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flyingscorpio@clevo 2022-02-28 09:04:34 +01:00
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algebre-non-lineaire/main.tex
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@ -872,4 +872,29 @@
\end{align*}
Donc $d_A$ est calculé à partir de $p$ et $q$.
\subsection{Signature de message}
\begin{center}
\begin{tikzpicture}
\node[rectangle,thick,draw,red] (alice) at (0,0) {
\parbox{4cm}{\centering Alice \\ $d_A$}
};
\node[rectangle,thick,draw,blue] (bob) at (10,0) {
\parbox{4cm}{\centering Bob \\ $d_B$ \\ preuve d'identité}
};
\node (alicekey) at (3.5,1) {$(n_A,e_A)$};
\node (bobkey) at (6.5,1) {$(n_B,e_B)$};
\node (bobcode) at (6,0) {$AC_B$};
\draw[-latex] (alice) -- (alicekey);
\draw[-latex] (bob) -- (bobkey);
\draw[-latex] (bob) -- (bobcode);
\end{tikzpicture}
\end{center}
Bob publie un code d'authentification $AC_B$.
Il calcule $AC_B^{d_B}[n_B] = V$.
Cette valeur $V$, seul Bob est capable de la produire.
Il la chiffre avec la clé d'Alice~: $(n_A, e_A)$.
On a donc $(S_B\rightarrow A)$, la signature de Bob à Alice.
\end{document}