X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/16dcc.git/blobdiff_plain/66ea391a3e386a6ed3b47d5011977ca136f65ad2..e1fe6e435ee452003a7135763d26e2320756398c:/stopping.tex?ds=sidebyside diff --git a/stopping.tex b/stopping.tex index d72f8bb..409dd83 100644 --- a/stopping.tex +++ b/stopping.tex @@ -1,11 +1,6 @@ - - - -Let thus be given such kind of map. -This article focuses on studying its iterations according to -the equation~(\ref{eq:asyn}) with a given strategy. -First of all, this can be interpreted as walking into its iteration graph -where the choice of the edge to follow is decided by the strategy. +This section considers functions $f: \Bool^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}} $ +issued from an hypercube where an Hamiltonian path has been removed +as described in previous section. Notice that the iteration graph is always a subgraph of ${\mathsf{N}}$-cube augmented with all the self-loop, \textit{i.e.}, all the edges $(v,v)$ for any $v \in \Bool^{\mathsf{N}}$. @@ -43,50 +38,19 @@ P=\dfrac{1}{6} \left( \end{xpl} -% % Let us first recall the \emph{Total Variation} distance $\tv{\pi-\mu}$, -% % which is defined for two distributions $\pi$ and $\mu$ on the same set -% % $\Bool^n$ by: -% % $$\tv{\pi-\mu}=\max_{A\subset \Bool^n} |\pi(A)-\mu(A)|.$$ -% % It is known that -% % $$\tv{\pi-\mu}=\frac{1}{2}\sum_{x\in\Bool^n}|\pi(x)-\mu(x)|.$$ - -% % Let then $M(x,\cdot)$ be the -% % distribution induced by the $x$-th row of $M$. If the Markov chain -% % induced by -% % $M$ has a stationary distribution $\pi$, then we define -% % $$d(t)=\max_{x\in\Bool^n}\tv{M^t(x,\cdot)-\pi}.$$ -% Intuitively $d(t)$ is the largest deviation between -% the distribution $\pi$ and $M^t(x,\cdot)$, which -% is the result of iterating $t$ times the function. -% Finally, let $\varepsilon$ be a positive number, the \emph{mixing time} -% with respect to $\varepsilon$ is given by -% $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$ -% It defines the smallest iteration number -% that is sufficient to obtain a deviation lesser than $\varepsilon$. -% Notice that the upper and lower bounds of mixing times cannot -% directly be computed with eigenvalues formulae as expressed -% in~\cite[Chap. 12]{LevinPeresWilmer2006}. The authors of this latter work -% only consider reversible Markov matrices whereas we do no restrict our -% matrices to such a form. - - - - - - - -This section considers functions $f: \Bool^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}} $ -issued from an hypercube where an Hamiltonian path has been removed. A specific random walk in this modified hypercube is first -introduced. We further detail -a theoretical study on the length of the path -which is sufficient to follow to get a uniform distribution. +introduced (See section~\ref{sub:stop:formal}). We further +theoretical study this random walk to +provide a upper bound of fair sequences +(See section~\ref{sub:stop:bound}). +We finally complete these study with experimental +results that reduce this bound (Sec.~\ref{sub:stop:stop}). Notice that for a general references on Markov chains -see~\cite{LevinPeresWilmer2006} -, and particularly Chapter~5 on stopping times. - +see~\cite{LevinPeresWilmer2006}, +and particularly Chapter~5 on stopping times. +\subsection{Formalizing the Random Walk}\label{sub:stop:formal} First of all, let $\pi$, $\mu$ be two distributions on $\Bool^{\mathsf{N}}$. The total variation distance between $\pi$ and $\mu$ is denoted $\tv{\pi-\mu}$ and is @@ -104,6 +68,13 @@ $$d(t)=\max_{X\in\Bool^{\mathsf{N}}}\tv{P^t(X,\cdot)-\pi}.$$ and $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$ + +Intuitively speaking, $t_{\rm mix}$ is a mixing time +\textit{i.e.}, is the time until the matrix $X$ of a Markov chain +is $\epsilon$-close to a stationary distribution. + + + One can prove that $$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(\frac{1}{4})$$ @@ -112,13 +83,13 @@ $$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}( % It is known that $d(t+1)\leq d(t)$. \JFC{references ? Cela a-t-il -% un intérêt dans la preuve ensuite.} +% un intérêt dans la preuve ensuite.} %and % $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$ -% One can prove that \JFc{Ou cela a-t-il été fait?