Questions:

\begin{itemize}

	\item Can we generalise the proof to other translationally invariant spaces, like the torus?

	\item In view of this proof, can we better characterise $a_k^{(k+1)}$?

	\item Why did Mario's and Tom's simulation show that for fixed $C$ the contribution coefficients have constant sign? Is it relevant for proving \ref{it:pos}-\ref{it:geq}?

\end{itemize} 

	%I think the same arguments would translate to the torus and other translationally invariant spaces, so we could go higher dimensional as Mario suggested. Then I think one would need to replace $|S_{><}|$ by the minimal number $k$ such that there is a $C$ set for which $S\cup C$ is connected. I am not entirely sure how to generalise Lemma~\ref{lemma:probIndep} though, which has key importance in the present proof.

\newpage

\section{General graphs proof}

We consider $R^{(n)}(p)$ as a power series in $p$ and our main aim in this section is to show that $R^{(n)}(p)$ and $R^{(n+k)}(p)$ are the same up to order $n-1$.

The proof will consider variations of the Markov Chain:

\begin{itemize}

    \item $\P^{(n)}$ refers to the original process on the length-$n$ cycle.

    \item $\P^{[a,b]}$ or $\P^{[n]}$ refers to a similar Markov Chain but on a finite chain ($[a,b]$ or $[1,n]$).

\end{itemize}

The process on the finite chain has the following modification at the boundary: if a boundary site is resampled, it can only resample itself and its single neighbour so it draws only two new bits. 

We use the notation $\E^{(n)}$,$\E^{[a,b]}$ and $\E^{[n]}$ similarly for denoting expectations.

%Note that an \emph{event} is a subset of all possible paths of the Markov Chain.

\begin{definition}[Events conditioned on starting state] \label{def:conditionedevents}

    For any state $b\in\{0,1\}^n$, define $\start{b}$ as the event that the starting state of the chain is the state $b$. For any event $A$ and any $v\in[n]$, define

    \begin{align*}

        \P^{(n)}_b(A) &= \P^{(n)}(A \;|\; \start{b}) \\

        \P^{[n]}_{b_v=1}(A) &= \P^{[n]}(A \;|\; v\text{ is initialized to }1) \\

        \P^{[n]}_{b_v=b_w=1}(A) &= \P^{[n]}(A \;|\; v\text{ and }w\text{ are initialized to }1) ,

    \end{align*}

    The last two probabilities are not conditioned on any other bits of the starting state.

\end{definition}

%Note that we have $\P^{(n)}(\start{b}) = (1-p)^{|b|}p^{n-|b|}$ by definition of our Markov Chain.

\begin{definition}[Vertex visiting event] \label{def:visitingResamplings}

    Denote by $\mathrm{Z}^{(v)}$ the event that site $v$ becomes zero at any point in time before the Markov Chain terminates. Denote the complement by $\mathrm{NZ}^{(v)}$, i.e. the event that site $v$ does \emph{not} become zero before it terminates. Furthermore define $\mathrm{NZ}^{(v,w)} := \mathrm{NZ}^{(v)} \cap \mathrm{NZ}^{(w)}$, i.e. the event that \emph{both} $v$ and $w$ do not become zero before termination.

\end{definition}

%\begin{figure}

%	\begin{center}

%    	\includegraphics{diagram_groups.pdf}

%    \end{center}

%    \caption{\label{fig:separatedgroups} Illustration of setup of Lemma \ref{lemma:eventindependence}. Here $b_1,b_2\in\{0,1\}^n$ are bitstrings such that all zeroes of $b_1$ and all zeroes of $b_2$ are separated by two indices $v,w$.}

%\end{figure}

\begin{wrapfigure}[7]{r}{0.25\textwidth} % The first [] argument is number of lines that are narrowed

    \centering

    \includegraphics{diagram_groups.pdf}

    \caption{\label{fig:separatedgroups} Lemma \ref{lemma:eventindependence}.}

\end{wrapfigure}

The following lemma considers two vertices $v,w$ that are never ``crossed'' so that two halves of the cycle become independent.\begin{lemma}[Conditional independence] \label{lemma:eventindependence} \label{claim:eventindependence}

