On the Partition Dimension and the Twin Number of a Graph ^{†}^{†}thanks: Research partially supported by grants MINECO MTM201563791R, Gen. Cat. DGR 2014SGR46 and MTM201460127P.
Abstract
A partition of the vertex set of a connected graph is a locating partition of if every vertex is uniquely determined by its vector of distances to the elements of . The partition dimension of is the minimum cardinality of a locating partition of . A pair of vertices of a graph are called twins if they have exactly the same set of neighbors other than and . A twin class is a maximal set of pairwise twin vertices. The twin number of a graph is the maximum cardinality of a twin class of .
In this paper we undertake the study of the partition dimension of a graph by also considering its twin number. This approach allows us to obtain the set of connected graphs of order having partition dimension . This set is formed by exactly 15 graphs, instead of 23, as was wrongly stated in the paper ”Discrepancies between metric dimension and partition dimension of a connected graph” (Disc. Math. 308 (2008) 5026–5031).
Key words: locating set; locating partition; metric dimension; partition dimension; twin number
1 Introduction
All the graphs considered are undirected, simple, finite and connected. The vertex set and edge set of a graph are denoted by and . Let be a vertex of . The open neighborhood of is , and the closed neighborhood of is . The degree of is . If (resp. ), then is called universal (resp. a leaf). Let be a subset of vertices of a graph . The open neighborhood of is , and the closed neighborhood of is . The subgraph of induced by , denoted by , has as vertex set and . The complement of , denoted by , is the graph on the same vertices as such that two vertices are adjacent in if and only if they are not adjacent in . Let , be two graphs having disjoint vertex sets. The (disjoint) union is the graph such that and . The join is the graph such that and .
The distance between vertices is denoted by , or if the graph is clear from the context. The diameter of is . The distance between a vertex and a set of vertices , denoted by is the minimum of the distances between and the vertices of , that is to say, . Undefined terminology can be found in [4].
A vertex resolves a pair of vertices if . A set of vertices is a locating set of , if every pair of distinct vertices of are resolved by some vertex in . The metric dimension of is the minimum cardinality of a locating set. Locating sets were first defined by [11] and [16], and they have since been widely investigated (see [2, 12] and their references).
Let be a graph of order . If is a partition of , we denote by the vector of distances between a vertex and the elements of , that is, . The partition is called a locating partition of if, for any pair of distinct vertices , . Observe that to prove that a given partition is locating, it is enough to check that the vectors of distances of every pair of vertices belonging to the same part are different. The partition dimension of is the minimum cardinality of a locating partition of . Locating partitions were introduced in [5], and further studied in [1, 3, 6, 7, 8, 9, 10, 14, 15, 17, 18, 19]. Next, some known results involving the partition dimension are shown.
Theorem 1 ([5]).
Let be a graph of order and diameter

.

. Moreover, this bound is sharp.

if and only if is isomorphic to either the star , or the complete split graph , or the graph .
In [17], its author approached the characterization of the set of graphs of order having partition dimension , presenting a collection of 23 graphs (as a matter fact there are 22, as the socalled graphs and are isomorphic). Although employing a different notation (see Table 1), the characterization given in this paper is the following.
Theorem 2 ([17]).
Figure 1  

Paper [17] 
Thus, in particular, and according to [17], for every , all the graphs displayed in Figure 1 satisfy . However, for all of them, it holds that , as we will prove in this paper (see Corollaries 3, 5 and 6). As a matter of example, we show next that , for every . First, notice that . Next, if , and , then consider the partition . Finally, observe that is a locating partition of since , for every .
The main contribution of this work is, after showing that the theorem of characterization presented in [17] is far for being true, finding the correct answer to this problem. Motivated by this objective, we introduce the socalled twin number of a connected graph , and present a list of basic properties, some of them directly related to the partition dimension .
The rest of the paper is organized as follows. Section 2 is devoted to introduce the notions of twin class and twin number, and to show some basic properties. In Section 3, subdivided in three subsections, a number of results involving both the twin number and the partition dimension of a graph are obtained. Finally, Section 4 includes a theorem of characterization presenting, for every , which graphs of order satisfy .
2 Twin number
A pair of vertices of a graph are called twins if they have exactly the same set of neighbors other than and . A twin set of is any set of pairwise twin vertices of . If , then they are called true twins, and otherwise false twins. It is easy to verify that the socalled twin relation is an equivalence relation on , and that every equivalence class is either a clique or a stable set. An equivalence class of the twin relation is referred to as a twin class.
Definition 1.
The twin number of a graph , denoted by , is the maximum cardinality of a twin class of . Every twin set of cardinality will be referred to as a set.
As a direct consequence of these definitions, the following list of properties hold.
Proposition 1.
Let be a graph of order . Let be a twin set of . Then

