modif presentation b

[16dcc.git] / generating.tex

diff --git a/generating.tex b/generating.tex
index 0837fdaec746214beea9f8734d672881c96634cc..b4839db9b84ad75d5322da432dfb81e0d88778e0 100644 (file)
--- 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 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
 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


@@ -19,11 +17,11 @@ 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 
 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}
 
 has been removed.
 \end{xpl}
 

-We first have proven the following result, which 
+We  have first proven the following result, which 

 states that the ${\mathsf{N}}$-cube without one
 Hamiltonian cycle 
 has the awaited property with regard to the connectivity.
 states that the ${\mathsf{N}}$-cube without one
 Hamiltonian cycle 
 has the awaited property with regard to the connectivity.


@@ -31,10 +29,10 @@ 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 
 \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}
 
 \end{thrm}
 

-Moreover, if all the transitions have the same probability ($\frac{1}{n}$),
+Moreover, when all the transitions have the same probability ($\frac{1}{n}$),

 we have proven the following results:
 \begin{thrm}
 The Markov Matrix $M$ resulting from the ${\mathsf{N}}$-cube in
 we have proven the following results:
 \begin{thrm}
 The Markov Matrix $M$ resulting from the ${\mathsf{N}}$-cube in


@@ -45,11 +43,11 @@ cycle is removed, is doubly stochastic.
 Let us consider now a ${\mathsf{N}}$-cube where an Hamiltonian 
 cycle is removed.
 Let $f$ be the corresponding function.
 Let us consider now a ${\mathsf{N}}$-cube where an Hamiltonian 
 cycle is removed.
 Let $f$ be the corresponding function.

-The question which remains to solve is:
+The question which remains to be solved is:

 \emph{can we always find $b$ such that $\Gamma_{\{b\}}(f)$ is strongly connected?}
 
 \emph{can we always find $b$ such that $\Gamma_{\{b\}}(f)$ is strongly connected?}
 

-The answer is indeed positive. We furthermore have the following strongest 
-result.
+The answer is indeed positive. Furthermore, we have the following results which are stronger
+than previous ones. 

 \begin{thrm}
 There exists $b \in \Nats$ such that $\Gamma_{\{b\}}(f)$ is complete.
 \end{thrm}
 \begin{thrm}
 There exists $b \in \Nats$ such that $\Gamma_{\{b\}}(f)$ is complete.
 \end{thrm}


@@ -64,10 +62,10 @@ 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.
 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 
+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 
 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 
+$2^{\mathsf{N}-1}$ times whereas the other 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 a uniform output.  
 $\left\lfloor \dfrac{2^{\mathsf{N-1}}}{\mathsf{N}-1} \right\rfloor$ times.
 It is clear that the function that is built from such a code would
 not provide a uniform output.