Add \section{Travaux en relation}

[these_qian.git] / ThefamilyofCIPRNG.tex

\chapter{The family of CIs PRNG}
\label{ThefamilyofCIPRNG}
\minitoc
\section{Mapping matrix}

Chaotic iterations introduced above can be described by using the mapping matrix defined bellow.

\begin{definition}
Let $f:\mathds{B}^{\mathsf{N}}\longrightarrow \mathds{B}^{\mathsf{N}}$
be an iteration function, then its associated \emph{mapping matrix}
$\mathsf{f}$ is the matrix of size $\mathsf{N} \times 2^\mathsf{N}$ whose element  $\mathsf{f}_{p,q}$ is the integer having the following binary decomposition:  $q_\mathsf{N}, \hdots, q_{\mathsf{N}-p}, f(q)_{\mathsf{N}-p+1}, q_{\mathsf{N}-p+2}, \hdots, q_1  $, where $q_i$ (resp. $f(q)_i$) is the $i-$th binary digit of $q$ (resp. of $f(q)$).
%$x_1, \hdots, x_\mathsf{N}$, where $x_i$ that is:
%\begin{itemize}
%\item $x_i$ is the $p-$th binary digit of $f(q)$, if $i=p$,
%\item the $i-$th binary digit of $q$, else.
%\end{itemize}
%$$\mathsf{f} =
%\left(
%\begin{array}{cccc}
%f(0)_1x_2... x_\mathsf{N}       &f(1)_1x_2...x_\mathsf{N}       &\ldots &f(2^\mathsf{N})_1 x_2...x_\mathsf{N} \\
%x_1f(0)_2...x_\mathsf{N}        &x_1f(1)_2...x_\mathsf{N}       &\ldots &x_1f(2^\mathsf{N})_2...x_\mathsf{N} \\
%\vdots                          &\vdots                         &\ddots~&\vdots~~~          ~~~~\\
%x_1x_2... f(0)_\mathsf{N}       &x_1x_2... f(1)_\mathsf{N}      &\ldots &x_1x_2...
%f(2^\mathsf{N})_\mathsf{N} \\
%\end{array}
%\right)
%$$
%where the value that lies in the $p$-th row and the $q$-th column of the
%matrix $\mathsf{f}$ is referred to $\mathsf{f}_{p,q}$.
\end{definition}

The relation between $\mathsf{f}$ and chaotic iterations of $f$ can be
understood as follows. If the current state of the system is $q$ and the
strategy is $p$, then the next state (under the chaotic iterations of $f
$) will be $\mathsf{f}_{p,q}$. Finally, the vector $\mathcal{F}(f)=(f(0),f(1),\ldots,f(2^{\mathsf{N}}-1)) \in \llbracket 0 ; 2^{\mathsf{N}}-1 \rrbracket^{2^{\mathsf{N}}}$ is called \emph{vector of images}.
An example is shown for the vectorial Boolean negation $f_0 (x_1, \hdots, x_\mathsf{N}) = (\overline{x_1}, \hdots, \overline{x_\mathsf{N}})$ in Table~
\ref{negation_output}.
\begin{table*}[!t]
\renewcommand{\arraystretch}{1.2}
\caption{The matrix $\mathsf{f}$ associated to $f_0$}
\label{negation_output}
\centering
\begin{tabular}{c|c@{}c@{}c@{}c@{}c@{}c@{}c@{}c@{}c@{}c@{}c@{}c@{}c@{}c@{}c@{}c}%\hline
\backslashbox{p}{q} &~~0 & ~~1~~ & ~~2~~ & ~~3~~ & ~~4~~ & ~~5~~& ~~6~~ & ~~7~~ &~~ 8~~ & ~~9~~ & ~~10~~ & ~~11~~
& ~~12~~ & ~~13~~ & ~~14~~ &15 \\ \hline
 $(f_0(q)_1,q_2,q_3,q_4)$ &\multicolumn{1}{@{}r@{}}{\multirow{4}*{$\left(
\begin{array}{@{}c@{}}
8  \\
4 \\
2\\
1\\
\end{array}
\right.$}}&9&10&11&12&13&14&15&0&1&2&3&4&5&6&
\multicolumn{1}{@{}l@{}}{\multirow{4}*{$\left.
\begin{array}{@{}c@{}}
7  \\
11 \\
13\\
14\\
\end{array}
\right)$
}}\\
$(q_1,f_0(q)_2,q_3,q_4)$&&5&6&7&0&1&2&3&12&13&14&15&8&9&10& \\
$(q_1,q_2,f_0(q)_3,q_4)$&&3&0&1&6&7&4&5&10&11&8&9&14&15&12& \\
$(q_1,q_2,q_3,f_0(q)_4)$& &0&3&2&5&4&7&6&9&8&11&10&13&12&15&\\\hline
$\mathcal{F}(f_0)$ &(15,&14,&13,&12,&11,&10,&9,&8,&7,&6,&5,&4,&3,&2,&1,&0)\\
\hline
\end{tabular}
\end{table*}
\section{Characterizing and Computing Functions for PRNG}\label{sec:instantiating}
This section presents other functions that theoretically could replace the 
negation function $\neg$ in the previous algorithms. 

In this algorithm and from the graph point of view, 
iterating the function $G_f$ from a configuration $x^0$
and according to a strategy $(S^t)^{t \in \mathds{N}}$  
consists in traversing the directed iteration graph $\Gamma(f)$
from a vertex $x^0$ following the edge labelled with $S^0$, $S^1$, \ldots
Obviously, if some vertices cannot be reached from other ones,
their labels expressed as numbers cannot be output by the generator.
The \emph{Strongly connected component of $\Gamma(f)$}
(\textit{i.e.}, when there is a path from each vertex to every other one),
denoted by SCC in the following~\cite{ita09},
is then a necessary condition for the function $f$.   
The following result shows this condition is sufficient to make 
iterations of $G_f$ chaotic.

\begin{theorem}[Theorem III.6, p. 91 in~\cite{GuyeuxThese10}]
\label{Thchaotiques}  
Let  $f$ be a function from $\mathds{B}^{n}$ to  $\mathds{B}^{n}$. Then 
$G_f$ is chaotic  according to  Devaney iff the graph
$\Gamma(f)$ is strongly connected.
\end{theorem}

Any function such that the graph $\Gamma(f)$ is strongly connected
is then a candidate for being iterated in $G_f$
for pseudo random number generating. 
Thus, let us show how to compute a map $f$ 
with a strongly connected graph of iterations $\Gamma(f)$.

We first consider the negation function $\neg$. The iteration graph 
$\Gamma(\neg)$ is obviously strongly connected:
since each configuration $(x_1,\ldots, x_n)$ may reach one of its $n$ neighbors,
there is then a bit by bit path from any 
$(x_1,\ldots, x_n)$ to any $(x'_1,\ldots, x'_n)$.
Let then $\Gamma$ be a graph, initialized with $\Gamma(\neg)$, 
the algorithm iteratively does the two following stages: 
\begin{enumerate}
\item select randomly an edge of the current iteration graph $\Gamma$ and
\item check whether the current iteration graph without that edge 
  remains strongly connected (by a Tarjan algorithm~\cite{Tarjanscc72}, for instance). In the positive case the edge is removed from $G$,
\end{enumerate}
until a rate $r$ of removed edges is greater
than a threshold given by the user.

