\documentclass[11 pt,xcolor={dvipsnames,svgnames,x11names,table}]{beamer}
\usepackage[english]{babel}
\usepackage{polski}
\usepackage[skins,theorems]{tcolorbox}
\tcbset{highlight math style={enhanced,
colframe=red,colback=white,arc=0pt,boxrule=1pt}}
\usetheme[
bullet=circle, % Other option: square
bigpagenumber, % circled page number on lower right
topline=true, % colored bar at the top of the frame
shadow=false, % Shading for beamer blocks
watermark=BG_lower, % png file for the watermark
]{Flip}
%\logo{\kern+1.em\includegraphics[height=1cm]{SHiP-3_LightCharcoal}}
\usepackage[lf]{berenis}
\usepackage[LY1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{emerald}
\usefonttheme{professionalfonts}
\usepackage[no-math]{fontspec}
\usepackage{listings}
\defaultfontfeatures{Mapping=tex-text} % This seems to be important for mapping glyphs properly
\setmainfont{Gillius ADF} % Beamer ignores "main font" in favor of sans font
\setsansfont{Gillius ADF} % This is the font that beamer will use by default
% \setmainfont{Gill Sans Light} % Prettier, but harder to read
\setbeamerfont{title}{family=\fontspec{Gillius ADF}}
\input t1augie.fd
%\newcommand{\handwriting}{\fontspec{augie}} % From Emerald City, free font
%\newcommand{\handwriting}{\usefont{T1}{fau}{m}{n}} % From Emerald City, free font
% \newcommand{\handwriting}{} % If you prefer no special handwriting font or don't have augie
%% Gill Sans doesn't look very nice when boldfaced
%% This is a hack to use Helvetica instead
%% Usage: \textbf{\forbold some stuff}
%\newcommand{\forbold}{\fontspec{Arial}}
\usepackage{graphicx}
\usepackage[export]{adjustbox}
\usepackage{amsmath, amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{bm}
\usepackage{colortbl}
\usepackage{mathrsfs} % For Weinberg-esque letters
\usepackage{cancel} % For "SUSY-breaking" symbol
\usepackage{slashed} % for slashed characters in math mode
\usepackage{bbm} % for \mathbbm{1} (unit matrix)
\usepackage{amsthm} % For theorem environment
\usepackage{multirow} % For multi row cells in table
\usepackage{arydshln} % For dashed lines in arrays and tables
\usepackage{siunitx}
\usepackage{xhfill}
\usepackage{grffile}
\usepackage{textpos}
\usepackage{subfigure}
\usepackage{tikz}
\usepackage{hyperref}
%\usepackage{hepparticles}
\usepackage[italic]{hepparticles}
\usepackage{hepnicenames}
% Drawing a line
\tikzstyle{lw} = [line width=20pt]
\newcommand{\topline}{%
\tikz[remember picture,overlay] {%
\draw[crimsonred] ([yshift=-23.5pt]current page.north west)
-- ([yshift=-23.5pt,xshift=\paperwidth]current page.north west);}}
% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
\usepackage{tikzfeynman} % For Feynman diagrams
\usetikzlibrary{arrows,shapes}
\usetikzlibrary{trees}
\usetikzlibrary{matrix,arrows} % For commutative diagram
% http://www.felixl.de/commu.pdf
\usetikzlibrary{positioning} % For "above of=" commands
\usetikzlibrary{calc,through} % For coordinates
\usetikzlibrary{decorations.pathreplacing} % For curly braces
