On determining number and metric dimension of graphs

Transcription

1 On determining number and metric dimension of graphs José Cáceres Delia Garijo María L. Puertas Carlos Seara October 24, 27 Abstract This paper initiates a study on the problem of computing the difference between the metric dimension and the determining number of graphs. We provide new proofs and results on the determining number of trees and Cartesian products of graphs and establish some lower bounds on the difference between the two parameters. Introduction Let G be a connected graph. A set of vertices S is a determining set of a graph G if every automorphism of G is uniquely determined by its action on S. The determining number is the smallest size of a determining set. Determining sets of connected graphs were introduced by Boutin in [7], where ways of finding and verifying determining sets are described. The author also gives natural lower bounds on the determining number of some graphs, developing a complete study on Kneser graphs. Concretely, tight bounds for their determining numbers are obtained and all Kneser graphs with determining number 2, 3 or 4 are provided. Recently, Boutin [9] has developed a study on the determining number of Cartesian products of graphs, paying special attention to powers of prime connected graphs. Moreover, she computes the determining number of the hypercube Q n. Independently, Erwin and Harary [5] defined an equivalent set and an equivalent number that they called the fixing set and the fixing number, respectively. They determined the fixing number of a tree, and found necessary and sufficient conditions for a tree to have fixing number. Furthermore, recent papers from Albertson, Boutin, Collins, Klavžar and Departamento de Estadística y Matemática Aplicada, Universidad de Almería, Spain. Supported by projects MEC MTM C2-, PAI P6-FQM-649 and PAI FQM-35. address: jcaceres@ual.es Departamento de Matemática Aplicada I, Universidad de Sevilla, Spain. Supported by projects MEC MTM C2-, PAI P6-FQM-649 and PAI FQM-64. address: dgarijo@us.es Departamento de Estadística y Matemática Aplicada, Universidad de Almería, Spain. Supported by projects MEC MTM C2-, PAI P6-FQM-649 and PAI FQM-35. address: mpuertas@ual.es Departament de Matemàtica Aplicada II, Universitat Politècnica de Catalunya, Spain. Supported by projects MEC MTM and DURSI 25SGR692. address: carlos.seara@upc.edu Graphs in this paper are finite, undirected and simple. The vertex-set and edge-set of a graph G are denoted by V (G) and E(G), respectively. The order of G is the number of its vertices, written as V (G). The distance between vertices v, w V (G) is denoted by d(v, w). For more terminology we follow [29].

2 Zhu [2, 3, 4, 5, 6, 8, 22] show the usefulness of determining sets in finding distinguishing numbers of graphs. We also refer the reader to the works done by Collins and Laison [3], and Gibbons and Laison [6] for more results on determining sets or equivalently fixing sets. Determining sets are frequently used to identify the automorphism group of a graph. Furthermore, they are obtained by using its connection with another well-known parameter of graphs: the metric dimension or location number. A set of vertices S V (G) resolves a graph G, and S is a resolving set of G, if every vertex is uniquely determined by its vector of distances to the vertices of S. A resolving set S of minimum cardinality is a metric basis, and S is the metric dimension of G. Resolving sets in graphs were first defined by Slater [28], and Harary and Melter [8]. They have since been widely studied, arising in several areas including coin weighing problems, network discovery and verification, robot navigation, connected joins in graphs, and strategies for Mastermind game. The works developed by Cáceres et al. [] and Hernando et al. [9] provide recent results and an extensive bibliography on this topic. Besides the above-mentioned papers, we want to remark some results obtained by Khuller et al. in [2] and Chartrand et al. in [2], some of which will be of interest throughout this paper. In [2], a formula and a linear time algorithm for computing the metric dimension of a tree is provided. They also obtain the metric dimension of the d-dimensional grid, and show that to compute the metric dimension of an arbitrary graph is a NP-hard problem. On the other hand, Chartrand et al. in [2] characterize the graphs with metric dimension, n, and n 2. They also provide a new proof for the metric dimension of trees and unicycle graphs. See also [25] for tight bounds on the metric dimension of unicycle graphs. Some other important works related to the metric dimension have to do with wheels and Cartesian products. Shanmukha and Sooryanarayana in [27] compute this parameter for wheels, and for graphs constructed by joining, in a certain way, wheels with paths, complete graphs, etc. The metric dimension of Cartesian products of graphs has been studied independently by Peters-Fransen and Oellermann [24] and by Cáceres et al. []. We come now to our main question: Can the difference between the determining number and the metric dimension of a graph of order n be arbitrarily large? This question turns out to be interesting since an automorphism preserves distances and resolving sets are determining sets (see for instance [7, 5]). It arises first as an open problem in [7], and its answer has led us to a number of results on the determining number of some families of graphs in which the metric dimension is known. A brief plan of the paper is the following. Section 2 recalls some definitions and basic tools. In Section 3, we study the determining number of trees, providing a linear time algorithm for computing minimum determining sets and metric basis formed by leaves. We also show that there exist trees for which the difference between the determining number and the metric dimension, considered as a function on the number of vertices, is arbitrarily large. Section 4 focuses on computing the determining number of Cartesian products of graphs, also evaluating the difference between the two parameters. In Section 5, we provide the family of graphs which attains, until now, the best lower bound on the difference between the metric dimension and the determining number. Finally, Section 6 summarizes the main results obtained in this paper, and establishes some open problems. 2

3 2 Definitions and tools An automorphism of G, f : G G, is a bijective mapping on V (G) such that f(u)f(v) E(G) uv E(G). As usual Aut(G) denotes the automorphism group of G. The ideas of determining set and resolving set have already been introduced in the previous section. The following are the precise and more technical definitions provided in [7] and [8, 28] (see also [2, 2, 25]). Definition. [7] A subset S V (G) is said to be a determining set of G if whenever g, h Aut(G) so that g(s) = h(s) for all s S, then g(v) = h(v) for all v V (G). The smallest size of a determining set of G, denoted by Det(G), is called the determining number of G. An equivalent definition of determining set is provided by Boutin in [7] by using the concept of pointwise stabilizer of S as follows. For any S V (G), Stab(S) = {g Aut(G) g(v) = v, v S} = v S Stab(v). Proposition. [7] S V (G) is a determining set of G if and only if Stab(S) = {id}. We also recall from [7] some simple examples of graphs in which the determining number has been computed. A leaf is a minimum determining set of a path P n and so Det(P n ) =. Any pair of non-antipodal vertices is a determining set of a cycle, thus Det(C n ) = 2. A minimum determining set of the complete graph K n is any set containing all but one vertex, and hence Det(K n ) = n. The determining number of the multipartite complete graph K n,n 2,...,n s can be also easily computed, Det(K n,n 2,...,n s ) = (n + n n s ) s since a minimum determining set contains n j vertices of each of the s classes. Observe that every graph has a determining set. It suffices to consider any set containing all but one vertex. Thus, Det(G) V (G) and both bounds are tight, since as it was said before Det(P n ) = and Det(K n ) = n. At some points of this paper, we will use the terminology followed by Erwing and Harary in [5]. Let S be the trivial group in Aut(G). For every v V (G), the set {φ(v) : φ Aut(G)} is the orbit of v under Aut(G). An automorphism φ fixes a vertex v if φ(v) = v. The set of automorphisms that fix v is a subgroup of Aut(G) called the stabilizer of v, and denoted by Aut(G) v (G). An automorphism φ fixes the set S V (G) if for every v S, φ(v) = v. If S is a set of vertices for which Aut(G) S (G) = S then S is a fixing set of G. The minimum cardinality of a fixing set is the fixing number of G. In fact, the fixing number was introduced by Harary in [7], and clearly fixing sets and determining sets are equivalent concepts. The metric dimension or location number is formally defined as follows. Definition 2. [8, 28] A set of vertices S resolves a graph G if every vertex of G is uniquely determined by its vector of distances to the vertices in S, that is, d(u, s) d(v, s) for all s S and u, v V (G) with u v. The metric dimension of G, denoted by β(g), is the minimum cardinality of a resolving set of G. 3

4 The following result was independently proved by Boutin [7], and Erwing and Harary (Lemma 2 in [5] using fixing sets instead of determining sets). Proposition 2. [7, 5] If S V (G) is a resolving set of G then S is a determining set of G. In particular, Det(G) β(g). Given a graph G of order n, the set of all its vertices but one is both a resolving set and a determining set. Moreover, every graph G has both a minimum resolving set and a minimum determining set. Thus, β(p n ) = Det(P n ) = Det(G) β(g) β(k n ) = Det(K n ) = n. There are many examples where both parameters are equal. For any graph G of order n, it is clear that β(g) Det(G) n 2. In order to determine whether β(g) Det(G) can be arbitrarily large, we first compute the determining number of some specific families of graphs in which the metric dimension is known. More concretely, the two following sections are devoted to study these two parameters, and the difference between them for trees and for Cartesian products of graphs. Throughout this paper, β(g) Det(G) will be considered as a function on the order of G, denoted by n = V (G). 3 Trees In this section, we focus on computing determining sets and metric basis of trees, comparing the determining number and the metric dimension, and providing bounds on the difference between the two parameters. A vertex v of a graph G is a determining vertex if there exists a minimum determining set S V (G) so that v S. It is clear that any minimum determining set of G only contains determining vertices. Thus, throughout this section we will distinguish between determining vertices and non-determining vertices in the sense of being necessary to determine the action of an automorphism of G. Let T be a n-vertex tree, n 2. Assume, unless otherwise stated, that T is not the path P n. Clearly, Det(P n ) = and most of the results proved in this section are trivial in that case. The results that follow use the following definitions and notations recalled from [2]. Let v be a fixed vertex of V (T ). A partition of E(T ) into equivalence classes is obtained by the following equivalence relation: e v e 2 if and only if there is a path in T containing e and e 2 that does not have v as an internal vertex. The subgraphs induced by the edges of the equivalence classes of E(T ) are called the bridges of T relative to v. The legs at v are the bridges of T relative to v which are paths (see Figure (a)). The number of legs at v is denoted by l v. We next provide a necessary condition for being a determining vertex. Lemma. Let S V (T ) be a minimum determining set of T and S >. For every vertex u S it holds that, { if u is not a leaf, l u = if u is a leaf. 4

5 v l 4 v l 3 u l l 2... (a) T T 2 T k (b) Figure : (a) A tree with four legs at v, (b) a tree formed by k subtrees, which are not legs, hanging from a vertex u with l u =. Proof. Since Det(T ) = S > then we can assume that there exist both determining vertices and non-determining vertices in T. Moreover, T is not a path which implies that l u = for every leaf u V (T ). Consider now a vertex u S which is not a leaf. Suppose on the contrary that l u >. Assume first that all the legs at u have different length. Obviously, no automorphism f Aut(T ) can interchange two of those legs. Thus, u is not a necessary vertex to determine the action of any f on the legs at u, which leads to a contradiction since S is minimum. Suppose now that there exist two legs at u with the same length. We denote the two legs by l = (u, v, v 2,..., v k ) and l 2 = (u, w, w 2,..., w k ) where v i, w i V (T ). Clearly, there exists f Aut(T ) such that f(u) = u and f(v i ) = w i for all i =,..., k. Hence, u is not a determining vertex. Since S is minimum, then u / S and the desired contradiction is again obtained. Note that any set containing all but one leaf is a determining set of T. The following result states that a minimum determining set can also be obtained by only using leaves. This result has been independently established by Erwing and Harary (Theorem 29 in [5] using fixing sets instead of determining sets). Lemma 2. For every tree T, there exists a minimum determining set of T formed by leaves. Proof. Let S be a minimum determining set of T. Suppose that there exists u S whose degree is greater than. We can assume that Det(T ) >, since the case Det(T ) = is obvious. Thus, by Lemma we have that l u =. Observe that this condition implies that there are some subtrees hanging from u, which are not legs, and that there exists exactly one leg at u (see Figure (b)). Hence the vertex u can be replaced by the leaf of the leg, denoted by v. For every f Aut(T ) we have that f(u) = u if and only if f(v) = v. The result holds by following the same procedure for every u S that is not a leaf. Lemma 2 provides the key tool to obtain the following properties of Det(T ). 5

6 Theorem. The determining number of any tree T satisfies the following statements:. Det(T ) n 2. Moreover, both bounds are tight. 2. Det(T ) = if and only if T is either a path, or there exists at most one vertex v V (T ) which has exactly two isomorphic legs. 3. Given n, k N with k n 2 and k n 3, there exists a tree T with n vertices such that Det(T ) = k. 4. There exists a tree T with n vertices so that Det(T ) = n 3 if and only if n = 4. Proof. The four statements are proved one by one.. Obviously, Det(T ) and any subset of V (T ) containing all but one vertex is a determining set of T, that is, Det(T ) n. Furthermore, T contains at most n leaves which implies, by Lemma 2, that the size of a determining set of T is at most n 2. Hence, Det(T ) n 2 and both bounds are tight. A path and a star are two trees with determining numbers and n 2, respectively. 2. ( =) First suppose that T is either a path, or it has at most one vertex with exactly two isomorphic legs. For each vertex v V (T ), compute the number of isomorphic legs at v. By hypothesis this number is either or 2. If v has non-isomorphic legs then these legs do not contain determining vertices. Otherwise, there is a determining vertex at exactly one of the two isomorphic legs at v. Since this last situation occurs at most once, then T has at most one determining vertex. Hence Det(T ) =. (= ) Assume now that Det(T ) = and T is not a path (otherwise the result holds trivially). Clearly, T has at most one determining vertex which implies that there are at most two isomorphic legs at such vertex. Figure 2 illustrates two instances of trees with determining numbers and 2, respectively. v v (a) (b) Figure 2: (a) A tree T with Det(T ) = and two legs at v with the same length, (b) a tree T with Det(T ) = 2, and two vertices which have two legs with the same length. The set of squared vertices is a determining set. 3. Consider now n, k N with k n 2, and k n 3. Figure 3 shows two groups of leaves, denoted by {u,..., u n } and {v,..., v n2 } with n + n 2 = k + 2, hanging from a path P n k 2. Clearly, the set {u,..., u n, v,..., v n2 } is a minimum determining set of that tree. Hence its determining number is n + n 2 = k. 6

7 u u 2 u n P n k 2 { }} { u... v v v 2 v n2 Figure 3: A tree with determining number n + n 2 2 formed by two groups of leaves hanging from the endpoints of a path. 4. ( =) When n = 4, the result holds by considering the path P 4. (= ) Suppose now on the contrary that there exists a tree T with n 4 vertices so that Det(T ) = n 3. Thus, there is a minimum determining set of size n 3. Moreover, Lemma 2 implies that this minimum set is formed by leaves. Hence T has at least n 2 leaves, which leads to the trees illustrated in Figure 4. The contradiction follows from the determining numbers of those graphs. Figure 4(a) shows a tree with determining number n 4, and the star (Figure 4(b)) has determining number n 2. Therefore, the theorem follows. u u 2 v v 2 u u n u 4 u n u v v n n 2 u 2 u 3 (a) (b) Figure 4: The two possible n-vertex trees with at least n 2 leaves. A result related to item 2 in Theorem is given by Erwing and Harary in [5], in terms of orbits, as follows: Let G be a nonidentity graph. Then Det(G) = if and only if G has an orbit of cardinality Aut(G). 3. Algorithmic study In [2] Khuller et al. show a linear time algorithm for computing the metric dimension of a tree T, which consists on computing l v for each vertex. If l v > then (l v ) landmarks are placed on all but one of the leaves associated with the legs at v. Otherwise, they place no landmarks. A metric basis of T is formed by those leaves on which we have placed landmarks. Thus, β(t ) is given by the following formula. Theorem 2. [2] Let T be a tree which is not a path. Then β(t ) = (l v ). v V (T ),l v> Since any metric basis is a determining set, the question is how many of the leaves which belong to the metric basis, are not necessary in order to obtain a minimum determining 7

8 set. Thus, we design an algorithm which computes all the minimum determining sets formed by leaves, and whose original sets are metric basis (also formed by leaves). It also provides the two values Det(T ) and β(t ). Recall first that the center of a graph is the subgraph induced by the vertices of minimum eccentricity. The center of a tree is either one vertex v or one edge v v 2 (see [2, 29]). In the first case, v is the best candidate for being the root of T in order to compute the determining number. Indeed, T can be viewed as a structure formed by the subtrees hanging from v verifying that, the possible isomorphic structures at T are either the subtrees hanging from v, or they are contained in one subtree hanging from v. Procedure: Minimum-Determining-Set-Tree Input: V (T ), E(T ). Output: A minimum determining set of T formed by leaves, and the value Det(T ). Case. Assume first that T has exactly one center v. Let T be rooted by its center and let T,..., T k be the subtrees hanging from v.. To decide whether T i, T j are isomorphic subtrees for i, j k can be done in O(n) time by following the procedure described in []. Figure 5 illustrates the idea of this algorithm. Labels are assigned to the vertices of T, starting at level and working up towards the root. Level 6 (,,2,2,3,3,4) Level 5 3 (2) 4 (4) () () 3 (2) 2 (,3) 2 (,3) Level 4 2 (2,3) 4 (5) () () 2 (2,3) 3 (4) () 3 (4) () Level (,3) (,) (2) () () (,3) (,) (,) () (,) () Level 2 3 () () 2 (,,) 3 () () () () () () Level () () Level Figure 5: Procedure to decide whether the subtrees hanging from v are isomorphic. Assign to all the leaves at level and go to level. At this level, there are tuples of zeros associated with every vertex. We then assign a different integer to every different tuple following a lexicographic order. Inductively, at each level we consider the tuples formed by both, zeros (representing leaves) and the integers obtained in the previous level sorted by non-decreasing order. The end of this procedure is to assign a tuple (i, i 2,..., i k ) to the root. Isomorphic subtrees hanging from v are represented in the tuple as the same integer number repeated as many times as isomorphic subtrees hang from v. 2. Denote by Λ v the set of subtrees hanging from v. A partition of Λ v into equivalence 8

9 classes is obtained by the following equivalence relation: T i T j if and only if T i = Tj. Let Ω Ti be the equivalence class whose representant is T i, and t i = Ω Ti. 3. Consider the representant of each equivalence class described in Step 2, that is, the non-isomorphic subtrees hanging from v, say T,..., T m. Place landmarks on each of them, if necessary, proceeding as follows for i =,..., m. (a) Compute l v for each vertex v V (T i ). (b) Assign a tuple (l v, (j, l ),..., (j p, l p )) to every v V (T i ), where j s is the number of legs at v with length l s. This process is shown in Figure 6. (2,4) (,3) (,3) (,2) (,2) (2,(,),(,3)) (2,(,),(,2)) (,) (2,2) (,) (,2) (,) (3,) (,) (,) (,) Figure 6: Placing landmarks, if necessary, on the non-isomorphic subtrees hanging from v. The squared vertices belong to every minimum determining set of T. (c) Check if l v > for each v V (T i ). If so, then place (j ) + + (j p ) landmarks on the legs at v. Otherwise, no landmark is placed. All the leaves but one, contained in every group of j s legs at v with the same length l s, belong to a minimum determining set of T (see Figure 6). 4. For every representant T i so that t i = Ω Ti > proceed as follows. (a) If s landmarks have been already placed on T i by following the procedure described in step 3, then for each copy of T i place s landmarks reproducing the situation at T i. Thus, the determining number increases s (t i ) units. The subtree T 4 in Figure 7 shows an instance of this case. (b) If we have placed no landmarks on T i in step 3, then place exactly one landmark on all the isomorphic copies but one of such subtree hanging from v, that is, t i landmarks. The subtrees T and T 3 in Figure 7 illustrate this case. 5. For every representant T i such that t i = Ω Ti =, we have to place no more landmarks than those fixed (if any) in step 3. The subtree T 2 in Figure 7 is an example of this case. Case 2. Suppose now that T has two centers v and v 2 which are adjacent vertices. This case can be viewed as the previous one proceeding as follows. 9

10 T T 2 T 3 T 3 T T 4 T 4 Figure 7: The set of 6 squared vertices of the tree form a minimum determining set.. Add a vertex v adjacent to both v and v 2. Let T be rooted by v. 2. Analyze each of the two subtrees hanging from v, denoted by T and T 2, following the procedure described in case. 3. If there are no landmarks on both, T and T 2, then place one landmark on a leaf of one of them and Det(T ) =. Otherwise, there are s and s 2 landmarks on T and T 2, respectively (including the cases s = or s 2 = ). Hence, Det(T ) = s + s 2 and a minimum determining set is computed as in the described case. We next show the correctness and complexity of our arguments. Theorem 3. The problem of computing a minimum determining set of a tree T, and the determining number Det(T ) can be solved in linear time. Proof. Clearly, the above-described algorithm computes a minimum determining set of T formed by leaves, and therefore its determining number Det(T ). Steps 2, 3, and 4 ensure the minimum cardinality of the obtained determining set, since each of the steps selects the minimum number of landmarks needed to distinguish isomorphic sub-structures contained in the subtrees of T. Moreover, Lemma 2 implies that these landmarks can always be placed on the leaves. The complexity of the procedure is shown as follows. We first have to decide whether T has either one or two centers, and to compute them. A linear time algorithm to compute the centers of a tree is provided in []. We next analyze the complexity time of the process described in case, since the complexity of case 2 is obtained analogously. The algorithm designed in [] enables us to affirm that step can be done in linear time. The complexity of step 2 follows from the O(n) time algorithm provided in [2] to compute the metric dimension of a tree. There is a main difference between both procedures: the length of the legs is also taken into account, in order to place landmarks to obtain the determining number. This extra-condition makes no change in the complexity time. Remark. Observe that the algorithm can be slightly modified maintaining the same O(n)-time complexity in order to output β(t ) and a metric basis formed by leaves. The

