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 number of transitions of each element must differ
at most by 2.
-This uniformity is a global property on the cycle, \textit{i.e.}
+This uniformity is a global property on the cycle, \textit{i.e.},
a property that is established while traversing the whole cycle.
On the opposite side, when the objective is to follow a subpart
of the Gray code and to switch each element approximately the
and an algorithm that establishes locally balanced Gray codes is given.
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
-$b$ time to produce a pseudo random number, \textit{i.e.} a vertex in the
-$\mathsf{N}$ cube.
+$f:\Bool^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}}$ by removing an Hamiltonian
+cycle in the $\mathsf{N}$-cube. Such a function is going to be iterated
+$b$ times to produce a pseudorandom number,
+\textit{i.e.}, a vertex in the
+$\mathsf{N}$-cube.
Obviously, the number of iterations $b$ has to be sufficiently large
-to provide a uniform output distribution.
-To reduced the number of iterations, the provided Gray code
-should ideally possess the both balanced and locally balanced properties.
+to provide an uniform output distribution.
+To reduce the number of iterations, it can be claimed
+that the provided Gray code
+should ideally possess both balanced and locally balanced properties.
However, none of the two algorithms is compatible with the second one:
balanced Gray codes that are generated by state of the art works~\cite{ZanSup04,DBLP:journals/combinatorics/BhatS96} are not locally balanced. Conversely,
locally balanced Gray codes yielded by Igor Bykov approach~\cite{Bykov2016}
\subsection{Analysis of the Robinson-Cohn extension algorithm}
As far as we know three works,
namely~\cite{Robinson:1981:CS},~\cite{DBLP:journals/combinatorics/BhatS96},
-and~\cite{ZanSup04} have adressed the probem of providing an approach
+and~\cite{ZanSup04} have addressed the problem of providing an approach
to produce balanced gray code.
The authors of~\cite{Robinson:1981:CS} introduced an inductive approach
aiming at producing balanced Gray codes, provided the user gives
a special subsequence of the transition sequence at each induction step.
This work have been strengthened in~\cite{DBLP:journals/combinatorics/BhatS96}
-where the authors have explicitely shown how to construct such a subsequence.
+where the authors have explicitly shown how to construct such a subsequence.
Finally the authors of~\cite{ZanSup04} have presented
the \emph{Robinson-Cohn extension}
-algorithm. There rigourous presentation of this one
-have mainly allowed them to prove two properties.
+algorithm. Their rigorous presentation of this one
+has mainly allowed them to prove two properties.
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
Let be given a $\mathsf{N}-2$-bit Gray code whose transition sequence is
$S_{\mathsf{N}-2}$. What follows is the
\emph{Robinson-Cohn extension} method~\cite{ZanSup04}
-which produces a $n$-bits Gray code.
+which produces a $\mathsf{N}$-bits Gray code.
\begin{enumerate}
\item \label{item:nondet}Let $l$ be an even positive integer. Find
$u_1, u_2, \dots , u_{l-2}, v$ (maybe empty) subsequences of $S_{\mathsf{N}-2}$
such that $S_{\mathsf{N}-2}$ is the concatenation of
$$
-s_{i_1}, u_0, s_{i_2}, u_1, s_{i_3}, u_2, . . . , s_{i_l-1}, u_{l-2}, s_{i_l}, v
+s_{i_1}, u_0, s_{i_2}, u_1, s_{i_3}, u_2, \dots , s_{i_l-1}, u_{l-2}, s_{i_l}, v
$$
where $i_1 = 1$, $i_2 = 2$, and $u_0 = \emptyset$ (the empty sequence).
-\item Replace in $S_{\mathsf{N}-2}$ the sequences $u_0, u_1, u_2, \ldots, u_{l-2}$
+\item\label{item:u'}Replace in $S_{\mathsf{N}-2}$ the sequences $u_0, u_1, u_2, \ldots, u_{l-2}$
by
$\mathsf{N} - 1, u'(u_1,\mathsf{N} - 1, \mathsf{N}) , u'(u_2,\mathsf{N}, \mathsf{N} - 1), u'(u_3,\mathsf{N} - 1,\mathsf{N}), \dots, u'(u_{l-2},\mathsf{N}, \mathsf{N} - 1)$
respectively, where $u'(u,x,y)$ is the sequence $u,x,u^R,y,u$ such that
$u^R$ is $u$ in reversed order.
The obtained sequence is further denoted as $U$.
-\item Construct the sequences $V=v^R,\mathsf{N},v$, $W=\mathsf{N}-1,S_{\mathsf{N}-2},\mathsf{N}$, and let $W'$ be $W$ where the first
+\item\label{item:VW} Construct the sequences $V=v^R,\mathsf{N},v$, $W=\mathsf{N}-1,S_{\mathsf{N}-2},\mathsf{N}$, and let $W'$ be $W$ where the first
two elements have been exchanged.
\item The transition sequence $S_{\mathsf{N}}$ is thus the concatenation $U^R, V, W'$.
\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
yielded Gray code is balanced.
-Next section shows how to choose the sequence $l$ to have the balancy property.
+Next section shows how to choose the sequence $l$ to have the balance property.
\subsection{Balanced Codes}
-Let us first recall how to formalize the balancy property of a Gray code.
+Let us first recall how to formalize the balance property of a Gray code.
Let $L = w_1, w_2, \dots, w_{2^\mathsf{N}}$ be the sequence
of a $\mathsf{N}$-bits cyclic Gray code.
The transition sequence
\begin{xpl}
-Let $L^*=000,100,101,001,011,111,110,010$ be the Gray code that corresponds to
+Let $L^*=000,100,101,001,011,111,$ $110,010$ be the Gray code that corresponds to
the Hamiltonian cycle that has been removed in $f^*$.
Its transition sequence is $S=3,1,3,2,3,1,3,2$ and its transition count function is
$\textit{TC}_3(1)= \textit{TC}_3(2)=2$ and $\textit{TC}_3(3)=4$. Such a Gray code is balanced.
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
+$L^4=0000, 0010, 0110, 1110, 1111, 0111, 0011,$ $0001, 0101,0100,$ $1100, 1101, 1001, 1011, 1010, 1000$
+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
is thus totally balanced.
On the contrary, for the standard $4$-bits Gray code
-$L^{\textit{st}}=0000,0001,0011,0010,0110,0111,0101,0100,1100,$
-$1101,1111,1110,1010,1011,1001,1000$,
+$L^{\textit{st}}=0000,0001,0011,$
+$0010,0110,0111,0101,0100,1100,1101,1111,1110,1010,1011,1001,1000$,
we have $\textit{TC}_4(1)=8$ $\textit{TC}_4(2)=4$ $\textit{TC}_4(3)=\textit{TC}_4(4)=2$ and
the code is neither balanced nor totally balanced.
\end{xpl}
\begin{thrm}\label{prop:balanced}
-Let $\mathsf{N}$ in $\Nats$, and $a_{\mathsf{N}}$ be defined by
-$a_{\mathsf{N}}= 2 \lfloor \dfrac{2^{\mathsf{N}}}{2\mathsf{N}} \rfloor$.
+Let $\mathsf{N}$ in $\Nats^*$, and $a_{\mathsf{N}}$ be defined by
+$a_{\mathsf{N}}= 2 \left\lfloor \dfrac{2^{\mathsf{N}}}{2\mathsf{N}} \right\rfloor$.
There exists then a sequence $l$ in
step~(\ref{item:nondet}) of the \emph{Robinson-Cohn extension} algorithm
such that all the transition counts $\textit{TC}_{\mathsf{N}}(i)$
-The proof is done by induction on $\mathsf{N}$. Let us imadialty verify
-that it is established for both odd and even smallest values, \textit{i.e.}
+The proof is done by induction on $\mathsf{N}$. Let us immediately verify
+that it is established for both odd and even smallest values, \textit{i.e.},
$3$ and $4$.
-For the initial case where $\mathsf{N}=3$, \textit{i.e.} $\mathsf{N-2}=1$ we successively have: $S_1=1,1$, $l=2$, $u_0 = \emptyset$, and $v=\emptyset$.
+For the initial case where $\mathsf{N}=3$, \textit{i.e.}, $\mathsf{N-2}=1$ we successively have: $S_1=1,1$, $l=2$, $u_0 = \emptyset$, and $v=\emptyset$.
Thus again the algorithm successively produces
$U= 1,2,1$, $V = 3$, $W= 2, 1, 1,3$, and $W' = 1,2,1,3$.
Finally, $S_3$ is $1,2,1,3,1,2,1,3$ which obviously verifies the theorem.
- For the initial case where $\mathsf{N}=4$, \textit{i.e.} $\mathsf{N-2}=2$
+ For the initial case where $\mathsf{N}=4$, \textit{i.e.}, $\mathsf{N-2}=2$
we successively have: $S_1=1,2,1,2$, $l=4$,
$u_0,u_1,u_2 = \emptyset,\emptyset,\emptyset$, and $v=\emptyset$.
Thus again the algorithm successively produces
For the inductive case, let us first define some variables.
Let $c_{\mathsf{N}}$ (resp. $d_{\mathsf{N}}$) be the number of elements
-whose transistion count is exactly $a_{\mathsf{N}}$ (resp $a_{\mathsf{N}} +2$).
+whose transition count is exactly $a_{\mathsf{N}}$ (resp $a_{\mathsf{N}} +2$).
These two variables are defined by the system
-$$
+\[
\left\{
\begin{array}{lcl}
c_{\mathsf{N}} + d_{\mathsf{N}} & = & \mathsf{N} \\
c_{\mathsf{N}}a_{\mathsf{N}} + d_{\mathsf{N}}(a_{\mathsf{N}}+2) & = & 2^{\mathsf{N}}
\end{array}
\right.
-\qquad
\Leftrightarrow
-\qquad
\left\{
\begin{array}{lcl}
-d_{\mathsf{N}} & = & \dfrac{2^{\mathsf{N}} -n.a_{\mathsf{N}}}{2} \\
+d_{\mathsf{N}} & = & \dfrac{2^{\mathsf{N}} -\mathsf{N}.a_{\mathsf{N}}}{2} \\
c_{\mathsf{N}} &= &\mathsf{N} - d_{\mathsf{N}}
\end{array}
\right.
+\]
+
+Since $a_{\mathsf{N}}$ is even, $d_{\mathsf{N}}$ is an integer.
+Let us first prove that both $c_{\mathsf{N}}$ and $d_{\mathsf{N}}$ are positive
+integers.
+Let $q_{\mathsf{N}}$ and $r_{\mathsf{N}}$, respectively, be
+the quotient and the remainder in the Euclidean division
+of $2^{\mathsf{N}}$ by $2\mathsf{N}$, \textit{i.e.},
+$2^{\mathsf{N}} = q_{\mathsf{N}}.2\mathsf{N} + r_{\mathsf{N}}$, with $0 \le r_{\mathsf{N}} <2\mathsf{N}$.
+First of all, the integer $r$ is even since $r_{\mathsf{N}} = 2^{\mathsf{N}} - q_{\mathsf{N}}.2\mathsf{N}= 2(2^{\mathsf{N}-1} - q_{\mathsf{N}}.\mathsf{N})$.
+Next, $a_{\mathsf{N}}$ is $\frac{2^{\mathsf{N}}-r_{\mathsf{N}}}{\mathsf{N}}$. Consequently
+$d_{\mathsf{N}}$ is $r_{\mathsf{N}}/2$ and is thus a positive integer s.t.
+$0 \le d_{\mathsf{N}} <\mathsf{N}$.
+The proof for $c_{\mathsf{N}}$ is obvious.
+
+
+For any $i$, $1 \le i \le \mathsf{N}$, let $zi_{\mathsf{N}}$ (resp. $ti_{\mathsf{N}}$ and $bi_{\mathsf{N}}$)
+be the occurrence number of element $i$ in the sequence $u_0, \dots, u_{l-2}$
+(resp. in the sequences $s_{i_1}, \dots , s_{i_l}$ and $v$)
+in step (\ref{item:nondet}) of the algorithm.
+
+Due to the definition of $u'$ in step~(\ref{item:u'}),
+$3.zi_{\mathsf{N}} + ti_{\mathsf{N}}$ is the
+ number of element $i$ in the sequence $U$.
+It is clear that the number of element $i$ in the sequence $V$ is
+$2bi_{\mathsf{N}}$ due to step (\ref{item:VW}).
+We thus have the following system:
+$$
+\left\{
+\begin{array}{lcl}
+3.zi_{\mathsf{N}} + ti_{\mathsf{N}} + 2.bi_{\mathsf{N}} + \textit{TC}_{\mathsf{N}-2}(i) &= &\textit{TC}_{\mathsf{N}}(i) \\
+zi_{\mathsf{N}} + ti_{\mathsf{N}} + bi_{\mathsf{N}} & =& \textit{TC}_{\mathsf{N}-2}(i)
+\end{array}
+\right.
+\qquad
+\Leftrightarrow
$$
-Since $a_{\mathsf{N}$ is even, $d_{\mathsf{N}}$ is defined.
-Both $c_{\mathsf{N}}$ and $d_{\mathsf{N}}$ are obviously positves.
+\begin{equation}
+\left\{
+\begin{array}{lcl}
+zi_{\mathsf{N}} &= &
+\dfrac{\textit{TC}_{\mathsf{N}}(i) - 2.\textit{TC}_{\mathsf{N}-2}(i) - bi_{\mathsf{N}}}{2}\\
+ti_{\mathsf{N}} &= & \textit{TC}_{\mathsf{N}-2}(i)-zi_{\mathsf{N}}-bi_{\mathsf{N}}
+\end{array}
+\right.
+\label{eq:sys:zt1}
+\end{equation}
+
+In this set of 2 equations with 3 unknown variables, let $b_i$ be set with 0.
+In this case, since $\textit{TC}_{\mathsf{N}}$ is even (equal to $a_{\mathsf{N}}$
+or to $a_{\mathsf{N}}+2$), the variable $zi_{\mathsf{N}}$ is thus an integer.
+Let us now prove that the resulting system has always positive integer
+solutions $z_i$, $t_i$, $0 \le z_i, t_i \le \textit{TC}_{\mathsf{N}-2}(i)$
+and s.t. their sum is equal to $\textit{TC}_{\mathsf{N}-2}(i)$.
+This latter constraint is obviously established if the system has a solution.
+We thus have the following system.
+
+
+
+\begin{equation}
+\left\{
+\begin{array}{lcl}
+zi_{\mathsf{N}} &= &
+\dfrac{\textit{TC}_{\mathsf{N}}(i) - 2.\textit{TC}_{\mathsf{N}-2}(i) }{2}\\
+ti_{\mathsf{N}} &= & \textit{TC}_{\mathsf{N}-2}(i)-zi_{\mathsf{N}}
+\end{array}
+\right.
+\label{eq:sys:zt2}
+\end{equation}
+
+The definition of $\textit{TC}_{\mathsf{N}}(i)$ depends on the value of $\mathsf{N}$.
+When $3 \le N \le 7$, values are defined as follows:
+\begin{eqnarray*}
+\textit{TC}_{3} & = & [2,2,4] \\
+\textit{TC}_{5} & = & [6,6,8,6,6] \\
+\textit{TC}_{7} & = & [18,18,20,18,18,18,18] \\
+\\
+\textit{TC}_{4} & = & [4,4,4,4] \\
+\textit{TC}_{6} & = & [10,10,10,10,12,12] \\
+\end{eqnarray*}
+It is not hard to verify that all these instanciations verify the aformentioned contraints.
+
+When $N \ge 8$, $\textit{TC}_{\mathsf{N}}(i)$ is defined as follows:
+\begin{equation}
+\textit{TC}_{\mathsf{N}}(i) = \left\{
+\begin{array}{l}
+a_{\mathsf{N}} \textrm{ if } 1 \le i \le c_{\mathsf{N}} \\
+a_{\mathsf{N}}+2 \textrm{ if } c_{\mathsf{N}} +1 \le i \le c_{\mathsf{N}} + d_{\mathsf{N}}
+\end{array}
+\right.
+\label{eq:TCN:def}
+\end{equation}
+
+
+We thus have
+\[
+\begin{array}{rcl}
+\textit{TC}_{\mathsf{N}}(i) - 2.\textit{TC}_{\mathsf{N}-2}(i)
+&\ge&
+a_{\mathsf{N}} - 2(a_{\mathsf{N}-2}+2) \\
+&\ge&
+\frac{2^{\mathsf{N}}-r_{\mathsf{N}}}{\mathsf{N}}
+-2 \left( \frac{2^{\mathsf{N-2}}-r_{\mathsf{N-2}}}{\mathsf{N-2}}+2\right)\\
+&\ge&
+\frac{2^{\mathsf{N}}-2N}{\mathsf{N}}
+-2 \left( \frac{2^{\mathsf{N-2}}}{\mathsf{N-2}}+2\right)\\
+&\ge&
+\frac{(\mathsf{N} -2).2^{\mathsf{N}}-2N.2^{\mathsf{N-2}}-6N(N-2)}{\mathsf{N.(N-2)}}\\
+\end{array}
+\]
+
+A simple variation study of the function $t:\R \rightarrow \R$ such that
+$x \mapsto t(x) = (x -2).2^{x}-2x.2^{x-2}-6x(x-2)$ shows that
+its derivative is strictly positive if $x \ge 6$ and $t(8)=224$.
+The integer $\textit{TC}_{\mathsf{N}}(i) - 2.\textit{TC}_{\mathsf{N}-2}(i)$ is thus positive
+for any $\mathsf{N} \ge 8$ and the proof is established.
+
+
+% \begin{table}[ht]
+% $$
+% \begin{array}{|l|l|l|l|l|l|}
+% \hline
+% \mathsf{N} & 3 & 4 & 5 & 6 & 7\\
+% \hline
+% a_{\mathsf{N}} & 2 & 4 & 6 & 10 & 18\\
+% \hline
+% \end{array}
+% $$
+% \label{tab:an}
+% \caption{First values of $a_{\mathsf{N}}$}
+% \end{table}
+
+For each element $i$, we are then left to choose $zi_{\mathsf{N}}$ positions
+among $\textit{TC}_{\mathsf{N}}(i)$, which leads to
+${\textit{TC}_{\mathsf{N}}(i) \choose zi_{\mathsf{N}} }$ possibilities.
+Notice that all such choices lead to an Hamiltonian path.
+
+
+
-\subsection{Toward a local uniform distribution of switches}
+%%% Local Variables:
+%%% mode: latex
+%%% TeX-master: "main"
+%%% ispell-dictionary: "american"
+%%% mode: flyspell
+%%% End:
