update xD
authorlindenmannm
Tue, 22 Feb 2011 19:14:19 +0100
changeset 5675024f99bf0
parent 2 7b0f43733557
child 6 37644b2389bd
update xD
AT_Exzerpt.aux
AT_Exzerpt.bbl
AT_Exzerpt.blg
AT_Exzerpt.dvi
AT_Exzerpt.log
AT_Exzerpt.out.ps
AT_Exzerpt.tex
AT_Exzerpt.toc
code/AzyklischeNetzwerke.code
code/BelmanFord.code
code/dijkstra.code
code/editierdistanz.code
code/kmp.code
code/quicksort.code
code/spannbaeumeGreedy.code
code/spannbaeumeKruskal.code
code/spannbaeumePrim.code
code/spurgraph.code
comp.log
exzerpt.aux
exzerpt.bbl
exzerpt.blg
exzerpt.dvi
exzerpt.log
exzerpt.out.ps
exzerpt.tex
exzerpt.toc
img/dijkstraExT.eps
img/dijkstraExT.png
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     2.4 +\begin{thebibliography}{THC01}
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    2.12 +
    2.13 +\bibitem[PW02]{2}
    2.14 +\textsc{Peter~Widmayer}, Thomas~O.:
    2.15 +\newblock \emph{Algorithmen und Datenstrukturen}.
    2.16 +\newblock 4. Auflage.
    2.17 +\newblock Spektrum Akademischer Verlag, 2002. --
    2.18 +\newblock ISBN 978--3827410290
    2.19 +
    2.20 +\bibitem[THC01]{1}
    2.21 +\textsc{Thomas H.~Cormen}, Robert L. Rivest und Cliford~S. Charles
    2.22 +  E.~Leiserson~L. Charles E.~Leiserson:
    2.23 +\newblock \emph{Introduction to Algorithms}.
    2.24 +\newblock 2nd.
    2.25 +\newblock Prentice Hall India, 2001. --
    2.26 +\newblock ISBN 978--8120321410
    2.27 +
    2.28 +\end{thebibliography}
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   6.106 +/OUT pdfmark
   6.107 +[
   6.108 +/Title(Amortisierte Analyse \(TODO\))
   6.109 +/Count -3
   6.110 +/Action/GoTo/Dest(section.4)cvn
   6.111 +/OUT pdfmark
   6.112 +[
   6.113 +/Title(Binomial Heaps \(TODO\))
   6.114 +/Action/GoTo/Dest(subsection.4.1)cvn
   6.115 +/OUT pdfmark
   6.116 +[
   6.117 +/Title(Fibonacci Heaps \(TODO\))
   6.118 +/Action/GoTo/Dest(subsection.4.2)cvn
   6.119 +/OUT pdfmark
   6.120 +[
   6.121 +/Title(Union Find \(TODO\))
   6.122 +/Action/GoTo/Dest(subsection.4.3)cvn
   6.123 +/OUT pdfmark
   6.124 +[
   6.125 +/Title(Greedy)
   6.126 +/Count -2
   6.127 +/Action/GoTo/Dest(section.5)cvn
   6.128 +/OUT pdfmark
   6.129 +[
   6.130 +/Title(K\374rzeste \(billigste\) Wege)
   6.131 +/Count -1
   6.132 +/Action/GoTo/Dest(subsection.5.1)cvn
   6.133 +/OUT pdfmark
   6.134 +[
   6.135 +/Title(Single Source Shortest Paths)
   6.136 +/Action/GoTo/Dest(subsubsection.5.1.1)cvn
   6.137 +/OUT pdfmark
   6.138 +[
   6.139 +/Title(Spannb\344ume minimalen Gewichts)
   6.140 +/Count -6
   6.141 +/Action/GoTo/Dest(subsection.5.2)cvn
   6.142 +/OUT pdfmark
   6.143 +[
   6.144 +/Title(Das Wachsen von min. Spannb\344umen)
   6.145 +/Action/GoTo/Dest(subsubsection.5.2.1)cvn
   6.146 +/OUT pdfmark
   6.147 +[
   6.148 +/Title(Schnitte)
   6.149 +/Action/GoTo/Dest(subsubsection.5.2.2)cvn
   6.150 +/OUT pdfmark
   6.151 +[
   6.152 +/Title(Sichere Kanten)
   6.153 +/Action/GoTo/Dest(subsubsection.5.2.3)cvn
   6.154 +/OUT pdfmark
   6.155 +[
   6.156 +/Title(Der Graph GA)
   6.157 +/Action/GoTo/Dest(subsubsection.5.2.4)cvn
   6.158 +/OUT pdfmark
   6.159 +[
   6.160 +/Title(Algorithmus von Kruskal)
   6.161 +/Action/GoTo/Dest(subsubsection.5.2.5)cvn
   6.162 +/OUT pdfmark
   6.163 +[
   6.164 +/Title(Algorithmus von Prim)
   6.165 +/Action/GoTo/Dest(subsubsection.5.2.6)cvn
   6.166 +/OUT pdfmark
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   6.168 +/Title(Bin Packing)
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   6.172 +[
   6.173 +/Title(Online Verfahren)
   6.174 +/Count -3
   6.175 +/Action/GoTo/Dest(subsection.6.1)cvn
   6.176 +/OUT pdfmark
   6.177 +[
   6.178 +/Title(Next Fit \(NF\))
   6.179 +/Action/GoTo/Dest(subsubsection.6.1.1)cvn
   6.180 +/OUT pdfmark
   6.181 +[
   6.182 +/Title(First-Fit \(FF\))
   6.183 +/Action/GoTo/Dest(subsubsection.6.1.2)cvn
   6.184 +/OUT pdfmark
   6.185 +[
   6.186 +/Title(Best-Fit \(BF\))
   6.187 +/Action/GoTo/Dest(subsubsection.6.1.3)cvn
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   6.190 +/Title(Offline Verfahren)
   6.191 +/Count -2
   6.192 +/Action/GoTo/Dest(subsection.6.2)cvn
   6.193 +/OUT pdfmark
   6.194 +[
   6.195 +/Title(First Fit Decreasing \(FFD oder FFNI\))
   6.196 +/Action/GoTo/Dest(subsubsection.6.2.1)cvn
   6.197 +/OUT pdfmark
   6.198 +[
   6.199 +/Title(Best Fit Decreasing \(BFD\))
   6.200 +/Action/GoTo/Dest(subsubsection.6.2.2)cvn
   6.201 +/OUT pdfmark
   6.202 +[
   6.203 +/Title(Dynamische Programmierung)
   6.204 +/Count -3
   6.205 +/Action/GoTo/Dest(section.7)cvn
   6.206 +/OUT pdfmark
   6.207 +[
   6.208 +/Title(Matrixkettenprodukt \(TODO\))
   6.209 +/Action/GoTo/Dest(subsection.7.1)cvn
   6.210 +/OUT pdfmark
   6.211 +[
   6.212 +/Title(Optimale Suchb\344ume \(TODO\))
   6.213 +/Action/GoTo/Dest(subsection.7.2)cvn
   6.214 +/OUT pdfmark
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   6.216 +/Title(Editierdistanz)
   6.217 +/Action/GoTo/Dest(subsection.7.3)cvn
   6.218 +/OUT pdfmark
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   6.220 +/Title(Suche in Texten)
   6.221 +/Action/GoTo/Dest(section.8)cvn
   6.222 +/OUT pdfmark
   6.223 +[
   6.224 +/Title(Landau-Symbole)
   6.225 +/Action/GoTo/Dest(appendix.A)cvn
   6.226 +/OUT pdfmark
     7.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
     7.2 +++ b/AT_Exzerpt.tex	Tue Feb 22 19:14:19 2011 +0100
     7.3 @@ -0,0 +1,624 @@
     7.4 +\documentclass[a4paper,11pt,twoside]{article}
     7.5 +
     7.6 +\usepackage[ngerman]{babel} \usepackage[T1]{fontenc}
     7.7 +\usepackage[latin1]{inputenc}
     7.8 +\usepackage[colorlinks=true,linkcolor=black]{hyperref}
     7.9 +
    7.10 +\usepackage{vaucanson-g}
    7.11 +\usepackage{hyperref}
    7.12 +\usepackage{hypcap}
    7.13 +\usepackage{fancyhdr}
    7.14 +\usepackage{graphicx}
    7.15 +\usepackage{lastpage}
    7.16 +\usepackage[cm]{fullpage}
    7.17 +\usepackage{amsthm, amsmath, amsfonts}
    7.18 +\usepackage{float}
    7.19 +\usepackage{paralist}
    7.20 +\usepackage{listings}
    7.21 +\usepackage{color}
    7.22 +\usepackage{bookmark}
    7.23 +
    7.24 +\bibliographystyle{alphadin}
    7.25 +
    7.26 +\title{\underline{Exzerpt Algorithmentheorie WS10/11}}
    7.27 +\author{Markus Lindenmann}
    7.28 +\date{\today{} }
    7.29 +
    7.30 +\begin{document}
    7.31 +\maketitle
    7.32 +
    7.33 +\tableofcontents \newpage
    7.34 +
    7.35 +\section{Einleitung}\label{sec:intro}
    7.36 +Dozent:     Matthias Westermann\\
    7.37 +Website:    \url{http://lak.informatik.uni-freiburg.de}\\
    7.38 +Klausur:    09.03.2011\\
    7.39 +Hilfsmittel:    Ein auf beiden Seiten handbeschriebenes DIN-A4-Blatt
    7.40 +\subsection{Themen:}
    7.41 +\begin{itemize}
    7.42 +  \item[$\bullet$] Divide and Conquer
    7.43 +  \item[$\bullet$] Greedy Prinzip
    7.44 +  \item[$\bullet$] Dynamische Programmierung
    7.45 +  \item[$\bullet$] Randomisierung
    7.46 +  \item[$\bullet$] Amortisierte Analyse
    7.47 +\end{itemize}
    7.48 +\subsection{Problem-/Anwendungsbereiche:}
    7.49 +\begin{itemize}
    7.50 +  \item[$\bullet$] Geometrische Algorithmen
    7.51 +  \item[$\bullet$] Algebraische Algorithmen
    7.52 +  \item[$\bullet$] Graphenalgorithmen
    7.53 +  \item[$\bullet$] Datenstrukturen
    7.54 +  \item[$\bullet$] Internet-Algorithmen
    7.55 +  \item[$\bullet$] Optimierungs-Verfahren
    7.56 +  \item[$\bullet$] Zeichenkettenverarbeitung
    7.57 +\end{itemize}
    7.58 +\subsection{Literatur:}
    7.59 +\bibliography{literatur}
    7.60 +\nocite{*}
    7.61 +
    7.62 +\newpage
    7.63 +\section{Divide and Conquer}\label{sec:div&conq}
    7.64 +\begin{itemize}
    7.65 +  \item[$\bullet$] Quicksort
    7.66 +  \item[$\bullet$] Formulierung und Analyse des Prinzips
    7.67 +  \item[$\bullet$] Geometrisches Devide and Conquer
    7.68 +  \begin{itemize}
    7.69 +    \item[-] Closest-Pair
    7.70 +    \item[-] Segmentschnitt
    7.71 +  \end{itemize}
    7.72 +\end{itemize}
    7.73 +
    7.74 +\subsection{Formulierung des Divide and Conquer Prinzips}
    7.75 +D\&C Verfahren zur Lösung eines Problems der Größe $n$
    7.76 +\begin{itemize}
    7.77 +  \item[1.]   Divide\\$n>2$:  Teile das Problem in $k$ Teilprobleme der Größe $n_1,\ldots n_k$ ($k\geq2$)\\$n\leq c:$ Löse das Problem direkt
    7.78 +  \item[2.]   Conquer\\Löse die $k$ Teilprobleme auf dieselbe Art (rekursiv)
    7.79 +  \item[3.]   Merge\\Füge die berechneten Teillösungen zu einer Gesamtlösung zusammen
    7.80 +\end{itemize}
    7.81 +
    7.82 +\subsection{Quicksort}\label{ch:qs}
    7.83 +Wähle beliebiges Pivot Element $v$ (meist das letzte Element in der
    7.84 +Input-Folge!) und sortiere alle Elemente aus der Menge, die größer sind als $v$
    7.85 +rechts von $v$ ein, die anderen links. Verfahre im selben Muster mit den neuen
    7.86 +Teilmengen links und rechts des Pivot Elements. Wenn die zu sortierende Folge
    7.87 +$F$ nur ein Element beinhaltet, gebe $F$ zurück.
