Relecture Sylvain avec commentaires/questions

[16dcc.git] / chaos.tex

diff --git a/chaos.tex b/chaos.tex
index d5ab5f94edc376ad4b7a09529fb1dd0d6d7f5511..ec2723aef86663af75c4d60192d7e8fab1104041 100644 (file)
--- a/chaos.tex
+++ b/chaos.tex
@@ -245,9 +245,10 @@ $\check{u}^{v^0}$ (on $n$ digits), ..., $\check{u}^{\check{v}^0-1}$ (on $n$ digi
 
 
 
 
 
 

+\newcommand{\ns}{$\hspace{.1em}$}

 
 \begin{xpl}
 
 \begin{xpl}

-Consider for instance that $\mathsf{N}=13$, $\mathcal{P}=\{1,2,11\}$ (so $\mathsf{p}=3$), and that
+Consider for instance that $\mathsf{N}=13$, $\mathcal{P}=\{1,2,11\}$ (so $\mathsf{p}=2$), and that

 $s=\left\{
 \begin{array}{l}
 u=\underline{6,} ~ \underline{11,5}, ...\\
 $s=\left\{
 \begin{array}{l}
 u=\underline{6,} ~ \underline{11,5}, ...\\


@@ -262,7 +263,7 @@ $\check{s}=\left\{
 \end{array}
 \right.$.
 
 \end{array}
 \right.$.
 

-So $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) = 0.010004000000000000000000011005 ...$
+So $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) = 0.01\ns00\ns04\ns00\ns00\ns00\ns00\ns00\ns00\ns00\ns00\ns00\ns01\ns10\ns05 ...$

 Indeed, the $p=2$ first digits are 01, as $|v^0-\check{v}^0|=1$, 
 and we use $p$ digits to code this difference ($\mathcal{P}$ being $\{1,2,11\}$, this difference can be equal to 10). We then take the $v^0=1$ first terms of $u$, each term being coded in $n=2$ digits, that is, 06. As we can iterate
 at most $\max{(\mathcal{P})}$ times, we must complete this
 Indeed, the $p=2$ first digits are 01, as $|v^0-\check{v}^0|=1$, 
 and we use $p$ digits to code this difference ($\mathcal{P}$ being $\{1,2,11\}$, this difference can be equal to 10). We then take the $v^0=1$ first terms of $u$, each term being coded in $n=2$ digits, that is, 06. As we can iterate
 at most $\max{(\mathcal{P})}$ times, we must complete this


@@ -425,7 +426,7 @@ $\mathcal{P}=\{2,3\}$. The graphs of iterations are given in
 The \textsc{Figure~\ref{graphe1}} shows what happens when 
 displaying each iteration result.
 On the contrary, the \textsc{Figure~\ref{graphe2}} explicits the behaviors
 The \textsc{Figure~\ref{graphe1}} shows what happens when 
 displaying each iteration result.
 On the contrary, the \textsc{Figure~\ref{graphe2}} explicits the behaviors

-when always applying 2 or 3 modification and next outputing results. 
+when always applying either 2 or 3 modifications before generating results. 

 Notice that here, orientations of arcs are not necessary 
 since the function $f_0$ is equal to its inverse $f_0^{-1}$. 
 \end{xpl}
 Notice that here, orientations of arcs are not necessary 
 since the function $f_0$ is equal to its inverse $f_0^{-1}$. 
 \end{xpl}


@@ -479,7 +480,7 @@ Let $y=(e,((u^0, ..., u^{\sum_{l=0}^{k_1}v^l-1}, a_0^0, ..., a_0^{|a_0|}, a_1^0,
 Conversely, if $\Gamma_{\mathcal{P}}(f)$ is not strongly connected, then there are 
 2 vertices $e_1$ and $e_2$ such that there is no path between $e_1$ and $e_2$.
 That is, it is impossible to find $(u,v)\in \mathds{S}_{\mathsf{N},\mathcal{P}}$
 Conversely, if $\Gamma_{\mathcal{P}}(f)$ is not strongly connected, then there are 
 2 vertices $e_1$ and $e_2$ such that there is no path between $e_1$ and $e_2$.
 That is, it is impossible to find $(u,v)\in \mathds{S}_{\mathsf{N},\mathcal{P}}$

-and $n \mathds{N}$ such that $G_f^n(e,(u,v))_1=e_2$. The open ball $\mathcal{B}(e_2, 1/2)$
+and $n\in \mathds{N}$ such that $G_f^n(e,(u,v))_1=e_2$. The open ball $\mathcal{B}(e_2, 1/2)$

 cannot be reached from any neighborhood of $e_1$, and thus $G_f$ is not transitive.
 \end{proof}
 
 cannot be reached from any neighborhood of $e_1$, and thus $G_f$ is not transitive.
 \end{proof}
 


@@ -498,7 +499,7 @@ $$\left\{(e, ((u^0, ..., u^{v^{k_1-1}},U^0, U^1, ...),(v^0, ..., v^{k_1},V^0, V^
 $$\left.\forall i,j \in \mathds{N}, U^i \in \llbracket 1, \mathsf{N} \rrbracket, V^j \in \mathcal{P}\right\}
 \subset \mathcal{B}(x,\varepsilon),$$
 and $y=G_f^{k_1}(e,(u,v))$. $\Gamma_{\mathcal{P}}(f)$ being strongly connected,
 $$\left.\forall i,j \in \mathds{N}, U^i \in \llbracket 1, \mathsf{N} \rrbracket, V^j \in \mathcal{P}\right\}
 \subset \mathcal{B}(x,\varepsilon),$$
 and $y=G_f^{k_1}(e,(u,v))$. $\Gamma_{\mathcal{P}}(f)$ being strongly connected,

-there is at least a path from the Boolean state $y_1$ of $y$ and $e$.
+there is at least a path from the Boolean state $y_1$ of $y$ and $e$ \ANNOT{Phrase pas claire : "from ... " mais pas de "to ..."}.

 Denote by $a_0, \hdots, a_{k_2}$ the edges of such a path.
 Then the point:
 $$(e,((u^0, ..., u^{v^{k_1-1}},a_0^0, ..., a_0^{|a_0|}, a_1^0, ..., a_1^{|a_1|},..., 
 Denote by $a_0, \hdots, a_{k_2}$ the edges of such a path.
 Then the point:
 $$(e,((u^0, ..., u^{v^{k_1-1}},a_0^0, ..., a_0^{|a_0|}, a_1^0, ..., a_1^{|a_1|},..., 


@@ -521,12 +522,18 @@ and only if its iteration graph $\Gamma_{\mathcal{P}}(f)$ is strongly connected.
   In this context, $\mathcal{P}$ is the singleton $\{b\}$.
   If $b$ is even, any vertex $e$ of $\Gamma_{\{b\}}(f_0)$ cannot reach 
   its neighborhood and thus $\Gamma_{\{b\}}(f_0)$ is not strongly connected. 
   In this context, $\mathcal{P}$ is the singleton $\{b\}$.
   If $b$ is even, any vertex $e$ of $\Gamma_{\{b\}}(f_0)$ cannot reach 
   its neighborhood and thus $\Gamma_{\{b\}}(f_0)$ is not strongly connected. 

-  If $b$ is even, any vertex $e$ of $\Gamma_{\{b\}}(f_0)$ cannot reach itself 
+  If $b$ is odd, any vertex $e$ of $\Gamma_{\{b\}}(f_0)$ cannot reach itself 

   and thus $\Gamma_{\{b\}}(f_0)$ is not strongly connected.
 \end{proof}
 
   and thus $\Gamma_{\{b\}}(f_0)$ is not strongly connected.
 \end{proof}
 

-The next section shows how to generate functions and a iteration number $b$
+The next section recalls a general scheme to produce
+functions and a iteration number $b$

 such that $\Gamma_{\{b\}}$ is strongly connected.
 
 
 such that $\Gamma_{\{b\}}$ is strongly connected.
 
 

- 
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