On $k$-transitive closures of directed paths

In this paper we study the structure of $k$-transitive closures of directed paths and formulate several properties. Concept of $k$-transitive orientation generalize the traditional concept of transitive orientation of a graph.

We use the standard notation. By an edge we mean an unoriented pair of vertices, and by an arc we mean an oriented pair of vertices. For a given graph G, V (G) and E(G) denotes the set of its vertices and the set of its edges, respectivly. For a digraph G, we write A(G) for the set of its arcs. By an oriented graph we mean such a digraph that if (a, b) is an arc, then (b, a) is not. All graphs and digraphs in this paper are finite.

Motivation
Orientation of a graph G is called transitive if for every (a, b) ∈ A(G), (b, c) ∈ A(G), also (a, c) ∈ A(G). This concept was studied by many authors in numerous papers, see the survey [Kel85] for example. The concept of transitive orientation was generalized in several ways in [GJK88] and [Tuz94], [HC12], and other papers.
A digraph is called k-transitive if every directed path of the length k has a shortcut joining the beginning and the end of this path. In other words, if (v 0 , . . . , v k ) is a path in the digraph G, then (v 0 , v k ) ∈ A(G).
A k-transitive closure of an oriented graph G = (V, A) is an oriented graph and it has the minimal (by inclusion) set of arcs among all graphs with the above stated properties.
Observe that there are oriented graphs for which the k-transitive closure does not exist. For example in a cyclically oriented cycle C k+1 it is not possible to add arcs to fulfill the condition (3).
If the k-transitive closure does exist for some oriented graph, it is unique.
Note that this definition is a partial answer to the point (4) in [GJK88,p. 41].
The aim of this paper is to describe k-transitive closures of directed paths.
2. Structure of the k-transitive closure of the directed path Instead of T k C(P n−1 ) we write k : n to denote the k-transitive closure of an oriented path on n vertices. We label the vertices by natural numbers 1, 2, . . . , n and assume that (i, i + 1) ∈ A(P n−1 ) for 1 ≤ i < n.
Although the graph k : n is oriented, some of the properties will be stated for simple graphs obtained by "forgetting" the orientation. We belive that it is clear from the context, but to be precise, for the unoriented case we write k : n .
In this paper by a degree sequence of a graph k : n we mean a sequence (deg(1), . . . , deg(n)). (In/out)degree sequence of a graph k : n is defined in a similiar way.
Observe that 2 : n is just the complete graph K n , and 2 : n is the tournament on n vertices.
The starting point in a construction of the k-transitive closure of the path P n−1 is to add arcs (i, i + k). Then we add arcs (i, i + 2k − 1), at the next stage arcs (i, i + 3k − 2), and so on. This construction shows that for every k, n ∈ N, k : n is well defined.
The key observations are: 2.1. Fact. Adding one vertex to the path adds only arcs ending in this new vertex. In other words, k : n is an induced subgraph of k : n + 1 .
Proof. It follows directly from the construction described above that To show the other inclusion we use the induction on n. First observe that for n ≤ k all arcs are of the form (i, i + 1). Assume that all arcs in (k : n − 1) have the length 1 + l(k − 1) for some l ≥ 0. To obtain a k-shortcut in k : n we need k arcs, each of them of length 1 + l In Figure 1 we present the graph 5 : 11 as an example.

Some properties
From the observations mentioned above, we conclude several properties of graphs k : n and k : n .
3.1. Fact. For n ≤ k, T k C(P n−1 ) = P n−1 . So for n ≤ k, the graph k : n is just the path P n−1 .
We can observe the following block structure in indegree/outdegree sequences of graphs k : n : 3.2. Theorem. Let n = 2 + l(k − 1) + m for some l ∈ N and 0 ≤ m < k − 1. In the oriented graph k : n the indegree sequence is built from uniform "blocks" of length k − 1 and has the form Similarly, the outdegree sequence is built from uniform "blocks" of length k − 1 and has the form Proof. The proof follows from Facts 2.2 and 2.1. We prove the part concerning the indegree sequence. First note that the indegree of the first vertex is 0. For the next k − 1 vertices there is no arcs ending in them other than the arcs in the initial path, so their indegree is 1. First vertex of indegree 2 is the (k + 1)-th vertex. First vertex of indegree 3 is the 2k-th vertex, and first vertex of indegree j is the ((j − 1)(k − 1) + 2)-th vertex.
The proof for the outdegree sequence is similiar; we just start from the last vertex.
Proof. This is a consequence of Theorem 3.2; just observe that summing up the indegree and outdegree sequences gives the constant sequence (l + 1, . . . , l + 1).

3.4.
Corollary. For every l ∈ N, and for every 0 < m < k − 1, all vertices of the graph k : 2 + l(k − 1) + m has degree l + 1 or l + 2. Morover, if we put a = l + 1 and b = l + 2, the degree sequence is built from "blocks" of the form repeated to get the sequence of the length 2+l(k −1)+m. Note that the last "block" has the length m + 2 (mod(k − 1)).
Proof. This is another consequence of Theorem 3.2.
Recall that by degree of a vertex v in a digraph we mean a pair (indegree(v), outdegree(v)).
For oriented graphs k : n we can observe the following: 3.6. Corollary. Every constant subsequence in the degree sequence of the non regular graph k : n is also the constant subsequence in the degree sequence of the oriented graph k : n .

Recall that an oriented graph G is irregular if for every two vertices
Strightforward consequence of Corollary 3.6 is that graphs k : n for k > 3 are not irregular. The natural question is: are the graphs 3 : n irregular? The answer is: 3.7. Theorem. Oriented graphs 3 : n are irregular iff n is odd.
Proof. By Theorem 3.2, pairs of vertices 2i, 2i + 1 for 1 ≤ i < n/2 have the same indegree. Because for the even n the graph 3 : n is regular, so 3 : n is not irregular if n is even.
Also by Theorem 3.2, indegree(2i) = indegree(2i − 1) + 1 for 1 ≤ i < n/2. By Corollary 3.4, for the odd n the degree sequence of the graph 3 : n is of the form (a, b, a, . . . , b, a). So for n odd, if for some two vertices their total degrees are equal then its indegrees are different. Hence 3 : n is irregular if n is odd.
Recall that the tournament 2 : n is irregular for any n.

Density
By density of the graph G, |V (G)| = n, we mean ratio "number of edges in the graph G"/"number of edges in complete graph K n "; in symbols Dens(G) = E(G)/E(K n ) = 2E(G)/(n(n − 1)).

Open problems
The main open problem concerning k-transitive closures in general, is to state what properties of an oriented graph G guarantee the existence of T k C (G).
There are also some other special classes of oriented graphs, such as cycles (with different orientations) and trees, for which there is a chance to obtain interested properties for their k-transitive closures.