integration style IEEE

[16dcc.git] / hamilton.tex

diff --git a/hamilton.tex b/hamilton.tex
index 17d93803a0d523ddadc31cfe930449bc6a996f41..b2c6e54f20363f788ccc99a530be58f74bbb40f2 100644 (file)
--- a/hamilton.tex
+++ b/hamilton.tex
@@ -1,5 +1,5 @@
 Many approaches have been developed to solve the problem of building
 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 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, 


@@ -16,14 +16,15 @@ 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 
  
 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 
 $b$ 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.
 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
-should ideally possess the both balanced and locally balanced properties.
+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}
 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}


@@ -50,7 +51,7 @@ 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 
 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 
 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 


@@ -83,7 +84,7 @@ two elements have been exchanged.
 \end{enumerate} 
 
 It has been proven in~\cite{ZanSup04} that 
 \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 
 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 


@@ -112,15 +113,15 @@ $|\textit{TC}_{\mathsf{N}}(i) - \textit{TC}_{\mathsf{N}}(j)| \le  2$.
 
    
 \begin{xpl}
 
    
 \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  
 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  

-$L^4=0000, 0010, 0110, 1110, 1111, 0111, 0011, 0001, 0101,$
+$L^4=0000, 0010, 0110, 1110, 1111, 0111, 0011,$ $0001, 0101,$

 $0100, 1100, 1101, 1001, 1011, 1010, 1000$
 $0100, 1100, 1101, 1001, 1011, 1010, 1000$

-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  
 is thus totally balanced.
 
 On the contrary, for the standard $4$-bits Gray code  


@@ -133,7 +134,7 @@ the code is neither balanced nor totally balanced.
 
 \begin{thrm}\label{prop:balanced}
 Let $\mathsf{N}$ in $\Nats^*$, and $a_{\mathsf{N}}$ be defined by
 
 \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 $\textit{TC}_{\mathsf{N}}(i)$ 
 There exists then a sequence $l$ in 
 step~(\ref{item:nondet}) of the \emph{Robinson-Cohn extension} algorithm
 such that all the transition counts $\textit{TC}_{\mathsf{N}}(i)$ 


@@ -170,23 +171,21 @@ 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 
  
 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.
 \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 
 \Leftrightarrow 

-\qquad 

 \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.
 \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
 
 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


@@ -282,7 +281,7 @@ a_{\mathsf{N}}+2 \textrm{ if } c_{\mathsf{N}} +1 \le i \le c_{\mathsf{N}} + d_{\
 
 
 We thus  have
 
 
 We thus  have

-$$
+\[

 \begin{array}{rcl} 
 \textit{TC}_{\mathsf{N}}(i) - 2.\textit{TC}_{\mathsf{N}-2}(i) 
 &\ge& 
 \begin{array}{rcl} 
 \textit{TC}_{\mathsf{N}}(i) - 2.\textit{TC}_{\mathsf{N}-2}(i) 
 &\ge& 


@@ -294,9 +293,9 @@ a_{\mathsf{N}} - 2(a_{\mathsf{N}-2}+2) \\
 \frac{2^{\mathsf{N}}-2N}{\mathsf{N}}
 -2 \left( \frac{2^{\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& 

-\dfrac{(\mathsf{N} -2).2^{\mathsf{N}}-2N.2^{\mathsf{N-2}}-6N(N-2)}{\mathsf{N.(N-2)}}\\
+\frac{(\mathsf{N} -2).2^{\mathsf{N}}-2N.2^{\mathsf{N-2}}-6N(N-2)}{\mathsf{N.(N-2)}}\\

 \end{array}
 \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 
 
 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 


@@ -328,3 +327,9 @@ Notice that all such choices lead to a hamiltonian path.
 
 
 
 
 
 

+%%% Local Variables:
+%%% mode: latex
+%%% TeX-master: "main"
+%%% ispell-dictionary: "american"
+%%% mode: flyspell
+%%% End: