1 Introduction
“It is possible to go to a graph theory conference and to ask oneself, at the end of every talk, what is the fractional analogue?”  Scheinermann and Ullman [5] made this remark in the preface of their book on fractional graph theory. Considering the popularity of studying fractional versions of wellstudied problems, it is a wonder that the fractional version of oriented coloring is yet to be studied. We initiate it with this article.
An oriented graph is a directed graph without any directed cycle of length 1 or 2. In this article, a graph refers to a simple or an oriented graph while denotes the set of vertices of and or refers to its set of edges or arcs, respectively.
The notion of oriented coloring was introduced by Bruno Courcelle [1] in 1994 inspiring a considerable volume of work on the topic (see recent survey [6] for details). An oriented coloring of an oriented graph is a function from to a set of colors such that (i) for every arc , and (ii) implies for every . The oriented chromatic number is the minimium such that admits an oriented coloring. The oriented chromatic number for a family of oriented graphs is given by
Without further ado, let us now define the natural analogue of the fractional version of the oriented chromatic number. Let be a set of colors and let denote the set of all subsets of having cardinality . A fold oriented coloring is a mapping from to satisfying for every arc , and implies for every . The fold oriented chromatic number of is the minimum such that admits an fold oriented coloring.
The fractional oriented chromatic number of is given by
Observe that for all . Thus the above limit exists due to the Subadditivity Lemma (see Appendix A.4 [5]). Naturally, for a family of oriented graphs
Notice that the oriented coloring and chromatic number are equivalent to the fold oriented coloring and chromatic number , respectively.
The close relation between oriented chromatic number and (oriented) graph homomorphisms is wellknown [6]. On the other hand, the fractional version of the usual chromatic number has a famous equivalent formulation using homomorphism to Kneser graphs [5]. Thus, naturally we study the relation between the oriented fractional coloring and oriented graph homomorphisms. Alongside such a study, we explore some other basic properties of oriented fractional coloring.
A relevant related concept is the oriented analogues of clique and clique number. The analogue of clique and clique number for oriented graphs ramified into two notions: (i) oriented absolute clique and clique number, and (ii) oriented relative clique and clique number. The later is more relevant here. An oriented relative clique [3] of an oriented graph is a vertex subset satisfying under any homomorphism of , and for any distinct vertices . The oriented relative clique number [3] of is the cardinality of a largest oriented relative clique. The oriented relative clique number for a family of oriented graphs is the maximum where .
We will soon see that the parameter is sandwiched between the parameters oriented relative clique number (lower bound) and oriented chromatic number (upper bound). Therefore, it is interesting to examine the oriented fractional chromatic number of oriented graphs or families of oriented graphs for which the values of and are different.
The most ordinary such graphs one can think of are probably the directed cycles
of length . The study of fractional oriented chromatic number of these simple looking graphs turn out to be quite challenging. Moreover, the corresponding results are utterly surprising and have an interesting relation with prime numbers. To motivate the readers, we present an interesting example where the fractional oriented chromatic number is strictly less than the oriented chromatic number.Example 1.1.
Let be the directed 7cycle. We know that . However, Fig. 1 shows a fold oriented coloring of implying . Is this upper bound tight? We will answer that later in this article, and till then we encourage the readers to think about it.
One of the most important open problems in the domain of oriented coloring is determining the oriented chromatic number of the family of oriented planar graphs. In particular, improving the upper bound [4] seems to be especially challenging. Moreover, the related class of questions of determining the oriented chormatic number of the family of oriented planar graphs with girth (length of a smallest cycle) at least are also interesting. However, till date the only known exact bounds are for the cases for all .
In general, finding the exact value of for seems to be tough problem. Therefore, we wondered if the fractional version will be any easier or if the exact values of the parameter will remain the same. For this particular article, we focus more on the second question and find that indeed the fractional oriented chromatic number of can be less than for large values of , and, in fact, it gets arbitrarily close to as .
The organization of this article as follows. In Section 2, we present the preliminary notation, terminology and establish some basic results. In Sections 3 and 4 we study the fractional oriented chromatic number of directed graphs and sparse planar graphs, respectively. Finally, in Section 5, we share our concluding remarks.
2 Preliminaries and basic results
Let and be two oriented graphs. A function is a homomorphism of to if for each arc of , is an arc of . We use the notation to denote that admits a homomorphism to . This definition readily motivates the following result.
Proposition 2.1.
If , then .
Proof.
Let , and thus there exists a fold oriented coloring of . Assume that is a homomorphism. Notice that the composition is a fold oriented coloring of . ∎
Now the next natural question is whether there exists any equivalent definition of oriented fractional coloring and chromatic number using the notion of homomorphisms. To express such an equivalent formulation we need some definitions.
The Kneser graph is a graph having the subsets of cardinality of a set of cardinality as vertices, and two vertices are adjacent if and only if they are disjoint sets. A consistent suborientation of is an oriented graph , whose underlying graph is a subgraph of , and whose arcs are oriented in such a way that given two arcs and we have . That brings us to the equivalent formulation of oriented fractional coloring using the notion of homomorphisms.
Theorem 2.2.
An oriented graph satisfies if and only if where is a consistent suborientation of and is a constant positive integer.
To prove the above theorem, we will first prove a supporting result.
Theorem 2.3.
If for some consistent suborientation of , then there exists a consistent suborientation of satisfying for every positive integer .
Proof.
It is enough to show that there exists a consistent suborientation of isomorphic to . This can be obtained by replacing each vertex of with . ∎
Now we are ready to prove Theorem 2.2.
Proof of Theorem 2.2. Let . That means, admits a fold oriented coloring for some satisfying . Notice that, as we have not assumed a reduced form of , it is possible to have and . However, due to Theorem 2.3 it is enough to prove assuming a reduced form of .
Now let us construct an oriented graph having the colors used for the fold oriented coloring of as vertices. Moreover, two vertices and of have an arc between them if there exists an arc in satisfying and . Notice that, is a consistent suborientation of and is a homomorphism of to . This takes care of the “if” part.
On the other hand, if where is a consistent suborientation of , then itself is also a fold coloring of . This implies . This completes the “only if” part. ∎
Next we will move onto the sandwich theorem mentioned in the introduction. Before stating it, it is useful to recall a handy characterization of an oriented relative clique.
Proposition 2.4.
[3] A vertex subset of is an oriented relative clique if and only if any nonadjacent pair of vertices in is connected by a directed 2path.
Now we are ready to state and prove the sandwich theorem.
Theorem 2.5.
For any oriented graph , .
Proof.
Let be a relative oriented clique of with . Suppose is a fold oriented coloring of . Then, as the nonadjacent vertices of are connected by a directed path, we must have for all distinct . Thus, a total of colors have been used on the vertices of . Therefore, which implies that . Hence the first inequality.
The second inequality follows from the trivial observation . ∎
We will sign off the section by establishing a general lower bound for oriented fractional chromatic number. To do so, we will introduce an oriented analogue of independent sets. A vertex subset of an oriented graph is an oriented independent set if for each distinct pair of vertices , it is possible to find an oriented coloring of satisfying .
Proposition 2.6.
Given an oriented graph , a vertex subset is an oriented independent set if and only if any two vertices of are neither adjacent nor connected by a directed path.
Proof.
The “if” part directly follows from Proposition 2.4.
For the “only if” part, given distinct , define the following function
Notice that is a homomorphism of to . ∎
We also use this definition to introduce the oriented analogue of the parameter independence number. The oriented independence number is the maximum where is an oriented independent set of . Finally, we are ready to state the result that gives us a general lower bound of oriented fractional chromatic number.
Theorem 2.7.
Given any oriented graph we have .
Proof.
Let be a fold coloring of with . This means, there is a set of colors whose set of all subsets having cardinality is , and in particular is a function from to .
Note that, if and , and , then are independent. Thus for any is an oriented independent set and . Moreover, observe that every vertex is part of exactly such sets. Thus we have
This implies
and completes the proof. ∎
Notice that, we can prove the tightness of the bound easily using the above result. In the next section, we are going to provide the generalized tight values for , where denotes the directed cycle of length , for all .
3 Directed cycles
We use the following classification of prime numbers to present our result. A prime number is a typeA prime if , and is a typeB prime if . Let be a positive integer. If does not have a typeA prime factor, then . Otherwise, where is the least typeA prime factor of . For instance , , and .
Theorem 3.1.
Let be a directed cycle of length . Then

if ,

,

,

, if and .
The first three parts of the above theorem follow directly from known results. However, the main challenge is to prove the last part of it. That part is unexpected and gives us another indication of how difficult, counterintuitive and yet beautiful the theory of fractional oriented coloring may actually be.
Throughout this section we will assume that the set of vertices of is and the set of arcs of is
Proof of Theorem 3.1(a,b,c). We know that for all [6]. We also know that and [6]. Thus the result follows due to Proposition 2.5.
The proof of Theorem 3.1(d) is broken into several lemmas and observations presented in the following.
For the rest of the section we only deal with having and . All the ‘’ and ‘’ operations perfomed in the subscript of is computed modulo unless otherwise stated.
Lemma 3.2.
For all and we have, .
Any fold oriented coloring of having is a miser fold oriented coloring.
Let be a miser fold oriented coloring of using a set of colors. Note that as a directed 2path is an oriented clique, its vertices must receive disjoint sets of colors. As a consequence:
Lemma 3.3.
For all satisfying we have .
Thus assume that , and . By Lemma 3.3, and . Suppose that . Observe that . Notice that the definitions of and depends on the coloring and the vertices .
Now we introduce some notations to aid our proof. Note that for each . We will use the notation to denote the set of colors used on . However, at some point of time if we are sure that some of the sets among or has empty intersection with , then we can drop the corresponding name of the set along with its subscript to denote . Moreover, if we are sure that for some , then we can replace with in the above notation.
One more type of notation that we use is the following: if for some we know that , then we may denote it as . For instance if , then we are sure that has colors from but it may or may not have colors from the other sets. Furthermore, if and for some , then we may denote it as . Similarly, we may use three or four set names among for this notation.
For instance, if at some point of time we learn that , and , then we can write , , , or to denote the set of colors used on . We will choose to use the notation that suits our purpose.
Now we are going to list some observations needed for the proof.
Observation 3.4.
There exists no index with .
Proof.
It is not possible to have as , . ∎
The next observation directly follows from the second condition from the definition of fold oriented coloring and the fact that , , .
Observation 3.5.
The following implications hold:

