X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/prng_gpu.git/blobdiff_plain/016edf5f7ca01fdf537b4abcb6de87d75c604510..3a4d92d48d8e34ab9e636f7eb092235bcfa0215d:/prng_gpu.tex diff --git a/prng_gpu.tex b/prng_gpu.tex index 57526f2..41e628a 100644 --- a/prng_gpu.tex +++ b/prng_gpu.tex @@ -8,6 +8,7 @@ \usepackage{moreverb} \usepackage{commath} \usepackage{algorithm2e} +\usepackage{listings} \usepackage[standard]{ntheorem} % Pour mathds : les ensembles IR, IN, etc. @@ -33,66 +34,138 @@ \newcommand{\alert}[1]{\begin{color}{blue}\textit{#1}\end{color}} -\title{Efficient generation of pseudo random numbers based on chaotic iterations on GPU} +\title{Efficient Generation of Pseudo-Random Numbers based on Chaotic Iterations +on GPU} \begin{document} -\author{Jacques M. Bahi, Rapha\"{e}l Couturier, and Christophe Guyeux\thanks{Authors in alphabetic order}} +\author{Jacques M. Bahi, Rapha\"{e}l Couturier, and Christophe +Guyeux\thanks{Authors in alphabetic order}} \maketitle \begin{abstract} -This is the abstract + \end{abstract} \section{Introduction} +Random numbers are used in many scientific applications and simulations. On +finite state machines, as computers, it is not possible to generate random +numbers but only pseudo-random numbers. In practice, a good pseudo-random number +generator (PRNG) needs to verify some features to be used by scientists. It is +important to be able to generate pseudo-random numbers efficiently, the +generation needs to be reproducible and a PRNG needs to satisfy many usual +statistical properties. Finally, from our point a view, it is essential to prove +that a PRNG is chaotic. Concerning the statistical tests, TestU01 is the +best-known public-domain statistical testing package. So we use it for all our +PRNGs, especially the {\it BigCrush} which provides the largest serie of tests. +Concerning the chaotic properties, Devaney~\cite{Devaney} proposed a common +mathematical formulation of chaotic dynamical systems. + +In a previous work~\cite{bgw09:ip} we have proposed a new familly of chaotic +PRNG based on chaotic iterations (IC). We have proven that these PRNGs are +chaotic in the Devaney's sense. In this paper we propose a faster version which +is also proven to be chaotic. + +Although graphics processing units (GPU) was initially designed to accelerate +the manipulation of images, they are nowadays commonly used in many scientific +applications. Therefore, it is important to be able to generate pseudo-random +numbers inside a GPU when a scientific application runs in a GPU. That is why we +also provide an efficient PRNG for GPU respecting based on IC. + + + + Interet des itérations chaotiques pour générer des nombre alea\\ Interet de générer des nombres alea sur GPU -\alert{RC, un petit state-of-the-art sur les PRNGs sur GPU ?} -... +\section{Related works on GPU based PRNGs} + +In the litterature many authors have work on defining GPU based PRNGs. We do not +want to be exhaustive and we just give the most significant works from our point +of view. + +In \cite{Pang:2008:cec}, the authors define a PRNG based on cellular automata +which does not require high precision integer arithmetics nor bitwise +operations. There is no mention of statistical tests nor proof that this PRNG is +chaotic. Concerning the speed of generation, they can generate about 3200000 +random numbers per seconds on a GeForce 7800 GTX GPU (which is quite old now). + +In \cite{ZRKB10}, the authors propose different versions of efficient GPU PRNGs +based on Lagged Fibonacci, Hybrid Taus or Hybrid Taus. They have used these +PRNGs for Langevin simulations of biomolecules fully implemented on +GPU. Performance of the GPU versions are far better than those obtained with a +CPU and these PRNGs succeed to pass the {\it BigCrush} test of TestU01. There is +no mention that their PRNGs have chaos mathematical properties. + +To the best of our knowledge no GPU implementation have been proven to have chaotic properties. + \section{Basic Recalls} \label{section:BASIC RECALLS} -This section is devoted to basic definitions and terminologies in the fields of topological chaos and chaotic iterations. -\subsection{Devaney's chaotic dynamical systems} +This section is devoted to basic definitions and terminologies in the fields of +topological chaos and chaotic iterations. +\subsection{Devaney's Chaotic Dynamical Systems} -In the sequel $S^{n}$ denotes the $n^{th}$ term of a sequence $S$ and $V_{i}$ denotes the $i^{th}$ component of a vector $V$. $f^{k}=f\circ ...\circ f$ denotes the $k^{th}$ composition of a function $f$. Finally, the following notation is used: $\llbracket1;N\rrbracket=\{1,2,\hdots,N\}$. +In the sequel $S^{n}$ denotes the $n^{th}$ term of a sequence $S$ and $V_{i}$ +denotes the $i^{th}$ component of a vector $V$. $f^{k}=f\circ ...\circ f$ +is for the $k^{th}$ composition of a function $f$. Finally, the following +notation is used: $\llbracket1;N\rrbracket=\{1,2,\hdots,N\}$. -Consider a topological space $(\mathcal{X},\tau)$ and a continuous function $f : \mathcal{X} \rightarrow \mathcal{X}$. +Consider a topological space $(\mathcal{X},\tau)$ and a continuous function $f : +\mathcal{X} \rightarrow \mathcal{X}$. \begin{definition} -$f$ is said to be \emph{topologically transitive} if, for any pair of open sets $U,V \subset \mathcal{X}$, there exists $k>0$ such that $f^k(U) \cap V \neq \varnothing$. +$f$ is said to be \emph{topologically transitive} if, for any pair of open sets +$U,V \subset \mathcal{X}$, there exists $k>0$ such that $f^k(U) \cap V \neq +\varnothing$. \end{definition} \begin{definition} -An element $x$ is a \emph{periodic point} for $f$ of period $n\in \mathds{N}^*$ if $f^{n}(x)=x$.% The set of periodic points of $f$ is denoted $Per(f).$ +An element $x$ is a \emph{periodic point} for $f$ of period $n\in \mathds{N}^*$ +if $f^{n}(x)=x$.% The set of periodic points of $f$ is denoted $Per(f).$ \end{definition} \begin{definition} -$f$ is said to be \emph{regular} on $(\mathcal{X}, \tau)$ if the set of periodic points for $f$ is dense in $\mathcal{X}$: for any point $x$ in $\mathcal{X}$, any neighborhood of $x$ contains at least one periodic point (without necessarily the same period). +$f$ is said to be \emph{regular} on $(\mathcal{X}, \tau)$ if the set of periodic +points for $f$ is dense in $\mathcal{X}$: for any point $x$ in $\mathcal{X}$, +any neighborhood of $x$ contains at least one periodic point (without +necessarily the same period). \end{definition} -\begin{definition} -$f$ is said to be \emph{chaotic} on $(\mathcal{X},\tau)$ if $f$ is regular and topologically transitive. +\begin{definition}[Devaney's formulation of chaos~\cite{Devaney}] +$f$ is said to be \emph{chaotic} on $(\mathcal{X},\tau)$ if $f$ is regular and +topologically transitive. \end{definition} -The chaos property is strongly linked to the notion of ``sensitivity'', defined on a metric space $(\mathcal{X},d)$ by: +The chaos property is strongly linked to the notion of ``sensitivity'', defined +on a metric space $(\mathcal{X},d)$ by: \begin{definition} \label{sensitivity} $f$ has \emph{sensitive dependence on initial conditions} -if there exists $\delta >0$ such that, for any $x\in \mathcal{X}$ and any neighborhood $V$ of $x$, there exist $y\in V$ and $n > 0$ such that $d\left(f^{n}(x), f^{n}(y)\right) >\delta $. +if there exists $\delta >0$ such that, for any $x\in \mathcal{X}$ and any +neighborhood $V$ of $x$, there exist $y\in V$ and $n > 0$ such that +$d\left(f^{n}(x), f^{n}(y)\right) >\delta $. $\delta$ is called the \emph{constant of sensitivity} of $f$. \end{definition} -Indeed, Banks \emph{et al.} have proven in~\cite{Banks92} that when $f$ is chaotic and $(\mathcal{X}, d)$ is a metric space, then $f$ has the property of sensitive dependence on initial conditions (this property was formerly an element of the definition of chaos). To sum up, quoting Devaney in~\cite{Devaney}, a chaotic dynamical system ``is unpredictable because of the sensitive dependence on initial conditions. It cannot be broken down or simplified into two subsystems which do not interact because of topological transitivity. And in the midst of this random behavior, we nevertheless have an element of regularity''. Fundamentally different behaviors are consequently possible and occur in an unpredictable way. +Indeed, Banks \emph{et al.} have proven in~\cite{Banks92} that when $f$ is +chaotic and $(\mathcal{X}, d)$ is a metric space, then $f$ has the property of +sensitive dependence on initial conditions (this property was formerly an +element of the definition of chaos). To sum up, quoting Devaney +in~\cite{Devaney}, a chaotic dynamical system ``is unpredictable because of the +sensitive dependence on initial conditions. It cannot be broken down or +simplified into two subsystems which do not interact because of topological +transitivity. And in the midst of this random behavior, we nevertheless have an +element of regularity''. Fundamentally different behaviors are consequently +possible and occur in an unpredictable way. -\subsection{Chaotic iterations} +\subsection{Chaotic Iterations} \label{sec:chaotic iterations} @@ -102,23 +175,23 @@ Boolean \emph{state}. Having $\mathsf{N}$ Boolean values for these cells leads to the definition of a particular \emph{state of the system}. A sequence which elements belong to $\llbracket 1;\mathsf{N} \rrbracket $ is called a \emph{strategy}. The set of all strategies is -denoted by $\mathbb{S}.$ +denoted by $\llbracket 1, \mathsf{N} \rrbracket^\mathds{N}.$ \begin{definition} \label{Def:chaotic iterations} The set $\mathds{B}$ denoting $\{0,1\}$, let $f:\mathds{B}^{\mathsf{N}}\longrightarrow \mathds{B}^{\mathsf{N}}$ be -a function and $S\in \mathbb{S}$ be a strategy. The so-called +a function and $S\in \llbracket 1, \mathsf{N} \rrbracket^\mathds{N}$ be a ``strategy''. The so-called \emph{chaotic iterations} are defined by $x^0\in \mathds{B}^{\mathsf{N}}$ and -$$ +\begin{equation} \forall n\in \mathds{N}^{\ast }, \forall i\in \llbracket1;\mathsf{N}\rrbracket ,x_i^n=\left\{ \begin{array}{ll} x_i^{n-1} & \text{ if }S^n\neq i \\ \left(f(x^{n-1})\right)_{S^n} & \text{ if }S^n=i. \end{array}\right. -$$ +\end{equation} \end{definition} In other words, at the $n^{th}$ iteration, only the $S^{n}-$th cell is @@ -128,49 +201,59 @@ $\left(f(x^{n-1})\right)_{S^{n}}$ can be replaced by $\left(f(x^{k})\right)_{S^{n}}$, where $k0$. \medskip +\begin{itemize} +\item If $\varepsilon \geqslant 1$, we see that 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 $.\bigskip \newline +In conclusion, +$$ +\forall \varepsilon >0,\exists N_{0}=max(n_{0},n_{1},n_{2})\in \mathds{N}% +,\forall n\geqslant N_{0}, + d\left( G_{f}(S^n,E^n);G_{f}(S,E)\right) +\leqslant \varepsilon . +$$ +$G_{f}$ 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 $S'$: $$\tilde +S=(S_0,\dots,S_{t_1-1},S'_0,\dots,S'_{t_2-1},S_0,\dots,S_{t_1-1},S'_0,\dots,S'_{t_2-1},S_0,\dots).$$ 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>$32);\\ +%% x = x\textasciicircum (unsigned int)(t3$>>$32);\\ +%% x = x\textasciicircum (unsigned int)t2;\\ +%% x = x\textasciicircum (unsigned int)(t1$>>$32);\\ +%% x = x\textasciicircum (unsigned int)t3;\\ +%% return x;\\ +%% \} +%% \end{minipage} +%% } +%% \end{center} +%% \caption{sequential Chaotic Iteration PRNG} +%% \label{algo:seqCIprng} +%% \end{figure} + + + +\lstset{language=C,caption={C code of the sequential chaotic iterations based +PRNG},label=algo:seqCIprng} +\begin{lstlisting} +unsigned int CIprng() { + static unsigned int x = 123123123; + unsigned long t1 = xorshift(); + unsigned long t2 = xor128(); + unsigned long t3 = xorwow(); + x = x^(unsigned int)t1; + x = x^(unsigned int)(t2>>32); + x = x^(unsigned int)(t3>>32); + x = x^(unsigned int)t2; + x = x^(unsigned int)(t1>>32); + x = x^(unsigned int)t3; + return x; +} +\end{lstlisting} + + + + + +In listing~\ref{algo:seqCIprng} a sequential version of our chaotic iterations +based PRNG is presented. The xor operator is represented by \textasciicircum. +This function uses three classical 64-bits PRNG: the \texttt{xorshift}, the +\texttt{xor128} and the \texttt{xorwow}. In the following, we call them +xor-like PRNGSs. These three PRNGs are presented in~\cite{Marsaglia2003}. As +each xor-like PRNG used works with 64-bits and as our PRNG works with 32-bits, +the use of \texttt{(unsigned int)} selects the 32 least significant bits whereas +\texttt{(unsigned int)(t3$>>$32)} selects the 32 most significants bits of the +variable \texttt{t}. So to produce a random number realizes 6 xor operations +with 6 32-bits numbers produced by 3 64-bits PRNG. This version successes the +BigCrush of the TestU01 battery~\cite{LEcuyerS07}. + +\section{Efficient prng based on chaotic iterations on GPU} + +In order to benefit from computing power of GPU, a program needs to define +independent blocks of threads which can be computed simultaneously. In general, +the larger the number of threads is, the more local memory is used and the less +branching instructions are used (if, while, ...), the better performance is +obtained on GPU. So with algorithm \ref{algo:seqCIprng} presented in the +previous section, it is possible to build a similar program which computes PRNG +on GPU. In the CUDA [ref] environment, threads have a local identificator, +called \texttt{ThreadIdx} relative to the block containing them. + + +\subsection{Naive version for GPU} + +From the CPU version, it is possible to obtain a quite similar version for GPU. +The principe consists in assigning the computation of a PRNG as in sequential to +each thread of the GPU. Of course, it is essential that the three xor-like +PRNGs used for our computation have different parameters. So we chose them +randomly with another PRNG. As the initialisation is performed by the CPU, we +have chosen to use the ISAAC PRNG [ref] to initalize all the parameters for the +GPU version of our PRNG. The implementation of the three xor-like PRNGs is +straightforward as soon as their parameters have been allocated in the GPU +memory. Each xor-like PRNGs used works with an internal number $x$ which keeps +the last generated random numbers. Other internal variables are also used by the +xor-like PRNGs. More precisely, the implementation of the xor128, the xorshift +and the xorwow respectively require 4, 5 and 6 unsigned long as internal +variables. + +\begin{algorithm} + +\KwIn{InternalVarXorLikeArray: array with internal variables of the 3 xor-like +PRNGs in global memory\; +NumThreads: Number of threads\;} +\KwOut{NewNb: array containing random numbers in global memory} +\If{threadIdx is concerned by the computation} { + retrieve data from InternalVarXorLikeArray[threadIdx] in local variables\; + \For{i=1 to n} { + compute a new PRNG as in Listing\ref{algo:seqCIprng}\; + store the new PRNG in NewNb[NumThreads*threadIdx+i]\; + } + store internal variables in InternalVarXorLikeArray[threadIdx]\; +} + +\caption{main kernel for the chaotic iterations based PRNG GPU naive version} +\label{algo:gpu_kernel} +\end{algorithm} + +Algorithm~\ref{algo:gpu_kernel} presents a naive implementation of PRNG using +GPU. According to the available memory in the GPU and the number of threads +used simultenaously, the number of random numbers that a thread can generate +inside a kernel is limited, i.e. the variable \texttt{n} in +algorithm~\ref{algo:gpu_kernel}. For example, if $100,000$ threads are used and +if $n=100$\footnote{in fact, we need to add the initial seed (a 32-bits number)} +then the memory required to store internals variables of xor-like +PRNGs\footnote{we multiply this number by $2$ in order to count 32-bits numbers} +and random number of our PRNG is equals to $100,000\times ((4+5+6)\times +2+(1+100))=1,310,000$ 32-bits numbers, i.e. about $52$Mb. + +All the tests performed to pass the BigCrush of TestU01 succeeded. Different +number of threads, called \texttt{NumThreads} in our algorithm, have been tested +upto $10$ millions. + +\begin{remark} +Algorithm~\ref{algo:gpu_kernel} has the advantage to manipulate independent +PRNGs, so this version is easily usable on a cluster of computer. The only thing +to ensure is to use a single ISAAC PRNG. For this, a simple solution consists in +using a master node for the initialization which computes the initial parameters +for all the differents nodes involves in the computation. +\end{remark} + +\subsection{Improved version for GPU} + +As GPU cards using CUDA have shared memory between threads of the same block, it +is possible to use this feature in order to simplify the previous algorithm, +i.e., using less than 3 xor-like PRNGs. The solution consists in computing only +one xor-like PRNG by thread, saving it into shared memory and using the results +of some other threads in the same block of threads. In order to define which +thread uses the result of which other one, we can use a permutation array which +contains the indexes of all threads and for which a permutation has been +performed. In Algorithm~\ref{algo:gpu_kernel2}, 2 permutations arrays are used. +The variable \texttt{offset} is computed using the value of +\texttt{permutation\_size}. Then we can compute \texttt{o1} and \texttt{o2} +which represent the indexes of the other threads for which the results are used +by the current thread. In the algorithm, we consider that a 64-bits xor-like +PRNG is used, that is why both 32-bits parts are used. + +This version also succeed to the BigCrush batteries of tests. + +\begin{algorithm} + +\KwIn{InternalVarXorLikeArray: array with internal variables of 1 xor-like PRNGs +in global memory\; +NumThreads: Number of threads\; +tab1, tab2: Arrays containing permutations of size permutation\_size\;} + +\KwOut{NewNb: array containing random numbers in global memory} +\If{threadId is concerned} { + retrieve data from InternalVarXorLikeArray[threadId] in local variables\; + offset = threadIdx\%permutation\_size\; + o1 = threadIdx-offset+tab1[offset]\; + o2 = threadIdx-offset+tab2[offset]\; + \For{i=1 to n} { + t=xor-like()\; + shared\_mem[threadId]=(unsigned int)t\; + x = x $\oplus$ (unsigned int) t\; + x = x $\oplus$ (unsigned int) (t>>32)\; + x = x $\oplus$ shared[o1]\; + x = x $\oplus$ shared[o2]\; + + store the new PRNG in NewNb[NumThreads*threadId+i]\; + } + store internal variables in InternalVarXorLikeArray[threadId]\; +} + +\caption{main kernel for the chaotic iterations based PRNG GPU efficient +version} +\label{algo:gpu_kernel2} +\end{algorithm} + +\subsection{Theoretical Evaluation of the Improved Version} + +A run of Algorithm~\ref{algo:gpu_kernel2} consists in four operations having +the form of Equation~\ref{equation Oplus}, which is equivalent to the iterative +system of Eq.~\ref{eq:generalIC}. That is, four iterations of the general chaotic +iterations are realized between two stored values of the PRNG. +To be certain that we are in the framework of Theorem~\ref{t:chaos des general}, +we must guarantee that this dynamical system iterates on the space +$\mathcal{X} = \mathcal{P}\left(\llbracket 1, \mathsf{N} \rrbracket\right)^\mathds{N}\times\mathds{B}^\mathsf{N}$. +The left term $x$ obviously belongs into $\mathds{B}^ \mathsf{N}$. +To prevent from any flaws of chaotic properties, we must check that each right +term, corresponding to terms of the strategies, can possibly be equal to any +integer of $\llbracket 1, \mathsf{N} \rrbracket$. + +Such a result is obvious for the two first lines, as for the xor-like(), all the +integers belonging into its interval of definition can occur at each iteration. +It can be easily stated for the two last lines by an immediate mathematical +induction. + +Thus Algorithm~\ref{algo:gpu_kernel2} is a concrete realization of the general +chaotic iterations presented previously, and for this reason, it satisfies the +Devaney's formulation of a chaotic behavior. + +\section{Experiments} + +Different experiments have been performed in order to measure the generation +speed. +\begin{figure}[t] +\begin{center} + \includegraphics[scale=.7]{curve_time_gpu.pdf} +\end{center} +\caption{Number of random numbers generated per second} +\label{fig:time_naive_gpu} +\end{figure} + + +First of all we have compared the time to generate X random numbers with both +the CPU version and the GPU version. + +Faire une courbe du nombre de random en fonction du nombre de threads, +éventuellement en fonction du nombres de threads par bloc. \section{The relativity of disorder} \label{sec:de la relativité du désordre} +In the next two sections, we investigate the impact of the choices that have +lead to the definitions of measures in Sections \ref{sec:chaotic iterations} and \ref{deuxième def}. + \subsection{Impact of the topology's finenesse} Let us firstly introduce the following notations. \begin{notation} -$\mathcal{X}_\tau$ will denote the topological space $\left(\mathcal{X},\tau\right)$, whereas $\mathcal{V}_\tau (x)$ will be the set of all the neighborhoods of $x$ when considering the topology $\tau$ (or simply $\mathcal{V} (x)$, if there is no ambiguity). +$\mathcal{X}_\tau$ will denote the topological space +$\left(\mathcal{X},\tau\right)$, whereas $\mathcal{V}_\tau (x)$ will be the set +of all the neighborhoods of $x$ when considering the topology $\tau$ (or simply +$\mathcal{V} (x)$, if there is no ambiguity). \end{notation} \begin{theorem} \label{Th:chaos et finesse} -Let $\mathcal{X}$ a set and $\tau, \tau'$ two topologies on $\mathcal{X}$ s.t. $\tau'$ is finer than $\tau$. Let $f:\mathcal{X} \to \mathcal{X}$, continuous both for $\tau$ and $\tau'$. +Let $\mathcal{X}$ a set and $\tau, \tau'$ two topologies on $\mathcal{X}$ s.t. +$\tau'$ is finer than $\tau$. Let $f:\mathcal{X} \to \mathcal{X}$, continuous +both for $\tau$ and $\tau'$. -If $(\mathcal{X}_{\tau'},f)$ is chaotic according to Devaney, then $(\mathcal{X}_\tau,f)$ is chaotic too. +If $(\mathcal{X}_{\tau'},f)$ is chaotic according to Devaney, then +$(\mathcal{X}_\tau,f)$ is chaotic too. \end{theorem} \begin{proof} Let us firstly establish the transitivity of $(\mathcal{X}_\tau,f)$. -Let $\omega_1, \omega_2$ two open sets of $\tau$. Then $\omega_1, \omega_2 \in \tau'$, becaus $\tau'$ is finer than $\tau$. As $f$ is $\tau'-$transitive, we can deduce that $\exists n \in \mathds{N}, \omega_1 \cap f^{(n)}(\omega_2) = \varnothing$. Consequently, $f$ is $\tau-$transitive. +Let $\omega_1, \omega_2$ two open sets of $\tau$. Then $\omega_1, \omega_2 \in +\tau'$, becaus $\tau'$ is finer than $\tau$. As $f$ is $\tau'-$transitive, we +can deduce that $\exists n \in \mathds{N}, \omega_1 \cap f^{(n)}(\omega_2) = +\varnothing$. Consequently, $f$ is $\tau-$transitive. -Let us now consider the regularity of $(\mathcal{X}_\tau,f)$, \emph{i.e.}, for all $x \in \mathcal{X}$, and for all $\tau-$neighborhood $V$ of $x$, there is a periodic point for $f$ into $V$. +Let us now consider the regularity of $(\mathcal{X}_\tau,f)$, \emph{i.e.}, for +all $x \in \mathcal{X}$, and for all $\tau-$neighborhood $V$ of $x$, there is a +periodic point for $f$ into $V$. -Let $x \in \mathcal{X}$ and $V \in \mathcal{V}_\tau (x)$ a $\tau-$neighborhood of $x$. By definition, $\exists \omega \in \tau, x \in \omega \subset V$. +Let $x \in \mathcal{X}$ and $V \in \mathcal{V}_\tau (x)$ a $\tau-$neighborhood +of $x$. By definition, $\exists \omega \in \tau, x \in \omega \subset V$. -But $\tau \subset \tau'$, so $\omega \in \tau'$, and then $V \in \mathcal{V}_{\tau'} (x)$. As $(\mathcal{X}_{\tau'},f)$ is regular, there is a periodic point for $f$ into $V$, and the regularity of $(\mathcal{X}_\tau,f)$ is proven. +But $\tau \subset \tau'$, so $\omega \in \tau'$, and then $V \in +\mathcal{V}_{\tau'} (x)$. As $(\mathcal{X}_{\tau'},f)$ is regular, there is a +periodic point for $f$ into $V$, and the regularity of $(\mathcal{X}_\tau,f)$ is +proven. \end{proof} \subsection{A given system can always be claimed as chaotic} -Let $f$ an iteration function on $\mathcal{X}$ having at least a fixed point. Then this function is chaotic (in a certain way): +Let $f$ an iteration function on $\mathcal{X}$ having at least a fixed point. +Then this function is chaotic (in a certain way): \begin{theorem} -Let $\mathcal{X}$ a nonempty set and $f: \mathcal{X} \to \X$ a function having at least a fixed point. -Then $f$ is $\tau_0-$chaotic, where $\tau_0$ is the trivial (indiscrete) topology on $\X$. +Let $\mathcal{X}$ a nonempty set and $f: \mathcal{X} \to \X$ a function having +at least a fixed point. +Then $f$ is $\tau_0-$chaotic, where $\tau_0$ is the trivial (indiscrete) +topology on $\X$. \end{theorem} \begin{proof} -$f$ is transitive when $\forall \omega, \omega' \in \tau_0 \setminus \{\varnothing\}, \exists n \in \mathds{N}, f^{(n)}(\omega) \cap \omega' \neq \varnothing$. -As $\tau_0 = \left\{ \varnothing, \X \right\}$, this is equivalent to look for an integer $n$ s.t. $f^{(n)}\left( \X \right) \cap \X \neq \varnothing$. For instance, $n=0$ is appropriate. - -Let us now consider $x \in \X$ and $V \in \mathcal{V}_{\tau_0} (x)$. Then $V = \mathcal{X}$, so $V$ has at least a fixed point for $f$. Consequently $f$ is regular, and the result is established. +$f$ is transitive when $\forall \omega, \omega' \in \tau_0 \setminus +\{\varnothing\}, \exists n \in \mathds{N}, f^{(n)}(\omega) \cap \omega' \neq +\varnothing$. +As $\tau_0 = \left\{ \varnothing, \X \right\}$, this is equivalent to look for +an integer $n$ s.t. $f^{(n)}\left( \X \right) \cap \X \neq \varnothing$. For +instance, $n=0$ is appropriate. + +Let us now consider $x \in \X$ and $V \in \mathcal{V}_{\tau_0} (x)$. Then $V = +\mathcal{X}$, so $V$ has at least a fixed point for $f$. Consequently $f$ is +regular, and the result is established. \end{proof} @@ -346,17 +1028,26 @@ Let us now consider $x \in \X$ and $V \in \mathcal{V}_{\tau_0} (x)$. Then $V = \ \begin{theorem} Let $\mathcal{X}$ be a set and $f: \mathcal{X} \to \X$. -If $\X$ is infinite, then $\left( \X_{\tau_\infty}, f\right)$ is not chaotic (for the Devaney's formulation), where $\tau_\infty$ is the discrete topology. +If $\X$ is infinite, then $\left( \X_{\tau_\infty}, f\right)$ is not chaotic +(for the Devaney's formulation), where $\tau_\infty$ is the discrete topology. \end{theorem} \begin{proof} -Let us prove it by contradiction, assuming that $\left(\X_{\tau_\infty}, f\right)$ is both transitive and regular. +Let us prove it by contradiction, assuming that $\left(\X_{\tau_\infty}, +f\right)$ is both transitive and regular. -Let $x \in \X$ and $\{x\}$ one of its neighborhood. This neighborhood must contain a periodic point for $f$, if we want that $\left(\X_{\tau_\infty}, f\right)$ is regular. Then $x$ must be a periodic point of $f$. +Let $x \in \X$ and $\{x\}$ one of its neighborhood. This neighborhood must +contain a periodic point for $f$, if we want that $\left(\X_{\tau_\infty}, +f\right)$ is regular. Then $x$ must be a periodic point of $f$. -Let $I_x = \left\{ f^{(n)}(x), n \in \mathds{N}\right\}$. This set is finite because $x$ is periodic, and $\mathcal{X}$ is infinite, then $\exists y \in \mathcal{X}, y \notin I_x$. +Let $I_x = \left\{ f^{(n)}(x), n \in \mathds{N}\right\}$. This set is finite +because $x$ is periodic, and $\mathcal{X}$ is infinite, then $\exists y \in +\mathcal{X}, y \notin I_x$. -As $\left(\X_{\tau_\infty}, f\right)$ must be transitive, for all open nonempty sets $A$ and $B$, an integer $n$ must satisfy $f^{(n)}(A) \cap B \neq \varnothing$. However $\{x\}$ and $\{y\}$ are open sets and $y \notin I_x \Rightarrow \forall n, f^{(n)}\left( \{x\} \right) \cap \{y\} = \varnothing$. +As $\left(\X_{\tau_\infty}, f\right)$ must be transitive, for all open nonempty +sets $A$ and $B$, an integer $n$ must satisfy $f^{(n)}(A) \cap B \neq +\varnothing$. However $\{x\}$ and $\{y\}$ are open sets and $y \notin I_x +\Rightarrow \forall n, f^{(n)}\left( \{x\} \right) \cap \{y\} = \varnothing$. \end{proof} @@ -370,77 +1061,94 @@ As $\left(\X_{\tau_\infty}, f\right)$ must be transitive, for all open nonempty \subsubsection{Toward a topological semiconjugacy} -In what follows, our intention is to establish, by using a topological semiconjugacy, that chaotic iterations over $\mathcal{X}$ can be described as iterations on a real interval. To do so, we must firstly introduce some notations and terminologies. +In what follows, our intention is to establish, by using a topological +semiconjugacy, that chaotic iterations over $\mathcal{X}$ can be described as +iterations on a real interval. To do so, we must firstly introduce some +notations and terminologies. -Let $\mathcal{S}_\mathsf{N}$ be the set of sequences belonging into $\llbracket 1; \mathsf{N}\rrbracket$ and $\mathcal{X}_{\mathsf{N}} = \mathcal{S}_\mathsf{N} \times \B^\mathsf{N}$. +Let $\mathcal{S}_\mathsf{N}$ be the set of sequences belonging into $\llbracket +1; \mathsf{N}\rrbracket$ and $\mathcal{X}_{\mathsf{N}} = \mathcal{S}_\mathsf{N} +\times \B^\mathsf{N}$. \begin{definition} -The function $\varphi: \mathcal{S}_{10} \times\mathds{B}^{10} \rightarrow \big[ 0, 2^{10} \big[$ is defined by: -$$ -\begin{array}{cccl} -\varphi: & \mathcal{X}_{10} = \mathcal{S}_{10} \times\mathds{B}^{10}& \longrightarrow & \big[ 0, 2^{10} \big[ \\ - & (S,E) = \left((S^0, S^1, \hdots ); (E_0, \hdots, E_9)\right) & \longmapsto & \varphi \left((S,E)\right) +The function $\varphi: \mathcal{S}_{10} \times\mathds{B}^{10} \rightarrow \big[ +0, 2^{10} \big[$ is defined by: +\begin{equation} + \begin{array}{cccl} +\varphi: & \mathcal{X}_{10} = \mathcal{S}_{10} \times\mathds{B}^{10}& +\longrightarrow & \big[ 0, 2^{10} \big[ \\ + & (S,E) = \left((S^0, S^1, \hdots ); (E_0, \hdots, E_9)\right) & \longmapsto & +\varphi \left((S,E)\right) \end{array} -$$ -\noindent where $\varphi\left((S,E)\right)$ is the real number: +\end{equation} +where $\varphi\left((S,E)\right)$ is the real number: \begin{itemize} -\item whose integral part $e$ is $\displaystyle{\sum_{k=0}^9 2^{9-k} E_k}$, that is, the binary digits of $e$ are $E_0 ~ E_1 ~ \hdots ~ E_9$. -\item whose decimal part $s$ is equal to $s = 0,S^0~ S^1~ S^2~ \hdots = \sum_{k=1}^{+\infty} 10^{-k} S^{k-1}.$ +\item whose integral part $e$ is $\displaystyle{\sum_{k=0}^9 2^{9-k} E_k}$, that +is, the binary digits of $e$ are $E_0 ~ E_1 ~ \hdots ~ E_9$. +\item whose decimal part $s$ is equal to $s = 0,S^0~ S^1~ S^2~ \hdots = +\sum_{k=1}^{+\infty} 10^{-k} S^{k-1}.$ \end{itemize} \end{definition} -$\varphi$ realizes the association between a point of $\mathcal{X}_{10}$ and a real number into $\big[ 0, 2^{10} \big[$. We must now translate the chaotic iterations $\Go$ on this real interval. To do so, two intermediate functions over $\big[ 0, 2^{10} \big[$ must be introduced: +$\varphi$ realizes the association between a point of $\mathcal{X}_{10}$ and a +real number into $\big[ 0, 2^{10} \big[$. We must now translate the chaotic +iterations $\Go$ on this real interval. To do so, two intermediate functions +over $\big[ 0, 2^{10} \big[$ must be introduced: \begin{definition} \label{def:e et s} Let $x \in \big[ 0, 2^{10} \big[$ and: \begin{itemize} -\item $e_0, \hdots, e_9$ the binary digits of the integral part of $x$: $\displaystyle{\lfloor x \rfloor = \sum_{k=0}^{9} 2^{9-k} e_k}$. -\item $(s^k)_{k\in \mathds{N}}$ the digits of $x$, where the chosen decimal decomposition of $x$ is the one that does not have an infinite number of 9: +\item $e_0, \hdots, e_9$ the binary digits of the integral part of $x$: +$\displaystyle{\lfloor x \rfloor = \sum_{k=0}^{9} 2^{9-k} e_k}$. +\item $(s^k)_{k\in \mathds{N}}$ the digits of $x$, where the chosen decimal +decomposition of $x$ is the one that does not have an infinite number of 9: $\displaystyle{x = \lfloor x \rfloor + \sum_{k=0}^{+\infty} s^k 10^{-k-1}}$. \end{itemize} $e$ and $s$ are thus defined as follows: -$$ +\begin{equation} \begin{array}{cccl} e: & \big[ 0, 2^{10} \big[ & \longrightarrow & \mathds{B}^{10} \\ & x & \longmapsto & (e_0, \hdots, e_9) \end{array} -$$ -\noindent and -$$ -\begin{array}{cccl} -s: & \big[ 0, 2^{10} \big[ & \longrightarrow & \llbracket 0, 9 \rrbracket^{\mathds{N}} \\ +\end{equation} +and +\begin{equation} + \begin{array}{cccc} +s: & \big[ 0, 2^{10} \big[ & \longrightarrow & \llbracket 0, 9 +\rrbracket^{\mathds{N}} \\ & x & \longmapsto & (s^k)_{k \in \mathds{N}} \end{array} -$$ +\end{equation} \end{definition} -We are now able to define the function $g$, whose goal is to translate the chaotic iterations $\Go$ on an interval of $\mathds{R}$. +We are now able to define the function $g$, whose goal is to translate the +chaotic iterations $\Go$ on an interval of $\mathds{R}$. \begin{definition} $g:\big[ 0, 2^{10} \big[ \longrightarrow \big[ 0, 2^{10} \big[$ is defined by: -$$ -\begin{array}{cccl} +\begin{equation} +\begin{array}{cccc} g: & \big[ 0, 2^{10} \big[ & \longrightarrow & \big[ 0, 2^{10} \big[ \\ -& \\ & x & \longmapsto & g(x) \end{array} -$$ -\noindent where g(x) is the real number of $\big[ 0, 2^{10} \big[$ defined bellow: +\end{equation} +where g(x) is the real number of $\big[ 0, 2^{10} \big[$ defined bellow: \begin{itemize} -\item its integral part has a binary decomposition equal to $e_0', \hdots, e_9'$, with: -$$ +\item its integral part has a binary decomposition equal to $e_0', \hdots, +e_9'$, with: + \begin{equation} e_i' = \left\{ \begin{array}{ll} e(x)_i & \textrm{ if } i \neq s^0\\ e(x)_i + 1 \textrm{ (mod 2)} & \textrm{ if } i = s^0\\ \end{array} \right. -$$ +\end{equation} \item whose decimal part is $s(x)^1, s(x)^2, \hdots$ \end{itemize} \end{definition} @@ -448,28 +1156,43 @@ $$ \bigskip -In other words, if $x = \displaystyle{\sum_{k=0}^{9} 2^{9-k} e_k + \sum_{k=0}^{+\infty} s^{k} ~10^{-k-1}}$, then: $$g(x) = \displaystyle{\sum_{k=0}^{9} 2^{9-k} (e_k + \delta(k,s^0) \textrm{ (mod 2)}) + \sum_{k=0}^{+\infty} s^{k+1} 10^{-k-1}}.$$ +In other words, if $x = \displaystyle{\sum_{k=0}^{9} 2^{9-k} e_k + +\sum_{k=0}^{+\infty} s^{k} ~10^{-k-1}}$, then: +\begin{equation} +g(x) = +\displaystyle{\sum_{k=0}^{9} 2^{9-k} (e_k + \delta(k,s^0) \textrm{ (mod 2)}) + +\sum_{k=0}^{+\infty} s^{k+1} 10^{-k-1}}. +\end{equation} + \subsubsection{Defining a metric on $\big[ 0, 2^{10} \big[$} -Numerous metrics can be defined on the set $\big[ 0, 2^{10} \big[$, the most usual one being the Euclidian distance recalled bellow: +Numerous metrics can be defined on the set $\big[ 0, 2^{10} \big[$, the most +usual one being the Euclidian distance recalled bellow: \begin{notation} \index{distance!euclidienne} -$\Delta$ is the Euclidian distance on $\big[ 0, 2^{10} \big[$, that is, $\Delta(x,y) = |y-x|^2$. +$\Delta$ is the Euclidian distance on $\big[ 0, 2^{10} \big[$, that is, +$\Delta(x,y) = |y-x|^2$. \end{notation} \medskip -This Euclidian distance does not reproduce exactly the notion of proximity induced by our first distance $d$ on $\X$. Indeed $d$ is finer than $\Delta$. This is the reason why we have to introduce the following metric: +This Euclidian distance does not reproduce exactly the notion of proximity +induced by our first distance $d$ on $\X$. Indeed $d$ is finer than $\Delta$. +This is the reason why we have to introduce the following metric: \begin{definition} Let $x,y \in \big[ 0, 2^{10} \big[$. -$D$ denotes the function from $\big[ 0, 2^{10} \big[^2$ to $\mathds{R}^+$ defined by: $D(x,y) = D_e\left(e(x),e(y)\right) + D_s\left(s(x),s(y)\right)$, where: +$D$ denotes the function from $\big[ 0, 2^{10} \big[^2$ to $\mathds{R}^+$ +defined by: $D(x,y) = D_e\left(e(x),e(y)\right) + D_s\left(s(x),s(y)\right)$, +where: \begin{center} -$\displaystyle{D_e(E,\check{E}) = \sum_{k=0}^\mathsf{9} \delta (E_k, \check{E}_k)}$, ~~and~ $\displaystyle{D_s(S,\check{S}) = \sum_{k = 1}^\infty \dfrac{|S^k-\check{S}^k|}{10^k}}$. +$\displaystyle{D_e(E,\check{E}) = \sum_{k=0}^\mathsf{9} \delta (E_k, +\check{E}_k)}$, ~~and~ $\displaystyle{D_s(S,\check{S}) = \sum_{k = 1}^\infty +\dfrac{|S^k-\check{S}^k|}{10^k}}$. \end{center} \end{definition} @@ -480,21 +1203,35 @@ $D$ is a distance on $\big[ 0, 2^{10} \big[$. \begin{proof} The three axioms defining a distance must be checked. \begin{itemize} -\item $D \geqslant 0$, because everything is positive in its definition. If $D(x,y)=0$, then $D_e(x,y)=0$, so the integral parts of $x$ and $y$ are equal (they have the same binary decomposition). Additionally, $D_s(x,y) = 0$, then $\forall k \in \mathds{N}^*, s(x)^k = s(y)^k$. In other words, $x$ and $y$ have the same $k-$th decimal digit, $\forall k \in \mathds{N}^*$. And so $x=y$. +\item $D \geqslant 0$, because everything is positive in its definition. If +$D(x,y)=0$, then $D_e(x,y)=0$, so the integral parts of $x$ and $y$ are equal +(they have the same binary decomposition). Additionally, $D_s(x,y) = 0$, then +$\forall k \in \mathds{N}^*, s(x)^k = s(y)^k$. In other words, $x$ and $y$ have +the same $k-$th decimal digit, $\forall k \in \mathds{N}^*$. And so $x=y$. \item $D(x,y)=D(y,x)$. -\item Finally, the triangular inequality is obtained due to the fact that both $\delta$ and $\Delta(x,y)=|x-y|$ satisfy it. +\item Finally, the triangular inequality is obtained due to the fact that both +$\delta$ and $\Delta(x,y)=|x-y|$ satisfy it. \end{itemize} \end{proof} -The convergence of sequences according to $D$ is not the same than the usual convergence related to the Euclidian metric. For instance, if $x^n \to x$ according to $D$, then necessarily the integral part of each $x^n$ is equal to the integral part of $x$ (at least after a given threshold), and the decimal part of $x^n$ corresponds to the one of $x$ ``as far as required''. -To illustrate this fact, a comparison between $D$ and the Euclidian distance is given Figure \ref{fig:comparaison de distances}. These illustrations show that $D$ is richer and more refined than the Euclidian distance, and thus is more precise. +The convergence of sequences according to $D$ is not the same than the usual +convergence related to the Euclidian metric. For instance, if $x^n \to x$ +according to $D$, then necessarily the integral part of each $x^n$ is equal to +the integral part of $x$ (at least after a given threshold), and the decimal +part of $x^n$ corresponds to the one of $x$ ``as far as required''. +To illustrate this fact, a comparison between $D$ and the Euclidian distance is +given Figure \ref{fig:comparaison de distances}. These illustrations show that +$D$ is richer and more refined than the Euclidian distance, and thus is more +precise. \begin{figure}[t] \begin{center} - \subfigure[Function $x \to dist(x;1,234) $ on the interval $(0;5)$.]{\includegraphics[scale=.35]{DvsEuclidien.pdf}}\quad - \subfigure[Function $x \to dist(x;3) $ on the interval $(0;5)$.]{\includegraphics[scale=.35]{DvsEuclidien2.pdf}} + \subfigure[Function $x \to dist(x;1,234) $ on the interval +$(0;5)$.]{\includegraphics[scale=.35]{DvsEuclidien.pdf}}\quad + \subfigure[Function $x \to dist(x;3) $ on the interval +$(0;5)$.]{\includegraphics[scale=.35]{DvsEuclidien2.pdf}} \end{center} \caption{Comparison between $D$ (in blue) and the Euclidian distane (in green).} \label{fig:comparaison de distances} @@ -505,15 +1242,19 @@ To illustrate this fact, a comparison between $D$ and the Euclidian distance is \subsubsection{The semiconjugacy} -It is now possible to define a topological semiconjugacy between $\mathcal{X}$ and an interval of $\mathds{R}$: +It is now possible to define a topological semiconjugacy between $\mathcal{X}$ +and an interval of $\mathds{R}$: \begin{theorem} -Chaotic iterations on the phase space $\mathcal{X}$ are simple iterations on $\mathds{R}$, which is illustrated by the semiconjugacy of the diagram bellow: +Chaotic iterations on the phase space $\mathcal{X}$ are simple iterations on +$\mathds{R}$, which is illustrated by the semiconjugacy of the diagram bellow: \begin{equation*} \begin{CD} -\left(~\mathcal{S}_{10} \times\mathds{B}^{10}, d~\right) @>G_{f_0}>> \left(~\mathcal{S}_{10} \times\mathds{B}^{10}, d~\right)\\ +\left(~\mathcal{S}_{10} \times\mathds{B}^{10}, d~\right) @>G_{f_0}>> +\left(~\mathcal{S}_{10} \times\mathds{B}^{10}, d~\right)\\ @V{\varphi}VV @VV{\varphi}V\\ -\left( ~\big[ 0, 2^{10} \big[, D~\right) @>>g> \left(~\big[ 0, 2^{10} \big[, D~\right) +\left( ~\big[ 0, 2^{10} \big[, D~\right) @>>g> \left(~\big[ 0, 2^{10} \big[, +D~\right) \end{CD} \end{equation*} \end{theorem} @@ -522,7 +1263,8 @@ Chaotic iterations on the phase space $\mathcal{X}$ are simple iterations on $\m $\varphi$ has been constructed in order to be continuous and onto. \end{proof} -In other words, $\mathcal{X}$ is approximately equal to $\big[ 0, 2^\mathsf{N} \big[$. +In other words, $\mathcal{X}$ is approximately equal to $\big[ 0, 2^\mathsf{N} +\big[$. @@ -534,10 +1276,14 @@ In other words, $\mathcal{X}$ is approximately equal to $\big[ 0, 2^\mathsf{N} \ \begin{figure}[t] \begin{center} - \subfigure[ICs on the interval $(0,9;1)$.]{\includegraphics[scale=.35]{ICs09a1.pdf}}\quad - \subfigure[ICs on the interval $(0,7;1)$.]{\includegraphics[scale=.35]{ICs07a95.pdf}}\\ - \subfigure[ICs on the interval $(0,5;1)$.]{\includegraphics[scale=.35]{ICs05a1.pdf}}\quad - \subfigure[ICs on the interval $(0;1)$]{\includegraphics[scale=.35]{ICs0a1.pdf}} + \subfigure[ICs on the interval +$(0,9;1)$.]{\includegraphics[scale=.35]{ICs09a1.pdf}}\quad + \subfigure[ICs on the interval +$(0,7;1)$.]{\includegraphics[scale=.35]{ICs07a95.pdf}}\\ + \subfigure[ICs on the interval +$(0,5;1)$.]{\includegraphics[scale=.35]{ICs05a1.pdf}}\quad + \subfigure[ICs on the interval +$(0;1)$]{\includegraphics[scale=.35]{ICs0a1.pdf}} \end{center} \caption{Representation of the chaotic iterations.} \label{fig:ICs} @@ -548,8 +1294,10 @@ In other words, $\mathcal{X}$ is approximately equal to $\big[ 0, 2^\mathsf{N} \ \begin{figure}[t] \begin{center} - \subfigure[ICs on the interval $(510;514)$.]{\includegraphics[scale=.35]{ICs510a514.pdf}}\quad - \subfigure[ICs on the interval $(1000;1008)$]{\includegraphics[scale=.35]{ICs1000a1008.pdf}} + \subfigure[ICs on the interval +$(510;514)$.]{\includegraphics[scale=.35]{ICs510a514.pdf}}\quad + \subfigure[ICs on the interval +$(1000;1008)$]{\includegraphics[scale=.35]{ICs1000a1008.pdf}} \end{center} \caption{ICs on small intervals.} \label{fig:ICs2} @@ -557,49 +1305,78 @@ In other words, $\mathcal{X}$ is approximately equal to $\big[ 0, 2^\mathsf{N} \ \begin{figure}[t] \begin{center} - \subfigure[ICs on the interval $(0;16)$.]{\includegraphics[scale=.3]{ICs0a16.pdf}}\quad - \subfigure[ICs on the interval $(40;70)$.]{\includegraphics[scale=.45]{ICs40a70.pdf}}\quad + \subfigure[ICs on the interval +$(0;16)$.]{\includegraphics[scale=.3]{ICs0a16.pdf}}\quad + \subfigure[ICs on the interval +$(40;70)$.]{\includegraphics[scale=.45]{ICs40a70.pdf}}\quad \end{center} \caption{General aspect of the chaotic iterations.} \label{fig:ICs3} \end{figure} -We have written a Python program to represent the chaotic iterations with the vectorial negation on the real line $\mathds{R}$. Various representations of these CIs are given in Figures \ref{fig:ICs}, \ref{fig:ICs2} and \ref{fig:ICs3}. It can be remarked that the function $g$ is a piecewise linear function: it is linear on each interval having the form $\left[ \dfrac{n}{10}, \dfrac{n+1}{10}\right[$, $n \in \llbracket 0;2^{10}\times 10 \rrbracket$ and its slope is equal to 10. Let us justify these claims: +We have written a Python program to represent the chaotic iterations with the +vectorial negation on the real line $\mathds{R}$. Various representations of +these CIs are given in Figures \ref{fig:ICs}, \ref{fig:ICs2} and \ref{fig:ICs3}. +It can be remarked that the function $g$ is a piecewise linear function: it is +linear on each interval having the form $\left[ \dfrac{n}{10}, +\dfrac{n+1}{10}\right[$, $n \in \llbracket 0;2^{10}\times 10 \rrbracket$ and its +slope is equal to 10. Let us justify these claims: \begin{proposition} \label{Prop:derivabilite des ICs} -Chaotic iterations $g$ defined on $\mathds{R}$ have derivatives of all orders on $\big[ 0, 2^{10} \big[$, except on the 10241 points in $I$ defined by $\left\{ \dfrac{n}{10} ~\big/~ n \in \llbracket 0;2^{10}\times 10\rrbracket \right\}$. - -Furthermore, on each interval of the form $\left[ \dfrac{n}{10}, \dfrac{n+1}{10}\right[$, with $n \in \llbracket 0;2^{10}\times 10 \rrbracket$, $g$ is a linear function, having a slope equal to 10: $\forall x \notin I, g'(x)=10$. +Chaotic iterations $g$ defined on $\mathds{R}$ have derivatives of all orders on +$\big[ 0, 2^{10} \big[$, except on the 10241 points in $I$ defined by $\left\{ +\dfrac{n}{10} ~\big/~ n \in \llbracket 0;2^{10}\times 10\rrbracket \right\}$. + +Furthermore, on each interval of the form $\left[ \dfrac{n}{10}, +\dfrac{n+1}{10}\right[$, with $n \in \llbracket 0;2^{10}\times 10 \rrbracket$, +$g$ is a linear function, having a slope equal to 10: $\forall x \notin I, +g'(x)=10$. \end{proposition} \begin{proof} -Let $I_n = \left[ \dfrac{n}{10}, \dfrac{n+1}{10}\right[$, with $n \in \llbracket 0;2^{10}\times 10 \rrbracket$. All the points of $I_n$ have the same integral prat $e$ and the same decimal part $s^0$: on the set $I_n$, functions $e(x)$ and $x \mapsto s(x)^0$ of Definition \ref{def:e et s} only depend on $n$. So all the images $g(x)$ of these points $x$: +Let $I_n = \left[ \dfrac{n}{10}, \dfrac{n+1}{10}\right[$, with $n \in \llbracket +0;2^{10}\times 10 \rrbracket$. All the points of $I_n$ have the same integral +prat $e$ and the same decimal part $s^0$: on the set $I_n$, functions $e(x)$ +and $x \mapsto s(x)^0$ of Definition \ref{def:e et s} only depend on $n$. So all +the images $g(x)$ of these points $x$: \begin{itemize} -\item Have the same integral part, which is $e$, except probably the bit number $s^0$. In other words, this integer has approximately the same binary decomposition than $e$, the sole exception being the digit $s^0$ (this number is then either $e+2^{10-s^0}$ or $e-2^{10-s^0}$, depending on the parity of $s^0$, \emph{i.e.}, it is equal to $e+(-1)^{s^0}\times 2^{10-s^0}$). -\item A shift to the left has been applied to the decimal part $y$, losing by doing so the common first digit $s^0$. In other words, $y$ has been mapped into $10\times y - s^0$. +\item Have the same integral part, which is $e$, except probably the bit number +$s^0$. In other words, this integer has approximately the same binary +decomposition than $e$, the sole exception being the digit $s^0$ (this number is +then either $e+2^{10-s^0}$ or $e-2^{10-s^0}$, depending on the parity of $s^0$, +\emph{i.e.}, it is equal to $e+(-1)^{s^0}\times 2^{10-s^0}$). +\item A shift to the left has been applied to the decimal part $y$, losing by +doing so the common first digit $s^0$. In other words, $y$ has been mapped into +$10\times y - s^0$. \end{itemize} -To sum up, the action of $g$ on the points of $I$ is as follows: first, make a multiplication by 10, and second, add the same constant to each term, which is $\dfrac{1}{10}\left(e+(-1)^{s^0}\times 2^{10-s^0}\right)-s^0$. +To sum up, the action of $g$ on the points of $I$ is as follows: first, make a +multiplication by 10, and second, add the same constant to each term, which is +$\dfrac{1}{10}\left(e+(-1)^{s^0}\times 2^{10-s^0}\right)-s^0$. \end{proof} \begin{remark} -Finally, chaotic iterations are elements of the large family of functions that are both chaotic and piecewise linear (like the tent map). +Finally, chaotic iterations are elements of the large family of functions that +are both chaotic and piecewise linear (like the tent map). \end{remark} \subsection{Comparison of the two metrics on $\big[ 0, 2^\mathsf{N} \big[$} -The two propositions bellow allow to compare our two distances on $\big[ 0, 2^\mathsf{N} \big[$: +The two propositions bellow allow to compare our two distances on $\big[ 0, +2^\mathsf{N} \big[$: \begin{proposition} -Id: $\left(~\big[ 0, 2^\mathsf{N} \big[,\Delta~\right) \to \left(~\big[ 0, 2^\mathsf{N} \big[, D~\right)$ is not continuous. +Id: $\left(~\big[ 0, 2^\mathsf{N} \big[,\Delta~\right) \to \left(~\big[ 0, +2^\mathsf{N} \big[, D~\right)$ is not continuous. \end{proposition} \begin{proof} -The sequence $x^n = 1,999\hdots 999$ constituted by $n$ 9 as decimal part, is such that: +The sequence $x^n = 1,999\hdots 999$ constituted by $n$ 9 as decimal part, is +such that: \begin{itemize} \item $\Delta (x^n,2) \to 0.$ \item But $D(x^n,2) \geqslant 1$, then $D(x^n,2)$ does not converge to 0. @@ -613,16 +1390,25 @@ The sequential characterization of the continuity concludes the demonstration. A contrario: \begin{proposition} -Id: $\left(~\big[ 0, 2^\mathsf{N} \big[,D~\right) \to \left(~\big[ 0, 2^\mathsf{N} \big[, \Delta ~\right)$ is a continuous fonction. +Id: $\left(~\big[ 0, 2^\mathsf{N} \big[,D~\right) \to \left(~\big[ 0, +2^\mathsf{N} \big[, \Delta ~\right)$ is a continuous fonction. \end{proposition} \begin{proof} -If $D(x^n,x) \to 0$, then $D_e(x^n,x) = 0$ at least for $n$ larger than a given threshold, because $D_e$ only returns integers. So, after this threshold, the integral parts of all the $x^n$ are equal to the integral part of $x$. - -Additionally, $D_s(x^n, x) \to 0$, then $\forall k \in \mathds{N}^*, \exists N_k \in \mathds{N}, n \geqslant N_k \Rightarrow D_s(x^n,x) \leqslant 10^{-k}$. This means that for all $k$, an index $N_k$ can be found such that, $\forall n \geqslant N_k$, all the $x^n$ have the same $k$ firsts digits, which are the digits of $x$. We can deduce the convergence $\Delta(x^n,x) \to 0$, and thus the result. +If $D(x^n,x) \to 0$, then $D_e(x^n,x) = 0$ at least for $n$ larger than a given +threshold, because $D_e$ only returns integers. So, after this threshold, the +integral parts of all the $x^n$ are equal to the integral part of $x$. + +Additionally, $D_s(x^n, x) \to 0$, then $\forall k \in \mathds{N}^*, \exists N_k +\in \mathds{N}, n \geqslant N_k \Rightarrow D_s(x^n,x) \leqslant 10^{-k}$. This +means that for all $k$, an index $N_k$ can be found such that, $\forall n +\geqslant N_k$, all the $x^n$ have the same $k$ firsts digits, which are the +digits of $x$. We can deduce the convergence $\Delta(x^n,x) \to 0$, and thus the +result. \end{proof} -The conclusion of these propositions is that the proposed metric is more precise than the Euclidian distance, that is: +The conclusion of these propositions is that the proposed metric is more precise +than the Euclidian distance, that is: \begin{corollary} $D$ is finer than the Euclidian distance $\Delta$. @@ -631,9 +1417,12 @@ $D$ is finer than the Euclidian distance $\Delta$. This corollary can be reformulated as follows: \begin{itemize} -\item The topology produced by $\Delta$ is a subset of the topology produced by $D$. +\item The topology produced by $\Delta$ is a subset of the topology produced by +$D$. \item $D$ has more open sets than $\Delta$. -\item It is harder to converge for the topology $\tau_D$ inherited by $D$, than to converge with the one inherited by $\Delta$, which is denoted here by $\tau_\Delta$. +\item It is harder to converge for the topology $\tau_D$ inherited by $D$, than +to converge with the one inherited by $\Delta$, which is denoted here by +$\tau_\Delta$. \end{itemize} @@ -644,82 +1433,47 @@ This corollary can be reformulated as follows: \subsubsection{Chaos according to Devaney} -We have recalled previously that the chaotic iterations $\left(\Go, \mathcal{X}_d\right)$ are chaotic according to the formulation of Devaney. We can deduce that they are chaotic on $\mathds{R}$ too, when considering the order topology, because: +We have recalled previously that the chaotic iterations $\left(\Go, +\mathcal{X}_d\right)$ are chaotic according to the formulation of Devaney. We +can deduce that they are chaotic on $\mathds{R}$ too, when considering the order +topology, because: \begin{itemize} -\item $\left(\Go, \mathcal{X}_d\right)$ and $\left(g, \big[ 0, 2^{10} \big[_D\right)$ are semiconjugate by $\varphi$, -\item Then $\left(g, \big[ 0, 2^{10} \big[_D\right)$ is a system chaotic according to Devaney, because the semiconjugacy preserve this character. -\item But the topology generated by $D$ is finer than the topology generated by the Euclidian distance $\Delta$ -- which is the order topology. -\item According to Theorem \ref{Th:chaos et finesse}, we can deduce that the chaotic iterations $g$ are indeed chaotic, as defined by Devaney, for the order topology on $\mathds{R}$. +\item $\left(\Go, \mathcal{X}_d\right)$ and $\left(g, \big[ 0, 2^{10} +\big[_D\right)$ are semiconjugate by $\varphi$, +\item Then $\left(g, \big[ 0, 2^{10} \big[_D\right)$ is a system chaotic +according to Devaney, because the semiconjugacy preserve this character. +\item But the topology generated by $D$ is finer than the topology generated by +the Euclidian distance $\Delta$ -- which is the order topology. +\item According to Theorem \ref{Th:chaos et finesse}, we can deduce that the +chaotic iterations $g$ are indeed chaotic, as defined by Devaney, for the order +topology on $\mathds{R}$. \end{itemize} This result can be formulated as follows. \begin{theorem} \label{th:IC et topologie de l'ordre} -The chaotic iterations $g$ on $\mathds{R}$ are chaotic according to the Devaney's formulation, when $\mathds{R}$ has his usual topology, which is the order topology. +The chaotic iterations $g$ on $\mathds{R}$ are chaotic according to the +Devaney's formulation, when $\mathds{R}$ has his usual topology, which is the +order topology. \end{theorem} -Indeed this result is weaker than the theorem establishing the chaos for the finer topology $d$. However the Theorem \ref{th:IC et topologie de l'ordre} still remains important. Indeed, we have studied in our previous works a set different from the usual set of study ($\mathcal{X}$ instead of $\mathds{R}$), in order to be as close as possible from the computer: the properties of disorder proved theoretically will then be preserved when computing. However, we could wonder whether this change does not lead to a disorder of a lower quality. In other words, have we replaced a situation of a good disorder lost when computing, to another situation of a disorder preserved but of bad quality. Theorem \ref{th:IC et topologie de l'ordre} prove exactly the contrary. +Indeed this result is weaker than the theorem establishing the chaos for the +finer topology $d$. However the Theorem \ref{th:IC et topologie de l'ordre} +still remains important. Indeed, we have studied in our previous works a set +different from the usual set of study ($\mathcal{X}$ instead of $\mathds{R}$), +in order to be as close as possible from the computer: the properties of +disorder proved theoretically will then be preserved when computing. However, we +could wonder whether this change does not lead to a disorder of a lower quality. +In other words, have we replaced a situation of a good disorder lost when +computing, to another situation of a disorder preserved but of bad quality. +Theorem \ref{th:IC et topologie de l'ordre} prove exactly the contrary. -\section{Efficient prng based on chaotic iterations} -On parle du séquentiel avec des nombres 64 bits\\ - -Faire le lien avec le paragraphe précédent (je considère que la stratégie s'appelle $S^i$\\ - -In order to implement efficiently a PRNG based on chaotic iterations it is -possible to improve previous works [ref]. One solution consists in considering -that the strategy used $S^i$ contains all the bits for which the negation is -achieved out. Then instead of applying the negation on these bits we can simply -apply the xor operator between the current number and the strategy $S^i$. In -order to obtain the strategy we also use a classical PRNG. - -\begin{figure}[htbp] -\begin{center} -\fbox{ -\begin{minipage}{14cm} -unsigned int CIprng() \{\\ - static unsigned int x = 123123123;\\ - unsigned long t1 = xorshift();\\ - unsigned long t2 = xor128();\\ - unsigned long t3 = xorwow();\\ - x = x\textasciicircum (unsigned int)t1;\\ - x = x\textasciicircum (unsigned int)(t2$>>$32);\\ - x = x\textasciicircum (unsigned int)(t3$>>$32);\\ - x = x\textasciicircum (unsigned int)t2;\\ - x = x\textasciicircum (unsigned int)(t1$>>$32);\\ - x = x\textasciicircum (unsigned int)t3;\\ - return x;\\ -\} -\end{minipage} -} -\end{center} -\caption{sequential Chaotic Iteration PRNG} -\label{algo:seqCIprng} -\end{figure} - -In Figure~\ref{algo:seqCIprng} a sequential version of our chaotic iterations -based PRNG is presented. This version uses three classical 64 bits PRNG: the -\texttt{xorshift}, the \texttt{xor128} and the \texttt{xorwow}. These three -PRNGs are presented in~\cite{Marsaglia2003}. As each PRNG used works with -64-bits and as our PRNG works with 32 bits, the use of \texttt{(unsigned int)} -selects the 32 least significant bits whereas \texttt{(unsigned int)(t3$>>$32)} -selects the 32 most significants bits of the variable \texttt{t}. This version -sucesses the BigCrush of the TestU01 battery [P. L’ecuyer and - R. Simard. Testu01]. - -\section{Efficient prng based on chaotic iterations on GPU} - -On parle du passage du sequentiel au GPU - -\section{Experiments} -On passe le BigCrush\\ -On donne des temps de générations sur GPU/CPU\\ -On donne des temps de générations de nombre sur GPU puis on rappatrie sur CPU / CPU ? bof bof, on verra \section{Conclusion}