THE DIRECTED DISTANCE DIMENSION OF ORIENTED GRAPHS ...

the (directed distance) dimension dim(D) ofD. Of course, not every oriented graphhas a dimension. An oriented graph of dimension k is also called k-dimensional.To determine whether an ordered set W = {w 1 ,w 2 ,...,w k } of vertices in anoriented graph D is a resolving set, we need only show that the representations of∣the vertices of V (D) − W are distinct since r(w i W ) is the only representationwhose ith coordinate is 0.The directed distance dimension of an oriented graph is a natural analogue ofthe metric dimension of a graph that wasD¡introduced independently by Harary andMelter [2] and Slater [3], [4]. This concept was also investigated in [1] as a result ofstudying a problem in pharmaceutical chemistry.uvxywFigure 1. An oriented graph D with dimension 2In the oriented graph D of Figure 1, let W 1 = {u, v}. The five representationsof the vertices of D with respect to W 1 are r(u ∣ W 1 )=(0, 2), r(v ∣ W 1 )=(1, 0),r(w ∣ W1 )=(2, 1), r(x ∣ W1 )=(2, 1), and r(y ∣ W1 )=(1, 3). Since x and w havethe same representation, W 1 is not a resolving set for D.The five representations of the vertices of D with respect to W 2 = {u, v, w} arer(u ∣ W2 )=(0, 2, 2), r(v ∣ W2 )=(1, 0, 3), r(w ∣ W2 )=(2, 1, 0),r(x ∣ W2 )=(2, 1, 1), r(y ∣ W2 )=(1, 3, 3)Since these five 3-vectors are distinct, W 2 is a resolving set for D. However, W 2is not a basis for D. To see this, let W 3 = {x, y}. Then r(u ∣ W3 ) = (1, 3),r(v ∣ W3 )=(2, 1), r(w ∣ W3 )=(3, 1), r(x ∣ W3 )=(0, 2), and r(y ∣ W3 )=(2, 0),which are distinct as well. So W 3 is a resolving set for D. Since there is no 1-elementresolving set for D, it follows that W 3 is a basis and dim(D) =2.Now let T be the tournament shown in Figure 2. Table 1 gives all 2-elementchoices for W and shows that for each such choice, there exist two equal 2-vectors,thus showing that dim(T ) 3. However, dim(T )=3since{v 1 ,v 3 ,v 6 } is a basisfor T . Figure 3 shows an oriented graph D containing T as an induced subdigraph.The set W = {x, y} is a basis of D, sodim(D) =2. Hencewehavethepossibly156

unexpected property that the 3-dimensional tournament T is an induced subdigraphof the 2-dimensional oriented graph D.v 1v 7T¢v 2v 6v 5 v 4v 3Figure 2. The tournament Tx£v 1v 7v 2u 1 vv 3 u 26v 5 v 4yFigure 3. The digraph DThere is a fundamental question here—one whose answer is not known to us, butone which deserves further study. What is a necessary and sufficient condition forthe dimension of a digraph D to be defined? Certainly, if D is strong, then dim(D) isdefined. Also, if D is connected and contains a vertex such that D − v is strong, thendim(D) is defined. This last statement follows because if od v>0, then V (D) −{v}is a resolving set; while if id v>0, then V (D) is a resolving set. There are numerousother sufficient conditions for dim(D) to be defined.157

Wequivalent vectors{v 1 ,v 2 } r(v 5 | W )=r(v 7 | W )=(1, 2){v 1 ,v 3 } r(v 6 | W )=r(v 7 | W )=(1, 1){v 1 ,v 4 } r(v 5 | W )=r(v 7 | W )=(1, 2){v 1 ,v 5 } r(v 6 | W )=r(v 7 | W )=(1, 2){v 1 ,v 6 } r(v 2 | W )=r(v 3 | W )=(2, 2){v 1 ,v 7 } r(v 5 | W )=r(v 6 | W )=(1, 1){v 2 ,v 3 } r(v 1 | W )=r(v 6 | W )=(1, 1){v 2 ,v 4 } r(v 5 | W )=r(v 7 | W )=(2, 2){v 2 ,v 5 } r(v 1 | W )=r(v 5 | W )=(1, 2){v 2 ,v 6 } r(v 4 | W )=r(v 5 | W )=(2, 1){v 2 ,v 7 } r(v 4 | W )=r(v 5 | W )=(2, 1){v 3 ,v 4 } r(v 1 | W )=r(v 2 | W )=(1, 1){v 3 ,v 5 } r(v 6 | W )=r(v 7 | W )=(1, 2){v 3 ,v 6 } r(v 1 | W )=r(v 2 | W )=(1, 2){v 3 ,v 7 } r(v 2 | W )=r(v 6 | W )=(1, 1){v 4 ,v 5 } r(v 2 | W )=r(v 3 | W )=(1, 1){v 4 ,v 6 } r(v 1 | W )=r(v 2 | W )=(1, 2){v 4 ,v 7 } r(v 1 | W )=r(v 3 | W )=(1, 2){v 5 ,v 6 } r(v 2 | W )=r(v 3 | W )=(1, 2){v 5 ,v 7 } r(v 2 | W )=r(v 4 | W )=(1, 1){v 6 ,v 7 } r(v 4 | W )=r(v 5 | W )=(1, 1)Table 1.2. 1-dimensional oriented graphsIn this section we characterize those oriented graphs having dimension 1. We alsodescribe some properties of bases for 1-dimensional oriented graphs.Theorem 2.1. Let D be a nontrivial oriented graph of order n. Thendim(D) =1if and only if there exists a vertex v in D such that(i) D contains a hamiltonian path P with terminal vertex v such that id D v =1;and(ii) if the hamiltonian path P in (i) is of the formv n−1 ,v n−2 ,...,v 1 ,v,then, for each pair i, j of integers with 1 i

ÈÖÓÓ. Assume that dim(D) = 1. Let W = {v}, v ∈ V (D), be a basisof D. Then the distance d(u, v) fromu to v is defined for each vertex u in Dand the set {d(u, v); u ∈ V (D)} = {0, 1,...,n− 1}. Thus, we may assume thatV (D) ={v, v 1 ,v 2 ,...,v n−1 } where d(v i ,v)=i (1 i n − 1). Clearly, id v =1. Since d(v n−1 ,v) = n − 1, there exists a hamiltonian path in D, namely P :v n−1 ,v n−2 ,...,v 1 ,v, so (i) holds. Furthermore, if there exists a pair i, j of integers(1 i < j n − 1) such that the arc (v j ,v i ) is in D − E(P ), thenj ≠ i +1 andd(v j ,v)=d(v i+1 ,v) (shown in Figure 4). This contradicts the factthat {d(u, v); u ∈ V (D)} consists of n distinct integers, so (ii) holds.v j vv i v i−1 v i−2 v 1 vi+1¤Figure 4.Conversely, assume that there is a vertex v in D such that (i) and (ii) hold.We show that W = {v} is a resolving set of D. Since d(u, v) is defined for eachu ∈ V (D), it suffices to show that the set {d(v i ,v); 1 i n − 1} consists of n − 1distinct integers. Suppose that this is not the case. Then there exist integers i, j(1 ikand k 2 >k. It follows that at least one of these arcs is in D − E(P ), but thisis a contradiction to (ii).□We now present some facts concerning bases in 1-dimensional oriented graphs.Theorem 2.2. Let D be a digraph of order n with dim(D) =1.Furthermore,let v 1 and v 2 be distinct vertices of D with d(v 1 ,v 2 )=2such that both {v 1 } and159

{v 2 } are bases of D. Ifv is a vertex of D such that (v 1 ,v), (v, v 2 ) ∈ E(D), then{v}is also a basis of D.ÈÖÓÓ. To show that {v} is a basis of D, we show that for each u ∈ V (D), thedistance d(u, v) is defined and the set {d(u, v); u ∈ V (D)} consists of n distinctintegers.First notice that id v = 1, for otherwise there exist distinct vertices x and y of Dsuch that d(x, v) =d(y, v) = 1. Since id v 2 = 1, by Theorem 2.1, we haved(x, v 2 )=d(y, v 2 )=d(x, v)+1=2This contradicts the fact that {v 2 } is a basis of D.Furthermore, suppose that there exist vertices u, w in D such that d(u, v) =d(w, v). Since id v =1,eachu − v path contains the arc (v 1 ,v) as its terminalarc, as does each w − v path, sod(u, v 1 )=d(w, v 1 )=d(u, v) − 1Again, this contradicts the fact that {v 1 } is a basis of D.□We now have an immediate consequence of Theorem 2.2.Corollary 2.3. If D is a 1-dimensional oriented graph of order n 3 such that{v} is a basis of D for every vertex v in D, thenD is a directed cycle.ÈÖÓÓ. Let V (D) ={v 1 ,v 2 ,...,v n }. By Theorem 2.2, id v = 1 for every vertexv of D. Moreover,D contains a hamiltonian path P . We can assume thatNext, we show that D contains the cycleP : v n ,v n−1 ,...,v 2 ,v 1C n : v n ,v n−1 ,...,v 2 ,v 1 ,v nSince id v n = 1, there exists a unique vertex v such that (v, v n ) ∈ E(D). If v ≠ v 1 ,then (v i ,v n ) ∈ E(D) forsomei (2 i n − 1). Since {v n } is a basis of D, thereexists a hamiltonian path in D with terminal vertex v n . However, since every vertexhas indegree 1, the only possible path in D with v n as its terminal vertex isP ′ : v n−1 ,v n−2 ,...,v i+1 ,v i ,v nSince P ′ has length n−i, it is not a hamiltonian path. This contradicts the fact that{v n } is a basis. So D contains the cycle C n . Furthermore, since id v =1,D cannotcontain any arc except those in C n .SoD = C n .□160

We can improve Corollary 2.3 slightly.Corollary 2.4. If D is a 1-dimensional oriented graph of order n 3 such that|{v ; {v} is a basis of D}| n − 1then D is a directed cycle.ÈÖÓÓ. Let V (D) ={v, v 1 ,v 2 ,...,v n−1 }. Without loss of generality, we assumethat {v i } is a basis of D for 1 i n − 1. By Corollary 2.3, it suffices to show that{v} is a basis as well.We claim that od v>0. Suppose that this is not the case. Then for each vertexu (≠ v), the distance d(v, u) is not defined, which contradicts the fact that {u} isa basis of D. Hence, there is a vertex x (≠ v) such that (v, x) ∈ E(D). Since {x}is also a basis of D, then by Theorem 2.1(i), D contains a hamiltonian path withterminal vertex x and id x = 1. This implies that there exists a vertex y distinct fromx and v such that (y, v) ∈ E(D). It follows that d(y, x) =2andbyTheorem2.2,{v} is also a basis of D.□The bound in Corollary 2.4 cannot be improved in general. For example, considerthe oriented graph D of order n in Figure 6. Since {v i } is a basis for D for1 i n − 2, dim(D) = 1. However, neither {v n−1 } nor {v n } is a basis D. So|{v ; {v} is a basis of D}| = n − 2andD is not a directed cycle.¦v 3 v 4v 2v 1v nv n−1v n−2Figure 6. An oriented graph with (n − 2) 1-element basesThere is only one 1-dimensional oriented tree of every order.Theorem 2.5. For every oriented tree T , dim(T )=1or dim(T ) is undefined.Furthermore, if dim(T )=1,thenT is a directed hamiltonian path.161

ÈÖÓÓ. There are certainly oriented trees whose dimension is undefined, forexample, any orientation of a star K 1,t ,wheret 3. Now let T be an oriented treewhose dimension is defined. Since T contains no cycles, for every pair x, y of vertices,whenever d(x, y) is defined, d(y, x) is undefined. Thus dim(T )=1.If dim(T ) = 1, then, by Theorem 2.1, T contains a hamiltonian path P and soT = P .□3. On oriented graphs with large dimensionWe have characterized those oriented graphs with dimension 1. But how large canthe dimension of an oriented graph of order n be? In this section, we describe upperbounds for the dimension of a connected oriented graph in terms of lower bounds forthe outdegrees of its vertices. The outdegree of every vertex in the oriented graphD of Figure 7 is 2, yet dim(D) is undefined. Such examples exist regardless of theoutdegrees.D :§Figure 7. The oriented graph DTheorem 3.1. If D is a connected oriented graph of order n 3 with od v 1for all v ∈ V (D) such that dim(D) is defined, then dim(D) n − 2.ÈÖÓÓ. Let D be an oriented graph satisfying the hypothesis of the theorem.Certainly dim(D) n − 1. Assume, to the contrary, that dim(D) =n − 1. LetW = {v 1 ,v 2 ,...,v n−1 } be a basis for D and let V (D) − W = {x}. Since odx 1,assume, without loss of generality, that x is adjacent to v 1 . Also, since odv 1 1,∣we may assume that v 1 is adjacent to v 2 .Sincedim(D) =n − 1, r(v i W −{vi })=r(x ∣ W −{v i })for1 i n − 1. Since x is adjacent to v 1 , it follows that v 2 isadjacent to v 1 , but this contradicts the fact that D is an oriented graph. □We now describe a class of oriented graphs. For k 2, let D k be an orientedgraph with vertex setV (D k )={u, v, w 1 ,w 2 ,...,w k }and let E(D k ) consist of the arc (u, v) andthearcs(v, w j )and(w j ,u)for1 j k.The oriented graph D k is shown in Figure 8. Then D k has order n = k +2 andod v 1 for all v ∈ V (D k ). We claim that dim(D k )=n − 3.162

w kw 2w 1D k :¨uvFigure 8. The oriented graph D k with minimum outdegree 1First we show that dim(D k ) n − 3. Let W = {w 2 ,w 3 ,...,w k },wherethen|W | = n − 3. The distances d(u, w 2 )=2,d(v, w 2 ) = 1, and d(w 1 ,w 2 ) = 3 show thatW is a resolving set for D k and so dim(D k ) n − 3. On the other hand, at leastk − 1 of the vertices w 1 ,w 2 ,...,w k must belong to every resolving set of D k sincethe distance from any two of these vertices to every other vertex of D k is the same.Hence dim(D k ) n − 3 and so dim(D) =n − 3. Of course, this does not show thatsharpness of the bound in Theorem 3.1, except that if D 1 is the directed 3-cycle,then dim(D 1 )=1=n − 2.We can, however, improve the bound in Theorem 3.1 if we require that the outdegreeof every vertex is at least 2.Theorem 3.2. If D is a connected oriented graph of order n 5 with od v 2for all v ∈ V (D) such that dim(D) is defined, then dim(D) n − 3.ÈÖÓÓ. Suppose, to the contrary, that D contains a basis B of cardinality n − 2.Let B = {v 1 ,v 2 ,...,v n−2 },andV (D) −B = {x, y}. For each i (1 i n − 2),B−{v i } is not a resolving set. Hence for each such i, some two of the three verticesx, y, v i have the same representations with respect to B−{v i }. We consider twocases.Case 1: For some i (1 i n − 2), x and y have the same representations withrespect to B−{v i }. Assume, without loss of generality, that x and y have the samerepresentations with respect to W = B−{v n−2 }. Then x and y have the sameout-neighbors in W .Sincex and y have distinct representations with respect to B,exactly one of x and y is adjacent to v n−2 ; for if neither x nor y is adjacent to v n−2 ,then d(x, v n−2 )=d(y, v n−2 ). Therefore, we may assume that y is adjacent to v n−2 .Let W ′ = {v 1 ,v 2 ,...,v n−4 ,v n−2 }. Two of x, y, andv n−3 have the same representationswith respect to W ′ . However, y is adjacent to v n−2 and x is not, so xand y do not have the same representations with respect to W ′ .Thustherearetwopossibilities.163

Subcase 1.1: r(x | W ′ )=r(v n−3 | W ′ ). We claim that x isadjacenttoatmostoneof v 1 ,v 2 ,...,v n−2 . Suppose that this is not the case. Then we can assume withoutloss of generality that x is adjacent to v 1 and v 2 .Thenr(v 1 |B−{v 1 })=r(x |B−{v 1 })orr(v 1 |B−{v 1 })=r(y |B−{v 1 }). Similarly, r(v 2 |B−{v 2 })=r(x |B−{v 2 })or r(v 2 |B−{v 2 })=r(y |B−{v 2 }). Since the out-neighbors of y in W are the sameas the out-neighbors of x in W ,wehavethatv 2 is an out-neighbor of v 1 and that v 1 isan out-neighbor of v 2 .SinceD is an oriented graph, this is impossible, so, as claimed,x isadjacenttoatmostoneofv 1 ,v 2 ,...,v n−2 .Now,sinceodx 2, it follows that xis adjacent to y and exactly one vertex from v 1 ,v 2 ,...,v n−2 ,sayv 1 . However, sincefor 1 i n−3, r(v i |B−{v i })=r(x |B−{v i })orr(v i |B−{v i })=r(y |B−{v i }),it follows that v 1 is an out-neighbor of every vertex in the set {x, y, v 2 ,v 3 ,...,v n−3 },so od v 1 1, which contradicts the assumption that every vertex in D has out-degreeat least 2.Subcase 1.2: r(y | W ′ )=r(v n−3 | W ′ ). We first suppose that x is adjacent tosome vertex in W ′ ,sayv 1 . Because of the assumptions in Case 1 and Subcase 1.2,it follows that y and v n−3 are also adjacent to v 1 . However, since for 2 i n − 3,r(v i |B−{v i })=r(x |B−{v i })orr(v i |B−{v i })=r(y |B−{v i }), it follows that v 1is an out-neighbor of every vertex in the set {x, y, v 2 ,v 3 ,...,v n−3 ,v n−2 },soodv 1 =0,which is a contradiction. Therefore, x is not adjacent to any of v 1 ,v 2 ,...,v n−4 ,v n−2 .Thus, since od x 2, it follows that x must be adjacent to both y and v n−3 . Buty is adjacent to v n−3 as well, because x and y have the same representations withrespect to W . Since x is not adjacent to any of v 1 ,v 2 ,...,v n−4 , it follows that yis not adjacent to any of v 1 ,v 2 ,...,v n−4 . Now r(y | W ′ )=r(v n−3 | W ′ ), so itfollows that v n−3 is not adjacent to any of v 1 ,v 2 ,...,v n−4 . All of this implies thatod v n−3 = 1, which is a contradiction.Case 2: For every i (1 i n − 2), x and y have distinct representations withrespect to B−{v i }. We next prove that every vertex of B is an out-neighbor ofx or y but at most one vertex of B is an out-neighbor of both x and y. To provethis, we first show that among the out-neighbors y 1 ,y 2 ,...,y k of y in B, atmostone y i has the same representation as y with respect to B−{y i }. Suppose thatthis is not the case. Then we may assume that r(y 1 |B−{y 1 })=r(y |B−{y 1 })and that r(y 2 |B−{y 2 })=r(y |B−{y 2 }). The first equality tells us that y 2 isan out-neighbor of y 1 and the second equality tells us that y 1 is an out-neighbor ofy 2 , contradicting the fact that D is an oriented graph. Similarly, among the outneighborsx 1 ,x 2 ,...,x l of x in B, atmostonex j has the same representation as xwith respect to B−{x j }.Next, we show that for each i (1 i n−2), at least one of x and y is adjacent tov i . This follows from the fact that if neither x nor y is adjacent to v i ,thennoother164

