ajout de fichiers

[16dcc.git] / stopping.tex

diff --git a/stopping.tex b/stopping.tex
index d72f8bbac8747217a733d0d31587816e84127c14..bb95663c6ca82fee5c21a017b39f496a76d6165b 100644 (file)
--- 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 the 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 $v \in \Bool^{\mathsf{N}}$. 
 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 $v \in \Bool^{\mathsf{N}}$. 


@@ -15,7 +10,7 @@ interpreted as Markov chains.
 \begin{xpl}
 Let us consider for instance  
 the graph $\Gamma(f)$ defined 
 \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:
 $$
 p(e) \left\{
 the probability function $p$ defined on the set of edges as follows:
 $$
 p(e) \left\{


@@ -38,55 +33,24 @@ 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}
 
 

-% % 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 
 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.
-Notice that for a general references on Markov chains
-see~\cite{LevinPeresWilmer2006}
-, and particularly Chapter~5 on stopping times.  
-
+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 this study with experimental
+results that reduce this bound (Sec.~\ref{sub:stop:exp}).
+For a general reference on Markov chains,
+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
 
 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


@@ -96,14 +60,28 @@ $$\tv{\pi-\mu}=\frac{1}{2}\sum_{X\in\Bool^{\mathsf{N}}}|\pi(X)-\mu(X)|.$$ 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,  $t_{\rm mix}(\varepsilon)$ is the time/steps required
+to be sure to be $\varepsilon$-close to the stationary distribution, wherever
+the chain starts. 
+
+
+

 One can prove that
 
 $$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(\frac{1}{4})$$
 One can prove that
 
 $$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(\frac{1}{4})$$


@@ -112,13 +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?}

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


@@ -140,8 +118,10 @@ randomized stopping time (possibly depending on the starting position $X$),
 such that  the distribution of $X_\tau$ is $\pi$:
 $$\P_X(X_\tau=Y)=\pi(Y).$$
 
 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
 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}


@@ -157,8 +137,8 @@ In other words, $E$ is the set of all the edges in the classical
 ${\mathsf{N}}$-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 
 ${\mathsf{N}}$-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 $\llbracket 1, {\mathsf{N}} \rrbracket$ 
+$X \in \Bool^{\mathsf{N}}$ whose edge is removed in the Hamiltonian cycle,
+\textit{i.e.}, which bit in $\llbracket 1, {\mathsf{N}} \rrbracket$ 

 cannot be switched.
 
 
 cannot be switched.
 
 


@@ -184,7 +164,7 @@ P_h(X,Y)=\frac{1}{2{\mathsf{N}}} & \textrm{if $X\neq Y$ and $(X,Y) \in E_h$}
 We denote by $\ov{h} : \Bool^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}}$ the function 
 such that for any $X \in \Bool^{\mathsf{N}} $, 
 $(X,\ov{h}(X)) \in E$ and $X\oplus\ov{h}(X)=0^{{\mathsf{N}}-h(X)}10^{h(X)-1}$. 
 We denote by $\ov{h} : \Bool^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}}$ the function 
 such that for any $X \in \Bool^{\mathsf{N}} $, 
 $(X,\ov{h}(X)) \in E$ and $X\oplus\ov{h}(X)=0^{{\mathsf{N}}-h(X)}10^{h(X)-1}$. 

-The function $\ov{h}$ is said {\it square-free} if for every $X\in \Bool^{\mathsf{N}}$,
+The function $\ov{h}$ is said to be {\it square-free} if for every $X\in \Bool^{\mathsf{N}}$,

 $\ov{h}(\ov{h}(X))\neq X$. 
 
 \begin{lmm}\label{lm:h}
 $\ov{h}(\ov{h}(X))\neq X$. 
 
 \begin{lmm}\label{lm:h}


@@ -222,16 +202,16 @@ $$
 
 
 
 
 
 

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

 %\section{Stopping time}
 
 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 $h(X_j)\neq \ell$.
 %\section{Stopping time}
 
 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 $j$) 
 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 $l$.  
+and where the configuration $X_j$ allows to cross the edge $l$.  

  
 Let $\ts$ be the first time all the elements of $\llbracket 1, {\mathsf{N}} \rrbracket$
 are fair. The integer $\ts$ is a randomized stopping time for
  
 Let $\ts$ be the first time all the elements of $\llbracket 1, {\mathsf{N}} \rrbracket$
 are fair. The integer $\ts$ is a randomized stopping time for


@@ -250,11 +230,15 @@ $b=1$ 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 ($\frac{1}{2}$).
 $\frac{1}{2}$. 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
 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}


@@ -267,7 +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
 % $$\lambda_h=\max_{X,\ell} S_{X,\ell}.$$


@@ -282,7 +271,7 @@ $E[S_{X,\ell}]\leq 8{\mathsf{N}}^2$ is established.
 \end{lmm}
 
 \begin{proof}
 \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, 
 \frac{1}{4{\mathsf{N}}^2}$. 
 Let $X_0= X$.
 Indeed, 


@@ -314,8 +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,
 $$E[S_{X,\ell}]\leq 1+1+2
 \sum_{i=1}^{+\infty}\left(1-\frac{1}{4{\mathsf{N}}^2}\right)^i=2+2(4{\mathsf{N}}^2-1)=8{\mathsf{N}}^2,$$


@@ -364,25 +359,106 @@ 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
 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 $\P_X(\tau > t)\leq \frac{E[\tau]}{t}$.
 \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
 Theorem~\ref{thm-sst}, it follows that

-$t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$.
+$t_{\rm mix}(\frac{1}{4})\leq 32N^2+16N\ln (N+1)=O(N^2)$ and that 
+

 
 
 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 
-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 $\Theta(N\ln N)$. 
+The calculus doesn't consider (balanced) Hamiltonian cycles, which 
+are more regular and more binding than this constraint.
+Moreover, the bound
+is obtained using the coarse Markov Inequality. For the
+classical (lazy) 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
 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 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 
+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.
+
+\begin{algorithm}[ht]
+%\begin{scriptsize}
+\KwIn{a function $f$, an initial configuration $x^0$ ($\mathsf{N}$ bits)}
+\KwOut{a number of iterations $\textit{nbit}$}
+
+$\textit{nbit} \leftarrow 0$\;
+$x\leftarrow x^0$\;
+$\textit{fair}\leftarrow\emptyset$\;
+\While{$\left\vert{\textit{fair}}\right\vert < \mathsf{N} $}
+{
+        $ s \leftarrow \textit{Random}(\mathsf{N})$ \;
+        $\textit{image} \leftarrow f(x) $\;
+        \If{$\textit{Random}(1) \neq 0$ and $x[s] \neq \textit{image}[s]$}{
+            $\textit{fair} \leftarrow \textit{fair} \cup \{s\}$\;
+            $x[s] \leftarrow \textit{image}[s]$\;
+          }
+        $\textit{nbit} \leftarrow \textit{nbit}+1$\;
+}
+\Return{$\textit{nbit}$}\;
+%\end{scriptsize}
+\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 generated according to the 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. 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 to the graph of $x \mapsto 2x\ln(2x+8)$.
+
+
+
+
+
+% \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: