\draw[gray,dotted] (0,2) -- (10,2);

    \draw[gray] (0,2) arc (90:270:1);

    \draw[gray] (10,0) arc (-90:90:1);

    \foreach \x in {0,...,10} {

        \draw[fill] (\x,0) circle (0.04);

    }

    \draw[fill] (11,1) circle (0.04);

    \draw[fill] (-1,1) circle (0.04);

    \foreach \x in {1,2,4,7,9} {

        \draw[fill,red] (\x,0) circle (0.08);

    }

    \draw[<->] (1,-0.5) -- (9,-0.5);

    \draw (5,-0.9) node {$\mathcal{D}(C)$};

    \draw[<->] (4.9,0.3) -- (6.1,0.3);

    \draw (5.5,0.7) node {$\mathrm{gap}(C)$};

    \draw[fill,red] (1,-2) circle (0.08);

    \draw (2,-2) node {slots $C$};

\end{tikzpicture}

\end{document}

	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]

	For a slot configuration $C\subseteq[n]$, we define the diameter $\diam{C}$ to be the minimum size of an interval containing $C$ where the interval is also considered modulo $n$. In other words $\diam{C} = n - \max\{ j \vert \exists i : [i,i+j-1]\cap C = \emptyset \}$. Figure \ref{fig:diametergap} shows the diameter in a picture.

\end{definition}

\begin{figure}

	\begin{center}

    	\includegraphics{diagram_gap.pdf}

    \end{center}

\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}

Example: for $C_0=\{1,2,4,7,9\}$ (the configuration shown in Figure \ref{fig:diametergap}) we computed the quantity up to order $p^{20}$ in an infinite system:

\begin{align*}

\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 $I_{\min}:=\min(I)$.

	Then $P_{I}(V^{(0)})=P_{I'}(V^{(0)}) + O(p^{I_{\max}+1-|I|})$.

\end{lemma}

\begin{proof}

	The proof uses induction on $|I|$. For $|I|=1$ the statement is easy, since every resample sequence that resamples the $0$ vertex must produce at least $I_{\max}$ number of $0$-s during the resamplings.

	Induction step: For an event $A$ and $k>0$ let us denote by $A_k=A\cap$``Each vertex in $0,1,2,\ldots, k-1$ gets $0$ before termination (either by resampling or initialisation), but not $k$''. Observe that $V^{(0)}=\dot{\bigcup}_{k=1}^{\infty}V^{(0)}_k$.

	\begin{align*}

		P_{I}(V^{(0)})

		&=\sum_{k=1}^{\infty}P(V^{(0)}_k)

		=\sum_{k\in \mathbb{N}\setminus I}P(V^{(0)}_k)\\

		&=\sum_{k\in\mathbb{N}\setminus I}P_{I_{<k}}(V^{(0)}_k)\cdot P_{I_{>k}}(\overline{V^{(k)}}) \tag{by Claim~\ref{claim:eventindependence}}\\

		&=\sum_{k\in I_{><}}P_{I_{<k}}(V^{(0)}_k)\cdot P_{I_{>k}}(\overline{V^{(k)}})+\mathcal{O}(p^{I_{\max}+1-|I|})

		\tag{$k<I_{\min}\Rightarrow P_{I_{<k}}(V^{(0)}_k)=0$}\\

		&=\sum_{k\in I_{><}}P_{I'_{<k}}(V^{(0)}_k)\cdot P_{I_{>k}}(\overline{V^{(k)}})+\mathcal{O}(p^{I_{\max}+1-|I|})	

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

		&=\sum_{k\in I_{><}}P_{I'_{<k}}(V^{(0)}_k)\cdot

		\left(P_{I'_{>k}}(\overline{V^{(k)}})+\mathcal{O}(p^{I_{\max}-k+1-|I_{>k}|})\right) +\mathcal{O}(p^{I_{\max}+1-|I|})	\tag{by induction, since for $k>I_{\min}$ we have $|I_{<k}|<|I|$}\\

		&=\sum_{k\in I_{><}}P_{I'_{<k}}(V^{(0)}_k)\cdot