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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 \\

            15& 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 & \cellcolor{blue!25}1009.38 & 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 $\forall k\in\mathbb{N}, \forall n,m\geq \max(k,3) : a^{(n)}_k=a^{(m)}_k$ \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}

	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}

    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*}

    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{Quasiprobability method}

Let us first introduce notation for paths of the Markov Chain

\begin{definition}[Paths]

    We define a \emph{path} of the Markov Chain as a sequence of states and resampling choices $\xi=((b_0,r_0),(b_1,r_1),...,(b_k,r_k)) \in (\{0,1\}^n\times[n])^k$ indicating that at time $t$ Markov Chain was in state $b_t\in\{0,1\}^n$ and then resampled site $r_t$. We denote by $|\xi|$ the length $k$ of such a path, i.e. the number of resamples that happened, and by $\mathbb{P}[\xi]$ the probability associated to this path.

    We denote by $\paths{b}$ the set of all valid paths $\xi$ that start in state $b$ and end in state $\mathbf{1} := 1^n$.

\end{definition}

We can write the expected number of resamplings per site $R^{(n)}(p)$ as

\begin{align}

    R^{(n)}(p) &= \frac{1}{n}\sum_{b\in\{0,1\}^{n}} \rho_b \; R_b(p) \label{eq:originalsum} ,

\end{align}

where $R_b(p)$ is the expected number of resamplings when starting from configuration $b$

\begin{align*}

	R_b(p) &= \sum_{\xi \in \paths{b}} \mathbb{P}[\xi] \cdot |\xi| .

\end{align*}

We consider $R^{(n)}(p)$ as a power series in $p$ and show that many terms in (\ref{eq:originalsum}) cancel out if we only consider the series up to some finite order $p^k$. The main idea is that if a path samples a $0$ then $\mathbb{P}[\xi]$ gains a factor $p$ so paths that contribute to $p^k$ can't be arbitrarily long.\\

To see this, we split the sum in (\ref{eq:originalsum}) into parts that will later cancel out. The initial probabilities $\rho_b$ contain a factor $p$ for every $0$ and a factor $(1-p)$ for every $1$. When expanding this product of $p$s and $(1-p)$s, we see that the $1$s contribute a factor $1$ and a factor $(-p)$ and the $0$s only give a factor $p$. We want to expand this product explicitly and therefore we no longer consider bitstrings $b\in\{0,1\}^n$ but bitstrings $b\in\{0,1,1'\}^n$. We view this as follows: every site can have one of $\{0,1,1'\}$ with `probabilities' $p$, $1$ and $-p$ respectively. A configuration $b=101'1'101'$ now has probability $\rho_{b} = 1\cdot p\cdot(-p)\cdot(-p)\cdot 1\cdot p\cdot(-p) = -p^5$ in the starting state $\rho$. It should not be hard to see that we have

\begin{align*}

    R^{(n)}(p) &= \frac{1}{n}\sum_{b\in\{0,1,1'\}^{n}} \rho_{b} \; R_{\bar{b}}(p) ,

\end{align*}

where $\bar{b}$ is the bitstring obtained by changing every $1'$ in it back to a $1$. It is simply the same sum as (\ref{eq:originalsum}) but now every factor $(1-p)$ is explicitly split into $1$ and $(-p)$.

Some terminology: for any configuration we call a $0$ a \emph{particle} (probability $p$) and a $1'$ an \emph{antiparticle} (probability $-p$). We use the word \emph{slot} for a position that is occupied by either a paritcle or antiparticle ($0$ or $1'$). In the initial state, the probability of a configuration is given by $\pm p^{\mathrm{\#slots}}$ where the $\pm$ sign depends on the parity of the number of antiparticles.

We can further rewrite the sum over $b\in\{0,1,1'\}^n$ as a sum over all slot configurations $C\subseteq[n]$ and over all possible fillings of these slots.

\begin{align*}

	R^{(n)}(p) &= \frac{1}{n} \sum_{C\subseteq[n]} \sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R_{C(f)} ,

\end{align*}

where $C(f)\in\{0,1,1'\}^n$ denotes a configuration with slots on the sites $C$ filled with (anti)particles described by $f$. The non-slot positions are filled with $1$s.

\begin{definition}[Diameter and gaps] \label{def:diameter} \label{def:gaps}

    For a subset $C\subseteq[n]$, we define the \emph{diameter} $\diam{C}$ to be the minimum size of an integer interval $I$ containing $C$. Here we consider both $C$ and the interval modulo $n$. In other words $\diam{C} = \min\{ j \vert \exists i : C\subseteq [i,i+j-1] \}$. We define the \emph{gaps} of $C$, as $I\setminus C$ and denote this by $\gaps{C}$. Note that $\diam{C} = |C| + |\gaps{C}|$.  Define $\maxgap{C}$ as the size of the largest connected component of $\gaps{C}$. Figure \ref{fig:diametergap} illustrates these concepts with a picture. 

\end{definition}

\begin{figure}

	\begin{center}

    	\includegraphics{diagram_gap.pdf}

    \end{center}

    \caption{\label{fig:diametergap} Illustration of Definition \ref{def:diameter}. A set $C=\{1,2,4,7,9\}\subseteq[n]$ consisting of 5 positions is shown by the red dots. The smallest interval containing $C$ is $[1,9]$, so the diameter is $\diam{C}=9$. The blue squares denote the set $\gaps{C} = \{3,5,6,8\}$. The dotted line at the top depicts the rest of the cycle which may be much larger. The largest gap of $C$ is $\maxgap{C}=2$ which is the largest connected component of $\gaps{C}$.}

\end{figure}

\begin{claim}[Strong cancellation claim] \label{claim:strongcancel}

	The lowest order term in

    \begin{align*}

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

    \end{align*}

	is $p^{\diam{C}}$ when $n$ is large enough. All lower order terms cancel out.

\end{claim}

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

        \cdots

        \P_{b_1}[\xi_1] \; \P_{b_2}[\xi_2] \tag{definition of $\P_{b_i}[\xi_i]$} \\

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

        &= \P(\text{path in grid}) \; \P_{b_1}[\xi_1] \; \P_{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*}

        \mathbb{P}_b(\mathrm{NZ}^{(j_1,j_2)},A_1,A_2)

        &= \sum_{\substack{\xi\in\paths{b} \cap \\ \mathrm{NZ}^{(j_1,j_2)}\cap A_1\cap A_2}} \mathbb{P}[\xi] \\

        &= \sum_{\substack{\xi_1\in\paths{b_1} \cap \\ \mathrm{NZ}^{(j_1,j_2)}\cap A_1}} \;\;

          \sum_{\substack{\xi_2\in\paths{b_1} \cap \\ \mathrm{NZ}^{(j_1,j_2)}\cap A_2}}

        \mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2] \\

        &=

        \mathbb{P}_{b_1}(\mathrm{NZ}^{(j_1,j_2)},A_1)

        \; \cdot \;

        \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)},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*}

        \mathbb{P}_b(\mathrm{NZ}^{(j_1,j_2)},A_1,A_2) R_{b,\mathrm{NZ}^{(j_1,j_2)},A_1,A_2}

        &\equiv \sum_{\substack{\xi\in\paths{b}\\\xi \in \mathrm{NZ}^{(j_1,j_2)}\cap A_1\cap A_2}} \mathbb{P}[\xi] |\xi| \\

        &= \sum_{\substack{\xi_1\in\paths{b_1}\\\xi_1 \in \mathrm{NZ}^{(j_1,j_2)}\cap A_1}}

          \sum_{\substack{\xi_2\in\paths{b_2}\\\xi_2 \in \mathrm{NZ}^{(j_1,j_2)}\cap A_2}}

        \mathbb{P}[\xi_1]\mathbb{P}[\xi_2] (|\xi_1| + |\xi_2|) \\

        &=

        \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)},A_2) \mathbb{P}_{b_1}(\mathrm{NZ}^{(j_1,j_2)},A_1) R_{b_1,\mathrm{NZ}^{(j_1,j_2)},A_1} \\

        &\quad +

        \mathbb{P}_{b_1}(\mathrm{NZ}^{(j_1,j_2)},A_1) \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)},A_2) R_{b_2,\mathrm{NZ}^{(j_1,j_2)},A_2} .

    \end{align*}

    Dividing by $\mathbb{P}_b(\mathrm{NZ}_{(j_1,j_2)},A_1,A_2)$ and using the first equality gives the desired result.

\end{proof}

\begin{comment}

TEST: Although a proof of claim \ref{claim:expectationsum} was already given, I'm trying to prove it in an alternate way using claim \ref{claim:eventindependence}.

~

Assume that $b_1$ ranges up to site $0$, the gap ranges from sites $1,...,k$ and $b_2$ ranges from site $k+1$ and onwards. For $j=1,...,k$ define the ``partial-zeros'' event $\mathrm{PZ}_j = \mathrm{Z}_1 \cap \mathrm{Z}_2 \cap ... \cap \mathrm{Z}_{j-1} \cap \mathrm{NZ}_j$ i.e. the first $j-1$ sites of the gap become zero and site $j$ does not become zero. Also define the ``all-zeros'' event $\mathrm{AZ} = \mathrm{Z}_1 \cap ... \cap \mathrm{Z}_k$, where all sites of the gap become zero. Note that these events partition the space, so we have for all $b$ that $\sum_{j=1}^k \mathbb{P}_b(\mathrm{PZ}_j) = 1 - \mathbb{P}_b(\mathrm{AZ}) = 1 - \mathcal{O}(p^k)$.

~

Furthermore, if site $j$ becomes zero when starting from $b_1$ it means all sites to the left of $j$ become zero as well. Similarly, from $b_2$ it implies all the sites to the right of $j$ become zero.

Because of that, we have

\begin{align*}

    \mathbb{P}_{b_1}(\mathrm{PZ}_j) &= \mathbb{P}_{b_1}(\mathrm{Z}_{j-1} \cap \mathrm{NZ}_j) = \mathcal{O}(p^{j-1}) \\

    \mathbb{P}_{b_2}(\mathrm{NZ}_j) &= 1 - \mathbb{P}_{b_2}(\mathrm{Z}_j) = 1 - \mathcal{O}(p^{k-j+1})

\end{align*}

Following the proof of claim \ref{claim:eventindependence} we also have

\begin{align*}

    \mathbb{P}_b(\mathrm{PZ}_{j})

    &=

    \mathbb{P}_{b_1}(\mathrm{PZ}_{j})

    \; \cdot \;

    \mathbb{P}_{b_2}(\mathrm{NZ}_{j}) \\

    R_{b,\mathrm{PZ}_{j}}

    &=

    R_{b_1,\mathrm{PZ}_{j}}

    \; + \;

    R_{b_2,\mathrm{NZ}_{j}}

\end{align*}

Now observe that

\begin{align*}

    R_b &= \sum_{j=1}^k \mathbb{P}_b(\mathrm{PZ}_j) R_{b,\mathrm{PZ}_j} + \mathbb{P}_b(\mathrm{AZ}) R_{b,\mathrm{AZ}} \\

        &= \sum_{j=1}^k \mathbb{P}_{b_2}(\mathrm{NZ}_j)\mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}

        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{NZ}_j) R_{b_2,\mathrm{NZ}_j}

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

        &= \sum_{j=1}^k \mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}

        - \sum_{j=1}^k \mathbb{P}_{b_2}(\mathrm{Z}_j)\mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}

        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{NZ}_j) R_{b_2,\mathrm{NZ}_j}

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

        &= \sum_{j=1}^k \mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}

        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{NZ}_j) R_{b_2,\mathrm{NZ}_j}

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

        &= R_{b_1}

        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{NZ}_j) R_{b_2,\mathrm{NZ}_j}

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

        &\overset{???}{=} R_{b_1} + R_{b_2} + \mathcal{O}(p^k)

\end{align*}

\end{comment}

Consider the chain (instead of the cycle) for simplicity with vertices identified by $\mathbb{Z}$.

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

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

    Let $b_I$ be the initial state where everything is $1$, apart from the vertices corresponding to $I$, which are set $0$. Define $P_I(A)=P_{b_I}(A)$ where the latter is defined in Definition \ref{def:conditionedevents}, i.e. the probability of seeing a resample sequence from $A$ when the whole procedure started in state $b_I$. 

\end{definition}

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:probIndep}

	Suppose we have a finite set $I\subset\mathbb{N}_+$ of vertices.

    Let $I_{\max}:=\max(I)$ and $I':=I\setminus\{I_{\max}\}$, and similarly let $I_{\min}:=\min(I)$. These definitions are illustraded in Figure \ref{fig:lemmaillustration}.

\end{lemma}

\begin{proof}

\begin{figure}

	\begin{center}

    	\includegraphics{diagram_proborders.pdf}

    \end{center}

    \caption{\label{fig:lemmaillustration} Illustration of setup of Lemma \ref{lemma:probIndep}.}

\end{figure}

	The proof uses induction on $|I|$. For $|I|=1$ the statement is easy, since every resample sequence that resamples vertex $0$ to zero must produce at least $I_{\max}$ zeroes in-between.

    Let $I_{<k}:=I\cap[1,k-1]$ and similarly $I_{>k}:=I\setminus[1,k]$, finally let $I_{><}:=\{I_{\min}+1,I_{\max}-1]\}\setminus I$ (note that $I_{><} = \gaps{I}$ as shown in Figure \ref{fig:diametergap}). Suppose we have proven the claim up to $|I|-1$, then the induction step can be shown by

	\begin{align*}

		\tag{$k< I_{\max}\Rightarrow I_{<k}=I'_{<k}$}\\

	\end{align*}

\end{proof}

	~

Here, I (Tom) tried to set do the same Lemma but for the cycle instead of the infinite chain.

\begin{lemma}[Startingstate dependence] \label{lemma:probIndepCycle}

    Let $d(a,b)$ be the distance between $a,b\in[n]$ on the cycle, so $d(a,b)=\min(|a-b| , n-|a-b|)$. Let $\dist{s}(a,b)$ be the distance between $a,b$ when taking the path that does \emph{not} cross $s$. Let $I\subseteq [n]$ be a non-empty set of vertices. Let $i_* \in I$ and define $I' = I \setminus \{i_*\}$. Let $j,s\notin I$, with $j\neq s$ be any vertices not in $I$.

    Then

    \begin{align*}

        \P_{I}(\Z{j})        &= \P_{I'}(\Z{j})        + \mathcal{O}(p^{d(i_*,j) + 1 - |I|}) \\

        \P_{I}(\Z{j},\NZ{s}) &= \P_{I'}(\Z{j},\NZ{s}) + \mathcal{O}(p^{\min\left( \dist{s}(i_*,j), \dist{j}(i_*,s) \right) + 1 - |I|}) .

    \end{align*}

\end{lemma}

\begin{proof}

    Without loss of generality, we can assume that $j=0$ and  $0 < i_* < s < n$ (because we can shift $j$ to $0$ and switch the direction to get the correct ordering). Therefore, we have to prove:

    \begin{align*}

        \P_{I}(\Z{0})        &= \P_{I'}(\Z{0})        + \mathcal{O}(p^{d(i_*,0) + 1 - |I|}) \\

        \P_{I}(\Z{0},\NZ{s}) &= \P_{I'}(\Z{0},\NZ{s}) + \mathcal{O}(p^{\min\left( i_*, s-i_* \right) + 1 - |I|}) .

    \end{align*}

    We will prove both statements inductively on $|I|$. For $|I|=1$ we have $I=\{i_*\}$ and $I'=\emptyset$, so $\P_{I'}(\Z{0})=0$ and

    \begin{align*}

        \P_{I}(\Z{0})        &= \mathcal{O}(p^{d(i_*,0)}) \\

        \P_{I}(\Z{0},\NZ{s}) &= \mathcal{O}(p^{i_*}) = \mathcal{O}(p^{\min\left( i_*, s-i_* \right)})

    \end{align*}

    simply because a chain of zeroes has to be formed between $i_*$ and $0$, and in the second case this chain can not go through $s$ so the shortest path has length $i_*$. Now assume both statements hold up to $|I|-1$, then we prove them both for sets of size $|I|$.

