 # Chromatic Polynomials of Oriented Graphs

The oriented chromatic polynomial of a oriented graph outputs the number of oriented k-colourings for any input k. We fully classify those oriented graphs for which the oriented graph has the same chromatic polynomial as the underlying simple graph, closing an open problem posed by Sopena. We find that such oriented graphs admit a forbidden subgraph characterization, and such graphs can be both identified and constructed in polynomial time. We study the analytic properties of this polynomial and show that there exist oriented graphs which have chromatic polynomials have roots, including negative real roots, that cannot be realized as the root of any chromatic polynomial of a simple graph.

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## 1 The Oriented Chromatic Polynomial

Let be a mixed graph. That is, is a graph in which a subset (possibly empty) of the edges have been oriented to be arcs. We say that is an oriented colouring of when is an oriented colouring when is restricted to the arcs and a proper colouring when is restricted to the edges. Notice that if every pair of vertices is either adjacent or at the ends of a directed path of length , then every vertrex must receive a distinct colour in every colouring. We define the oriented chromatic polynomial of a mixed graph analogously to that of oriented graphs. In this section we observe that the oriented chromatic polynomial introduced by Sopena can be generalized as the oriented chromatic polynomial of mixed graphs. This generalization allows us to find an explicit expression for the third coefficient of the oriented chromatic polynomial of mixed graphs. As every oriented graph is a mixed graph with an empty arc set, this expression leads us to an explicit expression for the third coefficient of the chromatic polynomial of an oriented graph. The expression for the third coefficient provides us with a tool to study chromatically invariant oriented graphs.

We begin by providing a recursive formula for the oriented chromatic polynomial of mixed graphs. Let be a mixed graph. If every pair of vertices is either adjacent or at the ends of a directed path of length (a -dipath), then

 fo(G,λ)=i=n−1∏i=0(λ−i)

Otherwise, let and be a pair of vertices that are neither adjacent nor at the ends of a directed path of length two. In this case we have

 fo(G,λ)=fo(G+uv,λ)+fo(Guv,λ),

where

• is the mixed graph formed from by adding an edge between and ; and

• is the mixed graph formed from by identifying and into a single vertex, deleting all parallel arcs and edges, and deleting any edge that is parallel with an arc.

Following the usual convention of having the picture of a graph stand in for its polynomial, an example of this recursion is provided in Figure 2. Figure 2: Computing the oriented chromatic polynomial by way of reduction.

The correctness of this reduction follows from the proof of the reduction provided in . The appendix gives Maple code for generating the oriented chromatic polynomial of a mixed graph.

For a mixed graph , let be the set of pairs of vertices that are at the ends of an induced -dipath. In Figure 1 we have .

###### Theorem 1.1.

 For any mixed graph with vertices

1. is a polynomial of order in ;

2. the coefficient of is ;

3. has no constant term;

4. the coefficient of is ; and

5. if has an isolated vertex , then .

Let be an mixed graph with non-incident arcs . We say that the arcs are obstructing, when

1. and are not the ends of a -dipath;

2. and are not the ends of a -dipath and

3. .

Let denote the set of unordered pairs of obstructing arcs in , a mixed graph. In Figure 1 we have

 OG={{u1u2,u4u3},{u1u2,u4u5},{u1u2,u5u6},{u1u2,u7u6}}.

If or , then .

###### Proof.

Observe that every colouring of using colours is a colouring of . However if has an induced -dipath or a pair of obstructing arcs, then the converse does not hold. As such, there exists such that . ∎

Let be the coefficient of in .

###### Theorem 1.3.

For a mixed graph , we have

 c2(G,λ)=(|AG|+|DD|+|EG|2)−|TG|−|DG|−|OG|,

where is the set of induced subgraphs isomorphic to in .

###### Proof.

Let be a minimum counter-example with respect to number of vertices. Among all such counter examples, let be the one that maximizes . Note that we may further assume by adding an edge between every pair of vertices in . The resulting mixed graph has the same set of oriented colourings, and thus the same oriented chromatic polynomial.

The oriented chromatic polynomial of a mixed complete graph on is equal to the chromatic polynomial of a complete graph on vertices. The third coefficient of such a graph is given by . Therefore the claim holds for mixed complete graphs. As is a minimum counter example, is not a complete mixed graph. As such there exists such that and are not adjacent, nor at the ends of a -dipath. Therefore . By the choice of , the claim holds for both and .

Let be the set of common neighbours of and in . Each of these common neighbours forms a triangle in . And so . Further observe that in , the arcs/edges from and to a common neighbour becomes a single adjacency in . Therefore .

A pair of obstructing arcs in is not obstructing in if and only if and are the head and tail, in some order, of the pair of obstructing arcs. Let be the set of such arcs. No new obstructing arcs can be created by adding an edge. Therefore = . Also note that a pair of obstructing arcs in form a -dipath in with centre vertex . All other induced -dipaths in are retained in , as and are not the ends of a -dipath. Therefore .

 c2(G,λ)=c2(G+uv,λ)+c1(Guv,λ) (1)
 =(|AG|+|DG|+|EG|+12)−(|TG+uv|+|DG+uv|+|OG+uv|)
 −(|AGuv|+|EGuv|+|DGuv|)
 =(|AG|+|DG|+|EG|+12)−(|TG|−|C|+|DG|+|OG|−|OuvG|)
 −(|AG|+|EG|+|C|+|DG|+|OuvG|)
 =(|AG|+|DG|+|EG|+12)−(|AG|+|DG|+|EG|)−|TG|−|DG|−|OG|
 =(|AG|+|DG|+|EG|2)−|TG|−|DG|−|OG.|

Thus the claim holds for , contradicting the choice of as a minimum counter example. ∎

For the case , we arrive at the desired result for oriented graphs.

###### Corollary 1.4.

For an oriented graph , we have

 c2(G,λ)=(|AG|+|DG|2)−|TG|−|DG|−|OG|.

We further note that in the case , we arrive at the usual result for the third coefficient of the chromatic polynomial: .

## 2 Oriented Chromatic Equivalence

A folklore construction gives an orientation of so that the resulting oriented graph has chromatic number . This common example is used to convince the reader that the oriented chromatic number and the chromatic number of the underlying simple graph can be arbitrarily far apart. We note, however that the set of colourings of using colours is exactly that of colourings of using colours. And so though the chromatic number of differs greatly to , there is still a relationship between colourings of and colourings of some simple graph. In this section we find a set of sufficient conditions so that the -colourings of an oriented graph are exactly those of some simple graph . We conclude this section by using these sufficient conditions to compute the chromatic polynomial of orientations of stars.

We are interested in the following decision problems:

CHROM-INVAR
Instance: A graph .
Question: Is there an orientation such that ?

OCHROM-INVAR
Instance: An oriented graph .
Question: Does ?

OCHROM-EQUIV
Instance: An oriented graph .
Question: Is there a graph such that ?

Figure 3 gives an example of an oriented graph and a graph so that and are chromatically equivalent. Figure 3: An oriented graph and a graph with the same chromatic polynomial.

Let be an oriented graph. Let be the mixed graph resulting from by adding an edge between and whenever and are at the ends of an induced -dipath. We observe the following.

.

###### Proof.

is a subgraph of . Therefore . Every oriented colouring of using colours is also an oriented colouring of , therefore .

###### Theorem 2.2.

For , an oriented graph, if and only if .

###### Proof.

By Lemma 2.1 it suffices to show . Notice that has no obstructing arcs if and only if has no obstructing arcs. By Lemma 1.2 is suffices to show that if has no obstructing arcs, then every colouring of is an oriented colouring of . Let be a colouring of . Since is a proper colouring, if is not an oriented colouring of , then the second condition of oriented colouring as been violated. However, this not possible as has neither an induced -dipath nor a pair of obstructing arcs. ∎

Recall the result of Corollary 1.4. If has no pair of obstructing arcs, then in we have . And so . Notice that this is exactly the third coefficient of the chromatic polynomial of .

###### Corollary 2.3.

An oriented graph is chromatically invariant if and only if has no induced -dipath and is -free.

###### Proof.

Assume has no induced -dipath and that is -free. It follows directly that has no obstructing arcs and that . The conclusion follows by Theorem 2.2.

Let be a chromatically invariant oriented graph. Therefore every proper -colouring of is an oriented colouring of . Therefore has no induced -dipath, nor does contain a pair of obstructing arcs. Since contains no pair of obstructing arcs, if contains an induced copy of , say , then without loss of generality, there must be an induced -dipath between and . This contradicts that contains no induced -dipath. Therefore has no induced -dipath and is -free. ∎

Introduced by Ghouila-Houri, oriented graphs that contain no induced -dipath are called quasi-transitive.

###### Theorem 2.4.

 A graph admits a quasi-transitive orientation if and only if is a comparability graph.

Combining Corollary 2.3 and Theorem 2.4 yields the following results.

###### Theorem 2.5.

An oriented graph is chromatically invariant if and only if it is a quasi-transitive orientation of a -free comparability graph

Together, these two theorems fully classify chromatically invariant oriented graphs, as well as those graphs that admit a chromatically invariant orientation. This closes an open problem posed by Sopena in  As comparability graphs can be recognized in polynomial time , we arrive at the following classification of CHROM-INVAR and OCHROM-INVAR.

###### Corollary 2.6.

The decision problems CHROM-INVAR and OCHROM-INVAR are Polynomial.

We note that such an orientation of a -free comparability graph need not be unique (up to converse). There are many methods in the literature (for example see ) that give a quasi-transitive orientation of a comparability graph. A common element of these methods is the construction of an auxiliary graph, , so that a -colouring of corresponds to a quasi-transitive orientation. Such constructions imply that a comparability graph has a unique quasi-transitive orientation (up to converse) if and only if is connected. As such constructions can be carried out in polynomial time, we find that given a -free comparability graph , one may find in polynomial time an orientation of , , so that .

Comparability graphs yield a forbidden subgraph characterization . Let be the set of graphs given in Figure 4 together with . The set is the set of -free graphs that are also forbidden subgraphs of a comparability graph. By Theorem 2.5, together with are the family of forbidden subgraphs of those graphs that underlie chromatically invariant oriented graphs. Figure 4: Forbidden subgraphs of graphs underlying chromatically invariant oriented graphs.
###### Corollary 2.7.

A graph admits an chromatically invariant orientation if and only if is -free.

We now consider an application of Theorem 2.2 and find the oriented chromatic polynomial of the family of orientations of stars. Let be the orientation of a star on vertices, with centre vertex , so that has in-neigbours and out-neighbours.

###### Proof.

Observe that has no obstructing arcs. Further observe that consists of a copy of together with a universal vertex. By Theorem 2.2, we have .

Conversely, using the results from Section 1, one can find families of oriented graphs for which there is no chromatically equivalent graph. Let be an orientation of for some . Recalling the notation of the previous section we have

• ;

• ;

• ; and

• .

If there exists such that , then by Theorem 1.1 and Corollary 1.4, it must be that has vertices, edges and copies of . A simple counting argument implies that no such can exist.

###### Conjecture 2.9.

Let be an oriented graph. There is a graph such that and are chromatically equivalent if and only if has no obstructing pairs of arcs.

When restricted to oriented graphs that contain no pair of obstructing arcs, OCHROM-EQUIV is Polynomial – every instance is a YES instance. However, for arbitrary inputs is not clear if OCHROM-EQUIV is contained in NP, as constructing the chromatic polynomial of a graph is NP-hard. We conjecture, however, that those oriented graphs with no obstructing arcs are the only oriented graphs whose -colourings have a one-to-one correspondence with the -colourings of some graph . If Conjecture 2.9 is true, then OCHROM-EQUIV is Polynomial for arbitrary inputs.

Conjecture 2.9 is not true when we allow to be a mixed graph, even when we require . Let be the mixed graph formed from a pair of disjoint arcs by adding an edge between the heads and an edge between the tails. Using the reduction given in Section 1, we find . Notably, this is exactly the chromatic polynomial of the example, , given in Figure 2. The oriented graph has . Therefore . Further . And so by Theorem 2.2 we have . In particular, has a pair of obstructing arcs, but yet there is graph such that . From this example, one may generate an example on vertices for any by repeatedly adding universal vertices to and .

## 3 Roots of Oriented Chromatic Polynomials

The location of the roots of polynomials has been well studied for a variety of graph polynomials, such independence, domination, reliability and chromatic polynomials. In this section we provide a results regarding the roots of the oriented chromatic polynomial. Chromatic polynomials have roots that are dense in the complex plane . Their coefficients alternate in sign and hence have no negative real roots . We show the following:

###### Theorem 3.1.

For every integer , there exists an oriented graph so that has a root where .

###### Proof.

Let be the oriented graph on vertices that consists of a directed path with leaves, directed from to .

The oriented chromatic polynomial can be computed in the following way. Vertex has colour choices, has colour choices and has colour choices. Now for we have two options. If is the same colour as then the leaves have choices of colour, since it can not be the same colour at or . If is a different colour than the others in the path it has colour choices and the leaves have colour choices. This means

 fo(Dn,λ)=λ(λ−1)(λ−2)((λ−2)n−4+(λ−3)(λ−1)n−4)

The polynomial has real roots at . We show that we can obtain arbitrarily large negative real roots by showing that a real root exists between and for even.

Observe,

 fo(Dn,−n)=(−1)3n(n+1)(n+2)((−1)n−4(n+2)n−4+(−1)n−3(n+3)(n+1)n−4)

It can be shown that if

 (1+1n+1)n−4

The quantity is bounded above by , and for all values of , therefore for all even values of .

Now consider .

 fo(Dn,−ln(n)) = (−1)3(ln(n))(ln(n)+1)(ln(n)+2)∗ ((−1)n−4(ln(n)+2)n−4+(−1)n−3(ln(n)+3)(ln(n)+1)n−4)

Clearly when

 (1+1ln(n)+1)n−4>ln(n)+3.

Let . The derivative of this function is
.

It is the case that , thus there exists so that for all , is an increasing function, as the derivative of is positive and hence and for large values of .

It then follows by the intermediate value theorem that can have an arbitrarily large negative root. ∎

A chromatic polynomial cannot have root in the interval . We have shown that oriented chromatic polynomials can have negative real roots. In addition, there exist oriented chromatic polynomials in the interval , as has a root at . Open problems regarding the roots of oriented chromatic polynomials include: does there exist an oriented graph whose real roots lie in ? What is the closure of the complex roots for the oriented chromatic polynomial?

## 4 Conclusion

The study of oriented graphs often goes hand-in-hand with that of signed graphs. Though the methods contained herein will extend to the study of chromatic polynomials of signed graphs, there will be a marked difference in the classification of chromatically invariant signed graphs. For example, letting all the edges of be positive leads to a chromatically invariant signed graph. However, every possible orientation of leads to an oriented graph that is not chromatically invariant. Of course, with this approach, every graph can have edge signs trivially assigned so that the resulting signed graph has the same chromatic polynomial as the underlying graph. And so one may require that there is at least one edge of each sign. With this added restriction it is unclear if chromatically invariant signed graphs can be identified in polynomial time, as we expect the characterization to require that signs be given so that there is no where the edges have different signs. Similarly, the generalization of signed graphs and oriented graphs to -mixed graphs should yield a definition of a chromatic polynomial that obeys the reduction outlined in Section 1. Consequently we expect the results of Theorems 1.1 and 1.3 to generalize in the same manner. One may also ask, then, for which graphs is there an assignment of arcs, edges, and corresponding colours, so that the resulting -mixed graph is chromatically invariant. We have shown that for oriented graphs that must be a -free comparability graph.

## Acknowledgments

The authors thank Gary MacGillivray for discussions regarding the recursive construction of the oriented chromatic polynomial.

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## Appendix

See pages - of MapleCode