X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/16dcc.git/blobdiff_plain/31468b39be240f447015be22c252d515b2dcfdac..79f46c777f6ef3202c2fb25194822518cad3e32c:/generating.tex?ds=sidebyside

diff --git a/generating.tex b/generating.tex
index d512a98..377c811 100644
--- a/generating.tex
+++ b/generating.tex
@@ -3,9 +3,7 @@ It has been shown~\cite[Theorem 4]{bcgr11:ip} that
 if its iteration graph $\Gamma(f)$ 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.
-
-
+if and only if its Markov matrix is a doubly stochastic one.
 In~\cite[Section 4]{DBLP:conf/secrypt/CouchotHGWB14},
 we have presented a general scheme which generates 
 function with strongly connected iteration graph $\Gamma(f)$ and
@@ -19,7 +17,7 @@ is removed, an edge $(x,x)$ is added.
 For instance, the iteration graph $\Gamma(f^*)$ 
 (given in Figure~\ref{fig:iteration:f*}) 
 is the $3$-cube in which the Hamiltonian cycle 
-$000,100,101,001,011,111,110,010,000$ 
+$000,100,101,001,011,111,$ $110,010,000$ 
 has been removed.
 \end{xpl}
 
@@ -31,7 +29,7 @@ has the awaited property with regard to the connectivity.
 \begin{thrm}
 The iteration graph $\Gamma(f)$ issued from
 the ${\mathsf{N}}$-cube where an Hamiltonian 
-cycle is removed is strongly connected.
+cycle is removed, is strongly connected.
 \end{thrm}
 
 Moreover, if all the transitions have the same probability ($\frac{1}{n}$),
@@ -51,7 +49,7 @@ The question which remains to solve is:
 The answer is indeed positive. We furthermore have the following strongest 
 result.
 \begin{thrm}
-There exist $b \in \Nats$ such that $\Gamma_{\{b\}}(f)$ is complete.
+There exists $b \in \Nats$ such that $\Gamma_{\{b\}}(f)$ is complete.
 \end{thrm}
 \begin{proof}
 There is an arc $(x,y)$ in the 
@@ -64,14 +62,13 @@ Thus, there exists $b$ such that there is an arc between any $x$ and $y$.
 This section ends with the idea of removing a Hamiltonian cycle in the 
 $\mathsf{N}$-cube. 
 In such a context, the Hamiltonian cycle is equivalent to a Gray code.
-Many approaches have been proposed a way to build such codes, for instance 
-the Reflected Binary Code. In this one, one of the bits is switched 
-exactly $2^{\mathsf{N}-}$ \ANNOT{formule incomplète : $2^{\mathsf{N}-1}$ ??} for a $\mathsf{N}$-length cycle. 
-
-%%%%%%%%%%%%%%%%%%%%%
-
-The function that is built 
-from the \ANNOT{Phrase non terminée}
+Many approaches have been proposed as a way to build such codes, for instance 
+the Reflected Binary Code. In this one and 
+for a $\mathsf{N}$-length cycle, one of the bits is exactly switched 
+$2^{\mathsf{N}-1}$ times whereas the others bits are modified at most 
+$\left\lfloor \dfrac{2^{\mathsf{N-1}}}{\mathsf{N}-1} \right\rfloor$ times.
+It is clear that the function that is built from such a code would
+not provide an uniform output.  
 
 The next section presents how to build balanced Hamiltonian cycles in the 
 $\mathsf{N}$-cube with the objective to embed them into the