Let be an undirected graph. An orientation of is a directed graph obtained by assigning a direction to every edge in . A classical result of Gallai, Roy and Vitaver states that in any orientation of a graph , the number of vertices in the longest directed path is at least the chromatic number of . Lin  asked whether every graph can be oriented such that the number of vertices in its longest directed path is where is any number between the chromatic number of and the number of vertices in the longest undirected path in . Let be a parameter of oriented graphs. We define
We say the parameter has the interval property if for every graph and every such that , there is an orientation of such that . For a graph is the length of the longest path in Note that = (the number of vertices in the longest path in ) − 1. Similarly, is the length of the longest directed path in a directed graph . Lin’s question can be phrased as whether has the interval property. In Section 2, we answer Lin’s question in the affirmative by describing a method to construct an orientation of any given graph with where is an integer between and . In Section 3, we show that the diameter of directed graphs does not have the interval property. We construct an infinite family of graphs and prove that it is not possible to orient these graphs so that their diameters take all the values between the minimum and the maximum diameters.
2. Longest Path in Oriented Graphs
Theorem 1 (Gallai  , Roy  , Vitaver  , also in the book  ) If is an orientation of , then Furthermore, equality holds for some orientation of .
Theorem 1 provides a lower bound of for all orientations of and is a trivial upper bound of . The question of whether the parameter has the interval property is the same as the question of whether can take every value in the interval . This is a question raised in  by C. Lin. In this section we describe a method to construct an orientation for each such integer thus giving an affirmative answer to the question of Lin and proving that the parameter has the interval property.
Theorem 2 For a graph with and and an integer such that , there always exists an orientation of such that .
Proof. Suppose that is a path of vertices in . Let be a -coloring of such that the color classes are . We may permute the colors so that . We define a -coloring of recursively for . In the -th coloring, the color classes will be labeled . For and we let . For
, we define for and
. What this means is that in the -th step, we take the vertex in the path and color it with a new color while keeping the colors of all the other vertices unchanged. This is a proper coloring of . Let be the orientation of induced by this -coloring. That is, in , the edge is directed from to if , and .
It is easy to see that and according to Theorem 1,
. Thus we have . On the other hand, is a directed path in . Therefore . The key idea of the proof of the theorem is the following observation: When the orientation changes from to , the length of the longest directed path will increase by at most one.
Claim: for all .
Let be a longest directed path in . We prove this claim in two cases.
Case 1: does not contain the vertex . Since all the edges that are not incident with are directed the same way in as in , is also a directed path in . We have .
Case 2: contains the vertex . Since all the edges that are incident to are directed towards in , must be the last vertex in . Therefore is a directed path in . We have . This completes the proof of the claim.
To prove the main theorem by contradiction, we assume that there is a value between and such that for all , Let
Since and , such exists and is strictly less than . We have and , therefore This contradicts to the claim.
An efficient algorithm can easily be derived to find the orientation in Theorem 2 when a longest path and a -coloring of is given, even though finding a longest path is a well known NP-complete problem (  , Problem ND29).
3. Diameter of Oriented Graphs
The distance between two vertices in a graph , denoted , is the length of a shortest path between and . The diameter of is the largest distance over all the pairs of vertices in , denoted . That is,
Similarly, in a directed graph , is the length of a shortest directed path from to and
Distances and diameters of oriented graphs and their applications have been studied extensively (see e.g.  -  ). A directed graph has a finite diameter if and only if is strongly connected. A well known theorem of Robbins  states that a graph has a strongly connected orientation if and only if is 2-connected. For this reason, we will only consider 2-connected graphs and their strongly connected orientations in this section. Another result we will use in this section is a theorem of Gutin  .
Theorem 3 Let be a 2-connected graph. The largest diameter of all strongly connected orientations of is .
We define an infinite family of graphs for
Definition 4 Let the vertex set of be , and the edges be: and . (See Figure 1).
It is easy to see that and
Lemma 5 There is an orientation of such that
Proof. Let be the orientation of obtained by using the pairs in Definition 4 as ordered pairs. (See Figure 2). We show that for every vertex and any other vertex , there is a directed -path of length at most . By symmetry, we only need to consider two cases that either or , .
Case 1. . Let . If , there is a -path of length at most : . If , there is a -path of length at most : . If , there is a -path of length at most :
Case 2. and . Let . We consider the following four subcases.
Case 2.1 . is a -path of length at most .
Case 2.2 . is a path of length at most since this path at at most edges from to and at most edges from to .
Case 2.3 . is a path of length at most since this path has at at most edges from to , one edge from to and at most edges from to .
Case 2.4 . is a path of length at most since this path has at most edges from to , one edge from to one edge from to , and at most edges from to .
Combining all these cases, we see that for all vertices in and Therefore .
Figure 1. G3k.
Figure 2. D3k.
Lemma 6 There are at most different orientations of with finite diameters.
Proof. The vertices all have degree two. For the orientation to be strongly connected, the path must be a directed path, either from to or from to . Similarly, the paths and
have to be directed paths too. There are two possible directions for each of these three paths and two possible directions for each of the edges and . Therefore, there are at most orientations that are strongly connected thus with finite diameters.
By symmetry, many of the orientations have the same diameter. The number of distinct values of the diameters of orientations of is much less than 64. Using Gutin’s theorem (  ), we find that the largest diameter among all orientations of is . By Lemma 5, the smallest diameter is at most . There are values in the interval . Therefore, we have our main result of this section.
Theorem 7 Let be an integer greater than 64. It is not possible to orient the graph such that the diameter of the orientations take all the values between the smallest diameter and the largest diameter.
Corollary 8 The diameter of oriented graphs does not have the interval property.
The author wish to thank two anonymous referees for their helpful comments.