On the metric dimension of corona product graphs
aa r X i v : . [ m a t h . C O ] O c t On the metric dimension of corona productgraphs
I. G. Yero , D. Kuziak and J. A. Rodr´ıguez-Vel´azquez Departament d’Enginyeria Inform`atica i Matem`atiques,Universitat Rovira i Virgili, Av. Pa¨ısos Catalans 26, 43007 Tarragona, [email protected], [email protected] Faculty of Applied Physics and MathematicsGda´nsk University of Technology, ul. Narutowicza 11/12 80-233 Gda´nsk, [email protected]
November 7, 2018
Abstract
Given a set of vertices S = { v , v , ..., v k } of a connected graph G , the metric representation of a vertex v of G with respect to S is the vector r ( v | S ) = ( d ( v, v ) , d ( v, v ) , ..., d ( v, v k )), where d ( v, v i ), i ∈ { , ..., k } denotes the distance between v and v i . S is a resolvingset for G if for every pair of vertices u, v of G , r ( u | S ) = r ( v | S ). Themetric dimension of G , dim ( G ), is the minimum cardinality of anyresolving set for G . Let G and H be two graphs of order n and n , respectively. The corona product G ⊙ H is defined as the graphobtained from G and H by taking one copy of G and n copies of H and joining by an edge each vertex from the i th -copy of H with the i th -vertex of G . For any integer k ≥
2, we define the graph G ⊙ k H recursively from G ⊙ H as G ⊙ k H = ( G ⊙ k − H ) ⊙ H . We giveseveral results on the metric dimension of G ⊙ k H . For instance,we show that given two connected graphs G and H of order n ≥ n ≥
2, respectively, if the diameter of H is at most two, then dim ( G ⊙ k H ) = n ( n + 1) k − dim ( H ). Moreover, if n ≥ he diameter of H is greater than five or H is a cycle graph, then dim ( G ⊙ k H ) = n ( n + 1) k − dim ( K ⊙ H ) . Keywords:
Resolving sets, metric dimension, corona graph.
AMS Subject Classification Numbers:
The concepts of resolvability and location in graphs were described inde-pendently by Harary and Melter [10] and Slater [19], to define the samestructure in a graph. After these papers were published several authors de-veloped diverse theoretical works about this topic [3, 4, 5, 6, 7, 16, 18, 20].Slater described the usefulness of these ideas into long range aids to nav-igation [19]. Also, these concepts have some applications in chemistry forrepresenting chemical compounds [14, 15] or to problems of pattern recogni-tion and image processing, some of which involve the use of hierarchical datastructures [17]. Other applications of this concept to navigation of robotsin networks and other areas appear in [6, 12, 16]. Some variations on re-solvability or location have been appearing in the literature, like those aboutconditional resolvability [18], locating domination [11], resolving domination[1] and resolving partitions [5, 8, 9, 21]. In this article we study the metricdimension of corona product graphs.We begin by giving some basic concepts and notations. Let G = ( V, E )be a simple graph of order n = | V | . Let u, v ∈ V be two different verticesin G , the distance d G ( u, v ) between two vertices u and v of G is the lengthof a shortest path between u and v . If there is no ambiguity, we will usethe notation d ( u, v ) instead of d G ( u, v ). The diameter of G is defined as D ( G ) = max u,v ∈ V { d ( u, v ) } . Given u, v ∈ V , u ∼ v means that u and v areadjacent vertices. Given a set of vertices S = { v , v , ..., v k } of a connectedgraph G , the metric representation of a vertex v ∈ V with respect to S is thevector r ( v | S ) = ( d ( v, v ) , d ( v, v ) , ..., d ( v, v k )). We say that S is a resolvingset for G if for every pair of distinct vertices u, v ∈ V , r ( u | S ) = r ( v | S ). The metric dimension of G is the minimum cardinality of any resolving set for G ,and it is denoted by dim ( G ).Let G and H be two graphs of order n and n , respectively. The coronaproduct G ⊙ H is defined as the graph obtained from G and H by taking onecopy of G and n copies of H and joining by an edge each vertex from the2 th -copy of H with the i th -vertex of G . We will denote by V = { v , v , ..., v n } the set of vertices of G and by H i = ( V i , E i ) the copy of H such that v i ∼ v for every v ∈ V i . Notice that the corona graph K ⊙ H is isomorphic to thejoin graph K + H . For any integer k ≥
2, we define the graph G ⊙ k H recursively from G ⊙ H as G ⊙ k H = ( G ⊙ k − H ) ⊙ H . We also note thatthe order of G ⊙ k H is n ( n + 1) k . We begin by presenting the following useful facts.
Lemma 1.
Let G = ( V, E ) be a connected graph of order n ≥ and let H be a graph of order at least two. Let H i = ( V i , E i ) be the subgraph of G ⊙ H corresponding to the i th -copy of H . (i) If u, v ∈ V i , then d G ⊙ H ( u, x ) = d G ⊙ H ( v, x ) for every vertex x of G ⊙ H not belonging to V i . (ii) If S is a resolving set for G ⊙ H , then V i ∩ S = ∅ for every i ∈ { , ..., n } . (iii) If S is a resolving set for G ⊙ H of minimum cardinality, then V ∩ S = ∅ . (iv) If H is a connected graph and S is a resolving set for G ⊙ H , then forevery i ∈ { , .., n } , S ∩ V i is a resolving set for H i .Proof. (i) Let y = v i ∈ V . The result directly follows from the fact that d G ⊙ H ( u, x ) = d G ⊙ H ( u, y )+ d G ⊙ H ( y, x ) = d G ⊙ H ( v, y )+ d G ⊙ H ( y, x ) = d G ⊙ H ( v, x ).(ii) We suppose V i ∩ S = ∅ for some i ∈ { , ..., n } . Let x, y ∈ V i . By(i) we have d G ⊙ H ( x, u ) = d G ⊙ H ( y, u ) for every vertex u ∈ S , which is acontradiction.(iii) We will show that S ′ = S − V is a resolving set for G ⊙ H . Now let x, y be two different vertices of G ⊙ H . We have the following cases.Case 1: x, y ∈ V i . By (i) we conclude that there exist v ∈ V i ∩ S ′ suchthat d G ⊙ H ( x, v ) = d G ⊙ H ( y, v ).Case 2: x ∈ V i and y ∈ V j , i = j . Let v ∈ V i ∩ S ′ . Then we have d G ⊙ H ( x, v ) ≤ < ≤ d G ⊙ H ( y, v ).Case 3: x, y ∈ V . Let x = v i and let v ∈ V i ∩ S ′ . Then we have d G ⊙ H ( x, v ) = 1 < d G ⊙ H ( y, x ) = d G ⊙ H ( y, v ).3ase 4: x ∈ V i and y ∈ V . If x ∼ y , then y = v i . Let v j ∈ V , j = i , andlet v ∈ V j ∩ S ′ . Then we have d G ⊙ H ( x, v ) = 1 + d G ⊙ H ( y, v ) > d G ⊙ H ( y, v ).For x y = v l we take v ∈ V l ∩ S ′ and we obtain d G ⊙ H ( x, v ) = d G ⊙ H ( x, y ) + d G ⊙ H ( y, v ) > d G ⊙ H ( y, v ).Therefore, S ′ is a resolving set for G ⊙ H .(iv) Let S i = S ∩ V i . For x ∈ S i or y ∈ S i the result is straightforward.We suppose x, y ∈ V i − S i . Since S is a resolving set for G ⊙ H , we have r ( x | S ) = r ( y | S ). By (i), d G ⊙ H ( x, u ) = d G ⊙ H ( y, u ) for every vertex u of G ⊙ H not belonging to V i . So, there exists v ∈ S i such that d G ⊙ H ( x, v ) = d G ⊙ H ( y, v ). Thus, either ( v ∼ x and v y ) or ( v x and v ∼ y ). In the firstcase we have d G ⊙ H ( x, v ) = d H i ( x, v ) = 1 and d G ⊙ H ( y, v ) = 2 ≤ d H i ( y, v ).The case v x and v ∼ y is analogous. Therefore, S i is a resolving set for H i . Theorem 2.
Let G and H be two connected graphs of order n ≥ and n ≥ , respectively. Then, dim ( G ⊙ k H ) ≥ n ( n + 1) k − dim ( H ) . Proof.
Let S be a resolving set of minimum cardinality in G ⊙ H . FromLemma 1 (iii) we have that S ∩ V = ∅ . Moreover, by Lemma 1 (ii) we havethat for every i ∈ { , ..., n } there exist a nonempty set S i ⊂ V i such that S = S n i =1 S i . Now, by using Lemma 1 (iv) we have that S i is a resolving setfor H i . Hence, dim ( G ⊙ H ) = | S | = P n i =1 | S i | ≥ P n i =1 dim ( H ) = n dim ( H ).As a result, the lower bound follows. Theorem 3.
Let G be a connected graph of order n ≥ and let H be agraph of order n ≥ . If D ( H ) ≤ , then dim ( G ⊙ k H ) = n ( n + 1) k − dim ( H ) . Proof.
Let S i ⊂ V i be a resolving set for H i and let S = S n i =1 S i . We willshow that S is a resolving set for G ⊙ H . Let us consider two different vertices x, y of G ⊙ H . We have the following cases.Case 1: x, y ∈ V i . Since D ( H i ) ≤
2, we have that r ( x | S i ) = r ( y | S i ) leadsto r ( x | S ) = r ( y | S ).Case 2: x ∈ V i and y ∈ V j , i = j . Let v ∈ S i . Hence we have d ( x, v ) ≤ < ≤ d ( y, v ).Case 3: x, y ∈ V . Let x = v i . Then for every vertex v ∈ S i we have d ( x, v ) = 1 < d ( y, x ) + 1 = d ( y, v ). 4ase 4: x ∈ V i and y ∈ V . If x ∼ y , then let v ∈ S j , for some j = i . Sowe have d ( x, v ) = 1 + d ( y, v ) > d ( y, v ). Moreover, if x y = v j , for v ∈ S j we have d ( x, v ) = d ( x, y ) + d ( y, v ) > d ( y, v ).Thus, for every different vertices x, y of G ⊙ H , we have r ( x | S ) = r ( y | S ),as a consequence, dim ( G ⊙ H ) ≤ n dim ( H ). Therefore, we have dim ( G ⊙ k H ) ≤ n ( n + 1) k − dim ( H ). By Theorem 2 we conclude the proof.In order to show a consequence of the above theorem we present thefollowing well known result, where K t denotes a complete graph of order t , K s,t denotes a complete bipartite graph of order s + t and N t denotes anempty graph of order t . Lemma 4. [6]
Let G be a connected graph of order n ≥ . Then dim ( G ) = n − if and only if G = K s,t , ( s, t ≥ , G = K s + N t , ( s ≥ , t ≥ , or G = K s + ( K ∪ K t ) , ( s, t ≥ . Corollary 5.
