The Partition Dimension of a Graph

c Birkh¨auser Verlag, Basel, 2000 Aequationes Math. 59 (2000) 45–54 0001-9054/00/010045-10 $ 1.50+0.20/0 Aequationes Mathematicae

The partition dimension of a graph

Gary Chartrand, Ebrahim Salehi and Ping Zhang

Summary. For a vertex v of a connected graph G and a subset S of V (G), the distance between v and S is d(v, S) = min d(v, x) x S . For an ordered k-partition Π = S1,S2, ,S of { | ∈ } { ··· k} V (G), the representation of v with respect to Π is the k-vector r(v Π) = (d(v, S1),d(v, S2), | ,d(v, Sk)). The k-partition Π is a resolving partition if the k-vectors r(v Π), v V (G), are··· distinct. The minimum k for which there is a resolving k-partition of V (G)| is the partition∈ dimension pd(G)ofG. It is shown that the partition dimension of a graph G is bounded above by 1 more than its metric dimension. An upper bound for the partition dimension of a bipartite graph G is given in terms of the cardinalities of its partite sets, and it is shown that the bound is attained if and only if G is a complete bipartite graph. Graphs of order n having partition dimension 2, n,orn 1 are characterized. −

Mathematics Subject Classiﬁcation (1991). 05C12.

1. Introduction

For vertices u and v in a connected graph G, the distance d(u, v) is the length of a shortest path between u and v in G. For an ordered set W = w1,w2, ,wk of vertices in a connected graph G and a vertex v of G,thek-vector{ (ordered···k-tuple)}

r(v W )=(d(v, w1),d(v, w2), ,d(v, wk)) | ··· is referred to as the (metric) representation of v with respect to W .ThesetW is called a resolving set for G if the vertices of G have distinct representations. A resolving set containing a minimum number of vertices is called a minimum resolving set or a basis for G. The number of vertices in a basis for G is its (metric) dimension dim(G). For example, the graph G of Figure 1 has the basis W = u, z and so dim(G) = 2. The representations for the vertices of G with respect{ to W} are

r(u W )=(0,1) r(v W )=(2,1) r(x W )=(1,2) | | | r(y W )=(1,1) r(z W )=(1,0). | | Research supported in part by a Western Michigan University Faculty Research and Creative Activities Grant. 46 G. Chartrand, E. Salehi and P. Zhang AEM

x u y G: v z

Figure 1. The representations of the vertices of a graph. This topic is the subject of the papers [1], [4], [5], and [6]. A directed graph version of this topic is explored in [2]. In this paper, we provide representations of the vertices of a connected graph G by other means, namely, through partitions of V (G) and the distances between each vertex of G and the subsets in the partition. Let G be a connected graph. For a subset S of V (G) and a vertex v of G,the distance d(v, S) between v and S is deﬁned as

d(v, S)=min d(v, x) x S . { | ∈ } For an ordered k-partition Π = S1,S2, ,Sk of V (G) and a vertex v of G,the representation of v with respect{ to Π is··· deﬁned} as the k-vector

r(v Π) = (d(v, S1),d(v, S2), ,d(v, Sk)) . | ··· The partition Π is called a a resolving partition if the k-vectors r(v Π),v V(G), are distinct. The minimum k for which there is a resolving k-partition| of∈V (G)is the partition dimension pd(G)ofG. As an illustration of these concepts, again consider the graph G in Figure 1. Let Π1 = S1,S2,S3 ,whereS1= u, x ,S2 = v, y , and S3 = z . Then the ﬁve representations{ (3-vectors)} are { } { } { }

r(u Π1)=(0,1,1),r(vΠ1)=(1,0,1),r(xΠ1)=(0,1,2), | | | r(y Π1)=(1,0,1),r(zΠ1)=(1,1,0). | | Since r(v Π1)=(1,0,1) = r(y Π1), it follows that Π1 is not a resolving partition | | of G. Next let Π2 = S1,S2,S3,S4 ,whereS1= u, x ,S2= v ,S3= y ,and { } { } { } { } S4 = z . Then the ﬁve representations (4-vectors) are { }

r(u Π2)=(0,2,1,1),r(vΠ2)=(1,0,1,1),r(xΠ2)=(0,1,1,2), | | | r(y Π2)=(1,1,0,1),r(zΠ2)=(1,1,1,0). | | Since the ﬁve 4-vectors are distinct, Π2 is a resolving partition of G. However, Π2 is not a minimum resolving partition of G. To see this, let Π3 = S1,S2,S3 ,where { } S1 = x ,S2= u ,and S3 = y,z,v . Then the corresponding representations are { } { } { }

r(u Π3)=(1,0,1),r(vΠ3)=(1,2,0),r(xΠ3)=(0,1,1), | | | r(y Π3)=(1,1,0),r(zΠ3)=(2,1,0). | | Vol. 59 (2000) The partition dimension of a graph 47

So Π3 is a resolving partition of G. Moreover, since no 2-partition is a resolving partition of G, it follows that Π3 is a minimum resolving partition of G and so pd(G) = 3. The partition dimension and metric dimension are related, as we next show.

Theorem 1.1. If G is a nontrivial connected graph, then

pd(G) dim(G)+1. ≤

Proof. Let dim(G)=kand let W = w1,w2, ,wk be a basis of G.Weconsider { ··· } the ordered partition Π= S1,S2, ,Sk+1 of V (G), where Si = wi (1 i k) { ··· } { } ≤ ≤ and Sk+1 = V (G) W .Sincer(vΠ) = (d(v, w1),d(v, w2), ,d(v, wk), 0) for v V (G) W and W is− a resolving set| of G, it follows that the··· representations r(v Π),∈ − | v Sk+1, are distinct. Moreover, only the representation r(wi Π), 1 i k,has ∈ | ≤ ≤ the ith entry 0, which implies that r(v Π) = r(wi Π) for all v V (G) W and all i with 1 i k. Therefore, Π is| a resolving6 | (k + 1)-partition∈ of G− and so pd(G) dim(≤G)+1.≤ ≤

The upper bound in Theorem 1.1 is attainable for the graphs Pn,Cn,Kn, and K1,k. We can say more, however.

b Theorem 1.2. For every pair a, b of positive integers with 2 +1 a b+1, there exists a connected graph G such that pd(G)=aand dim(G)=≤b. ≤

Proof. Let G = Ks,t with s = a and t = b a + 2. It is routine to verify that − b dim(G)=s+t 2=a+(b a+2) 2=b.Since 2 +1 a b+ 1, it follows that s>t. By Theorem− 2.4,−pd(G)=−s=a. ≤ ≤ Theorem 1.2 immediately brings up another question. Is it the case that pd(G) dim(G) + 1 for every nontrivial connected graph G?Itwasnoted ≥ 2 in [3, p. 204]l that determiningm the metric dimension of a graph is an NP-complete problem.

2. Some basic results on the partition dimension of a graph

If G is a connected graph of order n 2, then certainly 2 pd(G) n.Foreach integer n 2, there is only one graph≥ of order n having partition≤ dimension≤ 2. ≥ Proposition 2.1. Let G be a connected graph of order n 2.Thenpd(G)=2if ≥ and only if G = Pn.

Proof. First, let Pn : v1,v2, ,vn and let Π = S1,S2 be the partition of V (Pn) ··· { } 48 G. Chartrand, E. Salehi and P. Zhang AEM with S1 = v1 and S2 = v2,v3, ,vn .Since { } { ··· }

r(v1Π) = (0, 1) and r(vi Π) = (i 1, 0) for 2 i n, | | − ≤ ≤ it follows that Π is a resolving partition of Pn and so pd(Pn)=2. Now we verify the converse. Let Π = S1,S2 be a resolving partition of a { } graph G of order n.SinceGis connected, there exist adjacent vertices u S1 ∈ and v S2. Since the representations r(w Π) = (0,d(w, S2)), w S1,and ∈ | ∈ r(wΠ) = (d(w, S1), 0), w S2, are distinct, u is the unique vertex in S1 that | ∈ is adjacent to a vertex in S2 and v is the unique vertex in S2 that is adjacent to a vertex in S1. We show that S1 and S2 are paths in G.SinceGis connected, if h i h i S1 u = , every vertex in S1 is adjacent to at least one vertex in S1. Moreover, −{ }6 ∅ the vertex u is adjacent to at most one vertex in S1,forifuis adjacent to two vertices u1,u2 S1,thenr(u1Π) = r(u2Π) = (0, 2), contradicting the fact that ∈ | Π is a resolving partition of V (G). So assume that w is the unique vertex of S1 that is adjacent to u. Similarly, w isadjacenttoatmostonevertexinS1 that is distinct from u. Continuing this procedure, we see that S1 is a path in G.By h i the same reasoning, S2 is also a path in G,andsoGitself is a path. h i At the other extreme, for each n 2, there is also exactly one graph of order n with partition dimension n. Before presenting≥ this result, we establish the following lemma.

Lemma 2.2. Let Π be a resolving partition of V (G) and u, v V (G).Ifd(u, w)= d(v, w) for all w V (G) u, v ,thenuand v belong to distinct∈ elements of Π. ∈ −{ }

Proof. Let Π = S1,S2, ,Sk ,whereuand v belong to the same element, say { ··· } Si,ofΠ.Thend(u, Si)=d(v, Si) = 0. Since d(u, w)=d(v, w) for all w ∈ V (G) u, v ,wealsohavethatd(u, Sj)=d(v, Sj) for all j,where1 j=i k. Therefore,−{ r(}u Π) = r(v π) and Π is a not resolving partition. ≤ 6 ≤ | | Proposition 2.3. Let G be a connected graph of order n.Thenpd(G)=nif and only if G = Kn.

Proof. By Lemma 2.2, pd(Kn)=n. For the converse, let G be a graph of order n with pd(G)=n,whereV(G)= v1,v2, ,vn . Assume, to the contrary, { ··· } that G = Kn. Then we may assume that d(v1,vn)=1andd(vn 1,vn)=2. 6 − Let Π = S1,S2, ,Sn 1 be the partition of V (G) with S1 = v1,vn and { ··· − } { } Si = vi for 2 i n 1. For each i (1 i n 1), only the i-th entry of { } ≤ ≤ − ≤ ≤ − r(vi Π) is 0. So the representations r(vi Π), 1 i n 1, are distinct. Since the | | ≤ ≤ − ﬁrst entry of r(vn Π) is 0, r(vn Π) diﬀers from all r(vi Π), where 2 i n 1. | | | ≤ ≤ − Also, since the (n 1)-st entry of r(vn Π) is 2 and the (n 1)-st entry of r(v1 Π) − | − | is 1, it follows that r(vn Π) = r(v1 Π). Thus Π is a resolving partition of G and pd(G) n 1, producing| a contradiction.6 | ≤ − Vol. 59 (2000) The partition dimension of a graph 49

By Propositions 2.1 and 2.3, if G is a connected graph of order n 4thatis neither a path nor a complete graph, then 3 pd(G) n 1. By Proposition≥ 2.1, to verify that a graph G that is not a path≤ has partition≤ − dimension 3, it suffices to establish the existence of a resolving 3-partition of V (G). For example, one can verify that the partition dimension of an n-cycle (n 3) is 3, the partition ≥ dimension of the 3-cube Q3 is 3, and the partition dimension of the Petersen graph is 4. It can be verified that pd(H K2) pd(H) + 1 for every nontrivial × ≤ connected graph H. Hence a simple inductive argument yields that pd(Qn) n+1 for n 2. But establishing the partition dimensions of the more general n≤-cubes (n ≥4) appears to be difficult. For bipartite graphs in general, we have the following.≥ Since the proof is straightforward, we omit it.

Theorem 2.4. Let G be a connected bipartite graph with partite sets V1 and V2 of cardinalities r and s, respectively. Then (1) pd(G) r +1,ifr=s,and (2) pd(G) ≤ max r, s ,ifr=s. Moreover, equality≤ holds{ in} (1) or6 (2), if and only if G is a complete bipartite graph.

3. Graphs with partition dimension n 1 − We have already seen that the complete graphs are the only graphs of order n with partition dimension n. In this section, we determine all graphs of order n with partition dimension n 1. First we provide upper and lower bounds for the partition dimension of a graph− in terms of its order and diameter. For integers n and d with n>d 2, we deﬁne g(n, d) as the least positive integer k for which ≥ (d +1)k n. ≥ Theorem 3.1. If G is a graph of order n ( 3) and diameter d,then ≥ g(n, d) pd(G) n d +1. ≤ ≤ −

Proof. First, we establish the upper bound. Let u and v be vertices of G for which d(u, v)=dand let u = v1,v2, ,vd+1 = v be a u v path of length d.Let ··· − V(G)= v1,v2,...,vd, ,vn . It is straightforward to show that the partition { ··· }

Π= S1,S2, ,Sn d+1 { ··· − } of V (G), where S1 = v1,v2,...,vd and Si = vi+d 1 for 2 i n d +1,is a resolving (n d + 1)-partition,{ verifying} the upper{ − bound.} ≤ ≤ − Next, we consider− the lower bound. Let pd(G)=kand Π be a resolving k- partition of V (G). Since each representation of a vertex of G is a k-vector, every 50 G. Chartrand, E. Salehi and P. Zhang AEM coordinate of which is a nonnegative integer not exceeding d and all n representations are distinct, it follows that (d +1)k n. Hence g(n, d) k = pd(G). ≥ ≤ We show next that the upper bound given in Theorem 3.1 can be attained. Let G be a graph obtained from a path P : v1,v2,v3,v4 by attaching the pendant edge v2x to P (see Figure 2). Then G has order n =5anddiamG=3.Since the partition Π = S1,S2,S3 ,whereS1= v1,v2,v3 ,S2 = v4 and S3 = x , is a resolving 3-partition{ of V}(G), it follows{ that pd(}G)=3={ }n diam G {+1.} Unfortunately, we know of no graphs where the lower bound is attained,− so an improved lower bound may very well exist.

v1 v2 v3 v4

Figure 2. A graph G with pd(G)=n diam G +1. − Since pd(Kn)=n, we have the following consequence of Theorem 3.1.

Corollary 3.2. If G is a graph of order n ( 2) and pd(G)=n 1,then diam G =2. ≥ −

We have already seen by Theorem 2.4 that the star K1,n 1 (n 3) has partition − ≥ dimension n 1. It is not difficult to see that for n 3, the graphs Kn e and − ≥ − K1 +(K1 Kn 1) have partition dimension n 1 as well. We now show that these are the only− graphs of order n with partition− dimension n 1. S − Theorem 3.3. Let G be a connected graph of order n 3.Thenpd(G)=n 1 ≥ − if and only if G is one of the graphs K1,n 1, Kn e, K1 +(K1 Kn 1). − − − Proof. We have already noted that each of the graphs mentioned inS the theorem has partition dimension n 1. For the converse, assume that G is a connected graph of order n 3. By Corollary− 3.2, it follows that the diameter of G is 2. Assume ≥ first that G is bipartite. Since the diameter of G is 2, it follows that G = Kr,s for some integers r and s with n = r + s 3. By Theorem 2.4, G = K1,n 1. HencewemayassumethatGis≥ not bipartite. Let Y be the vertex− set of a maximum clique of G. We show that Y 3. Since G is not bipartite, G | |≥ contains an odd cycle. Let C2`+1 be the smallest odd cycle in G. Since the diameter of G is 2, it follows that C2`+1 is C3 or C5. Suppose first that C2`+1 = C5 : v1,v2,v3,v4,v5,v1.LetΠ= S1,S2, ,Sn 2 ,whereS1 = v1,v2,v3 , { ··· − } { } Vol. 59 (2000) The partition dimension of a graph 51

S2 = v4 , S3 = v5 ,andSi (4 i n 2) contains a single vertex of { } { } ≤ ≤ − V (G) v1,v2,v3,v4,v5 .Sincer(v1Π) = (0, 2, 1, ),r(v2 Π) = (0, 2, 2, ), −{ } | ··· | ··· and r(v3 Π) = (0, 1, 2, ), it follows that Π is a resolving (n 2)-partition of | ··· − V (G), contradicting pd(G)=n 1. Therefore, C2`+1 = C3.SinceGcontains K3 as a subgraph, it follows that Y− 3. Let U = V (G) Y .Since| G|≥is not complete, U 1. Assume ﬁrst that − | |≥ U =1.ThenG=Ks+(K1 Kt) for some integers s and t.Nows 1, since | | ∪ ≥ G is connected and t 1, since G is not complete. Let V (Ks)= u1,u2, ,us , ≥ { ··· } V(Kt)= v1,v2, ,vt ,andV(K1)= w . We consider two cases. { ··· } { } Case 1. s t.LetΠ= S1,S2, ,Ss+1 ,whereSi = ui,vi (1 i t), ≥ { ··· } { } ≤ ≤ Si = ui (t +1 i s), and Ss+1 = w .Sinced(u, w)=1foru V(Ks) { } ≤ ≤ { } ∈ and d(v, w)=2forv V(Kt), it follows that Π is a resolving (s + 1)-partition of V (G). Hence pd(G) ∈s + 1. By Lemma 2.2, pd(G) s. However, pd(G) = s,for ≤ ≥ 6 otherwise s = n 1andG=Kn. Therefore, pd(G)=s+1. Since pd(G)=n 1, it follows that s−= n 2andt= 1. Therefore, − −

