text:
\section{Introduction}

Consider a network where each node of the network has a private observation and needs to produce an action. These actions should be coordinated with the observations; therefore some form of communication is necessary among the nodes. The fundamental limits of the required communication was originally studied by Cuff et al.~in  \cite{cuff} where the authors assumed that the observation of node $i$ ($1\leq i\leq m$) are  i.i.d.~repetitions of some random variable $X_i$. The joint distribution of $(X_1, X_2, ..., X_m)$ was assumed to be a given. Denoting the action of the $i$-th node by $Y_i$, coordination was then modeled as requiring the joint pmf of the outputs conditioned on the inputs to be very close to some given $p(y_1, y_2, ..., y_m|x_1, x_2, ...,x_m)$. Here the authors introduce two notions of empirical and strong coordination: in the strong coordination, memoryless repetitions of the channel $p(y_1, y_2, ..., y_m|x_1, x_2, ...,x_m)$ are simulated, whereas in empirical coordination, only the data histograms (or its joint type) is equal  $p(y_1, y_2, ..., y_m, x_1, x_2, ...,x_m)$.
The common theme is that the conditional pmf $p(y_1, y_2, ..., y_m|x_1, x_2, ...,x_m)$ is approximated \emph{asymptotically} as the number of i.i.d.~observations (the block length) goes to infinity. In this work, however, we are interested in exact zero error coordination, \emph{i.e.,} coordination should be achieved with probability one. In this way, our work is related to \cite[Sec IV]{exact}, \cite{Winter} on exact strong coordination capacity, however we adopt a different \emph{set coordination} criterion (which is closer to the empirical notion of coordination).

In our setup, for any observation vector $(x_1, x_2, ..., x_m)$ by the $m$ nodes of a network, we assume a permissible set of output actions $(y_1, y_2, ..., y_m)$. In other words, we are not directly interested in simulating a given $$p(y_1, y_2, ..., y_m|x_1, x_2, ...,x_m).$$ Rather, for every $(x_1, x_2, ..., x_m)$, we define a set $$\mathcal{A}_{x_1, x_2, ..., x_m}\subseteq \mathcal{Y}_1\times \mathcal{Y}_2\times\cdots\times \mathcal{Y}_m,$$
such that $(y_1, y_2, ..., y_m)\in\mathcal{A}_{x_1, x_2, ..., x_m}$. Here $\mathcal{Y}_i$ is the action set of node $i$. We call this a ``set coordination." In Section \ref{sec:set-coordination}, we compare set coordination with empirical coordination.
\begin{example} If $|\mathcal{A}_{x_1, x_2, ..., x_m}|=1$, the value of $(y_1, y_2, ..., y_m)$ will be uniquely specified and will be a deterministic function of $(x_1, x_2, ..., x_m)$. In this case, coordination reduces to distributed computation. Distributed computation is itself a more general problem that the message transmission problem, since the functions computed by the nodes can be taken to be the message of other nodes.\end{example}

Consider the special case of a network with two nodes with node one has input $X_1$ and node two producing output $Y_2$. We assume that the input of node two, $X_2$, and the output of node one, $Y_1$, are disabled, \emph{i.e.,} $|\mathcal{X}_2|=|\mathcal{Y}_1|=1$. Then for every $x_1$ we have a set $\mathcal{A}_{x_1}\subseteq \mathcal{Y}_2$. Assume a one-way communication link from node one to node two. The goal of the first user will be to send a message from node one to node two that will enable production of $y_2\in\mathcal{A}_{x_1}$ at node two. We show in Section \ref{sec:3} that this  special case of the coordination problem is equivalent with the  ``Hide and Seek" problem of McEliece and Posner \cite{b}.  McEliece and Posner define a source coding problem and a zero-sum ``Hide and Seek" game. Rather surprisingly, they illustrate that the optimal compression rate of the source coding problem can be expressed in terms of  the Nash equilibrium of the game. Additional insight was provided by Lov\'asz who provided an elegant combinatorial argument for the result of McEliece and Posner in \cite{c}. We review other related results and in particular connections with R\'enyi mutual information of order $\alpha$ is discussed in Section \ref{sec:renyi}.

We continue by two examples that illustrate connections with zero error rate distortion  (see  \cite[Ch. 2]{d}, \cite{Max}), and with graph entropy. Firstly, consider a non-negative distortion function satisfying $d(x_1, y_2)=0$ if and only if $y_2\in\mathcal{A}_{x_1}$. Then, coordination is equivalent with zero distortion in reconstruction. Secondly, consider the source coding problem for a source that is taking values in a set $\mathcal X_1$. We are given a graph $\mathcal G$ on $\mathcal X_1$, where two symbols $x_1$ and $x'_1$ are connected to each other if it is legitimate to reconstruct $x'_1$ when the source value is $x_1$. We can model this by assuming that $\mathcal Y_2=\mathcal X_1$ and $\mathcal{A}_{x_1}$ being equal to the set of all $x'_1$ that are connected to $x_1$. The message transmitted from node one to node two represents the compressed message. The minimum compression rate in this case is equal to the logarithm of the chromatic number of the complement of $\mathcal G$ in the one-shot case when only one instance of the source is given. After coloring the vertices of  the complement of $\mathcal G$, the message can be the identity of the color that is assigned to the source symbol.  In the asymptotic case when multiple instances of the source are observed, the answer is the logarithm of the fractional  chromatic number \cite[p. 2215]{zero-error-it}.

For the asymptotically vanishing error model, authors in \cite[Conjecture 1]{cuff}  conjectured that empirical coordination and strong coordination have the same rate regions when infinite common randomness is provided to the parties. Considering a special two node network, we observe connections in the zero error model. In \cite{Winter}, authors considers strong coordination with unlimited common randomness and arrives at expressions that match the one given by McEliece and Posner. However, the work of McEliece and Posner (or its follow up works) are not cited  in \cite{Winter} and the connection is not noted. See Section \ref{sec:strong-coordination} for more details.

This paper is organized as follows: in Section \ref{prelim} we set up the notation that we use. Section \ref{sec:set-coordination} defines set coordination and defines one-way coordination capacity for a two node problem. A general lower bound for this problem is given in Section \ref{sec:general-lower}. Section \ref{sec:3} provides a detailed treatment for the special case of two nodes and discusses its connections with various known results. Section \ref{sec:with-side} computes the coordination capacity when the side information of the second node is a function of the side information of the first node. Finally, in Section \ref{sec:linear}, we consider linear coordination and provide several new results. Extensions to MAC and BC setups are given in Section \ref{sec:extensions}.

\section{Notation and Preliminaries}\label{prelim}
We adopt the notation of \cite{elgamal-kim}. In particular, we show the set $\{1,2,...,m\}$ by $[m]$, and the set $\{k+1, k+2, ..., m\}$ by $[k+1:m]$. All random variables in this paper are finite discrete random variables. All the logarithms are in base 2 in this paper.

Given two graphs $\mathcal G_1, \mathcal G_2$, the tensor product $\mathcal G_1\otimes\mathcal G_2$ is a graph whose vertex set is the Cartesian product of the vertex sets of $\mathcal G_1$ and $\mathcal G_2$ defined as follows: two vertices $(u_1,u_2)$ and $(v_1,v_2)$ in  are adjacent in $\mathcal G_1\otimes\mathcal G_2$  if and only if $u_1$ is adjacent with $v_1$ in $\mathcal G_1$ and $u_2$ is adjacent with $v_2$  in $\mathcal G_1$ .

There are many definitions for R\'enyi mutual information (see \cite{Verdu} for a review). One definition for R\'enyi mutual information of a joint pmf $p(x,y)$ is as follows:
\begin{equation}
I_{\alpha}(X;Y)=\min_{q(y)} D_{\alpha}(p(x,y)||p(x)q(y)),\label{defIalpha}
\end{equation}
Where $D_\alpha$ is the R\'enyi divergence between two pmfs is defined as follows: $$D_\alpha(p\| q) = \frac{1}{\alpha -1} \log \left(\sum_x p(x)^{\alpha}q(x)^{1-\alpha}\right).$$
Note that as $\alpha$ converges to one, R\'enyi divergence and R\'enyi mutual information of order $\alpha$ tend to the KL divergence and Shannon's mutual information.

In \cite[Eq 13]{e}, it is shown that mutual information of order $\alpha$, as defined in equation \eqref{defIalpha}, is equal to:
\begin{equation}
I_{\alpha}(X;Y)=\frac{\alpha}{\alpha-1}\log \bigg(\sum_{y}\Big[\sum_{x}p(x)p(y|x)^\alpha\Big]^{1/\alpha}\bigg).
\end{equation}
R\'enyi mutual information $I_\alpha(X;Y)$ is a non-decreasing function of $\alpha$ for $\alpha\in[0,\infty]$ \cite[Thm 3]{renyi}.

We now provide explicit expressions for $I_\alpha(X;Y)$ when $\alpha=0$ and $\alpha=\infty$.
When $\alpha$ goes to zero, $\sum_x p(x)p(y|x)^{\alpha}$ rises to power ${1}/{\alpha}$ which goes to infinity; thus only the largest term is important. In fact, one can show that
\begin{align}
\begin{split}
I_0(X;Y)&=\lim_{\alpha \to 0} \frac{\alpha}{\alpha-1}\log \bigg(\max_{y}\Big[\sum_{x}p(x)p(y|x)^\alpha\Big]^{1/\alpha}\bigg)\\&=\lim_{\alpha \to 0}\frac{1}{\alpha-1}\log \Big(\max_{y}\big[\sum_{x}p(x)p(y|x)^\alpha\big]\Big) \\&=-\log(\max_{y}\sum_{p(y|x)>0} p(x)).
\end{split}
\end{align}
Similarly, one can show that as $\alpha$ tends to infinity, we have
\begin{align}I_\infty(X;Y)=\log\left(\sum_{y}\underset{x: p(x)>0}{\max}\;p(y|x)\right)\label{eqn:mutu0inf}.\end{align}
\section{System Model}
\subsection{Set Coordination}\label{sec:set-coordination}
\begin{definition}
Given an input pmf $p(x_1, \cdots, x_m)$ and action sets $\mathcal{A}_{x_1, x_2, ..., x_m}$, one-shot and asymptotic coordination are defined as follows: in one-shot coordination, the parties observe only one instance of $X_i$ and coordination is achieved if $(y_1, y_2, ..., y_m)\in\mathcal{A}_{x_1, x_2, ..., x_m}$ for any $(x_1,\cdots, x_m)$ where $p(x_1, \cdots, x_m)>0$.
In the asymptotic version, the parties observe $n$ i.i.d.~ repetitions of the sources $X_1^n, X_2^n, \cdots, X_m^n$. Coordination is achieved if  $(y_{1i}, y_{2i}, ..., y_{mi})\in\mathcal{A}_{x_{1i}, x_{2i}, ..., x_{mi}}$  for any $i\in[n]$ and any $(x^n_1,\cdots, x^n_m)$ where $p(x^n_1, \cdots, x^n_m)=\prod_{i=1}^np(x_{1i}, \cdots, x_{mi})>0.$
\end{definition}

We assume that the nodes have access to limited communication resources, as well as possibly private or common randomness. However, similar to empirical coordination, without loss of generality we can assume that the nodes are deterministic and do not use shared or private randomness. This is because  $(y_1, y_2, ..., y_m)\in\mathcal{A}_{x_1, x_2, ..., x_m}$ with probability one, and hence it has to hold for all possible values of the shared randomness variable.

Set coordination is related to empirical coordination. Take some arbitrary conditional pmf $$p(y_1, y_2, ..., y_m|x_1, x_2, ...,x_m)$$ such that $p(y_1, ..., y_m|x_1, ...,x_m)>0$ only if $(y_1, y_2, ..., y_m)\in\mathcal{A}_{x_1, x_2, ..., x_m}$. Then a zero-error empirical coordination code for $p(y_1, y_2, ..., y_m|x_1, x_2, ...,x_m)$ is also a zero-error set coordination code. However, set coordination is more relaxed in the asymptotic formulation. Take a set coordination code of block length $n$ and two sequence $(x_1^n, x_2^n, \cdots, x_m^n)$ and $(\tilde x_1^n, \tilde x_2^n, \cdots, \tilde x_m^n)$ of the same type. Let $(y_1^n, y_2^n, \cdots, y_m^n)$ and $(\tilde y_1^n, \tilde y_2^n, \cdots, \tilde y_m^n)$ denote the actions of the nodes in response to $(x_1^n, x_2^n, \cdots, x_m^n)$ and $(\tilde x_1^n, \tilde x_2^n, \cdots, \tilde x_m^n)$  respectively. Then it can be the case that the joint types of the sequences $(x_1^n, x_2^n, \cdots, x_m^n, y_1^n, y_2^n, \cdots, y_m^n)$ and $(\tilde x_1^n, \tilde x_2^n, \cdots, \tilde x_m^n, \tilde y_1^n, \tilde y_2^n, \cdots, \tilde y_m^n)$ are different. Therefore, one cannot assign a single empirical conditional type $p(y_1, ..., y_m|x_1, ...,x_m)$ to the set coordination code.

\begin{remark}
Despite the apparent difference between set coordination and empirical coordination, it would be interesting to study whether set coordination can be expressed in terms of empirical coordination (under either zero error, or asymptotically zero error criteria). As we will see in the proof of Theorem \ref{thm:t1}, the known converse techniques for empirical coordination extend to set coordination.
\end{remark}

\subsection{One-way coordination capacity}\label{sec:one-waycoordination}
A two nodes network is characterized by two alphabet sets $\mathcal{X}_1$ and $\mathcal{X}_2$ for inputs and two action sets $\mathcal{Y}_1$ and $\mathcal{Y}_2$. For each pair of inputs $(x_1,x_2)\in \mathcal{X}_1\times\mathcal{X}_2$ we have a permissible action set $A_{x_1,x_2}\subseteq \mathcal{Y}_1\times\mathcal{Y}_2$. We are given some $p(x_1, x_2)$ on the inputs. Assume that there is a one-way communication link of limited rate $R$ from node one to node two, as depicted in Fig.~\ref{fig:one-way}.
\begin{figure}
\begin{center}
\begin{tikzpicture}[node distance=3.5cm,auto,>=latex', scale=1.4]
    \draw[black, thick,rounded corners] (0,0) rectangle (1,0.6) node[font=\fontsize{9.5}{144}\selectfont,pos=.5,anchor=center] {Node 1};
	\draw[black, thick,->] (1,.3) -- (3,.3) node[font=\fontsize{9}{144}\selectfont,pos=.5,anchor=south] {$M\in [2^{nR}]$};
	\draw[black,thick,rounded corners](3,0) rectangle (4,0.6) node[font=\fontsize{9.5}{144}\selectfont,pos=.5,anchor=center]{Node 2};
	\draw[black, thick,->] (0.5,1) ->(0.5,0.6) node[font=\fontsize{8.5}{144}\selectfont,pos=.2,anchor=south]{$X^n_1$};
	\draw[black, thick,->] (0.5,0) -> (0.5,-0.4) node[font=\fontsize{8.5}{144}\selectfont,pos=.85,anchor=north] {$Y^n_1$};
	\draw[black, thick,->] (3.5,1) -> (3.5,0.6) node[font=\fontsize{8.5}{144}\selectfont,pos=.2,anchor=south] {$X^n_2$};
	\draw[black, thick,->] (3.5,0) -> (3.5,-.4) node[font=\fontsize{8.5}{144}\selectfont,pos=.85,anchor=north] {$Y^n_2$};
\end{tikzpicture}
\end{center}
\caption{Coordination using a one-way communication link.}\label{fig:one-way}
\end{figure}

\begin{definition}
Coordination is achievable with one-way communication rate $R$ with block length $n$ if there are encoding and decoding functions
\begin{align}
\nonumber
&\mathcal{E}:{\mathcal{X}_1}^n\mapsto[2^{nR}],\qquad\mathcal{D}_1:\mathcal{X}_1^n\mapsto \mathcal{Y}^n_1,\qquad \mathcal{D}_2:[2^{nR}]\times \mathcal{X}_2^n\mapsto \mathcal{Y}^n_2,
\end{align}
such that $(Y_1^n, Y_2^n)=(\mathcal{D}_1(X_1^n), \mathcal{D}_1(M, X_2^n))$ is coordinated with $(X_1^n, X_2^n)$ where the message $M=\mathcal{E}(X_1^n)$.

A rate $R$ is said to be one-shot achievable if it is achievable with a code of block length $n=1$, and is said to be asymptotically achievable if  it is achievable with a code with for some arbitrarily large block length $n$. We use $C$ to denote the maximum one-shot achievable rate, and $\bar{C}$ to denote the supremum of the  asymptotically achievable rates.
\end{definition}
\begin{remark}Computing the  one-way capacity $\bar C$ is in general a difficult problem. For instance, consider the special case of $Y_1$ being a constant random variable, i.e. $|\mathcal{Y}_1|=1$, and $\mathcal{Y}_2=\mathcal{X}_1$. Further assume $y_2\in A_{x_1, x_2}$ if and only if $y_2=x_1$; in other words, the task of node one will be to communicate $x_1$ to node two. Given some $p(x_1, x_2)$, the problem reduces to a zero-error version of the Slepian-Wolf problem. The minimum rate for  zero-error source coding with side information is known as Witsenhausen's zero-error rate and is open in general \cite{Witsenhausens}. Practical code designs for this problem can be found in \cite{Slepian1, Slepian2, Slepian3}.
\end{remark}

\section{A general lower bound}\label{sec:general-lower}
One can use the fact that zero-error coordination is a more stringent condition than vanishing error coordination to derive the following lower bound on the asymptotic coordination capacity. Let $\mathcal{P}$ be the class of all $p(y_1,y_2|x_1,x_2)$ such that $p(y_1,y_2|x_1,x_2)>0$ only if $(y_1, y_2)\in A_{x_1, x_2}$. Then we have:
\begin{theorem}\label{thm:t1}
If node one observes $X_1$ and node two observes $X_2$ and they want to produce $Y_1$ and $Y_2$ respectively then minimum rate required from node one to node two is bounded from below by:
\begin{equation}
\max_{\substack{q(x_1,x_2): \\q(x_1, x_2)=0 \text{ if }p(x_1, x_2)=0}} ~\min_{q(y_1,y_2|x_1,x_2)\in \mathcal{P}} \min_{\substack{ q(f|x_1,y_1): \\I_q(X_1,Y_1;Y_2|F,X_2)=0}} I(F;X_1|X_2).
\end{equation}
\end{theorem}
\begin{proof}[Proof of Theorem \ref{thm:t1}] Coordination capacity depends only on the support of $p(x_1, x_2)$, i.e.~$\{(x_1, x_2): p(x_1, x_2)>0\}$. Furthermore, if the support of a pmf $q(x_1, x_2)$ is smaller than the support of $p(x_1, x_2)$, the coordination capacity for $q(x_1, x_2)$ will be less than or equal to that for $p(x_1, x_2)$. Therefore,  it suffices to show that the coordination capacity for the input pmf $p(x_1, x_2)$ is bounded from below by
\begin{equation}
\min_{p(y_1,y_2|x_1,x_2)\in \mathcal{P}} \min_{\substack{p(f|x_1,y_1): \\I_p(X_1,Y_1;Y_2|F,X_2)=0}} I(F;X_1|X_2).
\end{equation}
Take an arbitrary set coordination code of length $n$. Consider the  converse given in \cite[Appendix G]{Yassaee} for interactive channel simulation under the empirical coordination constraint. Consider the special case of one round of communication $r=1$. This converse can be exactly mimicked for the set coordination code until equation (102). In equations  (102) onwards in \cite[Appendix G]{Yassaee}, it is argued that the joint pmf of $(Y_{1Q}, Y_{2Q}, X_{1Q}, X_{2Q})$, where $Q$ is the time sharing random variable, is close to the desired joint pmf $p(y_1,y_2,x_1,x_2)$ that we wish to simulate. By contrast, in set coordination, we have that $(Y_{1Q}, Y_{2Q})\in A_{X_{1Q}, X_{2Q}}$ with probability one. Therefore, $p(y_{1Q}, y_{2Q}|x_{1Q}, x_{2Q})\in\mathcal{P}$. Therefore, the last step in the proof can be completed, and we get that the transmission rate is greater than or equal to $I(F;X_{1Q}|X_{2Q})$. Since  $p(y_{1Q}, y_{2Q}|x_{1Q}, x_{2Q})\in\mathcal{P}$, we get that the coordination rate is greater than or equal to
\begin{equation}
\min_{p(y_1,y_2|x_1,x_2)\in \mathcal{P}} \min_{\substack{p(f|x_1,y_1): \\I_p(X_1,Y_1;Y_2|F,X_2)=0}} I(F;X_1|X_2).
\end{equation}
\end{proof}
\begin{corollary}\label{cor} If $X_2$ and $Y_1$ are constant random variables, random variable $F$ will satisfy $X_1\rightarrow F\rightarrow Y_2$ and we would like to minimize $I(X_1;F)$. The choice of $F=Y_2$ will be optimal and we get that the coordination capacity is greater than or equal to
$$\max_{q(x_1)}\min_{p(y_2|x_1)\in \mathcal{P}}I(X_1;Y_2).$$
We will see later that this lower bound is tight.
\end{corollary}

 \section{No side information at node two}\label{sec:3}
In this section we restrict to the special case of a two nodes network, with node two having no observation, \emph{i.e.,} $|\mathcal{X}_2|=1$. Without loss of generality we assume that $p(x_1)>0$ for all $x_1$ throughout this section. In this case we can make some simplifications as follows: we can drop $x_2$ from $A_{x_1x_2}$ and define $$A_{x_1}\triangleq \{y_2:\exists y_1\in \mathcal{Y}_1, (y_1,y_2)\in A_{x_1x_2}\}.$$ Notice that if node two chooses a proper action $y_2$ from $A_{x_1}$, by definition there will exist a proper action $y_1$ for node one.  Since node one knows $x_1$ and the message sent to node two as well as the decoding strategy of node two, action $y_2$ and hence $y_1$ can be found by this node.

\subsection{One-shot achievable rates}\label{one-shot}
We would like to compute the minimum size of the alphabet set of message $M$ sent from node one to node two, such that node two can choose an action $y_2\in\mathcal{Y}_2$ where $y_2 \in A_{x_1}$. In other words, assuming that $M\in \mathcal{M}$, we would like to minimize $|\mathcal{M}|$ as much as possible. The one-shot capacity is minimum of $\log|\mathcal{M}|$.

\textbf{Characterization in terms of graph covers:}
The one-shot capacity can be also expressed in terms of the size of minimum graph cover \cite{b}. Before proceeding, we need some definitions:

\begin{definition}[Coordination Graph]\label{coordinationgraph}
 Consider a bipartite graph $\mathcal G=(\mathcal X_1, \mathcal Y_2)$, where by this notation we mean that the vertices in one part being indexed by elements of $\mathcal X_1$ and the vertices of the other part indexed by elements of $\mathcal Y_2$. An edge is drawn between $x_1\in \mathcal X_1$ and $y_2\in \mathcal{Y}_2$ if and only if $y_2\in A_{x_1}$. We call this a \emph{coordination graph}. Both the one-shot and asymptotic coordination capacity are characterized by the coordination graph. Hence we sometime denote the one-shot and asymptotic coordination capacity by $C(\mathcal G)$ and $\bar C(\mathcal G)$ respectively.
\end{definition}

\begin{definition}[\cite{b}]
Consider a bipartite graph $\mathcal{G}=(\mathcal X_1, \mathcal Y_2)$. A cover set for $\mathcal X_1$ is a subset $\mathcal S\subseteq \mathcal Y_2$ such that for each $x_1\in \mathcal X_1$, there exist some $y_2\in \mathcal S$ such that $x_1$ and $y_2$ are connected.
\end{definition}
Observe that the minimum length of message from node one to node two, $|\mathcal{M}|$ is equal to the size of minimum cover for $\mathcal X_1$. This is because given a cover set $\mathcal S$, node one can simply send the index of $y_2$ in the cover set $\mathcal S$ that is connected to $x_1$. Conversely, given any strategy by node one, one can produce a cover set by putting together the outputs $y_2$ corresponding to different values of the message $M$.

Finding minimum cover set for graph $\mathcal{G}=(\mathcal X_1, \mathcal Y_2)$ is equivalent to solving following integer linear programming:
\begin{align}
\begin{split}
 \mathsf{IP} (\mathcal G)&= \min \sum_{y_2\in \mathcal{Y}_2} \alpha(y_2)  \\
 s.t. \hspace{3 mm} \forall x_1&\in \mathcal{X}_1 \hspace{2 mm} \sum_{y_2: y_2\sim x_1} \alpha(y_2) \geq 1 \hspace{3 mm} \\  &\alpha(y_2)\in \{0,1\}.
\end{split}\label{eqn:LP}
\end{align}
Here $C(\mathcal G)=\log_2 \mathsf{IP} (\mathcal G)$ is the minimum message length in terms of bits.
\subsection{Asymptotic achievable rates}
In the asymptotic version of the problem, we fix some block length $n$. The first node observes some sequence $x_1^n$ and needs to convey it to node two in a way it can produce $y_2^n$ such that $y_{2i}\in A_{x_{1i}}$ for $i\in [n]$. Then, one can see that the solution to this problem is size of minimum cover for a bipartite graph on $\mathcal{X}_1^n$ and $\mathcal{Y}_2^n$, with two sequences $x_1^n$ and $y_2^n$ connected to each other if and only if $y_{2i}\in A_{x_{1i}}$ for $i\in [n]$. This graph can be expressed as $\mathcal{G}^n$, \emph{i.e., }the tensor product of the graph for the one-shot case, $\mathcal{G}$, with itself by $n$ times. Then the minimum required rate for block length $n$ is $\log (\mathsf{IP}(\mathcal G^{\otimes n}))/n$, and the limit of this when $n$ goes to infinity is the asymptotic coordination capacity is equal to $\bar C(\mathcal G)$, \emph{i.e.,}
\begin{equation}
\bar C(\mathcal G)\triangleq \log\lim_{n \to \infty}\big( \mathsf{IP}(\mathcal G^{\otimes n})^{1/n}\big).
\end{equation}
The above limit exist by the Fekete's lemma because the sequence of $\log (\mathsf{IP}(\mathcal G^{\otimes n}))$ is superadditive \cite{b}. In fact, for any two arbitrary graphs $\mathcal G_1$ and  $\mathcal G_2$, we have
\begin{align}
\label{prodID}
\mathsf{IP}(\mathcal G_1\otimes \mathcal G_2)\leq \mathsf{IP}(\mathcal G_1)\mathsf{IP}(\mathcal G_2),
\end{align}
because one possible way to find a cover set for $\mathcal G_1\otimes \mathcal G_2$ is to first find a cover set for $\mathcal G_1$ and for $\mathcal G_2$, and then take the Cartesian product of these two cover sets as a cover set for $\mathcal G_1\otimes \mathcal G_2$.

Authors in \cite{b} also consider the above asymptotic version of the problem and show that $\bar C(\mathcal G)=\log_2\mathsf{LP}(\mathcal G)$,  where $\mathsf{LP}(\mathcal G)$ is the relaxation of linear programming given in equation \eqref{eqn:LP} as follows:
\begin{align}
\begin{split}
\mathsf{LP}(\mathcal G)&= \min \sum_{y_2\in \mathcal{Y}_2} \alpha(y_2)  \\
 s.t. \hspace{3 mm} \forall x_1&\in \mathcal{X}_1 \hspace{2 mm} \sum_{y_2: y_2\sim x_1} \alpha(y_2) \geq 1 \hspace{3 mm} \\  &\alpha(y_2)\geq 0.
\end{split}\label{eqn:LP2}
\end{align}
Motivated by the Hide and Seek problem of \cite{b}, Lov\'asz provides a combinatorial proof for the above relation in \cite{c}. In order to prove \begin{align}\bar C(\mathcal G)=\log\lim_{n \to \infty}\big( \mathsf{IP}(\mathcal G^{\otimes n})^{1/n}\big)=\log\mathsf{LP}(\mathcal G),\label{LovaszThem}\end{align} firstly, it is clear that the above is relaxation of the linear program given in equation \eqref{eqn:LP}. Therefore $\mathsf{LP}(\mathcal G)\leq  \mathsf{IP}(\mathcal G)$. By using the dual problem of linear programming, Lov\'asz shows that \begin{align}
\label{prodID2}\mathsf{LP}(\mathcal G_1\otimes \mathcal G_2)=\mathsf{LP}(\mathcal G_1)\mathsf{LP}(\mathcal G_2).\end{align} Thus $\mathsf{IP}(\mathcal G^{\otimes n})\geq \mathsf{LP}(\mathcal G^{\otimes n})=\mathsf{LP}(\mathcal G)^n$ for all $n$. Hence $\bar C(\mathcal G)\geq \log(\mathsf{LP}(\mathcal G))$. To show the other direction, $\log(\mathsf{LP}(\mathcal G))\geq \bar C(\mathcal G)$,  Lov\'asz provides an elegant combinatorial argument using a greedy algorithm and the dual of the linear program. We refer the readers to \cite{c} for the details.

\begin{remark}
While computing the minimum cover $C(\mathcal G)$ is an NP-complete problem (it is the sixth problem of Karp's 21 NP-complete problems \cite{NP}), $\bar C(\mathcal G)$ can be calculated in polynomial time, as it is the solution of a real linear program.

It is known that $C(\mathcal G)$ can be arbitrarily large while $\bar C(\mathcal G)$ is arbitrarily small \cite{Orl}. This happens for the mail-order problem of Slepian, Wolf, and Wyner \cite{Mail}. On the other hand,
$C(\mathcal G)=\bar C(\mathcal G)$ holds for two families of ``interval graphs and forests" that are defined in \cite{Max}.

\end{remark}

We now make the observation that this special case of the coordination problem is nothing but the ``Hide and Seek"  problem of McEliece and Posner \cite{b}. Consider the following source coding problem. Let $\mathcal{X}_1$ be a set of natural numbers. Alice observes some number $x_1\in\mathcal{X}_1$. We have a certain list of properties such as a number being even, being divisible by five, being a prime number, etc. Given some $x_1\in\mathcal{X}_1$, Alice can find a subset of properties that are satisfied by $x_1$, \emph{e.g.,} if $x_1=5$, it is both prime and divisible by 5. The goal of Alice is to inform Bob of \emph{at least one} of these valid properties; thus the goal is not to inform Bob of $x_1$, but one of its valid properties. The question is the minimum amount of communication needed from Alice to Bob to accomplish this task. It is not difficult to see that this problem is identical to the two node coordination problem with no side information at node two: $\mathcal{Y}_2$ can denote the set of properties and $A_{x_1}$ can contain the list of properties that $x_1$ has.

\textbf{Connection to game theory:}
The solution to the Hide and Seek problem can be expressed in terms of the Nash equilibrium of the following zero-sum game: let player one (hider) choose $x_1\in\mathcal{X}_1$ and player two (seeker) choose some $y_2\in \mathcal{Y}_2$. Player one has to give player two one dollar if $y_2\in \mathcal A_{x_1}$, otherwise the payoff is zero. This game is called Hide and Seek because player one hides and player two seeks player one. We refer the readers to \cite{b} for details.

The fact that the above source coding is related to this game may come as a surprise. This is due to the fact that the commonly used achievability proofs and converses in information theory are not based on the  Nash value of games. However, we observe that
\begin{itemize}
\item The key feature of capacity regions and upper bounds thereof are essentially \emph{additive} regions, meaning that they expand by a factor of $n$, when evaluated on $n$ independent repetitions of a problem (see \cite{duality} for a discussion). The Nash value of the above zero-sum repeated game also has the additivity property, as the expected value of the total payoff is equal to the sum of the expected value of the payoffs in the individual games. Therefore, it is quite possible that rates of a capacity region have characterizations in terms of Nash value of carefully constructed games.
\item One can draw simple \emph{operational} connections between the communication problem and the Hide and Seek game. Take an arbitrary code in the communication problem. In the communication problem, the seeker (player two) can ensure that $y_2\in \mathcal A_{x_1}$ if he receives $R$ bits from the hider. However, in the game setup, there is no communication link. Nonetheless, the seeker can still guess the $R$ message bits and win the game with probability at least $2^{-R}$, regardless of the value of $x_1$. Therefore, the seeker has a strategy that gives him a payoff of $2^{-R}$ regardless of the action of the hider. This gives a lower bound on the Nash value of the game in terms of minimum value of $R$. This lower bound is tight by the result of  \cite{b}.
\end{itemize}

\subsubsection{Connection to the Rate Distortion theory}
While elegant, Lov\'asz's proof is combinatorial. Fortunately, the asymptotic capacity $\bar C$ can be found using standard information theory arguments as in \cite{b}. A formal way to do so is to express the the problem in terms of a zero-error rate distortion problem. Consider a distortion measure $d:\mathcal{X}_1\times\mathcal{Y}_2\to\{0,1\}$ where $d(x_1,y_2)=0$ if and only if $x_1$ and $y_2$ be connected, \emph{i.e.,} $y_2\in A_{x_1}$. When  $x_1$ and $y_2$ are not connected, we can set the distortion to an arbitrary positive value, say $d(x_1,y_2)=1$.

Let $R(p(x_1), D)$ be the standard rate distortion function for distortion function $d(\cdot, \cdot)$ when the source $X_1$ has pmf $p(x_1)$. Furthermore, let $R_0(p(x_1), D)$ be the zero-error rate distortion function, which is the minimum (asymptotic) rate which can guaranty average distortion less than or equal to $D$ \emph{with probability one}, \emph{i.e., } with probability of excess distortion being zero. Then, it is easy to see that $R_0(p(x_1), D)$ at $D=0$ is equal to $\bar{C}_\mathcal{G}$ because $E[d(X_1^n,Y_2^n)]\leq 0$ implies that then $d(x_1^n,y_2^n)=0$ for all pairs $(x_1^n,y_2^n)$ where $p(x_1^n,y_2^n)>0$.

Observe that $R_0(p(x_1), D)$ depends only on the support of $p(x_1)$, \emph{i.e.,} the values of $x_1$ where $p(x_1)>0$, and not on the exact values of probabilities $p(x_1)$. Furthermore, it is clear that $R(p(x_1), D)\leq R_0(p(x_1), D)$ since in $R_0(p(x_1), D)$ we ask for exactly zero probability of exceeding the distortion, whereas in $R(p(x_1), D)$ we ask for an asymptotically vanishing probability of excess distortion. Therefore, assuming that $p(x_1)>0$ for all $x_1$, we have
\begin{equation*}
R_0(p(x_1), D)\geq \max_{q(x_1)}R(q(x_1), D).
\end{equation*}
Interestingly, the above inequality holds with equality \cite[Thm 4.2]{d}:
\begin{equation}
R_0(p(x_1), D)=\max_{q(x_1)}R(q(x_1), D)=\max_{q(x_1)}~\min_{p(y_2|x_1): \mathbb E d(X_1,Y_2)\leq D} I(X_1;Y_2).\label{eqn:43}
\end{equation}

Now, let us specialize this result to our coordination problem, when distortion $D=0$. Let $$\mathcal{P}=\{p(y_2|x_1): p(y_2|x_1)>0 \textit{ only if }y_2\in A_{x_1}\}.$$ Then, the asymptotic coordination capacity is equal to \begin{equation}
\bar{C}(\mathcal G)=\max_{q(x_1)}\min_{p(y_2|x_1)\in \mathcal{P}}I(X_1;Y_2).
\end{equation}

An alternative expression for $\bar{C}(\mathcal G)$ follows from \cite[Cor. 3.7]{d}, where it is shown that the standard rate-distortion function has the following characterization at zero distortion $D=0$:
\begin{equation}
R(p(x_1), 0)=-\min_{q(x_1)}\Big\{D(p(x_1)||q(x_1))+\max_{y_2\in \mathcal{Y}_2}\log\big(\sum_{x_1:d(x_1,y_2)=0}q(x_2)\big)\Big\}.
\end{equation}
Then from equation \eqref{eqn:43}, we have
\begin{align}
\bar{C}(\mathcal G)&=R_0(p(x_1), 0)\nonumber
\\&=-\min_{q(x_1)}\max_{y_2\in \mathcal{Y}_2}\log\big(\sum_{x_1:d(x_1,y_2)=0}q(x_1)\big)\nonumber
\\&=\max_{q(x_1)}-\max_{y_2\in \mathcal{Y}_2}\log\big(\sum_{x_1:d(x_1,y_2)=0}q(x_1)\big)\label{eqn:re1}
\\&=\max_{q(x_1)}\min_{p(y_2|x_1)\in \mathcal{P}}I_0(X_1;Y_2),\nonumber
\end{align}
Where $I_0(X_1;Y_2)$ is the R\'enyi mutual information of order zero, defined in Section \ref{prelim}. The last equality holds because for any $q(x_1)$, from the definition of  $I_0(X_1;Y_2)$ we have that $$\min_{p(y_2|x_1)\in \mathcal{P}}I_0(X_1;Y_2)=-\log\big(\max_{y_2\in \mathcal{Y}_2}\sum_{x_1: d(x_1, y_2)=0}q(x_1)\big).$$
\begin{remark}
Curiously,  equation \eqref{eqn:re1} is also equal to the zero-error feedback capacity of a point-to-point channel, when it is positive. This expression is also related to the sphere-packing bound for error exponents (see for instance \cite[Eq. (8)]{Dalai}). It would be interesting to find operational interpretations for these facts.
\end{remark}

\subsubsection{Characterization in terms of R\'enyi mutual information}\label{sec:renyi}
So far, we have mentioned three characterizations for $\bar{C}(\mathcal G)$: one in terms of a relaxed linear program by Lov\'asz, and two as follows:
\begin{align}
\bar{C}(\mathcal G)&=\max_{q(x_1)}\min_{p(y_2|x_1)\in \mathcal{P}}I_0(X_1;Y_2), \label{eqn:mm1}
\\&=\max_{q(x_1)}\min_{p(y_2|x_1)\in \mathcal{P}}I(X_1;Y_2).  \label{eqn:mm2}
\end{align}
It is desirable to provide connections between various characterizations. Our first result states that the expression of Lov\'asz's linear program can be also understood in terms of R\'enyi mutual information, which to best of our knowledge is new. In particular, we use that Lov\'asz's linear program to show the following theorem:
\begin{theorem}Assuming $p(x_1)>0$ for all $x_1$, we have \label{thm:thm1}
\begin{align}
\bar{C}(\mathcal G)&=\max_{q(x_1)}\min_{p(y_2|x_1)\in \mathcal{P}}I_\infty(X_1;Y_2). \label{eqn:mm3}
\end{align}
\end{theorem}
\begin{remark} Since  mutual information of order $\alpha$ is a non-decreasing function in $\alpha$, equations \eqref{eqn:mm1}-\eqref{eqn:mm3} imply that for any $\alpha\in[0,\infty]$ we have
\begin{align}
\bar{C}(\mathcal G)&=\max_{q(x_1)}\min_{p(y_2|x_1)\in \mathcal{P}}I_\alpha(X_1;Y_2). \label{eqn:mm4}
\end{align}
\end{remark}
\begin{proof}[Proof of Theorem \ref{thm:thm1}]
Using equation \eqref{eqn:mutu0inf}, we need to show that
\begin{align*}\bar{C}(\mathcal G)&= \max_{q(x_1)}\min_{p(y_2|x_1)\in \mathcal{P}}\log\left(\sum_{y_2}\underset{x_1: q(x_1)>0}{\max}\;p(y_2|x_1)\right)
\\&=\min_{p(y_2|x_1)\in \mathcal{P}}\log\left(\sum_{y_2}\underset{x_1}{\max}\;p(y_2|x_1)\right).\end{align*}
Let $\alpha({y_2})=\underset{x_1}{\max}\;p(y_2|x_1)$. Then, we would like to show that
\begin{align*}\bar{C}(\mathcal G)= \log\min_{p(y_2|x_1)\in \mathcal{P}}\sum_{y_2}\alpha({y_2}).\end{align*}
Thus, $\bar{C}(\mathcal G)=\log \mathsf{LP}_2(\mathcal G)$ where
\begin{align}
\begin{split}
\mathsf{LP}_2(\mathcal G)&= \min \sum_{y_2} \alpha(y_2)  \\
 s.t. \hspace{3 mm} \alpha(y_2)&\geq p(y_2|x_1),\qquad\forall x_1
\\\hspace{3 mm}\sum_{y_2} p(y_2|x_1)&=1, \qquad p(y_2|x_1)\geq 0
\\\hspace{3 mm}p(y_2|&x_1)=0 \text{ if } y_2\nsim x_1,
\end{split}\label{eqn:LP3n}
\end{align}
where by $y_2\nsim x_1$ we mean that $y_2\notin A_{x_1}$.
Now, recall Lov\'asz's linear programming formulation: $\bar{C}(\mathcal G)=\log \mathsf{LP}(\mathcal G)$ where
\begin{align}
\begin{split}
\mathsf{LP}(\mathcal G)&= \min \sum_{y_2} \alpha(y_2)  \\
 s.t. \hspace{3 mm} \forall x_1&\in \mathcal{X}_1 \hspace{2 mm} \sum_{y_2: y_2\sim x_1} \alpha(y_2) \geq 1 \hspace{3 mm} \\  &\alpha(y_2)\geq 0.
\end{split}\label{eqn:LP2n}
\end{align}
Thus, we only need to show that the two LPs are equivalent. The LP given in equation \eqref{eqn:LP3n} has more variables than the one given in equation \eqref{eqn:LP2n}. Using a standard Fourier-Motzkin elimination on variables $p(y_2|x_1)$, one can see that the former LP reduces to the latter LP. Hence we are done.

\end{proof}

Our aim was to show the connection between Lov\'asz's linear program and the R\'enyi mutual information. But it is possible to provide an algebraic proof for  Theorem \ref{thm:thm1} (see \cite{Max} for another algebraic proof).

\begin{proof}[Second proof of Theorem \ref{thm:thm1}] Let $$F_\alpha(q(x_1))= \min_{p(y_2|x_1)\in \mathcal{P}}I_\alpha(X_1;Y_2).$$ Since R\'enyi mutual information $I_\alpha(X;Y)$ is a non-decreasing function of $\alpha$, we have that $$F_1(q(x_1))\leq F_\infty(q(x_1))\qquad \forall q(x_1).$$
Furthermore, using the fact that $p(x_1)>0$ for all $x_1$ we have that
$$F_\infty(q(x_1))= \log\left(\sum_{y_2}\underset{x_1}{\max}\;p(y_2|x_1)\right)$$
does not depend on $q(x_1)$. Therefore, it suffices to show that at the $q(x_1)$ that maximizes $F_1(q(x_1))$, we have that $F_1(q(x_1))\geq F_\infty(q(x_1))$.

A theorem by Shannon \cite[Thm. 2]{ze} studies the pmf $q(x_1)$ that maximize $F_1(q(x_1))$. This theorem implies that the joint pmf that obtains $\max_{q(x_1)}F_1(q(x_1))$ satisfies $q(x,y)=c\cdot q(x)q(y)$ for some fixed $c$ when $q(x,y)>0$. This implies that $I(X;Y)=I_{\infty}(X;Y)=\log(c)$. Hence,
$$ F_\infty(q(x_1))= \min_{p(y_2|x_1)\in \mathcal{P}}I_\infty(X_1;Y_2)\leq \log(c)=\max_{q(x_1)}F_1(q(x_1)).$$
\end{proof}

\subsection{Connections with strong coordination}\label{sec:strong-coordination}
Take some arbitrary $p(y_2|x_1)$. Then, consider the following channel simulation problem: Alice observes a sequence $x_1^n$, unknown to Bob. The goal of Bob is to sample a sequence $y_2^n$ from the pmf $\prod_{i=1}^np(y_{2i}|x_{1i})$. To achieve this, Alice can send $nR$ noiseless bits to Bob, \emph{i.e.,} the communication rate is $R$. This is the problem of  simulating the memoryless channel $p(y_2|x_1)$ via a noiseless link of limited rate using the strong coordination criterion. Alice and Bob may share common randomness, independent of $x_1^n$, at some limited rate $R_0$.

The channel simulation problem implies set coordination if  $p(y_2|x_1)$ is such that $p(y_2|x_1)>0$ only if $y_2\in {A}_{x_1}$. As discussed in the introduction, strong coordination with infinite shared randomness $R_0=\infty$ is related to the empirical coordination (at least in the vanishing error formulation).  It turns out that the connection between strong and empirical coordiantion exists here in the zero error case as well. Observe that  Empirical coordination, itself, is related to set coordination.

The above zero-error strong coordination problem has been studied in \cite{Winter}, where it is shown that the minimum rate $R$, when $R_0=\infty$, is equal to  \cite[Theorem 24]{Winter}: $$\log(\sum_{y_2} \max_{x_1} p(y_2|x_1)).$$
Even though not mentioned by the authors of \cite{Winter}, the above relation is nothing but $I_\infty(X_1;Y_2)$ for a $p(x_1)$ where $p(x_1)>0, \forall x_1$. Also in Section \cite[III.G]{Winter} (the section on \emph{Weak simulation and reversibility}), formulas similar to the ones given in the first proof of Theorem \ref{thm:thm1}  (both of the linear programs) appear, even though Theorem \ref{thm:thm1} is considering a different problem. Although, the linear program of equation \eqref{eqn:LP2} appears but the works of McEliece and Posner, or Lov\'asz are not cited.

\section{Two nodes with side information}\label{sec:with-side}

Assume that node two has some inputs $X_2$, but that node one is aware of that, \emph{i.e., }$H(X_2|X_1)=0$.
As in the previous section, we can ignore $Y_1$ in our analysis. Let us define $$A_{x_1,x_2}\triangleq \{y_2:\exists y_1\in \mathcal{Y}_1, (y_1,y_2)\in A_{x_1x_2}\}.$$  If node two chooses a proper action $y_2$ from $A_{x_1,x_2}$, by definition there will exist a proper action $y_1$ for node one.  Since node one knows $x_2$ and the message sent to node two as well as the decoding strategy of node two, action $y_2$ and hence $y_1$ can be found by this node.

Without loss of generality we assume that $p(x_2)>0$ for all $x_2$, throughout this section.

\begin{definition}Take some arbitrary $x^*_2\in\mathcal{X}_2$ and consider the coordination problem where $X_2=x_2^*$ is fixed and known to everybody, \emph{i.e.,} instead of the joint pmf $p(x_1, x_2)$ we consider the joint pmf $q(x_1, x_2)=p(x_1|x_2^*)\textbf{1}[x_2=x_2^*]$. Since $X_2$ is assumed to be fixed, this problem falls into the class of problems considered in Section \ref{sec:3}, and using Definition \ref{coordinationgraph}, a bipartite graph can be associated to it. We denote this graph by $\mathcal G_{x_2^*}$, which defined on $(\tilde{\mathcal{X}_1}, \mathcal{Y}_2)$, where $\tilde{\mathcal{X}_1}=\{x_1: p(x_1|x_2^*)>0\}$.
\end{definition}

\iffalse
Assume that $\mathcal{X}_1=\{x_{1,1},\cdots,x_{1,p}\}$ and $\mathcal{X}_2=\{x_{2,1},\cdots,x_{2,m}\}$ and $\mathcal{Y}_2=\{y_1,y_2,\cdots,y_k\}$. Now let $\mathcal G=(\mathcal X,\mathcal Y_2)$ notice that the vertices of one part are indexed by elements of $\mathcal X=\mathcal X_1\times \mathcal X_2$. Now for a specific $s \in \mathcal{X}_2$, $\mathcal G_s=(\mathcal X',\mathcal Y_2)$ denote a sub-graph of $G$ where the second component of inputs are fixed and are equal to $s$.

In one shot case, if $X_2=s_i$ then by result of Section \rom{3} we need $\log (C(\mathcal G_{s_i}))$ bits, and since we have to support all inputs of node two $\max_{s} \log (C(G_s))$ is required.
\fi

\begin{theorem}\label{thm:side-inf}
The one-shot coordination capacity for joint pmf $p(x_1, x_2)$ where $H(X_2|X_1)=0$ is equal to
$$\max_{x_2}C(\mathcal G_{x_2}).$$
The asymptotic coordination capacity for joint pmf $p(x_1, x_2)$ where $H(X_2|X_1)=0$ is equal to
$$\max_{x_2}\bar C(\mathcal G_{x_2}).$$
\end{theorem}
\begin{remark}
This result implies a cut-set bound  for general coordination networks. If we divide the set of nodes into two groups, say, nodes $\{1,2,..., k\}$ in one group and nodes $\{k+1, k+2, ..., m\}$ in the second group, we can write a cut-set bound as follows: we assume two super-nodes that have access to $x_{[k]}$  and $x_{[k+1:m]}$, and need to make actions $y_{[k]}$  and $y_{[k+1:m]}$ respectively. We also assume that a genie provides the inputs of the second group $x_{[k+1:m]}$ to the first super-node. Then, the minimum total communication rate from super-node one to super-node two is bounded from below by the corresponding bound given in the above theorem for such a two node scenario.
\end{remark}

\begin{proof}[Proof of Theorem \ref{thm:side-inf}] The equation for one-shot case follows directly from the definition of $\mathcal G_{x_2}$, the fact that coordination needs to hold for all values of $x_2$, and $x_2$ is known by both the nodes.

It remains to show the result for the asymptotic case. Using the one-shot result and applying it to the $n$-letter version of the problem, the minimum coordination rate for codebooks of length $n$, will be equal to
$$\frac{1}{n}\max_{x^n_2}C(\mathcal G_{x^n_2}),$$
where $\mathcal G_{x^n_2}=\mathcal G_{x_{21}}\otimes\mathcal G_{x_{22}}\otimes...\otimes\mathcal G_{x_{2n}}$ is the tensor product of the graphs for indices $1$ to $n$. Therefore, the one-way communication capacity is equal to
\begin{equation}
\lim_{n\to \infty} \frac{1}{n}\max_{x_2^n} C(\mathcal G_{x_2^n}).
\end{equation}
Thus, we need to show that
$$\lim_{n\to \infty} \frac{1}{n}\max_{x_2^n} C(\mathcal G_{x_2^n})=\max_{x_2}\bar C(\mathcal G_{x_2}).$$
Using the fact that $\log C(\mathcal G)=\mathsf{IP}(\mathcal G)\geq \mathsf{LP}(\mathcal G)=\log\bar C(\mathcal G)$ and $\mathsf{LP}(\mathcal G_1\otimes \mathcal G_2)=\mathsf{LP}(\mathcal G_1)\mathsf{LP}(\mathcal G_2)$, we have
\begin{align}
\begin{split}
 \frac{1}{n}\max_{x_2^n} C(\mathcal G_{x_2^n}) &\geq \log\max_{x_2^n} \mathsf{LP}(\mathcal G_{x_2^n})^{1/n}\\
&= \log\max_{x_2^n}  \prod_{i=1}^{n} \mathsf{LP}(\mathcal G_{x_{2i}})^{1/n}\\
&= \log \max_{x_2}\mathsf{LP}(\mathcal G_{x_{2}})\\
&=\max_{x_2}\log\mathsf{LP}(\mathcal G_{x_{2}})\\
&=\max_{x_2}\bar{C}(\mathcal G_{x_{2}}).
\end{split}
\end{align}

For the reverse direction we prove that for a given $\epsilon>0$, there exist a natural number $N$ such that for all $n>N$, we have
\begin{equation}
\mathsf{IP}(\mathcal G_{x_2^n})\leq \mathsf{LP}(\mathcal G_{x_2^n})(1+\epsilon)^n, \qquad \forall x_2^n.\label{eqn23}
\end{equation}
This equation would then imply that
\begin{align}
\begin{split}
 \frac{1}{n}\max_{x_2^n} C(\mathcal G_{x_2^n}) &= \frac{1}{n}\max_{x_2^n} \log \mathsf{IP}(\mathcal G_{x_2^n})\\
 &\leq \frac{1}{n}\max_{x_2^n} \log \mathsf{LP}(\mathcal G_{x_2^n})+\log (1+\epsilon)\\
&= \max_{x_2}\bar{C}(\mathcal G_{x_{2}})+\log (1+\epsilon).
\end{split}
\end{align}
 It remains to show equation \eqref{eqn23}. Without loss of generality assume that $\mathcal{X}_2=\{1,2,..., r\}$ for some natural number $r$. Take an arbitrary sequence $x_2^n$. Let $n_i$ be the number of indices $i$ such that $x_{2i}=i$ for $i\in[r]$. Then $\sum_i n_i=n$ and the tuple $(n_1/n, n_2/n, ..., n_r/n)$ indicates the type of the sequence $x_2^n$. Because for any two graphs $\mathcal{G}_1$ and $\mathcal{G}_2$ we have $$\mathsf{IP}(\mathcal G_{1}\otimes\mathcal G_{2})=\mathsf{IP}(\mathcal G_{2}\otimes\mathcal G_{1}),$$
we get that
$\mathsf{IP}(\mathcal G_{x_2^n})=\mathsf{IP}(\Motimes_{i=1}^r \mathcal G_{i}^{\otimes n_i})$.
From equations \eqref{prodID} and  \eqref{prodID2}, we have that
\begin{align}
\mathsf{IP}(\mathcal G_{x_2^n})&\leq \prod_{i=1}^r \mathsf{IP}(\mathcal G_{i}^{\otimes n_i}),\\
\mathsf{LP}(\mathcal G_{x_2^n})&=\prod_{i=1}^r \mathsf{LP}(\mathcal G_{i}^{\otimes n_i}).
\end{align}
Taking logarithms from both sides and dividing by $n$, we get
\begin{align}
\frac{1}{n}\log\mathsf{IP}(\mathcal G_{x_2^n})&\leq \sum_{i=1}^r \frac{n_i}{n}\log (\mathsf{IP}(\mathcal G_{i}^{\otimes n_i})^{1/n_i}),\label{eqn:abv-1}\\
\frac{1}{n}\log\mathsf{LP}(\mathcal G_{x_2^n})&=\sum_{i=1}^r \frac{n_i}{n}\log (\mathsf{LP}(\mathcal G_{i}^{\otimes n_i})^{1/n_i})\nonumber
\\&= \sum_{i=1}^r \frac{n_i}{n}\log (\mathsf{LP}(\mathcal G_{i})),  \label{eqn:abv}
\end{align}
Where in equation \eqref{eqn:abv} we used equation \eqref{prodID2}.

From Lov\'asz's result in equation \eqref{LovaszThem}, we know that
$$\lim_{n_i\to \infty} \mathsf{IP}(\mathcal G^{\otimes n_i}_{i})^{1/n_i}=\mathsf{LP}(\mathcal G_{i}).$$
Thus, given any $\delta>0$, for each $i\in[r]$, if $n_i>N_i$ for some sufficiently large threshold $N_i$, we have that
$\log (\mathsf{IP}(\mathcal G_{i}^{\otimes n_i})^{1/n_i})$ is within $\delta$ interval of $\log (\mathsf{LP}(\mathcal G_{i}))$.
Compare equations \eqref{eqn:abv-1} and \eqref{eqn:abv}. If $n_i>N_i$, we have that the corresponding terms in the sum are within $\delta$ interval of each other. If $n_i<N_i$, by letting $n$ go to infinity, we can make $n_i/n\leq N_i/n$ as small as we want. This would complete the proof.
\end{proof}

\begin{remark}
We  show that the general lower bound given in Section \ref{sec:general-lower} is tight in this case. Consider that $X_2$ is a function of $X_1$. Then, similar to Corollary \ref{cor}, when $X_1\rightarrow F\rightarrow Y_2$ and $X_2$ is a function of $X_1$, we have
\begin{equation}
I(F;X_1|X_2)=I(FY_2;X_1|X_2)\geq I(Y_2;X_1|X_2).
\end{equation}
Therefore, we know that
\begin{equation}
\max_{q(x_1,x_2)} \min_{p(y_2|x_1,x_2)\in \mathcal{P}} I(Y_2;X_1|X_2)
\end{equation}
is a lower bound for the coordination capacity, where the maximum is over all $q(x_1, x_2)$ such that $q(x_1, x_2)=0$ if $p(x_1, x_2)=0$. Observe that
\begin{align}
\max_{q(x_1,x_2)} \min_{p(y_2|x_1,x_2)\in \mathcal{P}} I(Y_2;X_1|X_2)&\geq \max_{x_2} \max_{q(x_1|x_2)} \min_{p(y_2|x_1,x_2)\in \mathcal{P}} I(Y_2;X_1|X_2=x_2)
\\&= \max_{x_2} \bar C(\mathcal G_{x_2}),
\end{align}
where the maximum in the second equation is over $q(x_1|x_2)$ where $q(x_1|x_2)>0$ only if $p(x_1|x_2)>0$. Thus, $ \max_{x_2} \bar C(\mathcal G_{x_2})$ is a tight lower bound to the  coordination capacity.
\end{remark}

\section{Linear Coordination} \label{sec:linear}
In this section we will study a special case of set coordination problems, where actions and inputs are constrained by linear equations. More specifically in a network with $m$ nodes, we assume that the inputs $X_i$ and output $Y_i$ are all vectors (of possibly different lengths) in a given field $\mathbb{F}$. We say that the nodes are coordinated if
\begin{equation}
K_1X_1+K_2X_2+\cdots+K_mX_m+K_{m+1}Y_1+\cdots+K_{2m}Y_m=0,
\end{equation}
for some matrices $K_i$, $i\in[2m]$.

One motivation for this model comes from linear control systems. Suppose that nodes are controllers and inputs are disturbances to the system. Controllers should undo the disturbance by producing proper actions.

\begin{example}Linear coordination is a generalization of linear network coding. To see this, for instance consider a network of $m$ nodes, where  node one has message of $k$ bits for node $m-1$, and node two has message of length $\ell$ bits for node $m$.
Then, we can write this as a linear coordination problem consider that
\begin{equation}
K_1=-K_{2m-1}=\begin{bmatrix}
I_{k\times k}  & 0\\
0 & 0
\end{bmatrix}_{(\ell+k)\times (\ell+k)}, \hspace{4 mm}
K_2=-K_{2m}=\begin{bmatrix}
0 & 0\\
0 & I_{\ell\times \ell}
\end{bmatrix}_{(\ell+k)\times (\ell+k)},
\end{equation}
While other $K_i$'s are zero. First $k$ bits of $X_1$ are the message for node $m-1$, and other $\ell$ bits are zero. In first $X_2$ first $k$ bits are zero and rest of them are the message for node $m$.
\end{example}

\subsection{Linear coordination capacity}
For a linearly constrained coordination problem, we can define linear or non-linear codes.  In a linear code, all encoding and decoding operations  are linear, \emph{i.e.,} the transmitted messages are constructed linearly from the inputs, and the output actions are reconstructed linearly from the messages and the inputs. On the other hand, a non-linear code allows for non-linear encoder and decoders.

For simplicity, we only consider the two nodes network with one-way communication of Section \ref{sec:one-waycoordination}. We assume that the inputs of nodes are column vectors $X_1\in \mathbb{F}^{r_1}, X_2 \in \mathbb{F}^{r_2}$, distributed according to some joint distribution. The outputs of the two nodes are also assumed to be column vectors $Y_1\in \mathbb{F}^{s_1}, Y_2 \in \mathbb{F}^{s_2}$. The nodes are coordinated if
\begin{equation}
K_1X_1+K_2X_2+K_3Y_1+K_4Y_2=0,\label{eqn:Ks}
\end{equation}
for some fixed matrices $K_1,K_2,K_3$ and $K_4$. These four matrices are assumed to have the same number of $c$ rows. The number of columns of $K_1, K_2, K_3$ and $K_4$ are $r_1, r_2, s_1$ and $s_2$ respectively. Similar to Section \ref{sec:3}, we further make the simplifying assumption that node two has no input $(r_2=0)$, or equivalently $K_2=0$.

The one-shot (non-linear) coordination capacity of the communication link from node one to node two, $C(K_1, K_3, K_4)$, is defined as before for the permissible action sets $$A_{x_1}=\{(y_1, y_2): K_1x_1+K_3y_1+K_4y_2=0\}, \qquad x_1\in\mathcal{X}_1.$$
On the other hand, the one-shot linear coordination capacity $C_L(K_1, K_3, K_4)$ is defined as follows:
\begin{definition}[One-shot linear coordination capacity]
 The message $M$ generated from node one is assumed to be a linear combination of coordinates of $X_1$, \emph{i.e.,} $M=SX_1$ for some matrix $S$ in $\mathbb{F}^{t\times r_1}$, where $t$ is the number of symbols that are transmitted. Actions $Y_1$ and $Y_2$ are constructed linearly according to $Y_1=AX_1$ and $Y_2=BM$. The goal is to find matrices $S$, $A$ and $B$ such that
equation \eqref{eqn:Ks} holds, while $t$ (the number of rows of matrix $S$) is minimized. The minimum value of $t$ is called the one-shot linear coordination capacity and denoted by $C_{{L}}(K_1, K_3, K_4)$.
\end{definition}

Because linear codes are a special case of non-linear codes, we have that  $C\leq C_L\log|\mathbb{F}|$.

In an asymptotic coordination problem, we have multiple inputs and outputs $X_{1i}, Y_{1i}, Y_{2i}$ for $i\in[n]$. We say that the nodes are coordinated if
\begin{equation}
K_1X_{1i}+K_3Y_{1i}+K_4Y_{2i}=0,\qquad \forall i\in[n]\label{coordin;v}.
\end{equation}
Alternatively, if we make an overall vector of inputs $\mathbf{X}_1$ by concatenating column vectors $X_{11}, X_{12}, ..., X_{1n}$, and similarly for $\mathbf{Y}_1$ and $\mathbf{Y}_2$, we can express equation \eqref{coordin;v} as follows:
\begin{equation}
(I_{n\times n}\otimes K_1)\mathbf{X}_{1}+(I_{n\times n}\otimes K_3)\mathbf{Y}_{1}+(I_{n\times n}\otimes K_4)\mathbf{Y}_{2}=0,\qquad \forall i\in[n]\label{coordin;v1}.
\end{equation}
Therefore, we can make the following definitions:

\begin{definition}[Asymptotic linear coordination capacity]
 The asymptotic linear coordination capacity is defined as
$$\bar C_{{L}}(K_1, K_3, K_4)=\lim_{n\rightarrow \infty}\frac 1n C_{{L}}(I_{n\times n}\otimes K_1, I_{n\times n}\otimes K_3, I_{n\times n}\otimes K_4).$$
Similarly, the  asymptotic non-linear coordination capacity can be expressed as
$$\bar C(K_1, K_3, K_4)=\lim_{n\rightarrow \infty}\frac 1n C(I_{n\times n}\otimes K_1, I_{n\times n}\otimes K_3, I_{n\times n}\otimes K_4).$$
\end{definition}

\subsection{Linear one-way coordination capacity}
In this section we study one-shot and asymptotic linear capacity for the one-way communication setup.

\begin{theorem}\label{LTh} Let subspace $\mathscr{V}$ be the linear span of all vectors $x_1\in \mathbb{F}^{r_1}$ such that $p(x_1)>0$. Then we have
$C_L(K_1, K_3, K_4)=\bar C_L(K_1, K_3, K_4)=f(K_1,K_3,K_4)$ where
\begin{align}
\begin{split}
f(K_1,K_3,K_4)&=\min_{\mathscr{U}} \dim \mathscr{U}\\
\text{s.t.\hspace{3 mm}} K_1\mathscr{V}&\subseteq  \mathfrak{Im}(K_3)\oplus \mathscr{U}\\
\mathscr{U}&\subseteq  \mathfrak{Im}(K_4).
\end{split}\label{eqn:Tmh4}
\end{align}
Here the minimum is over linear subspaces $\mathscr{U}$ that satisfy the given constraints, $\oplus$ is the Minkowski sum and $\mathfrak{Im}(\cdot)$ is the image operator.
\end{theorem}
\begin{proof}[Proof of Theorem \ref{LTh}] We first prove that $C_L(K_1, K_3, K_4)=f(K_1, K_3, K_4)$ and then prove that $\bar C_L(K_1, K_3, K_4)=C_L(K_1, K_3, K_4)$.

\textbf{Proof of $C_L(K_1, K_3, K_4)=f(K_1, K_3, K_4)$:}
The coordination constraint can be written as
\begin{align}\nonumber
\begin{split}
K_1X_1+K_3Y_1+K_4Y_2&=K_1X_1+K_3AX_1+K_4BSX_1\\&=(K_1+K_3A+K_4BS)X_1=0
\end{split}
\end{align}
Thus, $(K_1+K_3A+K_4BS)x_1=0$ for all $x_1$ where $p(x_1)>0$. Therefore,
the necessary and sufficient condition for this equation to hold is that
\begin{equation} \label{LC}
(K_1+K_3A+K_4BS)\mathscr{V}=0.
\end{equation}
We have that  $K_3A\mathscr{V}\subseteq \mathfrak{Im}(K_3)$ and $K_4BS\mathscr{V}\subseteq \mathfrak{Im}(K_4B)\triangleq\mathscr{U}$. Then since $B$ has $t$ columns we have $\dim \mathscr{U} \leq t$. Therefore, equation \eqref{LC} implies that equation \eqref{eqn:Tmh4} holds for some appropriate $\mathscr{U}$ and $C_L(K_1,K_2,K_3) \geq f(K_1,K_2,K_3)$.

To show the other direction, assume that there exist a vector space $\mathscr{U}\subseteq  \mathfrak{Im}(K_4)$ such that $K_1\mathscr{V}\subseteq  \mathfrak{Im}(K_3) \oplus \mathscr{U}$ and $\dim \mathscr{U}=t$. We will find appropriate matrices $A,B$ and $S$. Let $\{v_i\}$ be a basis for $\mathscr{V}$. Equation \eqref{LC} is true if and only if $(K_1+K_3A+K_4BS)v_i=0$. Now let $q_i=K_1v_i$. Since $K_1\mathscr{V}\subseteq  \mathfrak{Im}(K_3) \oplus \mathscr{U}$ we can find vectors $w_i\in \mathfrak{Im}(K_3)$ and $r_i\in \mathscr{U}$ such that $q_i=w_i+r_i$.

Now we want to find matrix $A$ such that $w_i=K_3Av_i$ for all $i$. If we show $[v_1|\cdots|v_l]$ by $V$ and $[w_1|\cdots|w_l]$ by $W$. Then we should find $A$ such that $W=K_3AV$.
Since $w_i\in \mathfrak{Im}(K_3)$ we can find matrix $L$ such that $W=K_3L$. Thus we need $L=AV$. Notice that since $\{v_i\}$ are a basis for $\mathscr{V}$, matrix $V$ is full column rank. Therefore its rows span the full space. Let us denote the $i$-th row of $L$ by $l_i$, and $i$-th row of $A$ by $a_i$. We have to solve linear equation $l_i=a_iV$ where $a_i$ is a vector that we need to find. Now since span of rows of $V$ is the full space we can always find proper vectors $a_i$. Hence we can find $A$ such that $W=K_3AV$.

Now for finding matrices $B$ and $S$ (with $t$ rows); we should have $r_i=K_4BSv_i$. Take $R=[r_1|\cdots|r_l]$. We need $R=K_4BSV$.

Since $r_i\in \mathscr{U}$, we have that $rank(R)\leq t$.
 Since $r_i \in \mathfrak{Im}(K_4)$, we can find $L'$ such that $R=K_4L'$. Now, we want to prove that we can choose $L'$ such that $rank(L')\leq t$. Notice that each columns of $L'$ can be written as the sum of two vectors such that one of them lies in the kernel space of $K_4$ and the other one is perpendicular to this space. Therefore $L'=L'_1+L'_2$. We have that $K_4L'=K_4L'_1+K_4L'_2$.
The first part $K_4L'_1$ will vanish and $K_4L'=K_4L'_2$. Therefore, without loss of generality we can assume that $\mathfrak{Im}(L') \subseteq Ker(K_4)^{\perp}$ or $\mathfrak{Im}(L')\cap Ker(K_4)=0$. Next, it is known for arbitrary matrices $A$ and $B$ that $rank(AB)=rank(B) - \dim ( \mathfrak{Im}(B) \cap Ker(A) )$. Thus,
\begin{equation} \label{RankL}
rank(R)=rank(L')-\dim(\mathfrak{Im}(L') \cap Ker(K_4))=rank(L').
\end{equation}
 Thus $rank(L')=rank(R)\leq t$.

Now we should have $L'=BSV$. Again notice that $V$ is full column rank and rank of $L'$ is equal to $t$. Therefore there exist $t$ rows that can produce all rows of $L'$. Choose $S$ such that $t$ rows of $SV$ can produce $L'$. Now if we denote rows of $SV$ by $s_1,\cdots,s_t$ and rows of $L'$ by $l'_1,\cdots,l'_{s_2}$, since $l'_i$ is in the span of $s_1,\cdots,s_t$, we can find real numbers $b_{ij}$ such that $l'_i=\sum_j b_{ij}s_j$ if we take $B=[b_{ij}]$, then we have $L'=B(SV)$. This completes the proof.

\textbf{Proof of $C_L(K_1, K_3, K_4)=\bar C_L(K_1, K_3, K_4)$:} It suffices to show that
$$\frac 12 C_{{L}}(I_{2\times 2}\otimes K_1, I_{2\times 2}\otimes K_3, I_{2\times 2}\otimes K_4)=C_L(K_1, K_3, K_4).$$
This is because the above equality can be used inductively to show that for any $n$, which is a power of two, we have
$$\frac 1n C_{{L}}(I_{n\times n}\otimes K_1, I_{n\times n}\otimes K_3, I_{n\times n}\otimes K_4)=C_L(K_1, K_3, K_4).$$
Therefore $C_L(K_1, K_3, K_4)=\bar C_L(K_1, K_3, K_4)$.

Consider that node one observe two vectors $X_{11}$ and $X_{12}$ and then sends $M=S\begin{bmatrix}X_{11}\\X_{12}\end{bmatrix}$ to the second node where $$S=\begin{bmatrix}S_1&S_2\\S_3&S_4\end{bmatrix}.$$ Node one produces $$\begin{bmatrix}Y_{11}\\Y_{12}\end{bmatrix}=\begin{bmatrix}A_1&A_2\\A_3&A_4\end{bmatrix}\begin{bmatrix}X_{11}\\X_{12}\end{bmatrix}$$ and node two produces $$\begin{bmatrix}Y_{21}\\Y_{22}\end{bmatrix}=BM=\begin{bmatrix}B_1&B_2\\B_3&B_4\end{bmatrix}\begin{bmatrix}S_1&S_2\\S_3&S_4\end{bmatrix}\begin{bmatrix}X_{11}\\X_{12}\end{bmatrix}.$$ Coordination constraint for first letter gives us
\begin{equation}
(K_1+K_2A_1+K_3(B_1S_1+B_2S_3))X_{11}+(K_2A_2+K_3(B_1S_2+B_2S_4))X_{12}=0.
\end{equation}
A similar condition holds for the second letter. Since $X_{11}$ and $X_{12}$ are independent, we get
$$(K_1+K_2A_1+K_3(B_1S_1+B_2S_3))X_{11}=(K_2A_2+K_3(B_1S_2+B_2S_4))X_{12}=0.$$
Equivalently,
$$(K_1+K_2A_1+K_3(B_1S_1+B_2S_3))\mathscr{V}=(K_2A_2+K_3(B_1S_2+B_2S_4))\mathscr{V}=0.$$
We claim that without loss of generality, we can make the following two assumptions:
\begin{itemize}
\item
 We can assume that columns of $B_i$ are perpendicular to the kernel space of $K_3$. To see this, observe that each columns of $B_i$ can be written as the sum of two vectors such that one of them lies in the kernel space of $K_3$ and the other one is perpendicular to this space. This gives us a decomposition of matrix $B_i$ as $B_i=B_{i1}+B_{i2}$ where $K_3B_{i1}=0$. Since  $K_3B_{i1}$ vanishes, only $K_3B_{i2}$ remains and we may assume that columns of $B_i$ are perpendicular to the kernel space of $K_3$.

\item we can assume that $S_i z=0$ for $i=1,2,3,4$ and for all vectors $z \in \mathscr V^{\perp}$, where $\mathscr V^{\perp}$ is the linear subspace perpendicular to $\mathscr V$.

Notice that for every matrix $S$ we can find matrix $S'$ with same dimension such that $Sv=S'v$ for all $v \in \mathscr V$; and $S'z=0$ for all  $z \in \mathscr V^{\perp}$. This is because we can define a linear function like $S':\mathbb{F}^{r_1} \to \mathbb{F}^t$ by determining effect of this function on a basis. If $\{v_1,\cdots,v_l\}$ be an orthonormal basis for $\mathscr V$ and we expand this to an orthonormal basis for $\mathbb{F}^r_1$ like $\{v_1,\cdots,v_l,z_1,\cdots,z_{r_1-l}\}$, then we can define $S'(v_i)=Sv_i$ and $S'(z_i)=0$. It is only effect of matrices on $\mathscr V$ in important for us, and we can consider that $Sz=0$ if $z \in \mathscr V^{\perp}$.
\end{itemize}

Each column of $K_3$ can be written as the sum of two vectors such that one of them lies in the image of $K_2$ and the other one perpendicular to image of $K_2$. This gives us a decomposition of matrix $K_3$ as $K_3=K_{31}+K_{32}$, where all columns of $K_{31}$ are in image of $K_2$ and all columns of $K_{32}$ are perpendicular to this space. Now since node one knows $B$ and $S$ it can choose $A'_1$ and $A'_2$ such that
\begin{align*}
\begin{split}
K_2A'_1&=K_2A_1+K_{31}(B_1S_1+B_2S_3),\\
K_2A'_2&=K_2A_2+K_{31}(B_1S_2+B_2S_4).
\end{split}
\end{align*}
Using the above equations, we get that
\begin{align*}
\begin{split}
(K_1+K_2A'_1&+K_{31}(B_1S_1+B_2S_3))\mathscr{V}=0\\
(K_2A'_2&+K_{32}(B_1S_2+B_2S_4))\mathscr{V}=0
\end{split}
\end{align*}
The second equation implies that $(B_1S_2+B_2S_4)\mathscr{V}=0$ since there is no common vector except zero in images of $K_2$ and $K_{32}$ (all columns of $K_{32}$ were perpendicular to the image space of $K_2$). A similar argument for the second letter shows that $(B_3S_1+B_4S_3)\mathscr{V}=0$.

On the other hand, from the assumption that $S_1\mathscr{V}^{\perp}=S_2\mathscr{V}^{\perp}=S_3\mathscr{V}^{\perp}=S_4\mathscr{V}^{\perp}=0$, we have that
$(B_1S_2+B_2S_4)\mathscr{V}^{\perp}=(B_3S_1+B_4S_3)\mathscr{V}^{\perp}=0$. This fact, in conjunction with  $(B_1S_2+B_2S_4)\mathscr{V}=(B_3S_1+B_4S_3)\mathscr{V}=0$, implies that $B_1S_2+B_2S_4=B_3S_1+B_4S_3=0$.

Now
\begin{equation}\nonumber
BM=\begin{bmatrix}
B_1S_1+B_2S_3 & B_1S_2+B_2S_4\\ B_3S_1+B_4S_3&B_3S_2+B_4S_4\end{bmatrix}=\begin{bmatrix}
B_1S_1+B_2S_3 & 0\\ 0&B_3S_2+B_4S_4
\end{bmatrix}
\end{equation}
So $rank(BM)=rank(B_1S_1+B_2S_3)+rank(B_3S_2+B_4S_4)$. Besides, $rank(BM)\leq 2t$. Therefore either $rank(B_1S_1+B_2S_3)\leq t$ or $rank(B_3S_2+B_4S_4)\leq t$. Thus, one can do one letter coordination with rate less than or equal to $t$. Therefore
$$\frac 12 C_{{L}}(I_{2\times 2}\otimes K_1, I_{2\times 2}\otimes K_3, I_{2\times 2}\otimes K_4)\geq C_L(K_1, K_3, K_4).$$
 and block coding cannot help.

\end{proof}

\subsection{Linear and non-linear coordination capacities}
\begin{theorem}Assuming $\mathfrak{Im}(K_4) \subseteq \mathfrak{Im}(K_1)$, we have that $C=\bar C=C_L\log|\mathbb{F}|=\bar C_L\log|\mathbb{F}|$. In other words, linear strategies are optimal and block coding does not help decrease the non-linear coordination rate.
\end{theorem}
\begin{proof} From Theorem~\ref{LTh}, we know that $C_L=\bar C_L$. We know that $C_L\log|\mathbb{F}|\geq C\geq \bar{C}$. The asymptotic coordination capacity $\bar{C}$ is equal to $\log \mathsf{LP}$ for the linear program of equation \eqref{eqn:LP2}, on a graph $\mathcal{G}$ constructed as follows: $\mathcal{G}$ is a bipartite graph with nodes indexed by $\mathcal{X}_1$ on one part, and by $\mathcal{Y}_2$ on the other part. Vertex $x_1$ is connected to $y_2$ if and only if there is some $y_1$ such that $K_1x_1+K_3y_1+K_4y_2=0$.

Let $\mathsf{LP}^{\dagger}$ be the dual of the linear program given in equation \eqref{eqn:LP2}. This linear program is given in \cite[page 2]{c} and consists of variables in the interval $[0,1]$. While the original LP involved a minimization and was asking for a fractional covering, the dual linear program involves maximizing a linear expression and can be understood as a fractional packing linear program. If we restrict the variables of the dual program to integers in $\{0,1\}$, we get a lower bound on  $\mathsf{LP}^{\dagger}$. We denote the answer to this integer program by $\mathsf{IP}^{\dagger}$. It is shown in \cite[page 2]{c} that $\mathsf{IP}^{\dagger}$ is the maximum number of vertices in $\mathcal{X}_1$ whose neighbor sets in the bipartite graph are disjoint.

To sum this up, we always have the following chain of inequalities:
$$C_L\log|\mathbb{F}|\geq C\geq \bar{C}=\log\mathsf{LP}=\log\mathsf{LP}^{\dagger}\geq \log \mathsf{IP}^{\dagger}.$$

We show that \begin{align}\log \mathsf{IP}^{\dagger}\geq C_L\log|\mathbb{F}|\label{eqnlas},\end{align} which implies that all of the above inequalities are equality. Observe that both $C_L\log|\mathbb{F}|$ and $\log \mathsf{IP}^{\dagger}$ are one-shot expressions and can be computed from the graph $\mathcal{G}$ (rather than its tensor products).

Assume that two vertices $x_1$ and $x'_1$ have a common neighbor like $y_2$. Then there exist $y_1, y'_1$ such that
\begin{equation}
K_1x_1+K_3y_1+K_4y_2=K_1x'_1+K_3y'_1+K_4y_2=0.
\end{equation}
Hence $K_1(x_1-x'_1)\in  \mathfrak{Im}(K_3)$. Furthermore, $K_1x_1, K_1x'_1\in\mathfrak{Im}(K_1)$.  Therefore, to show that $\mathsf{IP}^{\dagger}\geq N$, it suffices to find vectors $v_1, ..., v_N$ such that
\begin{align}\label{NL}
\begin{split}
v_i&\in  \mathfrak{Im}(K_1),\\
v_i-v_k&\notin  \mathfrak{Im}(K_3),\qquad \forall i\neq k,
\end{split}
\end{align}
Let $\mathscr U$ be the vector space with minimum dimension in Theorem \eqref{LTh}.  The dimension of $\mathscr U$ is equal to $C_L$. Hence there are $|\mathbb{F}|^{C_L}$ distinct vectors in $\mathscr U$. We claim that the set of vectors in $\mathscr U$, satisfy  both conditions of equation \eqref{NL}. This would imply that $\mathsf{IP}^{\dagger}\geq |\mathbb{F}|^{C_L}$ and gives us equation \eqref{eqnlas}.  First, observe that $\mathscr U\subseteq \mathfrak{Im}(K_4)\subseteq  \mathfrak{Im}(K_1)$. Hence, the first condition of  \eqref{NL} is clearly satisfied. To show the second condition, observe that $\mathscr U$ is a vector space with minimum dimension such that $ \mathfrak{Im}(K_1)\subseteq  \mathfrak{Im}(K_3)\oplus \mathscr{U}$. We claim that this implies $\mathscr U\cap \mathfrak{Im}(K_3)=\{0\}$. Otherwise if $u\in \mathscr{U}\cap\mathfrak{Im}(K_3)$, we can expand $u$ to a basis for $\mathscr U$ like $\{u,v_1,\cdots,v_{C_L-1}\}$. Let $\mathscr U'$ be the linear span of the vectors $\langle v_1,\cdots,v_{C_L-1}\rangle$. Then $\mathscr U'$  is a subspace of $ \mathfrak{Im}(K_4)$ that satisfies $ \mathfrak{Im}(K_3)\oplus \mathscr{U} =  \mathfrak{Im}(K_3)\oplus \mathscr{U}'$. Therefore, we can decrease dimension of $\mathscr U$ which is a contradiction. This completes the proof.
\end{proof}

\subsection{Extensions to multiple-terminal}\label{sec:extensions}
It is possible to extend the result in linear coordination to certain multi-terminal scenarios. For instance, consider a network with a broadcast channel topology depicted in Fig.~\ref{fig:BC}, where node one observes $X_1\in\mathbb{F}^{r_1}$ and sends $SX_1$ to node two and $TX_1$ to node three. All three nodes produce outputs, but nodes two and three have no inputs. Coordination constraint requires that $K_1X_1+K_4Y_1+K_5Y_2+K_6Y_3=0$. In a linear code, we have that $Y_1=BX_1$, and the outputs of nodes two and three are constructed linearly from their received messages, \emph{i.e., } $Y_2=C(SX_1)$ and $Y_3=D(TX_1)$ for some matrices $C$ and $D$. This gives us the equation
\begin{equation}
(K_1+K_4B+K_5CS+K_6DT)X_1=0.
\end{equation}

\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[node distance=3.5cm,auto,>=latex', scale=1.4]
    	\draw[black, thick,rounded corners] (0,0) rectangle (1,0.6) node[font=\fontsize{9.5}{144}\selectfont,pos=.5,anchor=center] {Node 1};
	\draw[black, thick,->] (1,.3) -- (3,1.2) node[font=\fontsize{8.5}{144}\selectfont,pos=.45,anchor=south] {$SX_1$};
	\draw[black,thick,rounded corners](3,.9) rectangle (4,1.5) node[font=\fontsize{9.5}{144}\selectfont,pos=.5,anchor=center]{Node 2};
	\draw[black,thick,rounded corners](3,-.3) rectangle (4,-.9) node[font=\fontsize{9.5}{144}\selectfont,pos=.5,anchor=center]{Node 3};
	\draw[black, thick,->] (1,.3) -- (3,-.6) node[font=\fontsize{8.5}{144}\selectfont,pos=.6,anchor=south] {$TX_1$};
	\draw[black, thick,->] (0.5,1) ->(0.5,0.6) node[font=\fontsize{8.5}{144}\selectfont,pos=.2,anchor=south]{$X_1$};
	\draw[black, thick,->] (0.5,0) -> (0.5,-0.4) node[font=\fontsize{8.5}{144}\selectfont,pos=.85,anchor=north] {$Y_1$};
	\draw[black, thick,->] (3.5,.9) -> (3.5,.5) node[font=\fontsize{8.5}{144}\selectfont,pos=.85,anchor=north] {$Y_2$};
	\draw[black, thick,->] (3.5,-.9) -> (3.5,-1.3) node[font=\fontsize{8.5}{144}\selectfont,pos=.85,anchor=north] {$Y_3$};
\end{tikzpicture}
\end{center}
\caption{Linear coordination with broadcast topology.}\label{fig:BC}
\end{figure}

Thus, $(K_1+K_4B+K_5CS+K_6DT)x_1=0$ for all $x_1$ where $p(x_1)>0$. Therefore, similar to previous parts we define $\mathscr{V}$ to be the linear span of all vectors $x_1\in \mathbb{F}^{r_1}$ such that $p(x_1)>0$. Hence
\begin{equation}
(K_1+K_4B+K_5CS+K_6DT)\mathscr{V}=0.
\end{equation}

Now with an argument similar to the one given in the proof of Theorem \ref{LTh}, we have following region for sizes of $S_{t_1\times r_1}$ and $T_{t_2\times r_1}$. The pair of $(t_1,t_2)$ is valid in one-shot case if and only if
\begin{align}\nonumber
\begin{split}
&t_1\geq \dim \mathscr{U}_1 \hspace{3 mm} t_2\geq \dim \mathscr{U}_1 \hspace{3 mm}\\
& K_1\mathscr{V} \subseteq  \mathfrak{Im} K_4 \oplus \mathscr{U}_1 \oplus \mathscr{U}_2\\
&\mathscr U_1\subseteq  \mathfrak{Im}(K_5), \hspace{3 mm} \mathscr U_2\subseteq  \mathfrak{Im}(K_6).
\end{split}
\end{align}

Next, consider a MAC channel where we assume that node one and two observe two independent vectors $X_1$ and $X_2$ respectively (see Fig.~\ref{fig:MAC}).  These two nodes send messages $SX_1$ and $TX_2$ to node three. All three nodes produce outputs linearly from their observations and their received vectors. We want to find minimum number of rows of $S$ and $T$.
Coordination constraint is $K_1X_1+K_2X_2+K_4Y_1+K_5Y_2+K_6Y_3=0$. Assuming that $Y_1=AX_1$, $Y_2=BX_2$ and $Y_3=CSX_1+DTX_2$, we get that
\begin{equation}\nonumber
K_1X_1+K_2X_2+K_4Y_1+K_5Y_2+K_6Y_3=(K_1+K_4A+K_6CS)X_1+(K_2+K_5B+K_6DT)X_2=0.
\end{equation}

\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[node distance=3.5cm,auto,>=latex', scale=1.4]
    	\draw[black, thick,rounded corners] (0,1.1) rectangle (1,1.7) node[font=\fontsize{9.5}{144}\selectfont,pos=.5,anchor=center] {Node 1};
	\draw[black, thick,->] (1,1.4) -- (3,.3) node[font=\fontsize{8.5}{144}\selectfont, pos=.5,anchor=south] {$SX_1$};
	\draw[black,thick,rounded corners](0,-.6) rectangle (1,-1.2) node[font=\fontsize{9.5}{144}\selectfont,pos=.5,anchor=center]{Node 2};
	\draw[black,thick,rounded corners](3,0) rectangle (4,.6) node[font=\fontsize{9.5}{144}\selectfont,pos=.5,anchor=center]{Node 3};
	\draw[black, thick,->] (1,-.8) -- (3,.3) node[font=\fontsize{8.5}{144}\selectfont ,pos=.5,anchor=south] {$TX_2$};
	\draw[black, thick,->] (0.5,2.1) ->(0.5,1.7) node[font=\fontsize{8.5}{144}\selectfont,pos=.2,anchor=south]{$X_1$};
	\draw[black, thick,->] (0.5,1.1) -> (0.5,.7) node[font=\fontsize{8.5}{144}\selectfont,pos=.85,anchor=north] {$Y_1$};
	\draw[black, thick,->] (0.5,-.2) -> (0.5,-.6) node[font=\fontsize{8.5}{144}\selectfont,pos=.2,anchor=south] {$X_2$};
	\draw[black, thick,->] (.5,-1.2) -> (.5,-1.6) node[font=\fontsize{8.5}{144}\selectfont,pos=.85,anchor=north] {$Y_2$};
	\draw[black, thick,->] (3.5,0) -> (3.5,-.4) node[font=\fontsize{8.5}{144}\selectfont,pos=.85,anchor=north] {$Y_3$};
\end{tikzpicture}
\end{center}
\caption{Linear coordination in MAC.}\label{fig:MAC}
\end{figure}

Now since $X_1$ and $X_2$ are independent both parenthesis should be zero. And these terms  are exactly same as the ones for the two nodes case. Therefore, if we denote the number of rows of $S$ and $T$ by$(t_1,t_2)$, then such a pair is valid  in one-shot case if and only if
\begin{align}\nonumber
\begin{split}
t_1\geq\dim &\mathscr{U}_1 \hspace{3 mm} t_2\geq\dim \mathscr{U}_1 \hspace{3 mm}\\
K_1\mathscr{V} &\subseteq  \mathfrak{Im}(K_4) \oplus \mathscr{U}_1, \\
K_2\mathscr{V} &\subseteq  \mathfrak{Im}(K_5) \oplus \mathscr{U}_2,\\
\mathscr U_1,\mathscr U_2&\subseteq  \mathfrak{Im} K_6.
\end{split}
\end{align}

redpajama_set_name: RedPajamaArXiv


text:
X Man is an album by drummer Andrew Cyrille. It was recorded in May 1993 at Sear Sound in New York City, and was released by Soul Note in 1994. On the album, Cyrille is joined by flutist James Newton, guitarist Alix Pascal, and bassist Anthony Cox.

Reception

In a review for AllMusic, Scott Yanow wrote: "Considering the musicians who are involved, it is often surprising how gentle and melodic this music is. Guitarist Alix Pascal is a new voice with a quiet sound, while flutist James Newton, bassist Anthony Cox, and drummer Andrew Cyrille have long been known for their versatility, open-minded approach to improvising, and high musicianship. Each player contributed two songs apiece, and the soloing is pretty democratic, with each musician having their opportunity to shine both as a soloist and in the ensembles. This is a subtle set that grows more interesting with each listen."

The authors of The Penguin Guide to Jazz awarded the album 3½ stars, and commented: "Flute and guitar add a new spectrum to X Man, one that seems closer to Cyrille's basic understanding of melody. Newton's 'E-Squat' is a strong, clearly stated idea from a player who seems to understand Cyrille's intentions from the bottom up, and the drummer responds with some of his simplest and least cluttered playing on record. 'Simple Melody' is extraordinary, something out of a far-off place that yet seems as familiar and immediate as the most overworked standard."

Track listing

 "Answer Me" (Pascal) – 6:09
 "Novo" (Cox) – 5:46
 "A Simple Melody" (Cyrille) – 5:03
 "E - Squat" (Newton) – 5:44
 "5:05" (Cox) – 17:29
 "Lydia" (Pascal) – 7:02
 "X Man" (Newton) – 9:40
 "Akan" (Cyrille) – 4:07

Personnel
 Andrew Cyrille – drums
 James Newton – flute
 Alix Pascal – guitar (tracks 1, 3, 6, and 8)
 Anthony Cox – bass

References

1994 albums
Andrew Cyrille albums

redpajama_set_name: RedPajamaWikipedia


Google continues its expansion into as many territories of the digital world as possible with its recent announcement to offer wireless service. Check it out here.
Called Project Fi, the service follows Google's usual tack in offering early invitations to the program, starting on April 22. Initially, the service will only support the Nexus 6 smartphone, and requires a special SIM card.
Among Project Fi's unique characteristics is that Google will only charge customers for data used, rather than a monthly fee. The service will debut with assistance from T-Mobile and Sprint, which allows users to seamlessly move between both companies' LTE networks, as well as Wi-Fi, to maintain strong reception. Google also claimed that a user's Project Fi phone number will live in the cloud, allowing calls to be made from laptops and tablets.


The cost of accommodation for third level students is making the headlines once again.
The Union of Students in Ireland said the unaffordable rental sector is forcing students out of college and is calling on the Minister for Housing, Planning and Local Government, Simon Coveney and the Dáil's committee on housing and homelessness, to consider the student accommodation crisis.
Cork, Dublin, Galway and Limerick have seen an increase of up to 16% in the rental sector, according to the latest report by property website Daft.ie. Rents have gone over €1,000 a month nationwide for the first time since 2008.
USI president, Kevin Donoghue, said the escalating cost of rent across Ireland is because of a chronic supply shortage and an increase in demand, which is forcing some students to commit to a lease they can't afford, sleep on friends' couches, live in poor quality accommodation or endure long commutes.
"The lack of purpose-built on-campus accommodation, private sector rent increases and the shortage of rental properties continues to fuel a dropout culture in third level education." Mr Donoghue said.
"The accommodation crisis will deter students, especially those from rural areas, from going to college, and long commutes will negatively affect the quality of their college experience.
The Dáil's committee on housing and homelessness will hear from legal experts on the housing crisis today (Wednesday). USI will be making a submission to the committee and is calling on them to consider student accommodation.
USI has campaigned for student accommodation since the 1970s and works with the PRTB, Threshold and other organisations on the issue. In August 2014, USI launched homes.usi.ie as an online service that hosts accommodation advertising 'digs' to students all over Ireland as a short-term immediate proactive solution to the student accommodation crisis. Donoghue said anyone with a spare room should sign up and students looking for a short-term solution should consider digs, but emphasised that this is not a long-term solution.


text:
Venezuela's presidential election riddled with controversy
By Lyndsey KohMay 22, 2018
Venezuela (MNN) — President Nicolás Maduro won Venezuela's presidential elections for a second term Sunday night. His next presidential term will last for another six years.
While Venezuela reports Maduro won 68 percent of the votes, the nation saw a staggeringly low turnout with less than half of registered voters participating.
Venezuela President Nicolas Maduro (Photo courtesy of newsonline via Flickr under Creative Commons: https://goo.gl/mNS16i)
It's no secret that controversy surrounded this last election season in Venezuela. The European Union and several Latin nations exhorted that Venezuela's presidential election was proceeding unfairly. Other nations, including the United States, denounced the election outright as a sham.
Steve Shantz with Trans World Radio (TWR) says, "I think the situation that is really of concern is that even if one of the opposition candidates were allowed to win, would that really change Venezuela's fortunes? The economic woes that they are experiencing, would there be any relief for the citizens, the people of Venezuela who are really suffering these days? And the answer to that is probably not because they would face severe opposition from the military and organized crime in the country that very much back the socialist revolution that has happened there in Venezuela."
This volatile election season comes at a time when the country is facing a lot of other turmoil. Citizens are suffering under a failing healthcare system and an unprecedented economic crisis. Venezuela currently has the world's highest inflation rate at 13,000 percent.
"I heard of one guy who quit his job because he realized he just spent his entire salary on a pizza with his girlfriend. People don't have enough resources to live on anymore. The minimum wage so far in 2018 has already been raised 56 times to try to keep track with inflation. Pharmacies have no medicines. Supermarkets have huge lines. So life is just very, very difficult for people on an everyday basis."
The prevailing feeling is that, so far, nothing seems like it will change. And yet, everything has changed.
(Photo courtesy of Christian Aid Mission)
"People are very disheartened," Shantz shares. "A friend of mine, a Venezuelan, just went back to Venezuela to visit her family and said that the whole mood in the country has changed and people have changed and everybody is despondent and depressed because of the situation that they're in."
In an environment like this, hope is hard to come by. But that's where TWR comes in. Their radio programs are broadcasting the Gospel throughout Venezuela.
One of TWR's morning shows is called Despertar, which means 'Wake Up' — a fitting title for a community that desperately needs to be reinvigorated with a spiritual hope that goes deeper than politics or the economy.
"We're not trying to address the political situation. We're trying to bring hope into people's lives and bringing a message of hope to them through what the Bible teaches for how we should deal with these situations in our lives and also that there is hope in faith in God and in Jesus Christ."
However, TWR's programming hasn't been without difficulty as well. "Our offices are in Maracay and there has been electrical failure in the neighborhood where the offices are in. So we've been unable to produce radio programs because there is no electricity. Our staff are really struggling to continue to operate there, but we are broadcasting a daily message of hope into Venezuela."
Venezuela needs the Body of Christ's prayers on multiple fronts.
For TWR, Shantz shares, "Our staff has asked us to pray for them so they would be able to continue to be able to function on a daily basis, just to get to work and get home again. They have asked us to pray that the electricity would be restored so that they can continue to produce the radio programs because they are passionate about the ministry to their own countrymen."
Please also pray for Venezuela as a nation and the citizens who are suffering in the current circumstances. Shantz suggests as you pray, put yourself in their shoes.
"Pick somebody, an imaginary person in Venezuela who has the same job as you and picture how you would be able to function and carry on your daily life in the situations that they are facing. Pray for that person specifically. Pray for somebody like you in Venezuela, that God would meet their needs and that God would fix the problems in the country and that Venezuela would be ultimately returned one day to a more stable economic basis so that people's lives can go on."
Click here to learn more about TWR's ministry in Venezuela.
(Header photo courtesy of Wikimedia Commons)
#hopecrisiseconomyelectionelectionshealthcareinflationmaduropresidentpresidentialtwrvenezuela
Phone: 800-456-7TWR
300 Gregson Drive/ Box 8700
Put yourself in the shoes of someone in Venezuela, and pray for the people's needs as you sympathize.
Ask God to strengthen Venezuela's economy.
Pray for the political leaders in Venezuela to have wisdom.
Pray for the Church in Venezuela to boldly proclaim Christ's hope.
Thank the Lord for TWR's Gospel broadcast ministry and how they are able to serve the Venezuelan people in this uncertain time.
Misconceptions about the Chinese Church
Hollywood stuntman sings Slavic Gospel Association's praises

redpajama_set_name: RedPajamaCommonCrawl


Through facebook fan pages, tweets and blogs we came to know that recently Shahid Khan Afridi buys 2011 Toyota Corolla for his wife. We are not confirmed that this news is true or rumor, remember, we publish what we hear. Currentblips Cars is not responsible if this news is found rumor.


From small start-ups, to Government agencies and well-known companies like Telstra, 7-Eleven and Victoria Police – Ferntree Gully Motor group offers a proven fleet service no matter what your operating size or needs.
With 20 years' experience, we know exactly what you expect from your fleet agreement and we go out of our way to deliver. From initial consultation, to ownership and servicing – our proven methods ensure your fleet experience is smooth and hassle-free.
Not only do we provide competitive pricing and quality products, but you'll have the peace of mind of working with a dedicated account manager who'll act as your main point of contact over the lifetime of your fleet agreement.
Whether you're an employee, a Government agency or a corporate customer, our tailored fleet management plans will get you on the move, and keep you there.


You should sell a pack of cards of these or something, like index card sized prints as a collection.
I love your Halloween series and I am majorly impressed by your ability to get so many done in such a short amount of time. Kudos to you.
Just love the Halloween series. I cant wait to see what else comes.
I LOVE ALL THE HALLOWEEN COSTUMES FOR THE DISNEY CHARACTERS!!! BRILLIANT!!!!
The Disney stuff is amazing ! You can be proud of yourself ! Seriously ! This is all I ever wanted and more !!
Loving your Disney Halloween series. Will the villains be getting in on the action too, or just the heroines?
I love Your Disney Halloween art! I'd love to see some non Disney girls as well!


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redpajama_set_name: RedPajamaGithub


We have released My default markup color (1) feature within the Default proof settings section of the Proofing defaults tab located on the Personal settings page. The feature allows you to set your preferred color for the markups used within the ProofHQ Viewer.
To allow tracking activity for individual users we have added "Last login" column (2) to the Contacts page. This option is visible to the Billing Administrators and Administrators only.


Cilantro Lime Asparagus and Rice | Dizzy Busy and Hungry!
Chipotle Lime Asparagus is a perfect side dish for those busy evening where you need a quick and easy meal. You will love this tender asparagus cooked in a butter-lime sauce and tossed with rice and cilantro. It pairs perfectly with SeaPak Butterfly Shrimp or any entree!
If you are a busy mom like I am, you know how tough it can be to serve a meal that everyone in the family really enjoys. And it can be extra challenging on weekdays, when the afternoons and evenings are jam-packed with homework and activities!
Every once in a while, a meal just comes together and we are all able to sit and enjoy a few minutes eating as a family and catching up with each other. When I can make that happen, I consider it a #MomVictory. With the help of SeaPak shrimp, this dinner was quick and easy, which made achieving that #MomVictory a whole lot easier!
Even though it is super-easy, I am pretty sure that cereal is not the best meal choice out there, so one of the ways that I avoid the Cheerios dinner is by being prepared with another option! I like to have a backup plan already in mind with something simple and quick that I can just throw together when a more complex undertaking just isn't going to happen.
This dinner with SeaPak Butterfly Shrimp served alongside Cilantro Lime Asparagus and Rice is one of my new favorite backup plans! In 20 minutes, I was able to get this meal together and make everyone happy. While the entree was cooking in the oven, I prepared the flavorful asparagus and rice side dish, which perfectly complemented the golden, crunchy butterfly shrimp. #MomVictory! The entire dinner was a hit!
Tender asparagus cooked in a butter-lime sauce and tossed with rice and cilantro for a side dish that pairs perfectly with SeaPak Butterfly Shrimp or any entree!
Blanch the asparagus (place in in a large pot of boiling water for approximately 1 to 2 minutes, until the asparagus turns bright green. Then drain and run very cold water over the asparagus to stop it from cooking). Set aside.
Add the lime juice, salt, and lime zest. Stir until combined.
Add the asparagus and stir until well-coated. Cook until hot, about 3 to 4 minutes.
Combine the cilantro and rice in a large bowl.
Mix in the asparagus and lime juice mixture.
Disclosure: This post sponsored by SeaPak Shrimp and Seafood, however all opinions are my own.
Like rice? Try my Crock Pot Enchilada Bake or Zucchini Walnut Fried Rice!
Looking for unique side dishes? You might like this Chipotle Lime Cole Slaw or these Jalapeno Garlic Basil Pickles!
I love cilantro. Lime too. Adding this to my menu. Thanks for sharing at the To Grandma's House We Go DIY, Crafts, Recipes and More Wednesday Link Party! Have a great week and join us again!
Great looking recipe! I must try it! Thank you for sharing on Merry Monday! Hope to see ya next week!
I would love to try the Clam Strips. Your recipe looks so delicious!


"And on the seventh day I, God, ended my work, and all things which I had made; and I rested on the seventh day from all my work, and all things which I had made were finished, and I, God, saw that they were good; And I, God, blessed the seventh day, and sanctified it; because that in it I had rested from all my work which I, God, had created and made." (Moses 3:2-3) God created the heavens and the earth in 6 days and on the 7th day He rested… and so shall we also labor 6 days and rest on the 7th.
"…every plant of the field before it was in the earth, and every herb of the field before it grew. For I, the Lord God, created all things, of which I have spoken, spiritually, before they were naturally upon the face of the earth…" (Moses 3:5) So here we learn that God created the blueprint (Spiritually) and then Jesus Christ created the world under God's direction.
God goes on to explain how man was formed and how the Garden of Eden was created for man to dwell.
"…I, the Lord God, formed man from the dust of the ground, and breathed into his nostrils the breath of life; and man became a living soul, the first flesh upon the earth… And I, the Lord God, planted a garden eastward in Eden, and there I put the man whom I had formed. And out of the ground made I, the Lord God, to grow every tree, naturally, that is pleasant to the sight of man; and man could behold it… And I, the Lord God, caused a river to go out of Eden to water the garden; and from thence it was parted, and became into fourheads. And I, the Lord God, called the name of the first Pison, and it compasseth the whole land of Havilah, where I, the Lord God, created much gold; And the gold of that land was good, and there was bdellium and the onyx stone. And the name of the second river was called Gihon; the same that compasseth the whole land of Ethiopia. And the name of the third river was Hiddekel; that which goeth toward the east of Assyria. And the fourth river was the Euphrates." (Moses 3:7-7-14) God breathes life into man and places man inside of the Garden of Eden which had four rivers running around it; Pison, Gihon, Hiddekel, and Euphrates.
God gave Adam the ability to choose for himself, his obedience was based on his own choice. But God warned him of the consequences, that he would 'surely die'. This death would cast him out of the Garden of Eden and bring upon man the physical death.
"And out of the ground I, the Lord God, formed every beast of the field, and every fowl of the air; and commanded that they should come unto Adam, to see what he would call them; and they were also living souls; for I, God, breathed into them the breath of life… And Adam gave names to all cattle, and to the fowl of the air, and to every beast of the field; but as for Adam, there was not found an help meet for him." (Moses 3:19-20) God commands all of the animals to come to Adam to be named, and Adam names them, and whatever name he chose for them, they would be called. God also explains to Moses that every animal was also a 'living soul', meaning that Animals have souls like humans have souls.
"And I, the Lord God, said unto mine Only Begotten, that it was not good that the man should be alone; wherefore, I will make an help meet for him. And I, the Lord God, caused a deep sleep to fall upon Adam; and he slept, and I took one of his ribs and closed up the flesh in the stead thereof; And the rib which I, the Lord God, had taken from man, made I a woman, and brought her unto the man. And Adam said: This I know now is bone of my bones, and flesh of my flesh; she shall be called Woman, because she was taken out of man. Therefore shall a man leave his father and his mother, and shall cleave unto his wife; and they shall be one flesh. And they were both naked, the man and his wife, and were not ashamed." (Moses 3:18; 22-25) God asks His Only Begotten if it is good for Man to be without a Woman and He responds that it is not. So God creates a woman from the rib of man and commands that neither should be without the other. Man and Women, through marriage become 'one flesh'.
This entry was posted in Moses, Pearl of Great Price and tagged Adam, Animals, Creation, Do animals have souls?, Life, Living, Man, Physical, Soul, Spiritual. Bookmark the permalink.


text:
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Texas Mom Decapitates Daughter for 'Asking for Cereal,' Is Sentenced to Life
March 9, 2020 By Lisa René LeClair
Photo Credit: Hays County Sheriff's Office and Justice for Giovanna/Facebook
Five-year-old Giovanna Larae Hernandez had no way of knowing what kind of mood her mother was going to be in when she woke up on January 5, 2017, and asked for breakfast. According to a statement released by the Hays County District Attorney's office, the response she got was straight out of a horror film.
Krystle Concepcion Villanueva, 24, was sentenced to life in prison without the possibility of parole last week after admitting to beheading her daughter and stabbing her father-in-law in the back and head. Let that sink in for a minute.
Villanueva was living with her daughter, the child's father — Refugio Hernandez, Jr. — and his parents — Eustorgio and Nancy Arellano — at the time the incident took place.
The child's grandfather was the first to call 911, telling operators that Villanueva had attacked him from behind without warning before he was able to get out of the house.
While authorities were en route to surround the home, Villanueva also called 911, stating she killed her daughter because "she asked for cereal," and then stabbed her father-in-law.
Upon arriving at the scene, Arellano alerted SWAT team members and crisis negotiators that his five-year-old granddaughter was still inside the home, injured or possibly dead. Once inside, the squad found Villanueva standing in the front room, naked and freshly showered. The child, located in a bedroom, had been stabbed to death and decapitated.
Villanueva was restrained and taken into custody by Hays County Sheriff's officers. Blood testing, taken at a hospital shortly afterward, showed the presence of alcohol and marijuana still in her system.
"I've been in this line of work for over 43 years — 11 years as a homicide detective," Hays County Sheriff Gary Cutler said in a statement. "This is one of the worst cases I've ever seen and been a part of. … It's very sad."
Despite admitting to killing her daughter and stabbing her father-in-law (while under the influence), Villanueva's defense team pleaded with jurors to return a verdict of not guilty by reason of insanity. According to the release, at the time of the attacks, Villanueva said she was under the delusional belief that her daughter and her father-in-law were clones. She claimed she had to kill them to bring her real family members back.
Though the defense was able to prove that Villanueva had undergone treatment for mental problems previously, the prosecution argued that all such treatments coincided with prior drug abuse, adding that that marijuana use can exacerbate existing psychiatric conditions and interfere with treatment. Additionally, they argued, the symptoms described at trial had never been observed before or after her arrest. The jury denied their request, returning a verdict of guilty on both counts.
Judge Bill Henry, who presided over the case, sentenced the defendant to life without parole for the capital murder of a child under ten, and 20 years in prison for the aggravated assault with a deadly weapon against Mr. Arellano.
"A case like this leaves an unforgettable mark on everyone involved, especially the child's family." said Hays County Criminal District Attorney Wes Mau. "Every law enforcement member who worked on this case will forever be traumatized by what Ms. Villanueva did to her innocent daughter."
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Filed Under: News/Trending, Parenting Tagged With: Giovanna Larae Hernandez, Justice for Giovanna, Krystle Concepcion Villanueva, mental illness, moms who kill
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The Circuit de Barcelona-Catalunya is a motorsport racing track in Montmeló, Catalonia, Spain. Until 2013 the track was known only as the Circuit de Catalunya before a sponsorship deal with Barcelona City Council added Barcelona to the track's title.
The 4.655 km (2.892 mi) long track features 16 corners and some pretty long straights making it an all-rounder circuit. The track has grandstands with a capacity of 140,700. Besides the GP layout, the FIA Grade 1 Circuit de Barcelona-Catalunya also features a 4.627 km (2.875 mi) Motorcycle layout.