X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/16dcc.git/blobdiff_plain/705f6ebe2219797761eb1fd336de90f287787a29..79f46c777f6ef3202c2fb25194822518cad3e32c:/stopping.tex?ds=inline

diff --git a/stopping.tex b/stopping.tex
index e86176a..142da7f 100644
--- a/stopping.tex
+++ b/stopping.tex
@@ -10,7 +10,7 @@ interpreted as Markov chains.
 \begin{xpl}
 Let us consider for instance  
 the graph $\Gamma(f)$ defined 
-in \textsc{Figure~\ref{fig:iteration:f*}.} and 
+in Figure~\ref{fig:iteration:f*} and 
 the probability function $p$ defined on the set of edges as follows:
 $$
 p(e) \left\{
@@ -33,19 +33,19 @@ P=\dfrac{1}{6} \left(
 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 
+introduced (see Section~\ref{sub:stop:formal}). We further 
  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 this study with experimental
 results that reduce this bound (Sec.~\ref{sub:stop:exp}).
-Notice that for a general references on Markov chains
+For a general reference 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
 $$\tv{\pi-\mu}\leq \tv{\pi-\nu}+\tv{\nu-\mu}$$
 
-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
 $$d(t)=\max_{X\in\Bool^{\mathsf{N}}}\tv{P^t(X,\cdot)-\pi}.$$
 
+%\ANNOT{incohérence de notation $X$ : entier ou dans $B^N$ ?}
 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.
+
+Intuitively 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}
 
-
 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}
@@ -132,7 +138,7 @@ ${\mathsf{N}}$-cube.
 Let $h$ be a function from $\Bool^{\mathsf{N}}$ into $\llbracket 1, {\mathsf{N}} \rrbracket$.
 Intuitively speaking $h$ aims at memorizing for each node 
 $X \in \Bool^{\mathsf{N}}$ which edge is removed in the Hamiltonian cycle,
-\textit{i.e.} which bit in $\llbracket 1, {\mathsf{N}} \rrbracket$ 
+\textit{i.e.}, which bit in $\llbracket 1, {\mathsf{N}} \rrbracket$ 
 cannot be switched.
 
 
@@ -202,7 +208,7 @@ $$
 An integer $\ell\in \llbracket 1,{\mathsf{N}} \rrbracket$ is said {\it fair} 
 at time $t$ if there
 exists $0\leq j <t$ such that $Z_{j+1}=(\ell,\cdot)$ and $h(X_j)\neq \ell$.
-In other words, there exist a date $j$ before $t$ where 
+In other words, there exists a date $j$ before $t$ where 
 the first element of the random variable $Z$ is exactly $l$ 
 (\textit{i.e.}, $l$ is the strategy at date $j$) 
 and where the configuration $X_j$ allows to traverse the edge $l$.  
@@ -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
-$\ell$-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}
@@ -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
-$$S_{X,\ell}=\min \{t \geq 1\mid h(X_{t-1})\neq \ell\text{ and }Z_t=(\ell,.)\text{ and } X_0=X\}.$$
+\[
+\begin{array}{rcl}
+S_{X,\ell}&=&\min \{t \geq 1\mid h(X_{t-1})\neq \ell\text{ and }Z_t=(\ell,.) \\
+&& \qquad \text{ and } X_0=X\}.
+\end{array}
+\]
 
 %  We denote by
 % $$\lambda_h=\max_{X,\ell} S_{X,\ell}.$$
@@ -259,7 +271,7 @@ $E[S_{X,\ell}]\leq 8{\mathsf{N}}^2$ is established.
 \end{lmm}
 
 \begin{proof}
-For every $X$, every $\ell$, one has $\P(S_{X,\ell})\leq 2)\geq
+For every $X$, every $\ell$, one has $\P(S_{X,\ell}\leq 2)\geq
 \frac{1}{4{\mathsf{N}}^2}$. 
 Let $X_0= X$.
 Indeed, 
@@ -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
-$$E[S_{X,\ell}]=\sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq i)\leq
-\P(S_{X,\ell}\geq 1)+\P(S_{X,\ell}\geq 2)+2 \sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq 2i).$$
+\[
+\begin{array}{rcl}
+  E[S_{X,\ell}]&=&\sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq i)\\
+&\leq& 
+\P(S_{X,\ell}\geq 1) +\P(S_{X,\ell}\geq 2)\\
+&& \qquad +2 \sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq 2i).
+\end{array}
+\]
 Consequently,
 $$E[S_{X,\ell}]\leq 1+1+2
 \sum_{i=1}^{+\infty}\left(1-\frac{1}{4{\mathsf{N}}^2}\right)^i=2+2(4{\mathsf{N}}^2-1)=8{\mathsf{N}}^2,$$
@@ -341,12 +359,12 @@ 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}.
+direct application of Lemma~\ref{prop:lambda} and~\ref{lm:stopprime}.
 \end{proof}
 
 Now using Markov Inequality, one has $\P_X(\tau > t)\leq \frac{E[\tau]}{t}$.
-With $t=32N^2+16N\ln (N+1)$, one obtains:  $\P_X(\tau > t)\leq \frac{1}{4}$. 
-Therefore, using the defintion of $t_{\rm mix)}$ and
+With $t_n=32N^2+16N\ln (N+1)$, one obtains:  $\P_X(\tau > t_n)\leq \frac{1}{4}$. 
+Therefore, using the definition of $t_{\rm mix}$ and
 Theorem~\ref{thm-sst}, it follows that
 $t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$.
 
@@ -354,24 +372,24 @@ $t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$.
 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. 
-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
-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 $\Theta(N\ln N)$. 
+is obtained using the coarse Markov Inequality. For the
+classical (lazzy) random walk the  $\mathsf{N}$-cube, without removing any
+Hamiltonian cycle, 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 
+In this latter 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 
+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.
@@ -396,19 +414,19 @@ $\textit{fair}\leftarrow\emptyset$\;
 }
 \Return{$\textit{nbit}$}\;
 %\end{scriptsize}
-\caption{Pseudo Code of stoping time calculus }
+\caption{Pseudo Code of stopping time computation}
 \label{algo:stop}
 \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 $E[\ts]$
-is executed 10000 times with a random seed. The Figure~\ref{fig:stopping:moy}
+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. 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}.
+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 $x \mapsto 2x\ln(2x+8)$.
 
@@ -431,7 +449,7 @@ is realistic according the graph of $x \mapsto 2x\ln(2x+8)$.
 
 \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}