X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/rairo15.git/blobdiff_plain/a5af617983aac5c657a64b171efdd174585433db..8845f0ad0e1fda56b399e8c451712a1cd6b3afe4:/preliminaries.tex?ds=sidebyside

diff --git a/preliminaries.tex b/preliminaries.tex
index f90fdae..9441aab 100644
--- a/preliminaries.tex
+++ b/preliminaries.tex
@@ -9,8 +9,9 @@ to itself
 such that 
 $x=(x_1,\dots,x_n)$ maps to $f(x)=(f_1(x),\dots,f_n(x))$.
 Functions are iterated as follows. 
-At the $t^{th}$ iteration, only the $s_{t}-$th component is
-``iterated'', where $s = \left(s_t\right)_{t \in \mathds{N}}$ is a sequence of indices taken in $\llbracket 1;n \rrbracket$ called ``strategy''. Formally,
+At the $t^{th}$ iteration, only the $s_{t}-$th component is said to be
+``iterated'', where $s = \left(s_t\right)_{t \in \mathds{N}}$ is a sequence of indices taken in $\llbracket 1;n \rrbracket$ called ``strategy''. 
+Formally,
 let $F_f: \llbracket1;n\rrbracket\times \Bool^{n}$ to $\Bool^n$ be defined by
 \[
 F_f(i,x)=(x_1,\dots,x_{i-1},f_i(x),x_{i+1},\dots,x_n).
@@ -34,79 +35,85 @@ the graph $\Gamma(f)$ contains an arc from $x$ to $F_f(i,x)$.
 Let us consider for instance $n=3$.
 Let 
 $f^*: \Bool^3 \rightarrow \Bool^3$ be defined by
-
 $f^*(x_1,x_2,x_3) = 
 (x_2 \oplus x_3, \overline{x_1}\overline{x_3} + x_1\overline{x_2},
-\overline{x_1}\overline{x_3} + x_1x_2)$
-
-
+\overline{x_1}\overline{x_3} + x_1x_2)$.
 The iteration graph $\Gamma(f^*)$ of this function is given in 
 Figure~\ref{fig:iteration:f*}.
 
 \vspace{-1em}
 \begin{figure}[ht]
 \begin{center}
-\includegraphics[scale=0.5]{images/iter_f0c.eps}
+\includegraphics[scale=0.5]{images/iter_f0c}
 \end{center}
 \vspace{-0.5em}
 \caption{Iteration Graph $\Gamma(f^*)$ of the function $f^*$}\label{fig:iteration:f*}
 \end{figure}
 \end{xpl}
 
-\vspace{-0.5em}
-It is easy to associate a Markov Matrix $M$ to such a graph $G(f)$
-as follows:
 
-$M_{ij} = \frac{1}{n}$ if there is an edge from $i$ to $j$ in $\Gamma(f)$ and $i \neq j$;  $M_{ii} = 1 - \sum\limits_{j=1, j\neq i}^n M_{ij}$; and $M_{ij} = 0$ otherwise.
+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 
+$n$-cube augmented with all the self-loop, \textit{i.e.}, all the 
+edges $(v,v)$ for any $v \in \Bool^n$. 
+Next, if we add probabilities on the transition graph, iterations can be 
+interpreted as Markov chains.
 
 \begin{xpl}
-The Markov matrix associated to the function $f^*$ is 
-
+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 
 \[
-M=\dfrac{1}{3} \left(
+P=\dfrac{1}{6} \left(
 \begin{array}{llllllll}
-1&1&1&0&0&0&0&0 \\
-1&1&0&0&0&1&0&0 \\
-0&0&1&1&0&0&1&0 \\
-0&1&1&1&0&0&0&0 \\
-1&0&0&0&1&0&1&0 \\
-0&0&0&0&1&1&0&1 \\
-0&0&0&0&1&0&1&1 \\
-0&0&0&1&0&1&0&1 
+4&1&1&0&0&0&0&0 \\
+1&4&0&0&0&1&0&0 \\
+0&0&4&1&0&0&1&0 \\
+0&1&1&4&0&0&0&0 \\
+1&0&0&0&4&0&1&0 \\
+0&0&0&0&1&4&0&1 \\
+0&0&0&0&1&0&4&1 \\
+0&0&0&1&0&1&0&4 
 \end{array}
 \right)
 \]
-
-
-
- 
-
 \end{xpl}
 
 
-It is usual to check whether rows of such kind of matrices
-converge to a specific 
-distribution. 
-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 
-$\Omega$  by:
-$$\tv{\pi-\mu}=\max_{A\subset \Omega} |\pi(A)-\mu(A)|.$$ 
-% It is known that
-% $$\tv{\pi-\mu}=\frac{1}{2}\sum_{x\in\Omega}|\pi(x)-\mu(x)|.$$
-
-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
-$$d(t)=\max_{x\in\Omega}\tv{M^t(x,\cdot)-\pi}.$$
-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 $\varepsilon$.
+% % 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  
@@ -115,52 +122,4 @@ that is sufficient to obtain a deviation lesser than $\varepsilon$.
 
 
 
-Let us finally present the pseudorandom number generator $\chi_{\textit{14Secrypt}}$
-which is based on random walks in $\Gamma(f)$. 
-More precisely, let be given a Boolean map $f:\Bool^n \rightarrow \Bool^n$,
-a PRNG \textit{Random},
-an integer $b$ that corresponds to an awaited mixing time, and 
-an initial configuration $x^0$. 
-Starting from $x^0$, the algorithm repeats $b$ times 
-a random choice of which edge to follow and traverses this edge.
-The final configuration is thus outputted.
-This PRNG is formalized in Algorithm~\ref{CI Algorithm}.
-
-
-
-\vspace{-1em}
-\begin{algorithm}[ht]
-%\begin{scriptsize}
-\KwIn{a function $f$, an iteration number $b$, an initial configuration $x^0$ ($n$ bits)}
-\KwOut{a configuration $x$ ($n$ bits)}
-$x\leftarrow x^0$\;
-\For{$i=0,\dots,b-1$}
-{
-$s\leftarrow{\textit{Random}(n)}$\;
-$x\leftarrow{F_f(s,x)}$\;
-}
-return $x$\;
-%\end{scriptsize}
-\caption{Pseudo Code of the $\chi_{\textit{14Secrypt}}$ PRNG}
-\label{CI Algorithm}
-\end{algorithm}
-\vspace{-0.5em}
-This PRNG is a particularized version of Algorithm given in~\cite{BCGR11}.
-Compared to this latter, the length of the random 
-walk of our algorithm is always constant (and is equal to $b$) whereas it 
-was given by a second PRNG in this latter.
-However, all the theoretical results that are given in~\cite{BCGR11} remain
-true since the proofs do not rely on this fact. 
-
-Let $f: \Bool^{n} \rightarrow \Bool^{n}$.
-It has been shown~\cite[Th. 4, p. 135]{BCGR11}} that 
-if its iteration graph is strongly connected, then 
-the output of $\chi_{\textit{14Secrypt}}$ follows 
-a law that tends to the uniform distribution 
-if and only if its Markov matrix is a doubly stochastic matrix.
-  
-Let us now present  a method to
-generate  functions
-with Doubly Stochastic matrix and Strongly Connected iteration graph,
- denoted as DSSC matrix.   
 
