The chromatic polynomial for cycle graphs

TL;DR

This paper revisits the chromatic polynomial of cycle graphs, providing multiple proofs including the well-known inductive proof, to deepen understanding of its combinatorial properties.

Contribution

It offers three new proofs of the chromatic polynomial for cycle graphs, complementing the classic deletion-contraction recurrence proof.

Findings

Provides multiple proofs of the chromatic polynomial formula

Enhances understanding of cycle graph colorings

Reinforces the combinatorial significance of the polynomial

Abstract

Let $P(G,\lambda)$ denote the number of proper vertex colorings of $G$ with $\lambda$ colors. The chromatic polynomial $P(C_n,\lambda)$ for the cycle graph $C_n$ is well-known as $$P(C_n,\lambda) = (\lambda-1)^n+(-1)^n(\lambda-1)$$ for all positive integers $n\ge 1$. Also its inductive proof is widely well-known by the \emph{deletion-contraction recurrence}. In this paper, we give this inductive proof again and three other proofs of this formula of the chromatic polynomial for the cycle graph $C_n$.

Equations

$$P(C_n,\lambda )=(\lambda -1{)}^n+(-1{)}^n(\lambda -1)$$

$$P(G,\lambda )=P(G-e,\lambda )-P(G/e,\lambda ),$$

$$P(G,\lambda )$$

$$P(G-e,\lambda )$$

$$P(G/e,\lambda )$$

$$P(C_n,\lambda )=(\lambda -1{)}^n+(-1{)}^n(\lambda -1)$$

$$P(C_1,\lambda )$$

$$P(C_2,\lambda )$$

$$P(C_3,\lambda )$$

$$P(P_n,\lambda )=\lambda (\lambda -1{)}^{n-1}.$$

$$P(C_{n+1},\lambda )$$

$$=P(P_{n+1},\lambda )-P(C_n,\lambda )$$

$$=\lambda (\lambda -1{)}^n-\left((\lambda -1{)}^n+(-1{)}^n(\lambda -1)\right)$$

$$=(\lambda -1{)}^{n+1}+(-1{)}^{n+1}(\lambda -1).$$

$$\left|\bigcap _{i=1}^n\overline{A_i}\right|$$

$$=\left|U\right|-\sum _{i=1}^n\left|A_i\right|+\sum _{i<j}\left|A_i\cap A_j\right|-\cdots +(-1{)}^n\left|A_1\cap \cdots \cap A_n\right|$$

$$P(C_n,\lambda )$$

$$={\lambda }^n-\left(\genfrac{}{}{0ex}{}{n}{1}\right){\lambda }^{n-1}+\left(\genfrac{}{}{0ex}{}{n}{2}\right){\lambda }^{n-2}+\cdots +(-1{)}^{n-1}(\genfrac{}{}{0ex}{}{n}{n-1})\lambda +(-1{)}^n\lambda$$

$$=(\lambda -1{)}^n-(-1{)}^n+(-1{)}^n\lambda$$

$$=(\lambda -1{)}^n+(-1{)}^n(\lambda -1).$$

$$A=\left(a_{ij}\right)=\left(\begin{array}{ccccc}0& 1& \cdots & 1& 1\\ 1& 0& \cdots & 1& 1\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 1& 1& \cdots & 0& 1\\ 1& 1& \cdots & 1& 0\end{array}\right),$$

$$A=\left(\begin{array}{ccccc}0& 1& \cdots & 1& 1\\ 1& 0& \cdots & 1& 1\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 1& 1& \cdots & 0& 1\\ 1& 1& \cdots & 1& 0\end{array}\right)\sim \left(\begin{array}{ccccc}\lambda -1& 0& \cdots & 0& 0\\ 0& -1& \cdots & 0& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& \cdots & -1& 0\\ 0& 0& \cdots & 0& -1\end{array}\right),$$

$$tr(A^n)=\sum _{i=1}^{\lambda }{u}_i^n=(\lambda -1{)}^n+\underset{\text{λ−1 times}}{\underbrace{(-1{)}^n+\cdots +(-1{)}^n}}.$$

$${\sigma }_i=\{\begin{array}{ll}{\omega }_i-1,& \text{mbox}\text{if ωi is a cyclic descent}\\ {\omega }_i,& \text{mbox}\text{otherwise}.\end{array}$$

$$\phi (\omega )=\sigma =(1,1,1,2,2,2,1,3,1)\in [3{]}^9.$$

$${\sigma }_1={\sigma }_2=\cdots ={\sigma }_n$$

$${\overline{\sigma }}_i=\{\begin{array}{ll}{\sigma }_i+1,& \text{mbox}\text{if σi is a cyclic descent}\\ {\sigma }_i,& \text{mbox}\text{otherwise}.\end{array}$$

$${\overline{\sigma }}_i={\overline{\sigma }}_{i+1}=\cdots ={\overline{\sigma }}_{i+\ell -1}\ne {\overline{\sigma }}_{i+\ell },$$

$$k=2:$$

$$k=4:$$

$$k=9:$$

$$\psi (\sigma )=\omega =(1,2,1,3,2,3,1,4,2)\in X_9.$$

$$\omega$$

$$P(C_n,\lambda )$$

$$=\left[(\lambda -1{)}^n-(\lambda -1)\right]+2(\lambda -1).$$

$$P(C_n,\lambda )$$

$$=\left[(\lambda -1{)}^n-(\lambda -1)\right]+0.$$

Full text

[TABLE]

for all positive integers $n\geq 1$. Also its inductive proof is widely well-known by the deletion-contraction recurrence. In this paper, we give this inductive proof again and three other proofs of this formula of the chromatic polynomial for the cycle graph $C_{n}$.

$\dagger$ Corresponding author. This work was supported by the National Research Foundation of Korea(NRF) grant funded by the Korea government(MSIP) (No. 2017R1C1B2008269).

1. Introduction

The number of proper colorings of a graph with finite colors was introduced only for planar graphs by George David Birkhoff [Bir13] in 1912, in an attempt to prove the four color theorem, where the formula for this number was later called by the chromatic polynomial. In 1932, Hassler Whitney [Whi32] generalized Birkhoff’s formula from the planar graphs to general graphs. In 1968, Ronald Cedric Read [Rea68] introduced the concept of chromatically equivalent graphs and asked which polynomials are the chromatic polynomials of some graph, that remains open.

Chromatic polynomial

For a graph $G$, a coloring means almost always a (proper) vertex coloring, which is a labeling of vertices of $G$ with colors such that no two adjacent vertices have the same colors. Let $P(G,\lambda)$ denote the number of (proper) vertex colorings of $G$ with $\lambda$ colors and $\chi(G)$ the least number $\lambda$ satisfying $P(G,\lambda)>0$, where $P(G,\lambda)$ and $\chi(G)$ are called a chromatic polynomial and chromatic number of $G$, respectively.

In fact, it is clear that the number of $\lambda$-colorings is a polynomial in $\lambda$ from a deletion-contraction recurrence.

Proposition 1** (Deletion-contraction recurrence).**

For a given a graph $G$ and an edge $e$ in $G$, we have

[TABLE]

where $G-e$ is a graph obtained by deletion the edge $e$ and $G/e$ is a graph obtained by contraction the edge $e$.

Example*.*

The chromatic polynomials of graphs in Figure 1 are

[TABLE]

It is confirmed that (1) is true for the graph $G$ and the edge $e$ in Figure 1.

Cycle graph

A cycle graph $C_{n}$ is a graph that consists of a single cycle of length $n$, which could be drown by a $n$-polygonal graph in a plane. The chromatic polynomial for cycle graph $C_{n}$ is well-known as follows.

Theorem 2**.**

For a positive integer $n\geq 1$, the chromatic polynomial for cycle graph $C_{n}$ is

[TABLE]

Example*.*

For an integer $n\leq 3$, it is easily checked that the chromatic polynomials of $C_{n}$ are from (2) as follows.

[TABLE]

As shown in Figure 2, the cycle graph $C_{1}$ is a graph with one vertex and one loop and $C_{1}$ cannot be colored, that means $P(C_{1},\lambda)=0$. The cycle graph $C_{2}$ is a graph with two vertices, where two edges between two vertices, and $C_{2}$ can have colorings by assigning two vertices with different colors, that means $P(C_{2},\lambda)=\lambda(\lambda-1)$. The cycle graph $C_{3}$ is drawn by a triangle and $C_{3}$ can have colorings by assigning all three vertices with different colors, that means $P(C_{3},\lambda)=\lambda(\lambda-1)(\lambda-2)$.

2. Four proofs of Theorem 2

In this section, we show the formula (2) in four different ways.

2.1. Inductive proof

This inductive proof is widely well-known. A path graph $P_{n}$ is a connected graph in which $n-1$ edges connect $n$ vertices of vertex degree at most $2$, which could be drawn on a single straight line. The chromatic polynomial for path graph $P_{n}$ is easily obtained by coloring all vertices $v_{1},\dots,v_{n}$ where $v_{i}$ and $v_{i+1}$ have different colors for $i=1,\dots,n-1$.

Lemma 3**.**

For a positive integer $n\geq 1$, the chromatic polynomial for path graph $P_{n}$ is

[TABLE]

We use an induction on the number $n$ of vertices by the deletion-contraction recurrence and the above lemma for path graph: It is already shown that (2) is true for $n\leq 3$ by the example in Section 1. Assume that (2) is true for a positive integer $n$. Using (1) and (3), we have

[TABLE]

Thus, (2) is true for all positive integers $n\geq 1$.

2.2. Proof by inclusion-exclusion principle

The inclusion-exclusion principle is a technique of counting the size of the union of finite sets.

Proposition 4** (Inclusion-exclusion principle).**

Let $A_{1},A_{2},\dots,A_{n}$ be subsets of a finite set $U$. Then number of elements excluding their union is as follows

[TABLE]

where $\overline{A}$ is the complement of $A$ in $U$.

Considering every condition to assign different colors to two adjacent vertices, for each edge $e$, we define a finite sets of arbitrary (including improper) colorings to assign same color to two adjacent vertices by the edge $e$.

Let $A_{i}$ be a set of colorings such that two vertices $v_{i}$ and $v_{i+1}$ are of same color, where $v_{n+1}$ is regarded as $v_{1}$. Applying the inclusion-exclusion principle, we can write the following

[TABLE]

Thus, (2) is true for all positive integers $n\geq 1$.

2.3. Algebric proof

Let us consider a case of $n=5$ and $\lambda=4$, that is, to assign the vertices of $C_{5}$ in four colors: red, blue, yellow, and green. Also let us consider a complete graph $K_{4}$ with vertex names red, blue, yellow, and green, see Figure 4.

When red-blue-red-yellow-green is assigned in order from the vertex $v_{1}$ to the vertex $v_{5}$ in $C_{5}$, it is corresponding to a closed walk of length $5$ in $K_{4}$ which begins and ends at red, that is, it is red-blue-red-yellow-green-red in $K_{4}$. By generalizing it, we have a correspondence between $\lambda$-colorings of $C_{n}$ and closed walks of length $n$ in $K_{\lambda}$. By this correspondence, it is enough to count the number of closed walks of length $n$ in $K_{\lambda}$, instead of the number of $\lambda$-colorings of $C_{n}$.

For a graph $G$ with vertex set $\left\{v_{1},\dots,v_{n}\right\}$, the adjacency matrix of $G$ is an $n\times n$ square matrix $A$ such that its element $A_{ij}$ is one when there is an edge between two vertices $v_{i}$ and $v_{j}$, and zero when there is no edge between $v_{i}$ and $v_{j}$.

The following related to an adjacency matrix is well-known.

Proposition 5**.**

Let $A$ be the adjacency matrix of the graph $G$ on $n$ vertices $v_{1},\dots,v_{n}$. Then the $(i,j)$th entry of the matrix $A^{n}$ is the number of the walk of length $n$ beginning at $v_{i}$ and ending at $v_{j}$.

By Proposition 5, we can calculate the number of closed walk of length $n$ in the complete graph $K_{\lambda}$: Let $A$ be an adjacency matrix of $K_{\lambda}$. Then $A$ is a $\lambda\times\lambda$ matrix as follows

[TABLE]

where $a_{ij}=0$ if $i=j$, and otherwise $a_{ij}=1$. So the number of closed walks of length $n$ in $K_{\lambda}$ is enumerated by $tr(A^{n})$, which equals the sum of all eigenvalues of $A^{n}$. Also let all eigenvalues of the matrix $A$ be denoted by $u_{1},\dots,u_{\lambda}$, then all eigenvalues of the matrix $A^{n}$ are $u_{1}^{n},\dots,u_{\lambda}^{n}$.

[TABLE]

Since the matrix $A$ have $\lambda$ eigenvalues $u_{1}=\lambda-1$ and $u_{2}=\dots=u_{\lambda}=-1$, we have

[TABLE]

Thus, (2) is true for all positive integers $n\geq 1$.

2.4. Bijective proof

Let $X_{n}$ denote the set of $\lambda$-colorings of $C_{n}$ and $[\lambda-1]^{n}$ be the set of $n$-tuples of positive integers less than $\lambda$, where $[\lambda-1]$ means $\left\{1,\dots,\lambda-1\right\}$. We consider a mapping $\varphi$ from $\lambda$-colorings of $C_{n}$ in $X_{n}$ to $n$-tuples in $[\lambda-1]^{n}$.

A mapping $\varphi$ from $X_{n}$ to $[\lambda-1]^{n}$

The mapping $\varphi:X_{n}\to[\lambda-1]^{n}$ is defined as follows: Let $\omega$ be a $\lambda$-coloring of $C_{n}$ in $X_{n}$, we write $\omega=(\omega_{1},\dots,\omega_{n})$ where $\omega_{i}$ is the color of $v_{i}$ in $C_{n}$ and it is obvious that $\omega_{i}\neq\omega_{i+1}$ for $1\leq i\leq\lambda$, where $\omega_{n+1}$ is regarded as $\omega_{1}$. An entry $\omega_{i}$ is called a cyclic descent of $C$ if $\omega_{i}>\omega_{i+1}$ for $1\leq i\leq\lambda$. Then we define $\varphi(\omega)=\sigma=(\sigma_{1},\dots,\sigma_{n})$ with

[TABLE]

Given a $\lambda$-coloring $\omega$, if $\omega_{i}=\lambda$ then $\omega_{i+1}<\lambda$, so $\omega_{i}=\lambda$ should be a cyclic descent. Thus we have $\sigma_{i}<\lambda$ for all $1\leq i\leq n$ and $\varphi(\omega)$ belongs to $[\lambda-1]^{n}$.

For example, in a case of $n=9$ and $\lambda=4$, $\omega=(1,2,1,3,2,3,1,4,2)\in X_{9}$ is given as an example of $4$-colorings of $C_{9}$. Here $\omega_{2}=2$, $\omega_{4}=3$, $\omega_{6}=3$, $\omega_{8}=4$, and $\omega_{9}=2$ are cyclic descents of $\omega$. So we have

[TABLE]

A mapping $\psi$ as the inverse of $\varphi$

Let $Z_{n}$ be the set of $n$-tuples $\sigma=(\sigma_{1},\sigma_{2},\dots,\sigma_{n})$ in $[\lambda-1]^{n}$ with

[TABLE]

and it is obvious that the size of $Z_{n}$ is $\lambda-1$.

We would like to describe a mapping $\psi:\left([\lambda-1]^{n}\setminus Z_{n}\right)\to X_{n}$ in order to satisfy $\varphi\circ\psi$ is the identity on $[\lambda-1]^{n}\setminus Z_{n}$ as follows: Given a $\sigma\in[\lambda-1]^{n}\setminus Z_{n}$, we define $\overline{\sigma}=(\overline{\sigma}_{1},\dots,\overline{\sigma}_{n})$ with

[TABLE]

Since $\overline{\sigma}$ may have consecutive same entries, we define $\psi(\sigma)=\omega=(\omega_{1},\dots,\omega_{n})$ from $\overline{\sigma}$ with $\omega_{i}=\overline{\sigma}_{i}+1$ for any entry $\overline{\sigma}_{i}$ of $\overline{\sigma}$ with a finite positive even integer $\ell$ satisfying

[TABLE]

where $\overline{\sigma}_{n+k}$ is regarded as $\overline{\sigma}_{k}$ for $1\leq k\leq n$, and $\omega_{i}=\overline{\sigma}_{i}$, otherwise. Thus $\omega$ has no consecutive same entries and $1\leq\omega_{i}\leq\lambda$ for all $1\leq i\leq n$, so $\psi(\sigma)=\omega$ belongs to $X_{n}$. Moreover, it is obvious that $\sigma_{i}\leq\omega_{i}\leq\sigma_{i}+1$ for all $1\leq i\leq n$ and if $\omega_{i}=\sigma_{i}+1$ for some $1\leq i\leq n$ then $\omega_{i}$ is a cyclic descent in $\omega$. Hence $\varphi(\omega)=\sigma$ and $\sigma\in[\lambda-1]^{n}\setminus Z_{n}$ if and only if $\psi(\sigma)=\omega$.

In a previous example, $\sigma=(1,1,1,2,2,2,1,3,1)$ is denoted as an example of $9$-tuples in $[3]^{9}$. Here $\sigma_{6}=2$, $\sigma_{8}=3$ are cyclic descents of $\sigma$ and we obtain $\overline{\sigma}=(1,1,1,2,2,3,1,4,1)$. And then there exist only three entries $\overline{\sigma}_{2}$, $\overline{\sigma}_{4}$, and $\overline{\sigma}_{9}$ in $\overline{\sigma}$ satisfying the following

[TABLE]

so we get $\omega_{2}=\overline{\sigma}_{2}+1=2$, $\omega_{4}=\overline{\sigma}_{4}+1=3$, $\omega_{9}=\overline{\sigma}_{9}+1=2$, and

[TABLE]

Let $Y_{n}$ be the set of $\lambda$-colorings $\omega$ in $X_{n}$ with $\varphi(\omega)\in Z_{n}$. Since two mapping $\varphi$ and $\psi$ are bijections between $X_{n}\setminus Y_{n}$ and $[\lambda-1]^{n}\setminus Z_{n}$, the size of the set $X_{n}\setminus Y_{n}$ is same with the size of the $[\lambda-1]^{n}\setminus Z_{n}$, which is equal to $(\lambda-1)^{n}-(\lambda-1).$

When $n$ is even, for any $1\leq i\leq\lambda-1$, there exist only two $n$-tuples in $X_{n}$

[TABLE]

satisfying $\varphi(\omega)=(i,i,\dots,i)\in Z_{n}$. If $n$ is even, the size of $Y_{n}$ is equal to $2(\lambda-1)$ and we obtain

[TABLE]

When $n$ is odd, there is no $n$-tuples satisfying $\varphi(\omega)\in Z_{n}$ and the set $Y_{n}$ is empty. If $n$ is odd, we obtain

[TABLE]

Therefore, (2) yields from (4) and (5) for all positive integers $n\geq 1$.