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a/Lectures_my/MC_2016/Lecture6/mchrzasz.pdf +++ b/Lectures_my/MC_2016/Lecture6/mchrzasz.pdf Binary files differ diff --git a/Lectures_my/MC_2016/Lecture6/mchrzasz.synctex.gz b/Lectures_my/MC_2016/Lecture6/mchrzasz.synctex.gz index a0ba3c6..a678dfa 100644 --- a/Lectures_my/MC_2016/Lecture6/mchrzasz.synctex.gz +++ b/Lectures_my/MC_2016/Lecture6/mchrzasz.synctex.gz Binary files differ diff --git a/Lectures_my/MC_2016/Lecture6/mchrzasz.tex b/Lectures_my/MC_2016/Lecture6/mchrzasz.tex index 6400591..553c7dd 100644 --- a/Lectures_my/MC_2016/Lecture6/mchrzasz.tex +++ b/Lectures_my/MC_2016/Lecture6/mchrzasz.tex @@ -308,30 +308,309 @@ \end{align*} has a distribution with a \cdf~of $F$ \end{alertblock} + + +\begin{center} +\includegraphics[width=0.5\textwidth]{images/code.png} +\end{center} + \ARROW E6.3 Using the above example please generate the \pdf~accordingly to: + \begin{equation} + \mathcal{ P}(X=k) =p_k=A \sin (\dfrac{\pi}{10}\left[k+\dfrac{1}{2}\right])\nonumber + \end{equation} + where $A$ is the normalization constant (calculate it!). From the generated numbers make a histogram and compare to the exact function. + \end{footnotesize} + +\end{frame} + + + + +\begin{frame}\frametitle{Reverting the \cdf, pros and cons} + \begin{footnotesize} +\ARROW Pros: +\begin{itemize} +\item Very accurate method. +\item Fast and easy. +\item To generate one random number from \cdf~you need one random number from $\mathcal{U}(0,1)$. +\end{itemize} +\ARROW Cons: +\begin{itemize} +\item Usually we require that the \cdf~is revertible analytically( small number of functions). +\item If we use the numerical method of reverting the function, then it's much slower and less accurate. +\item Super hard fro multidimensional distributions. +\end{itemize} + +\ARROW Mathematical digression: Numerical reverting the \cdf : +\begin{itemize} +\item We look for the zero of the function: $F_u(X)\equiv F(X)-U$, where $U \in \mathcal{U}(0,1)$. +\item $X_0=F^{-1}(U)$ +\end{itemize} + + \end{footnotesize} + +\end{frame} + +\begin{frame}\frametitle{J. von Neumann elimination method, 1951} + \begin{footnotesize} + \ARROW Let $X$ be a random variable with a \pdf~ $f(x)$ on an interval $\left[a,b\right]$, such that +$=\infty p(x): & \text{miss}. +\end{cases} +\end{align*} + +\column{2.5in} +\includegraphics[width=0.75\textwidth]{images/result.png} + + + +\end{columns} +\ARROW In the end we managed to generate a \pdf : $f(x)=\cos(x)$. + \end{footnotesize} + +\end{frame} + +\begin{frame}\frametitle{Elimination method, example for the Buffon needle} + \begin{footnotesize} +\begin{center} +\includegraphics[width=0.65\textwidth]{images/result.png} +\end{center} +\ARROW We generate the points from $(U_1,U_2)$ from $\mathcal{U}(0,\dfrac{\pi}{2}) \times \mathcal{U}(0,1)$.\\ +\ARROW Points that are below the function we keep.\\ +\ARROW The ones above we reject.\\ +\ARROW The new variable will have the $f(x)$ \pdf. + + + + + \end{footnotesize} + +\end{frame} + +\begin{frame}\frametitle{Elimination method, multi dimension case} + \begin{footnotesize} +\ARROW Let $X=(X_1,X_2,...,X_m)$ be a $m$ dimensional random variable with a \pdf of $f(x_1,x_2,...,x_m)$ of the domain of $\Omega \in \mathcal{R}^m$ and $f(x_1,x_2,..x_m)0,~n>1.$\\ +We can write down the functions: $g_Y(x)=ye^{-xy}$ and $h(y)=ny^{-(n+1)}$. +\begin{enumerate} +\item The numbers $Y$ should be generated from the \pdf $h(y)$. One can use the Reverting \cdf~for example: $Y=(1-U)^{1/n}$, where $U \in \mathcal{U}(0,1)$. +\item The numbers $X$ of a \pdf $g_Y(x)$ are generated from the exponential \pdf~$E(0,\frac{1}{y}): +X=- \frac{1}{Y}\ln V$, where $V \in \mathcal{U}(0,1)$. +\end{enumerate} + + + + \end{footnotesize} + +\end{frame} + + +\begin{frame}\frametitle{Discrete superposition of distribution} + \begin{footnotesize} +\ARROW Let's: +\begin{align*} +f(x) = \sum_{i=1}^{\infty} p_i g_i(x) +\end{align*} +where $p_i$ discrete \pdf . $p_i\geq 0$ and $\sum_{i=1}^{\infty} p_1 =1$\\ +\ARROW The algorithm: +\begin{enumerate} +\item Generate the $i$ number accordingly to the $p_i$ \pdf . Using Reverting \cdf~method. +\item For a given $i$ generate the number $X$ from the $g_i(x)$. +\end{enumerate} +\ARROW Example: $f(x)=\dfrac{5}{12}\left[ 1+ (x-1)^4\right],~0\leq x \leq 2.$\\ +$\rightarrow~f(x)=\dfrac{5}{6}g_1(x)+1/6 g_2(x),$\\ +where $g_1(x)=\dfrac{1}{2}$ and $g_2(x)=\dfrac{5}{2}(x-1)^4$, so $p_1=\dfrac{5}{6}$ and $p_2=\dfrac{1}{6}$. +\begin{align*} +X= \begin{cases} +2 U_2~~~~{ \rm if~ } U_1 < \dfrac{5}{6}\\ +1+(2U_2-1)^{\frac{1}{5}}~~~~{ \rm if~ } U_1 < \dfrac{5}{6} +\end{cases} +\end{align*} +$U_1,U_2 \in \mathcal{U}(0,1)$. + + \end{footnotesize} + +\end{frame} + +\iffalse +\begin{frame}\frametitle{Discrete superposition of distribution, multidim case} + \begin{footnotesize} \fixme - - \end{footnotesize} \end{frame} +\fi - -\begin{frame}\frametitle{Example in C++} +\begin{frame}\frametitle{Combination of the superposition and elimination method} \begin{footnotesize} - - - - \end{footnotesize} -\end{frame} - -\begin{frame}\frametitle{Reverting the \cdf, prose and cones} - \begin{footnotesize} - +\ARROW Let's: +\begin{align*} +f(x) = \sum_{i=1}^{\infty} p_i f_i(x),~~~p_i\leq 0,~~\sum_{i=1}^n p_i=1 +\end{align*} +where $f_i(x)$ functions that depend on $i$.\\ +For each $f_i$ we find a weight function $g_i(x)$ and a constant $c_i>0$ such as: +\begin{align*} +f_i(x) \leq c_i g_i(x) \forall_x . +\end{align*} +\ARROW The algorithm: +\begin{enumerate} +\item Generate $i$ accordingly to the distribution $p_i$ +\item For a given $i$ generate $X$ from the \pdf~$g_i(x)$. +\item Generate a $U\in \mathcal{U}(0,1)$. +\item If $c_i U g_i(X) \leq f_i(X)$ then we accept the $X$ if not then we start over. +\end{enumerate} +\ARROW Alternative: +\begin{enumerate} +\item One can also calculate the weights: $w(X)=\dfrac{f_i(X)}{g_i(X)}$. +\item Find the maximal weight: $w_i^{max}$. +\item If $w_i(X) \geq U w_i^{max}$ we accept $X$. If not then start over +\end{enumerate} @@ -339,226 +618,128 @@ \end{frame} - -\begin{frame}\frametitle{Overlaping-pairs-sparse-occupancy} +\begin{frame}\frametitle{Combination, warnings, example} \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: +\ARROW We need a different maximal weight for each ''branch''. This can be done either numerically or analytically.\\ +\ARROW Example: +\begin{itemize} +\item Let our \pdf~be in a form of: +\end{itemize} \begin{align*} -(I_1,I_2),(I_2,I_3),...(I_{n-1}, I_n) +f(x)=\sum_{i=1}^n c_ix^i,~~0 \leq x \leq 1 \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: +but $\exists_{j \in \lbrace 1,...,n \rbrace}: c_j <0$. +We can do a transmigration: \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). +cj=c^+_j -c^-_j: c^+_j, c^-_j>0 \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 $\alpha$: +\ARROW So: \begin{align*} -\mathcal{P}[D_n0 -\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 $h$: +f(x)=\sum_{i=1}^n c_i x^i \leq \sum_{i=1}^n c_i^+ x^i = g(x) +\end{align*} + \ARROW Now we use the weight function: \begin{align*} - y=(x_1-x_3)^2 +(x_2-X_4)^2 + \bar{g}(x)=\frac{g(x)}{\sum_{i=1}^n c_i^+} \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{frame}\frametitle{Combination, example} \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:\\ +\ARROW The algorithm: \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. +\item Generate $X$ accordingly to $\bar{g}(x)$ \pdf. The same we did in the discrete superposition method. +\item Generate $U \in \mathcal{U}(0,1)$ +\item If $Ug(X) \leq f(X)$ then we accept the $X$. If not start over. \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: +\ARROW Example $N(0,1)$: + \begin{align*} + f(x) = \sqrt{\dfrac{2}{\pi}} e^{-\frac{x^2}{2}},~x \geq 0. + \end{align*} +\ARROW We write the function as: \begin{align*} -Y=\begin{cases} -=a X ~\in A\\ -=b X~ \in B -\end{cases} +f(x) \alpha_1 g_1(x)h_1(x)+\alpha_2 g_2(x)h_2(x) \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: +where: +\begin{columns} +\column{2.5in} \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 +\alpha_1=\sqrt{\frac{2}{\pi}} +\end{align*} +\begin{align*} +g_1=\begin{cases} +1,~0\leq x \leq 1\\ +0,~x \geq 1\\ \end{cases} \end{align*} +\column{2.5in} +\begin{align*} +\alpha_2=\sqrt{\frac{1}{2 \pi}} +\end{align*} +\begin{align*} +g_2=\begin{cases} +2e^{-2(x-1)},~x \geq 1\\ +0,~0\leq x \leq 1\\ +\end{cases} +\end{align*} + + +\end{columns} + \end{footnotesize} + \end{frame} - - - -\begin{frame}\frametitle{Statistic distributions test- poker test} +\begin{frame}\frametitle{Combination, example} \begin{footnotesize} -\ARROW The values of $X$ random variable we divide into $k$ identical sub samples: + +\begin{columns} +\column{2.5in} + \begin{align*} -0>} -\pgfsyspdfmark {pgfid5}{23867907}{17900937} +\pgfsyspdfmark {pgfid8}{23867907}{17900937} \@writefile{nav}{\headcommand {\slideentry {0}{0}{2}{2/2}{}{0}}} \@writefile{nav}{\headcommand {\beamer@framepages {2}{2}}} -\pgfsyspdfmark 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a/Lectures_my/MC_2016/Lecture7/mchrzasz.pdf b/Lectures_my/MC_2016/Lecture7/mchrzasz.pdf index 2ef368f..1fd5134 100644 --- a/Lectures_my/MC_2016/Lecture7/mchrzasz.pdf +++ b/Lectures_my/MC_2016/Lecture7/mchrzasz.pdf Binary files differ diff --git a/Lectures_my/MC_2016/Lecture7/mchrzasz.synctex.gz