1 Introduction
Let be a graph. We use the concepts of path and cycle as defined in [3]. We may think of a path or cycle as a subgraph of . The length of a path (respectively, cycle) is its number of edges. The order of a path , denoted by , is defined as its number of vertices, that is, . Similarly, the order of a cycle is its number of vertices. Let denote the graph isomorphic to a cycle of length and let denote the complement of . We also use the concepts of stable set and clique as defined in [3]. The cardinality of a maximum stable set (respectively, maximum clique) is denoted by (respectively, ). A (proper) coloring of a graph is a partition of into stable sets. The cardinality of a minimum coloring is denoted by .
Let be a digraph. For every concept for graphs, we may have an analogue for digraphs. The underlying graph of , denoted by , is the simple graph with vertex set such that and are adjacent in if and only if or . We borrow terminology from undirected graphs when dealing with a digraph by considering its underlying graph . For example, we say that a stable set of a digraph is a stable set of its underlying graph . Conversely, we may obtain a directed graph from a graph by replacing each edge of by an arc , or an arc , or both; such directed graph is called a superorientation of . A superorientation which does not contain a digon (a directed cycle of length two) is an orientation. A digraph is symmetric if is a superorientation of a graph in which every edge of is replaced by both arcs and .
If is an arc of , then we say that dominates and is dominated by . If is not dominated by any of its neighbors, then we say that is a source. Similarly, if does not dominate any of its neighbors in , then we say that is a sink. A directed path or directed cycle is an orientation of a path or cycle, respectively, in which each vertex dominates its successor in the sequence. Henceforth, when we say path of a digraph, we mean directed path (note that we do not use this convention for cycles). We denote by () the cardinality of a maximum path in a graph (digraph). When we say a cycle of a digraph, we mean either a superorientation of an undirected cycle with length at least three or a digon.
Let and be two disjoint subsets of . We use the notation to denote that every vertex of dominates every vertex of in and no vertex of dominates a vertex of in . If (respectively, ), we may denote (respectively, ). A path partition of is a collection of vertexdisjoint paths of that cover . Let denote the cardinality of a smallest path partition of . In 1960, Gallai and Milgram [8] showed that, for every digraph, the size of a minimum path partition is less than or equal to the size of a maximum stable set . Actually, Gallai and Milgram showed a stronger statement that implies GallaiMilgram’s Theorem. It uses the concept of orthogonality, defined next. Let be a path partition and let be a stable set of . We say that and are orthogonal if for every ; we also say that is orthogonal to or vice versa.
Theorem 1.1 (Gallai and Milgram [8]).
Let be a digraph. For every minimum path partition of , there is a stable set such that and are orthogonal. In particular, .
A straightforward application of GallaiMilgram’s Theorem is the following corollary.
Corollary 1.2.
Let be a digraph. Let be a path of and let . If is adjacent to every vertex of , then has a path such that .
A graph is perfect if for every induced subgraph of . It is easy to show that if is perfect, then cannot contain either an odd cycle of order at least five or its complement as an induced subgraph. Berge [1] conjectured that the converse was true as well. In 2006, Chudnovsky, Robertson, Seymour and Thomas [4] proved this long standing open conjecture and it became known as the Strong Perfect Graph Theorem:
Theorem 1.3 (Chudnovsky et al [4]).
A graph is perfect, if and only if, does not contain an odd cycle with five or more vertices or its complement as an induced subgraph.
Motivated by GallaiMilgram’s Theorem and looking for stronger properties in the relationship between stable sets and paths in digraphs, Berge [2] introduced in 1982 a new class of digraphs which he called diperfect digraphs. A digraph is diperfect if every induced subdigraph of has the following property: for every maximum stable set of , there exists a path partition of such that and are orthogonal. Berge [2] proved that every symmetric digraph as well as every digraph whose underlying graph is perfect is diperfect. However, he also showed that there are superorientations of odd cycles that are not diperfect. We say that a superorientation of an odd cycle is an antidirected odd cycle if with and each of is either a source or a sink in . Figure 1 shows examples of anti directed odd cycles.
Berge [2] proved the following characterization of superorientations of odd cycles with at least five vertices that are diperfect.
Theorem 1.4 (Berge [2]).
Let be a superorientation of a , with . Then, is diperfect if and only if is not an antidirected odd cycle.
Analogously to Theorem 1.3, Berge was interested in obtaining a characterization of the class of diperfect digraphs in terms of forbidden subdigraphs. In fact, he proposed the following conjecture.
Conjecture 1.5 (Berge [2]).
A digraph is diperfect if and only if does not contain an antidirected odd cycle as an induced subdigraph.
Motivated by Berge’s Conjecture, Sambinelli, Silva and Lee [9] proposed in 2018 a similar conjecture. Before we state it, we need some definitons. A digraph is BeginEnddiperfect or simply BEdiperfect if every induced subdigraph of satisfies the following properties: (i) for every maximum stable set of there is a path partition of orthogonal to and (ii) every path starts or ends at a vertex of . We say that a superorientation of an odd cycle is a blocking odd cycle if with and each of and is either a source or a sink in . Figure 2 shows examples of blocking odd cycles.
Conjecture 1.6 (Sambinelli, Silva and Lee [9]).
A digraph is diperfect if and only if does not contain a blocking odd cycle as an induced subdigraph.
Recently, Conjectures 1.5 and 1.6 were verified for some specific classes of digraphs (see [10], [7] and [5]).
In this paper, we show that Berge’s Conjecture (Conjecture 1.5) is false for arbitrary digraphs. We present and characterize superorientations of complements of odd cycles with at least five vertices that are not diperfect. One can easily check that a complement of an odd cycle with at least five vertices cannot contain an induced odd cycle with at least five vertices. Thus every superorientation of the complement of an odd cycle is free from antidirected odd cycles. On the other hand, all these counterexamples to Conjecture 1.5 contain blocking odd cycles as induced subdigraphs. So these digraphs are not counterexamples to Conjecture 1.6. In fact, it can be shown that a superorientation of the complement of an odd cycle with at least five vertices is BEdiperfect if and only if does not contain an blocking odd cycle as an induced subdigraph [6].
2 Diperfect superorientations of
We start this section by presenting a necessary condition for a digraph to be diperfect.
Lemma 2.1.
Let be a digraph and let be a maximum stable set of of size at least two. If is diperfect, then or , for every pair of distinct vertices .
Proof.
Let be a path partition orthogonal to . Clearly, at least one path of must have size at least . Since and belong to distinct paths of , the result follows.
Let be a graph isomorphic to , with and let be a superorientation of . Henceforth, we may assume that the vertices of (and of ) are labeled as so that the cycle is . So, the nonneighbors of are and , where the indexes are taken modulo , as depicted in Figure 3. Moreover, note that and each pair is a maximum stable set of (and hence, of ).
We say that is a if there is no arc , for . Figure 4 shows the digraphs and .


Our goal is to show that a superorientation of the complement of an odd cycle with at least five vertices has a path partition orthogonal to if and only if is not isomorphic to . (Theorem 2.4). Note that this implies that there exists an infinite family of counterexamples of Berge’s conjecture. One may verify that the superorientation of a depicted in Figure 4(a) does not have a path partition orthogonal to . Note that this digraph is also an antidirected odd cycle. Similarly, the superorientation of a depicted in Figure 4(b) does not have a path partition orthogonal to .
Let be a graph isomorphic to for . Let be the graph obtained from by deleting a pair of nonadjacent vertices and and the edge . Note that is isomorphic to because is the complement of the cycle . This observation allows us to obtain a path partition orthogonal to of a superorientation of when a certain subdigraph of admits a specific path partition.
Lemma 2.2.
Let be a superorientation of a with . Let be the superorientation of a obtained from by deleting the vertices and and the arc between and . If has a path partition orthogonal to , then has a path partition orthogonal to .
Proof.
Suppose that has a path partition orthogonal to . Without loss of generality, we may assume that and . Since the only nonneighbors of in are and , it follows that is adjacent, in , to every vertex of . So, by Corollary 1.2, there is a path in such that . Similarly, the only nonneighbors of are and . So, it follows that is adjacent, in , to every vertex of . By Corollary 1.2, there is a path in such that . Thus, is a path partition of orthogonal to .
By adjusting notation, we immediately have the following corollary.
Corollary 2.3.
Let be a superorientation of a for . Let be the superorientation of a obtained from by deleting the vertices and and the arc between and . If has a path partition orthogonal to , then has a path partition orthogonal to .
Theorem 2.4.
Let be a superorientation of a for . Then, has a path partition orthogonal to if and only if is not isomorphic to .
Proof.
(Necessity) Assume that is isomorphic to . We show that and , and the result follows immediately by Lemma 2.1. Note that, by the Principle of Directional Duality, it suffices to prove that . Let . Let be a longest path of . For each , let if . By definition, for , there is no arc . Hence, the sequence is strictly increasing. Towards a contradiction, suppose that . So there must exist such that and . However, this is a contradiction since is nonadjacent to and . Thus, the size of is at most .
(Sufficiency) Suppose that does not have a path partition orthogonal to . We show that is isomorphic to . The proof follows by induction on . If , then is a superorientation of a . By Theorem 1.4, is an antidirected odd cycle. It is easy to verify that is isomorphic to (see Figure (a)a). So suppose that . Let be the superorientation of a obtained from by deleting the vertices and and the arc between and . By Lemma 2.2, we may assume that has no path partition orthogonal to . Since the vertices of are labeled as , we may apply the induction hypothesis and assume that is isomorphic to . By the Principle of Directional Duality, we may assume that the following property holds:

for , there is no arc in (and hence, in ) (see Figure (a)a).




Similarly, let be the superorientation of a obtained from by deleting the vertices and and the arc between and . By Corollary 2.3, we may assume that has no path partition orthogonal to . Since the vertices of are labeled as , we may adjust notation and apply the induction hypothesis. So is also isomorphic to . Moreover, since (a) holds, then must satisfy the following property:

for , there is no arc in (and hence, in ) (see Figure (b)b).
Let . Note that and (see Figure (c)c). Since (a) and (b) hold, it suffices to prove that , and . Towards a contradiction, suppose that dominates . Then, is a path partition of orthogonal to , a contradiction. Similarly, suppose that dominates . Then, is a path partition of orthogonal to , a contradiction. Now suppose that dominates . Then is a path partition of orthogonal to , a contradiction. Similarly, suppose that dominates . Then, is a path partition of orthogonal to , a contradiction. Finally, suppose that dominates . Then, is a path partition of orthogonal to , a contradiction. Thus, is isomorphic to (see Figure (d)d).
One may ask the following natural question.
Question 2.5.
Let be a digraph which does not contain an induced antidirected odd cycle or an induced , with . Is it true that is diperfect?
References
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