graphs, Networks, 17 (1987), 55-64.[9] P. J. Slater, Locating dominating sets, in Y. Alavi and A. Schwenk ed. Graph Theory,Combinatorics, and Applications, Proc. Seventh Quad International Conference on thetheory and applications of Graphs. John Wiley & Sons, Inc. (1995), 1073-1079.[10] A.Sebo and E.Tannier, On metric generators of graphs, Math. Opr. Res., 29 (2004), No.2,383-393.Unique Metro Domination Number of Circulant Graphs 61[11] B.Shanmukha, B.Sooryanarayana and K.S.Harinath, Metric dimension of wheels, Far EastJ. Appl. Math., 8(3) (2002) 217-229.[12] B.Sooryanarayana, On the metric dimension of graph, Indian J. Pure and Appl. 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Unique Metro Domination Number of Circulant Graphs

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Unique Metro Domination Number of Circulant Graphs

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