vertex v j from B−{v i } can be adjacent to v i since r(v j |B−{v j })=r(x |B−{v j })or r(v j |B−{v j })=r(y |B−{v j }). Thus id v i = 0, which is impossible since d(z,v i )must be defined for all z ∈ V (D). Finally, x and y are simultaneously adjacent toat most one vertex v i (1 i n − 2), for if v a and v b are distinct out-neighbors ofboth x and y, thenv a and v b are out-neighbors of each other, which is impossible.This creates a natural partition of the vertices of B into either two or three subsets,depending on whether there exists a vertex to which x and y are simultaneouslyadjacent. We now consider these two subcases.Subcase 2.1: There exists a unique common out-neighbor of x and y.We assume, without loss of generality, that v n−2 is an out-neighbor of both xand y. Furthermore, we can assume, without loss of generality, that the set X ={v 1 ,v 2 ,...,v k } consists of the out-neighbors of x and not y, and that the set Y ={v k+1 ,v k+2 ,...,v n−3 } consists of the out-neighbors of y and not x. We furtherassume, without loss of generality, that the representations of y and v n−2 with respectto B−{v n−2 } are the same. Therefore, there is no vertex in v j ∈ Y for which therepresentations of y and v j with respect to B−{v j } are the same. Therefore, forevery v j ∈ Y , the representations of x and v j with respect to B−{v j } are the same.Since x is adjacent to every vertex in X, every vertex in Y is adjacent to everyvertex in X ∪{v n−2 }. Now, there is at most one v i ∈ X for which the representationsof x and v i are the same with respect to B−{v i }. Therefore, if |X| 2, there existsat least one vertex v i ∈ X for which the representations of y and v i with respect toB−{v i } are the same. Hence, such a vertex v i is adjacent to every vertex in Y , butthis implies that D is not an oriented graph since for any v j ∈ Y , there is an arcfrom v i to v j and an arc from v j to v i . Therefore, |X| 1. But if |X| =1,thenv 1is the only vertex that could possibly be an out-neighbor of v n−2 . This contradictsthe assumption that the out-degree of every vertex in D is at least 2, so |X| =0. Wehave already seen that every vertex in Y ∪{x} is adjacent to vertex v n−2 ,soevenif|X| =0,wehavethatodv n−2 = 0, which cannot occur.Subcase 2.2: No vertex is a common out-neighbor of x and y.We assume, without loss of generality, that the set X = {v 1 ,v 2 ,...,v k } consistsof the out-neighbors of x and not y, and that the set Y = {v k+1 ,v k+2 ,...,v n−2 }consists of the out-neighbors of y and not x. Recall that there is at most one v i ∈ Xsuch that the representations of v i and x with respect to B−{v i } are equal and atmost one v j ∈ Y such that the representations of v j and y with respect to B−{v j }are equal. This produces three possibilities to consider.Subcase 2.2.1: For every v i ∈ X and v j ∈ Y , the representations of v i and y withrespect to B−{v i } are the same and the representations of v j and x with respect to165

B−{v j } are the same. Then every vertex in Y is adjacent to every vertex in X,and every vertex in X is adjacent to every vertex in Y . This contradicts the factthat D is an oriented graph as long as X and Y are both nonempty. However, if Xor Y is empty, then od x 1orody 1, respectively, which is a contradiction.Subcase 2.2.2: There is exactly one v i ∈ X for which the representations of v iand x with respect to B−{v i } are equal and there is no v j ∈ Y for which v j and yhave the same representations with respect to B−{v j }. (Note that this subcase issymmetric to the case when there is exactly one v j ∈ Y for which the representationsof v j and y with respect to B−{v j } are equal and for which there is no v i ∈ X suchthat v i and x have the same representations with respect to B−{v i }.) Now everyvertex in Y has the same out-neighbors as x, namely the vertices in the set X. SoifY ≠ ∅, then every vertex in Y is adjacent to every vertex in X. Furthermore, everyvertex in X −{v i } has the same out-neighbors as y. Soif|X| 2, then there is atleast one vertex in X which is adjacent to every vertex in Y . But this produces acontradiction since D is an oriented graph. Note that if Y = ∅, theny is adjacent toat most one vertex, namely x, and this is a contradiction.Assume now that |X| 1(so|Y | 2). If |X| =1,thenv i = v 1 and since everyvertex in Y is adjacent to v i ,thevertexv i is adjacent to no vertex except possiblyy. Hence, od v i 1, which is a contradiction. If X = ∅, thenx has no out-neighborsexcept possibly for y, but this contradicts the assumption that the out-degree of xis at least 2.Subcase 2.2.3: There exists exactly one v i ∈ X for which the representations of v iand x with respect to B−{v i } are the same and exactly one v j ∈ Y for which therepresentations of v j and y with respect to B−{v j } are the same. First, supposethat |X| 2and|Y | 2. Then there exists at least one vertex v ∈ X for whichthe representations of v and y with respect to B−{v} are the same. Therefore, v isadjacent to every vertex in Y . Similarly, there is at least one vertex w ∈ Y for whichthe representations of w and x with respect to B−{w} are the same. Therefore, wis adjacent to every vertex in X. However, since v ∈ X and w ∈ Y , it follows thatv is adjacent to w and w is adjacent to v. This contradicts the fact that D is anoriented graph.Next suppose that |X| =1,that|Y | 2, and that X = {v 1 }. Then the outneighborsof x are y and v 1 . Furthermore, v 1 is an out-neighbor of every vertex inY −{v j }. The only possible out-neighbors of v 1 are y and v j . However, if v i isadjacent to v j ,thenx is adjacent to v j , which contradicts the fact that v j /∈ X.Therefore, od v i 1, contradicting the fact that every vertex in D has out-degree atleast 2. The case where |Y | =1and|X| 2 is similar.□166

