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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 \\

        \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: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:eventindependenceGen}, 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]

		In other words

		\begin{align*}

		\end{align*}

		\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.)

\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: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*}

\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

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

	\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]