,

,

if , then ,
In the above statement the case if is considered under the special additional condition .
Observation 3.6.
If and , then for any .
Proof.
This observation holds by the pigeonhole principle as and . ∎
The next observation is based on the fact that the color sets assigned to the three vertices belonging to a directed 2path must be distinct under any fold oriented coloring as a directed 2path is an oriented clique.
Observation 3.7.
For any index and for any we have,
Proof.
Note that the vertices and of induces a directed 2path, and hence is a an oriented relative clique. Thus . However, . This implies . ∎
The above observation shows that three consecutive vertices of cannot all avoid colors from a particular set . The following observation will show the opposite.
Observation 3.8.
For any index and for any we cannot have,
Proof.
Suppose the contrary.
Based on the above observations we are able to show stronger implications.
Observation 3.9.
The following implications hold:

and and ,

and and ,

, and and .
Proof.
If and , then due to Observation 3.5. Moreover, if , then due to Observation 3.5. This implies contradicting Observation 3.7. Also is not possible due to Observation 3.4. Thus and .
The proofs are similar. ∎
An automorphism of an oriented graph is a function such that if and only if . An oriented graph is dipath transitive if for any two directed paths and of there exists an automorphism such that , , and .
Now we are going to prove one of the key lemmas to aid the proof of the theorem. We need to introduce some notations to state the lemma. The set of colors is called the label of for any . A sequence of consecutive labels are the labels used on for some . A sequence of consecutive labels of the form () is a triple. The sequence of consecutive labels of the form () is a quad.
Lemma 3.10.
Any label is either part of a triple or a quad.
Proof.
As , , , the lemma is true for all having .
Now assume that the lemma is true for all having . Observe that due to our assumption. Now we want to show that is part of either a triple or a quad.
If , then it contradicts Observation 3.7. Thus . However if , then . On the other hand, as , we have . This contradicts the second condition of the definition of fold oriented coloring. Therefore, .
The labels in the entire cycle can thus be viewed as a circular arrangement of triples and quads.
Let be a triple. If we rename the vertices as for all and some constant , then the definitions of the sets and gets changed accordingly. However, we observe that a triple remains a triple and a quad remains a quad.
Lemma 3.11.
If we rename the vertices as for all and some constant where is a triple, then a triple remains a triple and a quad remains a quad.
Proof.
For convenience we will refer to every vertex as per their names before renaming throughout the proof.
After renaming, the definitions of the sets and changes. We will use the labels and for convenience.
Suppose the first triple/quad after that does not remain a triple/quad after renaming is . We consider the following two cases.
Case 1: Suppose before renaming was a triple.
Considering the situation before renaming, due to pigeonhole principle, we can say that .
On the other hand, considering the situation after renaming , a contradiction. Thus must be a triple even after renaming.
Case 2: Suppose before renaming was a quad.
Considering the situation before renaming, due to pigeonhole principle, we can say that .
On the other hand, considering the situation after renaming , a contradiction. Thus must be a quad even after renaming. ∎
If the last element of a triple/quad is and the first element of a triple/quad is , then and are consecutive where is before and is after .
Lemma 3.12.
There are no consecutive triples.
Proof.
As , there must be at least one quad. Thus if there are consecutive triples, then there exists a quad and two triples such that are, respectively, before and after .
As the sets and are defined based on the set of colors assigned to the vertices and , respectively, and as is dipath transitive, we may assume that . Therefore, and .
Note that the following constraints follow directly from the definition of fold oriented coloring. For any distinct
(1) 
Observe that by pigeonhole principle. This implies by (1).
Furthermore . This implies by (1).
Thus . This is a contradiction to Lemma 3.3 as . ∎
Therefore, there are some positive number of quads in between any two triples. In order to analyze the number of quads in between two triples, we construct a symmetric binary matrix of dimension such that
(2) 
Thus implies by (1). The row of the binary matrix is denoted by . Three consecutive bits along a row or a column cannot be all s or all s due to Observations 3.7 and 3.8.
A cyclic rightshift
is the operation of rearranging the entries in a vector by moving the final entry to the first position, while shifting all other entries to the next position.
Now we will present an important property of the binary matrix .
Lemma 3.13.
The vector is obtained by a cyclic rightshift on .
Proof.
We have to show that for any . Suppose the contrary, that is, for some .
First assume that and . This implies by (1). Moreover, to avoid three consecutive s in the row we have . This implies by (1). Now the column has three consecutive s contradicting Observation 3.7. Thus and is not possible.
Similarly it can be shown that it is not possible to have and . ∎
Let be a triple and after that there are quads after which there is a triple . Then we say that is quads after .
Lemma 3.14.
If is quads after and is quads after , then .
Proof.
Without loss of generality let , , and .
As , we have
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