Let G be a connected graph of order n ≥ and let H be agraph of order n ≥ and diameter D ( H ) ≤ . Then dim ( G ⊙ k H ) = n ( n + 1) k − ( n − if and only if H = K s,t , ( s, t ≥ ; H = K s + N t , ( s ≥ , t ≥ , or H = K s + ( K ∪ K t ) , ( s, t ≥ . We recall that the wheel graph of order n +1 is defined as W ,n = K ⊙ C n ,where K is the singleton graph and C n is the cycle graph of order n . Themetric dimension of the wheel W ,n was obtained by Buczkowski et. al. in[2]. Remark 6. [2]
Let W ,n be a wheel graph. Then dim ( W ,n ) = for n = 3 , , for n = 4 , , (cid:4) n +25 (cid:5) otherwise. The fan graph F n ,n is defined as the graph join N n + P n , where N n is the empty graph of order n and P n is the path graph of order n . Thecase n = 1 corresponds to the usual fan graphs. Notice that, for the metricdimension of fan graphs, it is possible to find an equivalent result to Remark6 which was obtained by Caceres et. al. in [4].5 emark 7. [4] Let F ,n be a fan graph. Then dim ( F ,n ) = for n = 1 , for n = 2 , , for n = 6 , (cid:4) n +25 (cid:5) otherwise. As a particular case of the Theorem 3 we obtain the following results.
Corollary 8.
Let G be a connected graph of order n ≥ . If H is a wheelgraph or a fan graph of order n ≥ , then dim ( G ⊙ k H ) = n ( n + 1) k − (cid:22) n (cid:23) . Theorem 9.
Let G be a connected graph of order n ≥ and let H be agraph of order n ≥ . Let α be the number of connected components of H of order greater than one and let β be the number of isolated vertices of H .Then dim ( G ⊙ k H ) ≤ n ( n + 1) k − ( n − α − for α ≥ and β ≥ , n ( n + 1) k − ( n − α ) for α ≥ and β = 0 , n ( n + 1) k − ( n − for α = 0 .Proof. We suppose α ≥ β ≥
1. Let A i be the set of vertices of G ⊙ H formed by all but one of the vertices per each of the α connected componentsof H i . If β ≥ B i to be the set of vertices of G ⊙ H formed byall but one of the isolated vertices of H i . If β = 1 we assume B i = ∅ . Letus show that S = ∪ n j =1 ( A j ∪ B j ) is a resolving set for G ⊙ H . Let x, y betwo different vertices of G ⊙ H . We suppose x, y / ∈ S . We have the followingcases.Case 1. x = v i ∈ V and y ∈ V i . For every vertex u ∈ V j ∩ S , j = i , weobtain d ( y, u ) = d ( y, x ) + d ( x, u ) > d ( x, u ).case 2. x = v i ∈ V and y V i . For every v ∈ S ∩ V i we have d ( x, v ) =1 < d ( y, v ).Case 3. x ∈ V i and y ∈ V j , j = i . For every u ∈ V i ∩ S we have d ( x, u ) ≤ < ≤ d ( y, u ). 6ase 4. x, y ∈ V i . We consider, without loss of generality, that x is notan isolated vertex in H i . Then there exists v ∈ V i ∩ S such that v ∼ x , so d ( x, v ) = 1 < d ( y, v ).Thus, for every two different vertices x, y of G ⊙ H , we obtain r ( x | S ) = r ( y | S ) and, as a consequence, dim ( G ⊙ H ) ≤ n ( n − α − . As above, if β = 0 then we take S = ∪ n j =1 A j and we obtain dim ( G ⊙ H ) ≤ n ( n − α ) and if α = 0, then we take S = ∪ n j =1 B j and we obtain dim ( G ⊙ H ) ≤ n ( n − . Note that if α = 0, then it is not necessary toconsider Case 4. Thus, the result follows. Corollary 10.
Let G be a connected graphs of order n ≥ and let H be anunconnected graph of order n ≥ . Then dim ( G ⊙ k H ) = n ( n + 1) k − ( n − if and only if H ∼ = N n .Proof. In [13] the authors showed that dim ( G ⊙ N n ) = n ( n − dim ( G ⊙ k N n ) = n ( n +1) k − ( n − H is unconnected and H = N n , then dim ( G ⊙ k H ) ≤ n ( n + 1) k − ( n − Theorem 11.
