-The problem of deciding whether classical feedforward ANNs are
-suitable to approximate topological chaotic iterations may then be
-reduced to evaluate such neural networks on iterations of functions
-with Strongly Connected Component (SCC)~graph of iterations. To
-compare with non-chaotic iterations, the experiments detailed in the
-following sections are carried out using both kinds of function
-(chaotic and non-chaotic). Let us emphasize on the difference between
-this kind of neural networks and the Chaotic Iterations based
-multilayer peceptron.
-
-We are then left to compute two disjoint function sets that contain
-either functions with topological chaos properties or not, depending
-on the strong connectivity of their iterations graph. This can be
-achieved for instance by removing a set of edges from the iteration
-graph $\Gamma(f_0)$ of the vectorial negation function~$f_0$. One can
-deduce whether a function verifies the topological chaos property or
-not by checking the strong connectivity of the resulting graph of
-iterations.
-
-For instance let us consider the functions $f$ and $g$ from $\Bool^4$
-to $\Bool^4$ respectively defined by the following lists:
-$$[0, 0, 2, 3, 13, 13, 6, 3, 8, 9, 10, 11, 8, 13, 14,
- 15]$$ $$\mbox{and } [11, 14, 13, 14, 11, 10, 1, 8, 7, 6, 5, 4, 3, 2,
- 1, 0] \enspace.$$ In other words, the image of $0011$ by $g$ is
-$1110$: it is obtained as the binary value of the fourth element in
-the second list (namely~14). It is not hard to verify that
-$\Gamma(f)$ is not SCC (\textit{e.g.}, $f(1111)$ is $1111$) whereas
-$\Gamma(g)$ is. The remaining of this section shows how to translate
-iterations of such functions into a model amenable to be learned by an
-ANN. Formally, input and output vectors are pairs~$((S^t)^{t \in
- \Nats},x)$ and $\left(\sigma((S^t)^{t \in
- \Nats}),F_{f}(S^0,x)\right)$ as defined in~Eq.~(\ref{eq:Gf}).
+On considère par exemple les deux fonctions $f$ and $g$ de0 $\Bool^4$
+dans $\Bool^4$ définies par:
+
+\begin{eqnarray*}
+f(x_1,x_2,x_3,x_4) &= &
+(x_1(x_2+x_4)+ \overline{x_2}x_3\overline{x_4},
+x_2,
+x_3(\overline{x_1}.\overline{x_4}+x_2x_4+x_1\overline{x_2}),
+x_4+\overline{x_2}x_3) \\
+g(x_1,x_2,x_3,x_4) &= &
+(\overline{x_1},
+\overline{x_2}+ x_1.\overline{x_3}.\overline{x_4},
+\overline{x_3}(x_1 + x_2+x_4),
+\overline{x_4}(x_1 + \overline{x_2}+\overline{x_3}))
+\end{eqnarray*}
+On peut vérifier facilement que le graphe $\textsc{giu}(f)$
+n'est pas fortement connexe car $(1,1,1,1)$ est un point fixe de $f$
+tandis que le graphe $\textsc{giu}(g)$ l'est.
+
+L'entrée du réseau est une paire de la forme
+$(x,(S^t)^{t \in \Nats})$ et sa sortie correspondante est
+de la forme $\left(F_{h_u}(S^0,x), \sigma((S^t)^{t \in
+ \Nats})\right)$ comme définie à l'équationà l'équation~(\ref{eq:sch:unaire}).
+
+