    7.88 +\subsubsection{Algorithmus}
    7.89 +\lstinputlisting[language=Java,label=lst:TTlookup,caption=Algorithmus Quicksort,
    7.90 +captionpos=b, numbers=left, numberstyle=\tiny, numbersep=5pt,
    7.91 +basicstyle=\footnotesize]{./code/quicksort.code}
    7.92 +
    7.93 +\subsubsection{Analyse}
    7.94 +$T(n)$ - maximale Anzahl von Schritten, um ein Problem der Größe $n$ zu lösen.
    7.95 +\[T(n)=\begin{cases}n\neq c&a \\ n>c&T(n_1+\ldots+T(n_k)+\text{Divide- und Mergeaufwand}\end{cases}\]
    7.96 +Best Case: $k=2, n_1=n_2=\frac{n}{2}$\\
    7.97 +Divide- und Mergeaufwand: $DM(n)$\\
    7.98 +$T(1)=a$\\
    7.99 +$T(n)=2T(\frac{n}{2})+DM(n)$
   7.100 +
   7.101 +\subsection{Nächste Paare}
   7.102 +Eingabe:    $n$ Punkte in Ebene\\
   7.103 +Ausgabe:    Punkt-Paar $q,r$ mit geringstem Abstand\\
   7.104 +Abstand:    $d(q,r)$ bechreibt Abstand zwischen Punkten $q$ und $r$
   7.105 +\subsubsection{Algorithmus}
   7.106 +\begin{itemize}
   7.107 +  \item[$\bullet$] sortiere Punktmenge nach x-Koordinate
   7.108 +  \item[$\bullet$] Teile Punktmenge in der Mitte L in die Mengen Q und R
   7.109 +  \item[$\bullet$] Löse rekursiv auf Q und R
   7.110 +  \item[$\bullet$] Sortiere Punkte nach y-Koordinate
   7.111 +  \item[$\bullet$] Teste alle Paare, deren Abstand in der
   7.112 +  Sortierung kleiner als 16 ist (siehe unten)
   7.113 +  \item[$\bullet$] Füge zusammen
   7.114 +\end{itemize}
   7.115 +
   7.116 +$\delta=$ Minimum der Teillösungen\\
   7.117 +Gibt es $q\in Q$ und $r\in R$ mit $d(q,r)<\delta$, dann sind sowohl $q$ als auch $r$ höchstens $\delta$ von $L$ (Mitte) entfernt.\\
   7.118 +\textbf{Problem:} alle Punkte können innerhalb der $\delta$-Zone liegen\\
   7.119 +\textbf{Lösung:}\begin{itemize}
   7.120 +\item[$\bullet$] Sortiere Punkte in $\delta$ Zone nach $y$-Abstand
   7.121 +\item[$\bullet$] Liegen in dieser Sortierung mehr als $c$ Punkte zwischen $q$ und $r$, dann kann $(q,r)$ nicht nächstes Paar sein
   7.122 +\end{itemize}
   7.123 +
   7.124 +Dies kann für $c=15$ bewiesen werden (vgl. Abbildung \ref{cl_pair}):
   7.125 +\begin{itemize}
   7.126 +  \item[$\bullet$] Seien min. 15 Punkte zwischen $q$ und $r$
   7.127 +  \item[$\bullet$] Dann sind mindestens 3 Reihen Quadrate zwischen $q,r$, weil in jedem Quadrat höchstens ein Punkt ist.
   7.128 +  \item[$\rightarrow$] Dann muss $d(q,r)$ mindestens $\frac{3}{2}\delta > \delta$ sein.
   7.129 +\end{itemize}
   7.130 +
   7.131 +\begin{figure}[H]
   7.132 +\centering
   7.133 +\includegraphics[scale=0.5]{img/cl_pair}
   7.134 +\caption{Nächste Paare: Prüf Distanz}
   7.135 +\label{cl_pair}
   7.136 +\end{figure}
   7.137 +
   7.138 +\subsubsection{Analyse}
   7.139 +\[T(n)=\begin{cases}n\leq3&a \\ n>3&2T(\frac{n}{2})+an\end{cases}\]
   7.140 +\[T(n)\leq an \log n \]
   7.141 +
   7.142 +\subsection{Segmentschnitt}
   7.143 +Eingabe:    Menge S bestehend aus vertikalen Segmenten und Endpunkten vn horizontalen Segmenten\\
   7.144 +Ausgabe:    Alle Schnittpunkte von vertikalen Segmenten mit horizontalen Segmenten, von denen mindesens ein Endpunkt in S ist
   7.145 +
   7.146 +\subsubsection{Algorithmus}
   7.147 +\textbf{Divide \& Conquer:}\\Fallunterscheidung
   7.148 +\begin{itemize}
   7.149 +  \item[1.] $\left| S\right| > 1$:\\ Teile S mittels einer vertikalen Gerade G in 2 gleichgroße Mengen $S_1$ (links von G) und $S_2$ (rechts von G). Führe rekursiv auf beiden Hälften aus.
   7.150 +  \item[2.] else: S enthält keine Schnitte
   7.151 +\end{itemize}
   7.152 +
   7.153 +\textbf{Merge}\\
   7.154 +Bilde L(S), R(S) und V(S) (Definition am Beispiel $i=1,2$ berechnet: $S=S_1\cup S_2$):
   7.155 +\begin{itemize}
   7.156 +  \item[$\bullet$] L(S): y-Koordinate aller linken Endpunkte in S, deren rechter Partner nicht in S\\$=L(S_2)\cup(L(S_1)-R(S_2))$
   7.157 +  \item[$\bullet$] R(S): y-Koordinate aller rechten Endpunkte in S, deren linker Partner nicht in S\\$=R(S_1)\cup(R(S_2)-L(S_1))$
   7.158 +  \item[$\bullet$] V(S): y-Intervalle der vertikalen Segmente in S\\$=V(S_1)\cup V(S_2)$
   7.159 +\end{itemize}
   7.160 +L,R sortiert nach steigender y-Koordinate\\
   7.161 +V sortiert nach steigendem unteren Endpunkt\\
   7.162 +\medskip
   7.163 +
   7.164 +\textbf{Basisfälle}\\
   7.165 +S enthält nur ein Element s. Fallunterscheidung:
   7.166 +\begin{itemize}
   7.167 +  \item[1.] $s=(x,y)$ ist ein linker Endpunkt:\\$L(S)=\{y\},\qquad R(S)=\emptyset,\qquad V(S)=\emptyset$
   7.168 +  \item[2.] $s=(x,y)$ ist ein rechter Endpunkt:\\$L(S)=\emptyset,\qquad R(S)=\{y\},\qquad V(S)=\emptyset$
   7.169 +  \item[3.] $s=(x,y_1,y_2)$ ist ein vertikales Segment:\\$L(S)=\emptyset,\qquad R(S)=\emptyset,\qquad V(S)=\{[y_1,y_2]\}$
   7.170 +\end{itemize}
   7.171 +
   7.172 +\subsubsection{Analyse}
   7.173 +Eingabe wird Anfangs sortiert und in Array gespeichert
   7.174 +\[T(n)=2T(\frac{n}{2}+an+\text{Größe der Ausgabe}\]
   7.175 +\[T(1)=O(1)\]
   7.176 +\[O(n \log n + k \mid k=\#Schnittpunkte\]
   7.177 +
   7.178 +\subsection{Polynomprodukt und Fast Fourier-Transformation (TODO)}
   7.179 +%TODO
   7.180 +
   7.181 +\newpage
   7.182 +\section{Randomisierung}
   7.183 +\begin{itemize}
   7.184 +  \item Klassen von randomisierten Algorithmen
   7.185 +  \begin{itemize}
   7.186 +    \item \textbf{Las Vegas}: \underline{immer} Korrekt, erwartete Laufzeit
   7.187 +    (randomisierter Qicksort)
   7.188 +    \item \textbf{Monte Carlo}: \underline{wahrscheinlich} (MC:mostly correct)
   7.189 +    Korrekt, garantierte Laufzeit (randomisierter Primzahltest)
   7.190 +  \end{itemize}
   7.191 +  \item Randomisierter Quicksort
   7.192 +  \item Randomisierter Primzahltest
   7.193 +  \item Kryptographie
   7.194 +\end{itemize}
   7.195 +\subsection{Randomisierter Quicksort}
   7.196 +  Refer to \ref{ch:qs} for basic description of the algorithm.\\
   7.197 +  Problem von Quicksort: bisher nimmt Quicksort immer das letzte Element in der
   7.198 +  Folge $\implies$ Worst-Case: sortierte Folge als Eingabe ($\Theta(n^2)$).
   7.199 +  Ziel: keine quadratische Laufzeit.\\
   7.200 +  Lösung: randomisierte Auswahl des Pivotelements an Stelle $i$. Anschließend
   7.201 +  tauschen von Position $i$ und $r$ (mit $r$=rechte Grenze des zu sortierenden
   7.202 +  Segments).
   7.203 +  \subsubsection{Analyse}
   7.204 +  \textbf{Analyse 1: erwartete Laufzeit}\\
   7.205 +  Mit Wahrscheinlichkeit $\frac{1}{n}$ wird das $i$-kleinste Element $S_i$
   7.206 +  gewählt wobei $n$ zu sortierenden Elemente vorhanden sind. Abhängig von der
   7.207 +  Wahl von $S_i$ ergeben sich Teilprobleme der Größe $i-1$ und $n-i$. Damit
   7.208 +  ergibt sich für die Laufzeit $T(n)$.
   7.209 +  \[T(n)=\frac{1}{n}(T(0)+T(n-1))+\frac{1}{n}(T(1)+T(n-2))+\ldots+\frac{1}{n}(T(k-1)+T(n-k))+\ldots+\]
   7.210 +  \[\frac{1}{n}(T(n-1)+T(0))+\Theta(n))\mid
   7.211 +  \Theta(n)\text{ für partition}\]
   7.212 +  \[=\frac{1}{n}\sum_{k=1}^n(T(k-1)+T(n-k))+\Theta(n)\]
   7.213 +  \[\frac{2}{n}\sum_{k=1}^nT(k-1)+\Theta(n)\]
   7.214 +  \[=O(n \log n)\]
   7.215 +  \textbf{Analys 2: Erwartete Anzahl der Vergleiche}\\
   7.216 +  $x_{ij}=
   7.217 +  \begin{cases}
   7.218 +  	1&\text{falls }S_i\text{ mit }S_j\text{ verglichen wird}\\
   7.219 +  	0&\text{sonst}
   7.220 +  \end{cases}$
   7.221 +  \[E\left[\sum_{i=1}^n\sum_{j>1}x_{ij}\right]=\sum_{i=1}^n\sum_{j>1}E[x_{ij}]\]
   7.222 +  $p_{ij} =$ Wahrscheinlichkeit $S_i$ wird mit $S_j$ verglichen
   7.223 +  \[E[x_{ij}=1\cdot p_{ij}+0\cdot(1-p_{ij}=p_{ij}\]
   7.224 +\subsection{Randomisierter Primzahltest (TODO)}
   7.225 +%TODO schnelle exponentiation
   7.226 +\subsection{RSA (TODO)}
   7.227 +%TODO
   7.228 +%erw. Euklid
   7.229 +\subsection{Treaps (TODO)}
   7.230 +%TODO haben wir nicht so im detail gemacht wie Elsässer!
   7.231 +\subsection{Hashing (TODO)}
   7.232 +%TODO
   7.233 +
   7.234 +\newpage
   7.235 +\section{Amortisierte Analyse (TODO)}
   7.236 +%TODO
   7.237 +\subsection{Binomial Heaps (TODO)}
   7.238 +%TODO
   7.239 +\subsection{Fibonacci Heaps (TODO)}
   7.240 +%TODO
   7.241 +\subsection{Union Find (TODO)}
   7.242 +%TODO
   7.243 +
   7.244 +\newpage
   7.245 +\section{Greedy}
   7.246 +Momentan beste Teillösung wählen. Die Summe dieser optimalen Teillösung ergibt
   7.247 +die gesamt Lösung. Möglicher Zustand der Gesamtlösung:
   7.248 +\begin{itemize}
   7.249 +  \item Wir erhalten die optimale Gesamtlösung.
   7.250 +  \item Wir erhalten eine Lösung, die zwar nicht immer optimal ist, aber vom
   7.251 +  Optimum stets nur wenig abweicht.
   7.252 +  \item Die berechnete Lösung kann beliebig schlecht werden.
   7.253 +\end{itemize}
   7.254 +\underline{Beispiele:}
   7.255 +\begin{itemize}
   7.256 +  \item Münzwechselproblem\\\textbf{Ziel:} Bezahlung eines Betrages in möglichst
   7.257 +  wenig Münzen und Banknoten.\\\textbf{Lösung (Greedy):} Wähle die maximale
   7.258 +  Anzahl von Banknoten und Münzen mit jeweils größtmöglichem Wert, bis der
   7.259 +  gewünschte Betrag erreicht ist. (Dies führt nur für bestimmte Werte der
   7.260 +  Zahlungsmittel zur optimalen Gesamtlösung!)
   7.261 +  \item Handlungsreisenden-Problem (Traveling Sales Man Problem,
   7.262 +  TSP)\\\textbf{Ziel:} Finden der billigsten Rundreise, diealle Orte genau
   7.263 +  einmal besucht.\\\textbf{Lösung (Greedy):} Gehe immer den günstigsten Weg vom
   7.264 +  aktuellen Knoten zu einem noch nicht besuchten Knoten. Wenn alle besucht
   7.265 +  wurden, kehre zum Anfang zurück. (Mit Greedy keine optimale Gesamtlösung!)
   7.266 +  \item Aktivitäten Auswahlproblem\\\textbf{Ziel:} möglichst viele Anfragen
   7.267 +  erfüllen. (Muss nicht bedeuten, dass die Ressource optimal
   7.268 +  genutzt wird!)\\\textbf{Lösung (Greedy):} Nimm die Anfrage, die am frühesten
   7.269 +  fertig ist. (optimale Gesamtlösung!)\\\textbf{Analyse:} $\Theta(n)$
   7.270 +\end{itemize}
   7.271 +
   7.272 +\subsection{Kürzeste (billigste) Wege}
   7.273 +Pfad $P=$ Folge von Knoten.
   7.274 +Kosten eines Pfades:\[c(P)=\sum_{i=0}^{P.length-1}c(v_i,v_{i+1}\]
   7.275 +Entfernung von $v$ nach $w$ sind die Kosten des
   7.276 +günstigsten Pfades:\[dist(v,w)=inf{c(P)\mid P\text{= Weg von v nach
   7.277 +w}}\]
   7.278 +Kosten eines unerreichbaren Knoten sind definiert als $\infty$. Kosten eines
   7.279 +Pfades in einem negativen Zyklus: $-\infty$. Es können also nur Graphen ohne
   7.280 +negative Zyklen sinnvoll betrachtet werden!
   7.281 +\subsubsection{Single Source Shortest Paths}
   7.282 +Kürzeste/Günstigste Pfade von einem Knoten $s$ zu allen anderen Knoten. Hierfür
   7.283 +nutzt man die Dreiecksungleichung über $dist(u,v)\mid (u,v)\in E$ und $s\in V$:
   7.284 +\[dist(s,v)\leq dist(s,u)+dist(u,v)\]
   7.285 +\textbf{Greedy-Ansatz}
   7.286 +\begin{itemize}
   7.287 +  \item Überschätze die dist-Funktion \[dist(s,v)=\begin{cases}0&\text{ falls }
   7.288 +  v=s\\\infty&\text{ falls }v\neq s\end{cases}\]
   7.289 +  \item Solange eine Kante $e=(u,v)$ existiert mit
   7.290 +  $dist(s,v)>dist(s,u)+c(u,v)$ setze $dist(s,v)=dist(s,u)+c(u,v)$
   7.291 +\end{itemize}
   7.292 +Durch die Überschätzung kann die Dreicksungleichung nur noch für Knoten $s$ und
   7.293 +seine direkten Nachfolger verletzt werden!\\Alle folgenden Verfahren basieren
   7.294 +auf diesem Prinzip!\\
   7.295 +%Elsässer führt einige Lemmas ein und beweist sie - diese beziehen sich auf den
   7.296 +% ersten optimimierten Algorithmus und diesen erachte ich in Relation zu
   7.297 +% Dijkstra als irrelevant!
   7.298 +\textbf{Effiziente Implementierungen}
   7.299 +Diese sind im Allgemeinen nicht bekannt, jedoch für wichtige Spezialfälle.
   7.300 +\begin{itemize}
   7.301 +  \item Nicht-negative Netzwerke: \underline{Dijkstra}
   7.302 +  \item Netzwerke ohne negative Zyklen: \underline{Bellman-Ford}
   7.303 +  \item Azyklische Netzwerke
   7.304 +\end{itemize}
   7.305 +\underline{Dijkstra}
   7.306 +\lstinputlisting[language=Java,label=lst:TTlookup,caption=Algorithmus von
   7.307 +Dijkstra, captionpos=b, numbers=left, numberstyle=\tiny, numbersep=5pt,
   7.308 +basicstyle=\footnotesize]{./code/dijkstra.code}
   7.309 +Dijkstra Beispiel:
   7.310 +\begin{figure}
   7.311 +	\centering
   7.312 +	\includegraphics[scale=0.6]{img/dijkstraExT}
   7.313 +	\caption{Tree for Dijkstra Example}
   7.314 +	\label{fig:dijExT}
   7.315 +\end{figure}
   7.316 +\begin{table}[ht]
   7.317 +    \centering
   7.318 +    \begin{tabular}{|c c c c c c c|c|}
   7.319 +        \hline
   7.320 +        \multicolumn{7}{|c|}{DIST[v]} & \\
   7.321 +        \hline
   7.322 +        s & a & b & c & d & e & f & U\\\hline\hline
   7.323 +		0 & $\infty$ & $\infty$ & $\infty$ & $\infty$ & $\infty$ & $\infty$ & s,a,b,c,d,e,f\\
   7.324 +		0* & $\infty$ & $\infty$ & $\infty$ & $\infty$ & $\infty$ & $\infty$ & a,b,c,d,e,f\\
   7.325 +		0* & 6 & 7 & 5* & $\infty$ & $\infty$ & $\infty$ & a,b,d,e,f\\
   7.326 +		0* & 6* & 7 & 5* & $\infty$ & $\infty$ & 14 & b,d,e,f\\
   7.327 +		0* & 6* & 7* & 5* & 9 & 10 & 14 & d,e,f\\
   7.328 +		0* & 6* & 7* & 5* & 9* & 10 & 11 & e,f\\
   7.329 +		0* & 6* & 7* & 5* & 9* & 10* & 11 & f\\
   7.330 +		0* & 6* & 7* & 5* & 9* & 10* & 11* & $\emptyset$\\\hline
   7.331 +    \end{tabular}
   7.332 +    \caption{Dijkstra Example}\label{tab:dijEx}
   7.333 +\end{table}
   7.334 +Komplexität:
   7.335 +\[O(n\cdot(T_{insert}+T_{Empty}+T_{DeleteMin})+m\cdot T_{DecreaseKey}+m+n)\]
   7.336 +Implementierung mit Fibonaci-Heaps:\\
   7.337 +$T_{Insert}:O(1)$\\
   7.338 +$T_{DeleteMin}:O(\log n)$ (amortisiert)\\
   7.339 +$T_{DecreaseKey}:O(1)$ (amortisiert)
   7.340 +\[O(n\log n+m)\]
   7.341 +\underline{Bellman-Ford}
   7.342 +\lstinputlisting[language=Java,label=lst:TTlookup,caption=Algorithmus von
   7.343 +Belman-Ford, captionpos=b, numbers=left, numberstyle=\tiny, numbersep=5pt,
   7.344 +basicstyle=\footnotesize]{./code/BelmanFord.code}
   7.345 +\underline{Azyklische Netzwerke}
   7.346 +\lstinputlisting[language=Java,label=lst:TTlookup,caption=Algorithmus für
   7.347 +Azyklische Netzwerke, captionpos=b, numbers=left, numberstyle=\tiny,
   7.348 +numbersep=5pt, basicstyle=\footnotesize]{./code/AzyklischeNetzwerke.code}
   7.349 +\subsection{Spannbäume minimalen Gewichts}
   7.350 +Ungerichteter Graph $G=(V,E)$, Kostenfunktion $c:E\rightarrow R$.\\Sei
   7.351 +$T\subseteq E$ ein Baum (zusammenhängender azyklischer Teilgraph) Kosten von T:
   7.352 +$c(T)=\sum_{(u,v)\in T}c(u,v)$.\\Ein minimaler Spannbaum ist ein Baum
   7.353 +$T\subseteq E$, der alle Knoten in $V$ verbindet minimale Kosten hat.
   7.354 +\subsubsection{Das Wachsen von min. Spannbäumen}
   7.355 +\textbf{Invariante:} Verwalte Menge $A\subseteq E$, die Teilmenge eines
   7.356 +minimalen Spannbaums ist.\\
   7.357 +\textbf{Definition:} Eine Kante $(u,v)\in E\backslash A$ ist sicher für $A$, wenn
   7.358 +$A\cup {(u,v)}$ Teilmenge eines minimalen Spannbaums ist.
   7.359 +\lstinputlisting[language=Java,label=lst:TTlookup,caption=Greedy Algorithmus für
   7.360 +Spannbäume, captionpos=b, numbers=left, numberstyle=\tiny,
   7.361 +numbersep=5pt, basicstyle=\footnotesize]{./code/spannbaeumeGreedy.code}
   7.362 +\subsubsection{Schnitte}
   7.363 +Ein Schnitt $(S,V\backslash S)$ ist eine Partition von $V$. Eine Kante $(u,v)$
   7.364 +schneidet $(S,V\backslash S)$, wenn ein Endpunkt in $S$ und der andere in
   7.365 +$V\backslash S$ liegt.\\Ein Schnitt "`respektiert $A$"', wenn keine Kante aus
   7.366 +$A$ den Schnitt schneidet.\\Eine Kante heißt minimal bzgl. eines Schnitts, wenn
   7.367 +sie den Schnitt schneidet und unter allen solchen Kanten minimale Kosten hat.
   7.368 +\subsubsection{Sichere Kanten}
   7.369 +\textbf{Satz:} Sei $A$ Teilmenge eines minimalen Spannbaums $T$ und
   7.370 +$(S,V\backslash S)$ ein Schnitt, der $A$ respektiert. Ist $(u,v)$ eine minimale
   7.371 +Kante bzgl. $(S,V\S)$, dann ist $(u,v)$ sicher für $A$.\\
   7.372 +\textbf{Beweis}: \begin{itemize}
   7.373 +  \item[1. Fall] $(u,v)\in T$ : ok
   7.374 +  \item[2. Fall] $(u,v)\not\in T$ : Wir konstruieren minimalen Spannbaum $T'$
   7.375 +  mit $(u,v)\in T'$ und $A\subseteq T'$.
   7.376 +\end{itemize}
   7.377 +\subsubsection{Der Graph $G_A$}
   7.378 +\[G_A=(V,A)\]
   7.379 +\begin{itemize}
   7.380 +  \item ist ein \underline{Wald}, d.h. eine Menge von Bäumen
   7.381 +  \item anfangs, wenn $A=\emptyset$, ist jeder Baum ein einzelner Knoten
   7.382 +  \item eine sichere Kante verbindet verschiedene Bäume
   7.383 +\end{itemize}
   7.384 +\textbf{Korollar:} Sei $B$ ein Baum in $G_A=(V,A)$. Ist $(u,v)$ eine Kante
   7.385 +minimaler Kosten, die $B$ und einen anderen Baum in $G_A$ verbindet, dann ist
   7.386 +$(u,v)$ sicher für A.
   7.387 +\subsubsection{Algorithmus von Kruskal}
   7.388 +Wähle stets eine Kante minimaler Kosten, die zwei Bäume $B_1$ und $B_2$ in $G_A$
   7.389 +verbindet.
   7.390 +\lstinputlisting[language=Java,label=lst:TTlookup,caption=Algorithmus von
   7.391 +Kruskal, captionpos=b, numbers=left, numberstyle=\tiny, numbersep=5pt,
   7.392 +basicstyle=\footnotesize]{./code/spannbaeumeKruskal.code}
   7.393 +Laufzeit: $O(m\alpha(n)+m+m\log n)$
   7.394 +\subsubsection{Algorithmus von Prim}
   7.395 +$A$ ist immer ein einziger Baum. Starte mit einem beliebigen Wurzelknoten $w$.
   7.396 +Füge in jedem Schritt eine minimale Kante hinzu, die einen Knoten in $A$ mit
   7.397 +einem Knoten in $V\backslash A$ verbindet.
   7.398 +\textbf{Implementierung}\\$Q$: Prioritätenwarteschlange, die alle Knoten $v\in
   7.399 +V\backslash A$ enthält.\\Schlüssel von $v$: minimale Kosten einer Kante, die $v$
   7.400 +mit einem Knoten aus $A$ verbindet.\\Für einen Knoten $v$ ist $p[v]$ der
   7.401 +Elter-Knoten von $v$ in dem Baum.\[A={(v,p[v]):v\in V-{w}-Q}\]
   7.402 +\lstinputlisting[language=Java,label=lst:TTlookup,caption=Algorithmus von
   7.403 +Prim, captionpos=b, numbers=left, numberstyle=\tiny, numbersep=5pt,
   7.404 +basicstyle=\footnotesize]{./code/spannbaeumePrim.code}
   7.405 +Laufzeit: $O(n\log n+m)$
   7.406 +
   7.407 +\newpage
   7.408 +\section{Bin Packing}
   7.409 +\textbf{Problembeschreibung:}\\
   7.410 +Man hat $n$ Objekte verschiedener Größe mit $0<s_i\leq1\mid 1\leq i\leq n$. Gesucht ist die kleinst mögliche Anzahl an Kisten (Bins) der Größe 1, mit der alle Objekte verpackt werden können.
   7.411 +\subsection{Online Verfahren}
   7.412 +Jedes (ankommende) Objekt muss verpackt sein, bevor das nächste
   7.413 +Objekt betrachtet wird. Ein Objekt verbleibt in derjenigen Kiste,
   7.414 +in die es zuerst gepackt wird.\\ Kein Online Bin Packing Verfahren
   7.415 +kann stets eine optimale Lösung finden!\\ \textbf{Satz 1}: Es gibt
   7.416 +Eingaben, die jeden Online Bin Packing Algorithmus zwingen,
   7.417 +wenigstens $\frac{4}{3} OPT$ Bins zu verwenden, wobei $OPT$ die
   7.418 +minimal mögliche Binzahl ist.\\ \textbf{Beweis} Annahme: Online Bin
   7.419 +Packing Algorithmus benötigt stets weniger als $\frac{4}{3}OPT$
   7.420 +Bins.\\ Es wird eine Eingabe mit $2m$ Objekten angenommen. Die
   7.421 +ersten $m$ sind um $\epsilon$ kleiner als $\frac{1}{2}$, die
   7.422 +zweiten $m$ um $\epsilon$ größer. Die optimale Lösung sind $m$
   7.423 +Kisten, da immer ein großes mit einem kleinen Paket kombiniert
   7.424 +werden kann. Nun teilt man die Analyse in zwei Zeitpunkte: nach $m$
   7.425 +Objekten und nach $2m$ Objekten. Laut Annahme gilt nach nach $m$
   7.426 +Objekten, dass die Anzahl der Bins $b$ nur maximal $\frac{4}{3}
   7.427 +OPT$ groß sein darf. Mit $OPT=\frac{m}{2}$ gilt somit:
   7.428 +$b=\frac{4}{3}\cdot\frac{m}{2}=\frac{2}{3}m$. Sei $b=b_1+b_2\mid
   7.429 +b_1=\#$Bins mit einem Objekt, $b_2=\#$Bins mit zwei Objekten, so
   7.430 +folgt $b_1+2b_2=m\implies b_1=m-2b_2\implies b=b_1+b_2=m-b_2$. Nach
   7.431 +$2m$ Objekten ist $OPT=m$, wie zuvor beschrieben. $b_1$ Bins können
   7.432 +zu diesem Zeitpunkt mit Objekten des zweiten Bereichs gefüllt
   7.433 +werden, für die verbleibenden muss man neue Bins öffnen. Die Anzahl
   7.434 +der neu zu öffnenden beträgt damit genau $b_2\implies \#$Bins $\geq
   7.435 +b+m-b_1=m+b_2$. Die Annahme besagt $m+b_2\leq\#$Bins $<\frac{4}{3}m
   7.436 +\wedge b_2<\frac{m}{3}$. Dies ist nur möglich, wenn
   7.437 +$b=m-b_2>\frac{2}{3}m$ - Dies führt zu einem
   7.438 +\underline{Widerspruch}, da $b$ zum Zeitpunkt 1, nach $m$ Objekten
   7.439 +$<\frac{2}{3}m$ sein muss.
   7.440 +
   7.441 +\subsubsection{Next Fit (NF)}
   7.442 +Verpacke das nächste Objekt in dieselbe Kiste wie das
   7.443 +vorherige, wenn es dort noch hineinpasst, sonst öffne eine neue
   7.444 +Kiste und verpacke es dort.\\ \textbf{Satz 2}: (a) Für alle
   7.445 +Inputfolgen $I$ gilt: $NF(I)\leq2OPT(I)$. (b) Es gibt
   7.446 +Inputfolgen I mit $NF(I)\geq2OPT(I)-2$.\\ \textbf{Beweis}\\ (a)
   7.447 +Betrachte zwei beliebige aufeinanderfolgende Kisten $B_{2k-1},
   7.448 +B_{2k}$. Man weiß, dass die beinhalteten Objekte eine
   7.449 +Summengröße $>1$ haben, sonst wären selbige in einer Kiste.
   7.450 +Daraus folgt, man kann nicht mehr als die doppelte Anzahl an
   7.451 +Kisten als beim Optimum erreichen!\\ (b) Inputfolge: $0.5,
   7.452 +\frac{2}{n}, 0.5, \frac{2}{n}, 0.5, \frac{2}{n}, \ldots 0.5,
   7.453 +\frac{2}{n}$.\\Next-Fit liefert nun immer Bins gefüllt mit $0.5
   7.454 ++ \frac{2}{n}$. Darausfolgt $NF(I)=\frac{n}{2}$ und
   7.455 +$OPT=\frac{n}{4}+1$ (weil $\frac{n}{2}$ mal 0.5er Objekte in
   7.456 +$=\frac{n}{4}$ Kisten passen und $\frac{n}{2}$ $\frac{2}{n}$er
   7.457 +Objekte in eine Kiste passen)\\ Laufzeit für Inputfolgen der
   7.458 +Länge $n$: NF $O(n)$
   7.459 +\subsubsection{First-Fit (FF)}
   7.460 +Packe nächstes Objekt in die erste Kiste, in die es noch
   7.461 +hineinpasst, wenn es eine solche Kiste gibt, sonst in eine neue
   7.462 +Kiste.\\ \textbf{Beobachtung}: Zu jedem Zeitpunkt kann es
   7.463 +höchstens eine Kiste geben, die weniger als halb voll ist.
   7.464 +($\implies FF(I) \leq 2OPT(I)$)\\ \textbf{Satz 3}: (a) Für alle
   7.465 +Inputfolgen $I$ gilt $FF(I)\leq \lceil\frac{17}{10}
   7.466 +OPT(I)\rceil$. (b) Es gibt Inputfolgen $I$ mit
   7.467 +$FF(I)\geq\frac{17}{10}(OPT(I)-1)$. (b') Es gibt Inputfolgen
   7.468 +$I$ mit $FF(I)=\frac{10}{6}OPT(I)$.\\ \textbf{Beweis}\\ (b')
   7.469 +Inputfolge der Länge $3\cdot6m$. Dabei je $6m$ Objekte der
   7.470 +Größen $\frac{1}{7}+\epsilon, \frac{1}{3}+\epsilon,
   7.471 +\frac{1}{2}+\epsilon$, welche in gegebener Reihenfolge in
   7.472 +Gruppen eingegeben werden. FF liefert hierbei $m$ Kisten
   7.473 +gefüllt mit den Objekten der Größe $\frac{1}{7}+\epsilon$
   7.474 +(jeweils 6 pro Kiste) gefolgt von $\frac{6m}{2}=3m$ Kisten
   7.475 +gefüllt mit Objekten der Größe $\frac{1}{3}+\epsilon$ (jeweils
   7.476 +2 pro Kiste) und schlussendlich $6m$ Kisten mit je einem Objekt
   7.477 +der Größe $\frac{1}{2}+\epsilon$. Das macht in der Summe $10m$
   7.478 +Kisten. Die optimale Lösung packt je ein Objekt jeder Größe in
   7.479 +eine Kiste und benötigt somit nur $6m$ Kisten.\\ Laufzeit für
   7.480 +Inputfolgen der Länge $n$: FF $O(n^2) \rightarrow O(n \log n)$
   7.481 +\subsubsection{Best-Fit (BF)}
   7.482 +Verpacke das nächste Objekt in diejenige Kiste, in die es am
   7.483 +besten passt (d.h. den geringsten Platz ungenutzt lässt).
   7.484 +Verhalten von BF ähnlich zu FF. Laufzeit für Inputfolgen der
   7.485 +Länge $n$: BF $O(n^2) \rightarrow O(n \log n)$
   7.486 +\subsection{Offline Verfahren}
   7.487 +\underline{NP-schwieriges Problem! Entscheidungsproblem ist NP-vollständig!}\\
   7.488 +Zunächst wird die Anzahl $n$ und alle $n$ Objekte vorgegeben. Dann
   7.489 +beginnt die Verpackung.\\ $n$ und $s_1, ..., s_n$ sind gegeben,
   7.490 +bevor die Verpackung beginnt!\\ $\rightarrow$ \underline{Optimale
   7.491 +Verpackung} kann durch erschöpfende Suche bestimmt werden. ABER:
   7.492 +Benötigt exponentielle Zeit! Daher Approximimationsalgorithmen, die
   7.493 +schneller sind aber dafür unter Umständen nicht optimal!\\
   7.494 +Idee für Offline Approximationsalgorithmen: Sortiere die Objekte
   7.495 +zunächst nach abnhemender Größe und verpacke größere Objekte zuerst!
   7.496 +\subsubsection{First Fit Decreasing (FFD oder FFNI)}
   7.497 +\textbf{Lemma 1}: Sei $I$ eine Folge von $n$ Objekten mit
   7.498 +Größen $s_1\geq s_2\geq\ldots\geq s_n$ und sei $m=OPT(I)$. Dann
   7.499 +haben alle von FFD in den Bins
   7.500 +$B_{m+1},B_{m+2},\ldots,B_{FFD(I)}$ verpackten Objekte eine
   7.501 +Größe von höchstens $\frac{1}{3}$.
   7.502 +\\ \textbf{Beweis}:
   7.503 +%TODO:Elsässer hat das in seiner VL gemacht - aber hässlich lang
   7.504 +\textbf{Lemma 2}: Sei $I$ eine Folge von $n$ Objekten mit
   7.505 +Größen $s_1\geq s_2\geq\ldots\geq s_n$ und sei $m=OPT(I)$. Dann
   7.506 +ist die Anzahl der Objekte, die FFD in die Kisten
   7.507 +$B_{m+1},B_{m+2},\ldots,B_{FFD(I)}$ verpackt, höchstens
   7.508 +$m-1$.\\ \textbf{Beweis}: Annahme: Es gibt mehr als $m-1$
   7.509 +Objekte. $x_1,\ldots,x_m$ in $I$, die FFD in extra Kisten
   7.510 +verpacken. Dies führt zu einem Widerspruch.
   7.511 +%TODO:Warum?
   7.512 +\textbf{Satz}: Für alle Inputfolgen I
   7.513 +gilt $FFD(I)\leq(4 OPT(I)+\frac{1}{3})$.\\ \textbf{Satz}:
   7.514 +\begin{itemize}
   7.515 +  \item[1.] Für alle Inputfolgen $I$ gilt: $FFD(I)\leq \frac{11}{9} OPT(I)+4$
   7.516 +  \item[2.] Es gibt Inputfolgen $I$ mit: $FFD(I)=\frac{11}{9}OPT(I)$.
   7.517 +\end{itemize}
   7.518 +\textbf{Beweis}: %TODO
   7.519 +\subsubsection{Best Fit Decreasing (BFD)}
   7.520 +%TODO Nicht im Script???
   7.521 +
   7.522 +\newpage
   7.523 +\section{Dynamische Programmierung}
   7.524 +\textbf{Idee:} Speichern von zwischen Ergebnissen, welche sonst mehrfach
   7.525 +berechnet werden müssten.\\
   7.526 +\textbf{Vorgehen:}
   7.527 +\begin{itemize}
   7.528 +  \item Rekursive Beschreibung des Problems P
   7.529 +  \item Bestimmung einer Menge T, die alle Teilprobleme von P enthält, auf die
   7.530 +  bei der Lösung von P zugegriffen wird.
   7.531 +  \item Bestimmung einer Reihenfolge $T_0,\ldots T_k$ der Probleme in T, so dass
   7.532 +  bei der Lösung von $T_i$ nur auf Probleme $T_j$ mit $j<i$ zurückgegriffen
   7.533 +  wird.
   7.534 +  \item Sukzessive Berechnung und Speicherung von Lösungen für $T_0,\ldots T_k$.
   7.535 +\end{itemize}
   7.536 +Wird im Script anhand der Fibonacci Zahlen demonstriert aber ist trivial \ldots
   7.537 +\subsection{Matrixkettenprodukt (TODO)}
   7.538 +%TODO
   7.539 +\subsection{Optimale Suchbäume (TODO)}
   7.540 +%TODO
   7.541 +\subsection{Editierdistanz}
   7.542 +Berechne für zwei gegebene Zeichenfolgen $A$ und $B$ möglichst effizient die
   7.543 +Editier-Distanz $D(A,B)$ und eine minimale Folge von Editieroperationen, die
   7.544 +$A$ in $B$ überführt.\\
   7.545 +\textbf{Editieroperationen:}
   7.546 +\begin{itemize}
   7.547 +  \item Ersetzen eines Zeichens von $A$ durch ein Zeichen von $B$
   7.548 +  \item Löschen eines Zeichens von $A$
   7.549 +  \item Einfügen eines Zeichens von $B$
   7.550 +\end{itemize}
   7.551 +\textbf{Einheitskostenmodell:}
   7.552 +\[c(a,b)=\begin{cases}1 & \text{falls } a\neq b\\0 & \text{falls } a=b\end{cases}\]
   7.553 +$a=\epsilon, b=\epsilon$ möglich Dreiecksungleichung soll für $c$ im
   7.554 +allgemeinen gelten:
   7.555 +\[c(a,c)\leq c(a,b)+c(b,c)\]
   7.556 +$\rightarrow$ ein Buchstabe wird höchstens einmal
   7.557 +verändert.\\
   7.558 +Rekursionsgleichung, falls $m,n\geq 1$ 
   7.559 +\[D_{m,n}=min\left.\begin{cases}D_{m-1,n-1}+c(a_m,b_n)\\D_{m-1,n}+1\\D_{m,n-1}+1\end{cases}\right\}\]
   7.560 +Anfangsbedingung:\\
   7.561 +\[D_{0,0}=D(\epsilon,\epsilon)=0\]
   7.562 +\[D_{0,j}=D(\epsilon,B_j)=j\]
   7.563 +\[D_{i,0}=D(A_i,\epsilon)=i\]
   7.564 +\lstinputlisting[language=Java,label=lst:TTlookup,caption=Algorithmus zur
   7.565 +Editierdistanz, captionpos=b, numbers=left, numberstyle=\tiny, numbersep=5pt,
   7.566 +basicstyle=\footnotesize]{./code/editierdistanz.code}
   7.567 +\textbf{Spurgraph:}\\
   7.568 +Übersicht über alle möglichen Spuren zur Transformation von $A$ in $B$,
   7.569 +gerichtete Kanten von Knoten $(i,j)$ zu $(i+1,j)$, $(i,j+1)$ und $(i+1,j+1)$.
   7.570 +Gewichtung der Kanten entspricht den Editierkosten. Kosten nehmen entlang eines
   7.571 +optimalen Weges monoton zu. Jeder Weg mit monotin wachsenden Kosten von der
   7.572 +linken oberen Ecke zur rechten unteren Ecke entspricht einer optimalen Spur. Zur
   7.573 +Berechnung der Spur der minimalen Editieroperationen (es gibt mehrere
   7.574 +Lösungen), kann folgender Algorithmus verwendet werden:
   7.575 +\lstinputlisting[language=Java,label=lst:TTlookup,caption=Algorithmus zum Spurgraphen, captionpos=b, numbers=left, numberstyle=\tiny, numbersep=5pt, basicstyle=\footnotesize]{./code/spurgraph.code}
   7.576 +
   7.577 +\section{Suche in Texten}
   7.578 +\underline{Gegeben:} Text T $t_1t_2t_3\ldots t_n\in\Sigma^n$ und
   7.579 +rin Muster P $p_1p_2p_3\ldots p_m\in\Sigma^m$\\
   7.580 +\underline{Gesucht:} Ein oder alle Vorkommen des Musters im Text, d.h.
   7.581 +Verschiebungen $i$ mit $0\leq i\leq n-m$.\\
   7.582 +\textbf{Naives Verfahren:} iteriere über den Text und suche das Muster. Aufwand:
   7.583 +$m\cdot(n-m+1)$ Vergleiche $\implies \Omega(n\cdot m)$
   7.584 +\textbf{Verfahren nach Knuth-Morris-Pratt (KMP):} 
   7.585 +Präffix=Suffix Verfahren. Daraus resultiert eine Shiftmöglichkeit des Musters
   7.586 +über den Text. Der Shift geschieht um $next[j]$=Länge des längsten Präfix von
   7.587 +$P$, das echtes Suffix von $P_{1\ldots j}$ ist, wobei $j=$Position des letzten
   7.588 +Vergleiches.\\
   7.589 +\underline{Erstellung des $next[j]$ arrays über P.}\\
   7.590 +$P=0101101011$
   7.591 +\begin{table}[H]
   7.592 +	\centering
   7.593 +	\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
   7.594 +		1&2&3&4&5&6&7&8&9&10\\\hline
   7.595 +		0&1&0&1&1&0&1&0&1&1\\\hline
   7.596 +		 & &0& & & & & & & \\
   7.597 +		 & &0&1& & & & & & \\
   7.598 +		 & & & & &0& & & & \\
   7.599 +		 & & & & &0&1& & & \\
   7.600 +		 & & & & &0&1&0& & \\
   7.601 +		 & & & & &0&1&0&1& \\
   7.602 +		 & & & & &0&1&0&1&1\\
   7.603 +	\end{tabular}
   7.604 +	\caption{next[j] Beispiel}
   7.605 +	\label{tab:nextJ}
   7.606 +\end{table}
   7.607 +$\rightarrow next[j]=[0,0,1,2,0,1,2,3,4,5]$\\
   7.608 +\underline{Algorithmus:}
   7.609 +\lstinputlisting[language=Java,label=lst:TTlookup,caption=KMP Algorithmus,
   7.610 +captionpos=b, numbers=left, numberstyle=\tiny, numbersep=5pt, basicstyle=\footnotesize]{./code/kmp.code}
   7.611 +\underline{Analyse:}
   7.612 +inkrement $j$ = inkrement $i$ und dekrement $j$ = inkrement $j$!
   7.613 +$\rightarrow O(n)$ hinzu kommt noch die Berechnung des next-arrays mit $O(m)$.\\
   7.614 +KMP kann also mit $O(n+m)$ berechnet werden. (i.e. Untere Schranke
   7.615 +$\Omega(m+n/m)$)
   7.616 +
   7.617 +\newpage
   7.618 +\begin{appendix}
   7.619 +\section{Landau-Symbole}
   7.620 +\begin{figure}[H]
   7.621 +\centering
   7.622 +\includegraphics[scale=0.45]{img/landau_sym}
   7.623 +\caption{Definition der Landau Symbole}
   7.624 +\label{fig:landau_sym}
   7.625 +\end{figure}
   7.626 +\end{appendix}
   7.627 +\end{document}
   7.628 \ No newline at end of file
     8.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
     8.2 +++ b/AT_Exzerpt.toc	Tue Feb 22 19:14:19 2011 +0100
     8.3 @@ -0,0 +1,52 @@
     8.4 +\select@language {ngerman}
     8.5 +\contentsline {section}{\numberline {1}Einleitung}{3}{section.1}
     8.6 +\contentsline {subsection}{\numberline {1.1}Themen:}{3}{subsection.1.1}
     8.7 +\contentsline {subsection}{\numberline {1.2}Problem-/Anwendungsbereiche:}{3}{subsection.1.2}
     8.8 +\contentsline {subsection}{\numberline {1.3}Literatur:}{3}{subsection.1.3}
     8.9 +\contentsline {section}{\numberline {2}Divide and Conquer}{4}{section.2}
    8.10 +\contentsline {subsection}{\numberline {2.1}Formulierung des Divide and Conquer Prinzips}{4}{subsection.2.1}
    8.11 +\contentsline {subsection}{\numberline {2.2}Quicksort}{4}{subsection.2.2}
    8.12 +\contentsline {subsubsection}{\numberline {2.2.1}Algorithmus}{4}{subsubsection.2.2.1}
    8.13 +\contentsline {subsubsection}{\numberline {2.2.2}Analyse}{4}{subsubsection.2.2.2}
    8.14 +\contentsline {subsection}{\numberline {2.3}N\"achste Paare}{5}{subsection.2.3}
    8.15 +\contentsline {subsubsection}{\numberline {2.3.1}Algorithmus}{5}{subsubsection.2.3.1}
    8.16 +\contentsline {subsubsection}{\numberline {2.3.2}Analyse}{5}{subsubsection.2.3.2}
    8.17 +\contentsline {subsection}{\numberline {2.4}Segmentschnitt}{6}{subsection.2.4}
    8.18 +\contentsline {subsubsection}{\numberline {2.4.1}Algorithmus}{6}{subsubsection.2.4.1}
    8.19 +\contentsline {subsubsection}{\numberline {2.4.2}Analyse}{6}{subsubsection.2.4.2}
    8.20 +\contentsline {subsection}{\numberline {2.5}Polynomprodukt und Fast Fourier-Transformation (TODO)}{6}{subsection.2.5}
    8.21 +\contentsline {section}{\numberline {3}Randomisierung}{7}{section.3}
    8.22 +\contentsline {subsection}{\numberline {3.1}Randomisierter Quicksort}{7}{subsection.3.1}
    8.23 +\contentsline {subsubsection}{\numberline {3.1.1}Analyse}{7}{subsubsection.3.1.1}
    8.24 +\contentsline {subsection}{\numberline {3.2}Randomisierter Primzahltest (TODO)}{7}{subsection.3.2}
    8.25 +\contentsline {subsection}{\numberline {3.3}RSA (TODO)}{7}{subsection.3.3}
    8.26 +\contentsline {subsection}{\numberline {3.4}Treaps (TODO)}{7}{subsection.3.4}
    8.27 +\contentsline {subsection}{\numberline {3.5}Hashing (TODO)}{7}{subsection.3.5}
    8.28 +\contentsline {section}{\numberline {4}Amortisierte Analyse (TODO)}{8}{section.4}
    8.29 +\contentsline {subsection}{\numberline {4.1}Binomial Heaps (TODO)}{8}{subsection.4.1}
    8.30 +\contentsline {subsection}{\numberline {4.2}Fibonacci Heaps (TODO)}{8}{subsection.4.2}
    8.31 +\contentsline {subsection}{\numberline {4.3}Union Find (TODO)}{8}{subsection.4.3}
    8.32 +\contentsline {section}{\numberline {5}Greedy}{9}{section.5}
    8.33 +\contentsline {subsection}{\numberline {5.1}K\"urzeste (billigste) Wege}{9}{subsection.5.1}
    8.34 +\contentsline {subsubsection}{\numberline {5.1.1}Single Source Shortest Paths}{9}{subsubsection.5.1.1}
    8.35 +\contentsline {subsection}{\numberline {5.2}Spannb\"aume minimalen Gewichts}{11}{subsection.5.2}
    8.36 +\contentsline {subsubsection}{\numberline {5.2.1}Das Wachsen von min. Spannb\"aumen}{11}{subsubsection.5.2.1}
    8.37 +\contentsline {subsubsection}{\numberline {5.2.2}Schnitte}{12}{subsubsection.5.2.2}
    8.38 +\contentsline {subsubsection}{\numberline {5.2.3}Sichere Kanten}{12}{subsubsection.5.2.3}
    8.39 +\contentsline {subsubsection}{\numberline {5.2.4}Der Graph $G_A$}{12}{subsubsection.5.2.4}
    8.40 +\contentsline {subsubsection}{\numberline {5.2.5}Algorithmus von Kruskal}{12}{subsubsection.5.2.5}
    8.41 +\contentsline {subsubsection}{\numberline {5.2.6}Algorithmus von Prim}{13}{subsubsection.5.2.6}
    8.42 +\contentsline {section}{\numberline {6}Bin Packing}{14}{section.6}
    8.43 +\contentsline {subsection}{\numberline {6.1}Online Verfahren}{14}{subsection.6.1}
    8.44 +\contentsline {subsubsection}{\numberline {6.1.1}Next Fit (NF)}{14}{subsubsection.6.1.1}
    8.45 +\contentsline {subsubsection}{\numberline {6.1.2}First-Fit (FF)}{14}{subsubsection.6.1.2}
    8.46 +\contentsline {subsubsection}{\numberline {6.1.3}Best-Fit (BF)}{15}{subsubsection.6.1.3}
    8.47 +\contentsline {subsection}{\numberline {6.2}Offline Verfahren}{15}{subsection.6.2}
    8.48 +\contentsline {subsubsection}{\numberline {6.2.1}First Fit Decreasing (FFD oder FFNI)}{15}{subsubsection.6.2.1}
    8.49 +\contentsline {subsubsection}{\numberline {6.2.2}Best Fit Decreasing (BFD)}{15}{subsubsection.6.2.2}
    8.50 +\contentsline {section}{\numberline {7}Dynamische Programmierung}{16}{section.7}
    8.51 +\contentsline {subsection}{\numberline {7.1}Matrixkettenprodukt (TODO)}{16}{subsection.7.1}
    8.52 +\contentsline {subsection}{\numberline {7.2}Optimale Suchb\"aume (TODO)}{16}{subsection.7.2}
    8.53 +\contentsline {subsection}{\numberline {7.3}Editierdistanz}{16}{subsection.7.3}
    8.54 +\contentsline {section}{\numberline {8}Suche in Texten}{17}{section.8}
    8.55 +\contentsline {section}{\numberline {A}Landau-Symbole}{19}{appendix.A}
     9.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
     9.2 +++ b/code/AzyklischeNetzwerke.code	Tue Feb 22 19:14:19 2011 +0100
     9.3 @@ -0,0 +1,12 @@
     9.4 +Sortiere G=(V,E,c) topologisch;
     9.5 +DIST[s]=0;
     9.6 +forall(v:V\{s}){
     9.7 +    DIST[v]=infty;
     9.8 +}
     9.9 +U={v|v:V mit num(v)<n};
    9.10 +while(!U.Empty) {
    9.11 +    Wähle und streiche Knoten u:U mit kleinstem num-Wert;
    9.12 +    forall(e=(u,v):E) {
    9.13 +        if(DIST[v]>DIST[u]+c(u,v)) {
    9.14 +            DIST[v]=DIST[u]+c(u,v);
    9.15 +}}}
    9.16 \ No newline at end of file
    10.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
    10.2 +++ b/code/BelmanFord.code	Tue Feb 22 19:14:19 2011 +0100
    10.3 @@ -0,0 +1,17 @@
    10.4 +DIST[s]=0;
    10.5 +Z[s]=0;
    10.6 +forall(v:V\{s}) {
    10.7 +    DIST[v]=infty;
    10.8 +    Z[v]=0;
    10.9 +}
   10.10 +U={s};
   10.11 +while(!U.Empty) {
   10.12 +    Wähle und streiche Knoten u am Kopf von U;
   10.13 +    Z[u]=Z[u]+1;
   10.14 +    if Z[u]>n
   10.15 +        return "negativer Zyklus";
   10.16 +    forall(e=(u,v):E) {
   10.17 +        if(DIST[v]>DIST[u]+c(u,v)) {
   10.18 +            DIST[v]=DIST[u]+c(u,v);
   10.19 +            U=[U,{v}];
   10.20 +}}}
   10.21 \ No newline at end of file
    11.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
    11.2 +++ b/code/dijkstra.code	Tue Feb 22 19:14:19 2011 +0100
    11.3 @@ -0,0 +1,13 @@
    11.4 +DIST[s]=0;
    11.5 +Insert(U,0,s);
    11.6 +forall(v:V\{s}) {
    11.7 +	DIST[v]=infty;
    11.8 +	Insert(U, infty, v);
    11.9 +}
   11.10 +while !Empty(U) {
   11.11 +    (d,u)=DeleteMin(U);
   11.12 +    forall(e=(u,v):E) {
   11.13 +        if(DIST[v]>DIST[u]+c(u,v)){
   11.14 +            DIST[v]=DIST[u]+c(u,v);
   11.15 +            DecreaseKey(U,v,DIST[v]);
   11.16 +}}}
   11.17 \ No newline at end of file
    12.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
    12.2 +++ b/code/editierdistanz.code	Tue Feb 22 19:14:19 2011 +0100
    12.3 @@ -0,0 +1,10 @@
    12.4 +Input: Zwei Zeichenketten A und B
    12.5 +Output: Matrix D=(Dij)
    12.6 +D[0,0]:= 0
    12.7 +for i := 1 to m do D[i,0] = i
    12.8 +for j := 1 to n do D[0,j] = j
    12.9 +for i := 1 to m do
   12.10 +    for j := 1 to n do
   12.11 +         D[i,j] := min(D[i - 1,j] + 1,
   12.12 +                       D[i,j - 1] + 1,
   12.13 +                       D[i-1,j-1]+c(ai,bj))
   12.14 \ No newline at end of file
    13.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
    13.2 +++ b/code/kmp.code	Tue Feb 22 19:14:19 2011 +0100
    13.3 @@ -0,0 +1,34 @@
    13.4 +KMP := proc (T::string, P::string)
    13.5 +# Input: Text T und Muster P
    13.6 +# Output: Liste L mit Verschiebungen i,
    13.7 +#         an denen P in T vorkommt
    13.8 +    n := length (T); m := length(P);
    13.9 +    L := []; next := KMPnext(P);
   13.10 +    j := 0;
   13.11 +    for i from 1 to n do
   13.12 +        whilej>0and T[i]!=P[j+1]
   13.13 +            do j := next [j] od;
   13.14 +        if T[i] = P[j+1] then j := j+1 fi;
   13.15 +        if j = m then
   13.16 +            L := [L[] , i-m];
   13.17 +            j := next [j]
   13.18 +        fi;
   13.19 +    od;
   13.20 +    RETURN (L);
   13.21 +end;
   13.22 +
   13.23 +KMPnext := proc (P : : string)
   13.24 +#Input : Muster P
   13.25 +#Output : next-Array für P
   13.26 +    m := length (P);
   13.27 +    next := array (1..m);
   13.28 +    next [1] := 0;
   13.29 +    j := 0;
   13.30 +    for i from 2 to m do
   13.31 +        while j > 0 and P[i] <> P[j+1]
   13.32 +            do j := next [j] od;
   13.33 +        if P[i] = P[j+1] then j := j+1 fi;
   13.34 +        next [i] := j
   13.35 +    od;
   13.36 +    RETURN (next);
   13.37 +end;
   13.38 \ No newline at end of file
    14.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
    14.2 +++ b/code/quicksort.code	Tue Feb 22 19:14:19 2011 +0100
    14.3 @@ -0,0 +1,10 @@
    14.4 +Input: unsortierter Bereich [p, r] in Array A
    14.5 +Output: sortierter Bereich [p, r] in Array A
    14.6 +  if r > p then
    14.7 +    wähle Pivotelement x = A[r]
    14.8 +    m = partition(A, p , r)
    14.9 +    /* Teile A bzgl. x auf:
   14.10 +     * A[p],...,A[m-1] <= x <= A[m+1],...,A[r]
   14.11 +     */
   14.12 +  Quicksort(A, p , m - 1)
   14.13 +  Quicksort (A, m + 1, r)
   14.14 \ No newline at end of file
    15.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
    15.2 +++ b/code/spannbaeumeGreedy.code	Tue Feb 22 19:14:19 2011 +0100
    15.3 @@ -0,0 +1,6 @@
    15.4 +Algorithmus Generischer-MST(G,c);
    15.5 +A = EmptySet;
    15.6 +while(A ist kein Spannbaum) {
    15.7 +    Finde sichere Kante (u,v) für A;
    15.8 +    A = [A, {(u,v)}];
    15.9 +}
   15.10 \ No newline at end of file
    16.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
    16.2 +++ b/code/spannbaeumeKruskal.code	Tue Feb 22 19:14:19 2011 +0100
    16.3 @@ -0,0 +1,12 @@
    16.4 +A = EmptySet;
    16.5 +forall(v:V)
    16.6 +    Bv={v};
    16.7 +Erzeuge eine Liste L der Kanten in E, welche gemäß
    16.8 +nicht-fallenden Kantenkosten sortiert ist;
    16.9 +forall (u,v):L {
   16.10 +    B1=FIND(u);
   16.11 +    B2=FIND(v);
   16.12 +    if(B1!=B2)
   16.13 +        A=[A,{(u,v)}];
   16.14 +        UNION(B1, B2, B1);
   16.15 +}
   16.16 \ No newline at end of file
    17.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
    17.2 +++ b/code/spannbaeumePrim.code	Tue Feb 22 19:14:19 2011 +0100
    17.3 @@ -0,0 +1,12 @@
    17.4 +forall(v:V)
    17.5 +    Insert(Q, infty, v);
    17.6 +Wähle einen Knoten w:V als Wurzel;
    17.7 +DecreaseKey(Q, 0, w);
    17.8 +p[w]=nil;
    17.9 +while(!Empty(Q)) {
   17.10 +    (d,u)=DeleteMin(Q);
   17.11 +    forall((u,v):E)
   17.12 +        if(v:Q && c(u,v)<v.Key) {
   17.13 +            DecreaseKey(Q, c(u,v), v);
   17.14 +            p[v]=u;
   17.15 +        }
   17.16 \ No newline at end of file
    18.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
    18.2 +++ b/code/spurgraph.code	Tue Feb 22 19:14:19 2011 +0100
    18.3 @@ -0,0 +1,12 @@
    18.4 +Input: Berechnete Matrix D
    18.5 +Output: Sequenz der Editieroperationen
    18.6 +if i==0 & j==0 then return
    18.7 +if i!=0 and D[i,j]==D[i-1,j]+1
    18.8 +    then Editieroperationen(i-1,j)
    18.9 +    "lösche a[i]"
   18.10 +else if j!=0 and D[i,j]==D[i,j-1]+1
   18.11 +    then Editieroperationen(i,j-1)
   18.12 +    "füge b[j] ein"
   18.13 +else /* D[i,j]=D[i-1,j-1 ]+c(a[i],b[j]) */
   18.14 +    Editieroperationen(i-1,j-1)
   18.15 +    "ersetze a[i] durch b[j]"
   18.16 \ No newline at end of file
    19.1 --- a/comp.log	Tue Feb 22 18:58:21 2011 +0100
    19.2 +++ /dev/null	Thu Jan 01 00:00:00 1970 +0000
    19.3 @@ -1,126 +0,0 @@
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    21.1 --- a/exzerpt.bbl	Tue Feb 22 18:58:21 2011 +0100
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    21.3 @@ -1,25 +0,0 @@
    21.4 -\begin{thebibliography}{THC01}
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   21.14 -\textsc{Peter~Widmayer}, Thomas~O.:
   21.15 -\newblock \emph{Algorithmen und Datenstrukturen}.
   21.16 -\newblock 4. Auflage.
   21.17 -\newblock Spektrum Akademischer Verlag, 2002. --
   21.18 -\newblock ISBN 978--3827410290
   21.19 -
   21.20 -\bibitem[THC01]{1}
   21.21 -\textsc{Thomas H.~Cormen}, Robert L. Rivest und Cliford~S. Charles
   21.22 -  E.~Leiserson~L. Charles E.~Leiserson:
   21.23 -\newblock \emph{Introduction to Algorithms}.
   21.24 -\newblock 2nd.
   21.25 -\newblock Prentice Hall India, 2001. --
   21.26 -\newblock ISBN 978--8120321410
   21.27 -
   21.28 -\end{thebibliography}
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    26.1 --- a/exzerpt.tex	Tue Feb 22 18:58:21 2011 +0100
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    26.4 -\documentclass[a4paper,11pt,twoside]{article}
    26.5 -
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   26.10 -
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   26.24 -
   26.25 -% Festlegung Art der Zitierung - Havardmethode: Abkuerzung Autor + Jahr
   26.26 -\bibliographystyle{alphadin}
   26.27 -
   26.28 -\title{\underline{Exzerpt Algorithmentheorie WS10/11}}
   26.29 -\author{Markus Lindenmann}
   26.30 -\date{\today{} }
   26.31 -
   26.32 -\begin{document}
   26.33 -    \maketitle
   26.34 -    
   26.35 -    \tableofcontents
   26.36 -    \newpage
   26.37 -
   26.38 -    \section{Einleitung}\label{sec:intro}
   26.39 -        Dozent:     Matthias Westermann\\
   26.40 -        Website:    \url{http://lak.informatik.uni-freiburg.de}\\
   26.41 -        Klausur:    09.03.2011\\
   26.42 -        Hilfsmittel:    Ein auf beiden Seiten handbeschriebenes DIN-A4-Blatt
   26.43 -        \subsection{Themen:}
   26.44 -            \begin{itemize}
   26.45 -                \item[$\bullet$] Divide and Conquer
   26.46 -                \item[$\bullet$] Greedy Prinzip
   26.47 -                \item[$\bullet$] Dynamische Programmierung
   26.48 -                \item[$\bullet$] Randomisierung
   26.49 -                \item[$\bullet$] Amortisierte Analyse
   26.50 -            \end{itemize}
   26.51 -        \subsection{Problem-/Anwendungsbereiche:}
   26.52 -            \begin{itemize}
   26.53 -                \item[$\bullet$] Geometrische Algorithmen
   26.54 -                \item[$\bullet$] Algebraische Algorithmen
   26.55 -                \item[$\bullet$] Graphenalgorithmen
   26.56 -                \item[$\bullet$] Datenstrukturen
   26.57 -                \item[$\bullet$] Internet-Algorithmen
   26.58 -                \item[$\bullet$] ptimierungs-Verfahren
   26.59 -                \item[$\bullet$] Zeichenkettenverarbeitung
   26.60 -            \end{itemize}
   26.61 -        \subsection{Literatur:}
   26.62 -            \bibliography{literatur}
   26.63 -            \nocite{*}
   26.64 -    
   26.65 -    \newpage
   26.66 -    \section{Divide and Conquer}\label{sec:div&conq}
   26.67 -        \begin{itemize}
   26.68 -            \item[$\bullet$] Quicksort
   26.69 -            \item[$\bullet$] Formulierung und Analyse des Prinzips
   26.70 -            \item[$\bullet$] Geometrisches Devide and Conquer
   26.71 -            \begin{itemize}
   26.72 -                \item[-] Closest-Pair
   26.73 -                \item[-] Segmentschnitt
   26.74 -            \end{itemize}
   26.75 -        \end{itemize}
   26.76 -        
   26.77 -        \subsection{Formulierung des Divide and Conquer Prinzips}
   26.78 -            D\&C Verfahren zur Lösung eines Problems der Größe $n$
   26.79 -            \begin{itemize}
   26.80 -                \item[1.]   Divide\\$n>2$:  Teile das Problem in $k$ Teilprobleme der Größe $n_1,\ldots n_k$ ($k\geq2$)\\$n\leq c:$ Löse das Problem direkt
   26.81 -                \item[2.]   Conquer\\Löse die $k$ Teilprobleme auf dieselbe Art (rekursiv)
   26.82 -                \item[3.]   Merge\\Füge die berechneten Teillösungen zu einer Gesamtlösung zusammen
   26.83 -            \end{itemize}
   26.84 -        
   26.85 -        \subsection{Quicksort}
   26.86 -            Wähle beliebiges Pivot Element $v$ und sortiere alle Elemente aus der Menge, die größer sind als $v$ rechts von $v$ ein, die anderen links. Verfahre im selben Muster mit den neuen Teilmengen links und rechts des Pivot Elements. Wenn die zu sortierende Folge $F$ nur ein Element beinhaltet, gebe $F$ zurück.
   26.87 -			
   26.88 -            \subsubsection{Analyse}
   26.89 -                $T(n)$ - maximale Anzahl von Schritten, um ein Problem der Größe $n$ zu lösen.
   26.90 -                \[T(n)=\begin{cases}n\neq c&a \\ n>c&T(n_1+\ldots+T(n_k)+\text{Divide- und Mergeaufwand}\end{cases}\]
   26.91 -                Best Case: $k=2, n_1=n_2=\frac{n}{2}$\\
   26.92 -                Divide- und Mergeaufwand: $DM(n)$\\
   26.93 -                $T(1)=a$\\
   26.94 -                $T(n)=2T(\frac{n}{2})+DM(n)$
   26.95 -        
   26.96 -        \subsection{Nächste Paare}
   26.97 -            Eingabe:    $n$ Punkte in Ebene\\
   26.98 -            Ausgabe:    Punkt-Paar $q,r$ mit geringstem Abstand\\
   26.99 -            Abstand:    $d(q,r)$ bechreibt Abstand zwischen Punkten $q$ und $r$
  26.100 -            \subsubsection{Algorithmus}
  26.101 -                \begin{itemize}
  26.102 -                    \item[$\bullet$] sortiere Punktmenge nach x-Koordinate
  26.103 -                    \item[$\bullet$] Teile Punktmenge in der Mitte L in die Mengen Q und R
  26.104 -                    \item[$\bullet$] Löse rekursiv auf Q und R
  26.105 -                    \item[$\bullet$] Sortiere Punkte nach y-Koordinate
  26.106 -                    \item[$\bullet$] Teste alle Paare, deren Abstand in der Sortierung kleiner als 16 ist (siehe unten)
  26.107 -                    \item[$\bullet$] Füge zusammen
  26.108 -                \end{itemize}
  26.109 -            
  26.110 -                $\delta=$ Minimum der Teillösungen\\
  26.111 -                Gibt es $q\in Q$ und $r\in R$ mit $d(q,r)<\delta$, dann sind sowohl $q$ als auch $r$ höchstens $\delta$ von $L$ (Mitte) entfernt.\\
  26.112 -                \textbf{Problem:} alle Punkte können innerhalb der $\delta$-Zone liegen\\
  26.113 -                \textbf{Lösung:}\begin{itemize}
  26.114 -                    \item[$\bullet$] Sortiere Punkte in $\delta$ Zone nach $y$-Abstand
  26.115 -                    \item[$\bullet$] Liegen in dieser Sortierung mehr als $c$ Punkte zwischen $q$ und $r$, dann kann $(q,r)$ nicht nächstes Paar sein
  26.116 -                \end{itemize}
  26.117 -                
  26.118 -                Dies kann für $c=15$ bewiesen werden (vgl. Abbildung \ref{cl_pair}):
  26.119 -                \begin{itemize}
  26.120 -                    \item[$\bullet$] Seien min. 15 Punkte zwischen $q$ und $r$
  26.121 -                    \item[$\bullet$] Dann sind mindestens 3 Reihen Quadrate zwischen $q,r$, weil in jedem Quadrat höchstens ein Punkt ist.
