X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/rairo15.git/blobdiff_plain/6537ad6b39c8648e4c3597b469e79e505193e358..f7f2b1f98aa13a31b26d9cdccd417f4f1c44bc8b:/stopping.tex

diff --git a/stopping.tex b/stopping.tex
index 9664549..3a07e06 100644
--- a/stopping.tex
+++ b/stopping.tex
@@ -1,26 +1,105 @@
-This section considers functions $f: \Bool^n \rightarrow \Bool^n $ 
+
+
+
+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.
+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 $v \in \Bool^{\mathsf{N}}$. 
+Next, if we add probabilities on the transition graph, iterations can be 
+interpreted as Markov chains.
+
+\begin{xpl}
+Let us consider for instance  
+the graph $\Gamma(f)$ defined 
+in \textsc{Figure~\ref{fig:iteration:f*}.} and 
+the probability function $p$ defined on the set of edges as follows:
+$$
+p(e) \left\{
+\begin{array}{ll}
+= \frac{2}{3} \textrm{ if $e=(v,v)$ with $v \in \Bool^3$,}\\
+= \frac{1}{6} \textrm{ otherwise.}
+\end{array}
+\right.  
+$$
+The matrix $P$ of the Markov chain associated to the function $f^*$ and to its probability function $p$ is 
+\[
+P=\dfrac{1}{6} \left(
+\begin{array}{llllllll}
+4&1&1&0&0&0&0&0 \\
+1&4&0&0&0&1&0&0 \\
+0&0&4&1&0&0&1&0 \\
+0&1&1&4&0&0&0&0 \\
+1&0&0&0&4&0&1&0 \\
+0&0&0&0&1&4&0&1 \\
+0&0&0&0&1&0&4&1 \\
+0&0&0&1&0&1&0&4 
+\end{array}
+\right)
+\]
+\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.
- 
+Notice that for a general references on Markov chains
+see~\cite{LevinPeresWilmer2006}
+, and particularly Chapter~5 on stopping times.  
 
 
 
 
-First of all, let $\pi$, $\mu$ be two distributions on $\Bool^n$. The total
+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
 defined by
-$$\tv{\pi-\mu}=\max_{A\subset \Bool^n} |\pi(A)-\mu(A)|.$$ It is known that
-$$\tv{\pi-\mu}=\frac{1}{2}\sum_{X\in\Bool^n}|\pi(X)-\mu(X)|.$$ Moreover, if
-$\nu$ is a distribution on $\Bool^n$, one has
+$$\tv{\pi-\mu}=\max_{A\subset \Bool^{\mathsf{N}}} |\pi(A)-\mu(A)|.$$ It is known that
+$$\tv{\pi-\mu}=\frac{1}{2}\sum_{X\in\Bool^{\mathsf{N}}}|\pi(X)-\mu(X)|.$$ Moreover, if
+$\nu$ 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 $\Bool^n$. $P(X,\cdot)$ is the
+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
-$$d(t)=\max_{X\in\Bool^n}\tv{P^t(X,\cdot)-\pi}.$$
+$$d(t)=\max_{X\in\Bool^{\mathsf{N}}}\tv{P^t(X,\cdot)-\pi}.$$
 
 and
 
@@ -44,10 +123,10 @@ $$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(
 
 
 
-Let $(X_t)_{t\in \mathbb{N}}$ be a sequence of $\Bool^n$ valued random
+Let $(X_t)_{t\in \mathbb{N}}$ be a sequence of $\Bool^{\mathsf{N}}$ valued random
 variables. A $\mathbb{N}$-valued random variable $\tau$ is a {\it stopping
   time} for the sequence $(X_i)$ if for each $t$ there exists $B_t\subseteq
-(\Bool^n)^{t+1}$ such that $\{\tau=t\}=\{(X_0,X_1,\ldots,X_t)\in B_t\}$. 
+(\Bool^{\mathsf{N}})^{t+1}$ such that $\{\tau=t\}=\{(X_0,X_1,\ldots,X_t)\in B_t\}$. 
 In other words, the event $\{\tau = t \}$ only depends on the values of 
 $(X_0,X_1,\ldots,X_t)$, not on $X_k$ with $k > t$. 
  
