On Metric Dimensions of Hypercubes
Aleksander Kelenc, Aoden Teo Masa Toshi, Riste Skrekovski, Ismael G. Yero
aa r X i v : . [ m a t h . C O ] F e b On Metric Dimensions of Hypercubes
Aleksander Kelenc , , Aoden Teo Masa Toshi , Riste ˇSkrekovski , ,and Ismael G. Yero University of Maribor, FERI, 2000 Maribor, Slovenia Institute of Mathematics, Physics and Mechanics, 1000 Ljubljana, Slovenia Independent researcher, Singapore University of Ljubljana, FMF, 1000 Ljubljana, Slovenia Faculty of Information Studies, 8000 Novo Mesto, Slovenia Universidad de C´adiz, Departamento de Matem´aticas, Algeciras, Spain
February 23, 2021
Abstract
The metric (resp. edge metric or mixed metric) dimension of a graph G , is the cardinalityof the smallest ordered set of vertices that uniquely recognizes all the pairs of distinct vertices(resp. edges, or vertices and edges) of G by using a vector of distances to this set. In this notewe show two unexpected results on hypercube graphs. First, we show that the metric and edgemetric dimension of Q d differ by only one for every integer d . In particular, if d is odd, thenthe metric and edge metric dimensions of Q d are equal. Second, we prove that the metric andmixed metric dimensions of the hypercube Q d are equal for every d ≥ . We conclude thepaper by conjecturing that all these three types of metric dimensions of Q d are equal when d is large enough. Keywords:
Edge metric dimension; mixed metric dimension; metric dimension; hypercubes.
AMS Subject Classification numbers:
The metric dimension of connected graphs was introduced about 50 years ago in [6, 21], in con-nection with modeling navigation systems in networks, although this notion was already known by [email protected], [email protected], [email protected],and [email protected] G and two vertices u, v ∈ V ( G ) ,the distance d G ( u, v ) between these two vertices is the length of a shortest path connecting v and u . The vertices u, v are distinguished ( determined or resolved ) by a vertex x ∈ V ( G ) if d G ( u, x ) = d G ( v, x ) . A given set of vertices S is a metric generator for the graph G , if every twovertices of G are distinguished by a vertex of S . The cardinality of the smallest possible metricgenerator for G is the metric dimension of G , which is denoted by dim( G ) . The terminology ofmetric generators was introduced in [11], since the previous two works referred to such sets as resolving sets and locating sets , respectively. A metric generator for G of cardinality dim( G ) iscalled a metric basis . Although the classical metric dimension is an old topic in graph theory, thereare still several open problems that remain unsolved. Recent investigations on this concern are[3, 4, 8, 15].In order to uniquely identify the edges of a graph, by using vertices, the edge metric dimensionof connected graphs was introduced in [10] as follows. Let G be a connected graph and let uv bean edge of G such that u, v ∈ V ( G ) . The distance between a vertex x ∈ V ( G ) and the edge uv isdefined as, d G ( uv, x ) = min { d G ( u, x ) , d G ( v, x ) } . It is said that two distinct edges e , e ∈ E ( G ) are distinguished or determined by a vertex v ∈ V ( G ) if d G ( e , v ) = d G ( e , v ) . A set S ⊂ V ( G ) is called an edge metric generator for G if and only if for every pair of edges e , e ∈ E ( G ) , thereexists an element of S which distinguishes the edges. The cardinality of a smallest possible edgemetric generator of a graph is known as the edge metric dimension , and is denoted by edim( G ) .After the seminal paper [10], a significant number of researches on such parameter have appeared.Among them, some of the most recent ones are [3, 12–14, 18]. It is natural to consider comparingthe metric and edge metric dimensions of graphs. However, as first proved in [10], and continuedin [13, 14], both parameters are not in general comparable existing connected graphs G for which edim( G ) < dim( G ) , edim( G ) = dim( G ) or edim( G ) > dim( G ) .In order to combine the unique identification of vertices and of edges, in only one scheme, themixed metric dimension of graphs was introduced in [9]. For a connected graph G , a vertex w ∈ V ( G ) and an edge uv ∈ E ( G ) are distinguished or determined by a vertex x ∈ V ( G ) if d G ( w, x ) = d G ( uv, x ) . A set S ⊂ V ( G ) is called an mixed metric generator for G if and only if for every pairof elements of the graphs (vertices or edges) e, f ∈ E ( G ) ∪ V ( G ) , there exists a vertex of S whichdistinguishes them. The cardinality of a smallest possible mixed metric generator of G is knownas the mixed metric dimension of G , and is denoted by mdim( G ) . Some recent studies on mixedmetric dimension of graphs are [19, 20]. Clearly, every mixed metric generator must be a metricgenerator as well as an edge metric generator, and so, mdim( G ) ≥ max { dim( G ) , edim( G ) } ,for any connected graph G . In consequence (and due to the not comparability between dim( G ) and edim( G ) ), several situations relating these three parameters can be found. That is, graphs G with mdim( G ) >> max { dim( G ) , edim( G ) } , mdim( G ) = dim( G ) >> edim( G ) , mdim( G ) =edim( G ) >> dim( G ) , or mdim( G ) = dim( G ) = edim( G ) .The metric dimension of hypercube graphs has attracted the attention of several researchersfrom long ago. For instance, the work of Lindstr¨om [16] is probably one of the oldest ones, and forsome recent ones we suggest the works [7, 17, 22]. Surprisingly, for other related invariants, therehas been comparatively little research on hypercube graphs, although one can find some interesting2ecent results on this topic, such as those that appear in [5, 7]. It is our goal to present some resultson the closed connections that exist among the metric, edge metric and mixed metric dimensionsof hypercube graphs.The d -dimensional hypercube, denoted by Q d , with d ∈ N , is a graph whose vertices arerepresented by d -dimensional binary vectors, i.e. , u = ( u , . . . , u ) ∈ V ( Q d ) where u i ∈ { , } for every i ∈ { , . . . , d } . Two vertices are adjacent in Q d if their vectors differ in exactly onecoordinate. Hypercubes can be also seen as the d times Cartesian product of the graph P , thatis, Q d ∼ = P (cid:3) P (cid:3) · · · (cid:3) P , or recursively, Q d ∼ = Q d − (cid:3) P . The distance between two vertices in Q d represents the total number of coordinates in which their vectors differ. The hypercube Q d isbipartite, and has d vertices and d · d − edges. We remark that for instance, Q is the cycle C ,and that Q can be also seen as the torus graphs C (cid:3) C . Our first contribution is to relate the metric generators with the edge metric generators of bipartitegraphs.
Lemma 1.
Let G be a connected bipartite graph. Then, every metric generator for G is also anedge metric generator.Proof. Let S be an arbitrary metric generator for G . We will show that S is an edge metricgenerator as well.Let e = x y and e = x y be two arbitrary distinct edges of G . It is easy to choose u ∈ { x , y } and v ∈ { x , y } such that u and v are distinct and on even distance. In case e and e share a vertex, say y = y , then let u = x and v = x ; as these vertices are on distance 2, therequirements are satisfied. And, in case e and e have no common vertex, then simply let u = x and observe that precisely one of x and y is on even distance from u as G is a bipartite graph and y and y are adjacent. Without lost of generality assume d ( y , u ) is even, and finally set v = y .Now, as u and v are distinct, there must be a vertex s ∈ S that distinguishes them, i.e. d ( s, u ) = d ( s, v ) . We may assume that d ( s, u ) + 1 ≤ d ( s, v ) . Since u and v are on even distance, it followsthat distances d ( s, u ) and d ( s, v ) are of same parity, otherwise we encounter a closed walk of oddlength in G , which is not possible in a bipartite graph. This implies d ( s, u ) + 2 ≤ d ( s, v ) , and nowwe easily derive d ( e , s ) ≤ d ( u, s ) < d ( v, s ) − ≤ d ( e , s ) . In particular, d ( e , s ) < d ( e , s ) implies that e , e are distinguished by s ∈ S . Since the choice ofthese two edges was arbitrary, we conclude that S is also an edge metric generator.It is then natural to think in the opposite direction with regard to the result above. In particular,we ask if an edge metric generator for a bipartite graph is also a metric generator. In contrast withthe result above, achieving this seems to be a challenging task. However, we have at least managedto show a weaker result for an infinite family of bipartite graphs, namely the hypercubes Q d . That3s, when d is odd, every edge metric generator for Q d is indeed a metric generator, and when d iseven, every edge metric generator is “almost” a metric generator.From now on we denote by α i the vector of dimension d whose i th -coordinate is 1, and theremaining coordinates are 0. Also, by “ ⊕ ” we represent the standard (binary) XOR operation.Notice that, for any vertex u ∈ V ( Q d ) , u ⊕ α i means switching the i th -coordinate of u from 0 to1, or vice versa. Lemma 2.
Let S be an edge metric generator of Q d . If u and v are two vertices not distinguishedby S , then they must be antipodal in Q d and d is even. In particular, S is a metric generator of Q d for d being odd.Proof. Suppose that u = ( u , u , . . . , u d ) and v = ( v , v , . . . , v d ) are not antipodal. Then, u i = v i for some i . Let Q d − and Q d − be the half-cubes regarding the dimension i . Notice that u and v belongs to a same half-cube, say Q d − . Let e u and e v be the edges corresponding to the component i (in Q d ) incident with u and v , respectively. In other words, as u ⊕ α i and v ⊕ α i are the neighboursof u and v in Q d − , we have e u = ( u, u ⊕ α i ) and e v = ( v, v ⊕ α i ) . We claim that the edges e u and e v are not distinguished by S . To see this, observe that if s ∈ S belongs to Q d − , then d ( s, e u ) = d ( s, u ) = d ( s, v ) = d ( s, e v ) . Also, if s ∈ S belongs to Q d − , then d ( s, e u ) = d ( s, u ⊕ α i ) = d ( s, v ⊕ α i ) = d ( s, e v ) . We hence derive that edges e u and e v are not distinguished by S , which is a contradiction.Based on the above arguments, we conclude that u and v are antipodal, i.e. d ( u, v ) = d .Hence, every vertex x of S satisfies d ( u, x ) + d ( x, v ) = d . As every vertex s ∈ S must be equallydistanced from u and v , we conclude that d ( u, s ) = d ( s, v ) = d/ , and consequently, d must beeven. This establishes the main claim.Finally, observe that if d is odd, then no vertex is equally distanced from two antipodal verticesof Q d , and therefore, S is a metric generator of Q d .Next lemma will ensure that enlarging an edge metric generator of Q d with one chosen ele-ment, we get a metric generator of Q d . Lemma 3.
Let S be an edge metric generator of Q d and let s be an arbitrary element of S . Then, S ∪ { s ⊕ α } is a metric generator of Q d .Proof. If S is a metric generator of Q d , then S ∪ { s ⊕ α } is so too, and we are done. Thus, weassume that S is not a metric generator of Q d . Then, by Lemma 2, d is even and there must existantipodal vertices u and v such that d ( u, x ) = d ( v, x ) = d/ for every x ∈ S . This will not be acase for s ⊕ α , as | d ( u, s ⊕ α ) − d ( v, s ⊕ α ) | = 2 . Therefore, we conclude that S ∪ { s ⊕ α } isa metric generator of Q d .Since Q d is a bipartite graph, the two previous lemmas give us the following consequence.4 heorem 4. For every d ≥ it holds edim( Q d ) ≤ dim( Q d ) ≤ edim( Q d ) + 1 , with the second inequality being tight only if d is even.Proof. The lower bound holds by Lemma 1. The upper bound and its possible tightness (for morethan one case) follows by Lemmas 2 and 3.Notice that the upper bound dim( Q d ) ≤ edim( Q d ) + 1 is indeed tight for the case Q , since Q ) = edim( Q ) + 1 , as proved in [10].We now turn our attention to relating the metric dimension with the mixed metric dimensionof hypercubes. To this end, we will need the following two results. We must remark that the firstof next two lemmas already appeared in [17]. We include its proof here by completeness of ourpaper. Lemma 5. If S is a resolving set (metric basis) of Q d and s ∈ S , then ( S \ { s } ) ∪ { s ′ } is also aresolving set (metric basis) of Q d , where s ′ ∈ V ( Q d ) is the antipodal vertex of s .Proof. If s ∈ S distinguishes some pair of vertices x and y of Q d , then s ′ distinguishes such pairas well, since d ( x, s ′ ) = d − d ( x, s ) and d ( y, s ′ ) = d − d ( y, s ) . This also means that no metricbasis of Q d contains two antipodal vertices. Thus, if S is a metric generator (or a metric basis) of Q d , then S \ { s } ∪ { s ′ } is so too. Lemma 6. If S is a metric generator of Q d , then there is at most one index i ∈ { , . . . , d } suchthat all the vertices from S have the same value at the i th coordinate.Proof. Suppose that there exist two different indices i and j such that all vertices from S have thesame value at the i th and j th coordinates. In the proof, we proceed with the case when there arezeroes at such coordinates. The other cases can be shown by using similar arguments. Now, let x ∈ V ( Q d ) be a vertex having zeroes at all coordinates, except at the i th , and let y be a vertexhaving zeroes at all positions except at the j th . Then, d ( x, s ) = d ( y, s ) for any vertex s ∈ S , acontradiction.The mixed metric dimension of hypercubes Q and Q are and , respectively. This can bederived from results for paths and cycles from [9]. This gives us that dim( Q d ) < mdim( Q d ) , for d ∈ { , } . For all higher dimensions the mixed metric dimension is equal to the metric dimensionas we next show. Theorem 7.
For every d ≥ it holds, dim( Q d ) = mdim( Q d ) . roof. First, { (1 , , , (0 , , , (0 , , } and { (1 , , , , (0 , , , , (0 , , , , (0 , , , } aremixed metric bases for Q and Q , respectively. Thus, the equality follows for these cases since dim( Q ) = 3 and dim( Q ) = 4 . It remains to check the equality for d ≥ .Let S be a metric basis for Q d with d ≥ . By Lemma 1, S is an edge metric generator of Q d .In this sense, we only need to distinguish in Q d , those pairs of elements, one of them being a vertexand the other one, an edge. For this, let u be an arbitrary vertex and let e = xy be an arbitrary edgeof Q d .Suppose first that u is not a vertex of e . As d ( u, x ) and d ( u, y ) are of different parity, we mayassume that u and x are on even distance. Now, let s i be a vertex from S that distinguishes u and x . Similarly, as in Lemma 1, notice that d ( s i , u ) and d ( s i , x ) are of the same parity, and as they aredifferent, we have that | d ( s i , u ) − d ( s i , x ) | ≥ . So, if d ( s i , u ) < d ( s i , x ) , then we derive d ( s i , u ) < d ( s i , u ) + 1 ≤ d ( s i , x ) − ≤ d ( s i , e ) , and if d ( s i , x ) < d ( s i , u ) , then we have d ( s i , e ) ≤ d ( s i , x ) < d ( s i , u ) . Thus, in both cases e and u are distinguished by a vertex from S .So all the pairs of elements (vertices and edges) considered in the upper part are distinguishedby an arbitrary metric basis. To conclude the proof, we need to construct a metric basis of cardi-nality | S | that will also distinguish incident vertices and edges.Suppose now that u is an endpoint of e , say u = x . To distinguish u and e there needs to be avertex s ∈ S which is from the half-cube Q d − that contains vertex y and does not contain vertex x . To distinguish all such pairs there must be at least one vertex from the mixed metric generatorin every half-cube Q d − . For any index i ∈ { , . . . , d } , there exists a vertex from a mixed metricgenerator having on the i th coordinate, and a vertex from a mixed metric generator having onthe i th coordinate. In other words, a mixed metric basis does not have a column of zeroes or acolumn of ones at an arbitrary index i (if we arrange all vectors of the mixed metric basis as amatrix with such vectors as the rows of such matrix).We have started from an arbitrary metric basis S . Since Q d is a vertex transitive graph, wemay assume that the vertex s = (0 , , . . . , (all coordinates equal to 0) is in S . If S does notcontain a column of zeroes, then S is also a mixed metric basis. Otherwise, by Lemma 6, thereexists only one such column, say at index i . By Lemma 5, we know that we can replace any of thevertices from the set S with its antipodal vertex and the incurred set S ′ = S \ { s } ∪ { s ′ } is a metricbasis too, since the column at index i (all zeroes) ensures that no two vertices in S are antipodalto each other. Moreover, in view of Lemma 1, S is an edge metric generator as well.There exist at least four different vertices s = (0 , , . . . , , s , s and s in the set S , since dim( Q d ) ≥ , for d ≥ . We construct four sets S ′ i in the following way: S ′ = ( S \ { s } ) ∪ { s ′ } , S ′ = ( S \ { s , s } ) ∪ { s ′ , s ′ } ,S ′ = ( S \ { s } ) ∪ { s ′ } , S ′ = ( S \ { s , s } ) ∪ { s ′ , s ′ } , (I): If S ′ is not a mixed metric generator, then there is a column of ones in S ′ at some index i . (II): If S ′ is not a mixed metric generator, then there is a column of zeroes in S ′ at some index i . (III): If S ′ is not a mixed metric generator, then there is a column of zeroes in S ′ at some index i . (IV): If S ′ is not a mixed metric generator, then there is a column of ones in S ′ at some index i .Observe that all these indices i , i , i , i , and i are different. If none of the four sets S ′ i defined above is a mixed metric generator, then the set initial S looks as follows. i i i i i . . . s : s : s : s : s | S | : i , i , i and i . Let v be a vertex having zeroes at allpositions except at i and i and let v be a vertex having zeroes at all positions except at i and i .Then, d ( v , s ) = d ( v , s ) , for any vertex s ∈ S , a contradiction. Therefore, at least one of the sets S ′ i has to be a mixed metric generator, and therefore, the equality mdim( Q d ) = dim( Q d ) followssince any mixed metric basis is also a metric basis.Theorems 4 and 7 give us the following consequences, by using also the asymptotical resultfor the metric dimension of hypercubes known from [2]. Corollary 8.
For every d ≥ it holds dim( Q d ) − ≤ edim( Q d ) ≤ dim( Q d ) = mdim( Q d ) . Corollary 9.
For every d ≥ it holds mdim( Q d ) ∼ edim( Q d ) ∼ dim( Q d ) ∼ d log d . We conclude this little paper with the following conjecture.
Conjecture 1. If d is large enough, then edim( Q d ) = dim( Q d ) . As the above conjecture does not hold for d = 4 , d must be at least 5.7 cknowledgements A. Kelenc has been partially supported by the Slovenian Research Agency by the projects No. J1-1693 and J1-9109. R. ˇSkrekovski acknowledges the Slovenian research agency ARRS, programno. P1–0383 and project no. J1-1692. Ismael G. Yero has been partially supported by the SpanishMinistry of Science and Innovation through the grant PID2019-105824GB-I00.
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