11 idea is to consider the same leaves than those fixed for the minimum determining set, plus the (possible) extra leaves which play no role in the sense of distinguishing isomorphic structures but they do to distinguish vectors of distances to the metric basis. 3.2 Lower bounds on β(t ) Det(T ) There exist some examples of trees T for which both parameters β(t ) and Det(T ) are equal: β(p n ) = Det(P n ) =, β(k,n ) = Det(K,n ) = n 2. The following result shows a construction in which β(t ) Det(T ) is Ω( n). Nevertheless, this bound is a first approximation to the possible value n 2. Proposition 3. There exists a tree T of order n for which the difference β(t ) Det(T ) is Ω( n). Proof. Consider the tree T formed by connecting a single vertex u to k paths denoted by P m, P m+,..., P m+k with lengths m, m +,..., m + k, respectively (see Figure 8). Thus P i Pj = {u} for all i j. u v v 2 q t P m P m+ P m+2 v k- P m+k- Figure 8: Tree T formed by connecting a single vertex u to k paths. Since all the paths have different length then Det(T ) =. In fact, Aut(T ) = {id}. Therefore, any vertex is a determining set of T. On the other hand, all the distances are distinguished when we fix at least one vertex in exactly k paths of the k paths of T, say v i P m+i with u v i. Thus, the set S = {v,..., v k } is a resolving set, where the vectors of distances are as follows (see Figure 8).. For every vertex q P m such that d(q, u) = s we have, (d(q, v ), d(q, v 2 ),..., d(q, v j ), d(q, v j+ ),..., d(q, v k )) = = (s +, s +,..., s, s +,..., s + ). 2. For every vertex t P m+j such that d(t, u) = s 2 it follows that, (d(t, v ), d(t, v 2 ),..., d(t, v j ), d(t, v j+ ),..., d(t, v k )) = = (s +, s +,..., s +, s +,..., s + ). 3. The vector of distances (d(u, v ), d(u, v 2 ),..., d(u, v j ), d(u, v j+ ),..., d(u, v k )) is equal to (,,...,,,..., ).

12 Furthermore, S is a minimum resolving set since there is at most one path P j with no landmark. Indeed, suppose on the contrary that there exist two paths P j, P l with no landmarks. Clearly, for every q j P j and q l P l such that d(q j, u) = d(q l, u) = s we have that d(q j, v i ) = d(q l, v i ) = s + which leads to the desired contradiction. Hence, β(t ) = k. When m =, the order of T is given by, V (T ) = n = k2 + k Therefore β(t ) is Ω( n) and so β(t ) Det(T ) is also Ω( n). 4 Cartesian products of graphs This section arises as a natural consequence of the connection between determining number and metric dimension, and the studies developed by Boutin in [9] and Cáceres et al. in []. Our first purpose is to compute the determining number of some well-known Cartesian products of graphs, for which the metric dimension is known. The Cartesian product of graphs G and H, denoted by G H, is the graph with vertex set {(u, v) u V (G), v V (H)} where (u, v ) is adjacent to (u 2, v 2 ) whenever u = u 2 and v v 2 E(H), or v = v 2 and u u 2 E(G). Let S be a subset of V (G H). The projection of S onto G is the set of vertices u V (G) for which there exists a vertex (u, v) S. The projection of S onto H is defined analogously. A column of G H is the set of vertices {(u, v) v V (H)} for some vertex u V (G). A row of G H is the set of vertices {(u, v) u V (G)} for some vertex v V (H). Thus, each column and each row of G H induce a copy of H and G, respectively. Lemma 3. Let G,..., G p be connected finite graphs and G = G G p. For every determining set S of G, the projections of S onto G,..., G p are determining sets of G,..., G p, respectively. In particular, Det(G G p ) max{det(g ),..., Det(G p )}. Proof. Let S V (G) be a determining set of G. By Proposition, we have that Stab(S) = {id}. Consider the projections of S onto G i, written as S Gi, with i p. To prove that S Gi is a determining set of G i it suffices to show that, Stab(S G ) Stab(S Gp ) Stab(S) = {id}. Given f i Aut(G i ), the Cartesian product f f p Aut(G). Assume that f i (u) = u for all u S Gi. Then (f f p )(s) = s for all s S G S Gp = S. Hence f f p = id, and we conclude that Stab(SGi ) = {id} for every i =,..., p. A straightforward consequence of the above-proved result is that, Det(G G p ) max{det(g ),..., Det(G p )} since the determining sets of the projections of G have at most the same order than a minimum determining set of G. 2

13 We now recall some definitions from Boutin [9] and Sabidusi [26]. The unit graph, denoted by U, is the trivial graph given by V (U) = {u} and E(U) =. A graph G is prime with respect to the Cartesian product if it cannot be written as a Cartesian product of two smaller graphs, that is, G is not isomorphic to the unit graph, and if G = Z Z implies Z = U or Z = U. Two graphs G and G are said to be relative primes with respect to the Cartesian product if and only if G = H Z and G = H Z imply Z = U, that is, their prime factor decomposition share no common factor. The following theorem provides the determining number of the Cartesian product of pairwise relative prime graphs. Theorem 4. Let G,..., G p be connected graphs, and m,..., m p N. If G m,..., Gm p p are pairwise relative prime with respect to the Cartesian product, then Det(G m Gmp p ) = max{det(g m ),..., Det(Gmp p )}. Proof. Let S,..., S p be minimum determining sets of G m,..., Gm p p, respectively. Consider a minimum subset of V (G m Gmp p ), denoted by S, verifying that each set S i is the projection of S onto G m i i. Obviously, S = max{ S,..., S p } and hence it suffices to prove that S is a minimum determining set of G m Gmp p. Let f, g Aut(G m Gm p p ). Since G m,..., Gm p p are pairwise relative prime graphs then there exist f i, g i Aut(G m i i ) with i p such that f = f f p and g = g g p (see [26] for more details). Assume that f(s) = g(s) for all s S. Notice that s = (s,..., s p ) with s i S i. Hence, f(s) = g(s) (f (s ),..., f p (s p )) = (g (s ),..., g p (s p )) f i (s i ) = g i (s i ). Since the set S i is a minimum determining set of G m i i, it follows that f i (v) = g i (v) for all v V (G m i i ). Thus, we conclude that f(v) = g(v) for all v V (G m Gm p p ). Observe that every connected finite graph H has an unique prime factor decomposition with respect to the Cartesian product [26]. Thus, H can be written uniquely, up to order, as H = H m Hp mp where the graphs H i are connected, prime, and distinct. Moreover, H m i i is the Cartesian product of m i copies of H i. Since H m H m p p is a maximal decomposition of H into relative prime factors, then Det(H) = max{det(h m ),..., Det(H m p p )}. We want to stress that the above-mentioned consequence of Theorem 4 is equivalent to Theorem in [9], although the arguments of the proofs are different. To prove Theorem 4, we consider the automorphism group of the Cartesian product and some of its properties described by Sabidusi in [26]. On the other hand, the study developed by Boutin in [9] is based on characteristic matrices. Nevertheless, similar ideas appear implicitly in both proofs. We also refer the reader to the work done by Boutin in [9] for results on the determining number of Cartesian powers of prime connected graphs. Proposition 4. [5] Let P t, P m be two paths with t, m 2. It holds that, { 2 if m = t = 2 or 3, Det(P t P m ) = otherwise. 3

14 The following result provides the metric dimension of P t P m. Proposition 5. Let P t, P m be two paths with t, m 2. Then β(p t P m ) = 2. Proof. To prove that β(p t P m ) = 2 we can use that β(g) β(g P m ) β(g) + (see []), β(p m ) =, and so β(g P m ) 2. Obviously, two vertices are necessary and sufficient to obtain a metric basis of these graphs (see Figure 9). P 2 P 3 P 4 P 2 P3 P 5 Figure 9: Metric basis of P t P m. The connection between determining number and metric dimension, and Theorem 4 enable us to compute the determining number of the following Cartesian products. Proposition 6. Let K t, C t, and P t be the complete, cycle, and path graphs, respectively.. For every graph H and t 2β(H) + it holds that Det(K t H) = t. 2. Det(P t C m ) = 2 whenever t 2 and m For every t 2 and m 2 the following holds: 2 if t = m = 2, Det(K t P m ) = if t = 2 and m 2, t if t For every t 2 and m 3 the following holds: 2 if t 3 and m 3, Det(K t C m ) = 3 if t = m = 3, t if t For every t, m 3 we have { 3 if t = m = 3 or t = m = 4, Det(C t C m ) = 2 otherwise. Proof. We proceed by cases:. By Lemma 3 and taking into account that β(k t H) = t if t 2β(H) + (see Theorem 5.3 in []) we have, t = max{det(k t ), Det(H)} Det(K t H) β(k t H) = t therefore Det(K t H) = t. 4

15 2. It is a consequence of Theorem 4, using that Det(P t ) = and Det(C m ) = For t 3, it is a consequence of item, since t 2β(P m ) + = 3 and then Det(K t P m ) = t. For t = 2, we apply Proposition 5 obtaining that Det(P 2 P m ) is equal to 2 if m = 2 and equal to if m It is a straightforward consequence of Theorem 4, using that Det(K t ) = t and Det(C m ) = 2. It only remains to prove the case Det(K 3 C 3 ) which is shown in the next item 5 as Det(C 3 C 3 ). 5. Theorem 4 let us conclude that, Det(C t C m ) = 2 if t m, Det(C m C t ) 2 if t = m. It remains to prove that Det(C m C m ) = 2 whenever m 5, and Det(C m C m ) = 3 if m = 3 or m = 4. We next show that the set S = {(, ), (m 2, 2)} is a minimum determining set of C m C m whenever m 5. Suppose on the contrary that there exists a nontrivial f Aut(C m C m ) such that f(, ) = (, ), and f(m 2, 2) = (m 2, 2). Clearly, f preserves both, adjacency and distance between vertices. Consider the set A of vertices adjacent to (m 2, 2), that is, A = {(m 2, ), (m, 2), (m 2, 3), (m 3, 2)}. It is straightforward to prove that for m 5, the distances between those vertices and (, ) verify: d((, ), (m 2, )) = d((, ), (m, 2)) = 3; d((, ), (m 2, 3)) 5; and 4 d((, ), (m 3, 2)) < d((, ), (m 2, 3)) (see Figure (a)). (3,3) f (3,3) (2,2) (3,2) (4,2) f (2,2) f (3,2) f (3,) (3,) f (4,2) (,) f (,) (a) (b) Figure : (a) The set A of vertices adjacent to (3, 2) in C 5 C 5, (b) the action of a nontrivial automorphism f on A. Since f is nontrivial, then there are at most two vertices in A fixed by f. Thus, we have that f(m 2, ) = (m, 2), f(m, 2) = (m 2, ), f(m 2, 3) = (m 2, 3) and f(m 3, 2) = (m 3, 2) (see Figure (b)). Therefore, the vertices (m 3, 2) and (m, 2) have two adjacent vertices in common which leads to the desired contradiction whenever m 5. Thus, f is trivial and so S is a minimum determining set of C m C m. 5

16 Figure shows that the above reasoning can not be applied for m = 4. It is easy to prove that, for every u, v V (C 4 C 4 ) there always exists a nontrivial automorphism f Aut(C 4 C 4 ) so that f(u) = u, and f(v) = v. In Figure, the vertices (, ) and (2, 2) can play the roles of u and v, respectively. An analogous procedure can be followed to prove that Det(C 3 C 3 ) > 2. (,3) (,3) (2,3) (3,3) f (3,) f (3,3) f (2,3) f (2,) (,2) (,2) (2,2) (3,2) f (3,) f (3,2) f (2,2) f (2,) (,) (,) (2,) (3,) f (,) f (,2) f (,2) f (,) (,) (,) (2,) (3,) f (,) f (,3) f (,3) f (,) Figure : An automorphism f Aut(C 4 C 4 ) which fixes some vertices of C 4 C 4. In particular, f((, )) = (, ) and f((2, 2)) = (2, 2). On the other hand, the set of vertices S = {(, ), (, ), (, )} is a determining set of both C 3 C 3 and C 4 C 4, since every column and every row of the Cartesian product is fixed by the action of any automorphism on S. Thus, we conclude that Det(C m C m ) = 3 whenever m = 3 or m = 4. Therefore, the proposition follows. Notice that Theorem in [9] implies that Det(C m C n ) = 2 whenever m n. The above-proved result includes the case m = n. We next compute the determining number of K t K m. Observe that two vertices of this graph are adjacent if and only if they are in a common row or column. Fixed S V (K t K m ), a row or column is said to be empty if it contains no vertex in S. A lonely vertex v S is defined in [] as the only vertex of S in its row and in its column. Figure 2(a) shows instances of lonely and non-lonely vertices in K 7 K 7. Lemma 4. For t 2, a set S V (K t K t ) is a determining set of K t K t if and only if (a) there is at most one empty row and at most one empty column, and (b) there are at most S 2 lonely vertices. Proof. (= ) Assume that S V (K t K t ) is a determining set of K t K t. By Lemma 3, the projection of S onto K t is a determining set of K t. Since Det(K t ) = t then item (a) holds. Suppose now on the contrary that all the vertices of S are lonely vertices, which implies that S {(x, x) x t }. Thus, there exists a non-trivial automorphism f Aut(K t K t ) so that f(i, j) = (j, i). Therefore, S is not a determining set of K t K t which leads to the desired contradiction. 6

17 K7 K7 (a) (b) Figure 2: (a) The squared vertices form a set S V (K 7 K 7 ) in which there are two non-lonely vertices and one lonely vertex, (b) a minimum determining set of K 7 K 7. ( =) Assume now that S V (K t K t ) is a set verifying (a) and (b). Consider an automorphism f Aut(K t K t ) such that f(u) = u for all u S. Suppose on the contrary that f is non-trivial. Since there is at least one vertex of S in every row but one and in every column but one, then f neither interchanges two rows nor two columns of K t K t. Thus, f is a rotation which sends rows into columns. By item (b), we know that there is one row (or column, or both) in which we have two vertices of S fixed by f. The contradiction follows since such row is fixed by f and it can not be sent into a column. Proposition 7. For every t, m 2 the following holds: { max{t, m } if t m, Det(K t K m ) = t if t = m. Proof. The case t m is a straightforward consequence of Theorem 4, since K t and K m are relative prime graphs. Assume then that t = m 4 (the case K 3 K 3 was shown in item 5 of Proposition 6 as C 3 C 3 ). To prove that Det(K t K t ) = t, it suffices to show that t is the minimum number of vertices needed to satisfy items (a) and (b) of Lemma 4. Clearly, if we consider t vertices of K t K t and at least two of them are non-lonely vertices then there are either two empty rows or two empty columns, what contradicts item (a). Moreover, the set S = {(x, x) x t 2} {(, )} satisfies items (a) and (b) of Lemma 4 (see Figure 2(b)). Thus, Det(K t K t ) = t. Table summarizes the results obtained in this section on the determining number, and the corresponding known-results on the metric dimension of Cartesian products (taken from [, 24]). Notice that the case Det(C 3 C 3 ) = Det(K 3 K 3 ) = Det(K 2 3 ) = 3 matches the formula Det(K k 3 ) = log 3(2k + ) + obtained by Boutin in [9] for k = 2. Remark 2. The graph K t K m is the first instance of Cartesian product of graphs in which the difference between the metric dimension and the determining number is arbitrarily large whenever m t 2m. Indeed, β(k t K m ) Det(K t K m ) is Ω ( ) 2m t 3 where 7

18 Table : Summary of results. G Det(G) β(g) 2 if t = m = 2, 3 P t P m 2 otherwise 2 if m is odd P t C m 2 3 if m is even 3 if t = m = 3, 4 3 if t or m are odd C t C m 2 otherwise 4 if t and m are even K t P m t t 2 if t = 3 and m 3 3 if t = 4 and m is even K t C m 3 if t = m = 3 4 if t = 4 and m is odd t if t 4 t t 5 K t H t t t 2β(H) + max{t, m } if t m K t K m t if t = m 2 3 (t + m ) if m t 2m t if t 2m ( n ) V (K t K m ) = n = t m. In particular, for m = t, β(k t K t ) Det(K t K t ) is Ω 3. Table shows that in the rest of cases both parameters are equal or the difference is at most 2. To conclude this section, we want to point out the determining number and the metric dimension of the hypercube Q n. Recall that Q n is the graph whose vertices are the n- dimensional binary vectors, where two vertices are adjacent if they differ in exactly one coordinate. It is well-known that, Q n = K 2 K 2 K }{{} 2. n The determining number of the hypercube is given by Det(Q n ) = log 2 n + (see [9]). On the other hand, the works done by Erdős and Rényi [4], and Lindström [23] lead to lim β(q n) log n n n = 2. ) Therefore, β(q n ) Det(Q n ) is Ω for large enough values of n. ( 2n log 2 n log n 5 Lower bounds on β(g) Det(G) The studies developed in the two previous sections let us answer the question arose in [7]: Can the difference between the determining number and the metric dimension of a graph of order n be arbitrarily large? In Sections 3 and 4 we have shown that β(t ) Det(T ) = Ω( ( n ) n) for some trees, and β(k t K t ) Det(K t K t ) = Ω 3. The following result improves these lower bounds. Proposition 8. There exists a 2-connected graph G of order n such that ( ) 2n β(g) Det(G) = Ω. 5 8

