quelques corrections après remarques de Sylvain

[16dcc.git] / hamilton.tex

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 
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.}
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 
same amount of times,
local properties are wished.
For instance, the locally balanced property is studied in~\cite{Bykov2016} 
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
$b$ times to produce a pseudo random 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 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}
are not globally balanced.
This section thus shows how the non deterministic approach 
presented in~\cite{ZanSup04} has been automatized to provide balanced 
Hamiltonian paths such that, for each subpart, 
the number of switches of each element is as uniform as possible.

\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 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 explicitly shown how to construct such a subsequence.
Finally the authors of~\cite{ZanSup04} have presented 
the \emph{Robinson-Cohn extension} 
algorithm. There rigorous presentation of this one 
have 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 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 
(totally balanced) Gray code. 
What follows shows that this fact is established and first recalls the approach.


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.

\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 
$$
s_{i_1}, u_0, s_{i_2}, u_1, s_{i_3}, u_2, \dots , s_{i_l-1}, u_{l-2}, s_{i_l}, v
$$
where $i_1 = 1$, $i_2 = 2$, and $u_0 = \emptyset$ (the empty sequence).
\item\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\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 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 balance property.

\subsection{Balanced Codes}
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 
$S = s_1, s_2, \dots, s_{2^n}$, $s_i$, $1 \le i \le 2^\mathsf{N}$,
indicates which bit position changes between 
codewords at index $i$ and $i+1$ modulo $2^\mathsf{N}$. 
The \emph{transition count} function  
$\textit{TC}_{\mathsf{N}} : \{1,\dots, \mathsf{N}\} \rightarrow \{0, \ldots, 2^{\mathsf{N}}\}$ 
gives the number of times $i$ occurs in $S$,
\textit{i.e.}, the number of times 
the bit $i$ has been switched in $L$.   

The Gray code is \emph{totally balanced} if $\textit{TC}_{\mathsf{N}}$ 
is constant (and equal to $\frac{2^{\mathsf{N}}}{\mathsf{N}}$).
It is \emph{balanced} if for any two bit indices $i$ and $j$, 
$|\textit{TC}_{\mathsf{N}}(i) - \textit{TC}_{\mathsf{N}}(j)| \le  2$.


   
\begin{xpl}
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
is thus totally balanced.

On the contrary, for the standard $4$-bits Gray code  
$L^{\textit{st}}=0000,0001,0011,0010,0110,0111,0101,0100,1100,$
$1101,1111,1110,1010,1011,1001,1000$,
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
$a_{\mathsf{N}}= 2 \left\lfloor \dfrac{2^{\mathsf{N}}}{2\mathsf{N}} \right\rfloor$. 
There exists then a sequence $l$ in 
step~(\ref{item:nondet}) of the \emph{Robinson-Cohn extension} algorithm
such that all the transition counts $\textit{TC}_{\mathsf{N}}(i)$ 
are $a_{\mathsf{N}}$ or $a_{\mathsf{N}}+2$ 
for any $i$, $1 \le i \le \mathsf{N}$.
\end{thrm}





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 $v=\emptyset$.
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.} $\mathsf{N-2}=2$ 
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 
$U= 1,3,2,3,4,1,4,3,2$, $V = 4$, $W= 3, 1, 2, 1,2, 4$, and $W' = 1, 3, 2, 1,2, 4 $.
Finally, $S_4$ is 
$
2,3,4,1,4,3,2,3,1,4,1,3,2,1,2,4
$ 
such that $\textit{TC}_4(i) = 4$ and the theorem is established for 
odd and even initial values.


For the inductive case, let us first define some variables.
Let $c_{\mathsf{N}}$ (resp. $d_{\mathsf{N}}$) be the number of elements 
whose transition count is exactly $a_{\mathsf{N}}$ (resp $a_{\mathsf{N}} +2$).
These two variables are defined by the system 
 
\[
\left\{
\begin{array}{lcl}
c_{\mathsf{N}} + d_{\mathsf{N}} & = & \mathsf{N} \\
c_{\mathsf{N}}a_{\mathsf{N}} + d_{\mathsf{N}}(a_{\mathsf{N}}+2) & = & 2^{\mathsf{N}} 
\end{array}
\right.
\Leftrightarrow 
\left\{
\begin{array}{lcl}
d_{\mathsf{N}} & = & \dfrac{2^{\mathsf{N}} -\mathsf{N}.a_{\mathsf{N}}}{2} \\
c_{\mathsf{N}} &= &\mathsf{N} -  d_{\mathsf{N}}
\end{array}
\right.
\]

Since $a_{\mathsf{N}}$ is even, $d_{\mathsf{N}}$ is an integer.
Let us first proove 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 disvision
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 occurence 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:
$$
\left\{
\begin{array}{lcl}
3.zi_{\mathsf{N}} + ti_{\mathsf{N}} + 2.bi_{\mathsf{N}} + \textit{TC}_{\mathsf{N}-2}(i) &= &\textit{TC}_{\mathsf{N}}(i) \\
zi_{\mathsf{N}} + ti_{\mathsf{N}}  + bi_{\mathsf{N}} & =& \textit{TC}_{\mathsf{N}-2}(i)
\end{array}
\right.
\qquad 
\Leftrightarrow 
$$

\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 consraint 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 postive 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 a hamiltonian path.





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