The sharpness of the bound in Theorem 3.1 is not illustrated by the digraph D kshown in Figure 8 since the outdegrees of most vertices of D k are 1. We can, however,show that the upper bound in Theorem 3.2 is sharp. Let F k be an oriented graphwith vertex setV (F k )={u 1 ,u 2 ,v 1 ,v 2 ,w 1 ,w 2 ,...,w k }and let E(F k ) consist of (1) the arcs (u i ,v j )for1 i, j 2 and (2) the arcs (v i ,w j )and (w j ,u i )for1 i 2and1 j k. The oriented graph F k isshowninFigure9.Then F k has order n = k + 4 and the property that od v 2 for all v ∈ V (F k ). Weclaim that dim(F k )=n − 3.:©w kw 2w 1F ku 2 v 2u 1 v 1Figure 9. The oriented graph F k with minimum outdegree 2First we show that dim(F k ) n − 3. Let W = {u 1 ,v 1 ,w 2 ,w 3 ,...,w k },wherethen |W | = n − 3. The distances d(u 2 ,w 2 )=2,d(v 2 ,w 2 ) = 1, and d(w 1 ,w 2 )=3show that W is a resolving set for F k and so dim(F k ) n − 3. Next we show thatdim(F k ) n − 3. Let W be a resolving set for F k . Certainly at least k − 1ofthevertices w 1 ,w 2 ,...,w k must belong to W since the distance from any two of thesevertices to every other vertex of F k is the same. Moreover, at least one of u 1 andu 2 must belong to W since the distance from u 1 and u 2 every other vertex of F k isthe same. For the same reason, at least one of v 1 and v 2 must belong to W . Hencedim(F k ) n − 3 and so dim(F k )=n − 3.No additional restriction on the outdegrees of the vertices of an oriented graphyields an improved bound, however. Let r 2 be an integer. In the oriented graphof Figure 8, replace u 1 , u 2 by the r vertices u 1 ,u 2 ,..., u r and v 1 , v 2 by the r verticesv 1 ,v 2 ,...,v r and add the appropriate arcs. The resulting oriented graph H k hasod v r for all v ∈ V (H k ), but dim(H k )=n − 3.167

References[1] G. Chartrand, L. Eroh, M. Johnson, O. R. Oellermann: Resolvability in graphs and themetric dimension of a graph. Preprint.[2] F. Harary, R. A. Melter: On the metric dimension of a graph. Ars Combin. 2 (1976),191–195.[3] P.J. Slater: Leaves of trees. Congress. Numer. 14 (1975), 549–559.[4] P.J. Slater: Dominating and reference sets in graphs. J. Math. Phys. Sci. 22 (1988),445–455.Authors’ addresses: Gary Chartrand, Michael Raines, Ping Zhang, Department ofMathematics and Statistics Western Michigan University Kalamazoo, MI 49008, USA, e-mail: zhang@math-stat.wmich.edu.168