}

        \draw [->] (\x,6) -- (\x+0.9,6);

    }

    \foreach \y in {0,...,5}

        \draw [->] (8,\y) -- (8,\y+0.9);

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

    \draw[fill,red] (8,6) circle (0.08);

    \def\x{5};

    \def\y{3};

    \draw[fill,black] (\x,\y) circle (0.07);

    \draw[fill=white,draw=black] (\x+0.25,\y-0.26) rectangle +(0.5,0.5);

\newcommand{\BQP}{\textsf{BQP}}

\newcommand{\NP}{\textsf{NP}}

\newcommand{\SharpP}{\textsf{\# P}}

\newcommand{\diam}[1]{\mathcal{D}\left(#1\right)}

\newcommand{\paths}[1]{\mathcal{P}\left(#1\to\mathbf{1}\right)}

\long\def\ignore#1{}

\newtheorem{theorem}{Theorem}

\newtheorem{corollary}[theorem]{Corollary}%[theorem]

\newtheorem{lemma}[theorem]{Lemma}

			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. \\

	}

	\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:

		\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}

	\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}

    &\quad + 432.302 p^{17} + 862.926 p^{18} + 1662.05 p^{19} + 3112.9 p^{20} + \mathcal{O}(p^{21})

\end{align*}

and indeed the lowest order is $\diam{C}=9$.

~

\begin{claim}[Weak cancellation claim] \label{claim:weakcancel}

	For $C\subseteq[n]$ a configuration of slot positions, the lowest order term in

    \begin{align*}

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

    \end{align*}

\end{claim}

This weaker version would imply \ref{it:const} but for $\mathcal{O}(k^2)$ as opposed to $k+1$.

\newpage

The reason that claim \ref{claim:strongcancel} would prove \ref{it:const} is the following:

For a starting configuration that \emph{does} give a nonzero contribution, you can take that same starting configuration and translate it to get $n$ other configurations that give the same contribution. Therefore the coefficient in the expected number of resamplings is a multiple of $n$ which Andr\'as already divided out in the definition of $R^{(n)}(p)$. To show \ref{it:const} we argue that this is the \emph{only} dependency on $n$. This is because there are only finitely many (depending on $k$ but not on $n$) configurations where the $k$ slots are nearby regardless of the value of $n$. So there are only finitely many nonzero contributions after translation symmetry was taken out. For example, when considering all starting configurations with 5 slots one might think there are $\binom{n}{5}$ configurations to consider which would be a dependency on $n$ (more than only the translation symmetry). But since most of these configurations have a diameter larger than $k$, they do not contribute to $a_k$. Only finitely many do and that does not depend on $n$.

	We say two paths $\xi_i\in\paths{b_i}$ ($i=1,2$) of the Markov Chain are \emph{independent} if $\xi_1$ never resamples a site that was ever zero in $\xi_2$ and the other way around. It is allowed that $\xi_1$ resamples a $1$ to a $1$ that was also resampled from $1$ to $1$ by $\xi_2$ and vice versa. If the paths are not independent then we call the paths \emph{dependent}.

\end{definition}

The key ingredient of the proof is the following claim:

\begin{claim}[Sum of expectation values] \label{claim:expectationsum}

When $b=b_1\land b_2\in\{0,1\}^n$ is a state with two groups ($b_1\lor b_2 = 1^n$) of zeroes with $k$ $1$s inbetween the groups, then we have $R_b(p) = R_{b_1}(p) + R_{b_2}(p) + \mathcal{O}(p^{k})$ where $b_1$ and $b_2$ are the configurations where only one of the groups is present and the other group has been replaced by $1$s. To be precise, the sums agree up to and including order $p^{k-1}$.

\end{claim}

~

\begin{proof}

\[

\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

\[

    &= \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}

\begin{align*}

\end{align*}

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

    \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 $RS=\cup_{k\in\mathbb{N}}RS(k)$.

    %\end{definition}

    %\begin{definition}[Constrained resample sequence]\label{def:constrainedRes}