modif presentation b

[16dcc.git] / stopping.tex

diff --git a/stopping.tex b/stopping.tex
index 409dd83c971a5402faba645fedce5ebf85c777ae..9d7e74f374970e7d835203cef4d985973ec8c53a 100644 (file)
--- a/stopping.tex
+++ b/stopping.tex
@@ -1,6 +1,6 @@
 This section considers functions $f: \Bool^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}} $ 
 issued from an hypercube where an Hamiltonian path has been removed
 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 previous section.
+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 $v \in \Bool^{\mathsf{N}}$. 


@@ -10,7 +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\{


@@ -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}
 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 
 \]
 \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 
-(See section~\ref{sub:stop:bound}).
-We finally complete these study with experimental
-results that reduce this bound (Sec.~\ref{sub:stop:stop}).
-Notice that for a general references on Markov chains
+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.  
 
 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.
+
+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}
 
 
 \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}


@@ -131,8 +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.
 
 


@@ -158,7 +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}


@@ -202,10 +208,10 @@ $$
 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 $h(X_j)\neq \ell$.
 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


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


@@ -259,7 +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, 


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


@@ -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
 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
 $t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$.
 
 Theorem~\ref{thm-sst}, it follows that
 $t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$.
 


@@ -354,34 +372,28 @@ $t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$.
 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
-Hamiltonian cylce, the mixing time is in $\Theta(N\ln N)$. 
+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 $\Theta(N\ln
 N)$.
 
 
 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 $x^0$.
 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.
 
 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.
 

-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
-wœsmaller 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)}


@@ -389,36 +401,63 @@ wœsmaller 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 stopping time computation}

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

-
-
-
-\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}
+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: