From 4150953303213882b72e356be0b3e04495807f9b 2017-05-23 12:00:42 From: Tom Bannink Date: 2017-05-23 12:00:42 Subject: [PATCH] Update on Overleaf. --- diff --git a/Resample.bib b/Resample.bib new file mode 100644 index 0000000000000000000000000000000000000000..7dc6583b0f2fa3737a1ae13b85a34dab2fc0b9f4 --- /dev/null +++ b/Resample.bib @@ -0,0 +1,37 @@ +@inproceedings{ResampleLimit, + author = {Jan Dean Catarata and + Scott Corbett and + Harry Stern and + Mario Szegedy and + Tom{\'{a}}s Vyskocil and + Zheng Zhang}, + title = {The Moser-Tardos Resample algorithm: Where is the limit? (an experimental + inquiry)}, + booktitle = {Proceedings of the Ninteenth Workshop on Algorithm Engineering and + Experiments, {ALENEX} 2017, Barcelona, Spain, Hotel Porta Fira, January + 17-18, 2017.}, + pages = {159--171}, + year = {2017}, + crossref = {DBLP:conf/alenex/2017}, + url = {http://dx.doi.org/10.1137/1.9781611974768.13}, + doi = {10.1137/1.9781611974768.13}, + timestamp = {Fri, 27 Jan 2017 10:13:46 +0100}, + biburl = {http://dblp.uni-trier.de/rec/bib/conf/alenex/CatarataCSSVZ17}, + bibsource = {dblp computer science bibliography, http://dblp.org} +} + +@proceedings{DBLP:conf/alenex/2017, + editor = {S{\'{a}}ndor P. Fekete and + Vijaya Ramachandran}, + title = {Proceedings of the Ninteenth Workshop on Algorithm Engineering and + Experiments, {ALENEX} 2017, Barcelona, Spain, Hotel Porta Fira, January + 17-18, 2017}, + publisher = {{SIAM}}, + year = {2017}, + url = {http://dx.doi.org/10.1137/1.9781611974768}, + doi = {10.1137/1.9781611974768}, + isbn = {978-1-61197-476-8}, + timestamp = {Fri, 27 Jan 2017 10:08:19 +0100}, + biburl = {http://dblp.uni-trier.de/rec/bib/conf/alenex/2017}, + bibsource = {dblp computer science bibliography, http://dblp.org} +} diff --git a/coeffs_conv_radius.pdf b/coeffs_conv_radius.pdf new file mode 100644 index 0000000000000000000000000000000000000000..2e72b98b0946ccc71ee5d471a4cce542bdf5314f GIT binary patch literal 4867 zcma)A2|SeR_g5l~<+fa7jkjb;#_VIsRYSIGA6xdu%osB>t(=2~= zWe$b~PyjoC2{SPPkajc{ofib)K!^kb003#p2uU@NxRtjc0BCjc+E4ybT)@y$wl6o0|uYR^D@%9a4#qnVm_` ziM0I^6*V_TuTZV)4=}9VOkLskQnvGBVrOee5~`&ru2YzOKrQfGs>bLCyRsdJ-^~ox 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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 $p_c\approx0.62$, 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*} + + 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 $\rho*(M_{(n)}^1)*\mathbbm{1}, \ldots, \rho*(M_{(n)}^k)*\mathbbm{1}$. + +\section{Quasiprobability method} +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| \quad,\quad\text{where }|\xi|\text{ is the length of the path }\xi. +\end{align*} +Here $\paths{b}$ denotes the set of all possible ways (paths) to arrive in the all-one state (denoted by $\mathbf{1}$) and we take the expected length of these paths. Note that if a path samples a $0$ then $\mathbb{P}[\xi]$ gains a factor $p$.\\ + +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 $p^k$.\\ + +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 $p$. 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] + 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} + \caption{\label{fig:diametergap} A configuration $C=\{1,2,4,7,9\}\subseteq[n]$ consisting of 5 slots shown by the red dots. The dotted line at the top depicts the rest of the circle which may be much larger. The diameter of this configuration is $\diam{C}=9$ as shown and the largest gap of $C$ is $\mathrm{gap}(C)=2$. Note that we do not count the rest of the circle as a gap, we only consider gaps within the diameter of $C$.