    Let $b=b_1\land b_2\in\{0,1\}^n$ be a state with two separated groups of zeroes as in Figure \ref{fig:separatedgroups}. Let $v$, $w$ be any indices inbetween the groups, such that $b_1$ lies on one side of them and $b_2$ on the other, as shown in the figure. Furthermore, let $A_1$ be any event that depends only on the sites ``on the $b_1$ side of $v,w$'', and similar for $A_2$ (for example $\mathrm{Z}^{(i)}$ for an $i$ on the correct side). Then we have

    \begin{align*}

        \P^{(n)}_b(\mathrm{NZ}^{(v,w)}, A_1, A_2)

        &=

        \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)}, A_1)

        \; \cdot \;

        \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)}, A_2) \\

        \P^{(n)}_b(A_1, A_2 \mid \mathrm{NZ}^{(v,w)})

        &=

        \P^{(n)}_{b_1}(A_1 \mid \mathrm{NZ}^{(v,w)})

        \; \cdot \;

        \P^{(n)}_{b_2}(A_2 \mid \mathrm{NZ}^{(v,w)}) .%\\

        %R_{b,\mathrm{NZ}^{(v,w)},A_1,A_2}

        %&=

        %R_{b_1,\mathrm{NZ}^{(v,w)},A_1}

        %\; + \;

        %R_{b_2,\mathrm{NZ}^{(v,w)},A_2}

    \end{align*}

    %up to any order in $p$.

\end{lemma}

\begin{proof}

    From any path $\xi\in\start{b} \cap \mathrm{NZ}^{(v,w)}$ we can construct paths $\xi_1\in\start{b_1}\cap \mathrm{NZ}^{(v,w)}$ and $\xi_2\in\start{b_2}\cap\mathrm{NZ}^{(v,w)}$ as follows. Let us write the path $\xi$ as

    $$\xi=\left( (\text{initialize }b), (z_1, s_1, r_1), (z_2, s_2, r_2), ..., (z_{|\xi|}, s_{|\xi|}, r_{|\xi|}) \right)$$

    where $z_i\in[n]$ denotes the number of zeroes in the state before the $i$th step, $s_i\in [n]$ denotes the site that was resampled and $r_i\in \{0,1\}^3$ is the result of the three resampled bits. We have

    \begin{align*}

        \P^{(n)}_b[\xi] &= \P(\text{pick }s_1 | z_1) \P(r_1) \P(\text{pick }s_2 | z_2) \P(r_2) \cdots \P(\text{pick }s_{|\xi|} | z_{|\xi|}) \P(r_{|\xi|}) \\

                &= \frac{1}{z_1} \P(r_1) \frac{1}{z_2} \P(r_2) \cdots \frac{1}{z_{|\xi|}} \P(r_{|\xi|}) .

    \end{align*}

    To construct $\xi_1$ and $\xi_2$, start with $\xi_1 = \left( (\text{initialize }b_1) \right)$ and $\xi_2 = \left( (\text{initialize }b_2) \right)$. For each step $(z_i,s_i,r_i)$ in $\xi$ do the following: if $s_i$ is ``on the $b_1$ side of $v,w$'' then append $(z^{(1)}_i,s_i,r_i)$ to $\xi_1$ and if its ``on the $b_2$ side of $v,w$'' then append $(z^{(2)}_i,s_i,r_i)$ to $\xi_2$. Here $z^{(1)}_i$ is the number of zeroes that were on the $b_1$ side and $z^{(2)}_i$ is the number of zeroes on the $b_2$ side so we have $z_i = z^{(1)}_i + z^{(2)}_i$.

    %Let the resulting paths be

    %\begin{align*}

    %    \xi_1 &= \left( (z^{(1)}_{a_1}, s_{a_1}, r_{a_1}), (z^{(1)}_{a_2}, s_{a_2}, r_{a_2}), ..., (z^{(1)}_{a_{|\xi_1|}}, s_{a_{|\xi_1|}}, r_{a_{|\xi_1|}}) \right) \\

    %    \xi_2 &= \left( (z^{(2)}_{b_1}, s_{b_1}, r_{b_1}), (z^{(2)}_{b_2}, s_{b_2}, r_{b_2}), ..., (z^{(2)}_{b_{|\xi_1|}}, s_{b_{|\xi_1|}}, r_{b_{|\xi_1|}}) \right)

    %\end{align*}

    Now $\xi_1$ is a valid (terminating) path from $b_1$ to $\mathbf{1}$, because in the original path $\xi$, all zeroes ``on the $b_1$ side'' have been resampled by resamplings ``on the $b_1$ side''. Since the sites $v,w$ inbetween never become zero, there can not be any zero ``on the $b_1$ side'' that was resampled by a resampling ``on the $b_2$ side''.

    Vice versa, any two paths $\xi_1\in\start{b_1}\cap \mathrm{NZ}^{(v,w)}$ and $\xi_2\in\start{b_2}\cap\mathrm{NZ}^{(v,w)}$ also induce a path $\xi\in\start{b} \cap \mathrm{NZ}^{(v,w)}$ by simply interleaving the resampling positions. Note that $\xi_1,\xi_2$ actually induce $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths $\xi$ because of the possible orderings of interleaving the resamplings in $\xi_1$ and $\xi_2$.

    For a fixed $\xi_1,\xi_2$ we will now show the following:

    \begin{align*}

        \sum_{\substack{\xi\in\start{b} \cap \mathrm{NZ}^{(v,w)} \text{ s.t.}\\ \xi \text{ decomposes into } \xi_1,\xi_2 }} \P^{(n)}_b[\xi] &=

        \sum_{\text{interleavings of }\xi_1,\xi_2} \P(\text{interleaving}) \cdot \P^{(n)}_{b_1}[\xi_1] \cdot \P^{(n)}_{b_2}[\xi_2] \\

        &= \P^{(n)}_{b_1}[\xi_1] \cdot \P^{(n)}_{b_2}[\xi_2]

    \end{align*}

    where both sums are over $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ terms.

    This is best explained by an example. Lets consider the following fixed $\xi_1,\xi_2$ and an example interleaving where we choose steps from $\xi_2,\xi_1,\xi_1,\xi_2,\cdots$:

    \begin{align*}

        \xi_1 &= \left( (z_1, s_1, r_1), (z_2, s_2, r_2), (z_3, s_3, r_3), (z_4, s_4, r_4),\cdots  \right) \\

        \xi_2 &= \left( (z_1', s_1', r_1'), (z_2', s_2', r_2'), (z_3', s_3', r_3'), (z_4', s_4', r_4'),\cdots  \right) \\

        \xi   &= \left( (z_1 + z_1', s_1', r_1'), (z_1+z_2', s_1, r_1), (z_2+z_2', s_2, r_2), (z_3+z_2', s_2', r_2'), \cdots \right)

    \end{align*}

    The probability of $\xi_1$, started from $b_1$, is given by

    \begin{align*}

        \P^{(n)}_{b_1}[\xi_1] &= \P(\text{pick }s_1|z_1) \P(r_1) \P(\text{pick }s_2|z_2) \P(r_2) \cdots \P(\text{pick }s_{|\xi_1|}|z_{|\xi_1|}) \P(r_{|\xi_1|}) \\

                &= \frac{1}{z_1} \P(r_1) \frac{1}{z_2} \P(r_2) \cdots \frac{1}{z_{|\xi_1|}} \P(r_{|\xi_1|}) .

    \end{align*}

    and similar for $\xi_2$ but with primes.

    The following diagram illustrates all possible interleavings, and the red line corresponds to the particular interleaving $\xi$ in the example above.

    \begin{center}

        \includegraphics{diagram_paths2.pdf}

    \end{center}

    For the labels shown within the grid, define $p_{ij} = \frac{z_i}{z_i + z_j'}$.

