pch

[16dcc.git] / stopping.tex

diff --git a/stopping.tex b/stopping.tex
index 6632a23b05efdb558f76ede70dbff1ab25e2c1af..539d653dd3fa66c209cb9b7d11a49c05b32d6021 100644 (file)
--- a/stopping.tex
+++ b/stopping.tex
@@ -33,18 +33,18 @@ P=\dfrac{1}{6} \left(
 0&0&0&0&1&0&4&1 \\
 0&0&0&1&0&1&0&4 
 \end{array}
 0&0&0&0&1&0&4&1 \\
 0&0&0&1&0&1&0&4 
 \end{array}

-\right)
+\right).

 \]
 \end{xpl}
 
 
 A specific random walk in this modified hypercube is first 
 introduced (See section~\ref{sub:stop:formal}). We further 
 \]
 \end{xpl}
 
 
 A specific random walk in this modified hypercube is first 
 introduced (See section~\ref{sub:stop:formal}). We further 

-theoretical study this random walk to 
-provide a upper bound of fair sequences 
+ study this random walk in a theoretical way to 
+provide an upper bound of fair sequences 

 (See section~\ref{sub:stop:bound}).
 We finally complete these study with experimental
 (See section~\ref{sub:stop:bound}).
 We finally complete these study with experimental

-results that reduce this bound (Sec.~\ref{sub:stop:stop}).
+results that reduce this bound (Sec.~\ref{sub:stop:exp}).

 Notice that for a general references on Markov chains
 see~\cite{LevinPeresWilmer2006}, 
 and particularly Chapter~5 on stopping times.  
 Notice that for a general references on Markov chains
 see~\cite{LevinPeresWilmer2006}, 
 and particularly Chapter~5 on stopping times.  


@@ -60,18 +60,25 @@ $$\tv{\pi-\mu}=\frac{1}{2}\sum_{X\in\Bool^{\mathsf{N}}}|\pi(X)-\mu(X)|.$$ Moreov
 $\nu$ is a distribution on $\Bool^{\mathsf{N}}$, one has
 $$\tv{\pi-\mu}\leq \tv{\pi-\nu}+\tv{\nu-\mu}$$
 
 $\nu$ is a distribution on $\Bool^{\mathsf{N}}$, one has
 $$\tv{\pi-\mu}\leq \tv{\pi-\nu}+\tv{\nu-\mu}$$
 

-Let $P$ be the matrix of a Markov chain on $\Bool^{\mathsf{N}}$. $P(X,\cdot)$ is the
-distribution induced by the $X$-th row of $P$. If the Markov chain induced by
-$P$ has a stationary distribution $\pi$, then we define
+Let $P$ be the matrix of a Markov chain on $\Bool^{\mathsf{N}}$. For any
+$X\in \Bool^{\mathsf{N}}$, let $P(X,\cdot)$ be the distribution induced by the
+${\rm bin}(X)$-th row of $P$, where ${\rm bin}(X)$ is the integer whose
+binary encoding is $X$. If the Markov chain induced by $P$ has a stationary
+distribution $\pi$, then we define

 $$d(t)=\max_{X\in\Bool^{\mathsf{N}}}\tv{P^t(X,\cdot)-\pi}.$$
 
 $$d(t)=\max_{X\in\Bool^{\mathsf{N}}}\tv{P^t(X,\cdot)-\pi}.$$
 

+%\ANNOT{incohérence de notation $X$ : entier ou dans $B^N$ ?}

 and
 
 $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
 
 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.
+%% 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.
+
+Intutively speaking,  $t_{\rm mix}(\varepsilon)$ is the time/steps required
+to be sure to be $\varepsilon$-close to the stationary distribution, wherever
+the chain starts. 

 
 
 
 
 
 


@@ -83,13 +90,13 @@ $$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 
 
 
 % 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
 % $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
 
 
 
 %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?}

 % $$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(\frac{1}{4})$$
 
 
 % $$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(\frac{1}{4})$$
 
 


@@ -113,12 +120,11 @@ $$\P_X(X_\tau=Y)=\pi(Y).$$
 
 \subsection{Upper bound of Stopping Time}\label{sub:stop:bound}
 
 
 \subsection{Upper bound of Stopping Time}\label{sub:stop:bound}
 

-

 A stopping time $\tau$ is a {\emph strong stationary time} if $X_{\tau}$ is
 A stopping time $\tau$ is a {\emph strong stationary time} if $X_{\tau}$ is

-independent of $\tau$. 
+independent of $\tau$. The following result will be useful~\cite[Proposition~6.10]{LevinPeresWilmer2006},

 
 
 
 

-\begin{thrm}
+\begin{thrm}\label{thm-sst}

 If $\tau$ is a strong stationary time, then $d(t)\leq \max_{X\in\Bool^{\mathsf{N}}}
 \P_X(\tau > t)$.
 \end{thrm}
 If $\tau$ is a strong stationary time, then $d(t)\leq \max_{X\in\Bool^{\mathsf{N}}}
 \P_X(\tau > t)$.
 \end{thrm}


