Changeset - ec61b8b55280
[Not reviewed]
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Tom Bannink - 8 years ago 2017-09-08 14:32:50
tom.bannink@cwi.nl
Add event defs
1 file changed with 6 insertions and 6 deletions:
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main.tex
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@@ -40,48 +40,49 @@
 
\newcommand{\pvp}{\vec{p}{\kern 0.45mm}'}
 

	
 
\DeclarePairedDelimiter\bra{\langle}{\rvert}
 
\DeclarePairedDelimiter\ket{\lvert}{\rangle}
 
\DeclarePairedDelimiterX\braket[2]{\langle}{\rangle}{#1 \delimsize\vert #2}
 
\newcommand{\underflow}[2]{\underset{\kern-60mm \overbrace{#1} \kern-60mm}{#2}}
 

	
 
\def\Ind(#1){{{\tt Ind}({#1})}}
 
\def\Id{\mathrm{Id}}
 
\def\Pr{\mathrm{Pr}}
 
\def\Tr{\mathrm{Tr}}
 
\def\im{\mathrm{im}}
 
\newcommand{\bOt}[1]{\widetilde{\mathcal O}\left(#1\right)}
 
\newcommand{\bigO}[1]{\mathcal O\left(#1\right)}
 
\newcommand{\Res}[1]{\#\mathrm{Res}\left(#1\right)}
 

	
 
\newcommand{\QMAo}{\textsf{QMA$_1$}}
 
\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)}
 
\newcommand{\start}[1]{\textsc{start}\left(#1\right)}
 
\newcommand{\initone}[1]{\textsc{InitOne}\left(#1\right)}
 
\newcommand{\maxgap}[1]{\mathrm{maxgap}\left(#1\right)}
 
\newcommand{\gaps}[1]{#1_{\mathrm{gaps}}}
 
\renewcommand{\P}{\mathbb{P}}
 
\newcommand{\E}{\mathbb{E}}
 
\newcommand{\NZ}[1]{\mathrm{NZ}^{(#1)}}
 
\newcommand{\Z}[1]{\mathrm{Z}^{(#1)}}
 
%\newcommand{\dist}[1]{d_{\!\!\not\,#1}}
 
\newcommand{\dist}[1]{d_{\neg #1}}
 

	
 
\newcommand{\todo}[1]{{\color{red}\textbf{TODO:} #1}}
 

	
 
\long\def\ignore#1{}
 

	
 
\newtheorem{theorem}{Theorem}
 
\newtheorem{corollary}[theorem]{Corollary}%[theorem]
 
\newtheorem{lemma}[theorem]{Lemma}
 
\newtheorem{prop}[theorem]{Proposition}
 
\newtheorem{definition}[theorem]{Definition}
 
\newtheorem{claim}[theorem]{Claim}
 
\newtheorem{remark}[theorem]{Remark}
 

	
 
\newenvironment{proof}
 
{\noindent {\bf Proof. }}
 
{{\hfill $\Box$}\\	\smallskip}
 
@@ -572,56 +573,55 @@ The intuition of the following lemma is that the far right can only affect the z
 
		&= \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:eventindependenceNew}}\\
 
		&= \E^{[-N,N]}(\Res{1})+\bigO{p^{n}}.
 
		\end{align*}	
 
\end{comment}			
 

	
 
~
 

	
 
Questions:
 
\begin{itemize}
 
	\item Can we generalise the proof to other translationally invariant spaces, like the torus?
 
	\item In view of this proof, can we better characterise $a_k^{(k+1)}$?
 
	\item Why did Mario's and Tom's simulation show that for fixed $C$ the contribution coefficients have constant sign? Is it relevant for proving \ref{it:pos}-\ref{it:geq}?
 
\end{itemize} 
 

	
 
	%I think the same arguments would translate to the torus and other translationally invariant spaces, so we could go higher dimensional as Mario suggested. Then I think one would need to replace $|S_{><}|$ by the minimal number $k$ such that there is a $C$ set for which $S\cup C$ is connected. I am not entirely sure how to generalise Lemma~\ref{lemma:probIndep} though, which has key importance in the present proof.
 

	
 
\newpage
 
\section{General graphs proof}
 

	
 
We consider the following generalization of the Markov Chain.
 

	
 
Let $G=(V,E)$ be a graph with vertex set $V$ and edge set $E$. We define a Markov Chain $\mathcal{M}_G$ as the following process: initialize every vertex of $G$ independently to 0 with probability $p$ and 1 with probability $1-p$. Then at each step, select a uniformly random vertex that has value $0$ and resample it and its neighbourhood, all of them independently with the same probability $p$. The Markov Chain terminates when all vertices have value $1$. We use $\P^{G}$ to denote probabilities associated to this Markov Chain.
 

	
 
\begin{definition}[Events] \label{def:events}
 
    For any subset $S\subseteq V$ of vertices.
 
    For any state $b\in\{0,1\}^n$, define $\start{b}$ as the event that the starting state of the chain is the state $b$. For any event $A$ and any $v\in[n]$, define
 
    Let $S\subseteq V$ be any subset of vertices.
 
    Define $\Z{S}$ as the event that \emph{all} vertices in $S$ become zero at any point in time before the Markov Chain terminates.
 
    Define $\NZ{S}$ as the event that \emph{none} of the vertices in $S$ become zero at any point in time before the Markov Chain terminates.
 
    Define $\initone{S}$ as the event that all vertices in $S$ \emph{initially} get assigned the value 1, and define for any event $A$:
 
    \begin{align*}
 
        \P^{(n)}_b(A) &= \P^{(n)}(A \;|\; \start{b}) \\
 
        \P^{[n]}_{b_v=1}(A) &= \P^{[n]}(A \;|\; v\text{ is initialized to }1) \\
 
        \P^{[n]}_{b_v=b_w=1}(A) &= \P^{[n]}(A \;|\; v\text{ and }w\text{ are initialized to }1) ,
 
        \P^{G}_S(A) &= \P^{G}(A \;\mid\; \initone{S})
 
    \end{align*}
 
    The last two probabilities are not conditioned on any other bits of the starting state.
 
\end{definition}
 

	
 

	
 
$\NZ{S}$
 
$\Z{S}$
 

	
 
patch
 

	
 
$B(S;d)$
 

	
 

	
 

	
 
We consider $R^{(n)}(p)$ as a power series in $p$ and our main aim in this section is to show that $R^{(n)}(p)$ and $R^{(n+k)}(p)$ are the same up to order $n-1$.
 

	
 

	
 
%Note that we have $\P^{(n)}(\start{b}) = (1-p)^{|b|}p^{n-|b|}$ by definition of our Markov Chain.
 
\begin{definition}[Vertex visiting event] \label{def:visitingResamplings}
 
    Denote by $\mathrm{Z}^{(v)}$ the event that site $v$ becomes zero at any point in time before the Markov Chain terminates. Denote the complement by $\mathrm{NZ}^{(v)}$, i.e. the event that site $v$ does \emph{not} become zero before it terminates. Furthermore define $\mathrm{NZ}^{(v,w)} := \mathrm{NZ}^{(v)} \cap \mathrm{NZ}^{(w)}$, i.e. the event that \emph{both} $v$ and $w$ do not become zero before termination.
 
\end{definition}
 
%\begin{figure}
 
%	\begin{center}
 
%    	\includegraphics{diagram_groups.pdf}
 
%    \end{center}
 
%    \caption{\label{fig:separatedgroups} Illustration of setup of Lemma \ref{lemma:eventindependence}. Here $b_1,b_2\in\{0,1\}^n$ are bitstrings such that all zeroes of $b_1$ and all zeroes of $b_2$ are separated by two indices $v,w$.}
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