Non-chromatic-Adherence of the DP Color Function via Generalized Theta Graphs

Abstract

DP-coloring (also called correspondence coloring) is a generalization of list coloring that has been widely studied in recent years after its introduction by Dvořák and Postle in 2015. The chromatic polynomial of a graph is an extensively studied notion in combinatorics since its introduction by Birkhoff in 1912; denoted P(G, m), it equals the number of proper m-colorings of graph G. Counting function analogues of the chromatic polynomial have been introduced and studied for list colorings: $P_{\ell }$, the list color function (1990); DP colorings: $P_{DP}$, the DP color function (2019), and $P^*_{DP}$, the dual DP color function (2021). For any graph G and $m \in \mathbb {N}$, $P_{DP}(G, m) \le P_\ell (G,m) \le P(G,m) \le P_{DP}^*(G,m)$. A function f is chromatic-adherent if for every graph G, $f(G,a) = P(G,a)$ for some $a \ge \chi (G)$ implies that $f(G,m) = P(G,m)$ for all $m \ge a$. It is not known if the list color function and the DP color function are chromatic-adherent. We show that the DP color function is not chromatic-adherent by studying the DP color function of Generalized Theta graphs. The tools we develop along with the Rearrangement Inequality give a new method for determining the DP color function of all Theta graphs and the dual DP color function of all Generalized Theta graphs.

Appendices

A Proof of Lemma 12

In this section, we give the details of the remaining cases of this proof.

Continuing with (i), suppose $\varvec{x}_1$ is odd, $\varvec{x}_2$ is even, and $\varvec{x}_3$ is even. Then we have the following:

Continuing with (ii), we suppose $\varvec{x}_1$ is odd, $\varvec{x}_2$ is even, and $\varvec{x}_3$ is odd. Then we have:

Continuing with (iii), suppose $\varvec{x}_1$, $\varvec{x}_2$, and $\varvec{x}_3$ are all odd. Then we have the following:

$$\begin{aligned} x_{1, h_1(j)} = {\left\{ \begin{array}{ll} n_1 &{} \text {if } j \in [m] \\ n_1 + 1 &{} \text {if } j \in [m^2] - [m], \end{array}\right. } \end{aligned}$$

$$\begin{aligned} x_{2, h_2(j)} = {\left\{ \begin{array}{ll} n_2 &{} \text {if } j \in [2m] - [m] \\ n_2 + 1 &{} \text {if } j \in [m] \cup \left( [m^2] - [2m] \right) , \end{array}\right. } \end{aligned}$$

$$\begin{aligned} x_{3, h_3(j)} = {\left\{ \begin{array}{ll} n_3 &{} \text {if } j \in [3m] - [2m] \\ n_3 + 1 &{} \text {if } j \in [2m] \cup \left( [m^2] - [3m] \right) , \end{array}\right. } \\ x_{1, j} = {\left\{ \begin{array}{ll} n_1 &{} \text {if } j \in [m] \\ n_1 + 1 &{} \text {if } j \in [m^2] - [m], \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} x_{2, f(j)} = {\left\{ \begin{array}{ll} n_2 + 1 &{} \text {if } j \in [m(m - 1)] \\ n_2 &{} \text {if } j \in [m^2] - [m(m - 1)]. \end{array}\right. } \end{aligned}$$

We also have $N = m$, and so

$$\begin{aligned} x_{3, g(j)} = {\left\{ \begin{array}{ll} n_3 + 1 &{} \text {if } j \in [m] \cup \left( [m^2] - [2m] \right) \\ n_3 &{} \text {if } j \in [2m] - [m]. \end{array}\right. } \end{aligned}$$

Therefore, we obtain

$$\begin{aligned}&\sum _{j = 1}^{m^2} x_{1, h_1(j)} x_{2, h_2(j)} x_{3, h_3(j)} \\&\quad = \sum _{j = 1}^m n_1(n_2 + 1)(n_3 + 1) + \sum _{j = m + 1}^{2m} (n_1 + 1)n_2(n_3 + 1) + \sum _{j = 2m + 1}^{3m} (n_1 + 1)(n_2 + 1)n_3 \\&\qquad + \sum _{j = 3m + 1}^{m^2} (n_1 + 1)(n_2 + 1)(n_3 + 1) \\&\quad = \sum _{j = 1}^m n_1(n_2 + 1)(n_3 + 1) + \sum _{j = m + 1}^{2m} (n_1 + 1)(n_2 + 1)n_3 \\&\quad + \sum _{j = 2m + 1}^{m(m - 1)} (n_1 + 1)(n_2 + 1)(n_3 + 1) + \sum _{j = m(m - 1) + 1}^{m^2} (n_1 + 1)n_2(n_3 + 1) \\&\quad = \sum _{j = 1}^{m^2} x_{1, j} x_{2, f(j)} x_{3, g(j)}. \end{aligned}$$

B Proof of Lemma 13

We complete the proof of the claim that for each $k \in [3]$,

$$\begin{aligned} s_k = \frac{(m - 1)^{l_k} - (-1)^{l_k}}{m} + (-1)^{l_k} = \frac{(m - 1)^{l_k} + (-1)^{l_k}(m - 1)}{m} \end{aligned}$$

and

$$\begin{aligned} o_k = \frac{(m - 1)^{l_k} + (-1)^{l_k + 1}}{m} \end{aligned}$$

in the case $l_k$ is odd. We have $s_{k, (i, j)} = -1 + ((m - 1)^{l_k} - (-1)^{l_k})/m$ for exactly m choices of $(i, j) \in [m]^2$, and $s_{k, (i, j)} = ((m - 1)^{l_k} - (-1)^{l_k})/m$ for the remaining $m(m - 1)$ choices of $(i, j) \in [m]^2$. Therefore, we have $n_k = -1 + ((m - 1)^{l_k} - (-1)^{l_k})/m$, and

$$\begin{aligned} x_{k, \ell } = {\left\{ \begin{array}{ll} n_k &{} \text {if } \ell \in [m] \\ n_k + 1 &{} \text {if } \ell \in [m^2] - [m]. \end{array}\right. } \end{aligned}$$

It follows by definition that $\varvec{x}_k$ is odd, so that

$$\begin{aligned} s_k = n_k = \frac{(m - 1)^{l_k} - (-1)^{l_k}}{m} + (-1)^{l_k} \end{aligned}$$

and

$$\begin{aligned} o_k = n_k + 1 = \frac{(m - 1)^{l_k} - (-1)^{l_k}}{m}. \end{aligned}$$