If , then , for every vertex .

No two vertices of can belong to the same part of any locating partition.

induces either a complete graph or an empty graph.

Every vertex not in is either adjacent to all the vertices of or nonadjacent to any vertex of .

is a twin set of .

.

.

if and only if is the complete graph .

if and only if is the star .
It is a routine exercise to check all the results showed in Table 2 (see also [5] and the references given in [2]).
order  

1  2  
1  1  
2  3 
We conclude this section by characterizing the set of graphs such that .
Proposition 2.
Let be a graph of order . Then, if and only if is one of the following graphs (see Figure 2):

the complete split graph , obtained by removing an edge from the complete graph ;

the graph , obtained by attaching a leaf to the complete graph ;

the complete bipartite graph ;

the complete split graph .
Proof.
It is straightforward to check that the twin number of the four graphs displayed in Figure 2 is . Conversely, suppose that is a graph such that . Let such that is the set of . Since is connected, we may suppose without loss of generality that . We distinguish two cases.
Case 1: . If , then , and thus . If , then , as otherwise , a contradiction. Thus, .
Case 2: . If , then , and thus . If , then , as otherwise , a contradiction. Hence, . ∎
3 Twin number versus partition dimension
This section, consisting of 3 subsections, is devoted to obtain relations between the partition dimension and the twin number of a graph . In the first subsection, a realization theorem involving both parameters is presented, without any further restriction than the inequality . The second subsection is devoted to study the parameter , when is a graph of order with ”few” twin vertices, to be more precise, such that . Finally, the last subsection examines , whenever is a graph for which .
3.1 Realization Theorem for trees
A complete ary tree of height is a rooted tree whose internal vertices have children and whose leaves are at distance from the root. Let denote the complete ary tree of height 2. Suppose that is the root, are the children of , and are the children of for any (see Figure 3(a)).
Proposition 3.
For any integer , and .
Proof.
Certainly, , and thus . Suppose that and is a locating partition of size . In such a case, for every the vertices are twins, and thus each one belongs to a distinct part of . So, if for some pair , with , then , which is a contradiction. Hence, the vertices must belong to distinct parts of . We may assume that , for every . Thus, if belongs to the part , then , which is a contradiction. Hence, . Finally, consider the partition such that and, for any , . Then, for every and for every such that :
Therefore, is a locating partition, implying that . ∎
Proposition 4.
Let be integers such that and . Then, and .
Proof.
Certainly, . Let .
Next, we show that . Let be a locating partition of . If there exist two distinct pairs and such that the vertices are in the same part of and are in the same part, then , which is a contradiction. Notice that this tree contains pairs of vertices of the type and if , we achieve at most such pairs avoiding the preceding condition. Thus, . Moreover, if , then for every pair , there exists a pair such that and . So, by symmetry, we may assume without loss of generality that . Consider the vertices such that , and , . Then , which is a contradiction. Hence, .
To prove the equality , consider the partition such that:
Let . Then, for every , :
Therefore, if and . Moreover, it is straightforward to check that, if , then for every , , we have
Finally, for , we have
Therefore, is a locating partition, implying that . ∎
Theorem 3.
Let be integers such that . Then, there exists a tree such that and .
3.2 Twin number at most half the order
In this subsection, we approach the case when is a graph of order such that . Concretely, we prove that, in such a case, .
Lemma 1.
Let be a subset of vertices of size of a graph such that is neither complete nor empty. Then, there exist at least three different vertices such that and
Proof.
If is neither complete nor empty, then there is at least one vertex such that . Let (resp. ) be a a vertex adjacent (resp. nonadjacent) to . Then, satisfy the desired condition. ∎
Lemma 2.
If is a nontrivial graph of order with a vertex of degree , then .
Proof.
Let and and . Take the partition , where for . Observe that resolves the vertices of for . Therefore, is a locating partition of , impliying that . ∎
Corollary 1.
If is a graph of order with at least one vertex satisfying , then .
As a direct consequence of Theorem 1, we know that if is a graph such that , then . Next, we study the cases and .
Proposition 5.
Let be a graph of order . If and , then .
Proof.
By Corollary 1, and having also in mind that has no universal vertex, since its diameter is 3, we may suppose that, for every vertex , . Let be a vertex of eccentricity . Consider the nonempty subsets for . If at most one of these three subsets has exactly one vertex, then there exist five distinct vertices such that, for , and vertices and do not belong to the same set . Consider the partition , where and . Then, resolves every pair of vertices in and the vertices in . Therefore, is a locating partition, implying that .
Next, suppose that for exactly one value and for . We distinguish two cases.