G = Kn 2 +(K1 K1)=Kn e. − ∪ −

Case 2. s

x w

Figure 3. 52 G. Chartrand, E. Salehi and P. Zhang AEM

Let Π = S1, S2, , Sn 2 ,whereS1 = u,w,y ,S2 = x ,S3 = v ,and { ··· − } { } { } { } each set Si (4 i n 2) contains exactly one vertex from V (G) u, w, y, x, v . Since r(u Π)=(0≤ , 1≤, 2, − ), r(w Π)=(0, 2, 2, ), and r(y Π)=(0, 1−{, 1, ), it fol-} lows that| Π is a resolving··· (n |2)-partition··· of V (G), contradicting| the··· fact that pd(G)=n 1. Thus this subcase− cannot occur. − Subcase 1.2. Every vertex of Y is adjacent to either both u and w or to neither uorw. If u and w are adjacent to every vertex in Y v , then the vertices of (Y v ) u, w are pairwise adjacent, contradicting−{ the deﬁning} property of Y . Thus,−{ there} ∪{ exists} a vertex y Y such that y is distinct from v,andyis adjacent to neither u nor w. ∈ Since the diameter of G is 2, there is a vertex x of G that is adjacent to both u and v.ThusGcontains the subgraph shown in Figure 4.

Figure 4.

Let Π = S1, S2, , Sn 2 ,whereS1 = x,y,w ,S2 = u ,S3= v ,and { ··· − } { } { } { } each set Si (4 i n 2) contains only one vertex from V (G) u, w, y, x, v . Since r(x Π)=(0≤, 1≤, 1, −), r(y Π)=(0, 2, 1, ), and r(w Π)=(0, −{1, 2, ), it fol-} lows that| Π is a resolving··· (n | 2)-partition··· of V (G), contradicting| the··· fact that pd(G)=n 1. Thus Subcase− 1.2 and, in fact, Case 1 cannot occur. −

Case 2. There exist distinct vertices v and v0 in Y such that uv, wv0 / E(G). ∈ In fact, for each vertex y0 of Y , y0 is adjacent to at least one of u and w, for otherwise, we have the conditions of Case 1. Necessarily, vw, v0u E(G). Since ∈ Y 3, there exists a vertex y in Y distinct from v and v0. Also, at least one of the| |≥ edges yu and yw must be present in G,sayyu.ThusGcontains the subgraph shown in Figure 5. v u

v¢ w y

Figure 5.

Let Π = S1, S2, , Sn 2 ,whereS1 = u, w, y , S2 = v , S3 = v0 ,and { ··· − } { } { } { } each set Si (4 i n 2) contains only one vertex from V (G) u, w, y, v, v0 . Since r(u Π)=(0≤, 2≤, 1, − ), r(w Π)=(0, 1, 2, ), and r(y Π)=(0,−{1, 1, ), it fol-} | ··· | ··· | ··· Vol. 59 (2000) The partition dimension of a graph 53 lows that Π is a resolving (n 2)-partition of V (G), contradicting the fact that pd(G)=n 1. Therefore, U is− an independent set. Next we− claim the N(u)=N(w) for all u, w U. It suﬃces to show that if uv E(G), then uw E(G). Let uv E(G) for some∈ vertex v of G. Necessarily v ∈Y . Assume, to the∈ contrary, that∈wv / E(G). Since Y is the vertex set of a maximum∈ clique, there exists y Y such∈ that uy / E(G). Since G is connected and U is independent, w is adjacent∈ to some vertex of∈Y . We consider two subcases. Subcase 2.1. w is adjacent only to y.Sincewand y are not adjacent to u,it follows that d(w, u) = 3, which contradicts the fact that the diameter of G is 2. Subcase 2.2. There exists a vertex x in Y distinct from y such that wx E(G). Thus G contains the subgraph shown in Figure 6. ∈ y

x v

w u

Figure 6.

Let Π = S1,S2, ,Sn 2 ,whereS1 = u, w, x , S2 = v , S3 = y ,and { ··· − } { } { } { } each set Si (4 i n 2) contains only one vertex of V (G) u, w, x, v, y .Since r(uΠ)=(0, 1, 2≤, ≤), r−(w Π)=(0, 2, ),and r(x Π)=(0, 1, 1,−{), it follows} that Π is a| resolving (n···2)-partition| of V ···(G), contradicting| the fact··· that pd(G)=n 1. Therefore, V −(G)=Y U,where Y is complete, U is independent, Y −3, U 2, and N(u)=N(w∪) for all u, w,h iU. | |≥ | |≥Next we show that for each u U,∈ there exists at most one vertex of Y not contained in N(u). Suppose, to the∈ contrary, that there are two vertices x, y Y not in N(u). Let w be a vertex of U that is distinct from u.Thuswx, wy / E(∈G). Since G is connected, there exists z Y such that z N(u)=N(w).∈ Thus G contains the subgraph shown in Figure∈ 7. ∈ y

z x

w u

Figure 7.

Let Π = S1,S2, ,Sn 2 ,whereS1= y,z,w , S2 = u , S3 = x ,and { ··· − } { } { } { } each set Si (4 i n 2) contains only one vertex of V (G) y,z,w,u,x .Since r(yΠ)=(0, 2, 1≤, ≤), r−(z Π)=(0, 1, 1, ), and r(w Π)=(0, 2,−{2, ), it follows} that | ··· | ··· | ··· 54 G. Chartrand, E. Salehi and P. Zhang AEM

Π is a resolving (n 2)-partition of V (G), contradicting the fact that pd(G)=n 1. Now either N(−u)=Y or N(u)=Y v for some v Y .IfN(u)=−Y, −{ } ∈ then G = Ks + Kt for s = Y 3andt= U 2. If N(u)=Y v ,then | |≥ | |≥ −{ } G=Ks+(K1 Kt), where V (K1)= v ,s= Y 1 2, and t = U 2. ∪ { } | |− ≥ | |≥ However, Ks +(K1 Kt)=Ks+Kt+1.Ineithercase,G=Ks+Kt,wheret 3 ∪ ≥ and so s n 3. Let V (Ks)= u1,u2, ,us and V (Kt)= v1,v2, ,vt . We consider≤ three− cases. { ··· } { ··· }

Case 1. s = t.LetΠ= S1,S2, ,Ss+1 ,whereSi = ui,vi (1 i s 1), { ··· } { } ≤ ≤ − Ss = us ,andSs+1 = vs .Sinced(u, vs)=1(u V(Ks)) and d(v, vs)=2 { } { } ∈ (v V(Kt)), it follows that Π is a resolving (s + 1)-partition of V (G). Hence pd(∈G) s +1 n 3+1=n 2, which is a contradiction, and this case cannot occur. ≤ ≤ − −

Case 2. s>t.LetΠ= S1,S2, ,Ss+1 ,whereSi = ui,vi (1 i t 1), { ··· } { } ≤ ≤ − Si = ui (t +1 i s), and Ss+1 = vt .Sinced(u, vt)=1(u V(Ks)) and { } ≤ ≤ { } ∈ d(v, vt)=2(v V(Kt)), it follows that Π is a resolving (s + 1)-partition of V (G). Hence pd(G) ∈s +1 n 3+1=n 2, which is a contradiction, and this case too cannot occur.≤ ≤ − −

Case 3. s

References

[1] G. Chartrand, L. Eroh, M. Johnson and O. R. Oellermann, Resolvability in graphs and the metric dimension of a graph,preprint. [2] G. Chartrand, M. Raines and P. Zhang, The directed distance dimension of oriented graphs, Math. Bohem. (to appear). [3] M. R. Garey and D. S. Johnson, Computers and Intractability: A Guide to the Theory of NP-Completeness, Freeman, New York, 1979. [4] F. Harary and R. A. Melter, On the metric dimension of a graph,ArsCombin.2(1976), 191–195. [5] P. J. Slater, Leaves of trees,in:Proc. 6th Southeast Conf. Comb., Graph Theory, Comput.; Boca Raton, 14 (1975), 549–559. [6] P. J. Slater, Dominating and reference sets in graphs, J. Math. Phys. Sci. 22 (1988), 445–455.

G. Chartrand, P. Zhang E. Salehi Department of Mathematics and Statistics Department of Mathematical Sciences West Michigan University University of Nevada, Las Vegas Kalamazoo, MI 49008 Las Vegas, NV 89154 USA USA

Manuscript received: January 7, 1998 and, in ﬁnal form, June 6, 1999.