} +\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} + +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*} + \sum_{f\in\{0,1'\}^{|C_0|}} \rho_{C_0(f)} R_{C_0(f)} &= 0.0240278 p^{9} + 0.235129 p^{10} + 1.24067 p^{11} + 4.71825 p^{12} \\ + &\quad + 14.5555 p^{13} + 38.8307 p^{14} + 93.2179 p^{15} + 206.837 p^{16}\\ + &\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 $\diam{C}=9$. + +~ + +A weaker version of the claim is that if $C$ contains a gap of size $k$, then the sum is zero up to and including order $p^{k-1}$. +\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*} + is at least $p^{\mathrm{gap}(C)}$ when $n$ is large enough. Here $\mathrm{gap}(C)$ is defined as in Figure \ref{fig:diametergap}, its the size of the largest gap of $C$ within the diameter of $C$. All lower order terms cancel out. +\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 $n$. + +~ + +Section \ref{sec:computerb} shows how to compute $R_b$ (this is not relevant for showing the claim) and the section after that shows how to prove the weaker claim. + +\newpage +\subsection{Computation of $R_b$} \label{sec:computerb} + +By $R_{101}$ we denote $R_b(p)$ for a $b$ that consists of only $1$s except for a single zero. We compute $R_{101}$ up to second order in $p$. This requires the following transitions. +\begin{align*} + \framebox{$1 0 1$} &\to \framebox{$1 1 1$} & (1-p)^3 = 1-3p+3p^2-p^3\\ + \hline + \framebox{$1 0 1$} &\to + \begin{cases} + \framebox{$0 1 1$}\\ + \framebox{$1 0 1$}\\ + \framebox{$1 1 0$} + \end{cases} + & 3p(1-p)^2 = 3p-6p^2+3p^3\\ + \hline + \framebox{$1 0 1$} &\to \framebox{$0 1 0$} & p^2(1-p) = p^2-p^3\\ + \framebox{$1 0 1$} &\to + \begin{cases} + \framebox{$1 0 0$}\\ + \framebox{$0 0 1$} + \end{cases} + & 2p^2(1-p) = 2p^2 - 2p^3\\ + \hline + \framebox{$1 0 1$} &\to \framebox{$0 0 0$} & p^3 +\end{align*} +With this we can write a recursive formula for the expected number of resamples from $101$: +\begin{align*} + R_{101} &= (1-3p+3p^2 - p^3)(1) + (3p -6p^2 +3p^3) (1+R_{101}) \\ + &\quad + (p^2 - p^3) (1+R_{10101}) + (2p^2-2p^3) (1+R_{1001}) \\ + &= 1 + 3 p + 7 p^2 + 14.6667 p^3 + 29 p^4 + 55.2222 p^5 + 102.444 p^6 + 186.36 p^7 \\ + &\quad + 333.906 p^8 + 590.997 p^9 + 1035.58 p^{10} + 1799.39 p^{11} + 3104.2 p^{12} \\ + &\quad+ 5322.18 p^{13} + 9075.83 p^{14} + 15403.6 p^{15} + 26033.4 p^{16} + 43833.5 p^{17} \\ + &\quad+ 73555.2 p^{18} + 123053 p^{19} + 205290 p^{20} + 341620 p^{21} + 567161 p^{22} \\ + &\quad+ 939693 p^{23} + 1.5537\cdot10^{6} p^{24} + 2.56158\cdot10^{6} p^{25} + \mathcal{O}(p^{26}) +\end{align*} +where the recursion steps were done with a computer. This assumes $n$ to be much larger than the largest power of $p$ considered. + +Note: in the first line at the second term it uses that with probability $(3p-6p^2)$ the state goes to $\framebox{$101$}$ and then the expected number of resamplings is $1+R_{101}$. I (Tom) believe this requires the assumption $p_\mathrm{tot} := \sum_{\xi\in\paths{b}} \mathbb{P}[\xi] = 1$. To see why this is required, note that the actual term in the recursive formula should be $$(3p-6p^2)\cdot\left( \sum_{\xi\in\paths{101}} \mathbb{P}[\xi] \cdot \left( 1 + |\xi|\right) \right) = (3p-6p^2)\left( p_\mathrm{tot} + R_{101} \right)$$ +When there would be a non-zero probability of never stopping the resample process then $p_\mathrm{tot}$ (the probability of ever reaching $\mathbf{1}$) could be less than one. Therefore I assume that $R^{(n)}(p)$ is finite which implies that the probability of ever reaching $\mathbf{1}$ is 1. + +\newpage +\subsection{Cancellation of gapped configurations} + +Here we prove claim \ref{claim:weakcancel}, the weaker version of the claim. We require the following definition +\begin{definition}[Path independence] \label{def:independence} + 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} + +For example for $b_1 = 10111111$ and $b_2 = 11111000$ we have $b=10111000$ and $k=3$. The claim says that the expected time to reach $\mathbf{1}$ from $b$ is the time to make the first group $1$ plus the time to make the second group $1$, as if they are independent. When going up to order $p^{k}$ or higher, there will be terms where the groups interfere so they are no longer independent. + +~ + +\begin{proof} +Consider a path $\xi_1\in\paths{b_1}$ and a path $\xi_2\in\paths{b_2}$ such that $\xi_1$ and $\xi_2$ are independent (Definition \ref{def:independence}). The paths $\xi_1,\xi_2$ induce $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ different paths of total length $|\xi_1|+|\xi_2|$ in $\paths{b_1\land b_2}$. In the sums $R_{b_1}$ and $R_{b_2}$, the contribution of these paths are $\mathbb{P}[\xi_1]\cdot |\xi_1|$ and $\mathbb{P}[\xi_2]\cdot |\xi_2|$. The next diagram shows how these $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths contribute to $R_{b_1\land b_2}$. At every step one has to choose between doing a step of path 1 or a step of path 2. The number of zeroes in the current state determine probabilities with which this happens (aside from the probabilities associated to the two original paths already). The grid below shows that at every point one can choose to do a step of path 1 with probability $p_i$ or a step of path 2 with probability $1-p_i$. These $p_i$ could in principle be different at every point in this grid. The weight of such a new path is the weight of the path in the diagram below, multiplied by $\mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]$. By induction one can show that the sum over all $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths in the grid is $1$. Hence the contribution of all $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths together to $R_{b_1\land b_2}$ is given by +\[ +\mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]\cdot(|\xi_1|+|\xi_2|) = \mathbb{P}[\xi_2]\cdot\mathbb{P}[\xi_1]\cdot|\xi_1| \;\; + \;\; \mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]\cdot|\xi_2|. +\] +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 +\[ + R_{b_1\land b_2} = \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| + \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_1]\mathbb{P}[\xi_2]|\xi_2| + \sum_{\mathclap{\xi\;\mathrm{dependent}}} \mathbb{P}[\xi]|\xi|. +\] +The last sum only contains only terms of order $p^{k}$ or higher. Now for the first sum, note that +\[ + \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| + = \sum_{\xi_1\in\paths{b_1}} \sum_{\substack{\xi_2\in\paths{b_2}\\ \text{independent of }\xi_1}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| +\] +where the sum over independent paths could be empty. +\end{proof} + +\begin{center} +\includegraphics{diagram_paths.pdf} +\end{center} + +\textbf{Proof of claim \ref{claim:weakcancel}}: Say we have a group on the left with $l$ slots and a group on the right with $r$ slots, with enough space between the groups. Then on the left we have strings in $\{0,1'\}^l$ as possibilities and on the right we have strings in $\{0,1'\}^r$. The combined configuration can be described by strings $(a,b)\in\{0,1'\}^{l+r}$. Such a configuration has probability $(-1)^{|a|+|b|} p^{r+l}$ in $\rho$ and by claim \ref{claim:expectationsum} we know $R_{(a,b)} = R_a + R_b + \mathcal{O}(p^\mathrm{spacing})$. The total contribution of these configurations is therefore +\begin{align*} + \sum_{a\in\{0,1'\}^l} \sum_{b\in\{0,1'\}^r} (-1)^{|a|+|b|}p^{r+l} \left( R_a + R_b \right) + \mathcal{O}(p^\mathrm{spacing}) + &= p^{r+l}\sum_{a\in\{0,1'\}^l} (-1)^{|a|} R_a \sum_{b\in\{0,1'\}^r} (-1)^{|b|} \\ + &\quad + p^{r+l}\sum_{b\in\{0,1'\}^r} (-1)^{|b|} R_b \sum_{a\in\{0,1'\}^l} (-1)^{|a|} \\ + &\quad + \mathcal{O}(p^\mathrm{spacing})\\ + &= 0 + \mathcal{O}(p^\mathrm{spacing}) +\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} + % Let $C\subseteq[n]$ denote a slot configuration, and let $a\in\{\text{res},\neg\text{res}\}^{n-|C|}$, where the elements correspond to labels ``resampled" vs. ``not resampled" respectively. + % For $j\in[n-|C|]$ let $i_j$ denote the $j$-th index in $[n]\setminus C$. + % We define the set $A^{(C,a)}\subseteq RS$ as the set of resample sequences $(r_\ell)$ such that for all $j$ which has $a_j=\text{res}$ we have that $i_j$ appears in $(r_\ell)$ but for $j'$-s which have $a_{j'}=\neg\text{res}$ we have that $i_{j'}$ never appears in $(r_\ell)$. + %\end{definition} + \begin{definition}[Conditional expected number of resamples] + For a slot configuration $C\subseteq[n]$ and $a\in\{\!\text{ever},\text{ never}\}^{n-|C|}$ we define the event $A^{(C,a)}:=\bigwedge_{j\in[n-|C|]}\{i_j\text{ has }a_j\text{ become }0\text{ before reaching }\mathbf{1}\}$, + where $i_j$ is the $j$-th vertex of $[n]\setminus C$. + Then we also define + $$R^{(C,a)}_b:=\mathbb{E}[\#\{\text{resamplings when started from inital state }b\}|A^{(C,a)}].$$ + \end{definition} + + As in Mario's proof I use the observation that + \begin{align*} + R^{(n)}(p) &= \frac{1}{n}\sum_{b\in\{0,1,1'\}^{n}} \rho_b \; R_{\bar{b}}(p)\\ + &= \frac{1}{n}\sum_{C\subseteq [n]}\sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R_{C(f)}(p)\\ + &= \frac{1}{n}\sum_{C\subseteq [n]}\sum_{f\in\{0,1'\}^{|C|}}\sum_{a\in\{\!\text{ever},\text{ never}\}^{n-|C|}} \rho_{C(f)} R^{{(C,a)}}_{C(f)}(p)P_{C(f)}(A^{(C,a)})\\ + &= \frac{1}{n}\sum_{C\subseteq [n]}\sum_{a\in\{\!\text{ever},\text{ never}\}^{n-|C|}} \sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R^{{(C,a)}}_{C(f)}(p)P_{C(f)}(A^{(C,a)}), + \end{align*} + where we denote by $C\subseteq[n]$ a slot configuration, whereas $C(f)$ denotes the slots of $C$ filled with the particles described by $f$, while all other location in $[n]\setminus C$ are set to $1$. + When we write $R_{C(f)}$ we mean $R_{C(\bar{f})}$, i.e., replace $1'$-s with $1$-s. Since the notation is already heavy we dropped the bar from $f$, as it is clear from the context. Finally by $P_{C(f)}(A^{(C,a)})$ we denote the probability that the event $A^{(C,a)}$ holds. + + As in Definition for $j\in[n-|C|]$ let $i_j$ denote the $j$-th index in $[n]\setminus C$. + Suppose that $a$ is such that there are two indices $j_1\neq j_2$ such that + $a_{j_1}=\text{never}=a_{j_2}$, moreover the sets $\{i_{j_1}+1,\ldots, i_{j_2}-1\}$ and $\{i_{j_2}+1,\ldots, i_{j_1}-1\}$ partition $C$ non-trivially, and we denote by $C_l$,$C_r$ the corresponding partitions. + I wanted to prove that + \begin{equation}\label{eq:conditionalCancellation} + \sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R^{{(C,a)}}_{C(f)}(p)=0, + \end{equation} + based on the observation that for all $f\in\{0,1'\}^{|C|}$ we have + that + \begin{equation}\label{eq:keyIndependce} + R^{{(C,a)}}_{C(f)}(p)=R^{{(C_l,a_l)}}_{C_l(f_l)}(p)+R^{{(C_r,a_r)}}_{C_r(f_r)}(p), + \end{equation} + where $f_l\in\{0,1'\}^{|C_l|}$ is defined as taking only the indices (and values) of $f$ corresponding to vertices of $C_l$, also $a_l\in[n-|C_l|]$ is defined such that $a$ and $a_l$ agree on vertices where $a$ is defined, and on the vertices where $a$ is not defined, i.e., the vertices of $C_r$ we define $a_l$ to contain ``never". We define things analogously for $f_r$ and $a_r$. + + The reason why \eqref{eq:keyIndependce} holds is that as before the two halves of the cycle are conditionally independent because neither $i_{j_1}$ nor $i_{j_2}$ can become $0$. To be more precise each resample sequence $\left(C(f)\rightarrow \mathbf{1} \right)\in A^{(C,a)}$ can be uniquely decomposed to resample sequences $\left(C_l(f_l)\rightarrow \mathbf{1}\right)\in A^{(C_l,a_l)}$ and $\left(C_r(f_r)\rightarrow \mathbf{1}\right)\in A^{(C_r,a_r)}$. The sum of probabilities of the set of resample sequences $\{r\}$ which have decomposition $(r_l,r_r)$ have probability which is the product of the probabilities of $r_l$ and $r_r$ as shown in the proof of Claim~\ref{claim:expectationsum}. This proves that the set of all resample sequences $\left(C(f)\rightarrow \mathbf{1}\right)\in A^{(C,a)}$ for our purposes can be viewed as a product set with product probability distribution. Therefore the halves can be treated independently and so the expectation values just add up. + + From here I wanted to mimic Mario's proof: + \begin{align*} + \sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R^{{(C,a)}}_{C(f)}(p)&= + \sum_{f_l\in\{0,1'\}^{|C_l|}} \sum_{f_r\in\{0,1'\}^{|C_r|}} (-1)^{|f_l|+|f_r|}p^{|C_l|+|C_r|} \left( R^{{(C_l,a_l)}}_{C_l(f_l)}(p) + R^{{(C_r,a_r)}}_{C_r(f_l)}(p) \right)\\ + &= p^{|C|}\sum_{f_l\in\{0,1'\}^{|C_l|}} (-1)^{|f_l|} R^{{(C_l,a_l)}}_{C_l(f_l)}(p) \sum_{f_r\in\{0,1'\}^{|C_r|}} (-1)^{|f_r|} \\ + &\quad + p^{|C|}\sum_{f_r\in\{0,1'\}^{|C_r|}} (-1)^{|f_r|} R^{{(C_r,a_r)}}_{C_r(f_r)}(p) \sum_{f_l\in\{0,1'\}^{|C_l|}} (-1)^{|f_l|} \\ + &= 0. + \end{align*} + The nasty issue which I did not realise that the missing term $P_{C(f)}(A^{(C,a)})$ is non-constant: even though the event $A^{(C,a)}$ is independent of $f$ the probability $P_{C(f)}(A^{(C,a)})=P_{C(f_l)}(A^{(C_l,a_l)})\cdot P_{C(f_r)}(A^{(C_r,a_r)})$ is not and so the above breaks down. + + Observe that if \eqref{eq:conditionalCancellation} would hold for configurations that cut the slot configuration to two halves it would imply that the only non-zero contribution comes from pairs $(C,a)$ such that $C\cup\{i_j:a_j=\text{ever}\}$ is connected. This is because if this set is not connected, then either we can cut $C$ to two halves non-trivially along ``never" vertices, or there is an island of $\text{ever}$ vertices separated from any slots, and therefore from any $0$-s. This latter case has zero contribution since we cannot set these indices to $0$, without reaching them by some resamplings, and thereby building a path of $0$-s leading there. + + If $|C\cup\{i_j:a_j=\text{ever}\}|\geq k+1$ then all contribution has a power at least $k+1$ in $p$ since $(C,a)$ requires the prior appearance of at least $k+1$ particles. If $n\geq k+1$ than all $(C,a)$ such that $|C\cup\{i_j:a_j=\text{ever}\}|\leq k$ appears exactly $n$ times, since $(C,a)$ cannot be translationally invariant. Moreover the quantity $R^{{(C,a)}}_{C(f)}(p)$ is independent of $n$ due to the conditioning that every resampling happens on a connected component of length at most $k