Merge branch 'master' of ssh://bilbo.iut-bm.univ-fcomte.fr/16dcc

[16dcc.git] / stopping.tex

diff --git a/stopping.tex b/stopping.tex
index 989bb9eee1f7cb01b79d4b91bf71f0cc22f1e8f4..539d653dd3fa66c209cb9b7d11a49c05b32d6021 100644 (file)
--- a/stopping.tex
+++ b/stopping.tex
@@ -33,7 +33,7 @@ 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}
 
 \]
 \end{xpl}
 


@@ -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. 

 
 
 
 
 
 


@@ -113,9 +120,8 @@ $$\P_X(X_\tau=Y)=\pi(Y).$$
 
 \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}\label{thm-sst}
 
 
 \begin{thrm}\label{thm-sst}


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


@@ -345,7 +363,7 @@ direct application of lemma~\ref{prop:lambda} and~\ref{lm:stopprime}.
 \end{proof}
 
 Now using Markov Inequality, one has $\P_X(\tau > t)\leq \frac{E[\tau]}{t}$.
 \end{proof}
 
 Now using Markov Inequality, one has $\P_X(\tau > t)\leq \frac{E[\tau]}{t}$.

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


@@ -354,11 +372,11 @@ $t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$.
 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.
 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
+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 $\Theta(N\ln
 N)$.
 Hamiltonian cylce, the mixing time is in $\Theta(N\ln N)$. 
 We conjecture that in our context, the mixing time is also in $\Theta(N\ln
 N)$.


@@ -401,7 +419,7 @@ $\textit{fair}\leftarrow\emptyset$\;
 \end{algorithm}
 
 Practically speaking, for each number $\mathsf{N}$, $ 3 \le \mathsf{N} \le 16$, 
 \end{algorithm}
 
 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 $E[\ts]$
+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 $\mathsf{N}$.
 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 $\mathsf{N}$.


@@ -422,7 +440,7 @@ is realistic according the graph of $x \mapsto 2x\ln(2x+8)$.
 % \hline
 % \mathsf{N}  & 4 & 5 & 6 & 7& 8 & 9 & 10& 11 & 12 & 13 & 14 & 15 & 16 \\
 % \hline
 % \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 & 16 \\
+% \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}
 % $$
 % \hline
 % \end{array}
 % $$


@@ -431,7 +449,7 @@ is realistic according the graph of $x \mapsto 2x\ln(2x+8)$.
 
 \begin{figure}
 \centering
 
 \begin{figure}
 \centering

-\includegraphics[scale=0.5]{complexity}
+\includegraphics[width=0.49\textwidth]{complexity}

 \caption{Average Stopping Time Approximation}\label{fig:stopping:moy}
 \end{figure}
 
 \caption{Average Stopping Time Approximation}\label{fig:stopping:moy}
 \end{figure}