Let G and H be two connected graphs of order n ≥ and n ≥ , respectively. Then dim ( G ⊙ k H ) = n ( n + 1) k − ( n − if and only if H ∼ = K n . Moreover, if H = K n , then dim ( G ⊙ k H ) ≤ n ( n + 1) k − ( n − . Proof.
Since dim ( K n ) = n −
1, by Theorem 3 we conclude dim ( G ⊙ k K n ) = n ( n + 1) k − ( n − H = K n . Given a set X of vertices of H and a vertex v of H , N X ( v ) denotes the set of neighborsthat v has in X : N X ( v ) = { u ∈ X : u ∼ v } . Given two vertices a, b of H , let X a,b be the set formed by all vertices of H different from a and b . Since H isa connected graph and H = K n , there exist at least two vertices a, b of H such that N X a,b ( a ) = N X a,b ( b ). Let a i , b i be the vertices corresponding to a, b ,respectively, in the i th -copy H i = ( V i , E i ) of H . Let S = ∪ n i =1 ( V i − { a i , b i } ).We will show that S is a resolving set for G ⊙ H . Let x, y be two differentvertices of G ⊙ H such that x, y S . We have the following cases.7ase 1. x = a i and y = b i . Since N X a,b ( a ) = N X a,b ( b ) we have r ( x | S ) = r ( y | S ).Case 2. x = v i ∈ V and y ∈ V i . For every v ∈ V j − { a j , b j } , j = i , wehave d ( y, v ) = d ( y, x ) + d ( x, v ) > d ( x, v ). If x ∈ V i and y ∈ V j , j = i , thenfor every v ∈ V i − { a i , b i } we have d ( x, v ) ≤ < ≤ d ( y, v ).Case 3. x, y ∈ V . Say x = v i . Then for every v ∈ V i − { a i , b i } we have d ( x, v ) = 1 < d ( y, v ).Hence, for every two different vertices x, y of G ⊙ H , we obtain r ( x | S ) = r ( y | S ). Thus, dim ( G ⊙ H ) ≤ n ( n − . Therefore, the result follows.As we have shown in Corollary 5, the above bound is tight.
Theorem 12.
Let G be a connected graph of order n ≥ and let H be agraph of order n ≥ . Then dim ( G ⊙ k H ) ≤ n ( n + 1) k − dim ( K ⊙ H ) . Proof.
We denote by K ⊙ H i the subgraph of G ⊙ H , obtained by joining thevertex v i ∈ V with all vertices of H i . For every v i ∈ V , let B i be a resolvingset of minimum cardinality of K ⊙ H i and let B = S n i =1 B i . By Lemma 1 (iii)we have that v i does not belong to any resolving set of minimum cardinalityfor K ⊙ H i . So, B does not contain any vertex from G . We will show that B is a resolving set for G ⊙ H . Let x, y be two different vertices in G ⊙ H .We consider the following cases.Case 1: x, y ∈ V i . There exists u ∈ B i such that d K ⊙ H i ( x, u ) = d K ⊙ H i ( y, u ), which leads to d G ⊙ H ( x, u ) = d G ⊙ H ( y, u ).Case 2: x ∈ V i and y ∈ V j , i = j . Let v ∈ B i . We have d G ⊙ H ( x, v ) ≤ < ≤ d G ⊙ H ( y, v ).Case 3: x, y ∈ V . Suppose now that x is adjacent to the vertices of H i .Hence, for every vertex v ∈ B i we have d G ⊙ H ( x, v ) = 1 < d G ⊙ H ( y, x ) + 1 = d G ⊙ H ( y, v ).Case 4: x ∈ V i and y ∈ V . If x ∼ y , then for every vertex v ∈ B j , with j = i , we have d G ⊙ H ( x, v ) = 1+ d G ⊙ H ( y, v ) > d G ⊙ H ( y, v ). Now, let us assumethat x y . Hence, there exists v ∈ B j adjacent to y , with j = i . So, wehave d G ⊙ H ( x, v ) = d G ⊙ H ( x, y ) + 1 = d G ⊙ H ( x, y ) + d G ⊙ H ( y, v ) > d G ⊙ H ( y, v ).Thus, for every two different vertices x, y of G ⊙ H, we have r ( x | S ) = r ( y | S ) and, as a consequence, dim ( G ⊙ H ) ≤ n dim ( K ⊙ H ). Therefore,the result follows. 8 heorem 13. Let G be a connected graph of order n ≥ and let H be agraph of order n ≥ . If D ( H ) ≥ or H is a cycle graph, then dim ( G ⊙ k H ) = n ( n + 1) k − dim ( K ⊙ H ) . Proof.
Let S be a resolving set of minimum cardinality in G ⊙ H . By Lemma1 (iii) we have S ∩ V = ∅ , as a consequence, S = ∪ n i =1 S i , where S i ⊂ V i .Notice that, by Lemma 1 (ii), S i = ∅ for every i ∈ { , ..., n } . Now wedifferentiate two cases in order to show that r ( x | S i ) = (1 , ...,
1) for every x ∈ V i − S i .Case 1. H is a cycle graph of order n ≥
7. If r ( a | S i ) = (1 ,
1) for some a ∈ V i − S i , then, since n ≥
7, there exist two vertices x, y ∈ V i − S i suchthat d H i ( x, v ) > d H i ( y, v ) >
1, for every v ∈ S i . Hence, d G ⊙ H ( x, v ) = d G ⊙ H ( y, v ) = 2 for every v ∈ S i , which is a contradiction because, by Lemma1 (i), d G ⊙ H ( x, v ) = d G ⊙ H ( y, v ) for every vertex u of S not belonging to S i .Case 2. D ( H ) ≥
6. Let x, y ∈ V i − S i . Since S is a resolving set for G ⊙ H , we have r ( x | S ) = r ( y | S ). As we have noted before, by Lemma 1 (i) wehave that d G ⊙ H ( x, u ) = d G ⊙ H ( y, u ) for every vertex u of G ⊙ H not belongingto V i . So, there exists v ∈ S i such that d G ⊙ H ( x, v ) = d G ⊙ H ( y, v ) and, as aconsequence, either ( v ∼ x and v y ) or ( v x and v ∼ y ). Now we supposethat there exists a vertex a ∈ V i − S i such that r ( a | S i ) = (1 , , ... b ∈ V i − S i such that d H i ( b, u ) >
1, for every u ∈ S i , thenfor every w ∈ V i − ( S i ∪ { a, b } ), there exists v ∈ S i such that w ∼ v . Then D ( H i ) ≤
5. Moreover, if for every b ∈ V i − S i there exists v b ∈ S i such that v b ∼ b , then D ( H ) ≤
4. Therefore, if D ( H ) ≥
6, then r ( a | S i ) = (1 , , ... a ∈ V i − S i .Now, we denote by K ⊙ H i the subgraph of G ⊙ H , obtained by joiningthe vertex v i ∈ V with all vertices of the i th -copy of H . In both the abovecases we have r ( v i | S i ) = (1 , , ..., = r ( x | S i ) for every x ∈ V i − S i , so S i is a resolving set for K ⊙ H i . Hence, dim ( K ⊙ H i ) ≤ | S i | , for every i ∈ { , ..., n } . Thus, dim ( G ⊙ H ) ≥ n dim ( K ⊙ H i ) and, as a consequence, dim ( G ⊙ k H ) ≥ n ( n + 1) k − dim ( K ⊙ H ). We conclude the proof byTheorem 12. Corollary 14.
Let G be a connected graph of order n ≥ . (i) If n ≥ , then dim ( G ⊙ k C n ) = n ( n + 1) k − (cid:4) n +25 (cid:5) . (ii) If n ≥ , then dim ( G ⊙ k P n ) = n ( n + 1) k − (cid:4) n +25 (cid:5) . G ⊙ H for H of order at least two. Nowwe consider the case H ∼ = K . We obtain a general bound for dim ( G ⊙ k K )and, when G is a tree, we give the exact value for this parameter. Claim 15.
Let G be a simple graph. If v is a vertex of degree greater thanone in G , then for every vertex u adjacent to v there exists a vertex x = u, v of G , such that d ( v, x ) = d ( u, x ) + 1 . The following lemma obtained in [2] is useful to obtain the next result.
Lemma 16. [2] If G is a graph obtained by adding a pendant edge to anontrivial connected graph G , then dim ( G ) ≤ dim ( G ) ≤ dim ( G ) + 1 . Theorem 17.
For every connected graph G of order n ≥ , dim ( G ⊙ k K ) ≤ k − n − . Proof. If G ∼ = K , then dim ( K ⊙ K ) = dim ( P ) = 1. So, let us suppose G = K . Let us suppose, without loss of generality, that v n is a vertex ofdegree greater than one in G and let S = V − { v n } . For every i ∈ { , ..., n } ,let u i be the pendant vertex of v i in G ⊙ K . We will show that S is aresolving set for G ⊙ K . Let x, y be two different vertices of G ⊙ K . If x = u i and y = u j , i = j , then we have either i = n or j = n . Let ussuppose for instance i = n . So, we obtain that d ( x, v i ) = 1 = d ( y, v i ). Onthe other hand, if x = v n and y = u i , then let us suppose d ( x, v i ) = 1. Since v n is a vertex of degree greater than one in G , by Claim 15, there existsa vertex v j ∈ S such that d ( x, v j ) = d ( v i , v j ) + 1. So, we have d ( x, v j ) = d ( v i , v j ) + 1 = d ( v i , v j ) + d ( u i , v i ) = d ( y, v i ) + d ( v i , v j ) = d ( y, v j ) . Therefore,for every different vertices x, y of G ⊙ K we have r ( x | S ) = r ( y | S ) and, as aconsequence, dim ( G ⊙ K ) ≤ n −
1. Therefore, dim ( G ⊙ k K ) ≤ k − n − dim ( K n ⊙ K ) ≥ dim ( K n ) = n −
1. Thus, for k = 1 the above bound is achieved for the graph G = K n .To present the next result, we need additional definitions. A vertex ofdegree at least 3 in a graph G will be called a major vertex of G . Anyvertex u of degree one is said to be a terminal vertex of a major vertex v if d ( u, v ) < d ( u, w ) for every other major vertex w of G . The terminal degree of a major vertex v is the number of terminal vertices of v . A major vertex v is an exterior major vertex if it has positive terminal degree. Given a graph G , n ( G ) denotes the number of vertices of degree one and ex ( G ) denotesthe number of exterior major vertices of G .10 emma 18. [6, 10, 19] If T is a tree that is not a path, then dim ( T ) = n ( T ) − ex ( T ) . Theorem 19.
For any tree T of order n ≥ , dim ( T ⊙ k K ) = n ( T ) for k = 1 , k − n for k ≥ .Proof. If T is a path of order n ≥
3, then we have dim ( T ⊙ K ) = 2 = n ( T ).Now, if T is not a path, then by using Lemma 18, since T ⊙ K is a tree, n ( T ⊙ K ) = n and ex ( T ⊙ K ) = n − n ( T ), we obtain the result for k = 1.Since for every tree T of order n we have n ( T ⊙ K ) = n , we obtain theresult for k ≥ Acknowledgements
The research was partially done while the first author was at Gda´nsk Uni-versity of Technology, Poland, supported by “Fundaci´o Ferran Sunyer i Bal-aguer”, Catalunya, Spain. This work was partly supported by the SpanishMinistry of Education through projects TSI2007-65406-C03-01 “E-AEGIS”and CONSOLIDER INGENIO 2010 CSD2007-00004 “ARES”.
References [1] R. C. Brigham, G. Chartrand, R. D. Dutton, P. Zhang, Resolving dom-ination in graphs,
Mathematica Bohemica (1) (2003) 25–36.[2] P. S. Buczkowski, G. Chartrand, C. Poisson, P. Zhang, On k-dimensionalgraphs and their bases,
Periodica Mathematica Hungarica , (1) (2003),9–15.[3] J. Caceres, C. Hernando, M. Mora, I. M. Pelayo, M. L. Puertas, C.Seara, D. R. Wood, On the metric dimension of Cartesian product ofgraphs, SIAM Journal of Discrete Mathematics (2) (2007) 273–302.[4] J. Caceres, C. Hernando, M. Mora, I. M. Pelayo, M. L. Puertas, C.Seara, On the metric dimension of some families of graphs, ElectronicNotes in Discrete Mathematics (2005) 129–133.115] G. Chappell, J. Gimbel, C. Hartman, Bounds on the metric and parti-tion dimensions of a graph, Ars Combinatoria (2008) 349–366.[6] G. Chartrand, L. Eroh, M. A. Johnson, O. R. Oellermann, Resolvabil-ity in graphs and the metric dimension of a graph, Discrete AppliedMathematics (2000) 99–113.[7] G. Chartrand, C. Poisson, P. Zhang, Resolvability and the upper di-mension of graphs,
Computers and Mathematics with Applications (2000) 19–28.[8] G. Chartrand, E. Salehi, P. Zhang, The partition dimension of a graph, Aequationes Mathematicae (1-2) (2000) 45–54.[9] M. Fehr, S. Gosselin, O. R. Oellermann, The partition dimension ofCayley digraphs Aequationes Mathematicae (2006) 1–18.[10] F. Harary, R. A. Melter, On the metric dimension of a graph, Ars Com-binatoria (1976) 191–195.[11] T. W. Haynes, M. Henning, J. Howard, Locating and total dominatingsets in trees, Discrete Applied Mathematics (2006) 1293–1300.[12] B. L. Hulme, A. W. Shiver, P. J. Slater, A Boolean algebraic analysisof fire protection,
Algebraic and Combinatorial Methods in OperationsResearch (1984) 215–227.[13] H. Iswadi, E. T. Baskoro, R. Simanjuntak, A. N. M. Salman, The met-ric dimension of graph with pendant edges, Journal of CombinatorialMathematics and Combinatorial Computing , (2008) 139–145.[14] M. A. Johnson, Structure-activity maps for visualizing the graph vari-ables arising in drug design, Journal of Biopharmaceutical Statistics (1993) 203–236.[15] M. A. Johnson, Browsable structure-activity datasets, Advances inMolecular Similarity (R. Carb´o–Dorca and P. Mezey, eds.) JAI PressConnecticut (1998) 153–170.[16] S. Khuller, B. Raghavachari, A. Rosenfeld, Landmarks in graphs,
Dis-crete Applied Mathematics (1996) 217–229.1217] R. A. Melter, I. Tomescu, Metric bases in digital geometry, ComputerVision Graphics and Image Processing (1984) 113–121.[18] V. Saenpholphat, P. Zhang, Conditional resolvability in graphs: a sur-vey, International Journal of Mathematics and Mathematical Sciences (2004) 1997–2017.[19] P. J. Slater, Leaves of trees, Proceeding of the 6th Southeastern Con-ference on Combinatorics, Graph Theory, and Computing, CongressusNumerantium (1975) 549–559.[20] I. Tomescu, Discrepancies between metric and partition dimension of aconnected graph, Discrete Mathematics (2008) 5026–5031.[21] I. G. Yero and J. A. Rodr´ıguez-Vel´azquez. A note on the partition di-mension of Cartesian product graphs.