b/Lectures_my/MC_2016/Lecture7/mchrzasz.synctex.gz new file mode 100644 index 0000000..a678dfa --- /dev/null +++ b/Lectures_my/MC_2016/Lecture7/mchrzasz.synctex.gz Binary files differ diff --git a/Lectures_my/MC_2016/Lecture7/mchrzasz.tex b/Lectures_my/MC_2016/Lecture7/mchrzasz.tex index d936aca..553c7dd 100644 --- a/Lectures_my/MC_2016/Lecture7/mchrzasz.tex +++ b/Lectures_my/MC_2016/Lecture7/mchrzasz.tex @@ -46,6 +46,7 @@ \usepackage{graphicx} \usepackage[export]{adjustbox} +\usepackage{amsmath, amssymb} \usepackage{amsmath} \usepackage{amsfonts} @@ -122,6 +123,7 @@ \tikzstyle{block} = [draw, rectangle, minimum height=3em, minimum width=6em] +\def\ARROW{{\color{JungleGreen}{$\Rrightarrow$}}\xspace} @@ -203,12 +205,13 @@ \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[Random number generators and application]{Random number generators and application} +\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] @@ -220,7 +223,7 @@ \begin{center} \begin{columns} \begin{column}{0.9\textwidth} - \flushright\fontspec{Trebuchet MS}\bfseries \Huge {Applications of Monte Carlo methods} + \flushright\fontspec{Trebuchet MS}\bfseries \Huge {Arbitrary \pdf~generation} \end{column} \begin{column}{0.2\textwidth} %\includegraphics[width=\textwidth]{SHiP-2} @@ -242,419 +245,500 @@ \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}{Experimental Methods in Particle Physics, \\ 26 November, 2015} + \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 $F$. - - -\begin{frame}\frametitle{Markov Chain MC} -\begin{itemize} -\item Consider a finite possible states: $S_1$, $S_2$, ... -\item And the time steps of time, labelled as $1$, $2$, ... -\item At time $t$ the state is denoted $X_t$. -\item The conditional probability is defined as: -\end{itemize} -\begin{equation} -P(X_t=S_j \vert X_{t-1}=S_{j-1},..., X_{1}=S_{1}) \nonumber -\end{equation} -\begin{itemize} -\item The Markov chain is then if the probability depends only on previous step. -\end{itemize} -\begin{equation} -P(X_t=S_j \vert X_{t-1}=S_{j-1},..., X_{1}=S_{1}) = P(X_t=S_j \vert X_{t-1}=S_{j-1} )\nonumber -\end{equation} -\begin{itemize} -\item For this reason this reason MCMC is also knows as drunk sailor walk. -\item Very powerful method. Used to solve linear eq. systems, invert matrix, solve differential equations, etc. -\end{itemize} -\end{frame} - - - -\begin{frame}\frametitle{Linear Equations} -\begin{itemize} -\item Lets say we have a linear equation system: -\end{itemize} -\begin{equation} -\begin{array}{lcl} X & = & pY + (1-p) A \\ Y & = & qX + (1-q)B \end{array} \nonumber -\end{equation} -\begin{itemize} -\item We know $A,B,p,q$; $X$ and $Y$ are meant to be determined. -\item Algorithm: -\begin{enumerate} -\item We choose first element of the first equation with probability $p$ and second with probability $1-p$. -\item We we choose the second one, the outcome of this MCMC is $W=A$. -\item If we choose the first we go to second equation and choose the first element with probability $q$ and the second with $1-q$. -\item We we choose the second one, the outcome of this MCMC is $W=B$. -\item If we choose the first we go to the first equation back again. -\item We repeat the procedure. -\end{enumerate} -\item We can estimate the solution of this system: -\end{itemize} -\begin{equation} -\hat{X} = \dfrac{1}{N}\sum_{i=1} W_i{~}{~}{~}{~}{~} \hat{\sigma_X}=\dfrac{1}{\sqrt{N-1}}\sqrt{\dfrac{1}{N} \sum_{i=1}^N W_i^2-\hat{X}^2} \nonumber -\end{equation} + + \end{footnotesize} \end{frame} -\begin{frame}\frametitle{Neumann-Ulam method} +\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 Let's try apply the basic MCMC method to solve a simple linear equation system: -\end{itemize} -\begin{equation} -A \overrightarrow{x} = \overrightarrow{b} \nonumber -\end{equation} -\begin{itemize} -\item The above system can be (always, see linear algebra lecture) translated into system: -\end{itemize} -\begin{equation} -\overrightarrow{x} = \overrightarrow{a} + H \overrightarrow{x} \nonumber -\end{equation} -\begin{itemize} -\item For this method we assume that the norm of the matrix is: -\end{itemize} -\begin{equation} -\Vert H \Vert = \underset{1 \leq i \leq n}{max} \sum_{j=1}^n \vert h_{ij} \vert <1 \nonumber \end{equation} -\begin{itemize} -\item Which we can write in a form: -\end{itemize} -\begin{equation} -(1 -H)\overrightarrow{x}=\overrightarrow{a} \nonumber -\end{equation} -\end{frame} - - - - -\begin{frame}\frametitle{Neumann-Ulam method} -\begin{itemize} -\item The solution would be then: -\end{itemize} -\begin{equation} -\overrightarrow{x}_0=(1 -H)^{-1}\overrightarrow{a} \nonumber -\end{equation} -\begin{itemize} -\item We can Taylor expend this: -\end{itemize} -\begin{equation} -\overrightarrow{x}_0=(1 -H)^{-1}\overrightarrow{a} = \overrightarrow{a} + H \overrightarrow{a} + H^2 \overrightarrow{a} + H^3 \overrightarrow{a} +.... \nonumber -\end{equation} -\begin{itemize} -\item For the $i$-th component of the $\overrightarrow{x}$ vector: -\end{itemize} -\begin{equation} -x_0^i= a_i + \sum_{j=1}^n h_{ij} a_{j_1} + \sum_{j_1=1}^n \sum_{j_2=1}^n h_{ij_1} h_{ij_2} a_{j_2} + \sum_{j_1=1}^n \sum_{j_2=1}^n \sum_{j_3=1}^n h_{ij_1} h_{ij_2} h_{ij_3} a_{j_3} + ...\nonumber -\end{equation} -\begin{itemize} -\item One can construct probabilistic behaviour of a system that follows the path of equation above. -\end{itemize} +\item E 6.1 $\rho(X)= \dfrac{c}{X}$ on the interval $[a,b]$, where $0 0 \nonumber -\end{equation} -\begin{itemize} -\item The system has states: $\lbrace 0,1,2...,n\rbrace$ -\item State at $t$ time is denoted as $i_t$. -\item We make a random walk accordingly to to the following rules: -\begin{itemize} -\item At the begging of the walk ($t=0$) we are at $i_0$. -\item In the $t$ moment we are in the $i_t$ position then in $t+1$ time stamp we move to state $i_{t+1}$ with the probability $h_{i_t i_{t+1}}$. -\item We stop walking if we are in state $0$. -\end{itemize} -\item The path $X(\gamma) = (i_0, i_1, i_2, ..., i_k, 0)$ is called trajectory. -\item It can be proven that $x_i^0 =E \lbrace X (\gamma) \vert i_0=j \rbrace$. -\end{itemize} +\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} - -\end{frame} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Neumann-Ulam method, $\rm \color{RubineRed}{Lecture3/Markov}$} -\begin{itemize} -\item For example lets try to solve this equation system: -\end{itemize} -\begin{equation} -\overrightarrow{x} = -\left(\begin{array}{c} - 1.5 \\ --1.0\\ -0.7 \end{array} \right) -+ -\left(\begin{array}{ccc} -0.2 & 0.3 & 0.1 \\ -0.4 & 0.3 & 0.2 \\ -0.3 & 0.1 & 0.1 \end{array} \right) \overrightarrow{x} - \nonumber -\end{equation} -\begin{itemize} -\item The solution is $\overrightarrow{x}_0 = (2.154303, 0.237389, 1.522255)$. -\end{itemize} -\begin{columns} - -\column{0.1in} - -\column{2.5in} -\begin{itemize} -\item The propability matrix $h_{ij}$ has the shape: -\end{itemize} -\begin{tabular}{|c|cccc|} -\hline -$i/j$ & 1 & 2 & 3 & 4 \\ \hline -1 & 0.2 & 0.3 & 0.1 & 0.4 \\ -2 & 0.4 & 0.3 & 0.2 & 0.1 \\ -3 & 0.3 & 0.1 & 0.1 & 0.5 \\ \hline -\end{tabular} - -\column{2.5in} -\begin{itemize} -\item An example solution: -\end{itemize} -\includegraphics[width=0.95\textwidth]{images/mark.png} - -\end{columns} - - -\end{frame} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Neumann-Ulam dual method} -\begin{itemize} -\item The problem with Neumann-Ulam method is that you need to repeat it for each of the coordinates of the $\overrightarrow{x}_0$ vector. -\item The dual method calculates the whole $\overrightarrow{x}_0$ vector. -\item The algorithm: -\begin{itemize} -\item On the indexes: $\lbrace0,1,...,n\rbrace$ we set a probability distribution:\\ $q_1, q_2,..., q_n$, $q_i>0$ and $\sum_i=1^n q_i=1$. -\item The starting point we select from $q_i$ distribution. -\item If in $t$ time we are in $i_t$ state then with probability $p(i_{t+1} \vert i_t) = h_{i_{t+1},i_{t}}$ in $t+1$ we will be in state $i_1$. For $i_{t+1}=0$ we define the probability: $h_{0,i_{t}}=1-\sum_{j=1}^n h_{j,i_{t}}$. Here we also assume that $h_{j,i_{t}} > 0$. -\item NOTE: there the matrix is transposed compared to previous method: $H^{T}$. -\item Again we end our walk when we are at state $0$. -\item For the trajectory: $\gamma = (i_0, i_1,...,i_k, 0)$, we assign the vector: -\end{itemize} -\begin{equation} -\overrightarrow{Y}(\gamma) = \dfrac{a_{i_0}}{ q_{i_{0}} p(0 \vert i_k) } \widehat{e}_{i_{k}} \in \mathcal{R}^n \nonumber -\end{equation} -\item The solution will be : $\overrightarrow{x}^0 = \dfrac{1}{N} \sum \overrightarrow{Y}(\gamma)$ -\end{itemize} - - -\end{frame} - - - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Neumann-Ulam dual method, $\rm \color{RubineRed}{Lecture3/Markov2}$} -\begin{itemize} -\item Let's try to solve the equation system: -\end{itemize} -\begin{equation} -\overrightarrow{x} = -\left(\begin{array}{c} - 1.5 \\ --1.0\\ -0.7 \end{array} \right) -+ -\left(\begin{array}{ccc} -0.2 & 0.3 & 0.1 \\ -0.4 & 0.3 & 0.2 \\ -0.1 & 0.1 & 0.1 \end{array} \right) \overrightarrow{x} - \nonumber -\end{equation} -\begin{itemize} -\item The solution is: $\overrightarrow{x}_0 = (2.0, 0.0, 1.0)$. -\item Let's put the initial probability as constant: -\end{itemize} -\begin{equation} -q_1=q_2=q_3=\dfrac{1}{3} \nonumber -\end{equation} -\begin{columns} - -\column{0.1in} - -\column{2.5in} -\begin{itemize} -\item The propability matrix $h_{ij}$ has the shape: -\end{itemize} -\begin{tabular}{|c|cccc|} -\hline -$i/j$ & 1 & 2 & 3 & 4 \\ \hline -1 & 0.2 & 0.4 & 0.1 & 0.3 \\ -2 & 0.3 & 0.3 & 0.1 & 0.3 \\ -3 & 0.1 & 0.2 & 0.1 & 0.6 \\ \hline -\end{tabular} - -\column{2.5in} -\begin{itemize} -\item An example solution: -\end{itemize} -\includegraphics[width=0.95\textwidth]{images/mark2.png} - -\end{columns} -\end{frame} - - - - -\begin{frame}\frametitle{Look elsewhere effect, $\rm \color{RubineRed}{Lecture3/LEE}$} -\begin{itemize} -\item Look elsewhere effect addresses the following problem: -\begin{itemize} -\item Imagine you observed a $3\sigma$ deviation in one of the observable that you measured. -\item Before you get excited one needs to understand if given the fact that you had so many measurements this might happen! -\end{itemize} -\item Example: Let's say we have measured 50 observables. What is the probability to observed 1 that is $3\sigma$ away from theory prediction? -\item Let's simulate 50 Gaussian distribution centred at 0 and width of 1. We count how simulations where at least one of the 50 numbers have the absolute value $>3$. -\item More complicated example: what if you observed 3 in a row $2\sigma$ fluctuations among 50 measurements? -\item This kind of studies are the best solvable by MC simulations. -\end{itemize} - - -\end{frame} - - - - - - -\begin{frame}\frametitle{Travelling Salesman Problem} -\begin{itemize} -\item Salesman starting from his base has to visit $n-1$ other locations and return to base headquarters. The problem is to find the shortest way. -\item For large $n$ the problem can't be solver by brutal force as the complexity of the problem is $(n-1)!$ -\item There exist simplified numerical solutions assuming factorizations. Unfortunately even those require anonymous computing power. -\item Can MC help? YES :) -\item The minimum distance $l$ has to depend on 2 factors: $P$ the area of the city the Salesman is travelling and the density of places he wants to visit: $\dfrac{n}{P}$ -\item Form this we can assume: -\end{itemize} -\begin{equation} -l \sim P^a (\dfrac{n}{P})^b=P^{a-b}n^b. \nonumber -\end{equation} - -\end{frame} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% - - - - - - - -\begin{frame}\frametitle{Traveling Salesman Problem} -\begin{itemize} -\item From dimension analysis: -\end{itemize} -\begin{equation} -a-b=\dfrac{1}{2}. \nonumber -\end{equation} -\begin{itemize} -\item To get $l$ we need square root of area. -\item From this it's obvious: -\end{itemize} -\begin{equation} -l \sim P^a (\dfrac{n}{P})^b=P^{0.5}n^{a-0.5}. \nonumber -\end{equation} -\begin{itemize} -\item Now we can multiply the area by alpha factor that keeps the density constant then: -\end{itemize} -\begin{equation} -l \sim \alpha^0.5 \alpha6{a-0.5} = \alpha^a \nonumber -\end{equation} -\begin{itemize} -\item In this case the distance between the clients will not change, but the number of clients will increase by $\alpha$ so: -\end{itemize} -\begin{equation} -l \sim \alpha \nonumber -\end{equation} -\begin{itemize} -\item In the end we get: $a=1$ -\end{itemize} -\end{frame} - - - -\begin{frame}\frametitle{Traveling Salesman Problem} -\begin{itemize} -\item In total: -\end{itemize} -\begin{equation} -l \sim k (nP)^{0.5}\nonumber -\end{equation} -\begin{itemize} -\item Of course the k depends on the shape of the area and locations of client. However for large $n$ the k starts loosing the dependency. It's an asymptotically free estimator. -\item To use the above formula we need to somehow calculate k. -\item How to estimate this? Well make a TOY MC: take a square put uniformly $n$ points. Then we can calculate $l$. Then it's trivial: -\end{itemize} -\begin{equation} -k= l(nP)^{-0.5} \nonumber -\end{equation} -\end{frame} - - -\begin{frame}\frametitle{Traveling Salesman Problem} -\begin{itemize} -\item This kind of MC experiment might require large CPU power and time. The adventage is that once we solve the problem we can use the obtained k for other cases (it's universal constant!). -\item It turns out that: -\end{itemize} -\begin{equation} -k \sim \dfrac{3}{4} \nonumber -\end{equation} -\begin{itemize} -\item Ok, but in this case we can calculate $l$ but not the actual shortest way! Why the hell we did this exercise?! -\item Turns out that for most of the problems we are looking for the solution that is close to smallest $l$ not the exact minimum. -\end{itemize} -\end{frame} - -\begin{frame}\frametitle{War Games} - -\begin{itemize} -\item S. Andersoon 1966 simulated for Swedish government how would a tank battle look like. -\item Each of the sides has 15 tanks. that they allocate on the battle field. -\item The battle is done in time steps. -\item Each tank has 5 states: -\begin{itemize} -\item OK -\item Tank can only shoot -\item Tank can only move -\item Tank is destroyed -\item Temporary states -\end{itemize} -\item This models made possible to check different fighting strategies. - -\end{itemize} -\end{frame} - - - -\begin{frame} - \begin{center} -\begin{Huge} - Q \& A -\end{Huge} +\includegraphics[width=0.5\textwidth]{images/code.png} \end{center} - + \ARROW E6.3 Using the above example please generate the \pdf~accordingly to: + \begin{equation} + \mathcal{ P}(X=k) =p_k=A \sin (\dfrac{\pi}{10}\left[k+\dfrac{1}{2}\right])\nonumber + \end{equation} + where $A$ is the normalization constant (calculate it!). From the generated numbers make a histogram and compare to the exact function. + \end{footnotesize} \end{frame} +\begin{frame}\frametitle{Reverting the \cdf, pros and cons} + \begin{footnotesize} +\ARROW Pros: +\begin{itemize} +\item Very accurate method. +\item Fast and easy. +\item To generate one random number from \cdf~you need one random number from $\mathcal{U}(0,1)$. +\end{itemize} +\ARROW Cons: +\begin{itemize} +\item Usually we require that the \cdf~is revertible analytically( small number of functions). +\item If we use the numerical method of reverting the function, then it's much slower and less accurate. +\item Super hard fro multidimensional distributions. +\end{itemize} + +\ARROW Mathematical digression: Numerical reverting the \cdf : +\begin{itemize} +\item We look for the zero of the function: $F_u(X)\equiv F(X)-U$, where $U \in \mathcal{U}(0,1)$. +\item $X_0=F^{-1}(U)$ +\end{itemize} + + \end{footnotesize} + +\end{frame} + +\begin{frame}\frametitle{J. von Neumann elimination method, 1951} + \begin{footnotesize} + \ARROW Let $X$ be a random variable with a \pdf~ $f(x)$ on an interval $\left[a,b\right]$, such that +$=\infty p(x): & \text{miss}. +\end{cases} +\end{align*} + +\column{2.5in} +\includegraphics[width=0.75\textwidth]{images/result.png} + + + +\end{columns} +\ARROW In the end we managed to generate a \pdf : $f(x)=\cos(x)$. + \end{footnotesize} + +\end{frame} + +\begin{frame}\frametitle{Elimination method, example for the Buffon needle} + \begin{footnotesize} +\begin{center} +\includegraphics[width=0.65\textwidth]{images/result.png} +\end{center} +\ARROW We generate the points from $(U_1,U_2)$ from $\mathcal{U}(0,\dfrac{\pi}{2}) \times \mathcal{U}(0,1)$.\\ +\ARROW Points that are below the function we keep.\\ +\ARROW The ones above we reject.\\ +\ARROW The new variable will have the $f(x)$ \pdf. + + + + + \end{footnotesize} + +\end{frame} + +\begin{frame}\frametitle{Elimination method, multi dimension case} + \begin{footnotesize} +\ARROW Let $X=(X_1,X_2,...,X_m)$ be a $m$ dimensional random variable with a \pdf of $f(x_1,x_2,...,x_m)$ of the domain of $\Omega \in \mathcal{R}^m$ and $f(x_1,x_2,..x_m)0,~n>1.$\\ +We can write down the functions: $g_Y(x)=ye^{-xy}$ and $h(y)=ny^{-(n+1)}$. +\begin{enumerate} +\item The numbers $Y$ should be generated from the \pdf $h(y)$. One can use the Reverting \cdf~for example: $Y=(1-U)^{1/n}$, where $U \in \mathcal{U}(0,1)$. +\item The numbers $X$ of a \pdf $g_Y(x)$ are generated from the exponential \pdf~$E(0,\frac{1}{y}): +X=- \frac{1}{Y}\ln V$, where $V \in \mathcal{U}(0,1)$. +\end{enumerate} + + + + \end{footnotesize} + +\end{frame} + + +\begin{frame}\frametitle{Discrete superposition of distribution} + \begin{footnotesize} +\ARROW Let's: +\begin{align*} +f(x) = \sum_{i=1}^{\infty} p_i g_i(x) +\end{align*} +where $p_i$ discrete \pdf . $p_i\geq 0$ and $\sum_{i=1}^{\infty} p_1 =1$\\ +\ARROW The algorithm: +\begin{enumerate} +\item Generate the $i$ number accordingly to the $p_i$ \pdf . Using Reverting \cdf~method. +\item For a given $i$ generate the number $X$ from the $g_i(x)$. +\end{enumerate} +\ARROW Example: $f(x)=\dfrac{5}{12}\left[ 1+ (x-1)^4\right],~0\leq x \leq 2.$\\ +$\rightarrow~f(x)=\dfrac{5}{6}g_1(x)+1/6 g_2(x),$\\ +where $g_1(x)=\dfrac{1}{2}$ and $g_2(x)=\dfrac{5}{2}(x-1)^4$, so $p_1=\dfrac{5}{6}$ and $p_2=\dfrac{1}{6}$. +\begin{align*} +X= \begin{cases} +2 U_2~~~~{ \rm if~ } U_1 < \dfrac{5}{6}\\ +1+(2U_2-1)^{\frac{1}{5}}~~~~{ \rm if~ } U_1 < \dfrac{5}{6} +\end{cases} +\end{align*} +$U_1,U_2 \in \mathcal{U}(0,1)$. + + \end{footnotesize} + +\end{frame} + +\iffalse +\begin{frame}\frametitle{Discrete superposition of distribution, multidim case} + \begin{footnotesize} +\fixme + + \end{footnotesize} + +\end{frame} +\fi + + +\begin{frame}\frametitle{Combination of the superposition and elimination method} + \begin{footnotesize} + +\ARROW Let's: +\begin{align*} +f(x) = \sum_{i=1}^{\infty} p_i f_i(x),~~~p_i\leq 0,~~\sum_{i=1}^n p_i=1 +\end{align*} +where $f_i(x)$ functions that depend on $i$.\\ +For each $f_i$ we find a weight function $g_i(x)$ and a constant $c_i>0$ such as: +\begin{align*} +f_i(x) \leq c_i g_i(x) \forall_x . +\end{align*} +\ARROW The algorithm: +\begin{enumerate} +\item Generate $i$ accordingly to the distribution $p_i$ +\item For a given $i$ generate $X$ from the \pdf~$g_i(x)$. +\item Generate a $U\in \mathcal{U}(0,1)$. +\item If $c_i U g_i(X) \leq f_i(X)$ then we accept the $X$ if not then we start over. +\end{enumerate} +\ARROW Alternative: +\begin{enumerate} +\item One can also calculate the weights: $w(X)=\dfrac{f_i(X)}{g_i(X)}$. +\item Find the maximal weight: $w_i^{max}$. +\item If $w_i(X) \geq U w_i^{max}$ we accept $X$. If not then start over +\end{enumerate} + + + + \end{footnotesize} + +\end{frame} + +\begin{frame}\frametitle{Combination, warnings, example} + \begin{footnotesize} +\ARROW We need a different maximal weight for each ''branch''. This can be done either numerically or analytically.\\ +\ARROW Example: +\begin{itemize} +\item Let our \pdf~be in a form of: +\end{itemize} +\begin{align*} +f(x)=\sum_{i=1}^n c_ix^i,~~0 \leq x \leq 1 +\end{align*} +but $\exists_{j \in \lbrace 1,...,n \rbrace}: c_j <0$. +We can do a transmigration: +\begin{align*} +cj=c^+_j -c^-_j: c^+_j, c^-_j>0 +\end{align*} +\ARROW So: +\begin{align*} +f(x)=\sum_{i=1}^n c_i x^i \leq \sum_{i=1}^n c_i^+ x^i = g(x) +\end{align*} + \ARROW Now we use the weight function: + \begin{align*} + \bar{g}(x)=\frac{g(x)}{\sum_{i=1}^n c_i^+} + \end{align*} + + + \end{footnotesize} + +\end{frame} + +\begin{frame}\frametitle{Combination, example} + \begin{footnotesize} +\ARROW The algorithm: +\begin{itemize} +\item Generate $X$ accordingly to $\bar{g}(x)$ \pdf. The same we did in the discrete superposition method. +\item Generate $U \in \mathcal{U}(0,1)$ +\item If $Ug(X) \leq f(X)$ then we accept the $X$. If not start over. +\end{itemize} +\ARROW Example $N(0,1)$: + \begin{align*} + f(x) = \sqrt{\dfrac{2}{\pi}} e^{-\frac{x^2}{2}},~x \geq 0. + \end{align*} +\ARROW We write the function as: +\begin{align*} +f(x) \alpha_1 g_1(x)h_1(x)+\alpha_2 g_2(x)h_2(x) +\end{align*} +where: +\begin{columns} +\column{2.5in} +\begin{align*} +\alpha_1=\sqrt{\frac{2}{\pi}} +\end{align*} +\begin{align*} +g_1=\begin{cases} +1,~0\leq x \leq 1\\ +0,~x \geq 1\\ +\end{cases} +\end{align*} + +\column{2.5in} +\begin{align*} +\alpha_2=\sqrt{\frac{1}{2 \pi}} +\end{align*} +\begin{align*} +g_2=\begin{cases} +2e^{-2(x-1)},~x \geq 1\\ +0,~0\leq x \leq 1\\ +\end{cases} +\end{align*} + + +\end{columns} + + \end{footnotesize} + +\end{frame} +\begin{frame}\frametitle{Combination, example} + \begin{footnotesize} + +\begin{columns} +\column{2.5in} + +\begin{align*} +h_1(x)=e^{\frac{-x^2}{2}} +\end{align*} + +\column{2.5in} + +\begin{align*} +h_2(x)=e^{-\frac{(x-2)^2}{2}} +\end{align*} + +\end{columns} +{~}\\{~}\\ + +\ARROW The variable $X$ is chosen: +\begin{itemize} +\item With a probability: $\alpha_1/(\alpha_1+\alpha_2) =2/3$ from the $g_1$. Then we do the elimination accordingly to $h_1$ function, aka. $h_1(X) \geq U \in \mathcal{U}(0,1)$ +\item With a probability: $\alpha_2/(\alpha_1+\alpha_2) =2/3$ from the $g_2$. Then we do the elimination accordingly to $h_2$ function, aka. $h_2(X) \geq U \in \mathcal{U}(0,1)$ +\end{itemize} +\ARROW E6.6 Generate the above distribution using with the above method. Do a histogram and compare with desired \pdf. + + \end{footnotesize} + +\end{frame} + +\begin{frame}\frametitle{Acceptance - complement, Kornal \& Peterson 1981} + \begin{footnotesize} +\ARROW If $f=f_1+f_2$ is a \pdf~ of a random variable $X$ and $p=\int f_1(x) dx.$\\ +\ARROW $f_1/p_1$ and $f_2(1-p)$ are the \pdf of a discrete function.\\ +\ARROW Now if the functions $f_1$ and $f_2$ are easy to generate from then: +\begin{itemize} +\item We choose the numbers with the probability distribution $f_1/p_1$ from the \pdf~$f_1$. +\item We choose the numbers with the probability distribution $f_2/(1-p_1)$ from the \pdf~$f_2$. +\end{itemize} +\ARROW The method is super powerful if $f_2= { \rm const}$. + + + \end{footnotesize} + +\end{frame} + \backupbegin diff --git a/Lectures_my/MC_2016/Lecture8/Makefile b/Lectures_my/MC_2016/Lecture8/Makefile new file mode 100644 index 0000000..e4cf8e8 --- /dev/null +++ b/Lectures_my/MC_2016/Lecture8/Makefile @@ -0,0 +1,9 @@ +all: mchrzasz.pdf + +mchrzasz.pdf: mchrzasz.tex + xelatex mchrzasz.tex +# bibtex mchrzasz + xelatex mchrzasz.tex +# xelatex mchrzasz.tex +clean: + rm *.aux *.bbl *.blg *.log *.out *.pdf *.sh *.xml *.snm *.toc *.nav diff --git a/Lectures_my/MC_2016/Lecture8/beamercolorthemeFlip.sty b/Lectures_my/MC_2016/Lecture8/beamercolorthemeFlip.sty new file mode 100755 index 0000000..d6b9c50 --- /dev/null +++ b/Lectures_my/MC_2016/Lecture8/beamercolorthemeFlip.sty @@ -0,0 +1,179 @@ +% Color style file for the BEAMER FLIP THEME +% Copyright 2012 by Flip Tanedo +% This file may be distributed and/or modified +% 1. under the LaTeX Project Public License and/or +% 2. under the GNU Public License. + + + +\mode +\newif\ifbeamer@Flip@dark +\beamer@Flip@darkfalse +\DeclareOptionBeamer{dark}{\beamer@Flip@darktrue} +\ProcessOptionsBeamer + + +% These are a few colors that I like. + +\definecolor{crimsonred}{RGB}{153,0,0} % Neurtal red, good for dark or light bg +\definecolor{darkcharcoal}{RGB}{25,25,25} % Darker gray +\definecolor{charcoal}{RGB}{51,51,51} % Darker gray +\definecolor{ash}{RGB}{100,100,100} % medium gray +\definecolor{paleblue}{RGB}{0,102,102} % More of an `ocean' color +\definecolor{turtlegreen}{RGB}{51,153,0} % A more neutral green +\definecolor{paleale}{RGB}{204,204,102} % Only for dark BG +\definecolor{lager}{RGB}{140,110,10} % Use instead of pale ale for white BG +\definecolor{regal}{RGB}{90,0,120} % A more neutral purple +\definecolor{jeans}{RGB}{20,30,150} % A more neutral blue + + +% Define some "invariant colors" so that a color that doen't work with a +% light/dark background will automatically be substituted with one that does. +% These will be rewritten in beamercolorthemeFlipDark + +\ifbeamer@Flip@dark + \definecolor{FlipSand}{RGB}{204,204,102} % paleale, Only for dark BG + \definecolor{FlipGreen}{RGB}{0,255,0} % Neon-like green, only for dark BG +\else + \definecolor{FlipSand}{RGB}{140,110,10} % lager, Only for light BG + \definecolor{FlipGreen}{RGB}{51,153,0} % turtlegreen, works for both +\fi + +%% Future revision: It would be nice to define these in terms of the palette. + + +\ifbeamer@Flip@dark + \setbeamercolor{normal text}{fg=white, bg=black} + \setbeamercolor{structure}{fg=gray, bg=black} % fg controls bullets + \setbeamercolor*{footerlike}{fg=white,bg=black} % Controls the footerline +\else + % Default to black on white + \setbeamercolor*{footerlike}{fg=white,bg=charcoal} % for charcoal footer + \setbeamercolor{structure}{fg=charcoal, bg=white} % fg controls bullets + % \setbeamercolor{normal text}{fg=white, bg=black} +\fi + + + + + +% These defined colors can be used in \textcolor{colorname}{text in that color}. + +% Colors for Keynote background: DARK +\definecolor{keynotebottom}{rgb}{0.32,0.3,0.38} +\definecolor{keynotemiddle}{rgb}{0.08,0.08,0.16} +\definecolor{keynotetop}{rgb}{0,0,0} +%%% Example of usage: +%% \setbeamertemplate{background canvas}[vertical shading] +%% [bottom=keynotebottom, middle=keynotemiddle, top=keynotetop] + + + + +% ALERT AND COMMENT COLORS (just copies of the above colors) +% (I don't know how ot do this more elegantly) +\ifbeamer@Flip@dark + \definecolor{ALERT}{RGB}{153,0,0} % crimsonred + \definecolor{Alert}{RGB}{51,153,0} % turtlegreen + \definecolor{alert}{RGB}{204,204,102} % paleale + \definecolor{charcoal}{RGB}{51,51,51} % charcoal + \definecolor{comment}{RGB}{80,80,80} % light charcoal + \definecolor{Comment}{RGB}{100,100,100} % ash + \definecolor{COMMENT}{RGB}{80,20,120} % regal +\else +% \definecolor{ALERT}{RGB}{153,0,0} % crimsonred + \definecolor{ALERT}{RGB}{28,134,238} % DodgerBlue2 +% \definecolor{Alert}{RGB}{51,153,0} % turtlegreen + \definecolor{Alert}{RGB}{255,127,36} % chocolate1 +% \definecolor{alert}{RGB}{140,110,10} % lager + \definecolor{alert}{RGB}{0,205,102} % SpringGreen3 + \definecolor{charcoal}{RGB}{80,80,80} % light charcoal + \definecolor{comment}{RGB}{51,51,51} % charcoal + \definecolor{Comment}{RGB}{100,100,100} % ash + \definecolor{COMMENT}{RGB}{80,20,120} % regal +\fi + + +% \definecolor{crimsonred}{RGB}{153,0,0} % Neurtal red, good for dark or light bg +% \definecolor{charcoal}{RGB}{51,51,51} % Darker gray +% \definecolor{ash}{RGB}{100,100,100} % medium gray +% \definecolor{paleblue}{RGB}{0,102,102} % More of an `ocean' color +% \definecolor{turtlegreen}{RGB}{51,153,0} % A more neutral green +% \definecolor{paleale}{RGB}{204,204,102} % Only for dark BG +% \definecolor{lager}{RGB}{140,110,10} % Use instead of pale ale for white BG +% \definecolor{regal}{RGB}{90,0,120} % A more neutral purple +% \definecolor{jeans}{RGB}{20,30,150} % A more neutral blue + + +% A few useful colors +\setbeamercolor{alerted text}{fg=alert} + +%% I would like ot use these colors, but there seems to be a bug in Beamer +%% where the footnote color infects the normal text color +%% See: http://tex.stackexchange.com/questions/8264/beamer-changing-the-footnote-color-also-changes-the-normal-text-color/8268#8268 + +% \setbeamercolor{footnote}{parent=normal text, fg=FlipSand} +% \setbeamercolor{footnote mark}{parent=normal text, fg=FlipSand} + +% Blocks +\setbeamercolor{block title}{fg=gray!30!white,bg=charcoal} +\setbeamercolor{block body}{parent=normal text,bg=gray, fg=white} + +\setbeamercolor{block title example}{fg=turtlegreen!30!white,bg=turtlegreen!40!black} +\setbeamercolor{block body example}{parent=normal text,bg=turtlegreen, fg=white} +% +\setbeamercolor{block title alerted}{fg=crimsonred!30!white,bg=crimsonred!40!black} +\setbeamercolor{block body alerted}{parent=normal text,bg=crimsonred, fg=white} + + + +% These "setbeamercolor" commands are more standardized. + + +%% From the Beamer User Guide: +% Outer themes base the color of navigational elements and, possibly, also of other elements, on the +% four palette colors. The “primary” palette should be used for the most important navigational elements, +% which are usually the ones that change most often and hence require the most attention by the audience. +% The “secondary” and “tertiary” are less important, the “quaternary” one is least important. +% By default, the palette colors do not have a background and the foreground ranges from structure.fg to black. +% For the sidebar, there is an extra set of palette colors, see palette sidebar primary. + +\setbeamercolor*{palette primary}{fg=crimsonred,bg=charcoal} +\setbeamercolor*{palette secondary}{fg=paleale,bg=lager} +\setbeamercolor*{palette tertiary}{fg=turtlegreen,bg=jeans} +\setbeamercolor*{palette quaternary}{fg=paleblue,bg=regal} + +% \setbeamercolor*{palette primary}{fg=crimsonred,bg=white} +% \setbeamercolor*{palette secondary}{fg=charcoal,bg=white} +% \setbeamercolor*{palette tertiary}{fg=paleblue,bg=white} +% \setbeamercolor*{palette quaternary}{fg=turtlegreen,bg=white} + + + +%% These are some structures that I define for use in the outer theme. + +\setbeamercolor*{crimsontown}{fg=paleblue,bg=crimsonred} +\setbeamercolor*{bluetown}{fg=white,bg=DodgerBlue2} +%\setbeamercolor*{titlelike}{fg=crimsonred} +\setbeamercolor*{titlelike}{fg=DodgerBlue2} +\setbeamercolor*{topbarlike}{fg=white, bg=charcoal} + + + + + +%% Some old definitions that I might want to get back to +% +% \setbeamercolor*{frametitle}{bg=white,fg=charcoal} +% \setbeamercolor*{part title}{bg=white,fg=charcoal} +% \setbeamercolor*{item}{fg=charcoal, bg=white} +% +% \setbeamercolor*{separation line}{} +% \setbeamercolor*{fine separation line}{} + + + + + +\mode + diff --git a/Lectures_my/MC_2016/Lecture8/beamerinnerthemeFlip.sty b/Lectures_my/MC_2016/Lecture8/beamerinnerthemeFlip.sty new file mode 100755 index 0000000..9910f4d --- /dev/null +++ b/Lectures_my/MC_2016/Lecture8/beamerinnerthemeFlip.sty @@ -0,0 +1,95 @@ +% Inner style file for the BEAMER FLIP THEME +% Copyright 2010 by Flip Tanedo +% This file may be distributed and/or modified +% 1. under the LaTeX Project Public License and/or +% 2. under the GNU Public License. +% +% Based on: beamerinnerthemeFlip by Marco Barisione + +\mode + +% % Use alternative title page style. +% \DeclareOptionBeamer{alternativetitlepage}[true]{\def\beamer@Flip@alternativetitlepage{#1}} + +% Logo to use in the alternative title page. +% \def\beamer@Flip@titlepagelogo{} +% \DeclareOptionBeamer{titlepagelogo}{\def\beamer@Flip@titlepagelogo{#1}} + +% Bullet shape. +\DeclareOptionBeamer{bullet}{\def\beamer@Flip@bullet{#1}} +\DeclareOptionBeamer{shadow}[true]{\def\beamer@themerounded@shadow{#1}} +\ExecuteOptionsBeamer{shadow=false} + + +% \ExecuteOptionsBeamer{alternativetitlepage=false,bullet=square} +\ProcessOptionsBeamer + + + +% Margins. +\newlength{\beamer@Flip@normalmargin} +\setlength{\beamer@Flip@normalmargin}{.06\paperwidth} +\setbeamersize{text margin left=\beamer@Flip@normalmargin} +\setbeamersize{text margin right=\beamer@Flip@normalmargin} +\setlength\leftmargini{.6\beamer@Flip@normalmargin} +\setlength\leftmarginii{.6\beamer@Flip@normalmargin} +\setlength\leftmarginiii{.6\beamer@Flip@normalmargin} + +% Normal title page. +\defbeamertemplate*{title page normal}{Flip theme}[1][] +{ + \vbox{} + \vfill + \begin{centering} + \begin{beamercolorbox}[wd=\paperwidth,sep=8pt,center,#1]{title page header} + \usebeamerfont{title}\inserttitle\par% + \ifx\insertsubtitle\@empty% + \else% + \vskip0.25em% + {\usebeamerfont{subtitle}\usebeamercolor[fg]{subtitle}\insertsubtitle\par}% + \fi% + \end{beamercolorbox}% + \vskip1em\par + \begin{beamercolorbox}[sep=8pt,center,#1]{author} + \usebeamerfont{author}\insertauthor + \end{beamercolorbox} + \begin{beamercolorbox}[sep=8pt,center,#1]{institute} + \usebeamerfont{institute}\insertinstitute + \end{beamercolorbox} + \begin{beamercolorbox}[sep=8pt,center,#1]{date} + \usebeamerfont{date}\insertdate + \end{beamercolorbox}\vskip0.5em + {\usebeamercolor[fg]{titlegraphic}\inserttitlegraphic\par} + \end{centering} + \vfill +} + + +% Items. +\defbeamertemplate{itemize item}{squarealt}% +{\tiny\raise.5ex\hbox{\donotcoloroutermaths$\blacksquare$}} +\defbeamertemplate{itemize subitem}{squarealt}% +{\tiny\raise.4ex\hbox{\donotcoloroutermaths$\square$}} +\defbeamertemplate{itemize subsubitem}{squarealt}% +{\tiny\raise.3ex\hbox{\donotcoloroutermaths$\blacksquare$}} +% +\defbeamertemplate{itemize item}{circlealt}% +{\small\raise.2ex\hbox{\donotcoloroutermaths$\bullet$}} +\defbeamertemplate{itemize subitem}{circlealt}% +{\small\raise.1ex\hbox{\donotcoloroutermaths$\circ$}} +\defbeamertemplate{itemize subsubitem}{circlealt}% +{\scriptsize\raise.1ex\hbox{\donotcoloroutermaths$\bullet$}} +\setbeamertemplate{items}[circlealt] +% +\def\circletext{circle} +\ifx\beamer@Flip@bullet\circletext + \setbeamertemplate{items}[circlealt] +\else + \setbeamertemplate{items}[squarealt] +\fi + +\setbeamertemplate{blocks}[rounded][shadow=\beamer@themerounded@shadow] + + +\mode + diff --git a/Lectures_my/MC_2016/Lecture8/beamerouterthemeFlip.sty b/Lectures_my/MC_2016/Lecture8/beamerouterthemeFlip.sty new file mode 100755 index 0000000..d12f6d6 --- /dev/null +++ b/Lectures_my/MC_2016/Lecture8/beamerouterthemeFlip.sty @@ -0,0 +1,189 @@ +% Outer style file for the BEAMER FLIP THEME 212 +% Copyright 2012 by Flip Tanedo +% This file may be distributed and/or modified +% 1. under the LaTeX Project Public License and/or +% 2. under the GNU Public License. +% +% Based on: beamerouterthemesplit by Till Tantau + + +\mode + +% Watermark +\DeclareOptionBeamer{watermark}{\def\beamer@Flip@watermark{#1}} +\ExecuteOptionsBeamer{watermark=} % Default value is empty + + +\newif\ifbeamer@Flip@bigpagenumber +\beamer@Flip@bigpagenumberfalse +\DeclareOptionBeamer{bigpagenumber}{\beamer@Flip@bigpagenumbertrue} + + +\def\beamer@Flip@truetext{true} % Kludge for true/false so I can use ifx + +\DeclareOptionBeamer{topline}[true]{\def\beamer@Flip@topline{#1}} +\ExecuteOptionsBeamer{topline=false} % Default + + + +% \ExecuteOptionsBeamer{topline=} % Default value is empty + + +\ProcessOptionsBeamer + + +\usepackage{tikz} % For fancy decorations + + +% I should fix these up to be more robust against color changes +\setbeamercolor{section in head/foot}{parent=footerlike} +\setbeamercolor{subsection in head/foot}{parent=footerlike} +\setbeamercolor{author in head/foot}{parent=footerlike} +\setbeamercolor{title in head/foot}{parent=footerlike} + + +% \setbeamercolor{author in head/foot}{parent=section in head/foot} +% \setbeamercolor{title in head/foot}{parent=subsection in head/foot} + +%\setbeamercolor{topbar}{parent=crimsontown} +\setbeamercolor{topbar}{parent=bluetown} + + + + +\usesectionheadtemplate + {\hfill\insertsectionhead} + {\hfill\color{fg!50!bg}\insertsectionhead} + + +%%%%%%%%%%%%%%%%%% +% The footer bar % +%%%%%%%%%%%%%%%%%% + +\defbeamertemplate*{footline}{Flip theme} +{% + \leavevmode% +% OBSERVATION: the "%" symbols inside hbox are all very important here. +% The \hbox enviroment will insert spaces whenever there's whitespace +% Adding a % at the END of each line ensures that any hard returns aren't +% interpreted as white space. This allows the color boxes to be flush against +% one another. If a faint white line appears one could probably do something like +% hspace{-1px} and add an additional pixel-or-two wide beamercolorbox. + +\hbox{% +\begin{beamercolorbox}[wd=.4\paperwidth,ht=2.5ex,dp=1.125ex,leftskip=.3cm, rightskip=.3cm plus1fil]{author in head/foot}% + \usebeamerfont{author in head/foot}\insertshortauthor \end{beamercolorbox}% +% +\begin{beamercolorbox}[wd=.6\paperwidth,ht=2.5ex,dp=1.125ex,leftskip=.3cm,rightskip=.3cm plus1fil]{title in head/foot}% + \usebeamerfont{title in head/foot}{\textit \insertshorttitle}% + \hskip2ex plus1fill% + % \insertpagenumber\,/\,\insertpresentationendpage %% Want FRAME, not SLIDE number + \insertframenumber/\inserttotalframenumber + \end{beamercolorbox}% +}% + +% This is a complete kludge for the faint white line which appears +% between the beamer color boxes above. (It doesnt appear in presentation mode +% but sometimes looks funny when the pdfs are viewed in a window) +\begin{tikzpicture}[overlay] + \usebeamercolor[bg]{author in head/foot} + \draw[thick] (0,3.625ex) -- (\paperwidth,3.625ex); + \draw (.4\paperwidth,0) -- (.4\paperwidth,3.625ex); + \draw[thick] (0,0) -- (\paperwidth,0); +\end{tikzpicture} + + + +\ifbeamer@Flip@bigpagenumber + \begin{tikzpicture}[overlay] + \usebeamercolor[bg]{author in head/foot} + \draw[fill] (\paperwidth-5ex,3.625ex) circle (6ex); + \usebeamercolor[fg]{author in head/foot} + \draw[fill] (\paperwidth-5ex,3.625ex) circle (5ex); + \usebeamercolor[bg]{author in head/foot} + \node at (\paperwidth-5ex,4ex) {\small$^\text{\insertframenumber}/_\text{\inserttotalframenumber}$}; + \end{tikzpicture} +\fi +} + + + +\defbeamertemplate*{frametitle}{Flip theme}[1][left] +{ +% \ifbeamercolorempty[bg]{frametitle}{}{\nointerlineskip}% +\@tempdima=\textwidth% +\advance\@tempdima by\beamer@leftmargin% +\advance\@tempdima by\beamer@rightmargin% + +\vbox{}\vskip-3.5ex% +% +\begin{beamercolorbox}[sep=0.3cm,#1,wd=\the\@tempdima]{titlelike} + \usebeamerfont{frametitle}% + \vbox{}\vskip-1ex% + \if@tempswa\else\csname beamer@fte#1\endcsname\fi% + % \strut{\bf\insertframetitle}\strut\par% %% Include the \bf + \strut{\insertframetitle}\strut\par% + {% + \ifx\insertframesubtitle\@empty% + \else + {\usebeamerfont{framesubtitle} + \usebeamercolor[fg]{framesubtitle} + \insertframesubtitle\strut\par}% + \fi + }% + % + \vskip-1ex% + \if@tempswa\else\vskip-.3cm\fi% set inside beamercolorbox... evil here... +\end{beamercolorbox}% +%% + +% \def\beamer@fteright{\vskip0.35cm\advance\leftskip by 1.7cm\advance\rightskip by1.7cm} +} + + + + +% Sidebar right is important: anything placed here will be +% visible *behind* the main text. This is where to put any +% interesting watermarks. +% +% +\defbeamertemplate*{sidebar right}{Flip theme} +{ +% \llap{\includegraphics[width=50px]{footdecoration}} +% \llap{\includegraphics[width=\paperwidth,height=\paperheight]{upperleft_watermark_transp}} +% +% THIS WOULD BE A NICE WAY TO PUT THE PARTICLE BUBBLE CHAMBER BG +% ... make a png of the background with TRANSPARENT BG +% ... then put the image here (use \hfill to get it on the left) +% Now one can FURTHER apply a slight gradient in the BG +% +% +% % Top Line +\ifx\beamer@Flip@topline\beamer@Flip@truetext + \begin{tikzpicture}[overlay] + \usebeamercolor[fg]{titlelike} % alternately, topbarlike would be a good beamercolor +% \usebeamercolor[fg]{topbarlike} % alternately, topbarlike would be a good beamercolor + \draw[fill] (-\paperwidth,0) rectangle (0,-3pt); + \end{tikzpicture} +\fi +% +% + \vskip.2\beamer@leftmargin% + %\llap{\insertlogo\hskip.5\beamer@leftmargin}% + \llap{\insertlogo\hskip.08\beamer@leftmargin}% + \vfill% + \if \beamer@Flip@watermark\@empty + % This is kind of a kludge since if I put anything in here, I automatically get errors + % Somehow LaTeX wants to read the \else part! + \else% + \llap{\includegraphics[width=\paperwidth,height=\paperheight]{\beamer@Flip@watermark}} + \fi +} + + + +\mode +\endinput + + diff --git a/Lectures_my/MC_2016/Lecture8/beamerouterthemedecolines.sty b/Lectures_my/MC_2016/Lecture8/beamerouterthemedecolines.sty new file mode 100755 index 0000000..669e493 --- /dev/null +++ b/Lectures_my/MC_2016/Lecture8/beamerouterthemedecolines.sty @@ -0,0 +1,168 @@ +% Copyright 2007 by Marco Barisione +% +% This file may be distributed and/or modified +% +% 1. under the LaTeX Project Public License and/or +% 2. under the GNU Public License. + +\mode + +% String used between the current page and the total page count. +\def\beamer@decolines@pageofpages{/} +\DeclareOptionBeamer{pageofpages}{\def\beamer@decolines@pageofpages{#1}} + +% Show a line below the frame title. +\DeclareOptionBeamer{titleline}[true]{\def\beamer@decolines@titleline{#1}} + +% Image used for the watermark. +\def\beamer@decolines@watermarkorig{} +\DeclareOptionBeamer{watermark}{\def\beamer@decolines@watermarkorig{#1}} + +% Height of the watermark. +\def\beamer@decolines@watermarkheight{100px} +\DeclareOptionBeamer{watermarkheight}{\def\beamer@decolines@watermarkheight{#1}} + +% The original image height is watermarkheightmult * watermarkheight. +\def\beamer@decolines@watermarkheightmult{1} +\DeclareOptionBeamer{watermarkheightmult}{\def\beamer@decolines@watermarkheightmult{#1}} + +\ExecuteOptionsBeamer{titleline=false} +\ProcessOptionsBeamer + +% Enable/disable the watermark. +\def\watermarkon{% + \def\beamer@decolines@watermark{\beamer@decolines@watermarkorig}% +} +\def\watermarkoff{\def\beamer@decolines@watermark{}} + +% Initially enable the watermark. +\watermarkon + +% Colors. +\setbeamercolor*{lineup}{parent=palette primary} +\setbeamercolor*{linemid}{parent=palette secondary} +\setbeamercolor*{linebottom}{parent=palette tertiary} +\setbeamercolor*{page header}{parent=titlelike} + +% Lengths +\newlength{\headerheight} +\setlength{\headerheight}{.045\paperheight} +\newlength{\beamer@decolines@lineup} +\setlength{\beamer@decolines@lineup}{.025\paperheight} +\newlength{\beamer@decolines@linemid} +\setlength{\beamer@decolines@linemid}{.015\paperheight} +\newlength{\beamer@decolines@linebottom} +\setlength{\beamer@decolines@linebottom}{.01\paperheight} + +% The height of the watermark part below the 3 bottom lines. +\newlength{\beamer@decolines@watermarkheightbottom} +\addtolength{\beamer@decolines@watermarkheightbottom}{\beamer@decolines@lineup} +\addtolength{\beamer@decolines@watermarkheightbottom}{\beamer@decolines@linemid} +\addtolength{\beamer@decolines@watermarkheightbottom}{\beamer@decolines@linebottom} + +% The height of the watermark part over the 3 bottom lines before shrinking. +\newlength{\beamer@decolines@watermarkheightupperorig} +\setlength{\beamer@decolines@watermarkheightupperorig}{\beamer@decolines@watermarkheight} +\addtolength{\beamer@decolines@watermarkheightupperorig}{-\beamer@decolines@watermarkheightbottom} +\multiply\beamer@decolines@watermarkheightupperorig by \beamer@decolines@watermarkheightmult + +% % Footer. +% \defbeamertemplate*{footline}{decolines theme} +% { +% \leavevmode% +% % Page number. +% \hbox{% +% \begin{beamercolorbox}[wd=.2\paperwidth,ht=0ex,dp=0ex,center]{}% +% \usebeamerfont{palette primary}\insertframenumber{} \beamer@decolines@pageofpages{} \inserttotalframenumber% +% \end{beamercolorbox}% +% \begin{beamercolorbox}[wd=.8\paperwidth,ht=0ex,dp=0ex]{}% +% \end{beamercolorbox}% +% } % +% % First line. +% \hbox{% +% \begin{beamercolorbox}[wd=.2\paperwidth,ht=\beamer@decolines@lineup,dp=0pt]{}% +% \end{beamercolorbox}% +% \begin{beamercolorbox}[wd=.8\paperwidth,ht=\beamer@decolines@lineup,dp=0pt]{lineup}% +% \end{beamercolorbox}% +% } % +% % Second line. +% \hbox{% +% \begin{beamercolorbox}[wd=\paperwidth,ht=\beamer@decolines@linemid,dp=0pt]{linemid}% +% \end{beamercolorbox}% +% } % +% % Third line. +% \hbox{% +% \begin{beamercolorbox}[wd=.1\paperwidth,ht=\beamer@decolines@linebottom,dp=0pt]{}% +% \end{beamercolorbox}% +% \begin{beamercolorbox}[wd=.9\paperwidth,ht=\beamer@decolines@linebottom,dp=0pt]{linebottom}% +% \end{beamercolorbox}% +% }% +% % This seems to fix some alignment problems with the watermark. It has to be +% % always applied if you do not want to see the footer moving up and down when +% % moving from a page with watermark to a page without or vice versa. +% \vskip-.5px% +% % Watermark. +% \if\beamer@decolines@watermark\@empty\else% +% \vskip-\beamer@decolines@watermarkheightbottom% +% \llap{\includegraphics[height=\beamer@decolines@watermarkheightbottom,clip=true,% +% trim=0pt 0pt 0pt \beamer@decolines@watermarkheightupperorig]{\beamer@decolines@watermark}\hskip-\paperwidth}% +% \fi% +% } + +\defbeamertemplate*{headline}{decolines theme} +{ + \leavevmode% + \hbox{% + \begin{beamercolorbox}[wd=\paperwidth,ht=\headerheight,dp=0pt]{page header}% + \end{beamercolorbox}% + } % + \vskip0pt% +} + +\defbeamertemplate*{frametitle}{decolines theme}[1][left] +{ + \ifbeamercolorempty[bg]{frametitle}{}{\nointerlineskip}% + \@tempdima=\textwidth% + \advance\@tempdima by\beamer@leftmargin% + \advance\@tempdima by\beamer@rightmargin% + \vbox{}\vskip-.5\beamer@leftmargin% + \begin{beamercolorbox}[sep=\beamer@leftmargin,#1,wd=\the\@tempdima]{} + \usebeamerfont{frametitle}\usebeamercolor[bg]{framesubtitle}% + \vbox{}\vskip0ex% + \if@tempswa\else\csname beamer@fte#1\endcsname\fi% + \strut\insertframetitle\strut\par% + {% + \ifx\insertframesubtitle\@empty% + \else% + {\usebeamerfont{framesubtitle}\usebeamercolor[bg]{framesubtitle}\insertframesubtitle\strut\par}% + \fi + }% + \vskip-1ex% + \if@tempswa\else\vskip-\beamer@leftmargin\fi + \end{beamercolorbox}% + \def\beamer@decolines@truetext{true}% + \ifx\beamer@decolines@titleline\beamer@decolines@truetext% + \vskip-.5\beamer@leftmargin% + \begin{beamercolorbox}[wd=\textwidth,ht=.1ex,dp=0ex]{linemid}% + \end{beamercolorbox}% + \fi +} + +% Frame title continuations, default +\defbeamertemplate*{frametitle continuation}{decolines theme}{(\insertcontinuationcount)} + + + +\defbeamertemplate*{sidebar right}{decolines theme} +{ + \vskip.1\beamer@leftmargin% + \llap{\insertlogo\hskip.5\beamer@leftmargin}% + \vfill% + \if\beamer@decolines@watermark\@empty\else% + \llap{\includegraphics[height=\beamer@decolines@watermarkheight]{\beamer@decolines@watermark}}% + \vskip-\beamer@decolines@watermarkheightbottom% + \fi +} + +\mode + diff --git a/Lectures_my/MC_2016/Lecture8/beamerthemeFlip.sty b/Lectures_my/MC_2016/Lecture8/beamerthemeFlip.sty new file mode 100755 index 0000000..a615ff5 --- /dev/null +++ b/Lectures_my/MC_2016/Lecture8/beamerthemeFlip.sty @@ -0,0 +1,59 @@ +% Main style file for the BEAMER FLIP THEME 2012 +% Copyright 2012 by Flip Tanedo +% This file may be distributed and/or modified +% 1. under the LaTeX Project Public License and/or +% 2. under the GNU Public License. +% +% Comments: This is still a work in progress. +% A good way to present: http://www.cs.hmc.edu/~oneill/freesoftware/pdftokeynote.html + +% \usepackage{beamerthemesplit} +\mode + + +%% These are the options that are fed in through the driver file +%% Some of them get passed on to the other theme files +\DeclareOptionBeamer{bullet}{\PassOptionsToPackage{bullet=#1}{beamerinnerthemeFlip}} +\DeclareOptionBeamer{bigpagenumber}{\PassOptionsToPackage{bigpagenumber}{beamerouterthemeFlip}} +\DeclareOptionBeamer{topline}[true]{\PassOptionsToPackage{topline=#1}{beamerouterthemeFlip}} +\DeclareOptionBeamer{shadow}[false]{\PassOptionsToPackage{shadow=#1}{beamerinnerthemeFlip}} + +\DeclareOptionBeamer{watermark}{\PassOptionsToPackage{watermark=#1}{beamerouterthemeFlip}} + +\ProcessOptionsBeamer + +\useinnertheme{Flip} % Calls beamerinnerthemeFlip.sty +\useoutertheme{Flip} % Calls beamerouterthemeFlip.sty +\usecolortheme{Flip} % Calls beamercolorthemeFlip.sty + + + +%%%%%%%%%%%%%%%%%%%%%%%% +% Background Gradient % +%%%%%%%%%%%%%%%%%%%%%%%% + + +%% Usually I would leave this on +%% However, there seems to be problems with XeLaTeX +%% http://tex.stackexchange.com/questions/29497/xelatex-preventing-beamer-from-using-different-backgrounds +\setbeamertemplate{background canvas}[vertical shading][bottom=blue!.1, middle=white, top=white] + +%% HOWEVER: it seems to make it impossible to change this later on in the document + + +%% For a dark background: +% \setbeamertemplate{background canvas}[vertical shading][bottom=keynotebottom, middle=keynotemiddle, top=keynotetop] + + +%% An alternate way: \beamertemplateshadingbackground{blue!.1}{red!2} +%% In general: you want the gradient to be subtle! + + +%%%%%%%%%%%%%%%%%%%%%%% +% Navigation symbols % +%%%%%%%%%%%%%%%%%%%%%%% + +\setbeamertemplate{navigation symbols}{} % Turns off PDF navigation symbols + +\mode +\endinput \ No newline at end of file diff --git a/Lectures_my/MC_2016/Lecture8/images/BG_lower.png b/Lectures_my/MC_2016/Lecture8/images/BG_lower.png new file mode 100755 index 0000000..073034a --- /dev/null +++ b/Lectures_my/MC_2016/Lecture8/images/BG_lower.png Binary files differ diff --git a/Lectures_my/MC_2016/Lecture8/images/Focal_stability.png b/Lectures_my/MC_2016/Lecture8/images/Focal_stability.png new file 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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.}} + +\author{ {\fontspec{Trebuchet MS}Marcin Chrz\k{a}szcz} (Universit\"{a}t Z\"{u}rich)} +\institute{UZH} +\title[Random number generators and application]{Random number generators and application} +\date{\fixme} + + +\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 {Applications of Monte Carlo methods} + \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}{Experimental Methods in Particle Physics, \\ 26 November, 2015} +\end{center} +\end{frame} +} + + + + + +\begin{frame}\frametitle{Markov Chain MC} +\begin{itemize} +\item Consider a finite possible states: $S_1$, $S_2$, ... +\item And the time steps of time, labelled as $1$, $2$, ... +\item At time $t$ the state is denoted $X_t$. +\item The conditional probability is defined as: +\end{itemize} +\begin{equation} +P(X_t=S_j \vert X_{t-1}=S_{j-1},..., X_{1}=S_{1}) \nonumber +\end{equation} +\begin{itemize} +\item The Markov chain is then if the probability depends only on previous step. +\end{itemize} +\begin{equation} +P(X_t=S_j \vert X_{t-1}=S_{j-1},..., X_{1}=S_{1}) = P(X_t=S_j \vert X_{t-1}=S_{j-1} )\nonumber +\end{equation} +\begin{itemize} +\item For this reason this reason MCMC is also knows as drunk sailor walk. +\item Very powerful method. Used to solve linear eq. systems, invert matrix, solve differential equations, etc. +\end{itemize} +\end{frame} + + + +\begin{frame}\frametitle{Linear Equations} +\begin{itemize} +\item Lets say we have a linear equation system: +\end{itemize} +\begin{equation} +\begin{array}{lcl} X & = & pY + (1-p) A \\ Y & = & qX + (1-q)B \end{array} \nonumber +\end{equation} +\begin{itemize} +\item We know $A,B,p,q$; $X$ and $Y$ are meant to be determined. +\item Algorithm: +\begin{enumerate} +\item We choose first element of the first equation with probability $p$ and second with probability $1-p$. +\item We we choose the second one, the outcome of this MCMC is $W=A$. +\item If we choose the first we go to second equation and choose the first element with probability $q$ and the second with $1-q$. +\item We we choose the second one, the outcome of this MCMC is $W=B$. +\item If we choose the first we go to the first equation back again. +\item We repeat the procedure. +\end{enumerate} +\item We can estimate the solution of this system: +\end{itemize} +\begin{equation} +\hat{X} = \dfrac{1}{N}\sum_{i=1} W_i{~}{~}{~}{~}{~} \hat{\sigma_X}=\dfrac{1}{\sqrt{N-1}}\sqrt{\dfrac{1}{N} \sum_{i=1}^N W_i^2-\hat{X}^2} \nonumber +\end{equation} + +\end{frame} + +\begin{frame}\frametitle{Neumann-Ulam method} +\begin{itemize} +\item Let's try apply the basic MCMC method to solve a simple linear equation system: +\end{itemize} +\begin{equation} +A \overrightarrow{x} = \overrightarrow{b} \nonumber +\end{equation} +\begin{itemize} +\item The above system can be (always, see linear algebra lecture) translated into system: +\end{itemize} +\begin{equation} +\overrightarrow{x} = \overrightarrow{a} + H \overrightarrow{x} \nonumber +\end{equation} +\begin{itemize} +\item For this method we assume that the norm of the matrix is: +\end{itemize} +\begin{equation} +\Vert H \Vert = \underset{1 \leq i \leq n}{max} \sum_{j=1}^n \vert h_{ij} \vert <1 \nonumber \end{equation} +\begin{itemize} +\item Which we can write in a form: +\end{itemize} +\begin{equation} +(1 -H)\overrightarrow{x}=\overrightarrow{a} \nonumber +\end{equation} +\end{frame} + + + + +\begin{frame}\frametitle{Neumann-Ulam method} +\begin{itemize} +\item The solution would be then: +\end{itemize} +\begin{equation} +\overrightarrow{x}_0=(1 -H)^{-1}\overrightarrow{a} \nonumber +\end{equation} +\begin{itemize} +\item We can Taylor expend this: +\end{itemize} +\begin{equation} +\overrightarrow{x}_0=(1 -H)^{-1}\overrightarrow{a} = \overrightarrow{a} + H \overrightarrow{a} + H^2 \overrightarrow{a} + H^3 \overrightarrow{a} +.... \nonumber +\end{equation} +\begin{itemize} +\item For the $i$-th component of the $\overrightarrow{x}$ vector: +\end{itemize} +\begin{equation} +x_0^i= a_i + \sum_{j=1}^n h_{ij} a_{j_1} + \sum_{j_1=1}^n \sum_{j_2=1}^n h_{ij_1} h_{ij_2} a_{j_2} + \sum_{j_1=1}^n \sum_{j_2=1}^n \sum_{j_3=1}^n h_{ij_1} h_{ij_2} h_{ij_3} a_{j_3} + ...\nonumber +\end{equation} +\begin{itemize} +\item One can construct probabilistic behaviour of a system that follows the path of equation above. +\end{itemize} + +\end{frame} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\begin{frame}\frametitle{Neumann-Ulam method} +\begin{itemize} +\item To do so we add to our matrix an additional column of the matrix: +\end{itemize} +\begin{equation} +h_{i,0} = 1-\sum_{j=1}^n h_{ij} > 0 \nonumber +\end{equation} +\begin{itemize} +\item The system has states: $\lbrace 0,1,2...,n\rbrace$ +\item State at $t$ time is denoted as $i_t$. +\item We make a random walk accordingly to to the following rules: +\begin{itemize} +\item At the begging of the walk ($t=0$) we are at $i_0$. +\item In the $t$ moment we are in the $i_t$ position then in $t+1$ time stamp we move to state $i_{t+1}$ with the probability $h_{i_t i_{t+1}}$. +\item We stop walking if we are in state $0$. +\end{itemize} +\item The path $X(\gamma) = (i_0, i_1, i_2, ..., i_k, 0)$ is called trajectory. +\item It can be proven that $x_i^0 =E \lbrace X (\gamma) \vert i_0=j \rbrace$. +\end{itemize} + + + +\end{frame} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\begin{frame}\frametitle{Neumann-Ulam method, $\rm \color{RubineRed}{Lecture3/Markov}$} +\begin{itemize} +\item For example lets try to solve this equation system: +\end{itemize} +\begin{equation} +\overrightarrow{x} = +\left(\begin{array}{c} + 1.5 \\ +-1.0\\ +0.7 \end{array} \right) ++ +\left(\begin{array}{ccc} +0.2 & 0.3 & 0.1 \\ +0.4 & 0.3 & 0.2 \\ +0.3 & 0.1 & 0.1 \end{array} \right) \overrightarrow{x} + \nonumber +\end{equation} +\begin{itemize} +\item The solution is $\overrightarrow{x}_0 = (2.154303, 0.237389, 1.522255)$. +\end{itemize} +\begin{columns} + +\column{0.1in} + +\column{2.5in} +\begin{itemize} +\item The propability matrix $h_{ij}$ has the shape: +\end{itemize} +\begin{tabular}{|c|cccc|} +\hline +$i/j$ & 1 & 2 & 3 & 4 \\ \hline +1 & 0.2 & 0.3 & 0.1 & 0.4 \\ +2 & 0.4 & 0.3 & 0.2 & 0.1 \\ +3 & 0.3 & 0.1 & 0.1 & 0.5 \\ \hline +\end{tabular} + +\column{2.5in} +\begin{itemize} +\item An example solution: +\end{itemize} +\includegraphics[width=0.95\textwidth]{images/mark.png} + +\end{columns} + + +\end{frame} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\begin{frame}\frametitle{Neumann-Ulam dual method} +\begin{itemize} +\item The problem with Neumann-Ulam method is that you need to repeat it for each of the coordinates of the $\overrightarrow{x}_0$ vector. +\item The dual method calculates the whole $\overrightarrow{x}_0$ vector. +\item The algorithm: +\begin{itemize} +\item On the indexes: $\lbrace0,1,...,n\rbrace$ we set a probability distribution:\\ $q_1, q_2,..., q_n$, $q_i>0$ and $\sum_i=1^n q_i=1$. +\item The starting point we select from $q_i$ distribution. +\item If in $t$ time we are in $i_t$ state then with probability $p(i_{t+1} \vert i_t) = h_{i_{t+1},i_{t}}$ in $t+1$ we will be in state $i_1$. For $i_{t+1}=0$ we define the probability: $h_{0,i_{t}}=1-\sum_{j=1}^n h_{j,i_{t}}$. Here we also assume that $h_{j,i_{t}} > 0$. +\item NOTE: there the matrix is transposed compared to previous method: $H^{T}$. +\item Again we end our walk when we are at state $0$. +\item For the trajectory: $\gamma = (i_0, i_1,...,i_k, 0)$, we assign the vector: +\end{itemize} +\begin{equation} +\overrightarrow{Y}(\gamma) = \dfrac{a_{i_0}}{ q_{i_{0}} p(0 \vert i_k) } \widehat{e}_{i_{k}} \in \mathcal{R}^n \nonumber +\end{equation} +\item The solution will be : $\overrightarrow{x}^0 = \dfrac{1}{N} \sum \overrightarrow{Y}(\gamma)$ +\end{itemize} + + +\end{frame} + + + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\begin{frame}\frametitle{Neumann-Ulam dual method, $\rm \color{RubineRed}{Lecture3/Markov2}$} +\begin{itemize} +\item Let's try to solve the equation system: +\end{itemize} +\begin{equation} +\overrightarrow{x} = +\left(\begin{array}{c} + 1.5 \\ +-1.0\\ +0.7 \end{array} \right) ++ +\left(\begin{array}{ccc} +0.2 & 0.3 & 0.1 \\ +0.4 & 0.3 & 0.2 \\ +0.1 & 0.1 & 0.1 \end{array} \right) \overrightarrow{x} + \nonumber +\end{equation} +\begin{itemize} +\item The solution is: $\overrightarrow{x}_0 = (2.0, 0.0, 1.0)$. +\item Let's put the initial probability as constant: +\end{itemize} +\begin{equation} +q_1=q_2=q_3=\dfrac{1}{3} \nonumber +\end{equation} +\begin{columns} + +\column{0.1in} + +\column{2.5in} +\begin{itemize} +\item The propability matrix $h_{ij}$ has the shape: +\end{itemize} +\begin{tabular}{|c|cccc|} +\hline +$i/j$ & 1 & 2 & 3 & 4 \\ \hline +1 & 0.2 & 0.4 & 0.1 & 0.3 \\ +2 & 0.3 & 0.3 & 0.1 & 0.3 \\ +3 & 0.1 & 0.2 & 0.1 & 0.6 \\ \hline +\end{tabular} + +\column{2.5in} +\begin{itemize} +\item An example solution: +\end{itemize} +\includegraphics[width=0.95\textwidth]{images/mark2.png} + +\end{columns} +\end{frame} + + + + +\begin{frame}\frametitle{Look elsewhere effect, $\rm \color{RubineRed}{Lecture3/LEE}$} +\begin{itemize} +\item Look elsewhere effect addresses the following problem: +\begin{itemize} +\item Imagine you observed a $3\sigma$ deviation in one of the observable that you measured. +\item Before you get excited one needs to understand if given the fact that you had so many measurements this might happen! +\end{itemize} +\item Example: Let's say we have measured 50 observables. What is the probability to observed 1 that is $3\sigma$ away from theory prediction? +\item Let's simulate 50 Gaussian distribution centred at 0 and width of 1. We count how simulations where at least one of the 50 numbers have the absolute value $>3$. +\item More complicated example: what if you observed 3 in a row $2\sigma$ fluctuations among 50 measurements? +\item This kind of studies are the best solvable by MC simulations. +\end{itemize} + + +\end{frame} + + + + + + +\begin{frame}\frametitle{Travelling Salesman Problem} +\begin{itemize} +\item Salesman starting from his base has to visit $n-1$ other locations and return to base headquarters. The problem is to find the shortest way. +\item For large $n$ the problem can't be solver by brutal force as the complexity of the problem is $(n-1)!$ +\item There exist simplified numerical solutions assuming factorizations. Unfortunately even those require anonymous computing power. +\item Can MC help? YES :) +\item The minimum distance $l$ has to depend on 2 factors: $P$ the area of the city the Salesman is travelling and the density of places he wants to visit: $\dfrac{n}{P}$ +\item Form this we can assume: +\end{itemize} +\begin{equation} +l \sim P^a (\dfrac{n}{P})^b=P^{a-b}n^b. \nonumber +\end{equation} + +\end{frame} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + + + + + + + +\begin{frame}\frametitle{Traveling Salesman Problem} +\begin{itemize} +\item From dimension analysis: +\end{itemize} +\begin{equation} +a-b=\dfrac{1}{2}. \nonumber +\end{equation} +\begin{itemize} +\item To get $l$ we need square root of area. +\item From this it's obvious: +\end{itemize} +\begin{equation} +l \sim P^a (\dfrac{n}{P})^b=P^{0.5}n^{a-0.5}. \nonumber +\end{equation} +\begin{itemize} +\item Now we can multiply the area by alpha factor that keeps the density constant then: +\end{itemize} +\begin{equation} +l \sim \alpha^0.5 \alpha6{a-0.5} = \alpha^a \nonumber +\end{equation} +\begin{itemize} +\item In this case the distance between the clients will not change, but the number of clients will increase by $\alpha$ so: +\end{itemize} +\begin{equation} +l \sim \alpha \nonumber +\end{equation} +\begin{itemize} +\item In the end we get: $a=1$ +\end{itemize} +\end{frame} + + + +\begin{frame}\frametitle{Traveling Salesman Problem} +\begin{itemize} +\item In total: +\end{itemize} +\begin{equation} +l \sim k (nP)^{0.5}\nonumber +\end{equation} +\begin{itemize} +\item Of course the k depends on the shape of the area and locations of client. However for large $n$ the k starts loosing the dependency. It's an asymptotically free estimator. +\item To use the above formula we need to somehow calculate k. +\item How to estimate this? Well make a TOY MC: take a square put uniformly $n$ points. Then we can calculate $l$. Then it's trivial: +\end{itemize} +\begin{equation} +k= l(nP)^{-0.5} \nonumber +\end{equation} +\end{frame} + + +\begin{frame}\frametitle{Traveling Salesman Problem} +\begin{itemize} +\item This kind of MC experiment might require large CPU power and time. The adventage is that once we solve the problem we can use the obtained k for other cases (it's universal constant!). +\item It turns out that: +\end{itemize} +\begin{equation} +k \sim \dfrac{3}{4} \nonumber +\end{equation} +\begin{itemize} +\item Ok, but in this case we can calculate $l$ but not the actual shortest way! Why the hell we did this exercise?! +\item Turns out that for most of the problems we are looking for the solution that is close to smallest $l$ not the exact minimum. +\end{itemize} +\end{frame} + +\begin{frame}\frametitle{War Games} + +\begin{itemize} +\item S. Andersoon 1966 simulated for Swedish government how would a tank battle look like. +\item Each of the sides has 15 tanks. that they allocate on the battle field. +\item The battle is done in time steps. +\item Each tank has 5 states: +\begin{itemize} +\item OK +\item Tank can only shoot +\item Tank can only move +\item Tank is destroyed +\item Temporary states +\end{itemize} +\item This models made possible to check different fighting strategies. + +\end{itemize} +\end{frame} + + + +\begin{frame} + +\begin{center} +\begin{Huge} + Q \& A +\end{Huge} +\end{center} + + +\end{frame} + + + + + + + + +\backupbegin + +\begin{frame}\frametitle{Backup} + + +\end{frame} + +\backupend + +\end{document} diff --git a/Lectures_my/MC_2016/Lecture8/mchrzasz.toc b/Lectures_my/MC_2016/Lecture8/mchrzasz.toc new file mode 100644 index 0000000..429917f --- /dev/null +++ b/Lectures_my/MC_2016/Lecture8/mchrzasz.toc @@ -0,0 +1,5 @@ +\boolfalse {citerequest}\boolfalse {citetracker}\boolfalse {pagetracker}\boolfalse {backtracker}\relax +\defcounter {refsection}{0}\relax +\beamer@endinputifotherversion {3.24pt} +\defcounter {refsection}{0}\relax +\select@language {english} diff --git a/Lectures_my/MC_2016/Lecture8/tikzfeynman.sty b/Lectures_my/MC_2016/Lecture8/tikzfeynman.sty new file mode 100755 index 0000000..521fc65 --- /dev/null +++ b/Lectures_my/MC_2016/Lecture8/tikzfeynman.sty @@ -0,0 +1,43 @@ +% Tikz Feynman Diagrams +% by Flip Tanedo +% 4 January 2011, work in progress + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%% TIKZ - for drawing Feynman diagrams %%%% +%%% ... use with pdflatex %%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +\usepackage{tikz} +\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 +\usepackage[tikz]{bclogo} % For cute logo boxes +\usepackage{pgffor} % For repeating patterns + +\usetikzlibrary{decorations.pathmorphing} % For Feynman Diagrams +\usetikzlibrary{decorations.markings} +\tikzset{ + % >=stealth', %% Different kind of arrows + vector/.style={decorate, decoration={snake}, draw}, + fermion/.style={postaction={decorate}, + decoration={markings,mark=at position .55 with {\arrow{>}}}}, + fermionbar/.style={draw, postaction={decorate}, + decoration={markings,mark=at position .55 with {\arrow{<}}}}, + fermionnoarrow/.style={}, + gluon/.style={decorate, + decoration={coil,amplitude=4pt, segment length=5pt}}, + scalar/.style={dashed, postaction={decorate}, + decoration={markings,mark=at position .55 with {\arrow{>}}}}, + scalarbar/.style={dashed, postaction={decorate}, + decoration={markings,mark=at position .55 with {\arrow{<}}}}, + scalarnoarrow/.style={dashed,draw}, +% +%% Special vectors (when you need to fine-tune wiggles) + provector/.style={decorate, decoration={snake,amplitude=2.5pt}, draw}, + antivector/.style={decorate, decoration={snake,amplitude=-2.5pt}, draw}, +} +