A homomorphic polynomial for oriented graphs

In this article, we define a function that counts the number of (onto) homomorphisms of an oriented graph. We show that this function is always a polynomial and establish it as an extension of the notion of chromatic polynomials. We study algebraic properties of this function. In particular we show that the coefficients of these polynomials have alternating sign property and that the polynomials associated to the independent sets have relations with the Stirling numbers of the second kind.


Introduction
In 1994, Courcelle [4] extended the notion of vertex coloring for oriented graphs which inspired a number of works in this domain (see [20] for details). The way to view Courcelle's definition as a natural extension of vertex coloring for simple graphs is through homomorphisms.
A homomorphism f of a simple (oriented) graph G to a simple (oriented) target graph H is a function f : V (G) → V (H) such that f (u)f (v) is an edge (arc) if uv is an edge (arc). A homomorphism f with an image of cardinality k is a k-coloring (oriented k-coloring) of G and the image of a vertex is its color 1 . The minimum k such that G admits a k-coloring (oriented k-coloring) is the chromatic number χ(G) (oriented chromatic number χ o (G)) of G.
In 1994, Sopena [18] had defined a function, which turned out to be a polynomial, that counts the number of oriented colorings of a graph. He also noticed that this function lacks two very important properties, namely, the alternating signs of the coefficients property (present in chromatic polynomials) and the unimodal behavior of the absolute values of the coefficients property (present in chromatic polynomial, proved recently by Huh and Katz [9] through positively settling a long standing conjecture). Recently, Cox and Duffy [5] has made considerable progress on this topic by establishing several interesting properties of the chromatic polynomial. Looking back at the two decades of research on oriented coloring, it can be noted that, homomorphism, alongside coloring, is also an important aspect of oriented coloring.
Keeping that in mind, following the footsteps of Birkhoff [3] and Sopena [18], we define a function, which also turns out to be a polynomial, for counting the number of (onto) homomorphisms of an oriented graph to tournaments. This polynomial possesses the alternating signs of the coefficients property and, upon checking a number of examples, we conjecture it to have the unimodal behavier of the absolute values of the coefficients property as well.
We hope that these polynomials can become valuable tools for studying oriented chromatic number in the near future. We initiate the study by devising tools for studying these polynomials and by proving some of their interesting properties.
In this article, we motivate and introduce the notion of oriented homomorphic polynomials in Section 2 and justify its name in Section 3. In Section 4 we develop some useful tools to study the problem. In Section 5 we present interesting properties of the polynomials and prove most of them, while some of the proofs are moved to Section 6 to maintain the flow of reading. Finally in Section 7 we conclude the article.

Counting homomorphisms
Two functions f 1 and f 2 , having the same domain D, but possibly different co-domains, are equivalent if f 1 (v) = f 2 (v) for all v ∈ D, otherwise they are distinct.
The chromatic polynomial [7] χ(G, x) of a simple graph G denotes the number of ways to obtain a y-coloring of G, for all y ≤ x, using the colors {0, 1, · · · , x − 1}. The function χ(G, x) is known to be a polynomial in x, hence the name.
Observe that, to distinguish between two colorings we are only interested in the images of the vertices of G, not the structure of the target graph. That is not a concern though, as any y-coloring of G, for each y ≤ x, can be viewed as a homomorphism of G to the complete graph K x .
To be precise, let C x (G) be the set of distinct y-colorings (for all y ≤ x), using the colors {0, 1, · · · , x − 1}, of G and H x (G) be the set of distinct homomorphisms (not necessarily onto) of G to K x .
However, the scenario is completely different for oriented graphs. The major reason is that, unlike in the case of simple graphs, a maximal (with respect to number of edges) graph on x vertices is not unique for oriented graphs. To elaborate, while there is only one maximal graph K x (the complete graph) with set of vertices {0, 1, · · · , x − 1} in the   undirected case, there are 2 ( x 2 ) different maximal oriented graphs (the tournaments) with set of vertices {0, 1, · · · , x − 1}.
Therefore for an oriented graph G, the set OC x (G) of distinct oriented y-colorings, for all y ≤ x, using the colors {0, 1, · · · , x − 1} of G is not in a bijection with the set OH x (G) of all homomorphisms of G to tournaments having x vertices {0, 1, · · · , x − 1}.
Sopena [18] defined the oriented chromatic polynomial of an oriented graph G as where OC x (G) denote the set of distinct oriented y-colorings (for all y ≤ x) of G using the colors {0, 1, · · · , x − 1}. In the following we look at an example to better understand this function and some of its limitations for motivating our "oriented version" of the chromatic polynomial for oriented graphs.
An oclique [10] O is an oriented graph with χ o (O) = |V (O)|. It is known that an oriented graph O is an oclique if and only if each pair of its non-adjacent vertices are connected by a directed 2-path [10]. Also, it is known that the number of arcs of an oclique on n vertices may vary between (n log n − 3n 2 ) and n 2 [6]. However, the oriented chromatic polynomial for any oclique O on n vertices is the same [18]: Let P 2 and C 3 denote the directed 2-path and the directed cycle of order three having set of vertices V (P 2 ) = {a, b, c} and V (C 3 ) = {x, y, z}, respectively (as shown in Fig. 1). There are two non-isomorphic tournaments on 3 vertices, namely, the directed tournament DT 3 and the transitive tournament T T 3 .
Note that both P 2 and C 3 are ocliques on 3 vertices. Thus they both have the same oriented chromatic polynomial: In particular, from this example we can conclude that the oriented chromatic polynomial does not determine the number of arcs of the corresponding oriented graph.
Whereas, if we look at their set of homomorphisms to tournaments on 3 vertices, we observe some differences.
There are exactly three distinct homomorphisms of P 2 to DT 3 , namely, f 0 , f 1 and f 2 given by the following: for all i ∈ {0, 1, 2}, where the + operations are taken modulo 3. Similarly, there are exactly three distinct homomorphisms of C 3 to DT 3 , namely, g 0 , g 1 and g 2 given by the following: for all i ∈ {0, 1, 2}, where the + operations are taken modulo 3.
On the other hand, there is exactly one homomorphism of P 2 to T T 3 : In contrast, C 3 does not admit any homomorphism to T T 3 .
Remark 2.2. Notice that, while counting the oriented chromatic polynomial, the oriented colorings corresponding to the homomorphisms f 0 and f are counted as the same coloring.
Note that there are two distinct directed (labeled) tournaments DT 3 and six distinct transitive (labeled) tournaments T T 3 on set of vertices {0, 1, 2}. Thus |OH 3 (P 2 )| = 12 and |OH 3 (C 3 )| = 6. Hence note that, the function defined by |OH x (G)| is not the same for P 2 and C 3 . However, while counting 2 and we are having to make a lot of redundant counts due to the vertices which are not part of the image. Furthermore, due to the redundant counts, the function is not polynomial as per our example. To avoid these redundant counts and to, hopefully, obtain a better behaved function we define a simplified version of |OH x (G)| in the following.
2 For G = P 2 or C 3 , we can have x(x − 1)(x − 2) choices for the images of the vertices of G. Moreover, the directions of 2 or 3 arcs joining the vertices among the images of f (V (G)), respectively, will have fixed directions to satisfy the conditions of a homomorphism. That means, each of the remaining arcs will have two choices for direction. Notice that, there are, respectively, 1 or 0 such arcs joining the vertices among the images of f (V (G)), 3(x − 3) arcs with exactly one endpoint in f (V (G)), and x−3 2 arcs with both endpoints outside f (V (G)).