19 Proof. For n 5, let G = W,n be the wheel formed by connecting a single vertex v to all vertices of an (n )-cycle, denoted by C n. Every automorphism of W,n has to fix v, since its degree is at least 4 and the rest of the vertices have degree 3. Thus, it suffices to consider the action of f Aut(W,n ) on a minimum determining set of C n to determine the action of f on W,n. Hence, a set of two non-antipodal vertices of C n is a minimum determining set of W,n. Thus, the determining number of the wheel graphs is always 2 independently of the number of vertices. On the other hand, Shanmukha and Sooryanarayana in [27] show that β(w,n ) increases with the number of vertices: { 3 + 2k if x = 7 or 8, k N, β(w,x+5k ) = 4 + 2k if x = 9, or. Therefore the difference between the two parameters is: { ( 2k + = Ω 2n ) 5 β(w,x+5k ) Det(W,x+5k ) = 2k + 2 = Ω ( ) 2n 5 where n = V (W,x+5k ) = x + 5k +. if x = 7 or 8, if x = 9, or. We want to stress that we have also computed the metric dimension and the determining number of graphs formed by joining wheels in different ways using the explosion technique, that is, by joining the central vertex of a wheel W,m to any vertex of G as in [27]. As the graph G, it has been used P n, C n and K n. We have also considered the case of adding edges to the wheel graph W,n in different ways trying to increase the metric dimension, and maintaining the determining number to be a constant. Nevertheless, all these families of graphs give rise to at most the same Ω ( ) 2n 5 lower bound. 6 Concluding remarks and open questions We have shown that there exist graphs for which the difference between the metric dimension and the determining number is arbitrarily large, answering the question arose in [7]. In particular, we have proved that the wheel graphs are the family of graphs which attains, until now, the best lower bound for β(g) Det(G). Nevertheless, we have not been able to improve the Ω ( ) 2n 5 lower bound attained by the wheels. Thus, there is a gap between 2n 5 and n 2. The question is then whether we can find a graph G of order n verifying that Det(G) = k, β(g) = n k 2, and β(g) Det(G) = n k, where k, k, and k 2 are constants, closing the gap as much as possible. In otherwise case, can we prove an upper bound for β(g) Det(G) smaller than n 2? References [] A. V. Aho, J. E. Hopcroft, and J. D. Ullman. The design and analysis of computer algorithms. Addison-Wesley series in computer science and information processing, 974. [2] M. O. Albertson. Distinguishing Cartesian products of graphs. Electron. J. Combin., 2: Note 7 (electronic), 25. 9

20 [3] M. O. Albertson and D. L. Boutin. Automorphisms and distinguishing numbers of geometric cliques. Discrete and Computational Geometry, accepted. [4] M. O. Albertson and D. L. Boutin. Distinguishing geometric graphs. J. Graph Theory, 53 (26), [5] M. O. Albertson and D. L. Boutin. Using determining sets to distinguish Kneser graphs. Electron. J. Combin., 4: Research paper 4 (electronic), (27). [6] M. O. Albertson and K. L. Collins. Symmetry breaking in graphs. Electron. J. Combin., 3(): Research paper 8 (electronic), 996. [7] D. L. Boutin. Identifying graphs automorphisms using determining sets. The Electronic Journal of Combinatorics, 3, 26. [8] D. L. Boutin. Small label classes in 2-distinguishing labelings. Ars Mathematica Comtemporanea, submitted, 27. [9] D. L. Boutin. The determining number of a Cartesian product. SIAM Journal on Discrete Mathematics, submitted, 27. [] J. Cáceres, C. Hernando, M. Mora, I. M. Pelayo, M. L. Puertas, C. Seara, and D. R. Wood. On the metric dimension of Cartesian products of graphs. SIAM Journal in Discr. Math., 2, 423, 27. [] K. M. Chao and B. Y. Wu. A note on excentricities, diameters and radii. An excerpt from the book Spanning trees and optimization problems, Chapman and Hall/CRC Press, USA, 24. [2] G. Chartrand, L. Eroh, M. A. Johnson, and O. R. Oellermann. Resolvability in graphs and the metric dimension of a graph. Discrete Appl. Math., 5(-3), pp. 99 3, 2. [3] K. L. Collins and J. D. Laison. Fixing numbers of Kneser graphs. Manuscript. [4] P. Erdős and A. Rényi. On two problems of information theory. Magyar Tud. Akad. Mat. Kutató Int. Közl., 8, pp , 963. [5] D. Erwing and F. Harary. Destroying automorphisms by fixing nodes. Discrete Math., 36, pp , 26. [6] C. R. Gibbons and J. D. Laison. Fixing numbers of graphs and groups. Submitted, 26. [7] F. Harary. Methods of destroying the symmetries of a graph. Bull. Malasyan Math. Sc. Soc., 24(2) (2) pp [8] F. Harary and R. A. Melter. On the metric dimension of a graph. Ars Combinatoria, 2, pp. 9 95, 976. [9] C. Hernando, M. Mora, I. M. Pelayo, C. Seara, and D. R. Wood. Extremal graph theory for metric dimension and diameter. Electronic Notes in Discr. Math., 29 (27), pp

21 [2] C. Jordan. Sur les assemblages de lignes. J. Reine Angew. Math, 7, pp. 85 9, 869. [2] S. Khuller, B. Raghavachari, and A. Rosenfeld. Landmarks in graphs. Discrete Appl. Math., 7(3), pp , 996. [22] S. Klavžar and X. Zhu. Cartesian powers of graphs can be distinguished by two labels. European Journal of Combinatorics, 28 (27) pp [23] B. Lindström. On a combinatory detection problem. I. Magyar Tud. Akad. Mat. Kutató Int. Közl., 9, pp , 964. [24] J. Peters-Fransen and O. R. Oellermann. The metric dimension of Cartesian products of graphs. Util. Math., 69, pp 33 4, 26. [25] C. Poisson and P. Zhang. The metric dimension of unicycle graphs. J. Combin. Math. Combin. Comput. 4 (22) pp [26] G. Sabidussi. Graph multiplication. Math. Zeitschr., 72, pp , 96. [27] B. Shanmukha and B. Sooryanarayana. Metric dimension of wheels. Far East J. Appl. Math., 8(3), pp , 22. [28] P. J. Slater. Leaves of trees. Proc. 6th Southeastern Conf. on Combinatorics, Graph Theory and Computing, vol. 4 Congressus Numerantium, pp , 975. [29] D. B. West. Introduction to Graph Theory. Prentice Hall,