AENC/resampling_chain Changeset - 8a82b7e99663 · Centrum Wiskunde & Informatica (CWI)

@@ -259,8 +259,8 @@ Let $G=(V,E)$ be an undirected graph with vertex set $V$ and edge set $E$. We de

\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 $\Z{S}$ as the event that \emph{all} vertices in $S$ become zero at any point in time before the Markov Chain terminates.

        \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 $\NZ{S}$ 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)

@@ -268,7 +268,7 @@ Let $G=(V,E)$ be an undirected graph with vertex set $V$ and edge set $E$. We de

            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 boundary $\overline{\partial} S$ of $S$ as the set of vertices adjacent to $S$, excluding $S$ itself. In other words $\overline{\partial} S = B(S;1) \setminus S$.

        \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}

@@ -277,7 +277,7 @@ The following Lemma says that if a set $S$ splits the graph in two, then those t

    \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 any events that depends only on the sites in $X$ and $Y$ respectively. Then we have

    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)

&=

@@ -397,184 +397,60 @@ The following Lemma says that if a set $S$ splits the graph in two, then those t

    \end{align*}

\end{proof}

\todo{rewrite from here}

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

	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:splitting}, 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}

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{definition}[Connected patches]

		Let $P\subseteq V$ be a connected component of $G$. 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 $\patch{P}$ denote the event that one of the patches is equal to $P$.

		In other words

		\begin{align*}

		\patch{P} := \NZ{\overline{\partial}P} \cap \Z{P}.

		\end{align*}

		For a set of patches $\mathcal{P}$

		\begin{align*}

			\mathcal{P}'\in \mathcal{P} := \bigcup_{P\in \mathcal{P}'}\NZ{\overline{\partial}P} \cap \Z{P}.

		\end{align*}

	\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:probIndepNewGen}

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

\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!)

\end{lemma}

\begin{proof}

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

	We can assume without loss of generality, that $X\neq \emptyset\neq Y$, otherwise the statement is trivial.

	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:eventindependenceNewGen}}\\

	&=\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:eventindependenceNewGen}}\\

	&=\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*}

	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., $A^X=A^X\cap \Z{X}$.

	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 $A_{d+1}^X:=A^X\cap\bigcap_{j\in[d]}\Z{\overline{\partial}(X,j)}$,

	it is easy to see that it is partition $A^X=\dot\bigcup_{i\in [d+1]}A_i^X$.

	Also 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)}(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)}(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)}(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 using that $\P(A^X)=1-\P(B^X)$.

\end{proof}

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

	$\P^{[n]}(\Z{1})-\P^{[m]}(\Z{1}) = \bigO{p^{\min(n,m)}}$. (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:independenetSidesNewGen}

 		$$\P^{[k]}(\Z{1}\cap \Z{k})=\P^{[k]}(\Z{1})\P^{[k]}(\Z{k})+\bigO{p^{k}}=\left(\P^{[k]}(\Z{1})\right)^2+\bigO{p^{k}}.$$

 	\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:eventindependenceNewGen}}\\

 		&=\!\!\!\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:probIndepNewGen}}\\

 		&=\!\!\!\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:eventindependenceNewGen}}\\

 		&=\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

		$R^{(n)}-R^{(m)}=\bigO{p^{n}}$.

	\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} In the proof we identify the sites of the $n$-cycle with the$\mod n$ remainder classes.

		\vskip-3mm

	\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}\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:eventindependenceNewGen}}\\

			&= \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:independenetSidesNewGen}}\\

			&= \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:independenetSidesNewGen}}\\

			&= \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:eventindependenceNewGen}}\\[-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]

			&= \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*}

		\noindent Repeating the same argument with $m$ and comparing the results completes the proof.

		\vskip-7mm		></i>

	\end{proof}

\begin{comment}

		Let $N\geq \max(2n,2m)$, then

@@ -606,30 +482,7 @@ Questions:

\newpage

\section{Proving that $a_k^{(k+1)}=a_k^{(n)}$ for all $n>k$}

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.

	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.

	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}

@@ -639,46 +492,46 @@ We use the notation $\E^{(n)}$,$\E^{[a,b]}$ and $\E^{[n]}$ similarly for denotin

 		We proceed by induction on $k$. For $k=1,2$ the statement is trivial.

 		Now observe that:

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

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

 		$$\P^{[k]}(\Z{1})=\sum_{C\text{ connected}\,:\,1\in C}\P^{[k]}(P_{C})$$

 		$$\P^{[k]}(\Z{k})=\sum_{C\text{ connected}\,:\,k\in C}\P^{[k]}(P_{C})$$

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

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

 		\P^{[\ell+1]}_{\{\ell+1\}}(P_{[\ell]})

 		\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}}\\

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

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

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

 		\P^{[\ell+1]}_{\{\ell+1\}}(P_{[\ell]})

 		\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}}\\

 		\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]}([\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]}(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 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.

	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

		$R^{(n)}-R^{(m)}=\bigO{p^{n}}$.

@@ -689,15 +542,15 @@ We use the notation $\E^{(n)}$,$\E^{[a,b]}$ and $\E^{[n]}$ similarly for denotin

			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}} \\

			&= \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.