diff --git a/Algorithmen_und_Datenstrukturen_2/Uebung_1/Uebung_1.tex b/Algorithmen_und_Datenstrukturen_2/Uebung_1/Uebung_1.tex
new file mode 100644
index 0000000..9b6cfcc
--- /dev/null
+++ b/Algorithmen_und_Datenstrukturen_2/Uebung_1/Uebung_1.tex
@@ -0,0 +1,151 @@
+\documentclass[11pt]{scrartcl}
+\usepackage[utf8]{inputenc}
+\usepackage[ngerman]{babel}
+\usepackage{amsmath}
+%\usepackage{amssymb}
+\usepackage{multicol}
+%\usepackage{booktabs}
+%\usepackage{pstricks}
+%\usepackage{pst-node}
+
+\usepackage{listings}
+
+
+\lstset{
+ language=Java,
+ basicstyle=\ttfamily,
+ keywordstyle=\color{red},
+ commentstyle=\color{blue},
+ stringstyle=\color{darkgreen},
+ backgroundcolor=\color{lightgray},
+ morecomment=[s][\color{blue}]{/*}{*/},
+ emphstyle=\textbf,
+ frame=single,
+ showstringspaces=true,
+ breaklines=true,
+ numberstyle=\textcolor{grey},
+ numbers=left,
+ morekeywords={Merge,ExtractMinimum,Union,NIL,return,and},
+ tabsize=4,
+ captionpos=b
+}
+
+
+\usepackage[paper=a4paper,left=30mm,right=20mm,top=20mm,bottom =25mm]{geometry}
+\usepackage[
+ pdftitle={Uebungsblatt 1: Priority Queues - AlgoDS2},
+ pdfsubject={Algorithmen und Datenstrukturen 2; Uebungsblatt 1: Priority Queues},
+ pdfauthor={Thomas Battermann},
+ pdfkeywords={Uebungsblatt 1: Priority Queues - AlgoDS2},
+ pdfborder={0 0 0}
+]{hyperref}
+\usepackage{tabularx}
+\usepackage{graphicx}
+\usepackage[usenames,dvipsnames]{color}
+\definecolor{grey}{rgb}{0.5,0.5,0.5}
+\definecolor{lightgray}{rgb}{0.9,0.9,0.9}
+\definecolor{darkgreen}{rgb}{0.0,0.4,0.0}
+\usepackage{lastpage}
+\usepackage{fancyhdr}
+\setlength{\parindent}{0ex}
+\setlength{\parskip}{2ex}
+\setcounter{secnumdepth}{4}
+\setcounter{tocdepth}{4}
+\definecolor{darkgreen}{rgb}{0,0.5,0}
+\definecolor{darkblue}{rgb}{0,0,0.5}
+
+\pagestyle{fancy} %eigener Seitenstil
+\fancyhf{} %alle Kopf- und Fußzeilenfelder bereinigen
+\fancyhead[L]{Übungsblatt 1\\Priority Queues} %Kopfzeile links
+\fancyhead[C]{} %zentrierte Kopfzeile
+\fancyhead[R]{\textsc{Algorithmen \& Datenstrukturen 2}\\14. November 2011} %Kopfzeile rechts
+\renewcommand{\headrulewidth}{0.4pt} %obere Trennlinie
+\fancyfoot[C]{Seite \thepage\ von \pageref{LastPage}}
+\renewcommand{\footrulewidth}{0.4pt} %untere Trennlinie
+
+\newcommand{\spa}{\hspace*{4mm}}
+\newcommand{\defin}{\textcolor{darkgreen}{\textbf{Def.: }}}
+\newcommand{\rrfloor}{\right\rfloor}
+\newcommand{\llfloor}{\left\lfloor}
+
+
+\title{Übung Priority Queues -- AlgoDS2}
+\author{Thomas Battermann}
+\date{14. November 2011}
+
+\begin{document}
+\pagestyle{fancy}
+
+\section*{Aufgabe 1}
+\addcontentsline{toc}{section}{Aufgabe 1}
+
+%\includegraphics{bilder/aufgabe_1.eps}
+
+\begin{lstlisting}[frame=single]
+Merge(H1,H2):
+ head = NIL
+ l1 = head(H1)
+ l2 = head(H2)
+ while ( l1 != NIL and l2 != NIL ) do
+ if ( degree(l1) < degree(l2) ) then
+ y = l1
+ h = sibling(l1)
+ else
+ y = l2
+ l2 = sibling(l2)
+ if ( head == NIL ) then
+ head = y
+ x = y
+ if ( l1 != NIL ) then
+ if ( head == NIL ) then
+ head = l1
+ else
+ sibling(x) = h
+ if ( l2 != NIL ) then
+ if ( head != NIL ) then
+ head = l2
+ else
+ sibling(x) = l2
+ return head;
+\end{lstlisting}
+
+\newpage
+\section*{Aufgabe 2}
+\addcontentsline{toc}{section}{Aufgabe 2}
+
+\begin{lstlisting}[frame=single]
+ExtractMinimum(H):
+ x = head(H)
+ prev_x = NIL
+ min = x
+ prev_min = NIL
+ if ( x == NIL ) then
+ return NIL
+ prev_x = x
+ x = sibling(x)
+ while ( x != NIL ) do
+ if ( key(x) < key(min) ) then
+ min = x
+ prev_min = prev_x
+ prev_x = x
+ x = sibling(x)
+ if ( prev_min != NIL ) then
+ sibling(prev_min) = sibling(min)
+ else
+ head = sibling(min)
+ if ( child(min) != NIL ) then
+ H2 = InvertChildren(min)
+ Union(head,H2)
+ sibling(min) = NIL
+ parent(min) = NIL
+ child(min) = NIL
+ return min;
+\end{lstlisting}
+
+
+\section*{Aufgabe 3}
+\addcontentsline{toc}{section}{Aufgabe 3}
+
+Leider keine Lösung vorhanden :-(
+
+\end{document}
diff --git a/Algorithmen_und_Datenstrukturen_2/Uebung_2/Uebung_2.tex b/Algorithmen_und_Datenstrukturen_2/Uebung_2/Uebung_2.tex
new file mode 100644
index 0000000..9d88d7f
--- /dev/null
+++ b/Algorithmen_und_Datenstrukturen_2/Uebung_2/Uebung_2.tex
@@ -0,0 +1,159 @@
+\documentclass[11pt]{scrartcl}
+\usepackage[utf8]{inputenc}
+\usepackage[ngerman]{babel}
+\usepackage{ulem} % \sout
+\usepackage{marvosym} % \Lightning
+\usepackage{mathtools}
+\usepackage{multicol}
+\usepackage[paper=a4paper,left=30mm,right=20mm,top=20mm,bottom =25mm]{geometry}
+\usepackage[
+ pdftitle={Uebungsblatt 2: Hashing - AlgoDS2},
+ pdfsubject={Algorithmen und Datenstrukturen 2; Uebungsblatt 2: Hashing},
+ pdfauthor={Thomas Battermann},
+ pdfkeywords={Uebungsblatt 2: Hashing - AlgoDS2},
+ pdfborder={0 0 0}
+]{hyperref}
+\usepackage{tabularx}
+\usepackage{graphicx}
+\usepackage[usenames,dvipsnames]{color}
+\usepackage{lastpage}
+\usepackage{fancyhdr}
+\setlength{\parindent}{0ex}
+\setlength{\parskip}{2ex}
+
+\pagestyle{fancy} %eigener Seitenstil
+\fancyhf{} %alle Kopf- und Fußzeilenfelder bereinigen
+\fancyhead[L]{Übungsblatt 2\\Hashing}%Kopfzeile links
+\fancyhead[C]{} %zentrierte Kopfzeile
+\fancyhead[R]{\textsc{Algorithmen \& Datenstrukturen 2}\\1. Dezember 2011} %Kopfzeile rechts
+\renewcommand{\headrulewidth}{0.4pt} %obere Trennlinie
+\fancyfoot[C]{Seite \thepage\ von \pageref{LastPage}}
+\renewcommand{\footrulewidth}{0.4pt} %untere Trennlinie
+
+\title{Übung Hashing -- AlgoDS2}
+\author{Thomas Battermann}
+\date{1. Dezember 2011}
+
+\begin{document}
+\pagestyle{fancy}
+
+\section*{Aufgabe 1}
+\addcontentsline{toc}{section}{Aufgabe 1}
+
+ \subsection*{Mit der Formel}
+ \addcontentsline{toc}{subsection}{Mit der Formel}
+