is neither complete nor empty. Then by Lemma 1, there exist vertices such that and . Consider the sets and , where for , with the additional condition , which is possible since . Take the partition . Observe that resolves the vertices in and resolves every pair of vertices in . Therefore, is a locating partition, implying that .

is either complete or empty. We distinguish three cases, depending on for which , .

. Then, is a twin set with vertices, a contradiction as .

. Let be the (unique) vertex of . Then and are twin sets. If , then , a contradiction. If , then , again a contradiction.

. Let be the (unique) vertex of . Then, both and are twin sets. Notice that . We distinguish cases.

If (resp. ) , then (resp. ), a contradiction.

If , then . Let , , . Take the partition , where , and . Observe that resolves the vertices in and and resolves the vertices in . Therefore, is a locating partition, implying that .


∎
Proposition 6.
Let be a graph of order . If and , then .
Proof.
By Corollary 1, we may suppose that, for every vertex , . We distinguish three cases.

There exists a vertex of degree . Consider the subsets and . We distinguish two cases.

is neither complete nor empty. Then, Lemma 1, there exist three different vertices such that and . Consider two different vertices and let , , . Then, resolves the vertices in and in , and resolves the vertices in . Hence, is a locating partition of .

is either complete or empty. Then the subsets of , , and are twin sets. We distinguish cases.

If either or , then either or , in both cases a contradiction as .

If both and have degree at least , then , and . We distinguish cases, depending on the size of .

. Then, , a contradiction as .

. If , then , as is a (maximum) twin set of . Suppose that . Let , . Consider the partition , where , and . Observe that either or resolves the vertices of and . Notice also that resolves the vertices in . Hence, is a locating partition of .

. We may assume without loss of generality that and . Let , , . Consider the partition , where , and . Observe that resolves the vertices of and . Notice also that resolves the vertices in . Hence, is a locating partition of .




There exists at least one vertex of degree and there is no vertex of degree 2. In this case, the neighbor of is a universal vertex . Let be the set of vertices different from that are not leaves. Notice that there are at most two vertices of degree 1 in , as otherwise all vertices in would have degree between and , contradicting the assumption made at the beginning of the proof.
If there are exactly two vertices of degree 1, then . In such a case, induces a complete graph in , as otherwise the nonuniversal vertices in would have degree at most . So, is a twin set, implying that , a contradiction.
Suppose thus that is the only vertex of degree 1, which means that contains vertices, all of them of degree or . Consider the graph . Certainly, has vertices, all of them of degree or . Let denote the set of vertices of degree of , for . Observe that , since is a twin set in . Hence, , as and the size of must be even. We distinguish two cases, depending on the size of .

. Notice that . Let and such that . Consider the partition , where , and . Observe that , , . Hence, resolves the vertices in
