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
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}
-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.
\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 ($\frac{1}{n}$),
+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
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?}
-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 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
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 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.
The next section presents how to build balanced Hamiltonian cycles in the
$\mathsf{N}$-cube with the objective to embed them into the
