On the status sequences of trees

The status of a vertex $v$ in a connected graph is the sum of the distances from $v$ to all other vertices. The status sequence of a connected graph is the list of the statuses of all the vertices of the graph. In this paper we investigate the status sequences of trees. Particularly, we show that it is NP-complete to decide whether there exists a tree that has a given sequence of integers as its status sequence. We also present some results about trees whose status sequences are comprised of a few distinct numbers or many distinct numbers. In this direction, we provide a partial answer to a conjecture of Shang and Lin from 2011, showing that any status injective tree is unique among trees. Finally, we investigate how orbit partitions and equitable partitions relate to the status sequence.


Introduction
Sequences associated with a graph, such as the degree sequence, spectrum, and status sequence, contain useful information about the graph's structure and give a compact representation of the graph without using vertex adjacencies. Extracting and analyzing the information contained in such sequences is a crucial issue in many problems, such as graph isomorphism. In this paper, we study the status sequences of trees, and answer several questions about status realizability, uniqueness, and their relation to various graph partitions.
The graphs considered in this paper are finite, simple, loopless and connected. Let G = (V, E) be a graph. The status value (also called transmission index) of a vertex v in G, denoted s(v), is the sum of the distances between v and all other vertices, i.e., s(v) = u∈V d(v, u), where d(v, u) is the shortest path distance between v and u in G. This concept was introduced by Harary in 1959 [12]. The status sequence of G, denoted σ(G), is the list of the status values of all vertices arranged in nondecreasing order. Status sequences of graphs have recently attracted considerable attention, see, e.g., [17,18,20,21].
A connected graph is status injective if the status values of its vertices are all distinct. We denote by k(G) the number of different status values in σ(G). A graph is transmissionregular if k(G) = 1, i.e., if the status values of all its vertices are equal. Transmission-regular graphs have been studied by several authors (see, e.g., [1,2,13,14]). A sequence σ of integers is status realizable if there exists a graph G with σ(G) = σ . Let F be a family of connected graphs and G be a graph in F. A graph G is status unique in F if for any H ∈ F, σ(H) = σ(G) implies that H G, i.e., H and G are isomorphic. For example, paths are status unique in the family of all connected graphs, since a path of order n is the only graph of order n containing a vertex of status value n(n − 1)/2.
In general, compared to the adjacency list or the adjacency matrix of a graph, there is some loss of information in the status sequence. For example, one cannot obtain the distance degree sequence of a graph from its status sequence [18]. It is also well known that nonisomorphic graphs may have the same status sequence. Nevertheless, the status sequence contains important information about the graph, and the study of graphs with special status sequences has produced interesting results. For instance, it is known that spiders are status unique in trees [20]. In ( [3], p.185) the authors proposed the problem of finding status injective graphs. This problem was addressed by Pachter [18], who proved that for any graph G there exists a status injective graph H that contains G as an induced subgraph.
lems concerning distance concepts in graphs.
Problem 1.1 (Status sequence recognition). Characterize status sequences, i.e, find a characterization that determines whether a given sequence of positive integers is the status sequence of a graph.
Motivated by the above facts, in this paper we consider the following question raised by Shang and Lin in [20]: Are status injective trees status unique in all connected graphs?
The above problem was recently answered in the negative in [19], where the authors provide a construction of pairs of a tree and a nontree graph with the same status sequence. In this paper, we provide new results on the direction of the above problem for extremal trees. In particular, we show that status injective trees are status unique in trees.
Moreover, we also investigate the following special case of Problem 1.1: Recognition problems similar to 1.1 and 1.3 have been studied for some other graph sequences. For example, Erdös and Gallai [7] and Hakimi [11] gave conditions to determine whether a given sequence is the degree sequence of some graph. Sequences related to distances in a graph have also been studied, see, e.g., [4]. In those papers, the authors not only tackle the recognition problems, but also construct fast algorithms for finding graphs that realize given sequences. Following the same direction, we address the following question regarding status sequences: Problem 1.4 (Status realizability in trees). Given a sequence of integers σ , either construct a tree T such that σ(T ) = σ or report that such a tree does not exist.
We are also interested in the following problem related to the status uniqueness of a graph: Problem 1.5. What are necessary and/or sufficient conditions for a set of vertices of a graph to have the same status?
This paper is organized as follows. In Section 2, we prove that Tree status recognition is NP-complete. In Section 3, we present several polynomially solvable special cases of Status realizability in trees, specifically for symmetric and asymmetric trees. In Section 4, we explore how various well-known graph partitions relate to the status sequences.

Complexity of Tree status recognition
Let the depth of a tree T = (V, E) be the smallest number k such that there exists a vertex v ∈ V such that d(v, u) ≤ k for all u ∈ V . This is equivalent to saying that the diameter of the tree is at most 2k. In this section, we first study the complexity of Tree status recognition when the trees are restricted to have the depth of 3. We refer to the latter problem as SRT-D3.
P roof. SRT-D3 is clearly in NP, as the status sequence of a tree of depth 3 can be computed in linear time. We reduce the well-known strongly NP-complete problem 3-Partition to SRT-D3. 3-Partition (cf. [8]) reads: Given a multiset A = {a 1 , a 2 , . . . , a n } of positive integers, does there exist a partition of A into triplets such that all triplets have the same sum? Without loss of generality, assume n is divisible by 3 and let m = n/3, A = n i=1 a i and B = A/m. Since 3-Partition is strongly NP-complete, we may assume that the input of the problem is provided in unary encoding. Note that 3-Partition remains strongly NP-complete even if B/4 < a i < B/2 for all i ∈ {1, 2, . . . , n}.
Clearly, the size of the sequence σ is polynomial in the unary encoded size of I. We will show that I is a yes-instance of 3-Partition if and only if σ is a yes-instance of SRT-D3.
Suppose first that I is a yes-instance of 3-Partition, i.e., there is a partition of A into m triplets such that the sum of integers in each triplet equals B. Consider the following tree T . Take a root vertex with exactly m child-nodes which represent m triplets. For each child-node representing a triplet, create exactly three descendants representing the elements of the triplet. Finally, for each vertex representing an element a i of a triplet, create exactly a i descendants, which are the leaves of the tree. By construction the tree is of depth 3. See Figure 1 for an illustration.   The status of the root equals 1 · m + 2 · 3m + 3 · i a i = 3A + 7m. The status of any vertex representing a triplet equals 1 · 4 + 2 · (B + m − 1) There are m such vertices. The status of a vertex representing an element a i equals 1 · (a i + 1) Thus, tree T of depth 3 realizes σ .
Before tackling the opposite direction of the proof, we first recall some useful properties of the status values of the vertices in a tree. The median of a connected graph G is the set of vertices of G with the smallest status.
If v 1 is a vertex in the median of a tree T , v 1 , v 2 , . . . , v r is a path in T , and v 2 is not a median of T , then s(v 1 ) < s(v 2 ) < · · · < s(v r ).
Now, suppose, σ is a yes-instance of SRT-D3 and a tree T is its realization. Without loss of generality, we may assume A > 3B + 19m + 9, for otherwise we add to every element of the multiset an additive constant 3B + 19m + 9 preserving the hardness of the 3-partition to v equals the size of the tree after deletion of v together with all its x descendants minus the status difference between v and u and minus one for counting v itself. The size of the tree after deletion of v with all its descendants is Summarizing the findings, with necessity the following properties hold for T : 1. The median of T is the vertex of status 3A + 7m; 2. There are m vertices of status 4A − 2B + 11m − 7 adjacent to the median; 3. There are exactly B + 3 descendants for every vertex of status 4A − 2B + 11m − 7; Finally, consider a vertex u of status 4A − 2B + 11m − 7. As stated above, there are exactly B + 3 descendants of u. By assumption (in 3-Partition) that B/4 < a i < B/2 for all i ∈ {1, 2, . . . , n}, vertex u has exactly three immediate successors/descendants, say v i , v j , v k where i, j, k ∈ {1, 2, . . . , n}, and these vertices have exactly a i , a j , a k descendant leaves, respectively. Thus, a i + a j + a k = B, which completes the proof. P Theorem 2.1 straightforwardly implies the following corollary. P roof. The proof follows from Theorem 2.1 by a straightforward gap reduction. P

Polynomial special cases of Status realizability in trees
In this section, we investigate some special cases where status realizability in trees can be solved in polynomial time. We also address an extremal case of Problem 1.2, showing that status injective trees are status unique in trees.
We begin with a result about asymmetric trees. The following result shows that given an integer sequence with n distinct values, it can be determined in polynomial time whether or not this sequence is the status sequence of a tree T = (V, E) of order n. Note that such a tree, if realized, is highly asymmetric, as it cannot have any nontrivial automorphisms; see  P roof. The proof is constructive, i.e., we present an algorithm whose input is a sequence σ of n distinct positive integers and the output is either a unique tree T realizing σ or a conclusion that the sequence is not a status sequence of any tree.
Algorithm 1 attempts to construct a tree whose vertices v 1 , . . . , v n correspond to the status values in the input sequence σ (which by assumption are all distinct); the tree is specified by assigning a parent p i to each vertex v i , except the vertex v n which is treated as the root.
The algorithm also stores and updates the number of descendants of vertex v i (excluding v i itself) as the variable c i . In the main loop of the algorithm, the vertices are considered according to decreasing status values.
Algorithm 1: Status injective tree 1 Input: A sequence of integers σ = {a 1 , a 2 , . . . , a n }, with a 1 > a 2 > . . . > a n . 2 Output: A tree T whose status sequence is σ , or an affirmation that such a tree does not exist.
if such index j does not exist then return "A is not the status sequence of a tree" else We will show by induction that after iteration i of the algorithm, the parents of the vertices v 1 , . . . , v i are uniquely identified as p 1 , . . . , p i , and that the number of descendants of v i+1 In the first iteration of the algorithm, by Lemma 2.2, vertex v 1 (corresponding to status value a 1 ) must be a leaf. By Lemma 2.3, a 1 −a p 1 = (n−(c 1 +1))−(c 1 +1), so a p 1 = a 1 −n+2(c 1 +1).
If this value is not in σ , then σ is not the status sequence of a tree. Otherwise, the vertex   , and since k(T ) is an integer, it follows that The bound is tight, e.g., for stars and paths. P Now, we are ready to address polynomial solvability of SRT-D3 with a fixed number of distinct status values; these types of trees are highly symmetric. In a σ -realized tree T , consider two vertices, i ∈ I and an adjacent to it (non-root) vertex having status j, see Figure 3. For simplicity of the notation we refer to the latter vertex as j, though it can be any vertex of that status and there could be many such vertices. We The number of variables in the ILP is at most kδ. Thus, by Lenstra [16], the ILP can be Since a 1 = a 2 , we have that s(v 1 ) = s(v 2 ), s( 1 ) = s( 2 ), s(v 1 ) = s( 1 ), s(v 2 ) = s( 2 ), s(r) = s( 1 ), and s(r) = s( 2 ).
Suppose a 1 = t + b + a 2 + A − 1. Then, the status values of r, v 2 , 1 , and 2 are all distinct, a contradiction. Thus, all non-leaf neighbors of r must have the same number of leaf neighbors, say a. Suppose r has b ≥ 1 leaf neighbors, and let w be a leaf neighbor of r. Let v be any non-leaf neighbor of r, and let be any leaf neighbor of v. Then, The general case of graphs with status sequences having a unique value, that is, k(G) = 1, was studied in [1]. In that paper the authors obtain tight upper and lower bounds for the unique status value of the status sequence in terms of the number of vertices of the graph.

Equitable, orbit and status partititons
In this section we explore how the well-known concepts of equitable and orbit partitions relate to the status sequence of a graph.
Denote by P = {V 1 , . . . , V p } any partition of the vertex set V of a graph G. A status partition of a graph G, denoted by P s , is a partition of V (G) in which any two vertices in the same set have the same status value. An orbit partition of G, denoted P o , is a partition in which two vertices are in the same set if there is an automorphism which maps u to v, for some group of automorphisms of G. An equitable (or regular) partition of G, denoted by P e , is a partition of V into nonempty parts V 1 , . . . , V p such that for each i, j ∈ {1, . . . , p}, the number of neighbors in V j of a given vertex u in V i is determined by i and j, and is independent of the choice of u in V i . More precisely, let the distance matrix D of G be partitioned according to P = {V 1 , . . . , V p }, so that D i,j is a block of D formed by rows in V i and columns in V j .
The characteristic matrix S is the n × p matrix whose j th column is the characteristic vector of V j , 1 ≤ j ≤ p. The quotient matrix of D with respect to the partition P is the p × p matrix B = (b i,j ) whose entries are the average row sums of the blocks of D: where 1 is the all-one vector. The partition P is called equitable if each block D i,j of D has constant row (and column) sum, that is, SB = DS. See [10] for more details on orbit and equitable partitions.
Let v be a vertex of a graph G, and (v) denote the greatest distance between v and any other vertex of G. The distance partition of G with respect to v, denoted P d (v), is a partition The following result shows that an orbit partition is just a refinement of a status partition.
. Thus, P o is also a status partition of G. P The above elementary result can be applied to the highly asymmetric trees considered in The vertex a is adjacent to every vertex in A and b is adjacent to every vertex in B. The subgraph of G m induced by A is a cycle of length m 2 with The subgraph of G m induced by B consists of m disjoint cycles of length m with b i,j ∼ b i,j+1 , We define a matching between the vertices in A and those in B as follows: See Figure 5 for an illustration.  On the other hand, in the next result we find a necessary condition for a status partition to be an equitable partition. Suppose that the status partition P s is an equitable partition. Let u and v be any two vertices with the same status value and such that u, v ∈ V i for some i ∈ {1, . . . , k}. Then, since we assume P s is an equitable partition, it follows that u and v have the same number of neighbors in each V j for j = 1, . . . , k. Therefore the distance partitions of G with respect to u and v have the same quotient matrix.
Conversely, suppose that for any two vertices u, v with s(u) = s(v), it holds that the partitions P d (u) and P d (v) have the same quotient matrix. This implies that vertices u and v have the same number of neighbors at each possible distance. Hence it follows that P s is an equitable partition. P Note that the conditions of Proposition 4.5 can be verified efficiently. A simple approach to verify that two vertices with the same status value define a distance partition with the same quotient matrix is to build a breadth-first-search tree (in linear time) from each vertex, find the distance partition, and then compute the characteristic matrix and the quotient matrix as described at the beginning of the section.
Finally, we identify a large class of graphs for which a status partition is an equitable partition: distance mean-regular graphs. Distance mean-regular graphs were introduced in [5] as a generalization of both vertex-transitive and distance-regular graphs. Let G k (w) denote the number of vertices at distance k from a vertex w of a graph G. G = (V, E) is a distance mean-regular graph if, for a given u ∈ V , the averages of |G i (u) ∩ G j (v)|, computed over all vertices v at a given distance h ∈ {0, 1, . . . , diam(G)} from u, do not depend on u. Note that distance-mean regular graphs are super regular graphs, that is, the number of vertices at a fixed distance is the same for any vertex v i ∈ V . Note that for a super regular graph, it holds that s(v i ) is constant for any v i ∈ V . Corollary 4.6. If G is a distance-mean regular graph, then every status partition is an equitable partition. P roof. If G is a distance-mean regular graph, then by Proposition 2.2 in [5], the quotient matrix B with respect to the distance partition of every vertex is the same. This implies that the condition of Proposition 4.5 is satisfied and therefore it follows that every status partition in G is an equitable partition. P We conclude with two open problems. The first is related to status partitions.