Formally, if $r$ is close to $0\%$ (\textit{i.e.}, few edges are removed), 
there should remain about $n\times 2^n$ edges (let us recall that $2^n$ is the amount of nodes).
In the opposite case, if $r$ is close to $100\%$, there are left about $2^n$
edges.
In all the cases, this step returns the last graph $\Gamma$ 
that is strongly connected.  
It is not then obvious to return the function $f$ whose iteration
graph is $\Gamma$.

However, such an approach suffers from generating many functions with similar
behavior due to the similarity of their graph.    More formally, let us recall
the graph isomorphism definition that resolves this issue. 
Two directed graphs $\Gamma_1$ and $\Gamma_2$
 are \emph{isomorphic} 
if there exists a permutation $p$ from the vertices of
$\Gamma_1$ to the vertices of $\Gamma_2$ such that
there is an arc from vertex $u$ to vertex  $v$  in $\Gamma_1$ 
iff there is an arc from vertex $p(u)$ to vertex  $p(v)$  in 
$\Gamma_2$.


Then, let $f$ be a function, $\Gamma(f)$ 
be its iteration graph, and $p$ 
be a permutation of vertices of $\Gamma(f)$. 
Since $p(\Gamma(f))$ and $\Gamma(f)$ are isomorphic,   
then iterating  $f$ 
(\textit{i.e.}, traversing $\Gamma(f)$) from the initial configuration $c$
amounts to iterating the function whose iteration graph is $p(\Gamma(f))$
from the configuration $p(c)$.
Graph isomorphism being an equivalence relation, the sequel only 
consider the quotient set of functions with this relation over their graph.
In other words, two functions are distinct if and only if their iteration
graph are not isomorphic.



\begin{table}
\centering
\begin{tabular}{|c|c|c|}
\hline
Function $f$ & $f(x)$, for $x$ in $(0,1,2,\hdots,15)$ & Rate\\ 
\hline
$\neg$&(15,14,13,12,11,10,9,8,7,6,5,4,3,2,1,0)&0\%\\
\hline
$\textcircled{a}$&(15,14,13,12,11,10,9,8,7,6,7,4,3,2,1,0)&2.1\%\\
\hline
$\textcircled{b}$&(14,15,13,12,11,10,9,8,7,6,5,4,3,2,1,0)&4.1\%\\
\hline
$\textcircled{c}$&(15,14,13,12,11,10,9,8,7,7,5,12,3,0,1,0)&6.25\%\\
\hline
$\textcircled{d}$&(14,15,13,12,9,10,11,0,7,2,5,4,3,6,1,8)&16.7\%\\
\hline
$\textcircled{e}$&(11,2,13,12,11,14,9,8,7,14,5,4,1,2,1,9)&16.7\%\\
\hline
$\textcircled{f}$&(13,10,15,12,3,14,9,8,6,7,4,5,11,2,1,0)&20.9\%\\
\hline
$\textcircled{g}$&(13,7,13,10,11,10,1,10,7,14,4,4,2,2,1,0)&20.9\%\\
\hline
$\textcircled{h}$&(7,12,14,12,11,4,1,13,4,4,15,6,8,3,15,2)&50\%\\
\hline
$\textcircled{i}$&(12,0,6,4,14,15,7,15,11,1,14,2,7,4,7,9)&75\%\\
\hline
\end{tabular}
\caption{Functions with SCC graph of iterations\label{table:nc}}
\end{table}

Table~\ref{table:nc} presents generated functions 
that have been ordered by the rate of removed edges in their
graph of iterations compared to the iteration graph $\Gamma(\neg)$
of the boolean negation function $\neg$.

For instance let us consider  the function $\textcircled{g}$ from $\mathds{B}^4$ to $\mathds{B}^4$
defined by the following images: 
$[13,7,13,10,11,10,1,10,7,14,4,4,2,2,1,0]$.
In other words,  the image of $3 ~(0011)$ by $\textcircled{g}$ is $10 ~(1010)$: it is obtained
as  the  binary  value  of  the  fourth element  in  the  second  list
(namely~10).  It  is not  hard to verify  that $\Gamma(\textcircled{d})$ is  SCC.
Next section gives practical evaluations of these functions.

\section{Modifying the PRNG Algorithm}\label{sec:modif}
A coarse attempt could directly embed each function of table~\ref{table:nc}  
in the $\textit{iterate\_G}$ function defined in Algorithm~\ref{algo:it}.


\begin{algorithm}
\textbf{Input:} a function $f$, a PRNG $r$, 
  an iterations number $k$, a binary number $x^0$ ($n$ bits)\\
\textbf{Output:} a binary number $x$ ($n$ bits)
\begin{algorithmic}[1]
\STATE$x\leftarrow x^0$\;
\STATE S = $\textit{sample}(r,k,n)$\;
\FOR{$i=0,\dots,k-1$}
{
\STATE$s \leftarrow S[i]$\;
\STATE $x\leftarrow  F_f(s,x) $\;
}\ENDFOR
\STATE return $x$\;
\medskip
\caption{The \textit{iterate\_G} function. }
\label{algo:it}
\end{algorithmic}
\end{algorithm}

Let us show the drawbacks of this approach on a more simpler example.

Let us consider for instance $n$ is two, the negation function on $\mathds{B}^2$, and
the function $f$ defined by the list $[1,3,0,2]$ (i.e., $f(0,0) = (0,1), f(0,1) = (1,1), f(1,0) = (0,0),$ and $f(1,1)=(1,0)$) whose iterations graphs are represented 
in Fig.~\ref{fig:xplgraph}.
The two graphs are strongly connected and thus the vectorial negation function 
should theoretically  be replaced by the function $f$.



\begin{figure}[ht]
  \centering
  \subfigure[Negation]{
    \includegraphics[width=1.33cm]{images/neg2.eps}
    \label{fig:comp:n}
  }
  \subfigure[$(1,3,0,2)$]{
    \includegraphics[width=1.62cm]{images/f2.eps}
    \label{fig:comp:f}
  }
  \caption{Graphs of Iterations}
    \label{fig:xplgraph}
\end{figure}

%In what follows, we suppose that 
In the graph of iterations $\Gamma({\neg})$ (Fig.~\ref{fig:comp:n}), 
let us compute the probability $P^t_{\neg}(X)$ to reach the node $X$ in $t$ iterations 
from the node 00. Let $X_0$, $X_1$, $X_2$, $X_3$ be the nodes   
$00$, $01$, $10$ and $11$.
For $i\in \llbracket 0,3 \rrbracket$,  $P^1_{\neg}(X_i)$,  are respectively equal to 
0.0, 0.5, 0.0, 0.5. In two iterations   $P^2_{\neg}(X_i)$
are 0.5, 0.0, 0.5, 0.0.
It is obvious to establish that we have 
$P^{2t}(X_i) = P^{0}(X_i)$ and $P^{2t+1}(X_i) = P^{1}(X_i)$ for any $t\in \mathds{N}$.
Then in $k$ or $k+1$ iterations all these probabilities are equal to 0.25.  

Let us apply a similar reasoning for the function $f$ defined by $[1,3,0,2]$.
In its iterations graph $\Gamma(f)$ (Fig.~\ref{fig:comp:f}),
and with $X_i$ defined as above,
the probabilities $P^1_{f}(X_i)$ to reach the node $X_i$ 
in one iteration  from the node 00
are respectively equal to 
0.5, 0.5, 0.0, 0.0.  
Next, probabilities  $P^2_{f}(X)$  are 0.25, 0.5, 0.25, 0.0. 
Next, $P^3_{f}(X)$  are 0.125, 0.375, 0.375, 0.125.
For each iteration, we compute the average deviation rate $R^t$ 
with 0.25 as follows.
$$
R^t= \dfrac{ \Sigma_{i=0}^3 \mid P^t_{f}(X_i)-0.25 \mid}
{4}.
$$  
The higher is this rate, the less the generator may uniformly reach any $X_i$ from $00$.
For this example, it is necessary to iterate 14 times in order to
observe a deviation from 0.25 less 
than 1\%. 
A similar reasoning has been applied for all the functions listed in Table~\ref{table:nc}.
The table~\ref{tab:dev} summarizes their deviations with uniform distribution and gives the 
smallest iterations number the smallest deviation has been obtained. 

\begin{table}
\centering
\begin{tabular}{|c|r|r|}
\hline
Name & Deviation & Suff. number of it. \\

\hline
$\textcircled{a}$ &  8.1\% & 167 \\
\hline
$\textcircled{b}$ &  1\%  & 105 \\
\hline
$\textcircled{c}$ &  18\% & 58 \\
\hline
$\textcircled{d}$ &  1\% & 22  \\
\hline
$\textcircled{e}$ &  24\% & 19 \\
\hline
$\textcircled{f}$ &  1\%  & 14 \\
\hline
$\textcircled{g}$ &  20\% &  6 \\
\hline
$\textcircled{h}$ & 45.3\% & 7 \\
\hline
$\textcircled{i}$ & 53.2\%& 14 \\
\hline
\end{tabular}
\caption{Deviation with Uniform Distribution \label{tab:dev}}
\end{table}




With that material we present in Algorithm~\ref{CIs Algorithm}
the method that allows to take any chaotic function as 
the core of a PRNG.
Among the parameters, it takes the number $b$ of minimal iterations
that have to be executed to get a uniform like 
distribution. For our experiments $b$ is set with the value 
given in the third column of Table~\ref{tab:dev}.
\begin{algorithm}
\textbf{Input:} a function $f$, an iteration number $b$, an initial state $x^0$ ($n$ bits)\\
\textbf{Output:} a state $x$ ($n$ bits)
\begin{algorithmic}[1]
\STATE$x\leftarrow x^0$\;
\STATE$k\leftarrow b + (\textit{XORshift}() \mod 2)$\;
\FOR{$i=0,\dots,k-1$}
{
\STATE$s\leftarrow{\textit{XORshift}() \mod n}$\;
\STATE$x\leftarrow{F_f(s,x)}$\;
}\ENDFOR
\STATE return $x$\;
\medskip
\caption{modified PRNG with various functions}
\label{CIs Algorithm}
\end{algorithmic}
\end{algorithm}

% \begin{algorithm}
% \KwIn{a function $f$, an iteration number $b$, an initial state $x^0$ ($n$ bits)}
% \KwOut{a state $x$ ($n$ bits)}
% $x\leftarrow x^0$\;
% $k\leftarrow b + (\textit{XORshift}() \mod 2)$\;
% \For{$i=0,\dots,k-1$}
% {
% $s\leftarrow{\textit{XORshift}() \mod n}$\;
% $x\leftarrow{F_f(s,x)}$\;
% }
% return $x$\;
% \medskip
% \caption{modified PRNG with various functions}
% \label{CIs Algorithm}
% \end{algorithm}


Compared to the algorithm~\ref{Chaotic iteration1}
parameters of this one are 
the function $f$ to embed and 
the smallest number of time steps $G_f$ is iterated. 
First, the number of iterations is either $b$ or $b+1$ depending on the 
value of the \textit{XORshift} output (if the next value .
Next, a loop that iterates $G_f$ is executed.

In this example, $n$ and $b$ are equal to $4$ for easy understanding.
The initial state of the system $x^0$ can be seeded by the decimal part of the current time.
For example, the current time in seconds since the Epoch is 1237632934.484088,
so $t = 484088$. $x^0 = t \mod 16 $ in binary digits, then $x^0 = 0100$.
$m$ and $S$ can now be computed from \textit{XORshift}.
\begin{itemize}
\item $f$ = [14,15,13,12,11,10,9,8,7,6,5,4,3,2,1,0] 
\item $k$ = 4, 5, 4,\ldots
\item $s$ = 2,  4,  2,  3, ,  4,  1,  1,  4,  2, ,  0,  2,  3,  1,\ldots
\end{itemize}
Chaotic iterations are done with initial state $x^0$,
the mapping function $f$, and strategy $s^1$, $s^2$\ldots
The result is presented in Table \ref{table application example}. 
Let us recall that sequence $k$ gives the states $x^t$ to return: $x^4, x^{4+5}, x^{4+5+4}$\ldots
Successive stages are detailed in Table~\ref{table application example}.

\begin{table*}[t]
%\renewcommand{\arraystretch}{1.3}
\centering
\begin{tabular}{|c|c@{}c@{}c@{}c@{}c|c@{}c@{}c@{}c@{}c@{}c|c@{}c@{}c@{}c@{}c|}
\hline\hline
$k$ &  \multicolumn{5}{|c|}{4} &  \multicolumn{6}{|c|}{5} & \multicolumn{5}{|c|}{4}\\
\hline
$s$ & 2 & 4 & 2 & 3 & & 4 & 1 & 1 & 4 & 2 & & 0 & 2 & 3 & 1 &  \\ \hline
% [14,15,13,12,11,10,9,8,7,6,5,4,3,2,1,0] 
&$f(4)$&$f(0)$&$f(0)$&$f(4)$ &  &$f(6)$ &$f(7)$ &$f(15)$ &$f(7)$ &$f(7)$ & 
&$f(2)$ &$f(0)$ & $f(4)$& $f(6)$&  \\
%1ere ligne
\multirow{4}{*}{$f$} 
 & 1& 1& 1& 1&
 & 1& \textbf{1} & \textbf{0} &1 &1 &
 &1 & 1& 1&\textbf{1} &
  \\
%2eme ligne
 & \textbf{0} & 1& \textbf{1} & 0 &
 &0 &0& 0&0 & \textbf{0}&
 &1 &\textbf{1} & 0&0 & \\
%3eme ligne
 &1 & 1& 1& \textbf{1}&
 &0 &0 &0 &0 &0 &
& \textbf{0} & 1& \textbf{1} & 0 &
  \\
% 4eme ligne
 &1 &\textbf{0} &0 &1 &
 &\textbf{1} &0 &0 &\textbf{0} &0 &
 &1 &0 & 1&1 &
 \\\hline
$x^{0}$ & & & & & $x^{4}$ & & & & & & $x^{9}$ & & & & & $x^{13}$  \\
4 &0 &0 &4 &6&6 &7 &15 &7 &7 &7 &2&0  &4 &6 &14 &14   \\ 
& & &  & & && & & & & & &  & & &   \\
%1ere ligne
0 & & & & &
0 & & $\xrightarrow{1} 1$ & $\xrightarrow{1} 0$ & & &
0 & & & & $\xrightarrow{1} 1$ &
1  \\
%2eme ligne
1 & $\xrightarrow{2} 0$ & & $\xrightarrow{2} 1$ &  &
1 & & & & & $\xrightarrow{2} 0$ &
0 & & $\xrightarrow{2} 1$ & & & 
1 \\
%3eme ligne
0 & & & & $\xrightarrow{3} 1$ &
1 & & & & & &
1 & $\xrightarrow{3} 0$ & & $\xrightarrow{3} 1$ &  &
1 \\
% 4eme ligne
0 & & $\xrightarrow{4} 0$ & & &
0 &$\xrightarrow{4} 1$ & & & $\xrightarrow{4} 0$& &
0 & & & & &
0 \\
\hline\hline
\end{tabular}
% Binary Output: $x_1^{0}x_2^{0}x_3^{0}x_4^{0}x_5^{0}x_1^{4}x_2^{4}x_3^{4}x_4^{4}x_5^{4}x_1^{9}x_2^{9}x_3^{9}x_4^{9}x_5^{9}x_1^{13}x_2^{13}... = 0100011001110001...$
% Integer Output:
% $x^{0},x^{0},x^{4},x^{6},x^{8}... = 6,7,1...$
\caption{Application example}
\label{table application example}
\end{table*}

% So, in this example, the generated binary digits are: 0100011001110001... Or the integers are: 6, 2, 14\ldots




To illustrate the deviation, Figures~\ref{fig:f5} and~\ref{fig:f6} represent
the simulation outputs of 5120 executions with  $b$ equal to $40$
for $\textcircled{e}$ and $\textcircled{f}$ respectively.
In these two figures, the point $(x,y,z)$ can be understood as follows.
$z$ is the number of times the value $x$ has been succedded by the value $y$ in the 
considered generator.
These two figures explicitly confirm that outputs of functions $\textcircled{f}$ are 
more uniform that these of the function $\textcircled{e}$. 
In the former each number $x$ reaches about 20 times each number $y$ whereas
in the latter, results vary from 10 to more that 50. 

\begin{figure}
\centering
 \subfigure[Function $\textcircled{e}$]{
   \includegraphics[width=10cm]{images/f5.eps}
   \label{fig:f5}
 }
$\qquad$
\subfigure[Function $\textcircled{f}$]{
  \includegraphics[width=10cm]{images/f6.eps}
     \label{fig:f6}

}
\caption{Repartition of function outputs.} \label{fig:fs}
\end{figure}




\section{Experiments}


\begin{sidewaystable}
\renewcommand{\arraystretch}{1.3}
\caption{NIST SP 800-22 test results ($\mathbb{P}_T$)}
\label{The passing rate3}
\centering
  \begin{tabular}{|l||c|c|c|c|c|c|c|c|c|}
    \hline
Method &$\textcircled{a}$& $\textcircled{b}$ &  $\textcircled{c}$ & $\textcircled{d}$ & $\textcircled{e}$ & $\textcircled{f}$ & $\textcircled{g}$ & $\textcircled{h}$ & $\textcircled{i}$\\ \hline\hline

%$w^{j}$ & $\{1,..,8\}$ & $\{1,..,8\}$ & $\{1,..,8\}$ & $\{1,..,5\}$ & $\{1,..,5\}$ &$\{1,..,5\}$ \\ \hline \hline
Frequency (Monobit) Test                        &0.00000 &  0.45593 &  0.00000 &  0.38382 &  0.00000 &  0.61630 &  0.00000 &  0.00000 &    0.00000 \\ \hline
Frequency Test within a Block                   &0.00000 &  0.55442 &  0.00000 &  0.03517 &  0.00000 &  0.73991 &  0.00000 &  0.00000 &    0.00000 \\ \hline
Cumulative Sums (Cusum) Test*                   &0.00000 &  0.56521 &  0.00000 &  0.19992 &  0.00000 &  0.70923 &  0.00000 &  0.00000 &    0.00000 \\ \hline
Runs Test                                       &0.00000 &  0.59554 &  0.00000 &  0.14532 &  0.00000 &  0.24928 &  0.00000 &  0.00000 &    0.00000 \\ \hline                            
Test for the Longest Run of Ones in a Block     &0.20226 &  0.17186 &  0.00000 &  0.38382 &  0.00000 &  0.40119 &  0.00000 &  0.00000 &    0.00000 \\ \hline
Binary Matrix Rank Test                         &0.63711 &  0.69931 &  0.05194 &  0.16260 &  0.79813 &  0.03292 &  0.85138 &  0.12962 &    0.07571 \\ \hline
Discrete Fourier Transform (Spectral) Test      &0.00009 &  0.09657 &  0.00000 &  0.93571 &  0.00000 &  0.93571 &  0.00000 &  0.00000 &    0.00000 \\ \hline
Non-overlapping Template Matching Test*         &0.12009 &  0.52365 &  0.05426 &  0.50382 &  0.02628 &  0.50326 &  0.06479 &  0.00854 &    0.00927 \\ \hline
Overlapping Template Matching Test              &0.00000 &  0.73991 &  0.00000 &  0.55442 &  0.00000 &  0.45593 &  0.00000 &  0.00000 &    0.00000 \\ \hline
Universal Statistical Test                      &0.00000 &  0.71974 &  0.00000 &  0.77918 &  0.00000 &  0.47498 &  0.00000 &  0.00000 &    0.00000 \\ \hline
Approximate Entropy Test                        &0.00000 &  0.10252 &  0.00000 &  0.28966 &  0.00000 &  0.14532 &  0.00000 &  0.00000 &    0.00000\\ \hline
Random Excursions Test*                         &NaN &  0.58707 &NaN &  0.41184 &NaN &  0.25174 &NaN &NaN &NaN \\ \hline
Random Excursions Variant Test*                 &NaN &  0.32978 &NaN &  0.57832 &NaN &  0.31028 &NaN &NaN &NaN \\ \hline
Serial Test* (m=10)                             &0.11840 &  0.95107 &  0.01347 &  0.57271 &  0.00000 &  0.82837 &  0.00000 &  0.00000 &    0.00000 \\ \hline
Linear Complexity Test                          & 0.91141 &  0.43727 &  0.59554 &  0.43727 &  0.55442 &  0.43727 &  0.59554 &  0.69931 &    0.08558 \\ \hline
Success                                         &5/15&15/15&4/15&15/15&3/15&15/15&3/15&3/15&3/15  \\ \hline
Computational time                              &66.0507&47.0466&32.6808&21.6940&20.5759&19.2052&16.4945&16.8846&19.0256\\ \hline
  \end{tabular}
\end{sidewaystable}


Table~\ref{The  passing rate3} shows $\mathbb{P}_T$  of the  sequences  based on
discrete chaotic  iterations using  different ''iteration'' functions.  If there
are  at least  two statistical  values in  a test,  the test  is marked  with an
asterisk  and the  average value  is  computed to  characterize the  statistical
values. Here,  NaN means  a warning that  test is  not applicable because  of an
insufficient  number of cycles.  Time (in  seconds) is  related to  the duration
needed by each  algorithm to generate a $10^8$ bits long  sequence. The test has
been conducted using  the same computer and compiler  with the same optimization
settings for both algorithms, in order to make the test as fair as possible.

Firstly, the computational  time in  seconds has increased due to the
growth of the sufficient iteration numbers, as precised  in Table~\ref{tab:dev}.
For  instance, the fastest generator is $\textcircled{g}$ since each new 
number generation only requires 6 iterations.
Next, concerning the  NIST tests results, 
best situations are given by $\textcircled{b}$,  $\textcircled{d}$ and
$\textcircled{f}$. In the opposite, it can be observed that among the 15 tests,
less than 5 ones are a successful for  other functions. 
Thus,  we can draw a  conclusion that, $\textcircled{b}$, $\textcircled{d}$,
and $\textcircled{f}$ are qualified to be good PRNGs with chaotic property.
NIST tests results are not a surprise:
$\textcircled{b}$,  $\textcircled{d}$, and $\textcircled{f}$ have indeed a deviation less than 1\% with 
the uniform distribution as already precised in Table~\ref{tab:dev}.
The rate of removed edge in the graph $\Gamma(\neg)$ is then not a pertinent
criteria compared to the deviation with the uniform distribution property:
the function $\textcircled{a}$ whose graph $\Gamma(\textcircled{a})$ is $\Gamma(\neg)$ without the
edge $1010 \rightarrow 1000$ (\textit{i.e.}, with only one edge less than
$\Gamma(\neg)$) has dramatic results compared to the function 
$\textcircled{f}$ with many edges less.

Let us then try to give a characterization of convenient function. 
Thanks to a comparison with the other functions, we notice that 
$\textcircled{b}$,  $\textcircled{d}$, and $\textcircled{f}$ are composed of all the elements of
$\llbracket 0;15  \rrbracket$.
It means that $\textcircled{b}$,  $\textcircled{d}$, and $\textcircled{f}$, and even the vectorial  boolean
negation function are arrangements  of 
$\llbracket  0;2^n \rrbracket$  ($n=4$ in  this article)  into a
particular order.

\section{Description of the selection scheme}
\label{section:description}

In this section is explained how the iteration function $f_0$ can be replaced without losing chaos and randomness.

\subsection{Strong connectivity and chaos}


Let $f:\mathds{B}^\mathsf{N} \rightarrow \mathds{B}^\mathsf{N}$. Its
{\emph{iteration graph}} $\Gamma(f)$ is the directed graph defined as follows. 
The set of vertices is
$\mathds{B}^\mathsf{N}$, and $\forall x\in\mathds{B}^\mathsf{N}, \forall i\in \llbracket1;\mathsf{N}\rrbracket$,
$\Gamma(f)$ contains an arc labeled $i$ from $x = (x_1, \hdots, x_\mathsf{N})$ to $(x_1, \hdots, x_{i-1}, f(x)_i, x_{i+1}, \hdots, x_\mathsf{N})$. 
We have proven in~\cite{GuyeuxThese10} that:


\begin{theorem}
\label{IC_chaotiques}
The $CI_f(PRNG1,PRNG2)$ generator is chaotic according to Devaney if and only if the graph $\Gamma(f)$ is strongly connected.
\end{theorem}

Theorem \ref{IC_chaotiques} only focus on the topological chaos property.
However, it is possible to find chaotic sequences with bad statistical properties, in particular when the iteration function is unbalanced.
%
% \subsection{The balance property}
% \label{The Rule For Choosing Balance Mapping}
% \begin{tiny}
% \begin{table*}[!t]
% \renewcommand{\arraystretch}{1.2}
% \caption{The ratio of 0 in binary data for different functions}
% \label{ratio_0}
% \centering
% \begin{tabular}{cccccccccccccc}
% \hline
% iteration times & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 &13\\ \hline
% $\neg$ &0.500&0.500&0.500&0.500&0.500&0.500&0.500&0.500&0.500&0.500&0.500 &0.500&0.500\\ \hline
%
% $\textcircled{a}$ &0.496&0.494 &0.492 &0.491 &0.491 &0.490 &0.490 &0.490 &0.490 &0.490 &0.490 &0.490 &0.490\\ \hline
%
% $\textcircled{b}$ &0.500&0.500 &0.500 &0.500 &0.500 &0.500 &0.500 &0.500 &0.500 &0.500 &0.500 &0.500&0.500\\ \hline
%
% $\textcircled{c}$ &0.496&0.494 &0.492 &0.491 &0.490 &0.490 &0.490 &0.490 &0.489 &0.489 &0.489 &0.489 &0.489\\ \hline
%
% $\textcircled{d}$&0.500&0.500 &0.500 &0.500 &0.500 &0.500 &0.500 &0.500 &0.500 &0.500 &0.500 &0.500&0.500\\ \hline
%
% $\textcircled{e}$ &0.500&0.498 &0.497 &0.496 &0.496 &0.495 &0.495 &0.495 &0.495 &0.494 &0.494 &0.494 &0.494\\ \hline
%
% $\textcircled{f}$&0.500&0.500 &0.500 &0.500 &0.500 &0.500 &0.500 &0.500 &0.500 &0.500 &0.500 &0.500&0.500\\ \hline
%
% $\textcircled{g}$ &0.508&0.513 &0.516 &0.519 &0.521 &0.522 &0.523 &0.523 &0.524 &0.524 &0.524 &0.524 &0.524\\ \hline
%
% $\textcircled{h}$ &0.492&0.487 &0.483 &0.479 &0.475 &0.473 &0.471 &0.470 &0.469 &0.469 &0.468 &0.468 &0.468\\ \hline
%
% $\textcircled{i}$&0.484 &0.473 &0.459 &0.446 &0.434 &0.423 &0.413 &0.405 &0.398 &0.392 &0.387 &0.383 &0.380\\ \hline
%
% \hline
% \end{tabular}
% \end{table*}
% \end{tiny}
% By analyzing the sequences generated by the new functions which are randomly selected by using Graph with strongly connected components as a selection criterion, the authors have realized that the randomicity of
% some sequence are not very ideal.
%
% Let $\mathsf{N} = 4$, then the 10 functions are cited from ~\cite{}. Set initial values from 0 to 15, and at each iteration, every cell will be updated once to create one new input. The maps are Iterated 13 times. Table~\ref{ratio_0} shows the percentages of differences between zeros and ones for all output at each iterations. The balance property of illustrates that the sequence generated directly by
% $\neg$, $\textcircled{b}$,$\textcircled{d}$ and $\textcircled{f}$ fit strictly the uniform
% distribution. Therefore, effective selection criterion for chaotic iterate functions should be made to
% enhance the random statistical properties.
%
% Based on a large number of experiments, the authors propose the following method to select the chaotic
% iterate functions.

\subsection{Obtaining Balanced Maps}
\label{The generation of pseudo-random sequence}

% The design of the pseudo-random number generator based on discrete chaotic iterations is proposed
% in this section, its performance is evaluated in the next one.
%\subsection{The output distribution of negation mapping}

We now explain how to find balanced iterate functions.

\begin{theorem}
Let $j \in \llbracket 1; 2^\mathsf{N} \rrbracket$ and $F = \mathcal{F}(f_0)$ be the (balanced) vectorial Boolean negation: $F_{j}=2^\mathsf{N}-j$.

If $F' = \mathcal{F}(f)$, a vector of images of a \emph{balanced} iterate function $f$, is such that its $j-$th component differs from $F_j$ by only its $i-th$ bit (starting from the right), then $F'_{2^\mathsf{N}-F'_j}=2^\mathsf{N}-j$.
\end{theorem}

\begin{proof}
As $F'_j$ only differs from $F_j$ by its $i-th$ bit, we have: $F'_j=F_j-F_j\&2^{i-1}+(j-1)\&2^{i-1}.$
%Therefore, the value output using the iteration function $F'$ for its $i-th$ bit in $j-th$ element is computed as:
Therefore, the value $\mathsf{f}'_{i,j}$ of the mapping matrix of $F'$ can be computed as follows:
\begin{equation}
\label{eq2}
\begin{array}{l}
\mathsf{f}'_{i,j}=j_\mathsf{N}j_{\mathsf{N}-1}...f(j)_i...j_1 \\
=(j-1)-(j-1)\&2^{i-1}+F'_j\&2^{i-1}\\
=(j-1)-(j-1)\&2^{i-1}+F_j\&2^{i-1}-F_j\&2^{i-1}+(j-1)\&2^{i-1}\\
=(j-1)\\
\end{array}
\end{equation}
%In the original situation, the values in $\mathsf{f}$ are effectively distributed according to the standard uniform distribution. Now, there are two values of $j-1$ and no value of $\mathsf{f}_{i,j}$ in $i$-th row of the matrix $\mathsf{f}$. The balance of the outputs distribution has been broken, and they are not uniform any more. To fix it, the output value via $F$ equals $j-1$ should be found and changed it to $\mathsf{f}_{i,j}$, then:
The values in $\mathsf{f}$ are uniformly distributed.
However, in the new matrix $\mathsf{N}$, there are twice the value $j-1$ and no $\mathsf{f}_{i,j}$ in the $i$-th row: the uniform distribution is lost. 
To restore the balance, one of the two $j-1$ values must be found and replaced by $\mathsf{f}_{i,j}$. Let $k$ be a variable such that $\mathsf{f}_{i,k}=j-1$ and $\mathsf{f}'_{i,k}=\mathsf{f}_{i,j}$.
%\begin{equation}
%\mathsf{f}_{i,k}=j-1,
%\end{equation}
%\begin{equation}
%\label{eq4}
%\mathsf{f}'_{i,k}=\mathsf{f}_{i,j}
%\end{equation}

As the $i-$th bits in ${f}_{i,k}$ and ${f}'_{i,k}$ are equal, we have:
\begin{equation}
\label{eq5}
\mathsf{f}'_{i,k}\&(2^\mathsf{N}-1-2^{i-1})=\mathsf{f}_{i,j}\&(2^\mathsf{N}-1-2^{i-1}).
\end{equation}

We can thus transform the equation $\mathsf{f}_{i,k}=j-1$ as follows:
\begin{equation}
\label{eq6}
\begin{array}{lll}
\mathsf{f}_{i,k}&=&j-1\\
(k-1)-(k-1)\&2^{i-1}+F_k\&2^{i-1}& =&j-1\\
(k-1)\&(2^\mathsf{N}-1-2^{i-1})+F_k\&2^{i-1}&=&j-1.
\end{array}
\end{equation}

Moreover, from $F_k\&2^i = 2^\mathsf{N}-1-k$, we obtain:
\begin{equation}
\label{eq7}
\begin{array}{lll}
F_k\&2^{i-1}&=&(j-1)\&2^{i-1} \\
(2^\mathsf{N}-k)\&2^i&=&(j-1)\&2^{i-1}\\
(k-1)\&2^{i-1}&=&(k-1)\&2^{i-1}-(j-1)\&2^{i-1}.
\end{array}
\end{equation}

According to Equations (\ref{eq6}) and (\ref{eq7}), we have:
\begin{equation}
\label{eq8}
\begin{array}{ll}
k-1=(j-1)+(k-1)\&2^{i-1}-F_k\&2^{i-1}, \\
\text{where } F_k=2^\mathsf{N}-k \\
 =(j-1)+(k-1)\&2^{i-1}-2^{i-1}+(k-1)\&2^{i-1}\\
 =(j-1)-2^{i-1}+((k-1)\&2^{i-1})*2. \\
\text{But, due to Equation} \text{(\ref{eq8})}, \text{ we have:} \\
 =(j-1)-2^{i-1}+((k-1)\&2^{i-1}-(j-1)\&2^{i-1})*2 \\
 =(j-1)+2^{i-1}-((j-1)\&2^{i-1})*2 \\
 =(j-1)+(j-1)\&2^{i-1}+(2^\mathsf{N}-j)\&2^{i-1},\\
\text{where } F_j=2^\mathsf{N}-1-j \\
 =(j-1)+(j-1)\&2^{i-1}+F_j\&2^{i-1} \\
 =\mathsf{f}_{i,j}.
\end{array}
\end{equation}

As
\begin{equation}
\mathsf{f}'_{i,k}=(k-1)-(k-1)\&2^{i-1}+F'_k\&2^{i-1}\\
\end{equation}
and according to Equation (\ref{eq8}), we thus have:
\begin{equation}
\label{eq9}
\begin{array}{ll}
F'_k\&2^{i-1} &=(k-1)\&2^{i-1}.
\end{array}
\end{equation} 

Now, from Equation (\ref{eq5}), we can set that:
\begin{equation}
\label{eq10}
\begin{array}{lr}
\mathsf{f}'_{i,k}\&(2^\mathsf{N}-1-2^{i-1})=\mathsf{f}_{i,j}\&(2^\mathsf{N}-1-2^{i-1}) \\
\multicolumn{2}{l}{((k-1)-(k-1)\&2^{i-1}+F'_k\&2^{i-1})\&(2^\mathsf{N}-1-2^{i-1})=}\\
~~~~((k-1)-(k-1)\&2^{i-1}+F_k\&2^{i-1})\&(2^\mathsf{N}-1-2^{i-1}) \\
F'_k\&(2^\mathsf{N}-1-2^i)=F_k\&(2^\mathsf{N}-1-2^i).\\
\end{array}
\end{equation}

By using both Equations (\ref{eq9}) and (\ref{eq10}), we obtain:
\begin{equation}
\label{eq11}
\begin{array}{ll}
F'_k&=(k-1)\&2^{i-1}+F_k\&(2^\mathsf{N}-1-2^{i-1})\\
&=(k-1)\&2^{i-1}+F_k-F_k\&2^{i-1}\\
&=\mathsf{f}_{i,j} \& 2^{i-1}+(2^N-1-\mathsf{f}_{i,j})-(2^N-1-\mathsf{f}_{i,j}) \& 2^{i-1}\\
&=(2^N-1)-[\mathsf{f}_{i,j}+2^{i-1}-2 \times (\mathsf{f}_{i,j} \& 2^{i-1})] \\
&=(2^N-1)-((j-1)-(j-1)\&2^{i-1}+F_j\&2^{i-1}\\
&+2^{i-1}-2 \times [((j-1)-(j-1) \& 2^{i-1}+F_j \& 2^{i-1})\& 2^{i-1}\\
&=(2^N-1)-((j-1)-(j-1)\&2^{i-1}-F_j\&2^{i-1}+2^{i-1})\\
&=(2^N-1)-((j-1)+2^{i-1}\&(2^N-j)-F_j\&2^{i-1})\\
&=(2^N-1)-((j-1)+F_j\&2^{i-1}-F_j\&2^{i-1})\\
&=(2^N-1)-(j-1)\\
&=2^N-j.\\
\end{array}
\end{equation}


Finally, from Equation (\ref{eq8}), we can conclude that:
\begin{equation}
\label{eq12}
\begin{array}{ll}
k&=\mathsf{f}_{i,j}+1\\
&=(j-1)-(j-1)\&2^{i-1}+F_j\&2^{i-1} \\
&=2^N-(2^N-j)-(j-1)\&2^{i-1}+F_j\&2^{i-1}\\
&=2^N-(F_j-F_j\&2^i+(j-1)\&2^{i-1})  \\
&=2^N-F'_j.
\end{array}
\end{equation}
\end{proof}

With such equations (namely, Eq. (\ref{eq11}) and (\ref{eq12})), the balance of the new table can be obtained by computing the mapping values. In other words, there is a bijection from the set A of the inputs $x$ into the set B of $F'(x)$ values.

Let us give an example. In Table~\ref{negation_output} is given the mapping matrix for the vectorial Boolean negation, with $\mathsf{N}=4$. Obviously, the values in $\mathsf{f}$ are uniformly distributed: each integer from 0 to 15 occurs once per row. 
Now, if we desire to set $F'_1$ as 14, then $\mathsf{f}'_{4,1}=0$: there will be two $0$ and no $1$ in the fourth row of $\mathsf{f}'$. Due to the previous study, we know that $F'_2$ must be set to 15 too, which leads to $\mathsf{f}'_{4,2}=1$: the balance is recovered.

To sum up, we can determine whether the modification of a bit in the vector of images of the negation function preserves the balance of the outputs or not, by using the following rule (necessary condition):
\begin{itemize}
\item if $F'_j = C$,
\item then $C = F_j-F_j\&2^{i-1}+(j-1)\&2^{i-1}$,
\item and also $F_{2^N-C} = 2^N-j$.
\end{itemize}

This rule, we name it ``Balance Iteration Mapping Rule'', can be used as a criterion to find iterate functions leading to good CIs PRNGS, as it is depicted in Algorithm \ref{Chaotic iteration}.
Let us finally remark that, with such a process, it is possible to find new iteration functions by changing more than 1 couple of values in the vectorial Boolean negation $F$. Indeed it is obvious that 2, 3, 4, and even 8 couples of values can be changed using the Balance Iteration Mapping Rule.
For instance, Table~\ref{New vectors of images} contains 8 vectors of images obtained by using Algorithm \ref{Chaotic iteration} one or more times. All of these functions satisfy the hypothesis of Theorem \ref{IC_chaotiques} too, and thus their dynamical systems behave chaotically.
% that's all folks



%\subsection{Algorithm for selection}
% \begin{table*}[!t]
% \renewcommand{\arraystretch}{1.2}
% \caption{F}
% \label{negation_output}
% \centering
% \begin{tabular}{ccccccccccccccccc}
% \hline
% initial & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\ \hline
% 1st bit &14&15&11& 13& 10& 11& 8& 9& 6& 7& 4& 5& 2& 3& 0 &1\\
% 2nd bit &13&12&15& 14& 9 &8 &11& 10& 5 &4& 7& 6& 1& 0& 3& 2\\
% 3rd bit &11&10&9& 8& 15& 14& 13& 12& 3& 2& 1& 0& 7& 6& 5& 4\\
% 4th bit &7&6&5& 4& 3& 2& 1& 0& 15& 14& 13& 12& 11& 10& 9& 8\\ \hline
% \hline
% \end{tabular}
% \end{table*}
\begin{algorithm}
\textbf{Input:} a vector of images $F$\\
\textbf{Output:} a vector of images $r$ or 0
\begin{algorithmic}[1]
\FOR{$i=0,\dots,2^\mathsf{N}-1$}
{
\FOR{$j=0,\dots,N$}
{

\IF{$F(i+1) \neq F(i+1)-F(i+1) \& 2^j+i \& 2^j$}
{

\IF{$F(i+1) \neq 2^N-1-i$}
{
\STATE return 0;
}\ENDIF

}\ENDIF

}\ENDFOR

\IF{$F(2^N-F(i)) \neq 2^N-i$}
{
\STATE return 0;
}\ENDIF

}\ENDFOR
\STATE return $F$\;
\medskip
\caption{The Balance Iteration Mapping Rule.}
\label{Chaotic iteration}
\end{algorithmic}
\end{algorithm}

% \begin{algorithm}
% \SetAlgoLined
% \KwIn{a vector of images $F$}
% \KwOut{a vector of images $r$ or 0}
% \For{$i=0,\dots,2^\mathsf{N}-1$}
% {
% \For{$j=0,\dots,N$}
% {
% 
% \If{$F(i+1) \neq F(i+1)-F(i+1) \& 2^j+i \& 2^j$}
% {
% 
% \If{$F(i+1) \neq 2^N-1-i$}
% {
% return 0;
% }
% 
% }
% 
% }
% 
% \If{$F(2^N-F(i)) \neq 2^N-i$}
% {
% return 0;
% }
% 
% }
% return $F$\;
% \caption{The Balance Iteration Mapping Rule.}
% \label{Chaotic iteration}
% \end{algorithm}



\begin{table}
\centering
\begin{tabular}{|c|l|}
\hline
Name & Map \\
\hline
$F$&[15,14,13,12,11,10,9,8,7,6,5,4,3,2,1,0]\\
\hline
$F'1$&[14,15,13,12,11,10,9,8,7,6,5,4,3,2,1,0]\\
\hline
$F'2$&[14,15,13,12,9,10,11,8,7,6,5,4,3,2,1,0]\\
\hline
$F'3$&[14,15,9,4,11,8,13,10,7,6,5,12,3,2,1,0]\\
\hline
$F'4$&[14,15,9,12,3,8,13,10,7,6,5,4,11,2,1,0]\\
\hline
$F'5$&[14,15,9,4,11,8,13,10,7,6,5,12,3,2,0,1]\\
\hline
$F'6$&[14,15,9,4,11,8,13,10,3,6,5,12,7,2,0,1]\\
\hline
$F'7$&[14,15,9,4,3,8,13,10,5,2,7,12,11,6,1,0]\\
\hline
$F'8$&[14,15,5,8,9,2,11,12,3,4,13,6,7,10,0,1]\\
\hline
\end{tabular}
\caption{New vectors of images}
\label{New vectors of images}
\end{table}

\section{Statistical analysis}
\label{sec:nist}
% \subsection{NIST statistical test suite}
% Among the numerous standard tests for pseudo-randomness, a convincing way to prove the quality of the produced sequences is to confront them with the NIST (National Institute of Standards and Technology) Statistical Test Suite SP 800-22, %\footnote{A new version of the Statistical Test Suite (Version 2.0) has been released in August 25, 2008.}
% 
% 
% The NIST test suite, SP 800-22, is a statistical package consisting of 15 tests. They were developed to measure the randomness of (arbitrarily long) binary sequences produced by either hardware or software based cryptographic pseudorandom number generators. These tests focus on a variety of different types of non-randomness that could occur in such sequences. These 15 tests include in the NIST test suite are described in \cite{nist}.
% 
% 
% 
% \subsection{DieHARD battery of tests}
% The DieHARD battery of tests has been the most sophisticated standard for over a decade. Because of
% the stringent requirements in the DieHARD test suite, a generator passing this battery of
% tests can be considered good as a rule of thumb.
% 
% The DieHARD battery of tests consists of 18 different independent statistical tests. This collection
%  of tests is based on assessing the randomness of bits comprising 32-bit integers obtained from
% a random number generator. Each test requires $2^{23}$ 32-bit integers in order to run the full set
% of tests.
% Most of the tests in DieHARD return a p-value, which should be uniform on $[0,1)$ if the input file
% contains truly independent random bits.  Those $p-value$s are obtained by
% $p=F(X)$, where $F$ is the assumed distribution of the sample random variable $X$ \textendash often normal.
% But that assumed $F$ is just an asymptotic approximation, for which the fit will be worst
% in the tails. Thus occasional $p-value$s near 0 or 1, such as 0.0012 or 0.9983, can occur.
% An individual test is considered to be failed if $p$ value approaches 1 closely, for example $p>0.9999$.
% 
% \subsection{Test results}

We can conclude from Table \ref{The passing rate12} and Table \ref{The passing rate34}that all of the generators based on the new iterate functions have successfully passed both the NIST and DieHARD batteries of tests. 
These results show the good statistical properties of the proposed PRNGs, and thus the interest of the theoretical approach presented in this paper.

\begin{table*}[t]
\renewcommand{\arraystretch}{1.3}
\caption{Results through NIST SP 800-22 and DieHARD batteries of tests ($\mathbb{P}_T$ values)}
\label{The passing rate12}
\centering
  \begin{tabular}{|l@{}||c@{}|c@{}|c@{}|c@{}|}
    \hline
Method & $F'_1$ &  $F'_2$ & $F'_3$ & $F'_4$ \\ \hline\hline


%$w^{j}$ & $\{1,..,8\}$ & $\{1,..,8\}$ & $\{1,..,8\}$ & $\{1,..,5\}$ & $\{1,..,5\}$ &$\{1,..,5\}$ \\ \hline \hline
Frequency (Monobit) Test            &  0.102526 &  0.017912 &  0.171867 &  0.779188  \\ \hline
Frequency Test within a Block             &  0.085587 &  0.657933 &  0.779188 &  0.897763  \\ \hline
Cumulative Sums (Cusum) Test*             &  0.264576 &  0.185074 &  0.228927 &  0.736333 \\ \hline
Runs Test                    &  0.739918 &  0.334538 &  0.798139 &  0.834308  \\ \hline
Test for the Longest Run of Ones in a Block     &  0.678686 &  0.474986 &  0.637119 &  0.037566  \\ \hline
Binary Matrix Rank Test                & 0.816537 &  0.534146 &  0.249284 &  0.883171  \\ \hline
Discrete Fourier Transform (Spectral) Test     &   0.798139 &  0.474986 &  0.014550 &  0.366918  \\ \hline
Non-overlapping Template Matching Test*        &  0.489304 &  0.507177 &  0.477005 &  0.557597  \\ \hline
Overlapping Template Matching Test        &  0.514124 &  0.171867 &  0.162606 &  0.816537 \\ \hline
Maurer's ``Universal Statistical`` Test         &   0.249284 &  0.171867 &  0.096578 &  0.419021 \\ \hline
Approximate Entropy Test             & 0.236810 &  0.514124 &  0.262249 &  0.816537 \\ \hline
Random Excursions Test*                &  0.353142 &  0.403219 &  0.229832 &  0.481025 \\ \hline
Random Excursions Variant Test*            &  0.412987 &  0.369181 &  0.313171 &  0.513679  \\ \hline
Serial Test* (m=10)                &  0.304324 &  0.102735 &  0.270033 &  0.384058 \\ \hline
Linear Complexity Test                &0.759756 &  0.153763 &  0.883171 &  0.171867 \\ \hline
Success                     &15/15&15/15&15/15&15/15  \\ \hline\hline
Diehard Test               &pass&pass&pass&pass\\ \hline
  \end{tabular}
\end{table*}

\begin{table*}[t]
\renewcommand{\arraystretch}{1.3}
\caption{Results through NIST SP 800-22 and DieHARD batteries of tests ($\mathbb{P}_T$ values)}
\label{The passing rate34}
\centering
  \begin{tabular}{|l@{}|c@{}|c@{}|c@{}|c@{}|}
    \hline
Method & $F'_5$ & $F'_6$ & $F'_7$ & $F'_8$\\ \hline\hline


%$w^{j}$ & $\{1,..,8\}$ & $\{1,..,8\}$ & $\{1,..,8\}$ & $\{1,..,5\}$ & $\{1,..,5\}$ &$\{1,..,5\}$ \\ \hline \hline
Frequency (Monobit) Test            &    0.971699 &  0.275709 &  0.137282 &    0.699313 \\ \hline
Frequency Test within a Block             &   0.851383 &  0.383827 &  0.262249 &    0.122325 \\ \hline
Cumulative Sums (Cusum) Test*             &   0.462694 &  0.169816 &  0.391715 &    0.729111\\ \hline
Runs Test                    &    0.153763 &  0.719747 &  0.534146 &    0.262249 \\ \hline
Test for the Longest Run of Ones in a Block     &   0.366918 &  0.739918 &  0.236810 &    0.759756 \\ \hline
Binary Matrix Rank Test                &  0.739918 &  0.037566 &  0.798139 &    0.867692 \\ \hline
Discrete Fourier Transform (Spectral) Test     &     0.595549 &  0.115387 &  0.798139 &    0.153763 \\ \hline
Non-overlapping Template Matching Test*        &    0.452278 &  0.505673 &  0.541034 &    0.497140 \\ \hline
Overlapping Template Matching Test        &    0.319084 &  0.678686 &  0.534146 &    0.798139 \\ \hline
Maurer's ``Universal Statistical`` Test         &     0.171867 &  0.798139 &  0.115387 &    0.275709 \\ \hline
Approximate Entropy Test             &   0.474986 &  0.080519 &  0.000001 &    0.779188\\ \hline
Random Excursions Test*                &    0.317506 &  0.602978 &  0.362746 &    0.416274 \\ \hline
Random Excursions Variant Test*            &    0.274813 &  0.391166 &  0.454157 &    0.341012 \\ \hline
Serial Test* (m=10)                &    0.456684 &  0.125973 &  0.404429 &    0.253197 \\ \hline
Linear Complexity Test                & 0.366918 &  0.319084 &  0.678686 &    0.075719 \\ \hline
Success                     &15/15&15/15&15/15&15/15  \\ \hline\hline
Diehard Test               &pass&pass&pass&pass\\ \hline
  \end{tabular}
\end{table*}