@@ -196,7 +202,7 @@ $$
 
 
 
 
 
 

-%%%%%%%%%%%%%%%%%%%%%%%%%%%ù
+%%%%%%%%%%%%%%%%%%%%%%%%%%%ù

 %\section{Stopping time}
 
 An integer $\ell\in \llbracket 1,{\mathsf{N}} \rrbracket$ is said {\it fair} 
 %\section{Stopping time}
 
 An integer $\ell\in \llbracket 1,{\mathsf{N}} \rrbracket$ is said {\it fair} 


@@ -231,7 +237,8 @@ This probability is independent of the value of the other bits.
 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
 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
+$\ell$-th bit of $X_t$ is $0$ or $1$ with the same probability,  and
+independently of the value of the other bits, proving the

 lemma.\end{proof}
 
 \begin{thrm} \label{prop:stop}
 lemma.\end{proof}
 
 \begin{thrm} \label{prop:stop}


@@ -244,7 +251,12 @@ let $S_{X,\ell}$ be the
 random variable that counts the number of steps 
 from $X$ until we reach a configuration where
 $\ell$ is fair. More formally
 random variable that counts the number of steps 
 from $X$ until we reach a configuration where
 $\ell$ is fair. More formally

-$$S_{X,\ell}=\min \{t \geq 1\mid h(X_{t-1})\neq \ell\text{ and }Z_t=(\ell,.)\text{ and } X_0=X\}.$$
+\[
+\begin{array}{rcl}
+S_{X,\ell}&=&\min \{t \geq 1\mid h(X_{t-1})\neq \ell\text{ and }Z_t=(\ell,.) \\
+&& \qquad \text{ and } X_0=X\}.
+\end{array}
+\]

 
 %  We denote by
 % $$\lambda_h=\max_{X,\ell} S_{X,\ell}.$$
 
 %  We denote by
 % $$\lambda_h=\max_{X,\ell} S_{X,\ell}.$$


@@ -291,8 +303,14 @@ has, for every $i$, $\P(S_{X,\ell}\geq 2i)\leq
 since $S_{X,\ell}$ is positive, it is known~\cite[lemma 2.9]{proba}, that
 $$E[S_{X,\ell}]=\sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq i).$$
 Since $\P(S_{X,\ell}\geq i)\geq \P(S_{X,\ell}\geq i+1)$, one has
 since $S_{X,\ell}$ is positive, it is known~\cite[lemma 2.9]{proba}, that
 $$E[S_{X,\ell}]=\sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq i).$$
 Since $\P(S_{X,\ell}\geq i)\geq \P(S_{X,\ell}\geq i+1)$, one has

-$$E[S_{X,\ell}]=\sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq i)\leq
-\P(S_{X,\ell}\geq 1)+\P(S_{X,\ell}\geq 2)+2 \sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq 2i).$$
+\[
+\begin{array}{rcl}
+  E[S_{X,\ell}]&=&\sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq i)\\
+&\leq& 
+\P(S_{X,\ell}\geq 1) +\P(S_{X,\ell}\geq 2)\\
+&& \qquad +2 \sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq 2i).
+\end{array}
+\]

 Consequently,
 $$E[S_{X,\ell}]\leq 1+1+2
 \sum_{i=1}^{+\infty}\left(1-\frac{1}{4{\mathsf{N}}^2}\right)^i=2+2(4{\mathsf{N}}^2-1)=8{\mathsf{N}}^2,$$
 Consequently,
 $$E[S_{X,\ell}]\leq 1+1+2
 \sum_{i=1}^{+\infty}\left(1-\frac{1}{4{\mathsf{N}}^2}\right)^i=2+2(4{\mathsf{N}}^2-1)=8{\mathsf{N}}^2,$$


@@ -339,17 +357,31 @@ One can now prove Theorem~\ref{prop:stop}.
 \begin{proof}
 Since $\ts^\prime$ is the time used to obtain $\mathsf{N}-1$ fair bits.
 Assume that the last unfair bit is $\ell$. One has
 \begin{proof}
 Since $\ts^\prime$ is the time used to obtain $\mathsf{N}-1$ fair bits.
 Assume that the last unfair bit is $\ell$. One has

-$\ts=\ts^\prime+S_{X_\tau,\ell}$, and therefore
-$E[\ts] = E[\ts^\prime]+E[S_{X_\tau,\ell}]$. Therefore,
-Theorem~\ref{prop:stop} is a direct application of
-lemma~\ref{prop:lambda} and~\ref{lm:stopprime}.
+$\ts=\ts^\prime+S_{X_\tau,\ell}$, and therefore $E[\ts] =
+E[\ts^\prime]+E[S_{X_\tau,\ell}]$. Therefore, Theorem~\ref{prop:stop} is a
+direct application of lemma~\ref{prop:lambda} and~\ref{lm:stopprime}.

