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+\documentclass[a4paper,12pt]{scrartcl}
+\usepackage[ngerman]{babel}
+\usepackage{graphicx} %BIlder einbinden
+\usepackage{amsmath} %erweiterte Mathe-Zeichen
+\usepackage{amsfonts} %weitere fonts
+\usepackage[utf8]{inputenc} %Umlaute & Co
+\usepackage{hyperref} %Links
+\usepackage{ifthen} %ifthenelse
+\usepackage{enumerate}
+\usepackage{listings}
+\lstset{language=Python}
+
+\usepackage{algpseudocode} %Pseudocode
+\usepackage{dsfont} % schöne Zahlenräumezeichen
+\usepackage{amssymb, amsthm} %noch stärker erweiterte Mathe-Zeichen
+\usepackage{tikz} %TikZ ist kein Zeichenprogramm
+\usetikzlibrary{trees,automata,arrows,shapes}
+
+\pagestyle{empty}
+
+
+\topmargin-50pt
+
+\newcounter{aufgabe}
+\def\tand{&}
+
+\newcommand{\makeTableLine}[2][0]{%
+  \setcounter{aufgabe}{1}%
+  \whiledo{\value{aufgabe} < #1}%
+  {%
+    #2\tand\stepcounter{aufgabe}%
+  }
+}
+
+\newcommand{\aufgTable}[1]{
+  \def\spalten{\numexpr #1 + 1 \relax}
+  \begin{tabular}{|*{\spalten}{p{1cm}|}}
+    \makeTableLine[\spalten]{A\theaufgabe}$\Sigma$~~\\ \hline
+    \rule{0pt}{15pt}\makeTableLine[\spalten]{}\\
+  \end{tabular}
+}
+
+\def\header#1#2#3#4#5#6#7{\pagestyle{empty}
+\begin{minipage}[t]{0.47\textwidth}
+\begin{flushleft}
+{\bf #4}\\
+#5
+\end{flushleft}
+\end{minipage}
+\begin{minipage}[t]{0.5\textwidth}
+\begin{flushright}
+#6 \vspace{0.5cm}\\
+%                 Number of Columns    Definition of Columns      second empty line
+% \begin{tabular}{|*{5}{C{1cm}|}}\hline A1&A2&A3&A4&$\Sigma$\\\hline&&&&\\\hline\end{tabular}\\\vspace*{0.1cm}
+\aufgTable{#7}
+\end{flushright}
+\end{minipage}
+\vspace{1cm}
+\begin{center}
+{\Large\bf Sheet #1}
+
+{(Hand in #3)}
+\end{center}
+}
+
+
+
+%counts the exercisenumber
+\newcounter{n}
+
+%Kommando für Aufgaben
+%\Aufgabe{AufgTitel}{Punktezahl}
+\newcommand{\Aufgabe}[2]{\stepcounter{n}
+    \textbf{Exercise \arabic{n}: #1} (#2 Points)}
+
+
+\begin{document}
+    %\header{BlattNr}{Tutor}{Abgabedatum}{Vorlesungsname}{Namen}{Semester}{Anzahl Aufgaben}
+    \header{8}{}{2015-07-02}{Intelligent Systems I}{\textit{Maximus Mutschler}\\ \textit{Jan-Peter Hohloch}
+    }{SS 15}{4}
+    \vspace{1cm}
+    \Aufgabe{Kidney Stones}{25}
+        \begin{align*}
+            P(\text{large stone})&=\frac{343}{343+357}=0.49\\
+            P_A(R=1)&=\sum\limits_Z P(R=1|Z=z,T=A)P(Z=z)\\
+            &=0.73\cdot 0.49 + 0.93\cdot 0.51\approx 0.832\\
+            P_B(R=1)&=\sum\limits_Z P(R=1|Z=z,T=A)P(Z=z)\\
+            &=0.69\cdot 0.49 + 0.87\cdot 0.51\approx 0.782
+        \end{align*}
+        (Formula from the lecture)\\
+    \Aufgabe{PC}{25}
+        \begin{itemize}
+            \item Assume X and Y are directly connected.\\
+            There can't exists a set d-seperating them, because they are always linked over the direct connection.\checkmark
+            \item Assume X,Y are not directly connected, i.e. $X\not\in N(Y)\wedge Y\not\in N(X)$.\\
+            Any non-direct path between X and Y can be blocked:\begin{itemize}
+                \item directed paths are blocked by taking one of the intermediate nodes in the separation set
+                \item common ancestors can also be blocked
+                \item common descendants are not taken in the separation set
+            \end{itemize}
+            Because the graph is acyclic each path belongs clearly to one of the above categories and so can be blocked.
+        \end{itemize}
+    \Aufgabe{IC}{25}\\
+        \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto, node distance=2cm]
+                \node[circle, draw] (A) {A};
+                \node[circle, draw] (B) [right of=A] {B};
+                \node[circle, draw] (C) [below of=A] {C};
+                \node[circle, draw] (D) [below of=B] {D};
+
+                \draw (A) -- (B);
+                \draw (B) -- (D);
+                \draw (C) -- (D);
+                \draw (A) -- (C);
+            \end{tikzpicture}
+            \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto, node distance=2cm]
+                \node[circle, draw] (A) {A};
+                \node[circle, draw] (B) [right of=A] {B};
+                \node[circle, draw] (C) [below of=A] {C};
+                \node[circle, draw] (D) [below of=B] {D};
+
+                \draw (B) -- (A);
+                \draw (B) -- (D);
+                \draw (C) -- (D);
+                \draw (A) -- (C);
+            \end{tikzpicture}
+            \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto, node distance=2cm]
+                \node[circle, draw] (A) {A};
+                \node[circle, draw] (B) [right of=A] {B};
+                \node[circle, draw] (C) [below of=A] {C};
+                \node[circle, draw] (D) [below of=B] {D};
+
+                \draw (A) -- (B);
+                \draw (B) -- (D);
+                \draw (C) -- (D);
+                \draw (C) -- (A);
+            \end{tikzpicture}\\
+    \Aufgabe{IC Performance}{25}\\
+        \begin{itemize}
+            \item worst case:\\
+                All nodes are dependent of each other, so it has to check all possible subsets (the powerset) for all pairs of nodes:
+                $$\frac{p\cdot (p-1)}{2}\cdot 2^{p-2}\in\mathcal{O}(2^p)$$
+            \item best case:\\
+                first test shows independence, so we need only one test per pair:
+                $$\frac{p\cdot (p-1)}{2}\cdot 1\in\mathcal{O}(p^2)$$
+        \end{itemize}
+        %TODO Graphs
+\end{document}
+