    When we refer to an interval $[a,b]$ on the cycle we could be referring to two possible intervals because of the periodicity of the cycle. Define $[a,b]_j$ as the interval with vertex $j$ on the \emph{inside}. Furthermore by $-a$ we mean the vertex $n-a$, as one would expect modulo $n$.

 We will now consider intervals around vertex 0.

    For $l,r\geq 1$ and $l+r\leq n$, define the event ``zeroes patch'' $\mathrm{ZP}^{[-l,r]_0}$ as the event of getting zeroes inside the interval $[-l,r]_0$ but not on the boundary, i.e.

    $$\mathrm{ZP}^{[-l,r]_0} = \NZ{-l} \cap \Z{-l+1} \cap \cdots \cap \Z{0} \cap \cdots \cap \Z{r-1} \cap \NZ{r}$$

    Note that there are $r+l-1$ `zeroes' in this event, so $\P_{J}(\mathrm{ZP}^{[-l,r]_0}) = \mathcal{O}(p^{r+l-1-|J|})$ for $J\subseteq[-l,r]_0$ is a lower bound on the order of $p$.\\

    Claim:

    \begin{align*}

        \P_{I}(\mathrm{ZP}^{[-l,r]_0}) &= \P_{I'}(\mathrm{ZP}^{[-l,r]_0})

        + \mathcal{O}(p^{d(i_*,0)+1-|I|})

    \end{align*}

    If $r\geq i_*$ or $l\geq n-i_*$ then $\P_{I}(\mathrm{ZP}^{[-l,r]_0}) = \mathcal{O}(p^{d(i_*,0) + 1 - |I|})$ and also $\P_{I'}(\mathrm{ZP}^{[-l,r]_0}) = \mathcal{O}(p^{d(i_*,0) + 1 - |I|})$ so then the claim holds.

    If $-l\in I$ or $r\in I$ (and $-l,r$ are both not $i_*$ because of the previous point) then the probability of $\mathrm{ZP}^{[-l,r]_0}$ is zero for both $I$ and $I'$ so the claim holds.

    If $[-l,r]_0$ has no overlap with $I$ then both sides are also zero so it also holds. We are left with the case where: $-l,r,\notin I$ and $[-l,r]_0 \cap I \neq \emptyset$ and $i_*\notin[-l,r]_0$.

    The following diagram illustrates the situation

    \begin{center}

        \includegraphics{diagram_circle_lemma.pdf}

    \end{center}

    Note that by Claim~\ref{claim:eventindependence} we have

    \begin{align*}

        \P_{I}(\mathrm{ZP}^{[-l,r]_0}) = \P_{I \cap [-l,r]_0}(\mathrm{ZP}^{[-l,r]_0}) \;\cdot\; \P_{I\setminus [-l,r]_0}(\NZ{-l},\NZ{r})

    \end{align*}

    We have $i_*\in I \setminus[-l,r]_0$, and $I\cap[-l,r]_0 = I' \cap [-l,r]_0$. Define $J=I\setminus[-l,r]_0$ and $J'=I'\setminus[-l,r]_0$. We have $|J|<|I|$ so we can apply the induction hypothesis to $J$:

    \begin{align*}

        \P_{J}(\NZ{-l},\NZ{r})

        &=

        1

        - \P_{J}(\Z{-l},\NZ{r})

        - \P_{J}(\Z{r})

        \tag{partition of all events} \\

        &=

        1

        - \P_{J'}(\Z{-l},\NZ{r})

        - \P_{J'}(\Z{r}) \\

        &\quad + \mathcal{O}(p^{\min\left( \dist{r}(i_*,-l), \dist{-l}(i_*,r) \right) +1-|J|})

        + \mathcal{O}(p^{d(i_*,r)+1-|J|}) \\

        &=

        \P_{J'}(\NZ{-l},\NZ{b})