% http://www.math.ucla.edu/~getreuer/tikz.html
\usepackage{pgffor} % For repeating patterns
\usetikzlibrary{decorations.pathmorphing} % For Feynman Diagrams
\usetikzlibrary{decorations.markings}
\tikzset{
% >=stealth', %% Uncomment for more conventional arrows
vector/.style={decorate, decoration={snake}, draw},
provector/.style={decorate, decoration={snake,amplitude=2.5pt}, draw},
antivector/.style={decorate, decoration={snake,amplitude=-2.5pt}, draw},
fermion/.style={draw=gray, postaction={decorate},
decoration={markings,mark=at position .55 with {\arrow[draw=gray]{>}}}},
fermionbar/.style={draw=gray, postaction={decorate},
decoration={markings,mark=at position .55 with {\arrow[draw=gray]{<}}}},
fermionnoarrow/.style={draw=gray},
gluon/.style={decorate, draw=black,
decoration={coil,amplitude=4pt, segment length=5pt}},
scalar/.style={dashed,draw=black, postaction={decorate},
decoration={markings,mark=at position .55 with {\arrow[draw=black]{>}}}},
scalarbar/.style={dashed,draw=black, postaction={decorate},
decoration={markings,mark=at position .55 with {\arrow[draw=black]{<}}}},
scalarnoarrow/.style={dashed,draw=black},
electron/.style={draw=black, postaction={decorate},
decoration={markings,mark=at position .55 with {\arrow[draw=black]{>}}}},
bigvector/.style={decorate, decoration={snake,amplitude=4pt}, draw},
}
% TIKZ - for block diagrams,
% from http://www.texample.net/tikz/examples/control-system-principles/
% \usetikzlibrary{shapes,arrows}
\tikzstyle{block} = [draw, rectangle,
minimum height=3em, minimum width=6em]
\def\ARROW{{\color{JungleGreen}{$\Rrightarrow$}}\xspace}
\usetikzlibrary{backgrounds}
\usetikzlibrary{mindmap,trees} % For mind map
\newcommand{\degree}{\ensuremath{^\circ}}
\newcommand{\E}{\mathrm{E}}
\newcommand{\Var}{\mathrm{Var}}
\newcommand{\Cov}{\mathrm{Cov}}
\newcommand\Ts{\rule{0pt}{2.6ex}} % Top strut
\newcommand\Bs{\rule[-1.2ex]{0pt}{0pt}} % Bottom strut
\graphicspath{{images/}} % Put all images in this directory. Avoids clutter.
% SOME COMMANDS THAT I FIND HANDY
% \renewcommand{\tilde}{\widetilde} % dinky tildes look silly, dosn't work with fontspec
%\newcommand{\comment}[1]{\textcolor{comment}{\footnotesize{#1}\normalsize}} % comment mild
%\newcommand{\Comment}[1]{\textcolor{Comment}{\footnotesize{#1}\normalsize}} % comment bold
%\newcommand{\COMMENT}[1]{\textcolor{COMMENT}{\footnotesize{#1}\normalsize}} % comment crazy bold
\newcommand{\Alert}[1]{\textcolor{Alert}{#1}} % louder alert
\newcommand{\ALERT}[1]{\textcolor{ALERT}{#1}} % loudest alert
%% "\alert" is already a beamer pre-defined
\newcommand*{\Scale}[2][4]{\scalebox{#1}{$#2$}}%
\def\Put(#1,#2)#3{\leavevmode\makebox(0,0){\put(#1,#2){#3}}}
\usepackage{gmp}
\usepackage[final]{feynmp-auto}
\usepackage[backend=bibtex,style=numeric-comp,firstinits=true]{biblatex}
\bibliography{bib}
\setbeamertemplate{bibliography item}[text]
\makeatletter\let\frametextheight\beamer@frametextheight\makeatother
% suppress frame numbering for backup slides
% you always need the appendix for this!
\newcommand{\backupbegin}{
\newcounter{framenumberappendix}
\setcounter{framenumberappendix}{\value{framenumber}}
}
\newcommand{\backupend}{
\addtocounter{framenumberappendix}{-\value{framenumber}}
\addtocounter{framenumber}{\value{framenumberappendix}}
}
\definecolor{links}{HTML}{2A1B81}
%\hypersetup{colorlinks,linkcolor=,urlcolor=links}
% For shapo's formulas:
\def\lsi{\raise0.3ex\hbox{$<$\kern-0.75em\raise-1.1ex\hbox{$\sim$}}}
\def\gsi{\raise0.3ex\hbox{$>$\kern-0.75em\raise-1.1ex\hbox{$\sim$}}}
\newcommand{\lsim}{\mathop{\lsi}}
\newcommand{\gsim}{\mathop{\gsi}}
\newcommand{\wt}{\widetilde}
%\newcommand{\ol}{\overline}
\newcommand{\Tr}{\rm{Tr}}
\newcommand{\tr}{\rm{tr}}
\newcommand{\eqn}[1]{&\hspace{-0.7em}#1\hspace{-0.7em}&}
\newcommand{\vev}[1]{\rm{$\langle #1 \rangle$}}
\newcommand{\abs}[1]{\rm{$\left| #1 \right|$}}
\newcommand{\eV}{\rm{eV}}
\newcommand{\keV}{\rm{keV}}
\newcommand{\GeV}{\rm{GeV}}
\newcommand{\im}{\rm{Im}}
\newcommand{\disp}{\displaystyle}
\def\be{\begin{equation}}
\def\ee{\end{equation}}
\def\ba{\begin{eqnarray}}
\def\ea{\end{eqnarray}}
\def\d{\partial}
\def\l{\left(}
\def\r{\right)}
\def\la{\langle}
\def\ra{\rangle}
\def\e{{\rm e}}
\def\Br{{\rm Br}}
\def\fixme{{\color{red} FIXME!}}
\def\mc{{\color{Magenta}{MC}}}
\def\pdf{{\rm p.d.f.}}
\def\cdf{{\rm c.d.f.}}
\author{ {\fontspec{Trebuchet MS}Marcin Chrz\k{a}szcz} (Universit\"{a}t Z\"{u}rich)}
\institute{UZH}
\title[Arbitrary \pdf~generation]{Arbitrary \pdf~generation}
\date{\fixme}
\newcommand*{\QEDA}{\hfill\ensuremath{\blacksquare}}%
\newcommand*{\QEDB}{\hfill\ensuremath{\square}}%
\begin{document}
\tikzstyle{every picture}+=[remember picture]
{
\setbeamertemplate{sidebar right}{\llap{\includegraphics[width=\paperwidth,height=\paperheight]{bubble2}}}
\begin{frame}[c]%{\phantom{title page}}
\begin{center}
\begin{center}
\begin{columns}
\begin{column}{0.9\textwidth}
\flushright\fontspec{Trebuchet MS}\bfseries \Huge {Arbitrary \pdf~generation}
\end{column}
\begin{column}{0.2\textwidth}
%\includegraphics[width=\textwidth]{SHiP-2}
\end{column}
\end{columns}
\end{center}
\quad
\vspace{3em}
\begin{columns}
\begin{column}{0.44\textwidth}
\flushright \vspace{-1.8em} {\fontspec{Trebuchet MS} \Large Marcin Chrząszcz\\\vspace{-0.1em}\small \href{mailto:mchrzasz@cern.ch}{mchrzasz@cern.ch}}
\end{column}
\begin{column}{0.53\textwidth}
\includegraphics[height=1.3cm]{uzh-transp}
\end{column}
\end{columns}
\vspace{1em}
% \footnotesize\textcolor{gray}{With N. Serra, B. Storaci\\Thanks to the theory support from M. Shaposhnikov, D. Gorbunov}\normalsize\\
\vspace{0.5em}
\textcolor{normal text.fg!50!Comment}{Monte Carlo methods, \\ 7 April, 2016}
\end{center}
\end{frame}
}
\begin{frame}\frametitle{Reverting the \cdf}
\begin{footnotesize}
\begin{alertblock}{~}
\ARROW Let $U$ be a random variable from $\mathcal{U}(0,1)$\\
\ARROW Now let $F$ be a non decreasing function such that:
\begin{align*}
F(- \infty) =0~~~~F(\infty)=1
\end{align*}
then:
\begin{align*}
X=F^{-1}(U)
\end{align*}
has a \pdf~ distribution with a \cdf~function of F.
\end{alertblock}
\ARROW Prove:
\begin{align*}
F(x)=\mathcal{P}(U\leq F(x))=\mathcal{P}(F^{-1}(U) \leq x) = \mathcal{P}(X \leq x) ~~~~ \QEDB
\end{align*}
\ARROW So it looks very simple if $x_1,X_2,...,X_n$ are random variables from $\mathcal{U}(0,1)$ then: $\lbrace X_i=F^{-1}(x_i)\rbrace,i=1,...,n$ is the sequence that has a \cdf~distirbution of $F$.
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Reverting the \cdf, examples}
\begin{footnotesize}
\ARROW The exponential distribution: $E(0,1)$. \\
\ARROW The \pdf : $\rho(X)=e^{-X},~X \geq 0$.\\
\ARROW The \cdf : $F(x) = \int_0^x e^{-X} dX= 1- e^{-x}.$\\
\ARROW Now let $R \in \mathcal{U}(0,1):~R=F(X)=1-e^{-X} \longrightarrow X = -\ln(1-R)$\\
\ARROW Now we can play a trick: if $R \in \mathcal{U}(0,1)$ then $1-R$ also in $\mathcal{U}(0,1)$. \\
\ARROW In the end we get: $X=-\ln(R)$ \\
{~}\\
\ARROW Use the reverting to generate the following distributions:
\begin{itemize}
\item E 6.1 $\rho(X)= \dfrac{c}{X}$ on the interval $[a,b]$, where $0<a<b<\infty$.
\item E6.2 The Breit-Wigner function:
\begin{align*}
\rho_{\theta,\lambda}(X)=\dfrac{\lambda}{\pi} \dfrac{1}{(X-\theta)^2 + \lambda^2}
\end{align*}
Hit: First do $C(0,1)$ then transform the variables.
\end{itemize}
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Reverting the \cdf, general case}
\begin{footnotesize}
\begin{alertblock}
\ARROW Lets assume: $F$ - non decreasing function such that: $F(-\infty)=0$ and $F(\infty)=1$. Then a random variable $X$:
\begin{align*}
X=\inf \lbrace x: U \leq F(x) \rbrace
\end{align*}
has a distribution with a \cdf~of $F$
\end{alertblock}
\ARROW Prove:\\
One needs just to prove that if an event $\lbrace X \leq t \rbrace$ occurs then $\Leftrightarrow$ $\lbrace U \leq F(t)\rbrace $ occurs.\\
\fixme
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Example in C++}
\begin{footnotesize}
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Reverting the \cdf, prose and cones}
\begin{footnotesize}
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Overlaping-pairs-sparse-occupancy}
\begin{footnotesize}
\ARROW The OPSO (G.Marsaglia 1984)is an analysis of pairs obtained from random number generator.\\
\begin{exampleblock}{}
$X_1,X_2,...,X_n$ - $n$ random numbers obtained from generator. From each number we take $b$ bits from which we construct a second series: $I_1, I_2,...,I_n$, where $I_j \in \left[0,1,...,2^b-1 \right]$.\end{exampleblock}
\ARROW Next we create the pair series:
\begin{align*}
(I_1,I_2),(I_2,I_3),...(I_{n-1}, I_n)
\end{align*}
\ARROW $Y$ - number of pairs from : ${ (i,j):i,j=0,1,...,2^b-1}$, which DIDN'T occur in the above series.
\begin{center}
\includegraphics[width=0.35\textwidth]{images/test1.png}
\end{center}
\ARROW This kind of test can be exteded to triple-pairs, and quadro-pairs.\\
\ARROW See DIEHARD G.Marsgalia 1993 \href{http://stat.fsu.edu/pub/diehard/}{http://stat.fsu.edu/pub/diehard/}
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Kolomogorox-Smirnow}
\begin{footnotesize}
\ARROW The K-S test is used to check if a Random variable has pdf of a distribution $F$. The test is based on the difference between the two distributions:
\begin{align*}
D_n=\sup_{-\infty < x<\infty} \vert F_n(x) -F(x) \vert,~~~~~F_n =\dfrac{1}{n}\sum_{j=1}^n \Theta(x-X_j).
\end{align*}
\ARROW If the random generator is from the $F$ distribution then the $D_n \to 0$ with the probability 1.\\ \ARROW Large values of $D_n$ exclude the generator.
\ARROW The critical values of the test $D_n(\alpha)$ can be find in the mathematical tables for every $\alpha$:
\begin{align*}
\mathcal{P}[D_n<D_n(\alpha)]=\alpha
\end{align*}
\ARROW They do not depend on the $F$ function.\\
\ARROW For the $\mathcal{U}(0,1)$:
\begin{align*}
F(x)=x,~0<x<1
\end{align*}
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Kolomogorox-Smirnow in practice}
\begin{footnotesize}
\begin{exampleblock}{ Take note:}
Empirical CDF of $F_n$ is a step function and $\sup_{-\infty < x<\infty} \vert F_n(x) -F(x) \vert$ is achieved only in one point!
\end{exampleblock}
\ARROW In practice one should sort the numbers: $X_1,..., X_n$ and calculate the following:
\begin{align*}
D_n^{+}=\max_{1 \leq i\leq n} \left(\dfrac{i}{n} - F(X_{i:n} \right),~~~~D_n^{-}=\max_{1\leq i\leq n} \left(F(X_{i:n})- \dfrac{i-1}{n}\right)\\
D_n=\max\lbrace D_n^+, D_n^- \rbrace
\end{align*}
where $X_{i:n}$ is so-called position statistic: $X_{1:n}, X_{2:n},..., X_{i:n}$.\\
\ARROW The statistic $D_n$ asymptotically (in practice $n \geqslant 80$ ) is approaching the $\lambda$-Kolomogorows cdf:
\begin{align*}
\lim_{n \to \infty} \mathcal{P} \lbrace\sqrt{n}D_n \leqslant t \rbrace =K(t)= \sum_{j=-\infty}^{\infty} (-1)^j e^{-2j^2 t^2},~t>0
\end{align*}
for which the critical values $\lambda_{\alpha}(\mathcal{P}\lbrace\sqrt{n}D_n>\lambda_{\alpha}$ can be found in the mathematical tables.\\
\ARROW Commonly the $\lambda_{0.1}=1.224$, $\lambda_{0.05}=1.358$, $\lambda_{0.01}=1.628$ are used.
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Statistic distributions test- sum test}
\begin{footnotesize}
\ARROW The $h$ function has the form:
\begin{align*}
y=x_1+x_2+x_3...x_m.
\end{align*}
\ARROW the random variables form the new pdf:
\begin{align*}
g_m(y)=\begin{cases}
\dfrac{1}{m-1} \left[ y^{m-1} - {m \choose 1}(y-1)^{m-1} +{m \choose 2}(y-2)^{m-1}-.. \right]~~&{\rm for} 0\leq y\leq m,\\
0~~&{\rm else}
\end{cases}
\end{align*}
where you stop when $y-m$ is negative.
\ARROW For $m=2$ we have the triangle pdf:
\begin{align*}
g_2(y) =\begin{cases}
y,~{\rm for}~0 \leq y \leq 1\\
2-y,~{\rm for}~0 \leq y \leq 1\\
\end{cases}
\end{align*}
\ARROW For $m=3$ we have the triangle pdf:
\begin{align*}
g_3(y) =\begin{cases}
\dfrac{1}{2} y^2,~{\rm for}~0 \leq y \leq 1\\
\dfrac{1}{2} \left[y^2-3(y-1)^2\right],~{\rm for}~1 \leq y \leq 2\\
\dfrac{1}{2} \left[y^2-3(y-1)^2 3(y-2)^2 \right],~{\rm for}~2 \leq y \leq 3\\
\end{cases}
\end{align*}
\ARROW For large $m$ the $g_m$ approaches the normal distribution.
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Statistic distributions test- $d^2$}
\begin{footnotesize}
\ARROW for $m=4$ we define the $h$:
\begin{align*}
y=(x_1-x_3)^2 +(x_2-X_4)^2
\end{align*}
aka the square distance between $(x_1,x_2)$ and $(x_3,x_4)$.\\
\ARROW If the $X_1$, $X_2$, $X_3$, $X_4$ are from $\mathcal{U}(0,1)$ then:
\begin{align*}
d^2 = (X_1-X_3)^2+(X_2-X_4)^2
\end{align*}
had a pdf given by the following formula:
\begin{align*}
\mathcal{P}(d^2-y) =\begin{cases}
\pi y - \dfrac{8}{3}y^{\dfrac{3}{2}} + \dfrac{1}{2} y^2 ~~ &{\rm for }~0 \leq y \leq1 \\
-\dfrac{1}{2} y^2-4 {\rm arcsec} (y^{\dfrac{1}{2}})~~ &{\rm for}~ 1 \leq y \leq2 \\
\end{cases}
\end{align*}
\ARROW Test is to check if the generated numbers have the aforementioned distribution.
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Statistic distributions test- pair distance}
\begin{footnotesize}
\ARROW Generate $n$ points from $(0,1)^m$. We take ${ n \choose 2}$ pairs of points and we calculate the distance between them.\\
\ARROW If $D$ is the smallest distance between the pairs $\longmapsto$ for the $\mathcal{U}(0,1)^m$ the $T=n^2D^m/2$ has the exponential distribution with the mean $1/V_m$, where $V_m$ is the hiper volume of the unite ball.\\
\ARROW In Patrice:\\
\begin{itemize}
\item We generate $Nn$ points in the hipercube $(0,1)^m$, getting $N$ points in the $T$ statistics.
\item We compare the empirical distribution $T$ with the exponential distribution.
\item WARNING: the $N,n,m$ need to be choose smartly for the test to make sense.
\end{itemize}
\ARROW Linear generators usually fail this test!
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Statistic distributions test- series test}
\begin{footnotesize}
\ARROW Lets assume our numbers are generated with a CDF $F$. The values of $F$ we divide in two separated sub-samples: $A$ and $B$. \\
\ARROW Furthermore we define the new variables $Y$ such as:
\begin{align*}
Y=\begin{cases}
=a X ~\in A\\
=b X~ \in B
\end{cases}
\end{align*}
\ARROW The random number sequence we transform the $X_1,X_2,X_3,...,X_n$ into $Y_1,Y_2,Y_3,...,Y_n$. \\
\ARROW Next we make series: For example the $a,a,b,a,a, b,b,b, a$ will be grouped into $aa$, $b$, $aa$, $bbb$, $a$.\\
\ARROW Let $n_a$ be number of $a$ symbols in $Y_1,Y_2,Y_3,...,Y_n$. $n_b=N-n_a$. \\
\ARROW Distribution of number of series ($R$) is given by the equation:
\begin{align*}
\mathcal{P}(R=r, n_a,n_b)=\begin{cases}
2 {n_a- 1 \choose k-1}{n_b-1 \choose k-1}/{N \choose n_a}~{\rm if}~r=2k\\
[{n_a- 1 \choose k}{n_b-1 \choose k-1} +{n_a- 1 \choose k-1}{n_b-1 \choose k} ] /{N \choose n_a}~{\rm if}~r=2k+1
\end{cases}
\end{align*}
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Statistic distributions test- poker test}
\begin{footnotesize}
\ARROW The values of $X$ random variable we divide into $k$ identical sub samples:
\begin{align*}
0<a_1<...<a_k=1
\end{align*}
\ARROW For $X_1,X_2,...,X_n$ from $\mathcal{U}(0,1)$:
\begin{align*}
\mathcal{P}(a_{i-1}<X_j<a_i)=\dfrac{1}{k}.
\end{align*}
\ARROW We create the new variables $Y_1$ accordingly:
\begin{align*}
Y_j=i~{\rm if}~X_j\in(a_{i-1},a_i),~i=0,1,...k-1
\end{align*}
\ARROW Now we create ''the fives'':
\begin{align*}
(Y_1,Y_2,Y_3,Y_4,Y_5),(Y_6,...
\end{align*}
\ARROW There are couple of types of fives:
\begin{itemize}
\item[aabcd] pair
\item[aaabc] three
\item[aaaab] four
\item[aaaaa] five
\end{itemize}
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Statistic distributions test- poker test}
\begin{footnotesize}
\ARROW If the variables are independent then we can calculate the probability:
\begin{center}
\includegraphics[width=0.7\textwidth]{images/poker.png}
\end{center}
\ARROW In practice people choose: $k=2,8,10$\\
\ARROW The agreemnt of the distribution of different types of fives is check using the $\chi^2$ test.
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Conclusions}
\begin{small}
\ARROW There are infinite number of tests one can invent for the testing of the generators.\\
\ARROW All of the tests are in the same taste: invent a problem where you know the analytic solution, solve the problem and compare the results.\\
\ARROW Homework: Use one of the previously implemented random number generator and :
\begin{itemize}
\item E5.1 Test them with chi-square test k=10.
\item E5.2 Kolomorov-smirnon.
\item E5.3 Multidimensional test.
\end{itemize}
\end{small}
\end{frame}
\backupbegin
\begin{frame}\frametitle{Backup}
\end{frame}
\backupend
\end{document}
mchrzasz