@@ -61,72 +140,77 @@ randomized stopping time (possibly depending on the starting position $X$),
 such that  the distribution of $X_\tau$ is $\pi$:
 $$\P_X(X_\tau=Y)=\pi(Y).$$
 
+A stopping time $\tau$ is a {\emph strong stationary time} if $X_{\tau}$ is
+independent of $\tau$. 
 
-\begin{Theo}
-If $\tau$ is a strong stationary time, then $d(t)\leq \max_{X\in\Bool^n}
+
+\begin{thrm}
+If $\tau$ is a strong stationary time, then $d(t)\leq \max_{X\in\Bool^{\mathsf{N}}}
 \P_X(\tau > t)$.
-\end{Theo}
+\end{thrm}
 
 
 %Let $\Bool^n$ be the set of words of length $n$. 
 Let $E=\{(X,Y)\mid
-X\in \Bool^n, Y\in \Bool^n,\ X=Y \text{ or } X\oplus Y \in 0^*10^*\}$.
+X\in \Bool^{\mathsf{N}}, Y\in \Bool^{\mathsf{N}},\ X=Y \text{ or } X\oplus Y \in 0^*10^*\}$.
 In other words, $E$ is the set of all the edges in the classical 
-$n$-cube. 
-Let $h$ be a function from $\Bool^n$ into $\llbracket 1, n \rrbracket$.
+${\mathsf{N}}$-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 
-$X \in \Bool^n$ which edge is removed in the Hamiltonian cycle,
-\textit{i.e.} which bit in $\llbracket 1, n \rrbracket$ 
+$X \in \Bool^{\mathsf{N}}$ which edge is removed in the Hamiltonian cycle,
+\textit{i.e.} which bit in $\llbracket 1, {\mathsf{N}} \rrbracket$ 
 cannot be switched.
 
 
 
 We denote by $E_h$ the set $E\setminus\{(X,Y)\mid X\oplus Y =
-0^{n-h(X)}10^{h(X)-1}\}$. This is the set of the modified hypercube, 
-\textit{i.e.}, the $n$-cube where the Hamiltonian cycle $h$ 
+0^{{\mathsf{N}}-h(X)}10^{h(X)-1}\}$. This is the set of the modified hypercube, 
+\textit{i.e.}, the ${\mathsf{N}}$-cube where the Hamiltonian cycle $h$ 
 has been removed.
 
 We define the Markov matrix $P_h$ for each line $X$ and 
 each column $Y$  as follows:
-$$\left\{
+\begin{equation}
+\left\{
 \begin{array}{ll}
-P_h(X,X)=\frac{1}{2}+\frac{1}{2n} & \\
+P_h(X,X)=\frac{1}{2}+\frac{1}{2{\mathsf{N}}} & \\
 P_h(X,Y)=0 & \textrm{if  $(X,Y)\notin E_h$}\\
-P_h(X,Y)=\frac{1}{2n} & \textrm{if $X\neq Y$ and $(X,Y) \in E_h$}
+P_h(X,Y)=\frac{1}{2{\mathsf{N}}} & \textrm{if $X\neq Y$ and $(X,Y) \in E_h$}
 \end{array}
 \right.
-$$ 
+\label{eq:Markov:rairo}
+\end{equation} 
 
-We denote by $\ov{h} : \Bool^n \rightarrow \Bool^n$ the function 
-such that for any $X \in \Bool^n $, 
-$(X,\ov{h}(X)) \in E$ and $X\oplus\ov{h}(X)=0^{n-h(X)}10^{h(X)-1}$. 
-The function $\ov{h}$ is said {\it square-free} if for every $X\in \Bool^n$,
+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 $X\in \Bool^{\mathsf{N}}$,
 $\ov{h}(\ov{h}(X))\neq X$. 
 
-\begin{Lemma}\label{lm:h}
+\begin{lmm}\label{lm:h}
 If $\ov{h}$ is bijective and square-free, then $h(\ov{h}^{-1}(X))\neq h(X)$.
-\end{Lemma}
+\end{lmm}
 
 \begin{proof}
 Let $\ov{h}$ be bijective.
-Let $k\in \llbracket 1, n \rrbracket$ s.t. $h(\ov{h}^{-1}(X))=k$.
+Let $k\in \llbracket 1, {\mathsf{N}} \rrbracket$ s.t. $h(\ov{h}^{-1}(X))=k$.
 Then $(\ov{h}^{-1}(X),X)$ belongs to $E$ and 
-$\ov{h}^{-1}(X)\oplus X = 0^{n-k}10^{k-1}$.
+$\ov{h}^{-1}(X)\oplus X = 0^{{\mathsf{N}}-k}10^{k-1}$.
 Let us suppose $h(X) = h(\ov{h}^{-1}(X))$. In such a case, $h(X) =k$.
 By definition of $\ov{h}$, $(X, \ov{h}(X)) \in E $ and 
-$X\oplus\ov{h}(X)=0^{n-h(X)}10^{h(X)-1} = 0^{n-k}10^{k-1}$.
+$X\oplus\ov{h}(X)=0^{{\mathsf{N}}-h(X)}10^{h(X)-1} = 0^{{\mathsf{N}}-k}10^{k-1}$.
 Thus $\ov{h}(X)= \ov{h}^{-1}(X)$, which leads to $\ov{h}(\ov{h}(X))= X$.
 This contradicts the square-freeness of $\ov{h}$.
 \end{proof}
 
 Let $Z$ be a random variable that is uniformly distributed over
-$\llbracket 1, n \rrbracket \times \Bool$.
-For $X\in \Bool^n$, we
+$\llbracket 1, {\mathsf{N}} \rrbracket \times \Bool$.
+For $X\in \Bool^{\mathsf{N}}$, we
 define, with $Z=(i,b)$,  
 $$
 \left\{
 \begin{array}{ll}
-f(X,Z)=X\oplus (0^{n-i}10^{i-1}) & \text{if } b=1 \text{ and } i\neq h(X),\\
+f(X,Z)=X\oplus (0^{{\mathsf{N}}-i}10^{i-1}) & \text{if } b=1 \text{ and } i\neq h(X),\\
 f(X,Z)=X& \text{otherwise.} 
 \end{array}\right.
 $$
@@ -141,7 +225,7 @@ $$
 %%%%%%%%%%%%%%%%%%%%%%%%%%%ù
 %\section{Stopping time}
 
-An integer $\ell\in \llbracket 1,n \rrbracket$ is said {\it fair} 
+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 
@@ -149,14 +233,14 @@ 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 $l$.  
  
-Let $\ts$ be the first time all the elements of $\llbracket 1, n \rrbracket$
+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
 the Markov chain $(X_t)$.
 
 
-\begin{Lemma}
+\begin{lmm}
 The integer $\ts$ is a strong stationary time.
-\end{Lemma}
+\end{lmm}
 
 \begin{proof}
 Let $\tau_\ell$ be the first time that $\ell$ is fair. The random variable
@@ -173,57 +257,59 @@ 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
 lemma.\end{proof}
 
-\begin{Theo} \label{prop:stop}
+\begin{thrm} \label{prop:stop}
 If $\ov{h}$ is bijective and square-free, then
-$E[\ts]\leq 8n^2+ n\ln (n+1)$. 
-\end{Theo}
+$E[\ts]\leq 8{\mathsf{N}}^2+ 4{\mathsf{N}}\ln ({\mathsf{N}}+1)$. 
+\end{thrm}
 
-For each $X\in \Bool^n$ and $\ell\in\llbracket 1,n\rrbracket$, 
+For each $X\in \Bool^{\mathsf{N}}$ and $\ell\in\llbracket 1,{\mathsf{N}}\rrbracket$, 
 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
 $$S_{X,\ell}=\min \{t \geq 1\mid h(X_{t-1})\neq \ell\text{ and }Z_t=(\ell,.)\text{ and } X_0=X\}.$$
 
- We denote by
-$$\lambda_h=\max_{X,\ell} S_{X,\ell}.$$
-
+%  We denote by
+% $$\lambda_h=\max_{X,\ell} S_{X,\ell}.$$
 
-\begin{Lemma}\label{prop:lambda}
-If $\ov{h}$ is a square-free bijective function, then the inequality 
-$E[\lambda_h]\leq 8n^2$ is established.
 
-\end{Lemma}
+\begin{lmm}\label{prop:lambda}
+Let $\ov{h}$ is a square-free bijective function. Then
+for all $X$ and 
+all $\ell$, 
+the inequality 
+$E[S_{X,\ell}]\leq 8{\mathsf{N}}^2$ is established.
+\end{lmm}
 
 \begin{proof}
 For every $X$, every $\ell$, one has $\P(S_{X,\ell})\leq 2)\geq
-\frac{1}{4n^2}$. 
+\frac{1}{4{\mathsf{N}}^2}$. 
 Let $X_0= X$.
 Indeed, 
 \begin{itemize}
 \item if $h(X)\neq \ell$, then
-$\P(S_{X,\ell}=1)=\frac{1}{2n}\geq \frac{1}{4n^2}$. 
+$\P(S_{X,\ell}=1)=\frac{1}{2{\mathsf{N}}}\geq \frac{1}{4{\mathsf{N}}^2}$. 
 \item otherwise, $h(X)=\ell$, then
 $\P(S_{X,\ell}=1)=0$.
-But in this case, intutively, it is possible to move
+But in this case, intuitively, it is possible to move
 from $X$ to $\ov{h}^{-1}(X)$ (with probability $\frac{1}{2N}$). And in
 $\ov{h}^{-1}(X)$ the $l$-th bit can be switched. 
 More formally,
 since $\ov{h}$ is square-free,
 $\ov{h}(X)=\ov{h}(\ov{h}(\ov{h}^{-1}(X)))\neq \ov{h}^{-1}(X)$. It follows
 that $(X,\ov{h}^{-1}(X))\in E_h$. We thus have
-$P(X_1=\ov{h}^{-1}(X))=\frac{1}{2N}$. Now, by Lemma~\ref{lm:h},
+$P(X_1=\ov{h}^{-1}(X))=\frac{1}{2{\mathsf{N}}}$. Now, by Lemma~\ref{lm:h},
 $h(\ov{h}^{-1}(X))\neq h(X)$. Therefore $\P(S_{x,\ell}=2\mid
-X_1=\ov{h}^{-1}(X))=\frac{1}{2N}$, proving that $\P(S_{x,\ell}\leq 2)\geq
-\frac{1}{4N^2}$.
+X_1=\ov{h}^{-1}(X))=\frac{1}{2{\mathsf{N}}}$, proving that $\P(S_{x,\ell}\leq 2)\geq
+\frac{1}{4{\mathsf{N}}^2}$.
 \end{itemize}
 
 
 
 
-Therefore, $\P(S_{X,\ell}\geq 3)\leq 1-\frac{1}{4n^2}$. By induction, one
+Therefore, $\P(S_{X,\ell}\geq 3)\leq 1-\frac{1}{4{\mathsf{N}}^2}$. By induction, one
 has, for every $i$, $\P(S_{X,\ell}\geq 2i)\leq
-\left(1-\frac{1}{4n^2}\right)^i$.
+\left(1-\frac{1}{4{\mathsf{N}}^2}\right)^i$.
  Moreover,
 since $S_{X,\ell}$ is positive, it is known~\cite[lemma 2.9]{proba}, that
 $$E[S_{X,\ell}]=\sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq i).$$
@@ -232,40 +318,60 @@ $$E[S_{X,\ell}]=\sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq i)\leq
 \P(S_{X,\ell}\geq 1)+\P(S_{X,\ell}\geq 2)+2 \sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq 2i).$$
 Consequently,
 $$E[S_{X,\ell}]\leq 1+1+2
-\sum_{i=1}^{+\infty}\left(1-\frac{1}{4n^2}\right)^i=2+2(4n^2-1)=8n^2,$$
+\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,$$
 which concludes the proof.
 \end{proof}
 
-Let $\ts^\prime$ be the first time that there are exactly $n-1$ fair
-elements. 
+Let $\ts^\prime$ be the time used to get all the bits but one fair.
 
