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\title{Criticality of resampling on the cycle / in the evolution model}
%\author{?\thanks{QuSoft, CWI and University of Amsterdam, the Netherlands. \texttt{?@cwi.nl} }
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\begin{document}
	
	\maketitle

	\begin{abstract}
		The model we consider is the following~\cite{ResampleLimit}: We have a cycle of length $n\geq 3$. Initially we set each site to $0$ or $1$ independently at each site, such that we set it $0$ with probability $p$. After that in each step we select a random vertex with $0$ value and resample it together with its two neighbours assigning $0$ with probability $p$ to each vertex just as initially. The question we try to answer is what is the expected number of resamplings performed before reaching the all $1$ state. 
		
		We present strong evidence for a remarkable critical behaviour. We conjecture that there exists some $p_c\approx0.62$, such that for all $p\in[0,p_c)$ the expected number of resamplings is bounded by a $p$ dependent constant times $n$, whereas for all $p\in(p_c,1]$ the expected number of resamplings is exponentially growing in $n$.
	\end{abstract}
	%Let $R(n)$ denote this quantity for a length $n\geq 3$ cycle.
	
	We can think about the resampling procedure as a Markov chain. To describe the corresponding matrix we introduce some notation. For $b\in\{0,1\}^n$ let $r(b,i,(x_{-1},x_0,x_1))$ denote the bit string which differs form $b$ by replacing the bits at index $i-1$,$i$ and $i+1$ with the values in $x$, interpreting the indices $\!\!\!\!\mod n$. Also for $x\in\{0,1\}^k$ let $p(x)=p((x_1,\ldots,x_k))=\prod_{i=1}^{k}p^{(1-x_i)}(1-p)^{x_i}$. Now we can describe the matrix of the Markov chain. We use row vectors for the elements of the probability distribution indexed by bitstrings of length $n$. Let $M_{(n)}$ denote the matrix of the leaking Markov chain:
	$$
		M_{(n)}=\sum_{b\in\{0,1\}^n\setminus{\{1\}^n}}\sum_{i\in[n]:b_i=0}\sum_{x\in\{0,1\}^3}E_{(b,r(b,i,x))}\frac{p(x)}{n-|b|},
	$$
	where $E_{(i,j)}$ denotes the matrix that is all $0$ except $1$ at the $(i,j)$th entry.

	We want to calculate the average number of resamplings $R^{(n)}$, which we define as the expected number of resamplings divided by $n$. For this let $\rho,\mathbbm{1}\in[0,1]^{2^n}$ be indexed with elements of $\{0,1\}^n$ such that $\rho_b=p(b)$ and $\mathbbm{1}_b=1$. Then we use that the expected number of resamplings is just the hitting time of the Markov chain:
	\begin{align*}
		R^{(n)}:&=\mathbb{E}(\#\{\text{resampling before termination}\})/n\\
		&=\sum_{k=1}^{\infty}P(\text{at least } k \text{ resamplings are performed})/n\\
		&=\sum_{k=1}^{\infty}\rho M_{(n)}^k \mathbbm{1}/n\\
		&=\sum_{k=0}^{\infty}a^{(n)}_k p^k
	\end{align*}

	\begin{table}[]
	\centering
	\caption{Table of the coefficients $a^{(n)}_k$}
	\label{tab:coeffs}
	\resizebox{\columnwidth}{!}{%
		\begin{tabular}{c|ccccccccccccccccccccc}
			\backslashbox[10mm]{$n$}{$k$} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\		\hline
			3 &	0 & 1 & \cellcolor{blue!25}2 & 3+1/3 & 5.00 & 7.00 & 9.33 & 12.00 & 15.00 & 18.33 & 22.00 & 26.00 & 30.33 & 35.00 & 40.00 & 45.333 & 51.000 & 57.000 & 63.333 & 70.000 & 77.000 \\
			4 &	0 & 1 & 2 & \cellcolor{blue!25}3+2/3 & 6.16 & 9.66 & 14.3 & 20.33 & 27.83 & 37.00 & 48.00 & 61.00 & 76.16 & 93.66 & 113.6 & 136.33 & 161.83 & 190.33 & 222.00 & 257.00 & 295.50 \\
			5 &	0 & 1 & 2 & 3+2/3 & \cellcolor{blue!25}6.44 & 10.8 & 17.3 & 26.65 & 39.43 & 56.48 & 78.65 & 106.9 & 142.2 & 185.8 & 238.7 & 302.41 & 378.05 & 467.13 & 571.14 & 691.69 & 830.44 \\
			6 &	0 & 1 & 2 & 3+2/3 & 6.44 & \cellcolor{blue!25}11.0 & 18.5 & 30.02 & 47.10 & 71.68 & 106.0 & 152.9 & 215.4 & 297.4 & 403.1 & 537.21 & 705.25 & 913.31 & 1168.2 & 1477.4 & 1849.1 \\
			7 &	0 & 1 & 2 & 3+2/3 & 6.44 & 11.0 & \cellcolor{blue!25}18.7 & 31.21 & 50.83 & 80.80 & 125.3 & 189.7 & 280.8 & 407.0 & 578.6 & 808.13 & 1110.2 & 1502.6 & 2005.6 & 2643.2 & 3443.1 \\
			8 &	0 & 1 & 2 & 3+2/3 & 6.44 & 11.0 & 18.7 & \cellcolor{blue!25}31.44 & 52.08 & 84.95 & 136.0 & 213.6 & 328.9 & 496.5 & 735.6 & 1070.7 & 1532.5 & 2159.5 & 2998.8 & 4108.1 & 5556.7 \\
			9 &	0 & 1 & 2 & 3+2/3 & 6.44 & 11.0 & 18.7 & 31.44 & \cellcolor{blue!25}52.30 & 86.27 & 140.7 & 226.3 & 358.4 & 558.4 & 855.4 & 1289.0 & 1911.5 & 2791.4 & 4017.2 & 5701.4 & 7985.9 \\
			10&	0 & 1 & 2 & 3+2/3 & 6.44 & 11.0 & 18.7 & 31.44 & 52.30 & \cellcolor{blue!25}86.49 & 142.1 & 231.6 & 373.4 & 594.8 & 934.4 & 1447.1 & 2209.0 & 3324.6 & 4934.8 & 7226.9 & 10447. \\
            \vdots \\
            16& 0 & 1 & 2 & 3+2/3 & 6.44 & 11.08 & 18.76 & 31.45 & 52.31 & 86.49 & 142.33 & 233.31 & 381.17 & 621.02 & 1009.38 & \cellcolor{blue!25}1637.13 & % 2650.74 & 4285.68 & 6913.55 & 11171.2 & 18052.2
        \end{tabular}
	}
	\end{table}

	We observe that this is a power series in $p$. We discovered a very regular structure in this power series. It seems that for all $k\in\mathbb{N}$ and for all $n>k$ we have that $a^{(n)}_k$ is constant, this conjecture we verified using a computer up to $n=14$. 
	\newpage
	\noindent Based on our calculations presented in Table~\ref{tab:coeffs} and Figure~\ref{fig:coeffs_conv_radius} we make the following conjectures:
	\begin{enumerate}[label=(\roman*)]
		\item $\forall k\in\mathbb{N}, \forall n\geq 3 : a^{(n)}_k\geq 0$	\label{it:pos}	
        (A simpler version: $\forall k>0: a_k^{(3)}=(k+1)(k+2)/6$)
		\item $\forall k\in\mathbb{N}, \forall n>m\geq 3 : a^{(n)}_k\geq a^{(m)}_k$ \label{it:geq}		
		\item $\forall k\in\mathbb{N}, \forall n,m > \max(k,3) : a^{(n)}_k=a^{(m)}_k$ \label{it:const}		
  		\item $\exists p_c=\lim\limits_{k\rightarrow\infty}1\left/\sqrt[k]{a_{k}^{(k+1)}}\right.$ \label{it:lim}			
	\end{enumerate}
	\colorbox{red}{\ref{it:pos}-\ref{it:geq} is false since $a_{1114}^{(10)}<0$ -- needs to be double checked!}
	I figured this out by observing that $R^{(10)}(p)$ has a pole inside the disk of radius $0.96$. This also means that $R^{(10)}(p)=\sum_{k=0}^{\infty}a_k^{(10)}p^k$ is only true in an analytic sense, since for $p>0.96$ the right hand side does not converge.

	We also conjecture that $p_c\approx0.61$, see Figure~\ref{fig:coeffs_conv_radius}.

	\begin{figure}[!htb]\centering
	\includegraphics[width=0.5\textwidth]{coeffs_conv_radius.pdf}
	%\includegraphics[width=0.5\textwidth]{log_coeffs.pdf}	
	\caption{$1\left/\sqrt[k]{a_{k}^{(k+1)}}\right.$} %$\frac{1}{\sqrt[k]{a_k^{(k+1)}}}$
	\label{fig:coeffs_conv_radius}
	\end{figure}
    
    \newpage
    For reference, we also explicitly give formulas for $R^{(n)}(p)$ for small $n$. We also give them in terms of $q=1-p$ because they sometimes look nicer that way.
    \begin{align*}
    	R^{(3)}(p) &= \frac{1-(1-p)^3}{3(1-p)^3}
        			= \frac{1-q^3}{3q^3}\\
    	R^{(4)}(p) &= \frac{p(6-12p+10p^2-3p^3)}{6(1-p)^4}
                    = \frac{(1-q)(1+q+q^2+3q^3)}{6q^4}\\
        R^{(5)}(p) &= \frac{p(90-300p+435p^2-325p^3+136p^4-36p^5+6p^6)}{15(1-p)^5(6-2p+p^2)}\\
                   &= \frac{(1-q)(6+5q+6q^2+21q^3+46q^4+6q^6)}{15q^5(5+q^2)}
    \end{align*}
    For $n=3$ the system becomes very simple because regardless of the current state, the probability of going to $111$ is always equal to $(1-p)^3$. Therefore the expected number of resamplings is simply the expectation of a geometric distribution. This gives the formula for $R^{(3)}(p)$ as shown above. Note that the $k$-th coefficient of the powerseries of a function $f(p)$ is given by $\frac{1}{k!}\left.\frac{d^k f}{dp^k}\right|_{p=0}$, i.e. the $k$-th derivative to $p$ evaluated at $0$ divided by $k!$. For the function $R^{(3)}(p) =\frac{(1-p)^{-3} - 1}{3} $ this yields $a^{(3)}_k = (k+2)(k+1)/6$ for $k\geq 1$ and $a^{(3)}_0=0$.

    We can do the same for $n=4,5$, which gives, for $k\geq 1$ (with Mathematica):
    \begin{align*}
        a^{(3)}_k &= \frac{(k+2)(k+1)}{6}\\
        a^{(4)}_k &= \frac{1}{6}\left(2+\frac{(k+3)(k+2)(k+1)}{6}\right)\\
        a^{(5)}_k &= \frac{1}{15}\left(\frac{(k+4)(k+3)(k+2)(k+1)}{20} - \frac{(k+3)(k+2)(k+1)}{30} - \frac{(k+2)(k+1)}{50} + \frac{76(k+1)}{25}\right.\\
                  &  \qquad\quad \left. + \frac{626}{125} - \frac{4}{250}
                  \left( \left(\frac{1+i\sqrt{5}}{6}\right)^k(94-25\sqrt{5}i)+\left(\frac{1-i\sqrt{5}}{6}\right)^k(94+25\sqrt{5}i) \right)
                  \right)
    \end{align*}
    and from $n=6$ and onwards, the expression becomes complicated and Mathematica can only give expressions including roots of polynomials.

    ~

	If statements \ref{it:pos}-\ref{it:lim} are true, then we can define the function 
	$$R^{(\infty)}(p):=\sum_{k=0}^{\infty}a^{(k+1)}_k p^k,$$
	which would then have radius of convergence $p_c$, also it would satisfy for all $p\in[0,p_c)$ that $R^{(n)}(p)\leq R^{(\infty)}(p)$ and $\lim\limits_{n\rightarrow\infty}R^{(n)}(p)=R^{(\infty)}(p)$.
	It would also imply, that for all $p\in(p_c,1]$ we get $R^{(n)}(p)=\Omega\left(\left(\frac{p}{p_c}\right)^{n/2}\right)$.
	This would then imply a very strong critical behaviour. It would mean that for all $p\in[0,p_c)$ the expected number of resamplings is bounded by a constant $R^{(\infty)}(p)$ times $n$, whereas for all $p\in(p_c,1]$ the expected number of resamplings is exponentially growing in $n$.
	
	Now we turn to the possible proof techniques for justifying the conjectures \ref{it:pos}-\ref{it:lim}.
	First note that $\forall n\geq 3$ we have $a^{(n)}_0=0$, since for $p=0$ the expected number of resamplings is $0$.
	Also note that the expected number of initial $0$s is $p\cdot n$. If $p\ll1/n$, then with high probability there is a single $0$ initially and the first resampling will fix it, so the linear term in the expected number of resamplings is $np$, therefore $\forall n\geq 3$, $a^{(n)}_1=1$. 
	
	For the second order coefficients it is a bit harder to argue, but one can use the structure of $M_{(n)}$ to come up with a combinatorial proof. To see this, first assume we have a vector $e_b$ having a single non-zero, unit element indexed with bitstring $b$.
	Observe that $e_bM_{(n)}$ is a vector containing polynomial entries, such that the only indices $b'$ which have a non-zero constant term must have $|b'|\geq|b|+1$, since if a resampling produces a $0$ entry it also introduces a $p$ factor. Using this observation one can see that the second order term can be red off from $\rho M_{(n)}\mathbbm{1}+\rho M_{(n)}^2\mathbbm{1}$,
	which happens to be $2n$. (Note that it is already a bit surprising, form the steps of the combinatorial proof one would expect $n^2$ terms appearing, but they just happen to cancel each other.) Using similar logic one should be able to prove the claim for $k=3$, but for larger $k$s it seems to quickly get more involved.
	
	The question is how could we prove the statements \ref{it:pos}-\ref{it:lim} for a general $k$?
	
    \appendix
    
    \section{Lower bound on $R^{(n)}(p)$}
    Proof that \ref{it:pos} and \ref{it:lim} imply that for any fixed $p>p_c$ we have $R^{(n)}(p)\in\Omega\left(\left(\frac{p}{p_c}\right)^{n/2}\right)$. 
    
    By definition of $p_c = \lim_{k\to\infty} 1\left/ \sqrt[k]{a_k^{(k+1)}} \right.$ we know that for any $\epsilon$ there exists a $k_\epsilon$ such that for all $k\geq k_\epsilon$ we have $a_k^{(k+1)}\geq (p_c + \epsilon)^{-k}$. Now note that $R^{(n)}(p) \geq a_{n-1}^{(n)}p^{n-1}$ since all terms of the power series are positive, so for $n\geq k_\epsilon$ we have $R^{(n)}(p)\geq (p_c +\epsilon)^{-(n-1)}p^{n-1}$. Note that
    \begin{align*}
    	R^{(n)}(p)\geq(p_c+\epsilon)^{-(n-1)}p^{n-1}=\left(\frac{p}{p_c+\epsilon}\right)^{n-1} \geq \left(\frac{p}{p_c}\right)^{\frac{n-1}{2}},
    \end{align*}
    where the last inequality holds for $\epsilon\leq\sqrt{p_c}(\sqrt{p}-\sqrt{p_c})$.
    
    \section{Calculating the coefficients $a_k^{(n)}$}
    Let $\rho'\in\mathbb{R}[p]^{2^n}$ be a vector of polynomials, and let $\text{rank}(\rho')$ be defined in the following way: 
    $$\text{rank}(\rho'):=\min_{b\in\{0,1\}^n}\left( |b|+ \text{maximal } k\in\mathbb{N} \text{ such that } p^k \text{ divides } \rho'_b\right).$$
	Clearly for any $\rho'$ we have that $\text{rank}(\rho' M_{(n)})\geq \text{rank}(\rho') + 1$. Another observation is, that all elements of $\rho'$ are divisible by $p^{\text{rank}(\rho')-n}$.
    We observe that for the initial $\rho$ we have that $\text{rank}(\rho)=n$, therefore $\text{rank}(\rho*(M_{(n)}^k))\geq n+k$, and so $\rho*(M_{(n)}^k)*\mathbbm{1}$ is obviously divisible by $p^{k}$. This implies that $a_k^{(n)}$ can be calculated by only looking at $\rho*(M_{(n)}^1)*\mathbbm{1}, \ldots, \rho*(M_{(n)}^k)*\mathbbm{1}$.
    
\newpage
\section{General graphs proof}

We consider the following generalization of the Markov Chain.

Let $G=(V,E)$ be an undirected graph with vertex set $V$ and edge set $E$. We define a Markov Chain $\mathcal{M}_G$ as the following process: initialize every vertex of $G$ independently to 0 with probability $p$ and 1 with probability $1-p$. Then at each step, select a uniformly random vertex that has value $0$ and resample it and its neighbourhood, all of them independently with the same probability $p$. The Markov Chain terminates when all vertices have value $1$. We use $\P^{G}$ to denote probabilities associated to this Markov Chain and $\E^G$ to denote expectation values.

\begin{definition}[Events and notation] \label{def:events}
    Let $G=(V,E)$ be a graph. Let $S\subseteq V$ be any subset of vertices.
    \begin{itemize}
        \item Define $\NZ{S}$ as the event that \emph{none} of the vertices in $S$ become zero at any point in time before the Markov Chain terminates.    	
        \item Define $\Z{S}$ as the complement of $\NZ{S}$, i.e. the event that \emph{there exists} a vertex in $S$ that becomes zero at some point in time before the Markov Chain terminates.
        \item Define for any event $A$:
            \begin{align*}
                \P^{G}_S(A) &= \P^{G}(A \mid \text{All vertices in $S$ get initialized to }1)
            \end{align*}
            The condition does not apply to subsequent resamplings of vertices in $S$, it only specifies the initial assignment.
        \item Define $G\setminus S$ as the graph obtained by removing from $G$ all vertices in $S$ and edges adjacent to $S$.
        \item Define the $d$-neighbourhood $B^G(S;d)$ of $S$ as the set of vertices reachable from $S$ within $d$ steps.
        \item Define the distant-$k$ boundary $\overline{\partial}(S,k):=B(S,k)\setminus B(S,k-1)$ as the set of vertices lying at exactly distance $k$ from $S$.
    \end{itemize}
\end{definition}

The following Lemma says that if a set $S$ splits the graph in two, then those two parts become independent if the vertices in $S$ never become zero.
\begin{center}
    \includegraphics[scale=0.8]{diagram_splittinglemma.pdf}
\end{center}
\begin{lemma}[Splitting lemma] \label{lemma:splitting}
    Let $G=(V,E)$ be a graph. Let $S,X,Y\subseteq V$ be a partition of the vertices, such that $X$ and $Y$ are disconnected in the graph $G\setminus S$. Furthermore, let $A^X$ and $A^Y$ be local events on $X$ and $Y$ respectively. Then we have
    \begin{align*}
        \P^{G}_S(\NZ{S} \cap A^X \cap A^Y)
        &=
        \P^{G\setminus Y}_S(\NZ{S} \cap A^X)
        \; \cdot \;
        \P^{G\setminus X}_S(\NZ{S} \cap A^Y)
    \end{align*}
\end{lemma}

%\newcommand{\initone}[1]{\textsc{InitOne}_#1}
\begin{proof}
    We are considering three different Markov Chains, and the events $\NZ{S}$ in the different parts of the equation are events on different probability spaces. We will keep using the same notation for these events because it should be clear from the context which Markov Chain is being considered. We will consider paths (i.e. resampling sequences) and we will use a superscript to denote to which Markov Chain a path belongs. Let $\xi^G \in \NZ{S}$ be a path of the Markov Chain associated to the resample process on the graph $G$, that satisfies the event $\NZ{S}$. 
    From $\xi^G$ we will now construct paths $\xi^{G\setminus Y} \in \NZ{S}$ and $\xi^{G \setminus X} \in \NZ{S}$ of the other Markov Chains satisfying the corresponding events on those Markov Chains.
    Let us write the path $\xi^G$ as an initialization and a sequence of resamplings:
    \begin{align*}
        \xi^G=\left( (\text{initialize to }b), (z_1, v_1, r_1), (z_2, v_2, r_2), ..., (z_{|\xi^G|}, v_{|\xi^G|}, r_{|\xi^G|}) \right)
    \end{align*}
    where $b\in\{0,1\}^V$ is the initial state, $1 \leq z_i \leq |V|$ denotes the number of zeroes in the state before the $i$th step, $v_i\in V$ denotes the site that was resampled and $r_i\in \{0,1\}^{d(v_i)+1}$ is the result of the resampled bits. Here $d(v_i)$ is the degree of vertex $v_i$. By definition of the resample process, we have
    \begin{align*}
        \P^{G}_S(\xi^G) &=
        \P(\text{initialize }b \mid \text{initialize $S$ to }1)
        \P(\text{pick }v_1 \mid z_1) \P(r_1)
        \P(\text{pick }v_2 \mid z_2) \P(r_2) \cdots \\ 
        &= \frac{(1-p)^{|b|} p^{|V|-|b|}}{(1-p)^{|S|}} \cdot
        \frac{1}{z_1} \P(r_1) \cdot
        \frac{1}{z_2} \P(r_2) \cdots
        \frac{1}{z_{|\xi^G|}} \P(r_{|\xi^G|}) .
    \end{align*}
    Let $b|_{G\setminus X}$ be the restriction of $b$ to $G\setminus X$ and similar for $b|_{G\setminus Y}$.
    To construct $\xi^{G\setminus Y}$ and $\xi^{G\setminus X}$, start with $\xi^{G\setminus Y} = \left( (\text{initialize }b|_{G\setminus Y}) \right)$ and $\xi^{G\setminus X} = \left( (\text{initialize }b|_{G\setminus X}) \right)$. For each step $(z_i,v_i,r_i)$ in $\xi^G$ do the following: if $v_i \in X$ then append $(z^{X}_i,v_i,r_i)$ to $\xi^{G\setminus Y}$ and if $v_i \in Y$ then append $(z^{Y}_i,v_i,r_i)$ to $\xi^{G\setminus X}$.
    Here $z^{X}_i$ is the number of zeroes that were in $X$ and $z^{Y}_i$ is the number of zeroes in $Y$. We have $z_i = z^{X}_i + z^{Y}_i$ because $\xi^G\in\NZ{S}$ so there can not be any zero in $S$; they all have to be in $X$ or $Y$. Furthermore, this also makes sure that we always have either $v_i\in X$ or $v_i\in Y$ since only vertices that are zero can be chosen to resample.

    Now $\xi^{G\setminus Y}$ is a valid path of the Markov Chain associated to the graph $G\setminus Y$ (i.e. with vertices $X\cup S$), because in the original path $\xi^G$, all zeroes in $X$ have been resampled by resamplings in $X$. There can not be a vertex $v\in Y$ such that the resampling of $v$ changed a vertex in $X$, since $X$ and $Y$ are only connected through $S$ and we know $\xi^G\in\NZ{S}$.

    Vice versa, any two paths $\xi^{G\setminus Y}\in\NZ{S}$ and $\xi^{G\setminus X}\in\NZ{S}$ also induce a path $\xi^G\in\NZ{S}$ by simply interleaving the resampling positions. Note that $\xi^{G\setminus Y},\xi^{G\setminus X}$ actually induce $\binom{|\xi^{G\setminus Y}|+|\xi^{G\setminus X}|}{|\xi^{G\setminus Y}|}$ paths $\xi^G$ because of the possible orderings of interleaving the resamplings in $\xi^{G\setminus Y}$ and $\xi^{G\setminus X}$.
    For a fixed $\xi^{G\setminus Y},\xi^{G\setminus X}$ we will now show the following:
    \begin{align*}
        \sum_{\substack{\xi^G\in\NZ{S} \text{ s.t.}\\ \xi^G \text{ decomposes into } \xi^{G\setminus Y},\xi^{G\setminus X} }} \P^{G}_S(\xi^G) &=
        \sum_{\text{interleavings of }\xi^{G\setminus Y},\xi^{G\setminus X}} \P(\text{interleaving}) \cdot \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \cdot \P^{G\setminus X}_S(\xi^{G\setminus X}) \\
        &= \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \cdot \P^{G\setminus X}_S(\xi^{G\setminus X}) 
    \end{align*}
    where both sums are over $\binom{|\xi^{G\setminus Y}|+|\xi^{G\setminus X}|}{|\xi^{G\setminus Y}|}$ terms.
    This is best explained by an example. Lets consider the following fixed $\xi^{G\setminus Y},\xi^{G\setminus X}$ and an example interleaving where we choose vertices from $Y,X,X,Y,\cdots$:
    \begin{align*}
        \xi^{G\setminus Y} &= \left( (\text{initialize to }b^X\;1^S),
        (z^X_1, v^X_1, r^X_1),
        (z^X_2, v^X_2, r^X_2),
        (z^X_3, v^X_3, r^X_3),
        (z^X_4, v^X_4, r^X_4),
        \cdots  \right) \\
        \xi^{G\setminus X} &= \left( (\text{initialize to }1^S\;b^Y),
        (z^Y_1, v^Y_1, r^Y_1),
        (z^Y_2, v^Y_2, r^Y_2),
        (z^Y_3, v^Y_3, r^Y_3),
        (z^Y_4, v^Y_4, r^Y_4),
        \cdots  \right) \\
        \xi^G             &= \big( (\text{initialize to }b^X \; 1^S \; b^Y),
        (z^X_1+z^Y_1, v^Y_1, r^Y_1),
        (z^X_1+z^Y_2, v^X_1, r^X_1), \\
        &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
        (z^X_2+z^Y_2, v^X_2, r^X_2),
        (z^X_3+z^Y_2, v^Y_2, r^Y_2),
        \cdots \big)
    \end{align*}
    Here $b^X\in \{0,1\}^{X}$ and $b^Y\in\{0,1\}^Y$. Since we condition on the event that $S$ is initialized to ones, we know the initial state is of the form $b^X\;1^S$ in $\xi^{G\setminus Y}$. Similarly, since these paths satisfy the $\NZ{S}$ event, we know all the vertices $v_i$ resampled in $\xi^{G\setminus Y}$ are vertices in $X$, and the resampled bits $r_i$ are bits corresponding to vertices in $X$.
    In the newly constructed path $\xi^G$ the number of zeroes is the number of zeroes in $X$ and $Y$ together, so this starts as $z^X_1 + z^Y_1$. Then in this example, after the first step the number of zeroes is $z^X_1+z^Y_2$ since a step of $\xi^{G\setminus X}$ was done (so a vertex in $Y$ was resampled).
    The probability of $\xi^{G\setminus Y}$ is given by
    \begin{align*}
        \P^{G\setminus Y}_S(\xi^{G\setminus Y}) &=
        \P(\text{initialize }b^X\;1^S \mid \text{initialize $S$ to }1)
        \P(\text{pick }v^X_1 \mid z^X_1) \P(r^X_1)
        \P(\text{pick }v^X_2 \mid z^X_2) \P(r^X_2) \cdots \\ 
        &= (1-p)^{|b^X|} p^{|X|-|b^X|} \cdot
        \frac{1}{z^X_1} \P(r^X_1) \cdot
        \frac{1}{z^X_2} \P(r^X_2) \cdots
        \frac{1}{z^X_{|\xi^{G\setminus Y}|}} \P(r^X_{|\xi^{G\setminus Y}|}) .
    \end{align*}
    and similar for $\xi^{G\setminus X}$.
    Instead of choosing a step in $Y,X,X,Y,\cdots$ we could have chosen other orderings. The following diagram illustrates all possible interleavings, and the red line corresponds to the particular interleaving $Y,X,X,Y$ in the example above.
    \begin{center}
        \includegraphics{diagram_paths3.pdf}
    \end{center}
    For the labels shown within the grid, define $p_{ij} = \frac{z^X_i}{z^X_i + z^Y_j}$.
    The probability of this particular interleaving $\xi^G$ is given by
    \begin{align*}
        \P^{G}_S(\xi^{G})
        &= (1-p)^{|b^X\; b^Y|} p^{|X\cup Y|-|b^X\;b^Y|} \quad
        \frac{1}{z^X_1+z^Y_1} \P(r^Y_1) \cdot
        \frac{1}{z^X_1+z^Y_2} \P(r^X_1) \cdots \\
        &= (1-p)^{|b^X|} p^{|X|-|b^X|} \cdot (1-p)^{|b^Y|} p^{|Y|-|b^Y|} \\
        &\qquad \cdot
        \frac{z^Y_1}{z^X_1+z^Y_1} \frac{1}{z^Y_1} \P(r^Y_1) \;
        \frac{z^X_1}{z^X_1+z^Y_2} \frac{1}{z^X_1} \P(r^X_1) \;
        \frac{z^X_2}{z^X_2+z^Y_2} \frac{1}{z^X_2} \P(r^X_2)
        \cdots \tag{rewrite fractions}\\
        &=
        \frac{z^Y_1}{z^X_1+z^Y_1} 
        \frac{z^X_1}{z^X_1+z^Y_2} 
        \frac{z^X_2}{z^X_2+z^Y_2} 
        \cdots
        \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \; \P^{G\setminus X}_S(\xi^{G\setminus X})
        \tag{definition} \\
        &= (1-p_{1,1}) \; p_{1,2} \; p_{2,2} \; (1-p_{3,2}) \cdots \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \; \P^{G\setminus X}_S(\xi^{G\setminus X})
        \tag{definition of $p_{i,j}$} \\
        &= \P(\text{path in grid}) \; \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \; \P^{G\setminus X}_S(\xi^{G\setminus X})
    \end{align*}
    In the grid we see that at every point the probabilities sum to 1, and we always reach the end, so we know the sum of all paths in the grid is 1. This proves the required equality.
    We obtain
    \begin{align*}
        \P^{G}_S(\NZ{S} \cap A^X \cap A^Y)
        &= \sum_{\xi^G \in \NZ{S}\cap A^X \cap A^Y} \P^{G}_S(\xi^G) \\
        &= \sum_{\xi^{G\setminus Y} \in \NZ{S}\cap A^X}
           \sum_{\xi^{G\setminus X} \in \NZ{S}\cap A^Y}
            \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \cdot
            \P^{G\setminus X}_S(\xi^{G\setminus X}) \\
        &= \P^{G\setminus Y}_S(\NZ{S} \cap A^X) \; \cdot \; \P^{G\setminus X}_S(\NZ{S} \cap A^Y)
    \end{align*}
\end{proof}

The intuition of the following lemma is that if two sites have distance $d$ in the graph, then the only way they can affect each other is that an interaction chain is forming between them, meaning that every vertex should get resampled to $0$ at least once in between them.

\begin{lemma}\label{lemma:distancePower}
	Suppose $G=(V,E)$ is a graph, $X,Y\subseteq V$ and $A^X$ is a local event on $X$. Then
	$$\P^{G}(A^X)-\P^{G\setminus Y}(A^X)=\bigO{p^{d(X,Y)}}.$$
	(Should be true with $+1$ in the degree, when $d(X,Y)>0$!)
\end{lemma}
\begin{proof}
	We can assume without loss of generality, that $X\neq \emptyset\neq Y$, otherwise the statement is trivial. Also we can assume without loss of generality that $d(X,Y)\leq \infty$, i.e., $X,Y$ are in the same connected component of $G$, otherwise we can use Lemma~\ref{lemma:splitting} with $S=\emptyset$.
	
	The proof goes by induction on $d(X,Y)$. The statement is trivial for $d(X,Y)=0$, and is easy to check for $d(X,Y)=1$, by looking at resample sequences that reach the all $1$ state in at most $0$ step (which is simply the case when everything is sampled to $1$ initially).
	
	Now we show the inductive step, assuming we know the statement for $d$, and that $d(X,Y)=d+1$.
	First we assume, that $\NZ{X}\subseteq\overline{A^X}$, i.e., $A^X\subseteq \Z{X}$.
	
	For $i\in[d]$ we define $A_i^X:=A^X\cap{\NZ{\overline{\partial}(X,i)}}\cap\bigcap_{j\in[i-1]}\Z{\overline{\partial}(X,j)}$, 
	and define $A_{d+1}^X:=A^X\cap\bigcap_{j\in[d]}\Z{\overline{\partial}(X,j)}$,
	so that they form a partition $A^X=\dot\bigcup_{i\in [d+1]}A_i^X$. 
	It is easy to see that for all $i\in[d+1]$ we have $A_{i}^X\subseteq\Z{X}\cap\bigcap_{j\in[i-1]}\Z{\overline{\partial}(X,j)}$, and therefore 
	\begin{equation}\label{eq:AXorder}
		\P^G(A_{i}^X)=\bigO{p^{i}}.
	\end{equation}
	Now we use the Splitting lemma~\ref{lemma:splitting} to show that for all $i\in[d]$
	\begin{align}
		\P^G(A_{i}^X)
		&=\P^{G\cap B(X,i)}_{\overline{\partial}(X,i)}(A_{i}^X)\cdot \P^{G\setminus B(X,i-1)}(X,i)(\NZ{\overline{\partial}(X,i)}) \tag{by Lemma~\ref{lemma:splitting}}\\
		&=\P^{G\cap B(X,i)}_{\overline{\partial}(X,i)}(A_{i}^X)\cdot \left(\P^{G\setminus Y\setminus B(X,i-1)}(X,i)(\NZ{\overline{\partial}(X,i)})+\bigO{p^{d+1-i}}\right) \tag{by induction}\\
		&=\P^{G\cap B(X,i)}_{\overline{\partial}(X,i)}(A_{i}^X)\cdot \P^{G\setminus Y\setminus B(X,i-1)}(X,i)(\NZ{\overline{\partial}(X,i)})+\bigO{p^{d+1}} \tag{by equation \eqref{eq:AXorder}}\\
		&=\P^{G\setminus Y}(A_{i}^X)+\bigO{p^{d+1}} \tag{by Lemma~\ref{lemma:splitting}}\\
		&=\P^{G\setminus Y}(A_{i}^X)+\bigO{p^{d(Y,Y)}}. \label{eq:indStep}
	\end{align}
	Therefore 
	$$\P^G(A^X)
	\overset{\eqref{eq:AXorder}}{=}\sum_{i\in[d]}\P^G(A_i^X)+\bigO{p^{d(Y,Y)}}
	\overset{\eqref{eq:indStep}}{=}\sum_{i\in[d]}\P^{G\setminus Y}(A_i^X)+\bigO{p^{d(Y,Y)}}
	\overset{\eqref{eq:AXorder}}{=}\P^{G\setminus Y}(A^X)+\bigO{p^{d(Y,Y)}}.
	$$
	We finish the proof by observing that if $\NZ{X}\nsubseteq\overline{A^X}$,
	then we necessarily have $\NZ{X}\subseteq A^X$, and therefore we can use the above proof with $B^X:=\overline{A^X}$ and use that $\P(A^X)=1-\P(B^X)$.
\end{proof}
	
	\begin{theorem} If $2< 2m\leq n$ and $m\leq M$, then $R^{(n)}=\E^{[-M,M]}(\Res{0})+\bigO{p^{m}}$.
	\end{theorem}
	\begin{proof} 
		\vskip-12mm
		\begin{align*}
			R^{(n)}
			&= \E^{(n)}(\Res{0}) \tag{by translation invariance}\\
			&= \sum_{k=1}^{\infty}\P^{(n)}(\Res{0}\!\geq\! k) \\		
			&= \sum_{k=1}^{\infty}\P^{[-m+1,m-1]}(\Res{0}\!\geq\! k)+ \bigO{p^{m}} \tag{by Lemma~\ref{lemma:distancePower}}\\
			&= \sum_{k=1}^{\infty}\P^{[-M,M]}(\Res{0}\!\geq\! k)+ \bigO{p^{m}} \tag{by Lemma~\ref{lemma:distancePower}}\\	
			&=\E^{[-M,M]}(\Res{0})+\bigO{p^{m}}.
		\end{align*}  
		\vskip-7mm		
	\end{proof} 	
\begin{comment}
		Let $N\geq \max(2n,2m)$, then
		\begin{align*}
		R^{(n)}
		&= \E^{(n)}(\Res{1}) \tag{by translation invariance}\\
		&= \sum_{k=1}^{\infty}\P^{(n)}(\Res{1}\geq k) \\
		%&= \sum_{k=1}^{\infty}\sum_{\underset{\ell\geq r-1}{\ell,r\in[n]}}\P^{(n)}(\Res{1}\geq k\,\&\, [\ell+1,r-1]\in\mathcal{P}) \tag{partition}\\
		%&= \sum_{k=1}^{\infty}\sum_{\underset{\ell\geq r}{\ell,r\in[n]}}\P^{(n)}(\Res{1}\geq k\,\&\, [\ell+1,r-1]\in\mathcal{P})  +\bigO{p^{n}} \\	
		%&= \sum_{k=1}^{\infty}\sum_{\underset{\ell\geq r}{\ell,r\in[n]}}\P^{[l,r]}_{b_{\ell}=b_{r}=1}(\Res{1}\geq k\,\&\, [\ell+1,r-1]\in\mathcal{P}) \P^{[r,\ell]}(\NZ{\ell,r}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNewGen}}\\				
		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}\P^{(n)}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \tag{partition}\\
		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{(n)}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) +\bigO{p^{n}}\\
		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[P\cup \partial P]}_{b_{\partial P}=1}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \P^{[\overline{P}]}(\NZ{\partial P}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNewGen}}\\
		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[P\cup \partial P]}_{b_{\partial P}=1}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \left(\left(\P^{[|\overline{P}|]}(\NZ{1})\right)^2+\bigO{p^{|\overline{P}|}}\right) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:independenetSidesNewGen}}\\
		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[P\cup \partial P]}_{b_{\partial P}=1}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \left(\left(\P^{[N]}(\NZ{1})\right)^2+\bigO{p^{|\overline{P}|}}\right) +\bigO{p^{n}} \tag{by Corollary~\ref{cor:probIndepNewGen}}\\
		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[-N,N]}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNewGen}}\\
		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}\P^{[-N,N]}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNewGen}}\\
		&= \E^{[-N,N]}(\Res{1})+\bigO{p^{n}}.
		\end{align*}	
\end{comment}			

~

Questions:
\begin{itemize}
	\item Can we prove some upper bound of the coefficients in the difference, other than they are zero for small powers?
	\item In view of this proof, can we better characterise $a_k^{(k+1)}$?
\end{itemize} 

\newpage
\section{Proving that $a_k^{(k+1)}=a_k^{(n)}$ for all $n>k$}
	Let $$P_{C}:=\NZ{\overline{\partial}(C,1)}\cap\bigcap_{v\in C}\Z{\{v\}}$$ be the event that every points of $C$ gets to $0$ at some time, but not its boundary. If $P_{C}$ holds, we say $C$ is a patch of the $0$-s.
 	\begin{lemma}\label{lemma:independenetSidesNew}	
 		$$\P^{[k]}(\Z{1}\cap \Z{k})=\P^{[k]}(\Z{1})\P^{[k]}(\Z{k})+\bigO{p^{k}}=\left(\P^{[k]}(\Z{1})\right)^2+\bigO{p^{k}}.$$
 	\end{lemma}   
 	Note that using De Morgan's law and the inclusion-exclusion formula we can see that this is equivalent to saying:
 	$$\P^{[k]}(\NZ{1}\cap \NZ{k})=\P^{[k]}(\NZ{1})\P^{[k]}(\NZ{k})+\bigO{p^{k}}.$$
 	\begin{proof}
 		We proceed by induction on $k$. For $k=1,2$ the statement is trivial.
 		
 		Now observe that:
 		$$\P^{[k]}(\Z{1})=\sum_{C\text{ connected}\,:\,1\in C}\P^{[k]}(P_{C})$$
 		$$\P^{[k]}(\Z{k})=\sum_{C\text{ connected}\,:\,k\in C}\P^{[k]}(P_{C})$$
 		
 		Suppose we proved the statement up to $k-1$, then we proceed using induction similarly to the above
 		\begin{align*}
 		&\P^{[k]}(\Z{1}\cap \Z{k})=\\
 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!\P^{[k]}(P_{[\ell]}\cap P_{[r,k]})
 		+\P^{[k]}([k]\in\mathcal{P})\\
 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!\P^{[k]}(P_{[\ell]}\cap P_{[r,k]})
 		+\bigO{p^{k}} \tag*{$\left(\P^{[k]}([k]\in\mathcal{P})=\bigO{p^{k}}\right)$}\\	
 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!
 		\P^{[\ell+1]}_{\{\ell+1\}}(P_{[\ell]})
 		\P^{[\ell+1,r-1]}(\NZ{\ell+1}\cap \NZ{r-1})
 		\P^{[r-1,k]}_{\{r-1\}}(P_{[r,k]})
 		+\bigO{p^{k}} \tag{by Lemma~\ref{lemma:splitting}}\\
 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!
 		\P^{[\ell+1]}_{\{\ell+1\}}(P_{[\ell]})
 		\left(\P^{[\ell+1,r-1]}(\NZ{\ell+1})
		\P^{[\ell+1,r-1]}(\NZ{r-1})\right)
 		\P^{[r-1,k]}_{\{r-1\}}(P_{[r,k]})
 		+\bigO{p^{k}} \tag{by induction}\\
 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!
 		\P^{[\ell+1]}_{\{\ell+1\}}(P_{[\ell]})
 		\left(\P^{[\ell+1,k]}(\NZ{\ell+1})
 		\P^{[1,r-1]}_{\{r-1\}}(\NZ{r-1})\right)
 		\P^{[r-1,k]}(P_{[r,k]})
 		+\bigO{p^{k}} \tag{by Lemma~\ref{lemma:distancePower}}\\
 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!
 		\P^{[k]}(P_{[\ell]})
 		\P^{[k]}(P_{[r,k]})
 		+\bigO{p^{k}} \tag{by Lemma~\ref{lemma:splitting}}\\
 		&=\left(\sum_{\ell\in [k]}\P^{[k]}(P_{[\ell]})\right)
 		\left(\sum_{r\in [k]}\P^{[k]}(P_{[r,k]})\right)
 		+\bigO{p^{k}} \tag*{$\left(\P^{[k]}(P_{[\ell]})=\bigO{p^{\ell}}\right)$}\\	
 		&=\P^{[k]}(\Z{1})\P^{[k]}(\Z{k})
 		+\bigO{p^{k}}.	
 		\end{align*}
 	\end{proof}

	Again the intuition of the final theorem is simmilar to the previous lemma. A site can only realise the length of the cycle after an interaction chain was formed around the cycle, implying that every vertex was resampled to $0$ at least once.
 	
	\begin{theorem} $R^{(n)}=\E^{[-m,m]}(\Res{0})+\bigO{p^{n}}$ for all $m\geq n \geq 3$, thus
		$R^{(n)}-R^{(m)}=\bigO{p^{n}}$.
	\end{theorem}
	\begin{proof} In the proof we identify the sites of the $n$-cycle with the$\mod n$ remainder classes.
		\vskip-3mm
		\begin{align*}
			R^{(n)}
			&= \E^{(n)}(\Res{0}) \tag{by translation invariance}\\
			&= \sum_{k=1}^{\infty}\P^{(n)}(\Res{0}\!\geq\! k) \\		
			&= \sum_{k=1}^{\infty}\sum_{\underset{v+w\leq n+1}{v,w\in [n]}}\P^{(n)}(\Res{0}\!\geq\! k\,\cap\, \underset{P_{v,w}:=}{\underbrace{P_{[-v\!+\!1,w\!-\!1]}}}) \tag{partition}\\[-1mm]
			&= \sum_{k=1}^{\infty}\sum_{\underset{v+w\leq n}{v,w\in [n]}}\P^{(n)}(\Res{0}\!\geq\! k\,\cap\, P_{v,w}) +\bigO{p^{n}}\\[-1mm]
			&= \sum_{k=1}^{\infty}\smash{\sum_{\underset{v+w\leq n}{v,w\in [n]}}}\P^{[-v,w]}_{\{-v,w\}}(\Res{0}\!\geq\! k\,\cap\, P_{v,w}) \P^{[w,n-v]}(\NZ{w,n-v}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:splitting}}\\
			&= \sum_{k=1}^{\infty}\smash{\sum_{\underset{v+w\leq n}{v,w\in [n]}}}\P^{[-v,w]}_{\{-v,w\}}(\Res{0}\!\geq\! k\,\cap\, P_{v,w})  \left(\left(\P^{[w,n-v]}(\NZ{w})\right)^{\!\!2}\!+\!\bigO{p^{n-v-w+1}}\right) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:independenetSidesNew}}\\
			&= \sum_{k=1}^{\infty}\smash{\sum_{\underset{v+w\leq n}{v,w\in [n]}}}\P^{[-v,w]}_{\{-v,w\}}(\Res{0}\!\geq\! k\,\cap\, P_{v,w})  \left(\P^{[-m,-v]}(\NZ{-v})\P^{[w,m]}(\NZ{w})\!+\!\bigO{p^{n-v-w+1}}\right) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:independenetSidesNew}}\\	
			&= \sum_{k=1}^{\infty}\smash{\sum_{\underset{v+w\leq n}{v,w\in [n]}}}\P^{[-v,w]}_{\{-v,w\}}(\Res{0}\!\geq\! k\,\cap\, P_{v,w}) \P^{[-m,-v]}(\NZ{-v})\P^{[w,m]}(\NZ{w}) +\bigO{p^{n}} \tag{$|P_{v,w}|=v+w-1$}\\
			&= \sum_{k=1}^{\infty}\sum_{\underset{v+w\leq n}{v,w\in [n]}}\P^{[-m,m]}(\Res{0}\!\geq\! k\,\cap\, P_{v,w}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:splitting}}\\[-1mm]
			&= \sum_{k=1}^{\infty}\sum_{\underset{|C|<n}{C\text{ connected}:0\in C}}\P^{[-m,m]}(\Res{0}\!\geq\! k\,\cap\, P_C) +\bigO{p^{n}} \\[-1mm]
			&= \sum_{k=1}^{\infty}\sum_{C\text{ connected}:0\in C}\P^{[-m,m]}(\Res{0}\!\geq\! k\,\cap\, P_C) +\bigO{p^{n}} \\
			&= \E^{[-m,m]}(\Res{0})+\bigO{p^{n}}.\\[-3mm]										
		\end{align*}  
		\noindent Repeating the same argument with $m$ and comparing the results completes the proof.
	\end{proof} 	
\begin{comment}
		Let $N\geq \max(2n,2m)$, then
		\begin{align*}
		R^{(n)}
		&= \E^{(n)}(\Res{1}) \tag{by translation invariance}\\
		&= \sum_{k=1}^{\infty}\P^{(n)}(\Res{1}\geq k) \\
		%&= \sum_{k=1}^{\infty}\sum_{\underset{\ell\geq r-1}{\ell,r\in[n]}}\P^{(n)}(\Res{1}\geq k\,\&\, [\ell+1,r-1]\in\mathcal{P}) \tag{partition}\\
		%&= \sum_{k=1}^{\infty}\sum_{\underset{\ell\geq r}{\ell,r\in[n]}}\P^{(n)}(\Res{1}\geq k\,\&\, [\ell+1,r-1]\in\mathcal{P})  +\bigO{p^{n}} \\	
		%&= \sum_{k=1}^{\infty}\sum_{\underset{\ell\geq r}{\ell,r\in[n]}}\P^{[l,r]}_{b_{\ell}=b_{r}=1}(\Res{1}\geq k\,\&\, [\ell+1,r-1]\in\mathcal{P}) \P^{[r,\ell]}(\NZ{\ell,r}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\				
		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}\P^{(n)}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \tag{partition}\\
		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{(n)}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) +\bigO{p^{n}}\\
		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[P\cup \partial P]}_{b_{\partial P}=1}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \P^{[\overline{P}]}(\NZ{\partial P}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\
		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[P\cup \partial P]}_{b_{\partial P}=1}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \left(\left(\P^{[|\overline{P}|]}(\NZ{1})\right)^2+\bigO{p^{|\overline{P}|}}\right) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:independenetSidesNew}}\\
		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[P\cup \partial P]}_{b_{\partial P}=1}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \left(\left(\P^{[N]}(\NZ{1})\right)^2+\bigO{p^{|\overline{P}|}}\right) +\bigO{p^{n}} \tag{by Corollary~\ref{cor:probIndepNew}}\\
		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[-N,N]}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\
		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}\P^{[-N,N]}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\
		&= \E^{[-N,N]}(\Res{1})+\bigO{p^{n}}.
		\end{align*}	
\end{comment}			

~

\newpage
\section{Characterisation of $p_c$}
\textbf{Conjecture} for a fixed $p\in [0,1]$ the following are equivalent:
\begin{enumerate}
	\item $\lim_{n\to\infty}\P^{[-n,n]}_{\overline{\{0\}}}(\Z{\{n\}})>0$
	\item $\P^{[-\infty,\infty]}_{\overline{\{0\}}}(\text{Not reaching the all 1 state})>0$
	\item $\P^{[-\infty,\infty]}(\NZ{\{0\}})>0$
	\item $\P^{[0,\infty]}(\NZ{\{0\}})>0$
	\item $\lim_{n\to\infty}\P^{[0,n]}(\NZ{\{0\}})>0$		
	\item $\exists c,\lambda>0:\P^{[-\infty,\infty]}(\Z{[k]})<ce^{-\lambda k}$
	\item $\exists c,\lambda>0:\mathrm{Cov}^{[-\infty,\infty]}(A,B)<ce^{-\lambda d(A,B)}$
	\item $\exists c,\lambda>0\,\forall n\in\mathbb{N}:\mathrm{Cov}^{[n]}(A,B)<ce^{-\lambda d(A,B)}$	
	\item $R^{(\infty)}<\infty$
\end{enumerate}
\begin{proof}
	$1\Leftrightarrow 2:$
	\begin{align*}
		\P^{[-\infty,\infty]}_{\overline{\{0\}}}(\text{Not reaching the all 1 state})>0
		&=\P^{[-\infty,\infty]}_{\overline{\{0\}}}(\text{Resampling arbitrary far away})>0\\
		&=\P^{[-\infty,\infty]}_{\overline{\{0\}}}\left(\bigcap_{n=1}^{\infty}\Z{\{-n\}}\cup\Z{\{n\}}\right)>0\\
		&=\lim_{n\to\infty}\P^{[-\infty,\infty]}_{\overline{\{0\}}}(\Z{\{-n\}}\cup\Z{\{n\}})>0\\
		&=\lim_{n\to\infty}\P^{[-n,n]}_{\overline{\{0\}}}(\Z{\{-n\}}\cup\Z{\{n\}})>0
	\end{align*}
\end{proof}

    \textbf{New conjecture(s)}: The following statements are equivalent
    \begin{enumerate}
        \item $p<p_c$
        \item (requires continuous time) $\P^{[0,\infty]}(\NZ{0}) > 0$
        \item (requires continuous time) $\P^{[-\infty,\infty]}(\NZ{0}) > 0$
        \item (requires continuous time) $\exists c,\lambda$ such that $\P^{[-\infty,\infty]}(\Z{[0,k]}) \leq c \; e^{-\lambda \; k}$
        \item (requires continuous time) $\exists c,\lambda$ such that $\mathrm{COV}^{[-\infty,\infty]}(A,B) \leq c \; e^{-\lambda \; d(A,B)}$
        \item (requires continuous time) $\E^{(\infty)}(\text{\# resamples of }0) < \infty$
        \item (first lines requires continuous time)
            \begin{align*}
                \P^{[-\infty,\infty]}(\textsc{NonStop})
                &= \P^{[-\infty,\infty]}(\textsc{NonStop} \text{ and unbounded}) \\
                &= \P^{[-\infty,\infty]}(\bigcap_{n=1}^{\infty} \Z{[-n,n]}) \\
                &= \lim_{n\to\infty} \P^{[-\infty,\infty]}(\Z{[-n,n]}) \\
                &\overset{?}{=} \lim_{n\to\infty} \P^{[-n,n]}(\Z{[-n,n]})
            \end{align*}
        \item (does \emph{not} requires continuous time) $\P^{[-\infty,\infty]}(\textsc{NonStop} \mid \text{start with single 0 at 0})$
        \item $\lim_{n\to\infty} \P^{[0,n]}(\Z{n} \mid \text{start with single 0 at 0}) = 0$
        \item $\lim_{n\to\infty} \P^{[0,n]}(\NZ{0}) > 0$ (note: symbolically computable for small n)
        \item $\lim_{n\to\infty} \P^{[0,n]}(\Z{0}) < 1$ (note: symbolically computable for small n)
        \item For the non-terminating process (resample a random 1 in case there are only 1s available), the number of zeroes in the stationary distribution is $o(n)$.
    \end{enumerate}


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