Many approaches have been developed to solve the problem of building
-a Gray code in a $\mathsf{N}$ cube~\cite{Robinson:1981:CS,DBLP:journals/combinatorics/BhatS96,ZanSup04,Bykov2016}, according to properties
+a Gray code in a $\mathsf{N}$-cube~\cite{Robinson:1981:CS,DBLP:journals/combinatorics/BhatS96,ZanSup04,Bykov2016}, according to properties
the produced code has to verify.
For instance,~\cite{DBLP:journals/combinatorics/BhatS96,ZanSup04} focus on
balanced Gray codes. In the transition sequence of these codes,
The current context is to provide a function
$f:\Bool^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}}$ by removing a Hamiltonian
-cycle in the $\mathsf{N}$ cube. Such a function is going to be iterated
+cycle in the $\mathsf{N}$-cube. Such a function is going to be iterated
$b$ times to produce a pseudo random number,
\textit{i.e.} a vertex in the
-$\mathsf{N}$ cube.
+$\mathsf{N}$-cube.
Obviously, the number of iterations $b$ has to be sufficiently large
to provide a uniform output distribution.
To reduce the number of iterations, the provided Gray code
The former states that if
$\mathsf{N}$ is a 2-power, a balanced Gray code is always totally balanced.
The latter states that for every $\mathsf{N}$ there
-exists a Gray code such that all transition count numbers are
+exists a Gray code such that all transition count numbers
are 2-powers whose exponents are either equal
or differ from each other by 1.
However, the authors do not prove that the approach allows to build
\end{enumerate}
It has been proven in~\cite{ZanSup04} that
-$S_{\mathsf{N}}$ is transition sequence of a cyclic $\mathsf{N}$-bits Gray code
+$S_{\mathsf{N}}$ is the transition sequence of a cyclic $\mathsf{N}$-bits Gray code
if $S_{\mathsf{N}-2}$ is.
However, the step~(\ref{item:nondet}) is not a constructive
step that precises how to select the subsequences which ensures that
Let now
$L^4=0000, 0010, 0110, 1110, 1111, 0111, 0011, 0001, 0101,$
$0100, 1100, 1101, 1001, 1011, 1010, 1000$
-be a cyclic Gray code. Since $S=2,3,4,1,4,3,2,3,1,4,1,3,2,1,2,4$ $\textit{TC}_4$ is equal to 4 everywhere, this code
+be a cyclic Gray code. Since $S=2,3,4,1,4,3,2,3,1,4,1,3,2,1,2,4$, $\textit{TC}_4$ is equal to 4 everywhere, this code
is thus totally balanced.
On the contrary, for the standard $4$-bits Gray code
\begin{thrm}\label{prop:balanced}
Let $\mathsf{N}$ in $\Nats^*$, and $a_{\mathsf{N}}$ be defined by
-$a_{\mathsf{N}}= 2 \lfloor \dfrac{2^{\mathsf{N}}}{2\mathsf{N}} \rfloor$.
+$a_{\mathsf{N}}= 2 \left\lfloor \dfrac{2^{\mathsf{N}}}{2\mathsf{N}} \right\rfloor$.
There exists then a sequence $l$ in
step~(\ref{item:nondet}) of the \emph{Robinson-Cohn extension} algorithm
such that all the transition counts $\textit{TC}_{\mathsf{N}}(i)$
