text:
\subsection*{#1}
  \lstinputlisting{#2}
  \vspace{2em}
}

\begin{document}

\title{Non-chromatic-adherence of the DP Color Function via Generalized Theta Graphs}

\author{Manh Vu Bui$^1$, Hemanshu Kaul$^2$, Michael Maxfield$^1$, Jeffrey A. Mudrock$^1$, \\ Paul Shin$^1$, and Seth Thomason$^1$}

\footnotetext[1]{Department of Mathematics, College of Lake County, Grayslake, IL 60030.  E-mail:  {\tt {jmudrock@clcillinois.edu}}}

\footnotetext[2]{Department of Applied Mathematics, Illinois Institute of Technology, Chicago, IL 60616. E-mail: {\tt {kaul@iit.edu}}}
\maketitle

\begin{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 Dvo\v{r}\'{a}k 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 $m \in \mathbb{N}$, $P_{DP}(G, m) \leq P_\ell(G,m) \leq P(G,m) \leq P_{DP}^*(G,m)$.  A function $f$ is chromatic-adherent if for every graph $G$, $f(G,a) = P(G,a)$ for some $a \geq \chi(G)$ implies that $f(G,m) = P(G,m)$ for all $m \geq a$. 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.
\medskip

\noindent {\bf Keywords.} DP-coloring, correspondence coloring, chromatic polynomial, DP color function, rearrangement inequality.

\noindent \textbf{Mathematics Subject Classification.} 05C15, 05C30, 05A20, 05C69

\end{abstract}

\section{Introduction}\label{intro}

In this paper all graphs are nonempty, finite, simple graphs unless otherwise noted.  Generally speaking we follow West~\cite{W01} for terminology and notation.  The set of natural numbers is $\mathbb{N} = \{1,2,3, \ldots \}$.  For $m \in \mathbb{N}$, we write $[m]$ for the set $\{1, \ldots, m \}$.  We write \emph{AM-GM Inequality} for the inequality of arithmetic and geometric means.  Given a set $A$, $\mathcal{P}(A)$ is the power set of $A$.  If $G$ is a graph and $S, U \subseteq V(G)$, we use $G[S]$ for the subgraph of $G$ induced by $S$, and we use $E_G(S, U)$ to denote the subset of $E(G[S \cup U])$ with at least one endpoint in $S$ and at least one endpoint in $U$.  If $u$ and $v$ are adjacent in $G$, $uv$ or $vu$ refers to the edge between $u$ and $v$.  If $e \in E(G)$, we write $G \cdot e$ for the graph obtained from $G$ by contracting the edge $e$.

In this paper, we will use the following version of the Rearrangement Inequality (\cite{R52}) several times.
\begin{thm} [Rearrangement Inequality] \label{thm: RI}
For $k,n \in \mathbb{N}$, suppose that for each $(i,j) \in [k] \times [n]$, $x_{(i,j)}$ is a nonnegative real number so that $x_{(i,1)} \leq \cdots \leq x_{(i,n)}$ for each $i \in [k]$.  Let $\sigma_i$ be an arbitrary permutation of $[n]$ for each $i \in [k]$.  If $k=2$, then
$$\sum_{j=1}^n x_{(1, n+1-j)} x_{(2,j)} \leq \sum_{j=1}^n x_{(1, \sigma_1(j))} x_{(2,\sigma_2(j))}.$$
Furthermore, for any $k \in \mathbb{N}$,
$ \sum_{j=1}^n \prod_{i=1}^k x_{(i, \sigma_i(j))} \leq \sum_{j=1}^n \prod_{i=1}^k x_{(i, j)}.$
\end{thm}
Importantly, the first inequality in Theorem~\ref{thm: RI} doesn't hold when $k \geq 3$ (see e.g.,~\cite{W20}).  Throughout this paper whenever $l < j$, we take $\prod_{i=j}^l a_i$ to equal 1.

\subsection{DP-Coloring} \label{basic}

In the classical vertex coloring problem we wish to color the vertices of a graph $G$ with up to $m$ colors from $[m]$ so that adjacent vertices receive different colors, a so-called \emph{proper $m$-coloring}. The chromatic number of a graph $G$, denoted $\chi(G)$, is the smallest $m$ such that $G$ has a proper $m$-coloring.  List coloring is a well-known variation on classical vertex coloring which was introduced independently by Vizing~\cite{V76} and Erd\H{o}s, Rubin, and Taylor~\cite{ET79} in the 1970s.  For list coloring, we associate a \emph{list assignment} $L$ with a graph $G$ such that each vertex $v \in V(G)$ is assigned a list of available colors $L(v)$ (we say $L$ is a list assignment for $G$).  Then, $G$ is \emph{$L$-colorable} if there exists a proper coloring $f$ of $G$ such that $f(v) \in L(v)$ for each $v \in V(G)$ (we refer to $f$ as a \emph{proper $L$-coloring} of $G$).  A list assignment $L$ is called a \emph{$k$-assignment} for $G$ if $|L(v)|=k$ for each $v \in V(G)$.  The \emph{list chromatic number} of a graph $G$, denoted $\chi_\ell(G)$, is the smallest $k$ such that $G$ is $L$-colorable whenever $L$ is a $k$-assignment for $G$.  We say $G$ is \emph{$k$-choosable} if $k \geq \chi_\ell(G)$.  Note $\chi(G) \leq \chi_\ell(G)$, and this inequality may be strict since it is known that there are bipartite graphs with arbitrarily large list chromatic number (see~\cite{ET79}).

In 2015, Dvo\v{r}\'{a}k and Postle~\cite{DP15} introduced a generalization of list coloring called DP-coloring (they called it correspondence coloring) in order to prove that every planar graph without cycles of lengths 4 to 8 is 3-choosable. DP-coloring has been extensively studied  over the past 6 years (see e.g.,~\cite{B17, BK182, KM20, KO18, LL19, LLYY19, Mo18, M18}).  Intuitively, DP-coloring is a variation on list coloring where each vertex in the graph still gets a list of colors, but identification of which colors are different can change from edge to edge.  Following~\cite{BK17}, we now give the formal definition. Suppose $G$ is a graph.   A \emph{cover} of $G$ is a pair $\mathcal{H} = (L,H)$ consisting of a graph $H$ and a function $L: V(G) \rightarrow \mathcal{P}(V(H))$ satisfying the following four requirements:

\vspace{5mm}

\noindent(1) the set $\{L(u) : u \in V(G) \}$ is a partition of $V(H)$ of size $|V(G)|$; \\
(2) for every $u \in V(G)$, the graph $H[L(u)]$ is complete; \\
(3) if $E_H(L(u),L(v))$ is nonempty, then $u=v$ or $uv \in E(G)$; \\
(4) if $uv \in E(G)$, then $E_H(L(u),L(v))$ is a matching (the matching may be empty).

\vspace{5mm}

Suppose $\mathcal{H} = (L,H)$ is a cover of $G$.  We refer to the edges of $H$ connecting distinct parts of the partition $\{L(u) : u \in V(G) \}$ as \emph{cross-edges}.  An \emph{$\mathcal{H}$-coloring} of $G$ is an independent set in $H$ of size $|V(G)|$.  It is immediately clear that an independent set $I \subseteq V(H)$ is an $\mathcal{H}$-coloring of $G$ if and only if $|I \cap L(u)|=1$ for each $u \in V(G)$.  We say $\mathcal{H}$ is \emph{$m$-fold} if $|L(u)|=m$ for each $u \in V(G)$.  An $m$-fold cover $\mathcal{H}$ is a \emph{full cover} if for each $uv \in E(G)$, the matching $E_{H}(L(u),L(v))$ is perfect.  The \emph{DP-chromatic number} of $G$, $\chi_{DP}(G)$, is the smallest $m \in \mathbb{N}$ such that $G$ has an $\mathcal{H}$-coloring whenever $\mathcal{H}$ is an $m$-fold cover of $G$.

Suppose $\mathcal{H} = (L,H)$ is an $m$-fold cover of $G$.  We say that $\mathcal{H}$ has a \emph{canonical labeling} if it is possible to name the vertices of $H$ so that $L(u) = \{ (u,j) : j \in [m] \}$ and $(u,j)(v,j) \in E(H)$ for each $j \in [m]$ whenever $uv \in E(G)$.~\footnote{When $\mathcal{H}=(L,H)$ has a canonical labeling, we will always refer to the vertices of $H$ using this naming scheme.}  Now, suppose $\mathcal{H}$ has a canonical labeling and $G$ has a proper $m$-coloring.  Then, if $\mathcal{I}$ is the set of $\mathcal{H}$-colorings of $G$ and $\mathcal{C}$ is the set of proper $m$-colorings of $G$, the function $f: \mathcal{C} \rightarrow \mathcal{I}$ given by $f(c) = \{ (v, c(v)) : v \in V(G) \}$ is a bijection.  Also, given an $m$-assignment $L$ for a graph $G$, it is easy to construct an $m$-fold cover $\mathcal{H}'$ of $G$ such that $G$ has an $\mathcal{H}'$-coloring if and only if $G$ has a proper $L$-coloring (see~\cite{BK17}).  So, $\chi(G) \leq \chi_\ell(G) \leq \chi_{DP}(G)$.

\subsection{The DP Color Function and Dual DP Color Function}

In 1912, Birkhoff introduced the chromatic polynomial of a graph in hopes of using it to make progress on the four color problem.  For $m \in \mathbb{N}$, the \emph{chromatic polynomial} of a graph $G$, $P(G,m)$, is the number of proper $m$-colorings of $G$. It is well-known that $P(G,m)$ is a polynomial in $m$ of degree $|V(G)|$ (see~\cite{B12}).  For example, $P(K_n,m) = \prod_{i=0}^{n-1} (m-i)$, $P(C_n,m) = (m-1)^n + (-1)^n (m-1)$ whenever~\footnote{When considering graphs with multiple edges or loops, one should note that this formula for the chromatic polynomial of a cycle also works for $C_1$ and $C_2$.} $n \geq 3$, and $P(T,m) = m(m-1)^{n-1}$ whenever $T$ is a tree on $n$ vertices (see~\cite{W01}).

The notion of chromatic polynomial was extended to list coloring in the early 1990s. If $L$ is a list assignment for $G$, we use $P(G,L)$ to denote the number of proper $L$-colorings of $G$. The \emph{list color function} $P_\ell(G,m)$ is the minimum value of $P(G,L)$ where the minimum is taken over all possible $m$-assignments $L$ for $G$.  It is clear that $P_\ell(G,m) \leq P(G,m)$ for each $m \in \mathbb{N}$ since we must consider the $m$-assignment that assigns the same $m$ colors to all the vertices in $G$ when considering all possible $m$-assignments for $G$.  In general, the list color function can differ significantly from the chromatic polynomial for small values of $m$.  However, for large values of $m$, Wang, Qian, and Yan~\cite{WQ17} (improving upon results in~\cite{D92} and~\cite{T09}) showed:
If $G$ is a connected graph with $l$ edges, then $P_{\ell}(G,m)=P(G,m)$ whenever $m > (l-1)/\ln(1+ \sqrt{2})$.  It is also known that $P_{\ell}(G,m)=P(G,m)$ for all $m \in \mathbb{N}$ when $G$ is a cycle or chordal (see~\cite{KN16} and~\cite{AS90}).

A fundamental open question on the list color function asks whether the list color function of a graph and the corresponding chromatic polynomial stay the same after the first point at which they are both nonzero and equal.  Let $\mathcal{G}$ be the set of all finite, simple graphs.  We say a function $f: \mathcal{G} \times \mathbb{N} \rightarrow \mathbb{N}$ is \emph{chromatic-adherent} if for every graph $G$,  $f(G,a) = P(G,a)$ for some $a \geq \chi(G)$ implies that $f(G,m) = P(G,m)$ for all $m \geq a$.

\begin{ques} [\cite{KN16}] \label{ques: funlist}
Is $P_{\ell}$ chromatic-adherent?
\end{ques}

In 2019, two of the authors (Kaul and Mudrock in~\cite{KM19}) introduced a DP-coloring analogue of the chromatic polynomial in hopes of gaining a better understanding of DP-coloring and using it as a tool for making progress on some open questions related to the list color function~\cite{KM19}.  Since its introduction in 2019, the DP color function has received some attention in the literature (see e.g.,~\cite{BH21, DY21, HK21, KM21, M21, MT20}).   Suppose $\mathcal{H} = (L,H)$ is a cover of graph $G$.  Let $P_{DP}(G, \mathcal{H})$ be the number of $\mathcal{H}$-colorings of $G$.  Then, the \emph{DP color function} of $G$, $P_{DP}(G,m)$, is the minimum value of $P_{DP}(G, \mathcal{H})$ where the minimum is taken over all possible $m$-fold covers $\mathcal{H}$ of $G$.  A similar tool for studying enumerative aspects of DP coloring was recently introduced~\cite{M21}; specifically, the \emph{dual DP color function} of $G$, denoted $P^*_{DP}(G,m)$, is the maximum value of $P_{DP}(G, \mathcal{H})$ where the maximum is taken over all full $m$-fold covers $\mathcal{H}$ of $G$.~\footnote{We take $\mathbb{N}$ to be the domain of the DP color function and dual DP color function of any graph.} It is easy to show that for any graph $G$ and $m \in \mathbb{N}$,
$$P_{DP}(G, m) \leq P_\ell(G,m) \leq P(G,m) \leq P_{DP}^*(G,m).$$
Note that if $G$ is a disconnected graph with components: $H_1, H_2, \ldots, H_t$, then $P_{DP}(G, m) = \prod_{i=1}^t P_{DP}(H_i,m)$ (an analogous result holds for the dual DP color function).  So, we will only consider connected graphs from this point forward unless otherwise noted.

The list color function and DP color function of certain graphs behave similarly.  However, for some graphs there are surprising differences.  For example, similar to the list color function,  $P_{DP}(G,m) = P(G,m)$ for every $m \in \mathbb{N}$  whenever $G$ is chordal or an odd cycle~\cite{KM19}.  On the other hand, unlike the list color function, it is well-known that $P_{DP}(G,m)$ does not necessarily equal $P(G,m)$ for sufficiently large $m$.  Indeed Dong and Yang~\cite{DY21} recently generalized a result of Kaul and Mudrock~\cite{KM19} and showed that if $G$ is a simple graph that contains an edge $e$ such that the length of a shortest cycle containing $e$ is even, then there exists an $N \in \mathbb{N}$ such that $P_{DP}(G,m) < P(G,m)$ whenever $m \geq N$.  Nevertheless, while introducing the DP color function, Kaul and Mudrock asked the analogue of Question~\ref{ques: funlist} for the DP color function.
\begin{ques} [\cite{KM19}] \label{ques: funDP}
Is $P_{DP}$ chromatic-adherent?
\end{ques}
We will see below that the answer to Question~\ref{ques: funDP} is no.

\subsection{Summary of Results}

We answer Question~\ref{ques: funDP} by studying the DP color function of Generalized Theta graphs.   A \emph{Generalized Theta graph} $\Theta(l_1, \ldots, l_n)$ consists of a pair of end vertices joined by $n$ internally disjoint paths of lengths $l_1, \ldots, l_n \in \mathbb{N}$. When $n=3$, $\Theta(l_1, l_2, l_3)$ is simply called a \emph{Theta graph}.  From this point forward, we will always assume that the lengths of the paths of a Generalized Theta graph are listed such that: if $l_1, \ldots, l_n$ don't all have the same parity, then there exists some $r \in [n] - \{1\}$ such that $l_2, \ldots, l_r$ have the same parity which is different from that of $l_1, l_{r + 1}, \ldots, l_n$.

It is easy to prove that $\chi_{DP}(\Theta(l_1, \ldots, l_n))=3$ whenever $n \geq 2$.  It is also well-known (see~\cite{BH01}) that if $G = \Theta(l_1, \ldots, l_n)$, then for each $m \geq 2$,
\[P(G,m) = \frac{\prod_{i=1}^{n}((m-1)^{l_i+1}+(-1)^{l_i+1}(m-1))}{(m(m-1))^{n-1}} + \frac{\prod_{i=1}^n((m-1)^{l_i}+(-1)^{l_i}(m-1))}{m^{n-1}}.\]
Generalized Theta graphs have been widely studied for many graph theoretic problems (see e.g.,~\cite{BF16, CM15, ET79, LB16, LB19, MM21, SJ18}), and they are the main subject of two classical papers on the chromatic polynomial~\cite{BH01} and~\cite{S04} which include the celebrated result that the zeros of the chromatic polynomials of the Generalized Theta graphs are dense in the whole complex plane with the possible exception of the unit disc around the origin (by including the join of Generalized Theta graphs with $K_2$ this extends to all of the complex plane). Recently, exact formulas for the DP color function of all Theta graphs were determined, and it was shown that when $G = \Theta(l_1, \ldots, l_n)$, there is a polynomial $p(m)$ and $N \in \mathbb{N}$ such that $P_{DP}(G,m) = p(m)$ whenever $m \geq N$ (see~\cite{HK21}).

In Section~\ref{counter} we develop some elementary tools for analyzing the DP color function of a Generalized Theta graph.  These tools allow us to establish a sufficient condition for $P_{DP}(G,m)=P(G,m)$ when $G$ is a Generalized Theta graph.  This sufficient condition ultimately allows us to find two examples of graphs that demonstrate the answer to Question~\ref{ques: funDP} is no.
\begin{thm} \label{thm: negative}
If $G$ is $\Theta(2,3,3,3,2)$ or $\Theta(2,3,3,3,3,3,2,2)$, then $P_{DP}(G,3)=P(G,3)$ and there is an $N \in \mathbb{N}$ such that $P_{DP}(G,m) < P(G,m)$ for all $m \geq N$.
\end{thm}
Interestingly, Theorem~\ref{thm: negative} contains the only examples that we know of that demonstrate that the answer to Question~\ref{ques: funDP} is no.  So, the following question is natural.
\begin{ques} \label{ques: open}
For which graphs $G$ do there exist, $a,b \in \mathbb{N}$ with $\chi(G) \leq a < b$, $P_{DP}(G,a) = P(G,a)$, and $P_{DP}(G,b) < P(G,b)$?
\end{ques}
We do not even know whether there are infinitely many Generalized Theta graphs that satisfy the conditions of Question~\ref{ques: open}.  Since little is known about the enumerative aspects of DP coloring Generalized Theta graphs, we continue by studying the DP color function and dual DP color function of Generalized Theta graphs.

In Section~\ref{theta} we show how the tools we developed in Section~\ref{counter} along with the Rearrangement Inequality give us an elementary way to derive the formulas for the DP color function of all Theta graphs.
\begin{thm}\label{thm: generalTheta3}
Let $G = \Theta(l_1,l_2,l_3)$, where $l_1 = \min_{i \in [3]} l_i \geq 1$ and $l_i \geq 2$ for each $i \in \{2, 3\}$.

(i) If the parity of $l_1$ is different from both $l_2$ and $l_3$, then $P_{DP}(G, m) = P(G, m)$ for all $m \in \mathbb{N}$.

(ii) If the parity of $l_1$ is the same as $l_3$ and different from $l_2$, then for $m \geq 2$:

\noindent $\displaystyle P_{DP}(G, m) = \frac{1}{m} \left[ (m - 1)^{l_1 + l_2 + l_3} + (m - 1)^{l_1} - (m - 1)^{l_2} - (m - 1)^{l_3 + 1} + (-1)^{l_2 + 1}(m - 2) \right]$.

(iii) If $l_1$, $l_2$, and $l_3$ all have the same parity, then for $m \geq 3$:

\noindent $\displaystyle P_{DP}(G, m) = \frac{1}{m} \left[ (m - 1)^{l_1 + l_2 + l_3} - (m - 1)^{l_1} - (m - 1)^{l_2} - (m - 1)^{l_3} + 2(-1)^{l_1 + l_2 + l_3} \right]$.
\end{thm}

Finally, in Section~\ref{dual} we build on some of the ideas in Section~\ref{theta} and completely determine the dual DP color function of all Generalized Theta graphs.

\section{The DP Color Function is not Chromatic-adherent} \label{counter}

We begin this Section by establishing some conventions and notation that will be used for the remainder of this paper.  Whenever $\mathcal{H} = (L,H)$ is an $m$-fold cover of $G$ and $P \subseteq V(H)$, we let $N(P,\mathcal{H})$ be the number of $\mathcal{H}$-colorings containing $P \subseteq V(H)$.  Suppose $G=\Theta(l_1,\ldots,l_n)$.  We will always assume $n \geq 2$, $l_1 = \min_{i \in [n]} l_i \geq 1$, and $l_i \geq 2$ for each $i \in [n] - \{1\}$.  The endpoints of the paths that make up $G$ will always be called $u$ and $w$. When $\mathcal{H} = (L,H)$ is a full $m$-fold cover of $G$, we always suppose $L(v) = \{(v, j) : j \in [m]\}$ for each $v \in V(G)$. Furthermore, we let $G_i$ be the $u,w$-path of length $l_i$ used to form $G$ and $R_i = V(G_i)$. We let $\mathcal{H}_i = (L_i,H_i)$ where $L_i$ is $L$ with domain restricted to $R_i$ and $H_i$ is the graph defined by $H[\bigcup_{v \in R_i} L(v)] - E_H(L(u), L(w))$ if $l_1 = 1$ and $i \neq 1$, and $H[\bigcup_{v \in R_i} L(v)]$ otherwise. It is easy to see that $\mathcal{H}_i$ is a full $m$-fold cover of $G_i$.

We are now ready to present two lemmas that will be of fundamental importance throughout the paper.
\begin{lem}\label{lem: genThetaCount}
Let $G=\Theta(l_1,\ldots,l_n)$. Let $\mathcal{H} = (L,H)$ be a full $m$-fold cover of $G$ where $m \geq 2$. Then $$P_{DP}(G,\mathcal{H}) = \sum_{(i,j) \in [m]^2} \prod_{k=1}^{n} N(\{(u,i),(w,j)\},\mathcal{H}_k).$$
\end{lem}

\begin{proof}
Notice $P_{DP}(G,\mathcal{H}) = \sum_{(i,j) \in [m]^2} N(\{(u,i),(w,j)\},\mathcal{H})$. Let $A = \{(u,j_1),(w,j_2)\}$ where $j_1,j_2$ are fixed elements of $[m]$. We will show that $N(A,\mathcal{H}) = \prod_{i = 1}^n N(A,\mathcal{H}_i)$. Notice that if $(u,j_1)(w,j_2) \in E(H)$, then $N(A,\mathcal{H}) = N(A,\mathcal{H}_1) = 0$. So we can assume that $(u,j_1)(w,j_2) \notin E(H)$. Let $\mathcal{I}_i$ be the set of all $\mathcal{H}_i$-colorings of $G_i$ that contain $A$, and let $\mathcal{I}$ be the set of all $\mathcal{H}$-colorings of $G$ that contain $A$. If $\mathcal{I}_k$ is empty for some $k \in [n]$, then it is easy to see that $\mathcal{I}$ must also be empty; hence, $\prod_{i = 1}^n N(A, \mathcal{H}_i) = \prod_{i = 1}^n \lvert \mathcal{I}_i \rvert = 0 = \lvert \mathcal{I} \rvert = N(A, \mathcal{H})$. So, suppose that $\mathcal{I}_i$ is nonempty for each $i \in [n]$. Let $f: \prod_{i=1}^{n} \mathcal{I}_i \rightarrow \mathcal{I}$ be the function given by $f((I_1,\ldots,I_n)) = \bigcup_{i=1}^{n} I_i$. It is easy to check that $\bigcup_{i=1}^{n} I_i$ is an independent set of size $|V(G)|$ in $H$. Clearly, $f$ is a bijection. As such, $N(A,\mathcal{H}) = |\mathcal{I}| = \prod_{i=1}^{n} |\mathcal{I}_i| = \prod_{i=1}^{n} N(A,\mathcal{H}_i)$.
\end{proof}

\begin{lem}\label{lem: binaryThetaValues}
Let $G = \Theta(l_1,\ldots,l_n)$. Let $\mathcal{H} = (L,H)$ be a full $m$-fold cover of $G$ where $m \geq 2$. Let $(i,j) \in [m]^2$. For $k \in [n]$, if there is a path in $H_k$ connecting $(u,i)$ and $(w,j)$ consisting of only cross-edges of $\mathcal{H}_k$, then $$N(\{(u,i),(w,j)\},\mathcal{H}_k) = \frac{(m-1)^{l_k}+(-1)^{l_k}(m-1)}{m}.$$ Otherwise, $$N(\{(u,i),(w,j)\},\mathcal{H}_k) = \frac{(m-1)^{l_k}-(-1)^{l_k}}{m}.$$
\end{lem}

\begin{proof}
For $k \in [n]$, since $G_k$ is a tree, $\mathcal{H}_k$ has a canonical labeling. Let $r_k$ and $r_k'$ be the permutations of $[m]$ that produce a canonical labeling when $(u,i)$ is renamed to $(u,r_k(i))$ and $(w,j)$ is renamed to $(w,r_k'(j))$. Notice $r_k(i) = r_k'(j)$ if there is a path in $H_k$ connecting $(u,i)$ and $(w,j)$ consisting of only cross-edges of $\mathcal{H}_k$, and $r_k(i) \not= r_k'(j)$ otherwise. If $l_1 = 1$, let $e$ be an edge with endpoints $u$ and $w$ that is distinct from the edge $uw \in E(G_1)$; otherwise, let $e = uw$. For each $k \in [n]$, let $M_k$ be the graph with $V(M_k) = V(G_k)$ and $E(M_k) = E(G_k) \cup \{e\}$. Let $M_k' = M_k \cdot e$, where we do not remove multiple edges or loops upon contraction and where the vertex obtained from contracting $e$ is $v$. Notice $M_k = C_{l_k+1}$ and $M_k' = C_{l_k}$. When there is a path in $H_k$ connecting $(u,i)$ and $(w,j)$ consisting of only cross-edges of $\mathcal{H}_k$, $N(\{(u,i),(w,j)\},\mathcal{H}_k)$ is the number of proper $m$-colorings of $M_k'$ that color $v$ with $r_k(i)$. Thus, $N(\{(u,i),(w,j)\},\mathcal{H}_k) = P(C_{l_k},m)/m = ((m-1)^{l_k}+(-1)^{l_k}(m-1))/m$. When there is not a path in $H_k$ connecting $(u,i)$ and $(w,j)$ consisting of only cross-edges of $\mathcal{H}_k$, $N(\{(u,i),(w,j)\},\mathcal{H}_k)$ is the number of proper $m$-colorings of $M_k$ that color $u$ with $r_k(i)$ and $w$ with $r_k'(j)$. Thus, $N(\{(u,i),(w,j)\},\mathcal{H}_k) = P(C_{l_k+1},m)/(m(m-1)) = ((m-1)^{l_k}-(-1)^{l_k})/m$.
\end{proof}

Lemma~\ref{lem: binaryThetaValues} implies the possible values of $N(\{(u,i),(w,j)\},\mathcal{H}_k)$ are one apart; specifically, the possible values are: $\left((m-1)^{l_k}-(-1)^{l_k}\right)/m + (-1)^{l_k}$ and $\left((m-1)^{l_k}-(-1)^{l_k}\right)/m$.  Also, Lemmas~\ref{lem: genThetaCount} and~\ref{lem: binaryThetaValues} allow us to establish a sufficient condition for $P_{DP}(G,m)=P(G,m)$ when $G$ is a Generalized Theta graph.

\begin{lem} \label{lem: construct}
Let $G = \Theta(l_1, \ldots, l_n)$.  For any $m \geq 3$, if

$$P(G,m) =  \left \lceil m^2 \prod_{i=1}^n \left( \left(\frac{(m-1)^{l_i}+(-1)^{l_i}(m-1)}{(m-1)^{l_i} - (-1)^{l_i}} \right)^{1/m} \left(\frac{(m-1)^{l_i}-(-1)^{l_i}}{m} \right) \right)  \right \rceil,$$

then $P_{DP}(G,m) = P(G,m)$.
\end{lem}

\begin{proof}
For some $m \geq 3$, suppose that $\mathcal{H} = (L,H)$ is a full $m$-fold cover of $G$ such that $P_{DP}(G, \mathcal{H}) = P_{DP}(G,m)$.  By Lemma~\ref{lem: genThetaCount}, we know that \\ $P_{DP}(G,\mathcal{H}) = \sum_{(i,j) \in [m]^2} \prod_{k=1}^{n} N(\{(u,i),(w,j)\},\mathcal{H}_k)$. By the AM-GM inequality and the fact that $P_{DP}(G,m)$ is an integer, it follows that
$$P_{DP}(G,m) \geq \left \lceil m^2  \left(\prod_{k=1}^n \prod_{(i,j) \in [m]^2}  N(\{(u,i),(w,j)\},\mathcal{H}_k) \right)^{1/m^2} \right \rceil.$$
Now, suppose $l \in [n]$.  Let $H'_l$ be the spanning subgraph of $H_l$ that consists of only the cross-edges of $\mathcal{H}_l$.  Since $\mathcal{H}_l$ has a canonical labeling, we know that $H'_l$ is the disjoint union of $m$ paths of length $l_l$.  Consequently, by Lemma~\ref{lem: binaryThetaValues}, $N(\{(u,i),(w,j)\},\mathcal{H}_l) = ((m-1)^{l}+(-1)^{l}(m-1))/m$ for precisely $m$ ordered pairs $(i,j)$ in $[m]^2$, and $N(\{(u,i),(w,j)\},\mathcal{H}_l) = ((m-1)^{l}-(-1)^{l})/m$ for all other ordered pairs in $[m]^2$.  Thus,
\begin{align*}
&\left \lceil m^2  \left(\prod_{k=1}^n \prod_{(i,j) \in [m]^2}  N(\{(u,i),(w,j)\},\mathcal{H}_k) \right)^{1/m^2} \right \rceil \\
&= \left \lceil m^2 \left(\prod_{i=1}^n \left( \left(\frac{(m-1)^{l_i}+(-1)^{l_i}(m-1)}{m} \right)^m \left(\frac{(m-1)^{l_i}-(-1)^{l_i}}{m} \right)^{m(m-1)} \right) \right)^{1/m^2} \right \rceil.
\end{align*}
So, if the hypotheses of the Lemma are satisfied, we have $P_{DP}(G,m) \geq P(G,m)$, and the result follows.
\end{proof}

The final ingredient for the proof of Theorem~\ref{thm: negative} is a result from~\cite{HK21}~\footnote{Theorem~\ref{thm: generalized} is also implied by the main results in~\cite{DY21}.}.

\begin{thm} [\cite{HK21}] \label{thm: generalized}
	Suppose $G = \Theta(l_1, \ldots, l_n)$.

	\textit{(i)}  If there is a $j \in \{2, \ldots, n \}$ such that $l_1$ and $l_j$ have the same parity, then there is an $N \in \mathbb{N}$ such that $P_{DP}(G,m) < P(G,m)$ for all $m \geq N$.

	\textit{(ii)}  If $l_1$ and $l_j$ have different parity for each $j \in \{2, \ldots, n \}$, then there is an $N \in \mathbb{N}$ such that $P_{DP}(G,m) = P(G,m)$ for all $m \geq N$.
\end{thm}

We are now ready to prove Theorem~\ref{thm: negative}.

\begin{proof}
We will prove the result when $G= \Theta(2,3,3,3,2)$ (the proof in the other case is similar).  First, note that $P(G,3) = (6^2)(18^3)/6^4 + (6^2)(6^3)/3^4=258$.  Next, notice that if $(l_1, l_2, l_3, l_4, l_5) = (2,3,3,3,2)$, then
$$\left \lceil 9 \prod_{i=1}^5 \left( \left(\frac{(2)^{l_i}+(-1)^{l_i}(2)}{(2)^{l_i} - (-1)^{l_i}} \right)^{1/3} \left(\frac{(2)^{l_i}-(-1)^{l_i}}{3} \right) \right)  \right \rceil = \left \lceil 9 \left(\frac{6^5}{3^2 \cdot 9^3} \right)^{1/3} \left ( \frac{3^2 \cdot 9^3}{3^5} \right )   \right \rceil = 258.$$
So, by Lemma~\ref{lem: construct}, we have that $P_{DP}(G,3)=P(G,3)$.  Also, Theorem~\ref{thm: generalized} implies there is an $N \in \mathbb{N}$ such that $P_{DP}(G,m) < P(G,m)$ for all $m \geq N$.
\end{proof}

\section{Theta Graphs and the Rearrangement Inequality} \label{theta}

In this Section we show how we can use the Rearrangement Inequality along with the tools developed in Section~\ref{counter} to give an elementary proof of Theorem~\ref{thm: generalTheta3}.  We begin with a technical lemma that follows from the Rearrangement Inequality by an extremality argument.

\begin{lem}\label{lem: 3ListRearrangement}
Suppose $x_{i,j}$ is a non-negative integer for each $i \in [3]$ and $j \in [m^2]$ where $m \geq 3$. For each $i \in [3]$, suppose $n_i = x_{i,1} = \cdots = x_{i,m} \leq x_{i,m+1} = \cdots = x_{i,m(m-1)} \leq x_{i,m(m-1)+1} = \cdots = x_{i,m^2} = n_i + 1$ for some $n_i \geq 0$, and $n_1 \leq n_2,n_3$. Let $N = \lvert \{j \in [m^2] : x_{1, j} = n_1\} \rvert \in \{m, m(m - 1)\}$.
Suppose $f$ is a permutation of $[m^2]$ defined by $f(j) = m^2 + 1 - j$ and $g$ is a permutation of $[m^2]$ defined by
\[ g(j) = \begin{cases}
    m^2 + j - N & \text{if } j \in [N]\\
    j - N & \text{otherwise.}
   \end{cases}
\]
For any permutations $\sigma_1$ and $\sigma_2$ of $[m^2]$, $$\sum_{j=1}^{m^2} x_{1,j}x_{2,f(j)}x_{3,g(j)} \leq \sum_{j=1}^{m^2} x_{1,j}x_{2,\sigma_1(j)}x_{3,\sigma_2(j)}.$$
\end{lem

\begin{proof}
For any permutations $\sigma_1$ and $\sigma_2$ of $[m^2]$, let $M(\sigma_1,\sigma_2) = \lvert\{q\in[m^2]: x_{1,q}=n_1, x_{2,\sigma_1(q)}=n_2\}\rvert + \lvert\{q\in[m^2]: x_{1,q}=n_1, x_{3,\sigma_2(q)}=n_3\}\rvert$.

Among all permutations of $[m^2]$, $\sigma_1$ and $\sigma_2$, that minimize $\sum_{j=1}^{m^2} \left(x_{1,j}x_{2,\sigma_1(j)}x_{3,\sigma_2(j)}\right)$, choose $\gamma_1$ and $\gamma_2$ such that $M(\gamma_1,\gamma_2)$ is as small as possible.

Suppose there exist $a,b \in [m^2]$ such that $x_{1,a} = n_1$, $x_{1,b} = n_1+1$, $x_{2,\gamma_1(a)} = n_2$, and $x_{2,\gamma_1(b)} = n_2+1$. Then let $\gamma_1'$ be defined such that $\gamma_1'(a) = \gamma_1(b)$, $\gamma_1'(b) = \gamma_1(a)$, and $\gamma_1'$ equals $\gamma_1$ otherwise. We calculate
\begin{align*}
    &\sum_{j=1}^{m^2} \left(x_{1,j}x_{2,\gamma_1(j)}x_{3,\gamma_2(j)}\right) - \sum_{j=1}^{m^2} \left(x_{1,j}x_{2,\gamma_1'(j)}x_{3,\gamma_2(j)}\right)\\
    &= x_{1,a}x_{3,\gamma_2(a)}(x_{2,\gamma_1(a)} - x_{2,\gamma_1'(a)}) + x_{1,b}x_{3,\gamma_2(b)}(x_{2,\gamma_1(b)} - x_{2,\gamma_1'(b)})\\
    &= n_1 x_{3,\gamma_2(a)}(n_2 - n_2 - 1) + (n_1 + 1) x_{3,\gamma_2(b)}(n_2 + 1 - n_2)\\
    &= (n_1 + 1) x_{3,\gamma_2(b)} - n_1 x_{3,\gamma_2(a)} = n_1(x_{3,\gamma_2(b)} - x_{3,\gamma_2(a)}) + x_{3,\gamma_2(b)}\\
    &\geq n_1(n_3 - n_3 - 1) + n_3 \geq 0.
\end{align*}

By the choice of $\gamma_1$ and $\gamma_2$, $\sum_{j=1}^{m^2} \left(x_{1,j}x_{2,\gamma_1(j)}x_{3,\gamma_2(j)}\right) - \sum_{j=1}^{m^2} \left(x_{1,j}x_{2,\gamma_1'(j)}x_{3,\gamma_2(j)}\right)=0$. However, $M(\gamma_1',\gamma_2) = M(\gamma_1,\gamma_2)-1$ which is a contradiction.  Since a similar contradiction can be reached if we assume there are $a,b \in [m^2]$ such that $x_{1,a} = n_1$, $x_{1,b} = n_1+1$, $x_{3,\gamma_2(a)} = n_3$, and $x_{3,\gamma_2(b)} = n_3+1$, we know that if $N=m$ and $x_{1,a} = n_1$, then $x_{2,\gamma_1(a)} = n_2+1$ and $x_{3,\gamma_2(a)} = n_3 + 1$.  Similarly, we know that if $N=m(m-1)$ and $x_{2,\gamma_1(a)} = n_2+1$ or $x_{3,\gamma_2(a)} = n_3 + 1$, then $x_{1,a} = n_1$.  We will now prove the desired result when $N=m$ and when $N=m(m-1)$.

Suppose $N=m$.  Without loss of generality, assume $\gamma_1([m]) = \gamma_2([m]) = [m^2] - [m(m-1)]$. Then
\begin{align*}
    & \sum_{j=1}^{m^2} x_{1,j}x_{2,\gamma_1(j)}x_{3,\gamma_2(j)}
    = \left(\sum_{j=1}^{m} n_1(n_2 + 1)(n_3 + 1)\right) + \left(\sum_{j=m+1}^{m^2} x_{1,j}x_{2,\gamma_1(j)}x_{3,\gamma_2(j)}\right)\\
    &= \left(\sum_{j=1}^{m} n_1(n_2 + 1)(n_3 + 1)\right) + (n_1 + 1)\left(\sum_{j=m+1}^{m^2} x_{2,\gamma_1(j)}x_{3,\gamma_2(j)}\right)\\
    &= mn_1(n_2 + 1)(n_3 + 1) + (n_1 + 1)\left(\sum_{j=m+1}^{m^2} x_{2,\gamma_1(j)}x_{3,\gamma_2(j)}\right)\\
    &\geq mn_1(n_2 + 1)(n_3 + 1) + (n_1 + 1)\left(\sum_{j=1}^{m(m-1)} x_{2,m(m-1)+1-j}x_{3,j}\right) \text{ (Rearrangement Inequality)}\\
    &= mn_1(n_2 + 1)(n_3 + 1) + (n_1 + 1)\sum_{j=m+1}^{m^2} x_{2,f(j)}x_{3,g(j)} = \sum_{j=1}^{m^2} x_{1,j}x_{2,f(j)}x_{3,g(j)}
\end{align*}
as desired.

Now, suppose $N=m(m-1)$.  We know that if  $x_{1,a} = n_1 + 1$, then $x_{2,\gamma_1(a)} = n_2$ and $x_{3,\gamma_2(a)} = n_3$. Without loss of generality, assume $\gamma_1([m^2] - [m(m-1)]) = \gamma_2([m^2] - [m(m-1)]) = [m]$. Then a calculation along the lines of the previous case shows that $\sum_{j=1}^{m(m-1)} x_{1,j}x_{2,\gamma_1(j)}x_{3,\gamma_2(j)} + \sum_{j=m(m-1)+1}^{m^2} (n_1 + 1)n_2n_3 \ge n_1\left(\sum_{j=1}^{m(m-1)} x_{2,f(j)}x_{3,g(j)}\right) + m(n_1 + 1)n_2n_3$, which is the desired inequality.
\begin{comment}
\begin{align*}
    & \sum_{j=1}^{m^2} x_{1,j}x_{2,\gamma_1(j)}x_{3,\gamma_2(j)}\\
    &= \left(\sum_{j=1}^{m(m-1)} x_{1,j}x_{2,\gamma_1(j)}x_{3,\gamma_2(j)}\right) + \left(\sum_{j=m(m-1)+1}^{m^2} (n_1 + 1)n_2n_3\right)\\
    &= n_1\left(\sum_{j=1}^{m(m-1)} x_{2,\gamma_1(j)}x_{3,\gamma_2(j)}\right) + m(n_1 + 1)n_2n_3\\
    &\geq n_1\left(\sum_{j=1}^{m(m-1)} x_{2,m^2+1-j}x_{3,j+m}\right) + m(n_1 + 1)n_2n_3 \text{ (Rearrangement Inequality)}\\
    &= n_1\left(\sum_{j=1}^{m(m-1)} x_{2,f(j)}x_{3,g(j)}\right) + m(n_1 + 1)n_2n_3\\
    &= \sum_{j=1}^{m^2} x_{1,j}x_{2,f(j)}x_{3,g(j)}
\end{align*}
as desired.
\end{comment}
\end{proof}

We now introduce some terminology that will be useful in proving Theorem~\ref{thm: generalTheta3}.  Suppose that $\bm{x} = (x_1, \ldots, x_{m^2})$ satisfies $n = x_{1} = \cdots = x_{m} \leq x_{m+1} = \cdots = x_{m(m-1)} \leq x_{m(m-1)+1} = \cdots = x_{m^2} = n + 1$ for some $n \geq 0$. We say $\bm{x}$ is \emph{odd} if exactly $m$ of its coordinates are $n$, and we say $\bm{x}$ is \emph{even} if exactly $m$ of its coordinates are $n + 1$. These terms allow us to refer to the \emph{parity} of $\bm{x}$.

The following reformulation of Lemma~\ref{lem: 3ListRearrangement} will be useful in proving Theorem~\ref{thm: generalTheta3}.

\begin{lem} \label{lem: special}
Suppose $x_{i,j}$ is a non-negative integer for each $i \in [3]$ and $j \in [m^2]$ where $m \geq 3$. For each $i \in [3]$, suppose $n_i = x_{i,1} = \cdots = x_{i,m} \leq x_{i,m+1} = \cdots = x_{i,m(m-1)} \leq x_{i,m(m-1)+1} = \cdots = x_{i,m^2} = n_i + 1$ for some $n_i \geq 0$, and $n_1 \leq n_2,n_3$. Let $\bm{x}_i = (x_{i, 1}, \ldots, x_{i, m^2})$ for each $i \in [3]$. For each $i \in [3]$, if $\bm{x}_i$ is odd, let $s_i = n_i$ and $o_i = n_i + 1$, and if $\bm{x}_i$ is even, let $s_i = n_i + 1$ and $o_i = n_i$. Let $h_1$ be a permutation of $[m^2]$ such that $x_{1, h_1(j)} = s_1$ whenever $j \in [m]$ and $x_{1, h_1(j)} = o_1$ otherwise. We now define two more permutations $h_2$ and $h_3$ of $[m^2]$.

(i) If the parity of $\bm{x}_1$ is different from both $\bm{x}_2$ and $\bm{x}_3$, then for each $i \in \{2, 3\}$, let $h_i$ be a permutation of $[m^2]$ such that $x_{i, h_i(j)} = s_i$ whenever $j \in [m]$ and $x_{i, h_i(j)} = o_i$ otherwise.

(ii) If the parity of $\bm{x}_1$ is different from $\bm{x}_2$ and the same as $\bm{x}_3$, then let $h_2$ be a permutation of $[m^2]$ such that $x_{2, h_2(j)} = s_2$ whenever $j \in [m]$ and $x_{2, h_2(j)} = o_2$ otherwise, and let $h_3$ be a permutation of $[m^2]$ such that $x_{3, h_3(j)} = s_3$ whenever $j \in [2m] - [m]$ and $x_{3, h_3(j)} = o_3$ otherwise.

(iii) If $\bm{x}_1$, $\bm{x}_2$, and $\bm{x}_3$ have the same parity, then let $h_2$ be a permutation of $[m^2]$ such that $x_{2, h_2(j)} = s_2$ whenever $j \in [2m] - [m]$ and $x_{2, h_2(j)} = o_2$ otherwise, and let $h_3$ be a permutation of $[m^2]$ such that $x_{3, h_3(j)} = s_3$ whenever $j \in [3m] - [2m]$ and $x_{3, h_3(j)} = o_3$ otherwise.

Then, using the notation of Lemma~\ref{lem: 3ListRearrangement}, $$\sum_{j = 1}^{m^2} x_{1, h_1(j)} x_{2, h_2(j)} x_{3, h_3(j)} = \sum_{j=1}^{m^2} x_{1,j}x_{2,f(j)}x_{3,g(j)}.$$
In particular, for any permutations $\sigma_1, \sigma_2$ of $[m^2]$, $$ \sum_{j = 1}^{m^2} x_{1, h_1(j)} x_{2, h_2(j)} x_{3, h_3(j)} \leq \sum_{j = 1}^{m^2} x_{1, j} x_{2, \sigma_1(j)} x_{3, \sigma_2(j)}. $$
\end{lem}

\begin{proof}
Define $f$, $g$, and $N$ as in Lemma~\ref{lem: 3ListRearrangement}. We consider six cases corresponding to the parities of $\bm{x}_1$, $\bm{x}_2$, and $\bm{x}_3$ in (i), (ii), and (iii).

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

\noindent\begin{minipage}{.25\linewidth}
\begin{align*}
    x_{1, h_1(j)} =
    \begin{cases}
    n_1 + 1\\
    n_1
    \end{cases}
\end{align*}
\end{minipage}
\begin{minipage}{.25\linewidth}
\begin{align*}
    , \; x_{2, h_2(j)} =
    \begin{cases}
    n_2 \\
    n_2 + 1
    \end{cases}
\end{align*}
\end{minipage}
\begin{minipage}{.5\linewidth}
\begin{align*}
    , \; x_{3, h_3(j)} =
    \begin{cases}
    n_3 & \;\; \text{if } j \in [m] \\
    n_3 + 1 & \;\; \text{if } j \in [m^2] - [m];
    \end{cases}
\end{align*}
\end{minipage}

\noindent\begin{minipage}{.25\linewidth}
\begin{align*}
    x_{1, j} =
    \begin{cases}
    n_1  \\
    n_1 + 1
    \end{cases}
\end{align*}
\end{minipage}
\begin{minipage}{.25\linewidth}
\begin{align*}
    , \; x_{2, f(j)} =
    \begin{cases}
    n_2 + 1 \\
    n_2
    \end{cases}
\end{align*}
\end{minipage}
\begin{minipage}{.5\linewidth}
\begin{align*}
    , \; x_{3, g(j)} =
    \begin{cases}
    n_3 + 1 & \;\; \text{if } j \in [m(m - 1)] \\
    n_3 & \;\; \text{if } j \in [m^2] - [m(m - 1)].
    \end{cases}
\end{align*}
\end{minipage}
Therefore, we obtain
\begin{align*}
    \sum_{j = 1}^{m^2} x_{1, h_1(j)} x_{2, h_2(j)} x_{3, h_3(j)}
    &= \sum_{j = 1}^{m} (n_1 + 1)n_2n_3 + \sum_{j = m + 1}^{m^2} n_1(n_2 + 1)(n_3 + 1) \\
    &= m(n_1 + 1)n_2n_3 + m(m - 1)n_1(n_2 + 1)(n_3 + 1) \\
    &= \sum_{j = 1}^{m(m - 1)} n_1(n_2 + 1)(n_3 + 1) + \sum_{j = m(m - 1) + 1}^{m^2} (n_1 + 1)n_2n_3 \\
    &= \sum_{j = 1}^{m^2} x_{1, j} x_{2, f(j)} x_{3, g(j)}.
\end{align*}

The other possibility for (i) is that $\bm{x}_1$ is odd, $\bm{x}_2$ is even, and $\bm{x}_3$ is even. The details of the proof are similar to the previous case and are given in Appendix~\ref{A}.

For (ii), first we suppose $\bm{x}_1$ is even, $\bm{x}_2$ is odd, and $\bm{x}_3$ is even. Then we have:

\noindent\begin{minipage}{.25\linewidth}
\begin{align*}
    x_{1, h_1(j)} =
    \begin{cases}
    n_1 + 1 \\
    n_1
    \end{cases}
\end{align*}
\end{minipage}
\begin{minipage}{.5\linewidth}
\begin{align*}
    , \; x_{2, h_2(j)} =
    \begin{cases}
    n_2 & \;\; \text{if } j \in [m] \\
    n_2 + 1 & \;\; \text{if } j \in [m^2] - [m];
    \end{cases}
\end{align*}
\end{minipage}
\begin{align*}
    x_{3, h_3(j)} =
    \begin{cases}
    n_3 + 1 & \text{if } j \in [2m] - [m] \\
    n_3 & \text{if } j \in [m] \cup \left( [m^2] - [2m] \right);
    \end{cases}
\end{align*}

\noindent\begin{minipage}{.25\linewidth}
\begin{align*}
    x_{1, j} =
    \begin{cases}
    n_1 \\
    n_1 + 1
    \end{cases}
\end{align*}
\end{minipage}
\begin{minipage}{.5\linewidth}
\begin{align*}
    , \;\; x_{2, f(j)} =
    \begin{cases}
    n_2 + 1 & \; \text{if } j \in [m(m - 1)] \\
    n_2 & \; \text{if } j \in [m^2] - [m(m - 1)].
    \end{cases}
\end{align*}
\end{minipage}

We also have $N = m(m - 1)$, and so
\begin{align*}
    x_{3, g(j)} =
    \begin{cases}
    n_3 & \text{if } j \in [m(m - 2)] \cup \left( [m^2] - [m(m - 1)] \right) \\
    n_3 + 1 & \text{if } j \in [m(m - 1)] - [m(m - 2)].
    \end{cases}
\end{align*}
Therefore, we obtain
\begin{align*}
    &\sum_{j = 1}^{m^2} x_{1, h_1(j)} x_{2, h_2(j)} x_{3, h_3(j)} \\
    &= \sum_{j = 1}^m (n_1 + 1)n_2n_3 + \sum_{j = m + 1}^{2m} n_1(n_2 + 1)(n_3 + 1) + \sum_{j = 2m + 1}^{m^2} n_1(n_2 + 1)n_3 \\
    &= m(n_1 + 1)n_2n_3 + mn_1(n_2 + 1)(n_3 + 1) + m(m - 2)n_1(n_2 + 1)n_3 \\
    &= \sum_{j = 1}^{m(m - 2)} n_1(n_2 + 1)n_3 + \sum_{j = m(m - 2) + 1}^{m(m - 1)} n_1(n_2 + 1)(n_3 + 1) + \sum_{j = m(m - 1) + 1}^{m^2} (n_1 + 1)n_2n_3 \\
    &= \sum_{j = 1}^{m^2} x_{1, j} x_{2, f(j)} x_{3, g(j)}.
\end{align*}

The other possibility for (ii) is that $\bm{x}_1$ is odd, $\bm{x}_2$ is even, and $\bm{x}_3$ is odd.  The details of the proof are similar to the previous case and are given in Appendix~\ref{A}.

Finally, turning our attention to~(iii), we suppose $\bm{x}_1$, $\bm{x}_2$, and $\bm{x}_3$ are all even. Then we have the following:
\begin{align*}
    x_{1, h_1(j)} =
    \begin{cases}
    n_1 + 1 & \text{if } j \in [m] \\
    n_1 & \text{if } j \in [m^2] - [m];
    \end{cases}
\end{align*}
\begin{align*}
    x_{2, h_2(j)} =
    \begin{cases}
    n_2 + 1 & \text{if } j \in [2m] - [m] \\
    n_2 & \text{if } j \in [m] \cup \left( [m^2] - [2m] \right);
    \end{cases}
\end{align*}
\begin{align*}
    x_{3, h_3(j)} =
    \begin{cases}
    n_3 + 1 & \text{if } j \in [3m] - [2m] \\
    n_3 & \text{if } j \in [2m] \cup \left( [m^2] - [3m] \right);
    \end{cases}
\end{align*}
\begin{align*}
    x_{1, j} =
    \begin{cases}
    n_1 & \text{if } j \in [m(m - 1)] \\
    n_1 + 1 & \text{if } j \in [m^2] - [m(m - 1)];
    \end{cases}
\end{align*}
\begin{align*}
    x_{2, f(j)} =
    \begin{cases}
    n_2 + 1 & \text{if } j \in [m] \\
    n_2 & \text{if } j \in [m^2] - [m].
    \end{cases}
\end{align*}
We also have $N = m(m - 1)$, and so
\begin{align*}
    x_{3, g(j)} =
    \begin{cases}
    n_3 & \text{if } j \in [m(m - 2)] \cup \left( [m^2] - [m(m - 1)] \right) \\
    n_3 + 1 & \text{if } j \in [m(m - 1)] - [m(m - 2)].
    \end{cases}
\end{align*}
Therefore, we obtain
\begin{align*}
    &\sum_{j = 1}^{m^2} x_{1, h_1(j)} x_{2, h_2(j)} x_{3, h_3(j)} \\
    &= \sum_{j = 1}^m (n_1 + 1)n_2n_3 + \sum_{j = m + 1}^{2m} \mkern-10mu n_1(n_2 + 1)n_3 + \sum_{j = 2m + 1}^{3m} \mkern-16mu n_1n_2(n_3 + 1) + \sum_{j = 3m + 1}^{m^2} \mkern-16mu n_1n_2n_3 \\
    &= m(n_1 + 1)n_2n_3 + mn_1(n_2 + 1)n_3 + mn_1n_2(n_3 + 1) + m(m - 3)n_1n_2n_3 \\
    &= \sum_{j = 1}^m n_1(n_2 + 1)n_3 + \sum_{j = m + 1}^{m(m - 2)} \mkern-12mu n_1n_2n_3 + \sum_{j = m(m - 2) + 1}^{m(m - 1)} \mkern-20mu n_1n_2(n_3 + 1) + \sum_{j = m(m - 1) + 1}^{m^2} \mkern-20mu (n_1 + 1)n_2n_3 \\
    &= \sum_{j = 1}^{m^2} x_{1, j} x_{2, f(j)} x_{3, g(j)}.
\end{align*}

The other possibility for (iii) is that $\bm{x}_1$, $\bm{x}_2$, and $\bm{x}_3$ are all odd. The details of the proof are similar to the previous case and are given in Appendix~\ref{A}.
\end{proof}

We are now ready to show that each formula in Theorem~\ref{thm: generalTheta3} is a lower bound on the DP color function of the appropriate Theta graph.

\begin{lem} \label{lem: theta-dp-lower-bound}
Let $G = \Theta(l_1,l_2,l_3)$ and $\mathcal{H} = (L,H)$ be a full $m$-fold cover of $G$ where $m \geq 3$.

(i) If the parity of $l_1$ is different from both $l_2$ and $l_3$, then $P_{DP}(G, \mathcal{H}) \geq P(G, m)$.

(ii) If the parity of $l_1$ is different from $l_2$ and the same as $l_3$, then

\noindent $\displaystyle P_{DP}(G, \mathcal{H}) \geq \frac{1}{m} \left[ (m - 1)^{l_1 + l_2 + l_3} + (m - 1)^{l_1} - (m - 1)^{l_2} - (m - 1)^{l_3 + 1} + (-1)^{l_2 + 1}(m - 2) \right]$.

(iii) If $l_1$, $l_2$, and $l_3$ all have the same parity, then

\noindent $\displaystyle P_{DP}(G, \mathcal{H}) \geq \frac{1}{m} \left[ (m - 1)^{l_1 + l_2 + l_3} - (m - 1)^{l_1} - (m - 1)^{l_2} - (m - 1)^{l_3} + 2(-1)^{l_1 + l_2 + l_3} \right]$.
\end{lem}

\begin{proof}

We begin by using Lemmas~\ref{lem: genThetaCount} and~\ref{lem: binaryThetaValues} to find a formula for $P_{DP}(G, \mathcal{H})$ to which we can apply Lemma~\ref{lem: special}. For each $k \in [3]$ and $(i, j) \in [m]^2$, let $s_{k, (i, j)} = N(\{(u, i), (w, j)\}, \mathcal{H}_{k})$. Consider some $k \in [3]$ and $(i, j) \in [m]^2$. By Lemma~\ref{lem: binaryThetaValues}, we know that
\begin{align} \label{s_eq1}
    s_{k, (i, j)} = \frac{(m - 1)^{l_k} + (-1)^{l_k}(m - 1)}{m} = \frac{(m - 1)^{l_k} - (-1)^{l_k}}{m} + (-1)^{l_k}
\end{align}
if there is a path in $H_k$ from $(u, i)$ to $(w, j)$ consisting only of cross-edges of $\mathcal{H}_k$, and
\begin{align} \label{s_eq2}
    s_{k, (i, j)} = \frac{(m - 1)^{l_k} - (-1)^{l_k}}{m}
\end{align}
otherwise. In particular, notice that Equation~(\ref{s_eq1}) holds for exactly $m$ choices of $(i, j) \in [m]^2$; whereas, Equation~(\ref{s_eq2}) holds for the remaining $m(m - 1)$ choices of $(i, j) \in [m]^2$.

For each $k \in [3]$, let $n_k = \min_{(i, j) \in [m]^2} s_{k, (i, j)}$. Notice that, for each $(i, j) \in [m]^2$, either $s_{k, (i, j)} = n_k$ or $s_{k, (i, j)} = n_k + 1$. Moreover, we have $s_{k, (i, j)} = n_k$ for either $m$ or $m(m - 1)$ choices of $(i, j) \in [m]^2$, while $s_{k, (i, j)} = n_k + 1$ for the remaining $m(m - 1)$ or $m$ choices, respectively, of $(i, j) \in [m]^2$.

Now, let $\beta : [m]^2 \rightarrow [m^2]$ be the function defined by $\beta(i, j) = m(i - 1) + j$ for each $(i, j) \in [m]^2$. Notice that $\beta$ is bijective and hence has an inverse $\beta^{-1} : [m^2] \rightarrow [m]^2$. By Lemma~\ref{lem: genThetaCount},
\begin{align*}
    P_{DP}(G, \mathcal{H}) = \sum_{(i, j) \in [m]^2} s_{1, (i, j)} s_{2, (i, j)} s_{3, (i, j)} = \sum_{\ell = 1}^{m^2} s_{1, \beta^{-1}(\ell)} s_{2, \beta^{-1}(\ell)} s_{3, \beta^{-1}(\ell)}.
\end{align*}
If we let $s_{k, \ell} = s_{k, \beta^{-1}(\ell)}$ for each $k \in [3]$ and $\ell \in [m^2]$, then we can alternatively write
\begin{align*}
    P_{DP}(G, \mathcal{H}) = \sum_{\ell = 1}^{m^2} s_{1, \ell} s_{2, \ell} s_{3, \ell}.
\end{align*}

For each $k \in [3]$, let $\rho_k : [m^2] \rightarrow [m^2]$ be any permutation of $[m^2]$ such that for each $\ell_1, \ell_2 \in [m^2]$ with $\ell_1 < \ell_2$, we have $s_{k, \rho_k(\ell_1)} \leq s_{k, \rho_k(\ell_2)}$. Furthermore, let $x_{k, \ell} = s_{k, \rho_k(\ell)}$ for each $\ell \in [m^2]$. By definition, for each $\ell_1, \ell_2 \in [m^2]$ with $\ell_1 < \ell_2$, we have $x_{k, \ell_1} \leq x_{k, \ell_2}$. Moreover, notice that $x_{k, 1} = \cdots = x_{k, m} = n_k$ and $x_{k, m(m - 1) + 1} = \cdots = x_{k, m^2} = n_k + 1$. Finally, notice that $\rho_k$ is bijective, and so it has an inverse $\rho_k^{-1} : [m^2] \rightarrow [m^2]$. Using this notation, we may write
\begin{align*}
    P_{DP}(G, \mathcal{H}) &= \sum_{\ell = 1}^{m^2} s_{1, \rho_1(\ell)} s_{2, \rho_2(\rho_2^{-1}(\rho_1(\ell)))} s_{3, \rho_3(\rho_3^{-1}(\rho_1(\ell)))}
    = \sum_{\ell = 1}^{m^2} x_{1, \ell} x_{2, \rho_2^{-1}(\rho_1(\ell))} x_{3, \rho_3^{-1}(\rho_1(\ell))}.
\end{align*}
If we let $\sigma_1, \sigma_2 : [m^2] \rightarrow [m^2]$ be the permutations of $[m^2]$ defined by $\sigma_1(\ell) = \rho_2^{-1}(\rho_1(\ell))$ and $\sigma_2(\ell) = \rho_3^{-1}(\rho_1(\ell))$ for each $\ell \in [m^2]$, then we can alternatively write
\begin{align*}
    P_{DP}(G, \mathcal{H}) = \sum_{\ell = 1}^{m^2} x_{1, \ell} x_{2, \sigma_1(\ell)} x_{3, \sigma_2(\ell)}.
\end{align*}

Define $\bm{x}_k$, $h_k$, $s_k$, and $o_k$ for each $k \in [3]$ as in Lemma~\ref{lem: special}. By Lemma~\ref{lem: special},
\begin{align} \label{dp-lower-bound}
    P_{DP}(G, \mathcal{H}) \geq \sum_{\ell = 1}^{m^2} x_{1, h_1(\ell)} x_{2, h_2(\ell)} x_{3, h_3(\ell)}.
\end{align}

Now, we claim for each $k \in [3]$,
\begin{align} \label{sk-formula}
    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{align}
and
\begin{align} \label{ok-formula}
    o_k = \frac{(m - 1)^{l_k} + (-1)^{l_k + 1}}{m}.
\end{align}
To see why these formulas hold, consider the case where $l_k$ is even and the case where $l_k$ is odd separately.  Suppose that $l_k$ is even. Then we have
\begin{align*}
    s_{k, (i, j)} = \frac{(m - 1)^{l_k} - (-1)^{l_k}}{m} + 1
\end{align*}
for exactly $m$ choices of $(i, j) \in [m]^2$, and
\begin{align*}
    s_{k, (i, j)} = \frac{(m - 1)^{l_k} - (-1)^{l_k}}{m}
\end{align*}
for the remaining $m(m - 1)$ choices of $(i, j) \in [m]^2$. Therefore, we have
\begin{align*}
    n_k = \frac{(m - 1)^{l_k} - (-1)^{l_k}}{m}
\end{align*}
and
\begin{align*}
    x_{k, \ell} =
    \begin{cases}
    n_k & \text{if } \ell \in [m(m - 1)] \\
    n_k + 1 & \text{if } \ell \in [m^2] - [m(m - 1)].
    \end{cases}
\end{align*}
It follows by definition that $\bm{x}_k$ is even, so that
\begin{align*}
    s_k = n_k + 1 = \frac{(m - 1)^{l_k} - (-1)^{l_k}}{m} + (-1)^{l_k}\;\;
\text{and }
    o_k = n_k = \frac{(m - 1)^{l_k} - (-1)^{l_k}}{m}.
\end{align*}

The proof for the case $l_k$ is odd follows similarly. The details are given in Appendix~\ref{B}.

We are now ready to prove Statements~(i), (ii), and (iii). For Statement~(i), suppose that the parity of $l_1$ is different from that of both $l_2$ and $l_3$. Then, for each $k \in [3]$, we have
\begin{align*}
    x_{k, h_k(\ell)} =
    \begin{cases}
    s_k & \text{if } \ell \in [m] \\
    o_k & \text{if } \ell \in [m^2] - [m].
    \end{cases}
\end{align*}
So,
\begin{align*}
    P_{DP}(G, \mathcal{H}) &\geq \sum_{\ell = 1}^{m^2} x_{1, h_1(\ell)} x_{2, h_2(\ell)} x_{3, h_3(\ell)} = \sum_{\ell = 1}^m s_1s_2s_3 + \sum_{\ell = m + 1}^{m^2} o_1o_2o_3 \\
    &= ms_1s_2s_3 + m(m - 1)o_1o_2o_3.

\end{align*}

For Statement~(ii), suppose that the parity of $l_1$ is different from that of $l_2$ and the same as that of $l_3$. Then we have
\begin{align*}
    x_{k, h_k(\ell)} =
    \begin{cases}
    s_k & \text{if } \ell \in [m] \\
    o_k & \text{if } \ell \in [m^2] - [m]
    \end{cases}
\end{align*}
for $k \in [2]$ and
\begin{align*}
    x_{3, h_3(\ell)} =
    \begin{cases}
    s_k & \text{if } \ell \in [2m] - [m] \\
    o_k & \text{if } \ell \in [m] \cup \left( [m^2] - [2m] \right).
    \end{cases}
\end{align*}
So,
\begin{align*}
    P_{DP}(G, \mathcal{H}) &\geq \sum_{\ell = 1}^{m^2} x_{1, h_1(\ell)} x_{2, h_2(\ell)} x_{3, h_3(\ell)} = \sum_{\ell = 1}^m s_1 s_2 o_3 + \sum_{\ell = m + 1}^{2m} o_1 o_2 s_3 + \sum_{\ell = 2m + 1}^{m^2} o_1 o_2 o_3 \\
    &=ms_1s_2o_3 + mo_1o_2s_3 + m(m - 2)o_1o_2o_3.

\end{align*}

For Statement~(iii), suppose that $l_1$, $l_2$, and $l_3$ all have the same parity. Then, for each $k \in [3]$, we have
\begin{align*}
    x_{k, h_k(\ell)} =
    \begin{cases}
    s_k & \text{if } \ell \in [km] - [(k - 1)m] \\
    o_k & \text{if } \ell \in [(k - 1)m] \cup \left( [m^2] - [km] \right).
    \end{cases}
\end{align*}
So,
\begin{align*}
    P_{DP}(G, \mathcal{H}) &\geq \sum_{\ell = 1}^{m^2} x_{1, h_1(\ell)} x_{2, h_2(\ell)} x_{3, h_3(\ell)}\\
    &= \sum_{\ell = 1}^m s_1 o_2 o_3 + \sum_{\ell = m + 1}^{2m} o_1 s_2 o_3 + \sum_{\ell = 2m + 1}^{3m} o_1 o_2 s_3 + \sum_{\ell = 3m + 1}^{m^2} o_1 o_2 o_3 \\
    &= ms_1o_2o_3 + mo_1s_2o_3 + mo_1o_2s_3 + m(m - 3)o_1o_2o_3.

\end{align*}

Finally, substituting the formulas~\ref{sk-formula} and~\ref{ok-formula} for $s_k$ and $o_k$ into the three lower bounds that we just obtained yields the appropriate formula after some algebraic simplification.
\end{proof}

Having established the appropriate lower bounds, we are now ready to complete the proof of Theorem~\ref{thm: generalTheta3}.

\begin{proof}
Since $G$ contains a cycle, $P_{DP}(G, 1) = P_{DP}(G, 2) = 0$. So, the result holds when $m = 1, 2$. Therefore, throughout this proof we suppose that $m \geq 3$. Notice that if the parity of $l_1$ is different from both $l_2$ and $l_3$, then we know from Lemma~\ref{lem: theta-dp-lower-bound} that $P_{DP}(G, m) \geq P(G, m)$. Since we also know that $P_{DP}(G, m) \leq P(G, m)$, Statement~(i) follows.

For the remaining two statements, we will construct a full $m$-fold cover $\mathcal{H} = (L,H)$ of $G$ with an appropriate number of $\mathcal{H}$-colorings. For each $i \in [3]$, suppose the vertices of $R_i$ written in order are: $u,v_{i,1},\ldots,v_{i,l_i-1},w$. In the case $l_1 = 1$, $R_1$ has no internal vertices. Let $G' = G - \{v_{2,l_2-1}w,v_{3,l_3-1}w\}$. Suppose $\mathcal{H}' = (L,H')$ is an $m$-fold cover of $G'$ with a canonical labeling. Begin constructing edges of $H$ by including all the edges in $E(H')$. The remaining edges between $L(v_{2,l_2-1})$ and $L(w)$, and between $L(v_{3,l_3-1})$ and $L(w)$, will be specified below.

Let $\sigma : [m] \rightarrow [m]$ be the permutation of $[m]$ given by $\sigma(j) = (j \text{ (mod } m)) + 1$.  Suppose $l_3$ has the same parity as $l_1$ and $l_2$ has different parity than $l_1$. Complete the construction of $H$ by including $\{(v_{2,l_2 - 1},j)(w,j) : j \in [m]\} \cup \{(v_{3,l_3 - 1},j)(w,\sigma(j)) : j \in [m]\}$ in $E(H)$.
By Lemma~\ref{lem: binaryThetaValues}, for each $k \in [2]$ we have $N(\{(u,i),(w,j)\},\mathcal{H}_k) = ((m-1)^{l_k} + (-1)^{l_k}(m-1))/m$ when $i = j$ and $N(\{(u,i),(w,j)\},\mathcal{H}_k) = ((m-1)^{l_k} - (-1)^{l_k})/m$ otherwise.
Again by Lemma~\ref{lem: binaryThetaValues}, we have $N(\{(u,i),(w,j)\},\mathcal{H}_3) = ((m-1)^{l_3} + (-1)^{l_3}(m-1))/m$ when $j = \sigma(i)$ and $N(\{(u,i),(w,j)\},\mathcal{H}_3) = ((m-1)^{l_3} - (-1)^{l_3})/m$ otherwise. Thus, by Lemma~\ref{lem: genThetaCount},
\begin{align*}
    P_{DP}(G,\mathcal{H}) &= \sum_{(i,j) \in [m]^2} \prod_{k=1}^{3} N(\{(u,i),(w,j)\},\mathcal{H}_k)\\
    &= m\left(\frac{(m-1)^{l_1} + (-1)^{l_1}(m-1)}{m}\right)\left(\frac{(m-1)^{l_2} + (-1)^{l_2}(m-1)}{m}\right)\left(\frac{(m-1)^{l_3} - (-1)^{l_3}}{m}\right)\\
    &+ m\left(\frac{(m-1)^{l_1} - (-1)^{l_1}}{m}\right)\left(\frac{(m-1)^{l_2} - (-1)^{l_2}}{m}\right)\left(\frac{(m-1)^{l_3} + (-1)^{l_3}(m-1)}{m}\right)\\
    &+ (m^2 - 2m)\left(\frac{(m-1)^{l_1} - (-1)^{l_1}}{m}\right)\left(\frac{(m-1)^{l_2} - (-1)^{l_2}}{m}\right)\left(\frac{(m-1)^{l_3} - (-1)^{l_3}}{m}\right)\\
    &= \frac{1}{m}\left((m-1)^{l_1+l_2+l_3} + (m-1)^{l_1} - (m-1)^{l_2} - (m-1)^{l_3+1} - (-1)^{l_2}(m-2)\right).
\end{align*}

Suppose $l_2$ and $l_3$ have the same parity as $l_1$. Complete the construction of $H$ by including $\{(v_{2,l_2 - 1},j)(w,\sigma(j)) : j \in [m]\} \cup \{(v_{3,l_3 - 1},j)(w,\sigma^2(j)) : j \in [m]\}$ in $E(H)$ \footnote{Throughout this document, whenever $\sigma$ is a permutation of $[m]$ and $k \in \mathbb{N}$, we write $\sigma^{k}$ for $\sigma \circ \cdots \circ \sigma$, where $\sigma$ appears $k$ times. Moreover, we write $\sigma^0$ for the identity map on $[m]$.}.
By Lemma~\ref{lem: binaryThetaValues}, for each $k \in [3]$ we have $N(\{(u,i),(w,j)\},\mathcal{H}_k) = ((m-1)^{l_k} + (-1)^{l_k}(m-1))/m$ when $i = \sigma^{k - 1}(j)$ and $N(\{(u,i),(w,j)\},\mathcal{H}_k) = ((m-1)^{l_k} - (-1)^{l_k})/m$ otherwise.
Thus, by Lemma~\ref{lem: genThetaCount},
\begin{align*}
    P_{DP}(G,\mathcal{H}) &= \sum_{(i,j) \in [m]^2} \prod_{k=1}^{3} N(\{(u,i),(w,j)\},\mathcal{H}_k)\\
    &= m\left(\frac{(m-1)^{l_1} + (-1)^{l_1}(m-1)}{m}\right)\left(\frac{(m-1)^{l_2} - (-1)^{l_2}}{m}\right)\left(\frac{(m-1)^{l_3} - (-1)^{l_3}}{m}\right)\\
    &+ m\left(\frac{(m-1)^{l_1} - (-1)^{l_1}}{m}\right)\left(\frac{(m-1)^{l_2} + (-1)^{l_2}(m-1)}{m}\right)\left(\frac{(m-1)^{l_3} - (-1)^{l_3}}{m}\right)\\
    &+ m\left(\frac{(m-1)^{l_1} - (-1)^{l_1}}{m}\right)\left(\frac{(m-1)^{l_2} - (-1)^{l_2}}{m}\right)\left(\frac{(m-1)^{l_3} + (-1)^{l_3}(m-1)}{m}\right)\\
    &+ (m^2 - 3m)\left(\frac{(m-1)^{l_1} - (-1)^{l_1}}{m}\right)\left(\frac{(m-1)^{l_2} - (-1)^{l_2}}{m}\right)\left(\frac{(m-1)^{l_3} - (-1)^{l_3}}{m}\right)\\
    &= \frac{1}{m}\left((m-1)^{l_1+l_2+l_3} - (m-1)^{l_1} - (m-1)^{l_2} - (m-1)^{l_3} + 2(-1)^{l_1+l_2+l_3}\right).
\end{align*}
\end{proof}

\section{The Dual DP Color Function of Generalized Theta Graphs} \label{dual}

In this Section we show how the ideas we have developed thus far can be used to completely determine the dual DP color function of all Generalized Theta graphs.  In particular, we prove the following theorem.

\begin{thm} \label{thm: dual}
Let $G = \Theta(l_1, \ldots, l_n)$, where $n \geq 2$, $l_1 = \min_{i \in [n]} l_i \geq 1$, and $l_i \geq 2$ for each $i \in [n] - \{1\}$. If $l_1, \ldots, l_n$ do not all have the same parity, let $t = \max\{i \in [n] : (l_1 - l_i) \textup{ mod } 2 = 1\}$; otherwise, let $t = 1$. Let $m \geq 2$. For each $i \in [n]$, let
\begin{align*}
    n_i =
    \begin{cases}
    \displaystyle \frac{(m - 1)^{l_i} - (-1)^{l_i}}{m} & \text{if } l_i \text{ is even} \\[1em]
    \displaystyle \frac{(m - 1)^{l_i} + (-1)^{l_i}(m - 1)}{m} & \text{if } l_i \text{ is odd}.
    \end{cases}
\end{align*}
If $l_i$ is even, let $s_i = n_i + 1$ and $o_i = n_i$. If $l_i$ is odd, let $s_i = n_i$ and $o_i = n_i + 1$. Let $S = \prod_{i \in [n]} s_i$ and $O = \prod_{i \in [n]} o_i$. Then
\begin{align*}
    P_{DP}^*(G, m) = \frac{mS}{\prod_{i = 2}^t s_i} \prod_{i = 2}^t o_i + m(m - 2)O + \frac{mO}{\prod_{i = 2}^t o_i} \prod_{i = 2}^t s_i.
\end{align*}
\end{thm}

It is worth mentioning that Theorem~\ref{thm: dual} implies that when $t=1$, $P_{DP}^*(G, m) = P(G,m)$ (There is an analogous result for the DP color function; specifically, see Statement~(i) of Theorem~\ref{thm: generalized}.). The proof of Theorem~\ref{thm: dual} uses a lemma that follows immediately from the second part of the Rearrangement Inequality.

\begin{lem} \label{lem: rearrangementUpperBound}
Let $n, m \geq 2$. Suppose $x_{i, j}$ is a non-negative integer for each $i \in [n]$ and $j \in [m^2]$. For each $i \in [n]$, suppose $n_i = x_{i, 1} = \cdots = x_{i, m} \leq x_{i, m + 1} = \cdots = x_{i, m(m - 1)} \leq x_{i, m(m - 1) + 1} = \cdots = x_{i, m^2} = n_i + 1$ for some $n_i \geq 0$. Then for any permutations $\sigma_1, \ldots, \sigma_n$ of $[m^2]$,
\begin{align*}
    \sum_{j = 1}^{m^2} \prod_{i = 1}^n x_{i, \sigma_i(j)} \leq \sum_{j = 1}^{m^2} \prod_{i = 1}^n x_{i, j}.
\end{align*}
\end{lem}

We are now ready to show that each formula in Theorem~\ref{thm: dual} is an upper bound on the dual DP color function of the appropriate Generalized Theta graph.

\begin{lem} \label{lem: dual-dp-upper-bound}
Let $G = \Theta(l_1, \ldots, l_n)$ and $\mathcal{H}$ be a full $m$-fold cover of $G$. Then, using the same notation as Theorem~\ref{thm: dual},
\begin{align*}
    P_{DP}(G, \mathcal{H}) \leq \frac{mS}{\prod_{i = 2}^t s_i} \prod_{i = 2}^t o_i + m(m - 2)O + \frac{mO}{\prod_{i = 2}^t o_i} \prod_{i = 2}^t s_i.
\end{align*}
\end{lem}

\begin{proof}
We begin by using Lemmas~\ref{lem: genThetaCount} and~\ref{lem: binaryThetaValues} to find a formula for $P_{DP}(G, \mathcal{H})$ to which we can apply Lemma~\ref{lem: rearrangementUpperBound}.  For each $k \in [n]$ and $(i, j) \in [m]^2$ let $s_{k, (i, j)} = N(\{(u, i), (w, j)\}, \mathcal{H}_k)$. Consider some $k \in [n]$ and $(i, j) \in [m]^2$. By Lemma~\ref{lem: binaryThetaValues}, we know that
\begin{align} \label{s_eq4}
    s_{k, (i, j)} = \frac{(m - 1)^{l_k} + (-1)^{l_k}(m - 1)}{m} = \frac{(m - 1)^{l_k} - (-1)^{l_k}}{m} + (-1)^{l_k}
\end{align}
if there is a path in $H_k$ from $(u, i)$ to $(w, j)$ consisting only of cross-edges of $\mathcal{H}_k$, and
\begin{align} \label{s_eq5}
    s_{k, (i, j)} = \frac{(m - 1)^{l_k} - (-1)^{l_k}}{m}
\end{align}
otherwise. In particular, notice that Equation (\ref{s_eq4}) holds for exactly $m$ choices of $(i, j) \in [m]^2$, whereas Equation (\ref{s_eq5}) holds for the remaining $m(m - 1)$ choices of $(i, j) \in [m]^2$.

For each $k \in [n]$, let $n_k = \min_{(i, j) \in [m]^2} s_{k, (i, j)}$. Notice that for each $(i, j) \in [m]^2$, either $s_{k, (i, j)} = n_k$ or $s_{k, (i, j)} = n_k + 1$. Moreover, we have $s_{k, (i, j)} = n_k$ for either $m$ or $m(m - 1)$ choices of $(i, j) \in [m]^2$, while $s_{k, (i, j)} = n_k + 1$ for the remaining $m(m - 1)$ or $m$ choices, respectively, of $(i, j) \in [m]^2$.

Now, let $\beta : [m]^2 \rightarrow [m^2]$ be the function defined by $\beta(i, j) = m(i - 1) + j$ for each $(i, j) \in [m]^2$. Notice that $\beta$ is bijective and hence has an inverse $\beta^{-1} : [m^2] \rightarrow [m]^2$. By Lemma~\ref{lem: genThetaCount},
\begin{align*}
    P_{DP}(G, \mathcal{H}) = \sum_{(i, j) \in [m]^2} \prod_{k = 1}^n s_{k, (i, j)} = \sum_{\ell = 1}^{m^2} \prod_{k = 1}^n s_{k, \beta^{-1}(\ell)}.
\end{align*}
For each $k \in [n]$, let $\rho_k : [m^2] \rightarrow [m^2]$ be any permutation of $[m^2]$ such that for each $\ell_1, \ell_2 \in [m^2]$ with $\ell_1 < \ell_2$, we have $s_{k, \beta^{-1}(\rho_k(\ell_1))} \leq s_{k, \beta^{-1}(\rho_k(\ell_2))}$. Furthermore, let $x_{k, \ell} = s_{k, \beta^{-1}(\rho_k(\ell))}$ for each $\ell \in [m^2]$. By definition, for each $\ell_1, \ell_2 \in [m^2]$ with $\ell_1 < \ell_2$, we have $x_{k, \ell_1} \leq x_{k, \ell_2}$. Moreover, notice that $0 \leq n_k = x_{k, 1} = \cdots = x_{k, m} \leq x_{k, m + 1} = \cdots = x_{k, m(m - 1)} \leq x_{k, m(m - 1) + 1} = \cdots = x_{k, m^2} = n_k + 1$. By Lemma~\ref{lem: rearrangementUpperBound}, we have
\begin{align} \label{general-upper-bound-inequality}
    P_{DP}(G, \mathcal{H})
    &= \sum_{\ell = 1}^{m^2} \prod_{k = 1}^n s_{k, \beta^{-1}(\ell)} \leq \sum_{\ell = 1}^{m^2} \prod_{k = 1}^n x_{k, \ell}.
\end{align}

We now consider two cases: (1) $l_1$ is even and (2) $l_1$ is odd. First, consider case (1). If $t = 1$, then $l_1, \ldots, l_n$ are all even, and so we have $s_k = n_k + 1$, $o_k = n_k$, and
\begin{align*}
    x_{k, \ell} =
    \begin{cases}
    o_k & \text{if } \ell \in [m(m - 1)] \\
    s_k & \text{if } \ell \in [m^2] - [m(m - 1)]
    \end{cases}
\end{align*}
for each $k \in [n]$. Thus, Inequality (\ref{general-upper-bound-inequality}) becomes
\begin{align*}
    P_{DP}(G, \mathcal{H})
    &\leq \sum_{\ell = 1}^{m^2} \prod_{k = 1}^n x_{k, \ell} = \sum_{\ell = 1}^{m(m - 1)} \prod_{k = 1}^n x_{k, \ell} + \sum_{\ell = m(m - 1) + 1}^{m^2} \prod_{k = 1}^n x_{k, \ell}\\
    &= \sum_{\ell = 1}^{m(m - 1)} \prod_{k = 1}^n o_k + \sum_{\ell = m(m - 1) + 1}^{m^2} \prod_{k = 1}^n s_k
    = m(m - 1)O + mS.
\end{align*}
So, assume $t > 1$. Then, $l_1, l_{t + 1}, \ldots, l_n$ are even and $l_2, \ldots, l_t$ are odd. Hence, for each $k \in \{1, t + 1, \ldots, n\}$, we have $s_k = n_k + 1$, $o_k = n_k$, and
\begin{align*}
    x_{k, \ell} =
    \begin{cases}
    o_k & \text{if } \ell \in [m(m - 1)] \\
    s_k & \text{if } \ell \in [m^2] - [m(m - 1)].
    \end{cases}
\end{align*}
For each $k \in \{2, \ldots, t\}$, we have $s_k = n_k$, $o_k = n_k + 1$, and
\begin{align*}
    x_{k, \ell} =
    \begin{cases}
    s_k & \text{if } \ell \in [m] \\
    o_k & \text{if } \ell \in [m^2] - [m].
    \end{cases}
\end{align*}
Thus, Inequality (\ref{general-upper-bound-inequality}) becomes
\begin{align*}
    P_{DP}(G, \mathcal{H})
    &\leq \sum_{\ell = 1}^{m^2} \prod_{k = 1}^n x_{k, \ell}
    = \sum_{\ell = 1}^m \prod_{k = 1}^n x_{k, \ell} + \sum_{\ell = m + 1}^{m(m - 1)} \prod_{k = 1}^n x_{k, \ell} + \sum_{\ell = m(m - 1) + 1}^{m^2} \prod_{k = 1}^n x_{k, \ell} \\
    &= \sum_{\ell = 1}^m \left( o_1 \prod_{k = t + 1}^n o_k \prod_{k = 2}^t s_k \right) + \sum_{\ell = m + 1}^{m(m - 1)} \prod_{k = 1}^n o_k + \sum_{\ell = m(m - 1) + 1}^{m^2} \left( s_1 \prod_{k = t + 1}^n s_k \prod_{k = 2}^t o_k \right) \\
    &= \frac{mO}{\prod_{k = 2}^t o_k} \prod_{k = 2}^t s_k + m(m - 2)O + \frac{mS}{\prod_{k = 2}^t s_k} \prod_{k = 2}^t o_k.
\end{align*}

Next, we consider case (2). If $t = 1$, then $l_1, \ldots, l_n$ are all odd, and so we have $s_k = n_k$, $o_k = n_k + 1$, and
\begin{align*}
    x_{k, \ell} =
    \begin{cases}
    s_k & \text{if } \ell \in [m] \\
    o_k & \text{if } \ell \in [m^2] - [m]
    \end{cases}
\end{align*}
for each $k \in [n]$. Then substituting these values of $x_{k, \ell}$ into Inequality (\ref{general-upper-bound-inequality}), as done in the previous case, gives us the required expression.

\begin{comment}
Thus, Inequality (\ref{general-upper-bound-inequality}) becomes
\begin{align*}
    P_{DP}(G, \mathcal{H})
    &\leq \sum_{\ell = 1}^{m^2} \prod_{k = 1}^n x_{k, \ell}
    = \sum_{\ell = 1}^m \prod_{k = 1}^n x_{k, \ell} + \sum_{\ell = m + 1}^{m^2} \prod_{k = 1}^n x_{k, \ell} \\
    &= \sum_{\ell = 1}^m \prod_{k = 1}^n s_k + \sum_{\ell = m + 1}^{m^2} \prod_{k = 1}^n o_k
    = mS + m(m - 1)O.
\end{align*}
\end{comment}
So, assume $t > 1$. Then, $l_1, l_{t + 1}, \ldots, l_n$ are odd and $l_2, \ldots, l_t$ are even. Hence, for each $k \in \{1, t + 1, \ldots, n\}$, we have $s_k = n_k$, $o_k = n_k + 1$, and
\begin{align*}
    x_{k, \ell} =
    \begin{cases}
    s_k & \text{if } \ell \in [m] \\
    o_k & \text{if } \ell \in [m^2] - [m].
    \end{cases}
\end{align*}
For each $k \in \{2, \ldots, t\}$, we have $s_k = n_k + 1$, $o_k = n_k$, and
\begin{align*}
    x_{k, \ell} =
    \begin{cases}
    o_k & \text{if } \ell \in [m(m - 1)] \\
    s_k & \text{if } \ell \in [m^2] - [m(m - 1)].
    \end{cases}
\end{align*}

Then substituting these values of $x_{k, \ell}$ into Inequality (\ref{general-upper-bound-inequality}), as done in the previous case, gives us the required expression.\end{proof}

\begin{comment}
Thus, Inequality (\ref{general-upper-bound-inequality}) becomes
\begin{align*}
    P_{DP}(G, \mathcal{H})
    &\leq \sum_{\ell = 1}^{m^2} \prod_{k = 1}^n x_{k, \ell}
    = \sum_{\ell = 1}^m \prod_{k = 1}^n x_{k, \ell} + \sum_{\ell = m + 1}^{m(m - 1)} \prod_{k = 1}^n x_{k, \ell} + \sum_{\ell = m(m - 1) + 1}^{m^2} \prod_{k = 1}^n x_{k, \ell} \\
    &= \sum_{\ell = 1}^m \left( s_1 \prod_{k = 2}^t o_k \prod_{k = t + 1}^n s_k \right) + \sum_{\ell = m + 1}^{m(m - 1)} \prod_{k = 1}^n o_k + \sum_{\ell = m(m - 1) + 1}^{m^2} \left( o_1 \prod_{k = 2}^t s_k \prod_{k = t + 1}^n o_k \right) \\
    &= \frac{mS}{\prod_{k = 2}^t s_k} \prod_{k = 2}^t o_k + m(m - 2)O + \frac{mO}{\prod_{k = 2}^t o_k} \prod_{k = 2}^t s_k.
\end{align*}
\end{comment}

Having established the appropriate upper bounds, we are now ready to complete the proof of Theorem~\ref{thm: dual}.

\begin{proof}
First, notice that if $t=1$, then $mS + m(m-1)O = P(G,m) \leq P_{DP}^*(G,m)$.  Thus, the desired result holds when $t=1$, and we may now assume that $t > 1$.  We will construct a full $m$-fold cover $\mathcal{H} = (L, H)$ of $G$ with an appropriate number of $\mathcal{H}$-colorings. For each $k \in [n]$, suppose the vertices of $R_k$ written in order are $u, v_{k, 1}, \ldots, v_{k, l_k - 1}, w$. In the case $l_1 = 1$, $R_1$ has no internal vertices. Let $G' = G - \{v_{k, l_k - 1}w : k \in [n] - \{1\}\}$. Suppose $\mathcal{H}' = (L, H')$ is an $m$-fold cover of $G'$ with a canonical labeling. Begin the construction of $H$ by including all the edges in $E(H')$. The remaining edges between $L(v_{k, l_k - 1})$ and $L(w)$ for each $k \in [n] - \{1\}$ will be specified below.  Note that for each $k \in [n]$, $s_k = ((m - 1)^{l_k} + (-1)^{l_k}(m - 1)) / m$ and $o_k = ((m - 1)^{l_k} - (-1)^{l_k}) / m$.

\begin{comment}
We consider two cases: (1) $t = 1$ and (2) $t > 1$.\\
First, for case (1), complete the construction of $H$ by including $\bigcup_{k \in [n] - \{1\}} \{(v_{k, l_k - 1}, j)(w, j) : j \in [m]\}$ in $E(H)$. It follows from Lemma~\ref{lem: binaryThetaValues} that for each $k \in [n]$, $N(\{(u, i), (w, j)\}, \mathcal{H}_k) = ((m - 1)^{l_k} + (-1)^{l_k}(m - 1)) / m = s_k$ when $i = j$ and $N(\{(u, i), (w, j)\}, \mathcal{H}_k) = ((m - 1)^{l_k} - (-1)^{l_k}) / m = o_k$ otherwise. Thus, by Lemma~\ref{lem: genThetaCount},
\begin{align*}
    P_{DP}(G, \mathcal{H})
    &= \sum_{(i, j) \in [m]^2} \prod_{k = 1}^n N(\{(u, i), (w, j)\}, \mathcal{H}_k) \\
    &= \sum_{i \in [m]} \prod_{k = 1}^n N(\{(u, i), (w, i)\}, \mathcal{H}_k) + \sum_{\substack{i \in [m] \\ j \in [m] - \{i\}}} \prod_{k = 1}^n N(\{(u, i), (w, j)\}, \mathcal{H}_k) \\
    &= \sum_{i \in [m]} \prod_{k = 1}^n s_k + \sum_{\substack{i \in [m] \\ j \in [m] - \{i\}}} \prod_{k = 1}^n o_k
    = mS + m(m - 1)O.
\end{align*}
\end{comment}

Let $\sigma : [m] \rightarrow [m]$ be the permutation of $[m]$ given by $\sigma(j) = (j \text{ (mod } m)) + 1$. Complete the construction of $H$ by including $\bigcup_{k \in [n] - [t]} \{(v_{k, l_k - 1}, j)(w, j) : j \in [m]\}$ and \\ $\bigcup_{k \in [t] - \{1\}} \{(v_{k, l_k - 1}, j)(w, \sigma(j)) : j \in [m]\}$ in $E(H)$. It follows from Lemma~\ref{lem: binaryThetaValues} that for each $k \in \{1, t + 1, \ldots, n\}$ we have $N(\{(u, i), (w, j)\}, \mathcal{H}_k) = ((m - 1)^{l_k} + (-1)^{l_k}(m - 1)) / m = s_k$ when $i = j$ and $N(\{(u, i), (w, j)\}, \mathcal{H}_k) = ((m - 1)^{l_k} - (-1)^{l_k}) / m = o_k$ otherwise. Again by Lemma~\ref{lem: binaryThetaValues}, for each $k \in \{2, \ldots, t\}$ we have $N(\{(u, i), (w, j)\}, \mathcal{H}_k) = ((m - 1)^{l_k} + (-1)^{l_k}(m - 1)) / m = s_k$ when $j = \sigma(i)$ and $N(\{(u, i), (w, j)\}, \mathcal{H}_k) = ((m - 1)^{l_k} - (-1)^{l_k}) / m = o_k$ otherwise. Thus, by Lemma~\ref{lem: genThetaCount},
\begin{align*}
    P_{DP}(G, \mathcal{H})
    &= \sum_{(i, j) \in [m]^2} \prod_{k = 1}^n N(\{(u, i), (w, j)\}, \mathcal{H}_k) \\
    &= \sum_{i = 1}^m \prod_{k = 1}^n N(\{(u, i), (w, i)\}, \mathcal{H}_k) + \sum_{i = 1}^m \prod_{k = 1}^n N(\{(u, i), (w, \sigma(i))\}, \mathcal{H}_k) \\
    & + \sum_{\substack{i \in [m] \\ j \in [m] - \{i, \sigma(i)\}}} \prod_{k = 1}^n N(\{(u, i), (w, j)\}, \mathcal{H}_k) \\
    &= \sum_{i = 1}^m \left[ \prod_{k \in \{1, t + 1, \ldots, n\}} N(\{(u, i), (w, i)\}, \mathcal{H}_k) \prod_{k \in \{2, \ldots, t\}} N(\{(u, i), (w, i)\}, \mathcal{H}_k) \right] \\
    & + \sum_{i = 1}^m \left[ \prod_{k \in \{1, t + 1, \ldots, n\}} N(\{(u, i), (w, \sigma(i))\}, \mathcal{H}_k) \prod_{k \in \{2, \ldots, t\}} N(\{(u, i), (w, \sigma(i))\}, \mathcal{H}_k) \right] \\
    & + \sum_{\substack{i \in [m] \\ j \in [m] - \{i, \sigma(i)\}}} \prod_{k = 1}^n o_k \\
    &= \sum_{i = 1}^m \left( \prod_{k \in \{1, t + 1, \ldots, n\}} s_k \prod_{k \in \{2, \ldots, t\}} o_k \right) + \sum_{i = 1}^m \left( \prod_{k \in \{1, t + 1, \ldots, n\}} o_k \prod_{k \in \{2, \ldots, t\}} s_k \right)\\
		&+ m(m - 2)O \\
    &= \frac{mS}{\prod_{k = 2}^t s_k}\prod_{k = 2}^t o_k + \frac{mO}{\prod_{k = 2}^t o_k} \prod_{k = 2}^t s_k + m(m - 2)O.
\end{align*}
\end{proof}

{\bf Acknowledgment.}  This paper is a combination of research projects conducted with undergraduate students: Manh Bui, Michael Maxfield, Paul Shin, and Seth Thomason at the College of Lake County during the the spring and summer of 2021. The support of the College of Lake County is gratefully acknowledged.

redpajama_set_name: RedPajamaArXiv


text:
U.S.-Trained Syrian Insurgent Plans to Fight Assad, Citing Pentagon Training Program
A Free Syrian Army fighter in 2012 / AP
Morgan Chalfant • August 18, 2015 4:02 pm
A Syrian rebel trained by the Pentagon to fight the Islamic State plans to fight forces aligned with Bashar al-Assad's regime.
In an exclusive interview with CNN conducted via Skype from Syria, Abu Iskander cited the U.S. training program when justifying whether or not the pro-U.S. Syrian rebels should engage Assad's forces.
"The second rule in the training project is that we fight whoever fights us," Iskander explained. "The Assad regime is fighting us. We will control new areas from ISIS and we will have to face Assad, shall we stay sitting without fighting Assad?"
"Make a no fly zone in Syria, then we won't flee to Europe [as refugees], but will stay in our homes," he added, voicing a plea common among Syrian insurgents for a no fly zone over the northern part of Syria. "We don't want to cry [as victims] on your TV screens, instead we want the Assad regime to be stopped."
Iskander and the other 53 moderate Syrian insurgents who have received training and been vetted by the United States were required to sign a pledge that their priority will be to fight the Islamic State (IS, also known as ISIL or ISIS) and not Assad.
The program, which has cost $41 million in training and equipment, is expected to soon yield another 70 properly trained Syrian rebels to help the U.S. fight IS in the region.
No less than five of the U.S.-trained rebels–known collectively as the "New Syria Force"–were recently captured by al Nusra, an al Qaeda affiliate, days after they entered Syria upon completion of training. As such, only 40 members of the New Syria Force remain together, with others split up inside Syria.
Iskander expressed gratitude for the Pentagon program, labeling it a "dream," but nevertheless lamented its "slow" pace.
"Nearly 17,000 Syrian men want to join, but the training is very slow," he said. "We need it to be faster–30 days instead of 45 days. More trainees–for example, our training in Jordan did 85–we should have been 500 there and another 500 in Turkey. We are thankful but it needs to happen faster."
Iskander explained that his work involves providing targeting for U.S. airstrikes and that he wears GPS devices in order for U.S. forces to track him.
"I go to the front line against ISIS, and I give locations for the warplanes to bomb," he detailed. "We have developed communication devices using satellites that can target from any place on the front line whether we see it or not. There are daily drone flights and they're in the sky as I talk to you now. I speak to the Americans every hour, a total of four hours a day."
A spokesman for the U.S. military's Central Command refused to confirm whether or not Iskander is indeed member of New Syria Force "as a matter of policy."
"The Coalition does remain in contact with New Syrian Forces elements who have graduated from the program, who have expressed their eagerness to continue the counter-ISIL fight," Colonel Patrick Ryder said.
The desire among Syrian insurgents to fight Assad could pose a problem for the U.S. as it moves to implement its plan with Turkey to create a Islamic State free "safe zone" along the Syrian-Turkish border. The plan would require the U.S. to rely heavily on Syrian opposition fighters to clear and protect the safe zone.
President Obama has authorized U.S. forces to use air power to defend the Syrian insurgents in the event of an attack by Assad forces or other groups in Syria.
Published under: Bashar al-Assad, Department of Justice, Syria, Terrorism
Report: Taliban Threatened To Send 2,000 Suicide Bombers to Washington DC
Watchdog Flags Ethics Landmine for Biden Admin in Landmark Affirmative Action Case
Biden Administration Defends Vaccine Mandate To Supreme Court
Iranian 'Drone Armies' Step Up Attacks on US as Nuclear Talks Languish, GOP Lawmakers Say
Jake Sullivan's Wife Faces Growing Calls for Recusal From Durham Probe

redpajama_set_name: RedPajamaCommonCrawl


text:
Het zwart muskushert (Moschus fuscus) is een evenhoevige uit de familie van de muskusherten (Moschidae).

Kenmerken
Ze worden ongeveer 70-100 cm lang en wegen ongeveer 10-15 kg.

Voortplanting
Na een draagtijd van 185-195 dagen worden er 1-2 jongen geboren. Ze leven ongeveer tot 20 jaar in gevangenschap maar het is onbekend hoelang ze in het wild leven.

Verspreiding
Deze soort komt voor in Azië, vooral in de Himalaya en het gebied eromheen. Ze leven op een hoogte van 2600-3600 meter.

Dier uit het Palearctisch gebied
Muskusherten
IUCN-status bedreigd

redpajama_set_name: RedPajamaWikipedia


text:
Home and Learn - Free Java Course
Java and Inheritance
Another important concept in Object Oriented programming is Inheritance. Just what Inheritance is will become clearer with some programming examples. But essentially, it's having one class as a parent class (called a super class) and another class as a child of the parent (the sub class). The child class is said to be derived from the parent class. The reason to have a child class, though, is to keep information separate. The child can inherit all the methods and fields from its parent, but can then do its own thing.
As an example of inheritance, we'll create a sub class that handles information about certificates. If the student gets an "A" grade, we'll award a Certificate of Excellence; if the student gets a "B" or a "C" we'll award a Certificate of Achievement. Any other grade and no certificate will be awarded. But the point about the sub class is to keep the certificates data separate from the exam data. However, we may still want to access some of the information about the exam, such as which exam it was. We can even access the methods that turn a score into a grade, and all from the sub class.
So, create a new class by clicking File > New File from the NetBeans menu. When the dialogue box appears, select Java under the Categories heading, and Java Class under File Types. Click Next, and enter Certificates as the name of your new class. Click Finish.
When your new class is created, add a private String field and call it certificate. Your new class should then look like this:
To create a sub class (child) from a Java super class (parent), the keyword extends is used. You then follow the "extends" keyword with the parent class you want to extend. We want to create a sub class from the StudentResults class. The StudentResults class will be the super class and the Certificates class will be the sub class.
After "public class Certificates" in your code add "extends StudentResults". Your code should then look like this:
You have now created a sub class that inherits the code from the StudentResults class.
Just like the StudentResults class we can create a constructor for this new Certificates class. When we create an object from the class, Java will first of all call our constructor.
However, only one constructor can be called. If we call a new constructor from the Certificates class all those default values we set up for the StudentResults class fields won't get set. To get around this, there is a keyword called super. This makes a call to the constructor from the super class. Add the following constructor to your Certificates class:
The name of the constructor is the same as the name of the class: Certificates. The first line of the code between curly brackets is the super call (note the round brackets after super). When this line executes, all the default fields we set up in StudentResults will be set.
The second line of code in the constructor sets a default value for the String field called certificate. (You can actually set up more than one constructor. You'll see how to do this in a later section.)
To test out your new class, go back to your ExamDetails class, the one with the main method. Comment out any code you have so far. A quick way to do this is to highlight all the code, and then click the comments icon on the NetBeans toolbar:
To get rid of comments, highlight the code again and click the uncomment icon.
Now add the following line to create a new object from your class:
Certificates c1 = new Certificates();
Your code window should look something like this:
The object name is simply c1. The type of object is a Certificates object.
To check that you can access the methods from the StudentResults class, add this line below you new c1 object:
String exam = c1.examName("VB");
This is exactly the same as you did before: called the examName method of the StudentResults class. This time, however, you're using a Certificates object rather than a StudentResults object.
Add a print line and your code should look like this:
Run the programme to see the following Output window:
So the method from the parent class (the super class) has been called into action. We can now add a method to the child class (the sub class). Add the following method to your Certificates class, just below the constructor:
The method is called certificateAwarded, and is set up to return a String value. Inside the round brackets of the method, we're handing over an exam score.
The first line of the method is this:
String aGrade = examGrade(aScore);
The method examGrade is a method in the parent class. It's the one we set up in StudentResults. This method, remember, was set up to return a grade and some extra text, "Grade is A", "Grade is B", etc. We're now calling it from the child class. The IF Statement checks the value of the aGrade string to see what's in it. Depending on the value a new string is returned, awarding a particular certificate: Excellence, Achievement, or no certificate.
Click back on your ExamDetails class, and add the following line:
String award = c1.certificateAwarded(50);
This line calls the new method, and hands it a value of 50. The result is returned to the String we've called award.
Adapt the print line method in your code to this:
System.out.println( exam + " " + award );
Your ExamDetails class should look like this (we've deleted all the comments):
And here is the Output window when the programme is run:
So we've used Inheritance to keep the exam details separate from the certificate details. The child class (sub) accessed the methods of its parent (super). We were then able to output data from both.
Inheritance, then, is enhancing (extending) the usefulness of a parent class. You keep data separate by putting it in a child class. But the child is related in some way to the parent, and can access some or all of its code. Like children everywhere, however, it does its own thing!
In the next section, you'll learn how error handling in Java.
<-- Class Methods | Java Error Handling -->
Back to the Java Contents Page
Home and Learn Videos


text:
Q: microsoft.interop.selection text Before asking my question I would like to describe briefly background of my problem: I'm developing ms word COM addin on C# and I need to handle user's text selections. Now I'm able to catch selection event - it's look like
Microsoft.Interop.Word._Application app;
app = (Word._Application )Application; // Application object comes on addin's connection
app.Application.WindowSelectionChange+=
            new Word.ApplicationEvents4_WindowSelectionChangeEventHandler(selChange);

///
void selChange(Word.Selection selection){

   MessageBox.Show(selection.Text); // this is my problem, Text property is not available

}

// property Text doesn't exist,but documentation tells that it exists. I suspect, that this property is not available for ms word 2007 - in the documentation only 2003,2010 versions are mentioned. But how I can do something like selection.getSelectedText()? I tryed to play with selection.Rows, selection.Rows[0],selection.Words,selection.Words[0] - no success.

A: According to the documentation, the Selection.Text property should be available for Word 2007 as well. I made a small sample implementation of your case to test it, and I cannot make it fail on Word 2010 and 2013 at least:
var wordApplication = new Application() { Visible = true };
wordApplication.Documents.Add();
wordApplication.WindowSelectionChange += delegate(Selection mySelection) { Console.WriteLine(mySelection.Text); };

So, I suggest you check that you have included the right namespaces and that the Selection interface you are using are actually the one from the Microsoft.Office.Interop.Word namespace.

redpajama_set_name: RedPajamaStackExchange


text:
12/28/2011 - Clark Colborn Releases His Second Album
Clark Colborn Releases His Second Album
Again Debuts on January 2nd, 2012
December 28th, 2011, Rockford, IL – Clark Colborn's new album is entitled "Again," (his second) and it's due to be released on January 2nd. Colborn's first release came out under the name "Clark Plays Guitar." The new disc is a step forward from that debut.
The first time around Colborn earned some impressive critical accolades. Gregory Johnson of Beetcafe.com said, "Listening to the CD takes you back to the days when Randy Rhodes and Eddie Van Halen turned playing guitar into an art form." According to Wayne Klinger of Quintessence, "...Clark is definitely something quite different in terms of the way he writes and handles his music. The leads and rhythms are catchy and inspiring; just the whole damn CD is impressive." Michael Molenda, Editor in Chief, Guitar Player Magazine had this to say: "Clark Colborn's fat and massively effected tone comes on like Godzilla, and it shifts between slinky, silky, scary, spiky, and searing. ...his brilliantly schizoid technical/primitive approach to the guitar is pretty exciting. In short, he rocks."
With reviews like that some artists might be content to release a "more of the same" sequel. Rather than take the easy route, though, Colborn ups the ante with the follow up. The change can be noted by just looking at the credits as King's X's Ty Tabor serves as the mastering engineer on the release. But the difference doesn't stop there. Colborn admits that on the first disc he held back a bit, playing it a bit safe.
With Again, he says, "I let the songs take control and go where they wanted to go. So I ended up with longer songs (the obvious single, "Lie to Me," is six and a half minutes long!), more complex arrangements (but they still flow and rock!), and vocals. I always try to write music that will flow, and rock, and hold the attention of the listener. Sometimes getting too technical can ruin that flow, but for intelligent listeners not being technical enough can make the music boring. I think this album really strikes a good balance."
For more information you can check out Clark Colborn on the web at clarkplaysguitar.com and facebook.com/ClarkColborn. CDs will be sold at store.clarkplaysguitar.com, Amazon.com, CD Baby and other online retailers. Legitimate media representatives can get a review copy of the CD by sending an email to pr@musicstreetjournal.com. Use the same email address to set up interviews, show passes or to get any other information you need for features.
Contact: pr@musicstreetjournal.com
Clarkplaysguitar.com
facebook.com/ClarkColborn


Flu activity remained elevated across the U.S. for the week ending March 16, according to the CDC's most recent FluView report.
1. The percentage of outpatient visits for influenza-like illness was 4.4 percent for the week ending March 16, which is still higher than the national baseline of 2.2 percent. All 10 U.S. regions reported influenza-like illness at or above their region-specific baseline level.
2. The CDC confirmed 8,724 positive respiratory specimens for influenza A and 322 positive specimens for influenza B in the week ending March 16. The agency has confirmed 134,867 positive specimens since Sep. 30, 2018.
3. The overall flu-associated hospitalization rate increased from 41.3 per 100,000 for the week ending March 9 to 47.1 per 100,000 for the week ending March 16. The CDC reported 13,604 laboratory-confirmed flu-associated hospitalizations between Oct. 1, 2018, and March 16.
4. Twenty-six states experienced high influenza-like illness activity for the week ending March 16, down four states from a week prior. Twelve states reported moderate activity; eight states and Puerto Rico experienced low activity; and four states experienced minimal activity.
5. Eight pediatric flu deaths were reported to the CDC for the week ending March 16, bringing the total count to 76 for the 2018-19 season.
6. Forty-four states reported widespread flu activity for the week ending March 16. Four states and Puerto Rico reported regional flu activity, and the District of Columbia and two states reported local flu activity.


text:
Netflix increases price for US subscribers
MobileWorld January 16, 2019, 11:01 am January 16, 2019 0 405
Netflix is increasing monthly fees for its US subscribers by between 13 percent and 18 percent, the video streaming pioneer's first price increase since 2017 as it spends heavily on original content and international expansion.
This was disclosed in a statement by the company that prices for its popular standard plan, which allows streaming on two devices at the same time, will rise to $12.99 (R177.52) per month from $10.99.
Netflix shares rose 6.5 percent to close at $354.64, adding to their 30 percent rise so far this year, the company's top-tier plan, which allows streaming on four screens in high definition, will increase to $15.99 from $13.99 per month, while the fee for its basic plan will rise to $8.99 from $7.99.
In comparison, AT&T's HBO Now streaming service charges $14.99 per month, while Hulu's no-advertisements plan is priced at $11.99 per month.
The Pivotal Research Group analyst Jeff Wlodarczak, said, "It highlights that Netflix has pricing power and even after the increase it remains a very cheap entertainment alternative."
Netflix has been spending billions to bolster its original content, which boasts award-winning shows such as "The Crown," "Black Mirror" and "Wild Wild Country" to fend off intensifying competition from players such as Amazon.com's (AMZN.O) Prime Video service and Hulu. It was reported that the company had 137 million customers at the end of September.
The price hikes are expected to fetch Netflix hundreds of millions of dollars, ahead of the launch of streaming services from AT&T and Walt Disney.
While aggressive spending – a planned $8 billion in 2018 – has led to a surge in subscriber growth, its debt doubled to $6.50 billion in 2017 from $3.36 billion in 2016.
According to Daniel Morgan, senior portfolio manager at Synovus Trust Co, which owns 15,019 shares of Netflix, disclosed that the company is expected to have a debt level of $8.33 billion in 2018.
"With Netflix frequently tapping the debt markets on several recent occasions, the price hike could help ease concerns with a growing deficit on free cash flow to fund a likely continued escalation in Netflix's content spending, which likely topped $13 billion in 2018," CFRA analyst Tuna Amobi added.
However, Prices for Netflix will remain the same in South Africa.
Hot TopicNetflixPriceSubscribersUS
Previous ArticleABN Amro introduces contactless payments with wearablesNext ArticleWorldRemit launches in South Africa


HTF MI recently introduced Global Telecom IoT Market study with in-depth overview, describing about the Product / Industry Scope and elaborates market outlook and status to 2023. The market Study is segmented by key regions which is accelerating the marketization. At present, the market is developing its presence and some of the key players from the complete study are AT&T, China Mobile, Deutsche Telekom, Ericsson, Verizon Communications & Vodafone etc.
This report studies the Global Telecom IoT market size, industry status and forecast, competition landscape and growth opportunity. This research report categorizes the Global Telecom IoT market by companies, region, type and end-use industry.
Browse 100+ market data Tables and Figures spread through Pages and in-depth TOC on " Telecom IoT Market by Type (Connectivity Technology, Network Management Solution, Services & Others), by End-Users/Application (Smart Buildings and Home Automation, Capillary Network Management, Industrial Manufacturing and Automation, Vehicle Telematics, Transportation, Logistics Tracking and Traffic Management, Energy and Utilities, Smart Healthcare & Others), Organization Size, Industry, and Region - Forecast to 2023". Early buyers will receive 10% customization on comprehensive study.
The key players are highly focusing innovation in production technologies to improve efficiency and shelf life. The best long-term growth opportunities for this sector can be captured by ensuring ongoing process improvements and financial flexibility to invest in the optimal strategies. Company profile section of players such as AT&T, China Mobile, Deutsche Telekom, Ericsson, Verizon Communications & Vodafone includes its basic information like legal name, website, headquarters, its market position, historical background and top 5 closest competitors by Market capitalization / revenue along with contact information. Each player/ manufacturer revenue figures, growth rate and gross profit margin is provided in easy to understand tabular format for past 5 years and a separate section on recent development like mergers, acquisition or any new product/service launch etc.
The Global Telecom IoT Market has been divided into type, application, and region.
On The Basis Of Type: Connectivity Technology, Network Management Solution, Services & Others.
Telecom IoT (Volume, Value and Sales Price) table defined for each geographic region defined.


Tolu and his mum left for Chicago, leaving me alone in their house. So, I had to move to his sister's house and live there. Unfortunately, it was at a time I lost my admission and there was nothing to do. All I did was to sleep and wake up in his sister's house. Some people who knew my story had been going to my dad to complain; he didn't do anything to stop me or look for me.
It was after I left Tolu's house that my dad sent policemen to look for me and bring me home. Tolu had already travelled and the house was locked then. They left it for almost a year before they started looking for me. Couldn't believe it. Nice eeh uhmm.
When Tolu got to the US, he did all he could to bring me to the U.S. He paid an agent to process a visa for me, but I was eventually refused. After some few months, he came back to Nigeria on my birthday, and threw a surprise 21st birthday party for me. He got me a passport to enable me travel and told me to wait for him till. He said he would be back to take me to the US.
Three months later he came back with his sick mum, still sick and this time around, we all went to a distant village to seek spiritual help for his sick mum. This was not like the previous places; this was more like witchdoctor's house. When we got there, the man we met said we shouldn't have gotten to a particular place we stood, that we should have waited in the car for him to come and get us. This prompted him to do some spiritual arrangement for each person by getting some local eggs for sacrifice.
Tolu used the opportunity to convince me into taking an oath with him which I felt was okay. I was so blinded by love not knowing I was making a big mistake. He spoke with the witchdoctor and he asked us to come back to take the oath. His mum overheard us talking about it and refused to allow us to go. Tolu was upset and said we had to go, but because his mum did not support the idea, I decided not to go with him. However, after a while we went to Osun shrine in Osun State. All I could remember was, when we got to a bridge, the guy showing us around told us if we made any wish in the river flowing under the bridge, it will come to pass. We went to the river and we made a wish to be together forever. We even drank the water and did all sort of things.
Whenever I think about those days and where I am at the moment, I realise that if I had known I shouldn't have taken an oath with him. Shortly after, they left for the US again and he started behaving funny. I started hearing stories of him drinking and having girlfriends. I felt very bitter. I would call him several times but he wouldn't pick up. I also would call his mum but she won't say a word about his attitude. I felt really hurt and was determined to go to the U.S too.
I went to my dad one day, and showed him the passport Tolu gave me and I asked him to pay for my trip that I have got people in the US to look after me. I was 21 then. He told me if he paid for the travelling and I got deported, the money he would give me would be what I would inherit from him, so he would include it in his will. I accepted the inheritance pact.
My dad paid for the ticket and made arrangements for me to travel with the passport Tolu left for me and I left for Chicago.
I got to Chicago without telling Tolu, a friend picked me up from the airport. And getting to my friend's one bedroom house, she had four other boys and a lady living in the one bedroom flat with her. Staying there with them would have put the number at seven altogether. So, I decided to call one of Tolu's sisters who lived in Chicago. Immediately she came to pick me up.
She told me tales about their mum who she said was so mean. When they approached her about discouraging Tolu on having another girlfriend, she refused. She quoted what her mum said: 'If she (that's me) didn't take care of me, back in Nigeria, somebody else would have'. She was referring to me; can you imagine, after all I did. I lost my admission because of her and all I got was this from her. I felt really hurt, after all. I was very upset with her.
Note: The story continues next week.
Thanks so much for providing individuals with remarkably terrific opportunity to discover important secrets from this web site. It's always so excellent and also stuffed with a great time for me and my office fellow workers to search your website more than three times in a week to study the latest items you have got. And definitely, we are always motivated with all the special thoughts you give. Certain two ideas in this post are clearly the most impressive I have had.
Hi, Neat post. There is a problem along with your web site in web explorer, would check this IE still is the marketplace chief and a large component to folks will miss your excellent writing due to this problem.
I happen to be writing to let you be aware of of the magnificent encounter my cousin's child had reading yuor web blog. She picked up a wide variety of pieces, with the inclusion of what it is like to possess an excellent teaching character to let many people without difficulty know just exactly specified specialized subject matter. You really exceeded our own expectations. I appreciate you for showing the precious, trustworthy, explanatory and in addition unique tips on this topic to Lizeth.
Thank you for every other magnificent post. Where else may just anyone get that type of info in such an ideal way of writing? I have a presentation subsequent week, and I am on the look for such information.
naturally like your web-site but you have to test the spelling on quite a few of your posts. Many of them are rife with spelling issues and I to find it very troublesome to inform the reality however Iˇll surely come back again.
I like the valuable information you provide in your articles. I抣l bookmark your blog and check again here regularly. I am quite sure I will learn a lot of new stuff right here! Best of luck for the next!
What i don't realize is in truth how you're not really much more well-appreciated than you might be right now. You're very intelligent. You realize therefore significantly in terms of this topic, produced me in my view consider it from a lot of various angles. Its like men and women aren't interested until it is one thing to accomplish with Girl gaga! Your personal stuffs excellent. At all times handle it up!
Hi, for all time i used to check website posts here early in the morning, as i enjoy to learn more and more.
I have been surfing on-line greater than three hours nowadays, yet I by no means discovered any interesting article like yours. It is beautiful value enough for me. In my opinion, if all site owners and bloggers made just right content material as you did, the internet will probably be much more useful than ever before.
Hello to every body, it's my first go to see of this weblog; this webpage carries awesome and in fact excellent stuff designed for visitors.
thanks for visiting duchess international magazine,keep visiting, we always keep you updated.
Wow, superb weblog layout! How lengthy have you ever been running a blog for? you made running a blog look easy. The whole glance of your website is magnificent, as neatly as the content material!
Hello, Neat post. There is a problem along with your web site in web explorer, would check this IE nonetheless is the market leader and a good section of other people will miss your magnificent writing due to this problem.
Nice post. I was checking constantly this blog and I'm impressed! Very useful information particularly the last part 🙂 I care for such info much. I was seeking this certain information for a very long time. Thank you and good luck.
Thanks for every other informative web site. Where else may I get that type of information written in such an ideal means? I have a challenge that I am simply now running on, and I've been at the glance out for such information.