} +% One can prove that \JFc{Ou cela a-t-il été fait?} % $$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(\frac{1}{4})$$ @@ -140,6 +111,9 @@ randomized stopping time (possibly depending on the starting position $X$), such that the distribution of $X_\tau$ is $\pi$: $$\P_X(X_\tau=Y)=\pi(Y).$$ +\subsection{Upper bound of Stopping Time}\label{sub:stop:bound} + + A stopping time $\tau$ is a {\emph strong stationary time} if $X_{\tau}$ is independent of $\tau$. @@ -222,7 +196,7 @@ $$ -%%%%%%%%%%%%%%%%%%%%%%%%%%%ù +%%%%%%%%%%%%%%%%%%%%%%%%%%%ù %\section{Stopping time} An integer $\ell\in \llbracket 1,{\mathsf{N}} \rrbracket$ is said {\it fair} @@ -250,8 +224,11 @@ $b=1$ with probability $\frac{1}{2}$ and $b=0$ with probability $\frac{1}{2}$. Since $h(X_{\tau_\ell-1})\neq\ell$ the value of the $\ell$-th bit of $X_{\tau_\ell}$ is $0$ or $1$ with the same probability ($\frac{1}{2}$). +This probability is independent of the value of the other bits. + + - Moving next in the chain, at each step, +Moving next in the chain, at each step, the $l$-th bit is switched from $0$ to $1$ or from $1$ to $0$ each time with the same probability. Therefore, for $t\geq \tau_\ell$, the $\ell$-th bit of $X_t$ is $0$ or $1$ with the same probability, proving the @@ -378,11 +355,70 @@ Notice that the calculus of the stationary time upper bound is obtained under the following constraint: for each vertex in the $\mathsf{N}$-cube there are one ongoing arc and one outgoing arc that are removed. The calculus does not consider (balanced) Hamiltonian cycles, which -are more regular and more binding than this constraint. Moreover, the bound +are more regular and more binding than this constraint. +Moreover, the bound is obtained using Markov Inequality which is frequently coarse. For the classical random walkin the $\mathsf{N}$-cube, without removing any Hamiltonian cylce, the mixing time is in $\Theta(N\ln N)$. We conjecture that in our context, the mixing time is also in $\Theta(N\ln -N)$. -%In this later context, we claim that the upper bound for the stopping time -%should be reduced. +N)$. + + +In this later context, we claim that the upper bound for the stopping time +should be reduced. This fact is studied in the next section. + +\subsection{Practical Evaluation of Stopping Times}\label{sub:stop:exp} + +Let be given a function $f: \Bool^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}}$ +and an initial seed $x^0$. +The pseudo code given in algorithm~\ref{algo:stop} returns the smallest +number of iterations such that all elements $\ell\in \llbracket 1,{\mathsf{N}} \rrbracket$ are fair. It allows to deduce an approximation of $E[\ts]$ +by calling this code many times with many instances of function and many +seeds. + +Practically speaking, for each number $\mathsf{N}$,$ 3 \le \mathsf{N} \le 16$, +10 functions have been generaed according to method presented in section~\ref{sec:hamilton}. For each of them, the calculus of the approximation of $E[\ts]$ +is executed 10000 times with a random seed. The table~\ref{table:stopping:moy} +summarizes results. It can be observed that the approximation is largely +wœsmaller than the upper bound given in theorem~\ref{prop:stop}. + +\begin{algorithm}[ht] +%\begin{scriptsize} +\KwIn{a function $f$, an initial configuration $x^0$ ($\mathsf{N}$ bits)} +\KwOut{a number of iterations $\textit{nbit}$} + +$\textit{nbit} \leftarrow 0$\; +$x\leftarrow x^0$\; +$\textit{visited}\leftarrow\emptyset$\; + +\While{$\left\vert{\textit{visited}}\right\vert < \mathsf{N} $} +{ + $ s \leftarrow \textit{Random}(n)$ \; + $\textit{image} \leftarrow f(x) $\; + \If{$x[s] \neq \textit{image}[s]$}{ + $\textit{visited} \leftarrow \textit{visited} \cup \{s\}$ + } + $x[s] \leftarrow \textit{image}[s]$\; + $\textit{nbit} \leftarrow \textit{nbit}+1$\; +} +\Return{$\textit{nbit}$}\; +%\end{scriptsize} +\caption{Pseudo Code of the stoping time calculus} +\label{algo:stop} +\end{algorithm} + + + + +\begin{table} +$$ +\begin{array}{|*{15}{l|}} +\hline +\mathsf{N} & 3 & 4 & 5 & 6 & 7& 8 & 9 & 10& 11 & 12 & 13 & 14 & 15 & 16 \\ +\hline +\mathsf{N} & 3 & 10.9 & 5 & 17.7 & 7& 25 & 9 & 32.7& 11 & 40.8 & 13 & 49.2 & 15 & 16 \\ +\hline +\end{array} +$$ +\caption{Average Stopping Time}\label{table:stopping:moy} +\end{table}