+ \( m = 2^{12} = 4096; \quad A = \frac{\pi-2}{3} \)
+
+ \subsubsection*{a)}
+ \addcontentsline{toc}{subsubsection}{a)}
+
+ \begin{align*}
+ k &= 1000\\[4mm]
+ h &= \left( 1000 * \frac{\pi-2}{3} - \lfloor 1000 * \frac{\pi-2}{3} \rfloor \right) * 4096\\
+ &= 2174
+ \end{align*}
+
+ \subsubsection*{b)}
+ \addcontentsline{toc}{subsubsection}{b)}
+
+ \begin{align*}
+ k &= 2500\\[4mm]
+ h &= \left( 2500 * \frac{\pi-2}{3} - \lfloor 2500 * \frac{\pi-2}{3} \rfloor \right) * 4096\\
+ &= 1340
+ \end{align*}
+
+ \subsection*{Mit Ganzzahlarithmetik}
+ \addcontentsline{toc}{subsection}{Mit Ganzzahlarithmetik}
+
+ \textbf{Approximation}
+
+ \begin{align*}
+ &\left| \frac{S}{2^{16}} - A \right| &\text{ minimal}\\
+ &\frac{S}{2^{16}} - A = 0\\
+ &S = \lceil A * 2^{16} \rceil = 24939
+ \end{align*}
+
+ \subsubsection*{a)}
+ \addcontentsline{toc}{subsubsection}{a)}
+
+ \begin{align*}
+ k &= 1000\\[4mm]
+ r &= s * k \mod 2^{16}\\
+ &= 2493900 \mod 65536\\
+ &= 35320\\[4mm]
+ &\Rightarrow \underbracket{1000\,1001\,1111}\,1000\\
+ h(1000) &= 2207 = \lfloor \frac{35320}{2^4} \rfloor
+ \end{align*}
+
+ \subsubsection*{b)}
+ \addcontentsline{toc}{subsubsection}{b)}
+
+ \begin{align*}
+ k &= 2500\\[4mm]
+ r &= s * k \mod 2^{16}\\
+ &= 62347500 \mod 65536\\
+ &= 22764\\[4mm]
+ &\Rightarrow \underbracket{0101\,1000\,1110}\,1100\\
+ H(2500) &= 1422 = \lfloor \frac{22764}{2^4} \rfloor
+ \end{align*}
+
+
+\section*{Aufgabe 2}
+\addcontentsline{toc}{section}{Aufgabe 2}
+
+ \subsection*{a)}
+ \addcontentsline{toc}{subsection}{a)}
+
+ \includegraphics{bilder/aufgabe_2a.eps}
+
+ \subsection*{b)}
+ \addcontentsline{toc}{subsection}{b)}
+
+ \includegraphics{bilder/aufgabe_2b.eps}
+
+ \subsection*{c)}
+ \addcontentsline{toc}{subsection}{c)}
+
+ \includegraphics{bilder/aufgabe_2c.eps}
+
+ \begin{align*}
+ h(k,i) &= (k+\frac12i+\frac12i^2) \mod 16
+ \end{align*}
+
+ \subsection*{d)}
+ \addcontentsline{toc}{subsection}{d)}
+
+ \includegraphics{bilder/aufgabe_2d.eps}
+
+ \begin{align*}
+ h_1(k) &= k \mod 11\\
+ h_2(k) &= 1 + ( k \mod 10 )\\
+ h(k,i) &= \Big( h_1(k) + i * h_2(k) \Big) \mod 11
+ \end{align*}
+
+ \textbf{Sondierungsfrequenz für k=31}
+
+ \begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c}
+ $i$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\
+ \hline
+ $h(k,i)$ & 9 & 0 & 2 & 4 & 6 & 8 & 10 & 1 & 3 & 5 & 7
+ \end{tabular}
+
+ $<9,0,2,4,6,8,10,1,3,5,7>$
+
+\section*{Aufgabe 3}
+\addcontentsline{toc}{section}{Aufgabe 3}
+
+ Für alle $ 0 \le j < m $ gilt:\\
+ \[ \frac{n-j}{m-j} \le \frac n m \]
+
+ Annahme: Es gibt ein $j$ mit:\\
+ \begin{align*}
+ \frac{n-j}{m-j} &> \frac n m & | * (m-j) * m \\
+ (n-j)*m &> n * (m-j) \\
+ \sout{n*m} -j * m &> \sout{n*m} - j * m &| *\frac{1}{-j}\\[4mm]
+ m &< n & \text{\huge\color{Red}\Lightning}
+ \end{align*}
+
+\end{document}
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