-\begin{Lemma}\label{lm:stopprime}
-One has $E[\ts^\prime]\leq n \ln (n+1).$
-\end{Lemma}
+\begin{lmm}\label{lm:stopprime}
+One has $E[\ts^\prime]\leq 4{\mathsf{N}} \ln ({\mathsf{N}}+1).$
+\end{lmm}
 
 \begin{proof}
 This is a classical  Coupon Collector's like problem. Let $W_i$ be the
 random variable counting the number of moves done in the Markov chain while
-we had exactly $i-1$ fair bits. One has $\ts^\prime=\sum_{i=1}^{n-1}W_i$.
+we had exactly $i-1$ fair bits. One has $\ts^\prime=\sum_{i=1}^{{\mathsf{N}}-1}W_i$.
  But when we are at position $X$ with $i-1$ fair bits, the probability of
- obtaining a new fair bit is either $1-\frac{i-1}{n}$ if $h(X)$ is fair,
- or  $1-\frac{i-2}{n}$ if $h(X)$ is not fair. It follows that 
-$E[W_i]\leq \frac{n}{n-i+2}$. Therefore
-$$E[\ts^\prime]=\sum_{i=1}^{n-1}E[W_i]\leq n\sum_{i=1}^{n-1}
- \frac{1}{n-i+2}=n\sum_{i=3}^{n+1}\frac{1}{i}.$$
-
-But $\sum_{i=1}^{n+1}\frac{1}{i}\leq 1+\ln(n+1)$. It follows that
-$1+\frac{1}{2}+\sum_{i=3}^{n+1}\frac{1}{i}\leq 1+\ln(n+1).$
+ obtaining a new fair bit is either $1-\frac{i-1}{{\mathsf{N}}}$ if $h(X)$ is fair,
+ or  $1-\frac{i-2}{{\mathsf{N}}}$ if $h(X)$ is not fair. 
+
+Therefore,
+$\P (W_i=k)\leq \left(\frac{i-1}{{\mathsf{N}}}\right)^{k-1} \frac{{\mathsf{N}}-i+2}{{\mathsf{N}}}.$
+Consequently, we have $\P(W_i\geq k)\leq \left(\frac{i-1}{{\mathsf{N}}}\right)^{k-1} \frac{{\mathsf{N}}-i+2}{{\mathsf{N}}-i+1}.$
+It follows that $E[W_i]=\sum_{k=1}^{+\infty} \P (W_i\geq k)\leq {\mathsf{N}} \frac{{\mathsf{N}}-i+2}{({\mathsf{N}}-i+1)^2}\leq \frac{4{\mathsf{N}}}{{\mathsf{N}}-i+2}$.
+
+
+
+It follows that 
+$E[W_i]\leq \frac{4{\mathsf{N}}}{{\mathsf{N}}-i+2}$. Therefore
+$$E[\ts^\prime]=\sum_{i=1}^{{\mathsf{N}}-1}E[W_i]\leq 
+4{\mathsf{N}}\sum_{i=1}^{{\mathsf{N}}-1} \frac{1}{{\mathsf{N}}-i+2}=
+4{\mathsf{N}}\sum_{i=3}^{{\mathsf{N}}+1}\frac{1}{i}.$$
+
+But $\sum_{i=1}^{{\mathsf{N}}+1}\frac{1}{i}\leq 1+\ln({\mathsf{N}}+1)$. It follows that
+$1+\frac{1}{2}+\sum_{i=3}^{{\mathsf{N}}+1}\frac{1}{i}\leq 1+\ln({\mathsf{N}}+1).$
 Consequently,
-$E[\ts^\prime]\leq n (-\frac{1}{2}+\ln(n+1))\leq n\ln(n+1)$.
+$E[\ts^\prime]\leq 
+4{\mathsf{N}} (-\frac{1}{2}+\ln({\mathsf{N}}+1))\leq 
+4{\mathsf{N}}\ln({\mathsf{N}}+1)$.
 \end{proof}
 
 One can now prove Theorem~\ref{prop:stop}.
 
 \begin{proof}
-One has $\ts\leq \ts^\prime+\lambda_h$. Therefore,
+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}.
 \end{proof}
 
-
+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.
+In this later context, we claim that the upper bound for the stopping time 
+should be reduced.
