\usepackage{ctable}
\usepackage{tabularx}
\usepackage{multirow}
-
% Pour mathds : les ensembles IR, IN, etc.
\usepackage{dsfont}
\usepackage{graphicx}
% Pour faire des sous-figures dans les figures
\usepackage{subfigure}
+\usepackage{xr-hyper}
+\usepackage{hyperref}
+\externaldocument{prng_gpu}
+%\usepackage{hyperref}
\newtheorem{notation}{Notation}
-\title{Supplementary of ``Efficient and Cryptographically Secure Generation of Chaotic Pseudorandom Numbers on GPU''}
+\title{Annex document of ``Efficient and Cryptographically Secure Generation of Chaotic Pseudorandom Numbers on GPU''}
\begin{document}
\author{Jacques M. Bahi, Rapha\"{e}l Couturier, Christophe
\IEEEpeerreviewmaketitle
+\section{Proofs of Chaos of the General Formulation of the Chaotic Iterations}
+\label{deuxième def}
+Let us consider the discrete dynamical systems in chaotic iterations having
+the general form: $\forall n\in \mathds{N}^{\ast }$, $ \forall i\in
+\llbracket1;\mathsf{N}\rrbracket $,
+
+\begin{equation}
+ x_i^n=\left\{
+\begin{array}{ll}
+ x_i^{n-1} & \text{ if } i \notin \mathcal{S}^n \\
+ \left(f(x^{n-1})\right)_{S^n} & \text{ if }i \in \mathcal{S}^n.
+\end{array}\right.
+\label{general CIs}
+\end{equation}
+
+In other words, at the $n^{th}$ iteration, only the cells whose id is
+contained into the set $S^{n}$ are iterated.
+
+Let us now rewrite these general chaotic iterations as usual discrete dynamical
+system of the form $X^{n+1}=f(X^n)$ on an ad hoc metric space. Such a formulation
+is required in order to study the topological behavior of the system.
+
+Let us introduce the following function:
+\begin{equation}
+\begin{array}{cccc}
+ \chi: & \llbracket 1; \mathsf{N} \rrbracket \times \mathcal{P}\left(\llbracket 1; \mathsf{N} \rrbracket\right) & \longrightarrow & \mathds{B}\\
+ & (i,X) & \longmapsto & \left\{ \begin{array}{ll} 0 & \textrm{if }i \notin X, \\ 1 & \textrm{if }i \in X, \end{array}\right.
+\end{array}
+\end{equation}
+where $\mathcal{P}\left(X\right)$ is for the powerset of the set $X$, that is, $Y \in \mathcal{P}\left(X\right) \Longleftrightarrow Y \subset X$.
+
+Given a function $f:\mathds{B}^\mathsf{N} \longrightarrow \mathds{B}^\mathsf{N} $, define the function:
+$F_{f}: \mathcal{P}\left(\llbracket1;\mathsf{N}\rrbracket \right) \times \mathds{B}^{\mathsf{N}}
+\longrightarrow \mathds{B}^{\mathsf{N}}$
+\begin{equation*}
+\begin{array}{rll}
+ (P,E) & \longmapsto & \left( E_{j}.\chi (j,P)+f(E)_{j}.\overline{\chi(j,P)}\right) _{j\in \llbracket1;\mathsf{N}\rrbracket}%
+\end{array}%
+\end{equation*}%
+where + and . are the Boolean addition and product operations, and $\overline{x}$
+is the negation of the Boolean $x$.
+Consider the phase space:
+\begin{equation}
+\mathcal{X} = \mathcal{P}\left(\llbracket 1 ; \mathsf{N} \rrbracket\right)^\mathds{N} \times
+\mathds{B}^\mathsf{N},
+\end{equation}
+\noindent and the map defined on $\mathcal{X}$:
+\begin{equation}
+G_f\left(S,E\right) = \left(\sigma(S), F_f(i(S),E)\right), %\label{Gf} %%RAPH, j'ai viré ce label qui existe déjà avant...
+\end{equation}
+\noindent where $\sigma$ is the \emph{shift} function defined by $\sigma
+(S^{n})_{n\in \mathds{N}}\in \mathcal{P}\left(\llbracket 1 ; \mathsf{N} \rrbracket\right)^\mathds{N}\longrightarrow (S^{n+1})_{n\in
+\mathds{N}}\in \mathcal{P}\left(\llbracket 1 ; \mathsf{N} \rrbracket\right)^\mathds{N}$ and $i$ is the \emph{initial function}
+$i:(S^{n})_{n\in \mathds{N}} \in \mathcal{P}\left(\llbracket 1 ; \mathsf{N} \rrbracket\right)^\mathds{N}\longrightarrow S^{0}\in \mathcal{P}\left(\llbracket 1 ; \mathsf{N} \rrbracket\right)$.
+Then the general chaotic iterations defined in Equation \ref{general CIs} can
+be described by the following discrete dynamical system:
+\begin{equation}
+\left\{
+\begin{array}{l}
+X^0 \in \mathcal{X} \\
+X^{k+1}=G_{f}(X^k).%
+\end{array}%
+\right.
+\end{equation}%
+
+Once more, a shift function appears as a component of these general chaotic
+iterations.
+
+To study the Devaney's chaos property, a distance between two points
+$X = (S,E), Y = (\check{S},\check{E})$ of $\mathcal{X}$ must be defined.
+Let us introduce:
+\begin{equation}
+d(X,Y)=d_{e}(E,\check{E})+d_{s}(S,\check{S}),
+\label{nouveau d}
+\end{equation}
+\noindent where $ \displaystyle{d_{e}(E,\check{E})} = \displaystyle{\sum_{k=1}^{\mathsf{N}%
+ }\delta (E_{k},\check{E}_{k})}$ is once more the Hamming distance, and
+$ \displaystyle{d_{s}(S,\check{S})} = \displaystyle{\dfrac{9}{\mathsf{N}}%
+ \sum_{k=1}^{\infty }\dfrac{|S^k\Delta {S}^k|}{10^{k}}}$,
+%%RAPH : ici, j'ai supprimé tous les sauts à la ligne
+%% \begin{equation}
+%% \left\{
+%% \begin{array}{lll}
+%% \displaystyle{d_{e}(E,\check{E})} & = & \displaystyle{\sum_{k=1}^{\mathsf{N}%
+%% }\delta (E_{k},\check{E}_{k})} \textrm{ is once more the Hamming distance}, \\
+%% \displaystyle{d_{s}(S,\check{S})} & = & \displaystyle{\dfrac{9}{\mathsf{N}}%
+%% \sum_{k=1}^{\infty }\dfrac{|S^k\Delta {S}^k|}{10^{k}}}.%
+%% \end{array}%
+%% \right.
+%% \end{equation}
+where $|X|$ is the cardinality of a set $X$ and $A\Delta B$ is for the symmetric difference, defined for sets A, B as
+$A\,\Delta\,B = (A \setminus B) \cup (B \setminus A)$.
+
+
+\begin{proposition}
+The function $d$ defined in Eq.~\ref{nouveau d} is a metric on $\mathcal{X}$.
+\end{proposition}
+
+\begin{proof}
+ $d_e$ is the Hamming distance. We will prove that $d_s$ is a distance
+too, thus $d$, as being the sum of two distances, will also be a distance.
+ \begin{itemize}
+\item Obviously, $d_s(S,\check{S})\geqslant 0$, and if $S=\check{S}$, then
+$d_s(S,\check{S})=0$. Conversely, if $d_s(S,\check{S})=0$, then
+$\forall k \in \mathds{N}, |S^k\Delta {S}^k|=0$, and so $\forall k, S^k=\check{S}^k$.
+ \item $d_s$ is symmetric
+($d_s(S,\check{S})=d_s(\check{S},S)$) due to the commutative property
+of the symmetric difference.
+\item Finally, $|S \Delta S''| = |(S \Delta \varnothing) \Delta S''|= |S \Delta (S'\Delta S') \Delta S''|= |(S \Delta S') \Delta (S' \Delta S'')|\leqslant |S \Delta S'| + |S' \Delta S''|$,
+and so for all subsets $S,S',$ and $S''$ of $\llbracket 1, \mathsf{N} \rrbracket$,
+we have $d_s(S,S'') \leqslant d_e(S,S')+d_s(S',S'')$, and the triangle
+inequality is obtained.
+ \end{itemize}
+\end{proof}
+
+
+Before being able to study the topological behavior of the general
+chaotic iterations, we must first establish that:
+
+\begin{proposition}
+ For all $f:\mathds{B}^\mathsf{N} \longrightarrow \mathds{B}^\mathsf{N} $, the function $G_f$ is continuous on
+$\left( \mathcal{X},d\right)$.
+\end{proposition}
+
+
+\begin{proof}
+We use the sequential continuity.
+Let $(S^n,E^n)_{n\in \mathds{N}}$ be a sequence of the phase space $%
+\mathcal{X}$, which converges to $(S,E)$. We will prove that $\left(
+G_{f}(S^n,E^n)\right) _{n\in \mathds{N}}$ converges to $\left(
+G_{f}(S,E)\right) $. Let us remark that for all $n$, $S^n$ is a strategy,
+thus, we consider a sequence of strategies (\emph{i.e.}, a sequence of
+sequences).\newline
+As $d((S^n,E^n);(S,E))$ converges to 0, each distance $d_{e}(E^n,E)$ and $d_{s}(S^n,S)$ converges
+to 0. But $d_{e}(E^n,E)$ is an integer, so $\exists n_{0}\in \mathds{N},$ $%
+d_{e}(E^n,E)=0$ for any $n\geqslant n_{0}$.\newline
+In other words, there exists a threshold $n_{0}\in \mathds{N}$ after which no
+cell will change its state:
+$\exists n_{0}\in \mathds{N},n\geqslant n_{0}\Rightarrow E^n = E.$
+
+In addition, $d_{s}(S^n,S)\longrightarrow 0,$ so $\exists n_{1}\in %
+\mathds{N},d_{s}(S^n,S)<10^{-1}$ for all indexes greater than or equal to $%
+n_{1}$. This means that for $n\geqslant n_{1}$, all the $S^n$ have the same
+first term, which is $S^0$: $\forall n\geqslant n_{1},S_0^n=S_0.$
+
+Thus, after the $max(n_{0},n_{1})^{th}$ term, states of $E^n$ and $E$ are
+identical and strategies $S^n$ and $S$ start with the same first term.\newline
+Consequently, states of $G_{f}(S^n,E^n)$ and $G_{f}(S,E)$ are equal,
+so, after the $max(n_0, n_1)^{th}$ term, the distance $d$ between these two points is strictly less than 1.\newline
+\noindent We now prove that the distance between $\left(
+G_{f}(S^n,E^n)\right) $ and $\left( G_{f}(S,E)\right) $ is convergent to
+0. Let $\varepsilon >0$. \medskip
+\begin{itemize}
+\item If $\varepsilon \geqslant 1$, we see that the distance
+between $\left( G_{f}(S^n,E^n)\right) $ and $\left( G_{f}(S,E)\right) $ is
+strictly less than 1 after the $max(n_{0},n_{1})^{th}$ term (same state).
+\medskip
+\item If $\varepsilon <1$, then $\exists k\in \mathds{N},10^{-k}\geqslant
+\varepsilon > 10^{-(k+1)}$. But $d_{s}(S^n,S)$ converges to 0, so
+\begin{equation*}
+\exists n_{2}\in \mathds{N},\forall n\geqslant
+n_{2},d_{s}(S^n,S)<10^{-(k+2)},
+\end{equation*}%
+thus after $n_{2}$, the $k+2$ first terms of $S^n$ and $S$ are equal.
+\end{itemize}
+\noindent As a consequence, the $k+1$ first entries of the strategies of $%
+G_{f}(S^n,E^n)$ and $G_{f}(S,E)$ are the same ($G_{f}$ is a shift of strategies) and due to the definition of $d_{s}$, the floating part of
+the distance between $(S^n,E^n)$ and $(S,E)$ is strictly less than $%
+10^{-(k+1)}\leqslant \varepsilon $.
+
+In conclusion,
+%%RAPH : ici j'ai rajouté une ligne
+%%TOF : ici j'ai rajouté un commentaire
+%%TOF : ici aussi
+$
+\forall \varepsilon >0,$ $\exists N_{0}=max(n_{0},n_{1},n_{2})\in \mathds{N}
+,$ $\forall n\geqslant N_{0},$
+$ d\left( G_{f}(S^n,E^n);G_{f}(S,E)\right)
+\leqslant \varepsilon .
+$
+$G_{f}$ is consequently continuous.
+\end{proof}
+
+
+It is now possible to study the topological behavior of the general chaotic
+iterations. We will prove that,
+
+\begin{theorem}
+\label{t:chaos des general}
+ The general chaotic iterations defined on Equation~\ref{general CIs} satisfy
+the Devaney's property of chaos.
+\end{theorem}
+
+Let us firstly prove the following lemma.
+
+\begin{lemma}[Strong transitivity]
+\label{strongTrans}
+ For all couples $X,Y \in \mathcal{X}$ and any neighborhood $V$ of $X$, we can
+find $n \in \mathds{N}^*$ and $X' \in V$ such that $G^n(X')=Y$.
+\end{lemma}
+
+\begin{proof}
+ Let $X=(S,E)$, $\varepsilon>0$, and $k_0 = \lfloor log_{10}(\varepsilon)+1 \rfloor$.
+Any point $X'=(S',E')$ such that $E'=E$ and $\forall k \leqslant k_0, S'^k=S^k$,
+are in the open ball $\mathcal{B}\left(X,\varepsilon\right)$. Let us define
+$\check{X} = \left(\check{S},\check{E}\right)$, where $\check{X}= G^{k_0}(X)$.
+We denote by $s\subset \llbracket 1; \mathsf{N} \rrbracket$ the set of coordinates
+that are different between $\check{E}$ and the state of $Y$. Thus each point $X'$ of
+the form $(S',E')$ where $E'=E$ and $S'$ starts with
+$(S^0, S^1, \hdots, S^{k_0},s,\hdots)$, verifies the following properties:
+\begin{itemize}
+ \item $X'$ is in $\mathcal{B}\left(X,\varepsilon\right)$,
+ \item the state of $G_f^{k_0+1}(X')$ is the state of $Y$.
+\end{itemize}
+Finally the point $\left(\left(S^0, S^1, \hdots, S^{k_0},s,s^0, s^1, \hdots\right); E\right)$,
+where $(s^0,s^1, \hdots)$ is the strategy of $Y$, satisfies the properties
+claimed in the lemma.
+\end{proof}
+
+We can now prove the Theorem~\ref{t:chaos des general}.
+
+\begin{proof}[Theorem~\ref{t:chaos des general}]
+Firstly, strong transitivity implies transitivity.
+
+Let $(S,E) \in\mathcal{X}$ and $\varepsilon >0$. To
+prove that $G_f$ is regular, it is sufficient to prove that
+there exists a strategy $\tilde S$ such that the distance between
+$(\tilde S,E)$ and $(S,E)$ is less than $\varepsilon$, and such that
+$(\tilde S,E)$ is a periodic point.
+
+Let $t_1=\lfloor-\log_{10}(\varepsilon)\rfloor$, and let $E'$ be the
+configuration that we obtain from $(S,E)$ after $t_1$ iterations of
+$G_f$. As $G_f$ is strongly transitive, there exists a strategy $S'$
+and $t_2\in\mathds{N}$ such
+that $E$ is reached from $(S',E')$ after $t_2$ iterations of $G_f$.
+
+Consider the strategy $\tilde S$ that alternates the first $t_1$ terms
+of $S$ and the first $t_2$ terms of $S'$:
+%%RAPH : j'ai coupé la ligne en 2
+$$\tilde
+S=(S_0,\dots,S_{t_1-1},S'_0,\dots,S'_{t_2-1},S_0,$$$$\dots,S_{t_1-1},S'_0,\dots,S'_{t_2-1},S_0,\dots).$$ It
+is clear that $(\tilde S,E)$ is obtained from $(\tilde S,E)$ after
+$t_1+t_2$ iterations of $G_f$. So $(\tilde S,E)$ is a periodic
+point. Since $\tilde S_t=S_t$ for $t<t_1$, by the choice of $t_1$, we
+have $d((S,E),(\tilde S,E))<\epsilon$.
+\end{proof}
+
+
+
\section{Statistical Improvements Using Chaotic Iterations}
\label{The generation of pseudorandom sequence}
-Let us now explain why we have reasonable ground to believe that chaos
+Let us explain in this annex why we have reasonable ground to believe that chaos
can improve statistical properties.
We will show in this section that chaotic properties as defined in the
mathematical theory of chaos are related to some statistical tests that can be found
\begin{itemize}
- \item \textbf{Regularity}. As stated in Section~\ref{subsec:Devaney}, a chaotic dynamical system must
+ \item \textbf{Regularity}. As stated in Section~\ref{subsec:Devaney} of the main document, a chaotic dynamical system must
have an element of regularity. Depending on the chosen definition of chaos, this element can be the existence of
a dense orbit, the density of periodic points, etc. The key idea is that a dynamical system with no periodicity
is not as chaotic as a system having periodic orbits: in the first situation, we can predict something and gain a
\end{itemize}
-We have proven in our previous works~\cite{guyeux12:bc} that chaotic iterations satisfying Theorem~\ref{Th:Caractérisation des IC chaotiques} are, among other
+We have proven in our previous works~\cite{guyeux12:bc} that chaotic iterations satisfying Theorem~\ref{Th:Caractérisation des IC chaotiques} of the main document are, among other
things, strongly transitive, topologically mixing, chaotic as defined by Li and Yorke,
and that they have a topological entropy and an exponent of Lyapunov both equal to $ln(\mathsf{N})$,
where $\mathsf{N}$ is the size of the iterated vector.
+\section{Practical Security Evaluation}
+\label{sec:Practicak evaluation}
+
+Pseudorandom generators based on Eq.~\eqref{equation Oplus} of the main document are thus cryptographically secure when
+they are XORed with an already cryptographically
+secure PRNG. But, as stated previously,
+such a property does not mean that, whatever the
+key size, no attacker can predict the next bit
+knowing all the previously released ones.
+However, given a key size, it is possible to
+measure in practice the minimum duration needed
+for an attacker to break a cryptographically
+secure PRNG, if we know the power of his/her
+machines. Such a concrete security evaluation
+is related to the $(T,\varepsilon)-$security
+notion, which is recalled and evaluated in what
+follows, for the sake of completeness.
+
+Let us firstly recall that,
+\begin{definition}
+Let $\mathcal{D} : \mathds{B}^M \longrightarrow \mathds{B}$ be a probabilistic algorithm that runs
+in time $T$.
+Let $\varepsilon > 0$.
+$\mathcal{D}$ is called a $(T,\varepsilon)-$distinguishing attack on pseudorandom
+generator $G$ if
+
+\begin{flushleft}
+$\left| Pr[\mathcal{D}(G(k)) = 1 \mid k \in_R \{0,1\}^\ell ]\right.$
+\end{flushleft}
+
+\begin{flushright}
+$ - \left. Pr[\mathcal{D}(s) = 1 \mid s \in_R \mathds{B}^M ]\right| \geqslant \varepsilon,$
+\end{flushright}
+
+\noindent where the probability is taken over the internal coin flips of $\mathcal{D}$, and the notation
+``$\in_R$'' indicates the process of selecting an element at random and uniformly over the
+corresponding set.
+\end{definition}
+
+Let us recall that the running time of a probabilistic algorithm is defined to be the
+maximum of the expected number of steps needed to produce an output, maximized
+over all inputs; the expected number is averaged over all coin flips made by the algorithm~\cite{Knuth97}.
+We are now able to define the notion of cryptographically secure PRNGs:
+
+\begin{definition}
+A pseudorandom generator is $(T,\varepsilon)-$secure if there exists no $(T,\varepsilon)-$distinguishing attack on this pseudorandom generator.
+\end{definition}
+
+
+
+
+
+
+
+Suppose now that the PRNG of Eq.~\eqref{equation Oplus} of the main document will work during
+$M=100$ time units, and that during this period,
+an attacker can realize $10^{12}$ clock cycles.
+We thus wonder whether, during the PRNG's
+lifetime, the attacker can distinguish this
+sequence from a truly random one, with a probability
+greater than $\varepsilon = 0.2$.
+We consider that $N$ has 900 bits.
+
+Predicting the next generated bit knowing all the
+previously released ones by Eq.~\eqref{equation Oplus} of the main document is obviously equivalent to predicting the
+next bit in the BBS generator, which
+is cryptographically secure. More precisely, it
+is $(T,\varepsilon)-$secure: no
+$(T,\varepsilon)-$distinguishing attack can be
+successfully realized on this PRNG, if~\cite{Fischlin}
+\begin{equation}
+T \leqslant \dfrac{L(N)}{6 N (log_2(N))\varepsilon^{-2}M^2}-2^7 N \varepsilon^{-2} M^2 log_2 (8 N \varepsilon^{-1}M)
+\label{mesureConcrete}
+\end{equation}
+where $M$ is the length of the output ($M=100$ in
+our example), and $L(N)$ is equal to
+$$
+2.8\times 10^{-3} exp \left(1.9229 \times (N ~ln~ 2)^\frac{1}{3} \times (ln(N~ln~ 2))^\frac{2}{3}\right)
+$$
+is the number of clock cycles to factor a $N-$bit
+integer.
+
+
+
+
+A direct numerical application shows that this attacker
+cannot achieve his $(10^{12},0.2)$ distinguishing
+attack in that context.
+
+
+
+
+
\bibliographystyle{plain}
\bibliography{mabase}