        + \mathcal{O}(p^{\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right)+1-|J|})

    \end{align*}

    Note that the event $\mathrm{ZP}^{[-l,r]_0}$ contains $l+r-1$ zeroes, so $\P_{I \cap [-l,r]_0}(\mathrm{ZP}^{[-l,r]_0}) = \mathcal{O}(p^{l+r-1-|I\cap[-l,r]_0|})$. This means

    \begin{align*}

        \P_{I}(\mathrm{ZP}^{[-l,r]_0})

        &= \P_{I' \cap [-l,r]_0}(\mathrm{ZP}^{[-l,r]_0})

        \left( \P_{I' \setminus [-l,r]_0}(\NZ{a},\NZ{b}) + \mathcal{O}(p^{\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right)+1-|J|}) \right) \\

        &= \P_{I' \cap [-l,r]_0}(\mathrm{ZP}^{[-l,r]_0}) \;\cdot\; \P_{I'\setminus [-l,r]_0}(\NZ{a},\NZ{b}) \\

        &\qquad + \mathcal{O}(p^{\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right)+1-|J| + l+r-1-|I\cap[-l,r]_0|}) \\

        &= \P_{I'}(\mathrm{ZP}^{[-l,r]_0})

        + \mathcal{O}(p^{\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right)+l+r-|I|})

    \end{align*}

    Where we used Claim~\ref{claim:eventindependence} again.

    Case separation shows that

    $$\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right) + l +r \geq d(i_*,0) + 1$$

    for $l,r\geq 1$ which proves the claim.

    The first equality that we have to prove now follows from the fact that the ``zeroes patch'' events are a partition of $\Z{0}$:

    \begin{align*}

        \P_{I}(\Z{0})

        &=\sum_{\substack{l,r\geq 1\\l+r\leq n}}

        \P_I(\mathrm{ZP}^{[-l,r]_0})

        \tag{the events are a partition of $\Z{0}$}\\

        &=\sum_{\substack{l,r\geq 1\\l+r\leq n}}

        \P_{I'}(\mathrm{ZP}^{[-l,r]_0})

        + \mathcal{O}(p^{d(i_*,0)+1-|I|})

        \tag{by claim} \\

        &= \P_{I'}(\Z{0}) + \mathcal{O}(p^{d(i_*,0)+1-|I|})

    \end{align*}

    Similarly, we have

    \begin{align*}

        \P_{I}(\Z{0} , \NZ{s})

        &=\sum_{l=1}^{n-s}\sum_{r=1}^{s}

        \P_{I}(\mathrm{ZP}^{[-l,r]_0},\NZ{s})

        \tag{partition of $\Z{0}$}\\

        &=\sum_{l=1}^{n-s}\sum_{r=1}^{i_*-1}

        \P_{I}(\mathrm{ZP}^{[-l,r]_0},\NZ{s})

        +\mathcal{O}(p^{i_*+1-|I|}) \\

        &=\sum_{l=1}^{n-s}\sum_{r=1}^{i_*-1}

        \P_{I\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot

        \P_{I\setminus [s,r]_0}(\NZ{r},\NZ{s})

        +\mathcal{O}(p^{i_*+1-|I|})

        \tag{Claim~\ref{claim:eventindependence}}\\

        &=\sum_{l=1}^{n-s}\sum_{r=1}^{i_*-1}

        \P_{I'\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot

        \P_{I\setminus [s,r]_0}(\NZ{r},\NZ{s})

        +\mathcal{O}(p^{i_*+1-|I|})

        \tag{$i_*\in I \setminus[s,r]_0$}\\

        &=\sum_{l=1}^{n-s}\sum_{r=1}^{i_*-1}

        \P_{I'\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot

        \P_{I'\setminus [s,r]_0}(\NZ{r},\NZ{s}) \\

        &\qquad +\mathcal{O}(p^{\min\left( \dist{r}(i_*,s) , d(i_*,r)\right)+l+r-|I|})

        +\mathcal{O}(p^{i_*+1-|I|})

        \tag{same argument as before}\\

        &=\sum_{l=1}^{n-s}\sum_{r=1}^{i_*-1}

        \P_{I'\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot

        \P_{I'\setminus [s,r]_0}(\NZ{r},\NZ{s}) \\

        &\qquad

        +\mathcal{O}(p^{\min\left( i_* , s-i_* \right) +1-|I|})

        \tag{case separation}\\

        &= \P_{I'}(\Z{0} , \NZ{s})

        +\mathcal{O}(p^{\min\left( i_* , s-i_* \right) +1-|I|})

    \end{align*}

    This finishes the proof.

\end{proof}

    Similarly to Mario's proof I use the observation that 

    \begin{align*}

    R^{(n)} &= \frac{1}{n}\sum_{b\in\{0,1,1'\}^{n}} \rho_b \; R_{\bar{b}}(p)\\

    &= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} R_{S(f)}\\

    &= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)}

    \sum_{\mathcal{P}\text{ patches}} \mathbb{P}_{S(f)}(A^{(\mathcal{P})}) R^{(\mathcal{P})}_{S(f)} \\

    &= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)}

    \sum_{\mathcal{P}\text{ patches}} \mathbb{P}_{S(f)}(A^{\mathcal{P}}) \sum_{P\in\mathcal{P}} R^{(P,\mathcal{P})}_{S(f)}\\

    &= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} 

    \sum_{\mathcal{P}\text{ patches}} \mathbb{P}_{S(f)}(A^{\mathcal{P}}) \sum_{P\in\mathcal{P}} R^{(P)}_{S(f)\cap P}\tag{by Claim~\ref{claim:eventindependence}}\\ 

    &= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} 

    \sum_{P\text{ patch}} R^{(P)}_{S(f)\cap P}\sum_{\mathcal{P}:P\in\mathcal{P}}\mathbb{P}_{S(f)}(A^{\mathcal{P}})\\     

    &= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{P\text{ patch}}\sum_{f\in\{0,1'\}^{|S|}}

     \rho_{S(f)} R^{(P)}_{S(f)\cap P}\mathbb{P}_{S(f)}(A^{(P)}) \tag{by definition}\\        

    &= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{P\text{ patch}}\sum_{f\in\{0,1'\}^{|S|}}

    &= \frac{1}{n}\sum_{S\subseteq [n]} \sum_{P\text{ patch}} \sum_{f_P\in\{0,1'\}^{|S\cap P|}}

    \rho_{S(f_P)} R^{(P)}_{S(f_P)} \mathbb{P}_{S(f_P)}(A^{(P)})

	&= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{P\text{ patch}}\sum_{f_P\in\{0,1'\}^{|S\cap P|}}

	\rho_{S(f_P)}

        \sum_{f_{\overline{P}}\in\{0,1'\}^{|S\cap \overline{P}|}}\rho_{S(f_{\overline{P}})}\mathcal{O}(p^{|S_{><}|}) \tag{see below} \\

	&= \frac{1}{n}\sum_{S\subseteq [n]}\mathcal{O}(p^{|S|+|S_{><}|}).

    \end{align*}

\begin{figure}

	\begin{center}

    	\includegraphics{diagram_patches.pdf}

    \end{center}

    \caption{\label{fig:patches} Illustration of last steps of the proof.}

\end{figure}

    The penultimate inequality can be seen by case separation as follows: If $S\subseteq P$ then there is no splitting into $S\cap P$ and $S\setminus P$, and we already have $\mathbb{P}_{S(f_P)}(A^{(P)})=\mathcal{O}(p^{|S_{><}|})$ simply because the patch $P$ must be filled with zeroes that were not yet in $S$, so this is at least $|S_{><}|$ resampled zeroes. For the more general case, assume that $S$ is larger than $P$ on both sides of $P$. This is illustrated in Figure \ref{fig:patches}. We will focus on the following sum that was in the above equations:

    \begin{align*}

    \end{align*}

    By Lemma \ref{lemma:eventindependence} we can split this sum into two parts: the part to the left of $P$ and the part to the right of $P$. Define $S_\mathrm{left}=S\cap[S_\mathrm{min},P_{\mathrm{min}}-1]$ and $S_\mathrm{right}=S\cap[P_{\mathrm{max}}+1,S_\mathrm{max}]$, so that $S\cap\overline{P} = S_\mathrm{left} \cup S_\mathrm{right}$. These are also illustrated in Figure \ref{fig:patches}. Then we have

    \begin{align*}

    \end{align*}

    and hence we can split the sum. Without loss of generality we now only consider the `right' part of the sum:

    \begin{align*}

    \end{align*}

    Now further split this sum over the value of $f$ at position $S_\mathrm{max}$:

    \begin{align*}

        \sum_{f\in\{0,1'\}^{|S_\mathrm{right}\setminus\{S_\mathrm{max}\}|}} \sum_{f'\in\{0,1'\}}

    \end{align*}

    and we use the definition of $\rho$ for the sum over $f'$:

    \begin{align*}

         \sum_{f\in\{0,1'\}^{|S_\mathrm{right}\setminus\{S_\mathrm{max}\}|}}

    \end{align*}

    Now we recognize the setup of Lemma~\ref{lemma:probIndep} with $I=S_\mathrm{right}(f\,0)$ and $I'=S_\mathrm{right}(f\,1)$. The lemma yields

    \begin{align*}

    \end{align*}

    Entering this back into the sum gives

    \begin{align*}

         \sum_{f\in\{0,1'\}^{|S_\mathrm{right}\setminus\{S_\mathrm{max}\}|}}

        \rho_{S_\mathrm{right}(f)} \mathcal{O}(p^{S_\mathrm{max}-P_{\mathrm{max}}-|S_\mathrm{right}|+1})

         = \sum_{f\in\{0,1'\}^{|S_\mathrm{right}|}}

        \rho_{S_\mathrm{right}(f)} \mathcal{O}(p^{S_\mathrm{max}-P_{\mathrm{max}}-|S_\mathrm{right}|})

    \end{align*}

    One can do the same for the `left' part, which gives a term $\mathcal{O}(p^{P_\mathrm{min}-S_{\mathrm{min}}-|S_\mathrm{left}|})$. The part of $S$ that was within $P$ gives $\mathbb{P}_{S(f_P)}(A^{(P)})=\mathcal{O}(p^{P_\mathrm{max}-P_\mathrm{min}+1-|S\cap P|})$. Combining these three factors yields

    \begin{align*}

        (\textrm{left part})(P\textrm{ part})(\textrm{right part}) &=

\mathcal{O}(p^{P_\mathrm{min}-S_{\mathrm{min}}-|S_\mathrm{left}|}) \cdot \mathcal{O}(p^{P_\mathrm{max}-P_\mathrm{min}+1-|S\cap P|}) \cdot \mathcal{O}(p^{S_\mathrm{max}-P_{\mathrm{max}}-|S_\mathrm{right}|}) \\

        &= \mathcal{O}(p^{S_\mathrm{max}-S_\mathrm{min}+1-|S_\mathrm{left}\cup S_\mathrm{right}\cup (S\cap P)|})\\

        &= \mathcal{O}(p^{S_\mathrm{max}-S_\mathrm{min}+1-|S|})

        = \mathcal{O}(p^{|S_{><}|})

    \end{align*}

    as required. This finishes the proof.

\begin{comment}

    \subsection{Sketch of the (false) proof of the linear bound \ref{it:const}}

    Let us interpret $[n]$ as the vertices of a length-$n$ cycle, and interpret operations on vertices mod $n$ s.t. $n+1\equiv 1$ and $1-1\equiv n$.

    %\begin{definition}[Resample sequences]

    %	A sequence of indices $(r_\ell)=(r_1,r_2,\ldots,r_k)\in[n]^k$ is called resample sequence if our procedure performs $k$ consequtive resampling, where the first resampling of the procedure resamples around the mid point $r_1$ the second around $r_2$ and so on. Let $RS(k)$ the denote the set of length $k$ resample sequences, and let $RS=\cup_{k\in\mathbb{N}}RS(k)$.