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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 \\

        \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 $\forall k\in\mathbb{N}, \forall n,m > \max(k,3) : a^{(n)}_k=a^{(m)}_k$ \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}

	\colorbox{red}{\ref{it:pos}-\ref{it:geq} is false since $a_{1114}^{(10)}<0$ -- needs to be double checked!}

	I figured this out by observing that $R^{(10)}(p)$ has a pole inside the disk of radius $0.96$. This also means that $R^{(10)}(p)=\sum_{k=0}^{\infty}a_k^{(10)}p^k$ is only true in an analytic sense, since for $p>0.96$ the right hand side does not converge.

	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*}

    For $n=3$ the system becomes very simple because regardless of the current state, the probability of going to $111$ is always equal to $(1-p)^3$. Therefore the expected number of resamplings is simply the expectation of a geometric distribution. This gives the formula for $R^{(3)}(p)$ as shown above. Note that the $k$-th coefficient of the powerseries of a function $f(p)$ is given by $\frac{1}{k!}\left.\frac{d^k f}{dp^k}\right|_{p=0}$, i.e. the $k$-th derivative to $p$ evaluated at $0$ divided by $k!$. For the function $R^{(3)}(p) =\frac{(1-p)^{-3} - 1}{3} $ this yields $a^{(3)}_k = (k+2)(k+1)/6$ for $k\geq 1$ and $a^{(3)}_0=0$.

    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.

	Again the intuition of the final theorem is simmilar to the previous lemmas. A site can only realise the length of the cycle after an interaction chain was formed around the cycle, implying that every vertex was resampled to $0$ at least once.

	\begin{theorem} $R^{(n)}=\E^{[-m,m]}(\Res{0})+\bigO{p^{n}}$ for all $m\geq n \geq 3$, thus

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

	\end{theorem}

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

		\vskip-3mm

		\begin{align*}

			R^{(n)}

			&= \E^{(n)}(\Res{0}) \tag{by translation invariance}\\

			&= \sum_{k=1}^{\infty}\P^{(n)}(\Res{0}\!\geq\! k) \\		

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

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

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

		\end{align*}  

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

	\end{proof} 	

\begin{comment}

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

		\begin{align*}

		R^{(n)}

		&= \E^{(n)}(\Res{1}) \tag{by translation invariance}\\

		&= \sum_{k=1}^{\infty}\P^{(n)}(\Res{1}\geq k) \\

		%&= \sum_{k=1}^{\infty}\sum_{\underset{\ell\geq r-1}{\ell,r\in[n]}}\P^{(n)}(\Res{1}\geq k\,\&\, [\ell+1,r-1]\in\mathcal{P}) \tag{partition}\\

		%&= \sum_{k=1}^{\infty}\sum_{\underset{\ell\geq r}{\ell,r\in[n]}}\P^{(n)}(\Res{1}\geq k\,\&\, [\ell+1,r-1]\in\mathcal{P})  +\bigO{p^{n}} \\	

		%&= \sum_{k=1}^{\infty}\sum_{\underset{\ell\geq r}{\ell,r\in[n]}}\P^{[l,r]}_{b_{\ell}=b_{r}=1}(\Res{1}\geq k\,\&\, [\ell+1,r-1]\in\mathcal{P}) \P^{[r,\ell]}(\NZ{\ell,r}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNewGen}}\\				

		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}\P^{(n)}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \tag{partition}\\

		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{(n)}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) +\bigO{p^{n}}\\

		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[P\cup \partial P]}_{b_{\partial P}=1}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \P^{[\overline{P}]}(\NZ{\partial P}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNewGen}}\\

		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[P\cup \partial P]}_{b_{\partial P}=1}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \left(\left(\P^{[|\overline{P}|]}(\NZ{1})\right)^2+\bigO{p^{|\overline{P}|}}\right) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:independenetSidesNewGen}}\\

		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[P\cup \partial P]}_{b_{\partial P}=1}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \left(\left(\P^{[N]}(\NZ{1})\right)^2+\bigO{p^{|\overline{P}|}}\right) +\bigO{p^{n}} \tag{by Corollary~\ref{cor:probIndepNewGen}}\\

		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[-N,N]}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNewGen}}\\

		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}\P^{[-N,N]}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNewGen}}\\

		&= \E^{[-N,N]}(\Res{1})+\bigO{p^{n}}.

		\end{align*}	

\end{comment}			

~

Questions:

\begin{itemize}

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

\end{itemize} 

\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*}