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The Partition Dimension of a Graph

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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.
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