Let T x be the set of all tournaments whose set of vertices are subsets of {0, 1, · · · , x−1}. Given an oriented graph G and a tournament T , let OH T (G) be the set of all distinct onto homomorphisms of G to T . Now define OH * x (G) = ∪ T ∈Tx OH T (G) and consider the function χ o (G, x) = |OH * x (G)|. Reworking our example we obtain distinct polynomials 3 In fact, notice that Proposition 2.3. Given two oriented graphs G and H, |OH Let us prove this by induction on x. For the base case, observe that We will show that it is true for y = x as well. Notice that the number of homomorphisms in OH x (G) (resp., OH x (H)) whose cardinality of the image set is equal to y ≤ x can be expressed as x x−y · |OH * y (G)| (resp., x x−y · |OH * y (H)|). Therefore, using the induction hypothesis and our basic assumption For the converse, assume that |OH * That concludes the proof.
In this sense the new function |OH * x (·)| is truly a good modification of |OH x (·)|. Upon further study, in hindsight, the function χ o (·, ·) turns out to be very interesting. To reiterate, let T x be the set of all tournaments whose set of vertices are subsets of {0, 1, · · · , x − 1} for all integer x ≥ 0. We define as the homomorphic polynomial of an oriented graph G and devote this article to study this polynomial. The justification for the name is given in Theorem 3.3.

Remark 2.4.
Let K x be the set of all complete graphs whose set of vertices are subsets of {0, 1, · · · , x − 1}. Given a graph G and a complete graph . As any y-coloring of G, for each y ≤ x, corresponds to an onto homomorphism of G to K y , we have χ(G, x) = |H * x (G)|. Therefore, the homomorphic polynomial χ o (·, ·) is indeed an "oriented analogue" of the chromatic polynomial χ(·, ·).

Homomorphic polynomials
We start by computing the homomorphic polynomial of a tournament.
Proof. As a tournament T admit onto homomorphisms only to itself, it is enough to count the number of ways to label the vertices of T using {0, 1, · · · , x − 1}.
The next result proves a formula for computing χ o (G, x) of an oriented graph. The formula along with its derivation is similar to the deletion-contraction formula [7] for counting chromatic polynomials, even though it is not an exact analogue. For oriented chromatic polynomial, a similar formula is established by Duffy and Cox [5]. However, to prove that formula, it was necessary to extend the concept of oriented chromatic polynomials to mixed graphs.
Before presenting the formula we need to define a few notation. Let G be an oriented graph and let u, v be two non-adjacent vertices of G. Then G · uv denotes the oriented graph obtained by identifying the vertices u, v of G and G+uv denotes the oriented graph obtained by adding the arc uv to G.
The set of onto homomorphisms having f (u) = f (v) is in a one-to-one correspondence with the set of onto homomorphisms of G · uv.
Also the set of onto homomorphisms such that f (u)f (v) is an arc is in a one-to-one correspondence with the set of onto homomorphisms of G + uv.
Similarly, the set of onto homomorphisms such that f (v)f (u) is an arc is in a one-toone correspondence with the set of onto homomorphisms of G + vu.
Moreover, it is possible to have f (u) = f (v) if and only if u and v are neither adjacent, nor connected by a directed 2-path [11]. Thus the result follows. Now we are ready to prove that the function Proof. Let us introduce a partial order for oriented graphs to facilitate induction. Let G ≺ H if one of the following conditions hold: Now assume that G is a minimal (with respect to ≺) counter-example to the statement of the theorem. Thus, due to Proposition 3.1 G is not a tournament. Hence, there exists a pair of non-adjacent vertices u, v in G. Now apply Theorem 3.2 on G to obtain depending on whether u, v are connected by a directed 2-path or not. In any case, observe that G + uv ≺ G, G + vu ≺ G, and (if it exists) G · uv ≺ G. Thus, due the minimality of Therefore, we are done using Theorem 3.2.
The above result justifies the name homomorphic polynomial.

Associated (2, 3)-rooted tree and factorial form
Notice that Theorem 3.2 enables us to compute the homomorphic polynomial of any oriented graph. Using that idea we will describe a rooted (2, 3)-tree, not neccessarily unique, associated to each oriented graph G which will enable us to compute its homomorphic polynomial.
Given an oriented graph G, we describe the design of a rooted (2, 3)-tree T G as follows: the vertices of T G are oriented graphs with its root being G. Now take two non-adjacent vertices u, v of G. If u, v are connected by a directed 2-path, then there are two children G + uv and G + vu of G and if u, v are not connected by a directed 2-path, then there are three children G + uv, G + vu and G · uv of G. We recursively continue this process for each node of the tree unless that node is a tournament. Thus, finally, we obtain a rooted (2, 3)-tree T G whose leaves are tournaments. Observe that such a tree is not unique.
Note that if we stop the recursive process of obtaining a T G for an oriented graph G at any point and the leaves at that point of time are G 1 , G 2 , · · · , G t , then As the collection of leaves of the tree is a multiset of tournaments after finishing the whole process, the homomorphic polynomial χ o (G, x) of G can be expressed as a sum of homomorphic polynomials of some tournaments. In other words, let such a multiset L(G) have a i tournaments of order i for each i ∈ {1, 2, · · · , n}. Then the homomorphic polynomial of G is From now on, for convenience, we will denote such a multiset L(G) by {a n · T n , a n−1 · T n−1 , · · · , a 1 · T 1 }.
Observe that as the homomorphic polynomial of an oriented graph G is well defined, the multiset L(G) will not depend on the choice of the (2, 3)-rooted tree associated to G.

Properties
Throughout this section, the set of vertices and arcs of an oriented graph G is denoted by V (G) and A(G), respectively. Also we would like to suggest the master's thesis of Hubai Tamás [8] as a detailed survey on properties of chromatic polynomials for simple graphs.
As hinted in the introduction, the homomorphic polynomial carries information about the order (number of vertices) and size (number of arcs) of an oriented graph.
Theorem 5.1. Let G be an oriented graph with homomorphic polynomial χ o (G, x) = a n x n + a n−1 x n−1 + · · · + a 1 x + a 0 where a n ̸ = 0. Then |V (G)| = n and |A(G)| = n 2 − log 2 a n . Proof. Recall the partial order ≺ defined in the proof of Theorem 3.3 and assume that G is a minimal (with respect to ≺) counter-example to this theorem.
Notice that G cannot be a tournament due to Proposition 3.1. Hence, there exists a pair of non-adjacent vertices u, v in G. Now apply Theorem 3.2 on G to obtain depending on whether u, v are connected by a directed 2-path or not. In any case, observe that G + uv ≺ G, G + vu ≺ G, and (if it exists) G · uv ≺ G. Now, notice that the number of vertices in G is the same as the number of vertices in G + uv or G + vu and one more than G · uv. Due the minimality of G, the polynomials χ o (G + uv, x), χ o (G + vu, x), and χ o (G · uv, x) (if it exists) must have degree n, n, and (n − 1), respectively. Therefore, the polynomial χ o (G, x) also has degree n = |V (G)|. Now let us concentrate on the other part of the statement. Notice that, as χ o (G·uv, x) has degree (n−1), its coefficients does not contribute in the value of a n . To be precise, the leading coefficient a n is a sum of the leading coefficients a uv and a vu (say) of χ o (G + uv, x) and χ o (G + vu, x), respectively. Also, as both G + uv and G + vu has one arc more than the number of arcs in G, due to the minimality of G, we must have a uv = a vu . Hence, This completes the proof.
Clearly, χ o (G, x) is not a monic polynomial like the chromatic polynomial [7] and the oriented chromatic polynomial [18]. Anyways, it is possible to characterize all the monic homomorphic polynomials as a simple corollary of Theorem 5.1.
Recall the partial order ≺ defined in the proof of Theorem 3.3 and assume that the oclique G is a minimal (with respect to ≺) counter-example to this proposition.
Notice that G cannot be a tournament due to Proposition 3.1. Hence, there exists a pair of non-adjacent vertices u, v in G. Now apply Theorem 3.2 on G to obtain as u, v are connected by a directed 2-path due to G being an oclique. In any case, observe that G + uv ≺ G and G + vu ≺ G.
Observe that both G + uv and G + vu are ocliques as they are each obtained by adding an arc to an oclique. Thus, due to the minimality of G, the fact that |A(G + uv, x)| = |A(G + vu, x)| = (m + 1), we must have where t ′ = n 2 − (m + 1). Now using the above equation along with the relation where t = (t ′ + 1) = n 2 − m. Let us look at some other properties of the homomorphic polynomial. It is easy to show that the constant term of an homomorphic polynomial is 0. This is also a property of chromatic polynomials [7] and oriented chromatic polynomials [18]. Among the other standard graphs, we want to compute the homomorphic polynomial for the independent set I n of cardinality n. Surprisingly, it turns out to be a challenging problem. For convenience, we will figure out the factorial form of the polynomial instead of its standard form.
The proof of Theorem 5.5 is provided in Section 6.
Let n k denote the number of ways to partition a set of n labeled objects into k nonempty unlabeled subsets. These numbers are called the Stirling numbers of the second kind.
Using the polynomial χ o (I n , x), it is possible to prove the following inequality involving Stirling numbers of the second kind.
Theorem 5.6. For any k ≥ 1, Proof. To count the number of onto homomorphisms of I n to a fixed labeled tournament on i vertices, we partition the n vertices in i subsets in n i ways and can map a particular partition to separate vertices in i! ways. Furthermore, there are total 2 ( i 2 ) labeled tournaments on i vertices. Therefore, the coefficient of x (i) in the homomorphic polynomial of I n is 2 ( i 2 ) · i! · n i ·. Thus the result follows.
Furthermore, using Theorem 5.6 one can compute the Stirling numbers of the second kind using the polynomial χ o (I n , x). The forward difference operator [17] of a real function and that the operator commutes with the operation + as well as scalar multiplication.
Proof. Note that, by Theorem 5.6 we have Observe the following before continuing with the calculation steps. [ Therefore, The final equality is justified by applying the formula given in eq n (1) to the equality from Theorem 5.6. Thus we are done.
Computation of the chromatic polynomial of a disconnected undirected simple graph has a nice relation with the chromatic polynomials of its components.
Proposition 5.8. [7] Let G be an undirected simple graph with c connected components G 1 , G 2 , · · · , G c . Then However, the above condition does not hold at all in the oriented case. Thus, if we mandate an oriented analogue of the Tutte polynomial to retain the important property of factoring through components, then we can conclude that the homomorphic polynomial cannot be a valid candidate for it. However, it is worth studying in details about what can be a natural analogue of Tutte polynomial in the context of oriented coloring. On the other hand, in pursuit of a similar result for homomorphic polynomials, we found out a particular type of construction where such a result holds.
We will describe the construction before stating the result. Let G 1 , G 2 , · · · , G c be c oriented graphs. Let T be a tournament on c vertices {1, 2, · · · , c}. Now construct the oriented graph G as follows. Take the disjoint union of G 1 , G 2 , · · · , G c . For each arc ij of T , put an arc from each vertex of G i to each vertex of G j . The so obtained oriented graph is G. It is worth mentioning that this construction appears frequently while studying oriented coloring. For instance, such constructions are used to prove lower bounds for oriented chromatic number of several graph families, such as, outerplanar graphs [19], partial 2-trees [14], planar graphs [15], graphs on surfaces [2], etc.
Theorem 5.9. Let G be an oriented graph as described above. Then Proof. Observe that for any i ̸ = j, a vertex of G i is adjacent to a vertex of G j . Thus, the images of G i and G j are disjoint under any homomorphism. Hence, the formula described in Theorem 3.2 can be independently applied to G i and G j for any distinct i, j ∈ {1, 2, · · · , c}.
We are going to present a few more results in the same direction. But before presenting them, let us describe a graph construction.
Let G 1 and G 2 be two oriented graphs. First take the disjoint union of G 1 , G 2 and then add a new vertex v to it. After that for each vertex u of G 1 , add an arc uv and for each vertex w of G 2 , add an arc vw. Let us denote the so-obtained oriented graph by Theorem 5.10. For any non-negative integer k, the homomorphic polynomial is equal to where n(G, j) satisfies the recurrence relation with the initial condition n(G, 1) = χ o (G, 1).
Proof. We want to count the number of ways to choose an onto homomorphism f of G 1 ⋉ v G 2 to distinct tournaments on vertices labeled by subset of {0, 1, · · · , k}. Note that the vertex v will always have a distinct image to any other vertices of the graph with respect to any homomorphism. Thus we can choose the image f (v) of v from the set {0, 1, · · · , k} in (k + 1) different ways.
Also any vertex of G 1 will have a distinct image to that of any vertex of G 2 with respect to any homomorphism as they are end points of a directed 2-path whose middle vertex is v.
Let n(G i , k i ) denote the number of onto homomorphisms of G i to all possible tournaments on k i vertices. Notice that, n(G i , 1) = χ o (G i , 1) and Therefore, given a subset S i ⊆ {0, 1, · · · , k} having cardinality k i , there are exactly n(G i , k i ) onto homomorphisms of G i to tournaments of order k i with vertices labeled by elements of S for each i ∈ {1, 2}.
Furthermore, we can choose a subset S 1 of cardinality k 1 from the set {0, 1, · · · , k} \ {f (v)} in k k 1 ways and can choose a a subset S 2 of cardinality k 2 from the set {0, 1, · · · , k}\ ways. There can be k 1 k 2 different arcs between the vertices of S 1 and S 2 . Those arcs can be chosen in 2 k 1 k 2 ways. Thus Hence we are done.
In the construction of G 1 ⋉ v G 2 , if we choose G 2 (or G 1 ) to be the null graph, then the resultant graph we obtain is denoted by G + 1 (or G − 2 ). A weaker version of the above theorem is the following.
Corollary 5.11. Let G be an oriented graph. Then Above we tried to investigate the nature of the homomorphic polynomials of graphs with a dominating vertex. Now let us probe the opposite aspect.
Theorem 5.12. Let G v be the disjoint union of an oriented graph G and an isolated vertex v. If the factorial form of the homomorphic polynomial of G is χ o (G, x) = a n x (n) + a n−1 x (n−1) + · · · + a 1 x (1) , The proof of Theorem 5.12 is provided in Section 6. The chromatic polynomial for a simple graph has alternative signs of its coefficients [7]. Sopena [18] showed that this property does not hold for oriented chromatic polynomials. However, the above property holds for homomorphic polynomials.
Theorem 5.13. Let G be an oriented graph with homomorphic polynomial χ o (G, x) = a n x n + a n−1 x n−1 + · · · + a 1 x where a n ̸ = 0. Then a i ≥ 0 for all i ≡ n (mod 2) and a i ≤ 0, otherwise.
The proof of Theorem 5.13 is provided in Section 6.
An immediate corollary of the above result is the following: Corollary 5.14. A homomorphic polynomial does not have any negative real root.
Proof. Let G be an oriented graph with homomorphic polynomial χ o (G, x) = a n x n + a n−1 x n−1 + · · · + a 1 x where a n ̸ = 0 and let y be a negative real number.
Thus χ o (G, y) > 0 if n is even and χ o (G, y) < 0 if n is odd due to Theorem 5.13.
It is worth noting that oriented chromatic polynomials do not posses this property [5]. One other important property of chromatic polynomials that was not satisfied by oriented chromatic polynomials is the unimodal property [21]. That the coefficients of a chromatic polynomial have unimodal property was observed by Read [16], later conjectured by Nijenhuis and Wilf [13], and finally proved by Huh and Katz [9].
Even though we have not proved the unimodal property for homomorphic polynomials this property was observable without in many examples we have worked, which makes us propose the following conjecture. where c n ̸ = 0. Then there exists i < n such that |c n | ≤ |c n−1 | ≤ · · · ≤ |c n−i | ≥ |c n−i−1 | ≥ · · · |c 1 |.

The proofs
We open our proof section with an important lemma. Lemma 6.1. Let G be the union of a tournament T of order k and an isolated vertex v.
Proof. Assume that the vertices of T are v 1 , v 2 , · · · , v k . Now let us construct a (2, 3) rooted tree T T as follows. We will recursively use the formula given by Theorem 3.2 on the pair of vertices v and v i , where i ∈ {1, 2, · · · , k} starts with the initial value 1 and gets incremented in each step until it reaches k.
Let T i denote the mixed graph obtained by adding the edges vv i for all i ∈ {1, 2, · · · , i} in G. An orientation of T i refers to an oriented graph obtained by replacing its edges with arcs (in any possible direction). That means, there can be exactly 2 i distinct orientations of T i . We claim that the (i+1) th level of the rooted tree consists of 2 i distinct orientations of T i and the tournament T for all i ∈ {1, 2, · · · , k + 1}. We will use induction to prove the correctness of this claim.
For the base case, observe that the root of T T is G. It will have three children G + vv 1 , G + v 1 v and G · vv 1 , which will be the level 2 of the tree. Note that G · vv 1 is exactly the tournament T and G + vv 1 , G + v 1 v are the two distinct orientations of T 1 . Hence our claim is correct for i = 1.
For the induction step, let the claim be true for all j ≤ i. We want to prove it for j = (i + 1). For that, notice that among all distinct orientations of T i−1 , there exists exactly one in which v and v i are not connected by a directed 2-path. Thus for that orientation of T i−1 we will apply the part (ii) of the formula given in Theorem 3.2, while we will apply part (i) of the formula for the rest. The tournament T will be a child of the particular orientation of T i−1 where v and v i are not connected by a directed 2-path. The other children of it, and the children of the rest of the orientations of T i−1 will be all the orientations of T i . Finally, the tournament T in the i th level will not have any child. This proves our claim.
The construction of the tree will be complete after k steps, that is, at the (k + 1) th level. In the (k + 1) th level, we have 2 k distinct orientations of T k , which are tournaments on (k + 1) vertices. Moreover, every level, except for the 0 th one, contains a leaf which is the tournament T . This completes the proof. Now we are ready to prove Theorem 5.12.
Proof of Theorem 5.12. Note that to obtain a (2, 3)-rooted tree T Gv associated to G v , we can simply include an isolated vertex to each node of T G and continue the branching process as described in Section 4.
Thus using Lemma 6.1 we obtain This concludes the proof. □ For proving Theorem 5.5 we are going to introduce some notations. Let H k denote the set of all (labeled) tournaments of order k. Furthermore, let a · H k denote the multiset having a copies of each elements belonging to H k . Note that |H k | = 2 ( k 2 ) and |a · H k | = a · 2 ( k 2 ) . Now we are ready to prove Theorem 5.5.
Proof of Theorem 5.5. To prove the theorem, it is sufficient to prove that the multiset L(I n ) of leaves of a (2, 3)-rooted tree T In associated to I n is L(I n ) = {a n · T n , a n−1 · T n−1 , · · · , a 1 · T 1 } where a i 's are as in the statement of Theorem 5.5.
We will prove this by induction. For n ≤ 2, the result is trivially true. Assume that the statement is true for all n ≤ t for some fixed t. Then we want to show that the statement is true for n = t + 1 as well.
Note that to obtain a (2, 3)-rooted tree T I t+1 associated to I t+1 , we can simply include an isolated vertex to each node of T It and continue the branching process as described in Section 4.
Due to Lemma 6.1 b t+1 = 2 t · a t = 2 t · 2 ( t 2 ) = 2 ( t+1 2 ) . Furthermore, b 1 = 1 as there is only one tournament on 1 vertex and there is exactly one homomorphism of I t+1 to that tournament.
For an i ∈ {2, 3, · · · , t}, we have This concludes the proof. □ For proving Theorem 5.13 we need another lemma. Note that we can write where the coefficients C n,k are elementary symmetric polynomials in the first positive (n − 1) integers, given by For k < 0 or k ≥ n we will set C n,k = 0 The following lemma proves a recurrence relation on the coefficients C n,k .
Proof. The term C n,k can be written as sum of two expressions e 1 + e 2 , such that every additive term in e 1 has (n − 1) as a factor and every additive term in e 2 does not have (n − 1) as a factor. In e 1 , (n − 1) can be factored out to give (n − 1) · C n−1,k−1 and e 2 is C n−1,k by definition.
Now we are ready to prove Theorem 5.13.
Proof of Theorem 5.13. We will prove this by the method of strong induction on |V (G)|. For the base case, observe that the statement is correct for all |V (G)| ≤ 2. This is verifiable using Proposition 3.1 and Theorem 5.5. As induction hypothesis, assume that the statement holds for all |V (G)| ≤ (n − 1). Thus, if we can show that the statement is true for |V (G)| = n, we will be done. Let r = |V (G)| 2 − |E(G)| be the number of missing arcs in G and let p be the number of pairs of vertices in G that are neither adjacent nor connected by a directed 2-path. Note that p ≤ r. Call those p pairs as identifiable pairs.
Let {v 1 , v 2 , . . . , v d } be the d distinct vertices among the identifiable pairs. Moreover, let d i be the number of vertices identifiable with v i for each i ∈ {1, 2, · · · , d}. We know that χ o (G, x) = n i=1 a i x (i) where a i is the number of tournaments on i vertices in L(G) and |V (G)| = n.
Observe that while constructing an associated (2, 3)-rooted tree T G of G we can stop in between in such a way that a leaf is either a tournament of order n or a tournament of order (n − 1) or an oriented graph of order (n − 2). Therefore we can express the homomorphic polynomial of G as follows: χ o (G, x) = a n x (n) + a n−1 x (n−1) + W where W is a sum of some homomorphic polynomials of oriented graphs of order (n − 2). Notice that the polynomial W has alternating signs due to the the induction hypothesis. Hence if we can show that the polynomial a n x (n) + a n−1 x (n−1) has alternating signs, then we will be done.
For this to be true, the absolute value of the ith term in a n x (n) should be greater than that of the (i − 1)th term in a n−1 x (n−1) . In other words we need to show that a n C n,i ≥ a n−1 C n−1,i−1 .
While constructing an associated (2, 3)-rooted tree T G of G, if v i is identified with another vertex to obtain the oriented graph G ′ , then the number of missing arcs in G ′ is at most r − d i . Let T G ′ be the subtree of T G rooted at G ′ and let L(G ′ ) be the multiset of the leaves in T G ′ . Thus, tournaments of order n − 1 in L(G ′ ) is at most 2 r−d i .
There are d i such vertices with which v i can identify. Hence the number of tournaments having (n − 1) vertices that has v i identified with another vertex is at most d i · 2 r−d i . Moreover, d i · 2 r−d i ≤ 2 r−1 .
As there are d distinct vertices among the identifiable pairs in G, the number of tournaments having (n − 1) vertices is at most d 2 · 2 r−1 . On the other hand, a n = 2 r by Theorem 5.1. So the ratio a n−1 an is at most d 4 . With this bound the eq n (2) is satisfied proving that a n x (n) + a n−1 x (n−1) has alternating signs. □

Conclusions
In this article we introduced and studied homomorphic polynomials for oriented graphs.
To state a few highlights of this work, we would like to mention that we found an interesting relation between homomorphic polynomials for independent sets and Stirling numbers of the second kind. We also showed that the polynomials possesses the alternating signs of the coefficients property. Moreover, we conjecture homomorphic polynomials to be unimodal.
While working through several examples, one question has peeped into our mind: is homomorphic polynomial a refinement of oriented chromatic polynomial? A precise restatement of the question is the following. We have already noted that the converse is false.
Recently, Beaton, Cox, Duffy, and Zolkavich [1] has introduced and studied chromatic polynomials for 2-edge-colored graphs. One may also consider the analogue of homomorphic polynomials in that set up. It can be an interesting task to generalize these notions for (m, n)-colored mixed graphs [12].