 \end{proof}
 
 \end{proof}
 

+Now using Markov Inequality, one has $\P_X(\tau > t)\leq \frac{E[\tau]}{t}$.
+With $t_n=32N^2+16N\ln (N+1)$, one obtains:  $\P_X(\tau > t_n)\leq \frac{1}{4}$. 
+Therefore, using the defintion of $t_{\rm mix)}$ and
+Theorem~\ref{thm-sst}, it follows that
+$t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$.
+
+

 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. 
 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 
+The calculus doesn't consider (balanced) Hamiltonian cycles, which 

 are more regular and more binding than this constraint.
 are more regular and more binding than this constraint.

+Moreover, the bound
+is obtained using the coarse Markov Inequality. For the
+classical (lazzy) random walk 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. This fact is studied in the next section.
 
 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.
 


@@ -362,12 +394,6 @@ number of iterations such that all elements $\ell\in \llbracket 1,{\mathsf{N}} \
 by calling this code many times with many instances of function and many 
 seeds.
 
 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
-smaller 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)}
 \begin{algorithm}[ht]
 %\begin{scriptsize}
 \KwIn{a function $f$, an initial configuration $x^0$ ($\mathsf{N}$ bits)}


@@ -375,36 +401,63 @@ smaller than the upper bound given in theorem~\ref{prop:stop}.
 
 $\textit{nbit} \leftarrow 0$\;
 $x\leftarrow x^0$\;
 
 $\textit{nbit} \leftarrow 0$\;
 $x\leftarrow x^0$\;

-$\textit{visited}\leftarrow\emptyset$\;
-
-\While{$\left\vert{\textit{visited}}\right\vert < \mathsf{N} $}
+$\textit{fair}\leftarrow\emptyset$\;
+\While{$\left\vert{\textit{fair}}\right\vert < \mathsf{N} $}

 {
 {

-        $ s \leftarrow \textit{Random}(n)$ \;
+        $ s \leftarrow \textit{Random}(\mathsf{N})$ \;

         $\textit{image} \leftarrow f(x) $\;
         $\textit{image} \leftarrow f(x) $\;

-        \If{$x[s] \neq \textit{image}[s]$}{
-            $\textit{visited} \leftarrow \textit{visited} \cup \{s\}$
+        \If{$\textit{Random}(1) \neq 0$ and $x[s] \neq \textit{image}[s]$}{
+            $\textit{fair} \leftarrow \textit{fair} \cup \{s\}$\;
+            $x[s] \leftarrow \textit{image}[s]$\;

           }
           }

-        $x[s] \leftarrow \textit{image}[s]$\;

         $\textit{nbit} \leftarrow \textit{nbit}+1$\;
 }
 \Return{$\textit{nbit}$}\;
 %\end{scriptsize}
         $\textit{nbit} \leftarrow \textit{nbit}+1$\;
 }
 \Return{$\textit{nbit}$}\;
 %\end{scriptsize}

-\caption{Pseudo Code of the stoping time calculus}
+\caption{Pseudo Code of stoping time calculus }

 \label{algo:stop}
 \end{algorithm}
 
 \label{algo:stop}
 \end{algorithm}
 

+Practically speaking, for each number $\mathsf{N}$, $ 3 \le \mathsf{N} \le 16$, 
+10 functions have been generated 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 Figure~\ref{fig:stopping:moy}
+summarizes these results. In this one, a circle represents the 
+approximation of $E[\ts]$ for a given $\mathsf{N}$.
+The line is the graph of the function $x \mapsto 2x\ln(2x+8)$. 
+It can firstly 
+be observed that the approximation is largely
+smaller than the upper bound given in theorem~\ref{prop:stop}.
+It can be further deduced  that the conjecture of the previous section 
+is realistic according the graph of $x \mapsto 2x\ln(2x+8)$.

 
 
 
 
 
 

-\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}
+
+
+% \begin{table}
+% $$
+% \begin{array}{|*{14}{l|}}
+% \hline
+% \mathsf{N}  & 4 & 5 & 6 & 7& 8 & 9 & 10& 11 & 12 & 13 & 14 & 15 & 16 \\
+% \hline
+% \mathsf{N}  & 21.8 & 28.4 & 35.4 & 42.5 & 50 & 57.7 & 65.6& 73.5 & 81.6 & 90 & 98.3 & 107.1 & 115.7 \\
+% \hline
+% \end{array}
+% $$
+% \caption{Average Stopping Time}\label{table:stopping:moy}
+% \end{table}
+
+\begin{figure}
+\centering
+\includegraphics[width=0.49\textwidth]{complexity}
+\caption{Average Stopping Time Approximation}\label{fig:stopping:moy}
+\end{figure}
+
+
+
+%%% Local Variables:
+%%% mode: latex
+%%% TeX-master: "main"
+%%% ispell-dictionary: "american"
+%%% mode: flyspell
+%%% End: