X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/hdrcouchot.git/blobdiff_plain/eb86b87fcb8bd9aa66283fcea4d9efbe964179ff..2bf0bca226facdce323e96552021bf90952eefce:/15RairoGen.tex?ds=inline

diff --git a/15RairoGen.tex b/15RairoGen.tex
index e4f8858..aff664f 100644
--- a/15RairoGen.tex
+++ b/15RairoGen.tex
@@ -517,45 +517,76 @@ $\mathcal{X}_{\mathsf{N},\mathcal{P}} = \mathds{B}^\mathsf{N} \times \mathds{S}_
 oÃ¹ $s=(u,v)$ et $\check{s}=(\check{u},\check{v})$ sont dans $ \mathds{S}_{\mathsf{N},\mathcal{P}} = 
 \mathcal{S}_{\llbracket 1, \mathsf{N} \rrbracket} \times \mathcal{S}_\mathcal{P}$. 
 \begin{itemize}
-\item $e$ et $\check{e}$ sont des entiers appartenant Ã  $\llbracket 0, 2^{\mathsf{N}-1} \rrbracket$. The Hamming distance
-on their binary decomposition, that is, the number of dissimilar binary digits, constitutes the integral
-part of $d(X,\check{X})$.
-\item The fractional part is constituted by the differences between $v^0$ and $\check{v}^0$, followed by the differences
-between finite sequences $u^0, u^1, \hdots, u^{v^0-1}$ and  $\check{u}^0, \check{u}^1, \hdots, \check{u}^{\check{v}^0-1}$, followed by 
- differences between $v^1$ and $\check{v}^1$, followed by the differences
-between $u^{v^0}, u^{v^0+1}, \hdots, u^{v^1-1}$ and  $\check{u}^{\check{v}^0}, \check{u}^{\check{v}^0+1}, \hdots, \check{u}^{\check{v}^1-1}$, etc.
-More precisely, let $p = \lfloor \log_{10}{(\max{\mathcal{P}})}\rfloor +1$ and $n = \lfloor \log_{10}{(\mathsf{N})}\rfloor +1$.
+\item $e$ et $\check{e}$ sont des entiers appartenant Ã  $\llbracket 0, 2^{\mathsf{N}-1} \rrbracket$. 
+La distance de Hamming $d_{\mathds{B}^\mathsf{N}}$ sur entre les 
+dÃ©compositions binaires de $e$ et de $\check{e}$ (\textit{i.e.}, le 
+le nombre de bits qu'elles ont de diffÃ©rent) constitue 
+la partie entiÃ¨re de $d(X,\check{X})$.
+\item la partie dÃ©cimale est construite Ã  partir des diffÃ©rences entre 
+$v^0$ et $\check{v}^0$, suivie des diffÃ©rences entre les sÃ©quences finies
+$u^0, u^1, \hdots, u^{v^0-1}$ et  $\check{u}^0, \check{u}^1, \hdots, \check{u}^{\check{v}^0-1}$, suivie par les diffÃ©rences entre  $v^1$ et $\check{v}^1$, 
+suivie par les diffÃ©rences entre $u^{v^0}, u^{v^0+1}, \hdots, u^{v^1-1}$ et
+$\check{u}^{\check{v}^0}, \check{u}^{\check{v}^0+1}, \hdots, \check{u}^{\check{v}^1-1}$, etc.
+
+Plus prÃ©cisemment, soit 
+$p = \lfloor \log_{10}{(\max{\mathcal{P}})}\rfloor +1$ et 
+$n = \lfloor \log_{10}{(\mathsf{N})}\rfloor +1$.
 \begin{itemize}
-\item The $p$ first digits of $d(x,\check{x})$ is $|v^0-\check{v}^0|$ written in decimal numeration (and with $p$ digits).
-\item The next $n\times \max{(\mathcal{P})}$ digits aim at measuring how much $u^0, u^1, \hdots, u^{v^0-1}$ differs from $\check{u}^0, \check{u}^1, \hdots, \check{u}^{\check{v}^0-1}$. The $n$ first
-digits are $|u^0-\check{u}^0|$. They are followed by 
-$|u^1-\check{u}^1|$ written with $n$ digits, etc.
+\item Les $p$ premiers Ã©lÃ©ments de $d(x,\check{x})$ sont $|v^0-\check{v}^0|$ 
+  Ã©crits en base 10 et sur $p$ indices;
+\item les $n\times \max{(\mathcal{P})}$ Ã©lÃ©ments suivants servent 
+  Ã  Ã©valuer de combien $u^0, u^1, \hdots, u^{v^0-1}$ diffÃ¨re de 
+  $\check{u}^0, \check{u}^1, \hdots, \check{u}^{\check{v}^0-1}$. 
+  Les $n$ premiers Ã©lÃ©ments sont $|u^0-\check{u}^0|$. Il sont suivis de 
+$|u^1-\check{u}^1|$ Ã©crits Ã  l'aide de $n$ Ã©lÃ©ments, etc.
 \begin{itemize}
-\item If
-$v^0=\check{v}^0$, then the process is continued until $|u^{v^0-1}-\check{u}^{\check{v}^0-1}|$ and the fractional
-part of $d(X,\check{X})$ is completed by 0's until reaching
-$p+n\times \max{(\mathcal{P})}$ digits.
-\item If $v^0<\check{v}^0$, then the $ \max{(\mathcal{P})}$  blocs of $n$
-digits are $|u^0-\check{u}^0|$, ..., $|u^{v^0-1}-\check{u}^{v^0-1}|$,
-$\check{u}^{v^0}$ (on $n$ digits), ..., $\check{u}^{\check{v}^0-1}$ (on $n$ digits), followed by 0's if required.
-\item The case $v^0>\check{v}^0$ is dealt similarly.
+\item Si
+$v^0=\check{v}^0$,
+alors le processus se continue jusqu'Ã  $|u^{v^0-1}-\check{u}^{\check{v}^0-1}|$ et la 
+partie dÃ©cimale de $d(X,\check{X})$ est complÃ©tÃ©e par des 0
+jusqu'Ã  atteindre 
+$p+n\times \max{(\mathcal{P})}$ Ã©lÃ©ments.
+\item Si $v^0<\check{v}^0$, alors les $ \max{(\mathcal{P})}$  blocs de $n$
+Ã©lÃ©ments sont $|u^0-\check{u}^0|$, ..., $|u^{v^0-1}-\check{u}^{v^0-1}|$,
+$\check{u}^{v^0}$ (sur $n$ Ã©lÃ©ments), ..., $\check{u}^{\check{v}^0-1}$ (sur $n$ Ã©lÃ©ments), suivi par des 0, si besoin.
+\item Le cas $v^0>\check{v}^0$ est similaire, et donc omis
 \end{itemize}
-\item The next $p$ digits are $|v^1-\check{v}^1|$, etc.
+\item Les $p$ suivants sont $|v^1-\check{v}^1|$, etc.
 \end{itemize}
 \end{itemize}
 
 
+La fonction $d$ peut se formaliser comme suit:
+$$d(x,\check{x})=d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s})+d_{\mathds{B}^\mathsf{N}}(e,\check{e}),$$
+oÃ¹: % $p=\max \mathcal{P}$ and:
+\begin{itemize}
+\item $d_{\mathds{B}^\mathsf{N}}$ est la distance de Hamming,
+\item $\forall s=(u,v), \check{s}=(\check{u},\check{v}) \in \mathcal{S}_{\mathsf{N},\mathcal{P}}$,\newline 
+$$\begin{array}{rcl}
+ d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) &= &
+   \sum_{k=0}^\infty \dfrac{1}{10^{(k+1)p+kn\max{(\mathcal{P})}}} 
+   \bigg(|v^k - \check{v}^k|  \\
+   & & + \left| \sum_{l=0}^{v^k-1} 
+       \dfrac{u^{\sum_{m=0}^{k-1} v^m +l}}{ 10^{(l+1)n}} -
+       \sum_{l=0}^{\check{v}^k-1} 
+       \dfrac{\check{u}^{\sum_{m=0}^{k-1} \check{v}^m +l}}{ 10^{(l+1)n}} \right| \bigg)
+\end{array}
+$$ %\left| \sum_{l=0}^{v^k-1} \dfrac{u^{\sum_{m=0}^{k-1} v^m +l}}{ 10^{l}} - \sum_{l=0}^{\check{v}^k-1} \dfrac{\check{u}^{\sum_{m=0}^{k-1} \check{v}^m +l}}{ 10^{l}}\right|\right)}.$$
+\end{itemize}
+
 
 
 \begin{xpl}
-Consider for instance that $\mathsf{N}=13$, $\mathcal{P}=\{1,2,11\}$ (so $\mathsf{p}=3$), and that
+On considÃ¨re par exemple 
+$\mathsf{N}=13$, $\mathcal{P}=\{1,2,11\}$ ($\mathsf{p}$ vaut ainsi $3$), 
+et 
 $s=\left\{
 \begin{array}{l}
 u=\underline{6,} ~ \underline{11,5}, ...\\
 v=1,2,...
 \end{array}
 \right.$
-while
+avec 
 $\check{s}=\left\{
 \begin{array}{l}
 \check{u}=\underline{6,4} ~ \underline{1}, ...\\
@@ -563,65 +594,56 @@ $\check{s}=\left\{
 \end{array}
 \right.$.
 
-So $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) = 0.010004000000000000000000011005 ...$
-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
-value by some 0's in such a way that the obtained result
-has $n\times \max{(\mathcal{P})}=22$ digits, that is: 
-0600000000000000000000. Similarly, the $\check{v}^0=2$ first
-terms in $\check{u}$ are represented by 0604000000000000000000, and the absolute value of their
-difference is equal to 0004000000000000000000. These digits are concatenated to 01, and
-we start again with the remainder of the sequences.
+Ainsi $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) = 0.010004000000000000000000011005 ...$
+En effet, les $p=2$ premiers Ã©lÃ©ments sont  01, c'est-Ã -dire 
+$|v^0-\check{v}^0|=1$, 
+et on utilise $p$ Ã©lÃ©ments pour reprÃ©senter cette diffÃ©rence
+(Comme $\mathcal{P}=\{1,2,11\}$, cette diffÃ©rence peut valoir 10).
+On prend alors le $v^0=1$ premier terme de $u$, 
+chaque terme Ã©tant codÃ© sur $n=2$ Ã©lÃ©ments, soit 06.
+Comme on itÃ¨re au plus $\max{(\mathcal{P})}$ fois, 
+on complÃ¨te cette valeur par des 0 de sorte que 
+la chaine obtenue a $n\times \max{(\mathcal{P})}=22$ Ã©lÃ©ments, soit:
+0600000000000000000000. 
+De maniÃ¨re similaire, les $\check{v}^0=2$ premiers
+termes de $\check{u}$ sont reprÃ©sentÃ©s par 
+0604000000000000000000.
+LA valeur absolue de leur diffÃ©rence est Ã©gale Ã  
+0004000000000000000000.
+Ces Ã©lÃ©ments sont concatÃ©nÃ©s avec 01. On peut construire alors le reste de 
+la sÃ©quence.
 \end{xpl}
 
 
 \begin{xpl}
-Consider now that $\mathsf{N}=9$, and $\mathcal{P}=\{2,7\}$, and that
-
-$s=\left\{
+On considÃ¨re Ã  prÃ©sent que  $\mathsf{N}=9$, que $\mathcal{P}=\{2,7\}$ et que
+$$s=\left\{
 \begin{array}{l}
 u=\underline{6,7,} ~ \underline{4,2,} ...\\
 v=2,2,...
 \end{array}
-\right.$
-while
-$\check{s}=\left\{
+\right.$$
+avec
+$$\check{s}=\left\{
 \begin{array}{l}
 \check{u}=\underline{4, 9, 6, 3, 6, 6, 7,} ~ \underline{9, 8}, ...\\
 \check{v}=7,2,...
 \end{array}
-\right.$
+\right.
+$$
 
-So $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) = 0.5173633305600000...$, as $|v^0-\check{v}^0|=5$, $|4963667-6700000| = 1736333$, $|v^1-\check{v}^1|=0$,
-and $|9800000-4200000| = 5600000$.
+Ainsi $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) = 0.5173633305600000...$, 
+puisque 
+$|v^0-\check{v}^0|=5$, $|4963667-6700000| = 1736333$, $|v^1-\check{v}^1|=0$,
+et $|9800000-4200000| = 5600000$.
 \end{xpl}
 
 
 
-$d$ can be more rigorously written as follows:
-$$d(x,\check{x})=d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s})+d_{\mathds{B}^\mathsf{N}}(e,\check{e}),$$
-where: % $p=\max \mathcal{P}$ and:
-\begin{itemize}
-\item $d_{\mathds{B}^\mathsf{N}}$ is the Hamming distance,
-\item $\forall s=(u,v), \check{s}=(\check{u},\check{v}) \in \mathcal{S}_{\mathsf{N},\mathcal{P}}$,\newline 
-$$\begin{array}{rcl}
- d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) &= &
-   \sum_{k=0}^\infty \dfrac{1}{10^{(k+1)p+kn\max{(\mathcal{P})}}} 
-   \bigg(|v^k - \check{v}^k|  \\
-   & & + \left| \sum_{l=0}^{v^k-1} 
-       \dfrac{u^{\sum_{m=0}^{k-1} v^m +l}}{ 10^{(l+1)n}} -
-       \sum_{l=0}^{\check{v}^k-1} 
-       \dfrac{\check{u}^{\sum_{m=0}^{k-1} \check{v}^m +l}}{ 10^{(l+1)n}} \right| \bigg)
-\end{array}
-$$ %\left| \sum_{l=0}^{v^k-1} \dfrac{u^{\sum_{m=0}^{k-1} v^m +l}}{ 10^{l}} - \sum_{l=0}^{\check{v}^k-1} \dfrac{\check{u}^{\sum_{m=0}^{k-1} \check{v}^m +l}}{ 10^{l}}\right|\right)}.$$
-\end{itemize}
-
-
-Let us show that,
-\begin{prpstn}
-$d$ is a distance on $\mathcal{X}_{\mathsf{N},\mathcal{P}}$.
-\end{prpstn}
+On a la proposition suivante, qui est dÃ©montrÃ©e en annexes~\ref{anx:generateur}.
+\begin{lemma}
+$d$ est une distance sur $\mathcal{X}_{\mathsf{N},\mathcal{P}}$.
+\end{lemma}
 
 
 \subsection{Le graphe $\textsc{giu}_{\mathcal{P}}(f)$ Ã©tendant  $\textsc{giu}(f)$}