    The probability of $\xi$ is given by

    \begin{align*}

        \P^{(n)}_b[\xi] &= \frac{1}{z_1+z_1'} \P(r_1') \frac{1}{z_1+z_2'} \P(r_1) \frac{1}{z_2+z_2'} \P(r_2) \frac{1}{z_3+z_2'} \P(r_2') \cdots \tag{by definition}\\

        &=

        \frac{z_1'}{z_1+z_1'} \frac{1}{z_1'} \P(r_1') \;

        \frac{z_1 }{z_1+z_2'} \frac{1}{z_1 } \P(r_1 ) \;

        \frac{z_2 }{z_2+z_2'} \frac{1}{z_2 } \P(r_2 ) \;

        \frac{z_2'}{z_3+z_2'} \frac{1}{z_2'} \P(r_2')

        \cdots \tag{rewrite fractions}\\

        &=

        \frac{z_1'}{z_1+z_1'} \;

        \frac{z_1 }{z_1+z_2'} \;

        \frac{z_2 }{z_2+z_2'} \;

        \frac{z_2'}{z_3+z_2'}

        \cdots

        \P^{(n)}_{b_1}[\xi_1] \; \P^{(n)}_{b_2}[\xi_2] \tag{definition of $\P^{(n)}_{b_i}[\xi_i]$} \\

        &= (1-p_{1,1}) \; p_{1,2} \; p_{2,2} \; (1-p_{3,2}) \; \P^{(n)}_{b_1}[\xi_1] \; \P^{(n)}_{b_2}[\xi_2] \tag{definition of $p_{i,j}$} \\

        &= \P(\text{path in grid}) \; \P^{(n)}_{b_1}[\xi_1] \; \P^{(n)}_{b_2}[\xi_2]

    \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^{(n)}_b(\mathrm{NZ}^{(v,w)},A_1,A_2)

        &= \sum_{\substack{\xi\in\start{b} \cap \\ \mathrm{NZ}^{(v,w)}\cap A_1\cap A_2}} \P^{(n)}_b(\xi) \\

        &= \sum_{\substack{\xi_1\in\start{b_1} \cap \\ \mathrm{NZ}^{(v,w)}\cap A_1}} \;\;

          \sum_{\substack{\xi_2\in\start{b_1} \cap \\ \mathrm{NZ}^{(v,w)}\cap A_2}}

        \P^{(n)}_{b_1}(\xi_1)\cdot\P^{(n)}_{b_2}(\xi_2) \\

        &=

        \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)},A_1)

        \; \cdot \;

        \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)},A_2).

    \end{align*}

    The second equality follows directly from $\mathbb{P}(A\mid B)=\mathbb{P}(A,B)/\mathbb{P}(B)$ and setting $A_1,A_2$ to the always-true event.

\end{proof}

\begin{lemma}[Conditional independence 2] \label{lemma:eventindependenceNew}

	Let $v,w \in [n]$, and let $A$ be any event that depends only on the sites $[v,w]$ (meaning the initialization and resamples) and similarly $B$ an event that depends only on the sites $[w,v]$. (For example $\mathrm{Z}^{(s)}$ or ``$s$ has been resampled at least $k$ times'' for an $s$ on the correct interval). Then we have

	\begin{align*}

		\P^{(n)}(\mathrm{NZ}^{(v,w)}\cap A\cap B)

		=

		\P_{b_v=b_w=1}^{[v,w]}(\mathrm{NZ}^{(v,w)}\cap A)

		\; \cdot \;

		\P^{[w,v]}(\mathrm{NZ}^{(v,w)}\cap B),

	\end{align*}

	and similarly

	\begin{align*}

		\P^{[n]}(\mathrm{NZ}^{(v)}\cap A\cap B)

		=

		\P_{b_v=1}^{[v]}(\mathrm{NZ}^{(v)}\cap A)

		\; \cdot \;

		\P^{[v,n]}(\mathrm{NZ}^{(v)}\cap B)

	\end{align*}

	where there is no longer a condition on the starting state.

\end{lemma}

\begin{proof}

    We start by relating the different Markov Chains.

    If $b$ is a starting state where all the zeroes are inside an interval $[v,w]$ (not on the boundary) then we can switch between the cycle and the finite chain:

    \begin{align*}

        \P^{(n)}_{b} (\NZ{v,w} \cap A) = \P^{[v,w]}_b (\NZ{v,w}\cap A) .

    \end{align*}

    If vertex $v$ and $w$ never become zero, then the zeroes never get outside of the interval $[v,w]$ and we can ignore the entire circle and only focus on the process within $[v,w]$.

    We can apply this to the result of Lemma \ref{lemma:eventindependence}, to get

    \begin{align*}

        \P^{(n)}_b(\mathrm{NZ}^{(v,w)} \cap A \cap B)

        &=

        \P^{[v,w]}_{b|_{[v,w]}}(\mathrm{NZ}^{(v,w)} \cap A)

        \; \cdot \;

        \P^{[w,v]}_{b|_{[w,v]}}(\mathrm{NZ}^{(v,w)} \cap B)

    \end{align*}

    Note that this also holds if $b$ has zeroes on the boundary (i.e. $b_v=0$ or $b_w=0$), because then both sides of the equations are zero.

    For the starting state we have the expression $\P^{(n)}(\start{b}) = (1-p)^{|b|} p^{n-|b|}$ so it splits into a product

    \begin{align*}

        \P^{(n)}(\start{b}) = \P^{[v,w]}(\start{b|_{[v+1,w-1]}}) \;\; \P^{[w,v]}(\start{b|_{[w,v]}})

    \end{align*}

    where we have to be careful to count the boudary only once.

    We now have

    \begin{align*}

		\P^{(n)}(\mathrm{NZ}^{(v,w)}\cap A\cap B)

        &= \sum_{b\in\{0,1\}^n} \P^{(n)}_b(\mathrm{NZ}^{(v,w)}\cap A\cap B) \; \P^{(n)}(\start{b}) \\

        &= \sum_{b\in\{0,1\}^n}

            \P^{[v,w]}_{b|_{[v,w]}}(\mathrm{NZ}^{(v,w)}\cap A)

            \P^{[v,w]}(\start{b|_{[v+1,w-1]}})

            \\ &\qquad\qquad\quad\cdot

            \P^{[w,v]}_{b|_{[w,v]}}(\mathrm{NZ}^{(v,w)}\cap B)

            \P^{[w,v]}(\start{b|_{[w,v]}}) \\

        &= \left( \sum_{\substack{b_1\in\{0,1\}^{[v,w]}\\ b_v=b_w=1}}

            \P^{[v,w]}_{b_1}(\mathrm{NZ}^{(v,w)}\cap A)

            \P^{[v,w]}(\start{b_1}) \right)

            \\ &\qquad \cdot

           \left( \sum_{b_2\in\{0,1\}^{[w,v]}}

            \P^{[w,v]}_{b_2}(\mathrm{NZ}^{(v,w)}\cap B)

            \P^{[w,v]}(\start{b_2}) \right) \\

        &=  \P^{[v,w]}_{b_v=b_w=1}(\mathrm{NZ}^{(v,w)}\cap A) \cdot

            \P^{[w,v]}(\mathrm{NZ}^{(v,w)}\cap B)

    \end{align*}

    The second equality follows in a similar way.

\end{proof}

	\begin{definition}[Connected patches]

		Let $P$ be an interval $[a,b]$. We say that $P$ is a patch of a particular run of the process if $P$ is a maximal connected component of the vertices that have ever become $0$ before termination. We denote the set of patches of a run by $\mathcal{P}$. For a patch $P$ let $P\in \mathcal{P}$ denote the event that one of the patches is equal to $P$. 

		In other words

		\begin{align*}

		P\in\mathcal{P} := \NZ{a-1} \cap \Z{a} \cap \Z{a+1} \cap \cdots \cap \Z{b-1} \cap \Z{b} \cap \NZ{b+1} .

		\end{align*}

		(In the extreme case when $P$ covers the whole cycle $[n]$, then instead $P\in\mathcal{P}:= \bigcap_{v\in[n]}\Z{v}$.)

	\end{definition} 

	We are often going to use the observation that we can partition the event $\Z{v}$ using patches:

	\begin{align*}

	\Z{v} = \dot\bigcup_{P\text{ patch } : v\in P} (P\in\mathcal{P})

	\end{align*}

The intuition of the following lemma is that the far right can only affect the zero vertex if there is an interaction chain forming, which means that every vertex should get resampled to $0$ at least once.

\begin{lemma}\label{lemma:probIndepNew}

	$\forall n\in \mathbb{N}_+:\P^{[n]}(\Z{1})-\P^{[n+1]}(\Z{1}) = \bigO{p^{n}}$. (Should be true with $\bigO{p^{n+1}}$ as well.)

\end{lemma}

\begin{proof}

	The proof uses induction on $n$. For $n=1$ the statement is easy, since $\P^{[1]}(\Z{1})=p$ and $\P^{[2]}(\Z{1})=p+p^2+\bigO{p^{3}}$.

	Induction step: suppose we proved the claim for $n-1$, then

	\begin{align*}

	\P^{[n+1]}(\Z{1})

	&=\sum_{k=1}^{n+1}\P^{[n+1]}([k]\in\mathcal{P}) \tag{the events form a partition}\\

	&=\sum_{k=1}^{n-1}\P^{[n+1]}([k]\in\mathcal{P}) + \bigO{p^{n}}\tag*{$\left(\P^{[n+1]}([k]\in\mathcal{P})=O(p^{k})\right)$}\\	

	&=\sum_{k=1}^{n-1}\P^{[k+1]}_{b_{k+1}=1}([k]\in\mathcal{P})\cdot \P^{[n-k+1]}(\NZ{1})+ \bigO{p^{n}} \tag{by Claim~\ref{lemma:eventindependenceNew}}\\

	&=\sum_{k=1}^{n-1}\P^{[k+1]}_{b_{k+1}=1}([k]\in\mathcal{P})\cdot \left(\P^{[n-k]}(\NZ{1})+\bigO{p^{n-k}}\right)+ \bigO{p^{n}} \tag{by induction} \\	

	&=\sum_{k=1}^{n-1}\P^{[k+1]}_{b_{k+1}=1}([k]\in\mathcal{P})\cdot \P^{[n-k]}(\NZ{1})+ \bigO{p^{n}} \tag*{$\left(\P^{[k+1]}_{b_{k+1}=1}([k]\in\mathcal{P})=\bigO{p^{k}}\right)$}\\	

	&=\sum_{k=1}^{n-1}\P^{[n]}([k]\in\mathcal{P})+ \bigO{p^{n}} \tag{by Claim~\ref{lemma:eventindependenceNew}}\\

	&=\sum_{k=1}^{n}\P^{[n]}([k]\in\mathcal{P})+ \bigO{p^{n}} \tag*{$\left(\P^{[n]}([n]\in\mathcal{P})=\bigO{p^{n}}\right)$}\\	

	&=\P^{[n]}(\Z{1})	+ \bigO{p^{n}} 

	\end{align*}

\end{proof}

\begin{corollary}\label{cor:probIndepNew}

	$\P^{[n]}(\Z{1})-\P^{[m]}(\Z{1}) = \bigO{p^{\min(n,m)}}$. (Should be true with $\bigO{p^{\min(n,m)+1}}$ too.)

\end{corollary}

	The intuition of the following lemma is simmilar to the previous. The events on the two sides should be independent unless an interaction chain is forming, implying that every vertex gets resampled to $0$ at least once.

 	\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:

 		$$\P^{[k]}(\Z{1})=\sum_{P\text{ patch}\,:\,1\in P}\P^{[k]}(P\in\mathcal{P})$$

 		$$\P^{[k]}(\Z{k})=\sum_{P\text{ patch}\,:\,k\in P}\P^{[k]}(P\in\mathcal{P})$$

 		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]}([\ell],[r,k]\in\mathcal{P})

 		+\P^{[k]}([k]\in\mathcal{P})\\

 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!\P^{[k]}([\ell],[r,k]\in\mathcal{P})

 		+\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]}_{b_{\ell+1}=1}([\ell]\in\mathcal{P})

 		\P^{[\ell+1,r-1]}(\NZ{\ell+1}\cap \NZ{r-1})

 		\P^{[r-1,k]}_{b_{r-1}=1}([r,k]\in\mathcal{P})

 		+\bigO{p^{k}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\

 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!

 		\P^{[\ell+1]}_{b_{\ell+1}=1}([\ell]\in\mathcal{P})

 		\left(\P^{[\ell+1,r-1]}(\NZ{\ell+1})

		\P^{[\ell+1,r-1]}(\NZ{r-1})\right)

 		\P^{[r-1,k]}_{b_{r-1}=1}([r,k]\in\mathcal{P})

 		+\bigO{p^{k}} \tag{by induction}\\

 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!

 		\P^{[\ell+1]}_{b_{\ell+1}=1}([\ell]\in\mathcal{P})

 		\left(\P^{[\ell+1,k]}(\NZ{\ell+1})

 		\P^{[1,r-1]}_{b_{r-1}=1}(\NZ{r-1})\right)

 		\P^{[r-1,k]}([r,k]\in\mathcal{P})

 		+\bigO{p^{k}} \tag{by Corrolary~\ref{cor:probIndepNew}}\\

 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!

 		\P^{[k]}([\ell]\in\mathcal{P})

 		\P^{[k]}([r,k]\in\mathcal{P})

 		+\bigO{p^{k}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\

 		&=\left(\sum_{\ell\in [k]}\P^{[k]}([\ell]\in\mathcal{P})\right)

 		\left(\sum_{r\in [k]}\P^{[k]}([r,k]\in\mathcal{P})\right)

 		+\bigO{p^{k}} \tag*{$\left(\P^{[k]}([\ell]\in\mathcal{P})=\bigO{p^{\ell}}\right)$}\\	

 		&=\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 lemmas. 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\,\&\, \underset{P_{v,w}:=}{\underbrace{[-v\!+\!1,w\!-\!1]}}\in\mathcal{P}) \tag{partition}\\[-1mm]

			&= \sum_{k=1}^{\infty}\sum_{\underset{v+w\leq n}{v,w\in [n]}}\P^{(n)}(\Res{0}\!\geq\! k\,\&\, P_{v,w}\!\in\!\mathcal{P}) +\bigO{p^{n}}\\[-1mm]

			&= \sum_{k=1}^{\infty}\smash{\sum_{\underset{v+w\leq n}{v,w\in [n]}}}\P^{[-v,w]}_{b_{-v}=b_{w}=1}(\Res{0}\!\geq\! k\,\&\, P_{v,w}\!\in\!\mathcal{P}) \P^{[w,n-v]}(\NZ{w,n-v}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\

			&= \sum_{k=1}^{\infty}\smash{\sum_{\underset{v+w\leq n}{v,w\in [n]}}}\P^{[-v,w]}_{b_{-v}=b_{w}=1}(\Res{0}\!\geq\! k\,\&\, P_{v,w}\!\in\!\mathcal{P})  \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]}_{b_{-v}=b_{w}=1}(\Res{0}\!\geq\! k\,\&\, P_{v,w}\!\in\!\mathcal{P})  \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]}_{b_{-v}=b_{w}=1}(\Res{0}\!\geq\! k\,\&\, P_{v,w}\!\in\!\mathcal{P}) \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\,\&\, P_{v,w}\!\in\!\mathcal{P}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\[-1mm]

			&= \sum_{k=1}^{\infty}\sum_{\underset{|P|<n}{P\text{ patch}:0\in P}}\P^{[-m,m]}(\Res{0}\!\geq\! k\,\&\, P\in\mathcal{P}) +\bigO{p^{n}} \\[-1mm]

			&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:0\in P}\P^{[-m,m]}(\Res{0}\!\geq\! k\,\&\, P\in\mathcal{P}) +\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}			

~