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    R_{b_1} + R_{b_2} &= 3 + 8p + 17.67p^2+35.78p^3 + 69.26p^4 +\mathcal{O}(p^5)

\end{align*}

and indeed the sums agree up to order $p^{k-1}=p^2$. When going up to order $p^{k}$ or higher, there will be terms where the groups interfere so they are no longer independent.

~

\begin{proof}

    Consider a path $\xi_1\in\paths{b_1}$ and a path $\xi_2\in\paths{b_2}$ such that $\xi_1$ and $\xi_2$ are independent (Definition \ref{def:independence} or \ref{def:independence2}). The paths $\xi_1,\xi_2$ induce $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ different paths of total length $|\xi_1|+|\xi_2|$ in $\paths{b_1\land b_2}$. In the sums $R_{b_1}$ and $R_{b_2}$, the contribution of these paths are $\mathbb{P}[\xi_1]\cdot |\xi_1|$ and $\mathbb{P}[\xi_2]\cdot |\xi_2|$. The next diagram shows how these $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths contribute to $R_{b_1\land b_2}$. Point $(i,j)$ in the grid indicates that $i$ steps of $\xi_1$ have been done and $j$ steps of $\xi_2$ have been done. At every point (except the top and right edges of the grid) one has to choose between doing a step of $\xi_1$ or a step of $\xi_2$. The number of zeroes in the current state determine the probabilities with which this happens (beside the probabilities associated to the two original paths already). The grid below shows that at a certain point one can choose to do a step of $\xi_1$ with probability $p_i$ or a step of $\xi_2$ with probability $1-p_i$. These $p_i$ could in principle be different at every point in this grid. The weight of such a new path $\xi\in\paths{b_1\land b_2}$ is $p_\mathrm{grid}\cdot\mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]$ where $p_\mathrm{grid}$ is the weight of the path in the diagram. By induction one can show that the sum over the $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ different terms $p_\mathrm{grid}$ is $1$.

\begin{center}

\includegraphics{diagram_paths.pdf}

\end{center}

 Hence the contribution of all $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths together to $R_{b_1\land b_2}$ is given by

\[

\mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]\cdot(|\xi_1|+|\xi_2|) = \mathbb{P}[\xi_2]\cdot\mathbb{P}[\xi_1]\cdot|\xi_1| \;\; + \;\; \mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]\cdot|\xi_2|.

\]

Ideally we would now like to sum this expression over all possible paths $\xi_1,\xi_2$ and use $p_\mathrm{tot}:=\sum_{\xi\in\paths{b_i}} \mathbb{P}[\xi] = 1$ (which also holds up to arbitrary order in $p$). The above expression would then become $R_{b_1} + R_{b_2}$. However, not all paths in the sum would satisfy the independence condition so it seems we can't do this. We now argue that it works up to order $p^{k-1}$.

For all $\xi\in\paths{b_1\land b_2}$ we have that \emph{either} $\xi$ splits into two independent paths $\xi_1,\xi_2$ as above, \emph{or} it does not. In the latter case, when $\xi$ can not be split like that, we know $\mathbb{P}[\xi]$ contains a power $p^k$ or higher because there is a gap of size $k$  and the paths must have moved at least $k$ times `towards each other' (for example one path moves $m$ times to the right and the other path moves $k-m$ times to the left). So the total weight of such a combined path is at least order $p^k$. Therefore we have

\[

	R_{b_1\land b_2} = \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| + \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_1]\mathbb{P}[\xi_2]|\xi_2| + \sum_{\mathclap{\xi\;\mathrm{dependent}}} \mathbb{P}[\xi]|\xi|.

\]

where last sum only contains only terms of order $p^{k}$ or higher. Now for the first sum, note that

\[

	\sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|

    = \sum_{\xi_1\in\paths{b_1}} \sum_{\substack{\xi_2\in\paths{b_2}\\ \text{independent of }\xi_1}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|

\]

where the sum over independent paths could be empty for certain $\xi_1$. Now we replace this last sum by a sum over \emph{all} paths $\xi_2\in\paths{b_2}$. This will change the sum but only for terms where $\xi_1,\xi_2$ are dependent. For those terms we already know that $\mathbb{P}[\xi_1]\mathbb{P}[\xi_2]$ contains a factor $p^k$ and hence we have 

\begin{align*}

    \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|

    &= \sum_{\xi_1\in\paths{b_1}} \sum_{\xi_2\in\paths{b_2}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| + \mathcal{O}(p^k) \\

    &= \sum_{\xi_1\in\paths{b_1}} \mathbb{P}[\xi_1]|\xi_1| + \mathcal{O}(p^k) \\

    &= R_{b_1} + \mathcal{O}(p^k)

\end{align*}

we can do the same with the second term and this proves the claim.

\end{proof}

~\\

\textbf{Proof of claim \ref{claim:weakcancel}}: We can assume $C$ consists of a group on the left with $l$ slots and a group on the right with $r$ slots (so $r+l=|C|$), with a gap of size $k=\mathrm{gap}(C)$ between these groups. Then on the left we have strings in $\{0,1'\}^l$ as possibilities and on the right we have strings in $\{0,1'\}^r$. The combined configuration can be described by strings $f=(a,b)\in\{0,1'\}^{l+r}$. The initial probability of such a state $C(a,b)$ is $\rho_{C(a,b)} = (-1)^{|a|+|b|} p^{r+l}$ and by claim \ref{claim:expectationsum} we know $R_{C(a,b)} = R_{C(a)} + R_{C(b)} + \mathcal{O}(p^k)$ where $C(a)$ indicates that only the left slots have been filled by $a$ and the other slots are filled with $1$s. The total contribution of these configurations is therefore

\begin{align*}

    \sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R_{C(f)}

    &= \sum_{a\in\{0,1'\}^l} \sum_{b\in\{0,1'\}^r} (-1)^{|a|+|b|}p^{r+l} \left( R_{C(a)} + R_{C(b)} + \mathcal{O}(p^k) \right) \\

    &=\;\;\; p^{r+l}\sum_{a\in\{0,1'\}^l} (-1)^{|a|} R_{C(a)} \sum_{b\in\{0,1'\}^r} (-1)^{|b|} \\

    &\quad + p^{r+l}\sum_{b\in\{0,1'\}^r} (-1)^{|b|} R_{C(b)} \sum_{a\in\{0,1'\}^l} (-1)^{|a|}

        + \mathcal{O}(p^{r+l+k})\\

    &= 0 + \mathcal{O}(p^{|C|+k})

\end{align*}

where we used the identity $\sum_{a\in\{0,1\}^l} (-1)^{|a|} = 0$.

\newpage

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

    For any state $b\in\{0,1\}^n$, define $\textsc{start}(b)$ as the event that the starting state of the chain is the state $b$. For any event $A$, define

    \begin{align*}

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

        %R_{b,A} &= \mathbb{E}( \#resamples \;|\; A \; , \; \textsc{start}(b))

    \end{align*}

\end{definition}

%Note that we have $\P^{(n)}(\textsc{start}(b)) = (1-p)^{|b|}p^{n-|b|}$ 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{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 groups of zeroes that are separated by at least one site inbetween, 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}

The lemma says that conditioned on $v$ and $w$ not being crossed, the two halves of the cycle are independent. 

\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*}

    \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.

    %For the third equality, by the same reasoning we can decompose the paths

    %\begin{align*}

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

    %    &\equiv \sum_{\substack{\xi\in\start{b}\\\xi \in \mathrm{NZ}^{(v,w)}\cap A_1\cap A_2}} \P^{(n)}[\xi] |\xi| \\

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

    %      \sum_{\substack{\xi_2\in\start{b_2}\\\xi_2 \in \mathrm{NZ}^{(v,w)}\cap A_2}}

    %    \P^{(n)}[\xi_1]\P^{(n)}[\xi_2] (|\xi_1| + |\xi_2|) \\

    %    &=

    %    \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)},A_2) \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)},A_1) R_{b_1,\mathrm{NZ}^{(v,w)},A_1} \\

    %    &\quad +

    %    \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)},A_1) \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)},A_2) R_{b_2,\mathrm{NZ}^{(v,w)},A_2} .

    %\end{align*}

    %Dividing by $\P^{(n)}_b(\mathrm{NZ}_{(v,w)},A_1,A_2)$ and using the first equality gives the desired result.

\end{proof}

\begin{definition}[Starting state dependent probability distribution.]

	Let $I\subset\mathbb{Z}$ be a finite set of vertices.

\end{definition}

	\begin{align*}

	&=

	\; \cdot \;

	&=

	\; \cdot \;

	\end{align*}

\end{lemma}

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}

	$\forall n\in \mathbb{N}_+:\P^{[n]}(\Z{1})-\P^{[n+1]}(\Z{1}) = O(p^{n})$. (Should be true with $O(p^{n+1})$ 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 $\P^{[2]}(\Z{1})=p+p^2+O(p^{3})$.

	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]}([k]\in\mathcal{P}) \tag{the events are a partition}\\

	&=\sum_{k=1}^{n-1}\P^{[n]}([k]\in\mathcal{P}) + O(p^{n})\\

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

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

	&=\sum_{k=1}^{n-1}\P^{[k+1]}([k]\in\mathcal{P})\cdot \P^{[n-k]}(\NZ{1})/(1-p)+ O(p^{n}) \\	

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

	&=\sum_{k=1}^{n}\P^{[n]}([k]\in\mathcal{P})+ O(p^{n}) \\

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

	\end{align*}

\end{proof}

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

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

\end{corollary}

 	\begin{lemma}\label{lemma:independenetSidesNew}	

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

 	\end{lemma}   

 	Note that using De Morgan's law and the inclusion-exclusion formula we can see that this is equivalent to saying:

 	$$\P^{[k]}(\NZ{1}\cap \NZ{k})=\P^{[k]}(\NZ{1})\P^{[k]}(\NZ{k})+\mathcal{O}(p^{k}).$$

 	\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})

 		+\mathcal{O}(p^{k})\\

 		&\overset{Lemma~\ref{lemma:eventindependenceNew}}{=}\sum_{\ell, r\in [k]: \ell<r-1}

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

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