  26.122 -                    \item[$\rightarrow$] Dann muss $d(q,r)$ mindestens $\frac{3}{2}\delta > \delta$ sein.
  26.123 -                \end{itemize}
  26.124 -                
  26.125 -                \begin{figure}[H]
  26.126 -                    \centering
  26.127 -                    \includegraphics[scale=0.5]{img/cl_pair}
  26.128 -                    \caption{Nächste Paare: Prüf Distanz}
  26.129 -                    \label{cl_pair}
  26.130 -                \end{figure}
  26.131 -
  26.132 -            \subsubsection{Analyse}
  26.133 -                \[T(n)=\begin{cases}n\leq3&a \\ n>3&2T(\frac{n}{2})+an\end{cases}\]
  26.134 -                \[T(n)\leq an \log n \]
  26.135 -
  26.136 -        \subsection{Segmentschnitt}
  26.137 -            Eingabe:    Menge S bestehend aus vertikalen Segmenten und Endpunkten vn horizontalen Segmenten\\
  26.138 -            Ausgabe:    Alle Schnittpunkte von vertikalen Segmenten mit horizontalen Segmenten, von denen mindesens ein Endpunkt in S ist
  26.139 -            
  26.140 -            \subsubsection{Algorithmus}
  26.141 -                \textbf{Divide \& Conquer:}\\Fallunterscheidung
  26.142 -                \begin{itemize}
  26.143 -                    \item[1.] $\left| S\right| > 1$:\\ Teile S mittels einer vertikalen Gerade G in 2 gleichgroße Mengen $S_1$ (links von G) und $S_2$ (rechts von G). Führe rekursiv auf beiden Hälften aus.
  26.144 -                    \item[2.] else: S enthält keine Schnitte
  26.145 -                \end{itemize}
  26.146 -                
  26.147 -                \textbf{Merge}\\
  26.148 -                Bilde L(S), R(S) und V(S) (Definition am Beispiel $i=1,2$ berechnet: $S=S_1\cup S_2$):
  26.149 -                \begin{itemize}
  26.150 -                    \item[$\bullet$] L(S): y-Koordinate aller linken Endpunkte in S, deren rechter Partner nicht in S\\$=L(S_2)\cup(L(S_1)-R(S_2))$
  26.151 -                    \item[$\bullet$] R(S): y-Koordinate aller rechten Endpunkte in S, deren linker Partner nicht in S\\$=R(S_1)\cup(R(S_2)-L(S_1))$
  26.152 -                    \item[$\bullet$] V(S): y-Intervalle der vertikalen Segmente in S\\$=V(S_1)\cup V(S_2)$
  26.153 -                \end{itemize}
  26.154 -                L,R sortiert nach steigender y-Koordinate\\
  26.155 -                V sortiert nach steigendem unteren Endpunkt\\
  26.156 -                \medskip
  26.157 -                
  26.158 -                \textbf{Basisfälle}\\
  26.159 -                S enthält nur ein Element s. Fallunterscheidung:
  26.160 -                \begin{itemize}
  26.161 -                    \item[1.] $s=(x,y)$ ist ein linker Endpunkt:\\$L(S)=\{y\},\qquad R(S)=\emptyset,\qquad V(S)=\emptyset$
  26.162 -                    \item[2.] $s=(x,y)$ ist ein rechter Endpunkt:\\$L(S)=\emptyset,\qquad R(S)=\{y\},\qquad V(S)=\emptyset$
  26.163 -                    \item[3.] $s=(x,y_1,y_2)$ ist ein vertikales Segment:\\$L(S)=\emptyset,\qquad R(S)=\emptyset,\qquad V(S)=\{[y_1,y_2]\}$
  26.164 -                \end{itemize}
  26.165 -            
  26.166 -            \subsubsection{Analyse}
  26.167 -                Eingabe wird Anfangs sortiert und in Array gespeichert
  26.168 -                \[T(n)=2T(\frac{n}{2}+an+\text{Größe der Ausgabe}\]
  26.169 -                \[T(1)=O(1)\]
  26.170 -                \[O(n \log n + k \mid k=\#Schnittpunkte\]
  26.171 -        
  26.172 -        \subsection{Polynomprodukt und Fast Fourier-Transformation}
  26.173 -            
  26.174 -            
  26.175 -	  \newpage
  26.176 -    \section{Randomisierung}
  26.177 -        \subsection{Hashing}
  26.178 -    
  26.179 -    \newpage
  26.180 -    \section{Amortisierte Analyse}
  26.181 -        \subsection{Binomial Heaps}
  26.182 -        \subsection{Fibonacci Heaps}
  26.183 -        \subsection{Union Find}
  26.184 -    
  26.185 -    \newpage
  26.186 -    \section{Greedy}
  26.187 -        \subsection{Kürzeste (billigste) Wege}
  26.188 -    
  26.189 -    \newpage
  26.190 -    \section{Bin Packing}
  26.191 -        \textbf{Problembeschreibung:}\\
  26.192 -        Man hat $n$ Objekte verschiedener Größe mit $0<s_i\leq1\mid 1\leq i\leq n$. Gesucht ist die kleinst mögliche Anzahl an Kisten (Bins) der Größe 1, mit der alle Objekte verpackt werden können.
  26.193 -        \subsection{Online Verfahren}
  26.194 -            Jedes (ankommende) Objekt muss verpackt sein, bevor das nächste Objekt betrachtet wird. Ein Objekt verbleibt in derjenigen Kiste, in die es zuerst gepackt wird.\\
  26.195 -            Kein Online Bin Packing Verfahren kann stets eine optimale Lösung finden!\\            
  26.196 -            \textbf{Satz 1}: Es gibt Eingaben, die jeden Online Bin Packing Algorithmus zwingen, wenigstens $\frac{4}{3} OPT$ Bins zu verwenden, wobei $OPT$ die minimal mögliche Binzahl ist.\\
  26.197 -            \textbf{Beweis} Annahme: Online Bin Packing Algorithmus benötigt stets weniger als $\frac{4}{3}OPT$ Bins.\\
  26.198 -            Es wird eine Eingabe mit $2m$ Objekten angenommen. Die ersten $m$ sind um $\epsilon$ kleiner als $\frac{1}{2}$, die zweiten $m$ um $\epsilon$ größer. Die optimale Lösung sind $m$ Kisten, da immer ein großes mit einem kleinen Paket kombiniert werden kann. Nun teilt man die Analyse in zwei Zeitpunkte: nach $m$ Objekten und nach $2m$ Objekten. Laut Annahme gilt nach nach $m$ Objekten, dass die Anzahl der Bins $b$ nur maximal $\frac{4}{3} OPT$ groß sein darf. Mit $OPT=\frac{m}{2}$ gilt somit: $b=\frac{4}{3}\cdot\frac{m}{2}=\frac{2}{3}m$. Sei $b=b_1+b_2\mid b_1=\#$Bins mit einem Objekt, $b_2=\#$Bins mit zwei Objekten, so folgt $b_1+2b_2=m\implies b_1=m-2b_2\implies b=b_1+b_2=m-b_2$. Nach $2m$ Objekten ist $OPT=m$, wie zuvor beschrieben. $b_1$ Bins können zu diesem Zeitpunkt mit Objekten des zweiten Bereichs gefüllt werden, für die verbleibenden muss man neue Bins öffnen. Die Anzahl der neu zu öffnenden beträgt damit genau $b_2\implies \#$Bins $\geq b+m-b_1=m+b_2$. Die Annahme besagt $m+b_2\leq\#$Bins $<\frac{4}{3}m \wedge b_2<\frac{m}{3}$. Dies ist nur möglich, wenn $b=m-b_2>\frac{2}{3}m$ - Dies führt zu einem \underline{Widerspruch}, da $b$ zum Zeitpunkt 1, nach $m$ Objekten $<\frac{2}{3}m$ sein muss.
  26.199 -            
  26.200 -            \subsubsection{Next Fit (NF)}
  26.201 -                Verpacke das nächste Objekt in dieselbe Kiste wie das vorherige, wenn es dort noch hineinpasst, sonst öffne eine neue Kiste und verpacke es dort.\\
  26.202 -                \textbf{Satz 2}: (a) Für alle Inputfolgen $I$ gilt: $NF(I)\leq2OPT(I)$. (b) Es gibt Inputfolgen I mit $NF(I)\geq2OPT(I)-2$.\\
  26.203 -                \textbf{Beweis}\\
  26.204 -                (a) Betrachte zwei beliebige aufeinanderfolgende Kisten $B_{2k-1}, B_{2k}$. Man weiß, dass die beinhalteten Objekte eine Summengröße $>1$ haben, sonst wären selbige in einer Kiste. Daraus folgt, man kann nicht mehr als die doppelte Anzahl an Kisten als beim Optimum erreichen!\\
  26.205 -                (b) Inputfolge: $0.5, \frac{2}{n}, 0.5, \frac{2}{n}, 0.5, \frac{2}{n}, \ldots 0.5, \frac{2}{n}$.\\Next-Fit liefert nun immer Bins gefüllt mit $0.5 + \frac{2}{n}$. Darausfolgt $NF(I)=\frac{n}{2}$ und $OPT=\frac{n}{4}+1$ (weil $\frac{n}{2}$ mal 0.5er Objekte in $=\frac{n}{4}$ Kisten passen und $\frac{n}{2}$ $\frac{2}{n}$er Objekte in eine Kiste passen)\\
  26.206 -                Laufzeit für Inputfolgen der Länge $n$: NF $O(n)$
  26.207 -            \subsubsection{First-Fit (FF)}
  26.208 -                Packe nächstes Objekt in die erste Kiste, in die es noch hineinpasst, wenn es eine solche Kiste gibt, sonst in eine neue Kiste.\\
  26.209 -                \textbf{Beobachtung}: Zu jedem Zeitpunkt kann es höchstens eine Kiste geben, die weniger als halb voll ist. ($\implies FF(I) \leq 2OPT(I)$)\\
  26.210 -                \textbf{Satz 3}: (a) Für alle Inputfolgen $I$ gilt $FF(I)\leq \lceil\frac{17}{10} OPT(I)\rceil$. (b) Es gibt Inputfolgen $I$ mit $FF(I)\geq\frac{17}{10}(OPT(I)-1)$. (b') Es gibt Inputfolgen $I$ mit $FF(I)=\frac{10}{6}OPT(I)$.\\
  26.211 -                \textbf{Beweis}\\
  26.212 -                (b') Inputfolge der Länge $3\cdot6m$. Dabei je $6m$ Objekte der Größen $\frac{1}{7}+\epsilon, \frac{1}{3}+\epsilon, \frac{1}{2}+\epsilon$, welche in gegebener Reihenfolge in Gruppen eingegeben werden. FF liefert hierbei $m$ Kisten gefüllt mit den Objekten der Größe $\frac{1}{7}+\epsilon$ (jeweils 6 pro Kiste) gefolgt von $\frac{6m}{2}=3m$ Kisten gefüllt mit Objekten der Größe $\frac{1}{3}+\epsilon$ (jeweils 2 pro Kiste) und schlussendlich $6m$ Kisten mit je einem Objekt der Größe $\frac{1}{2}+\epsilon$. Das macht in der Summe $10m$ Kisten. Die optimale Lösung packt je ein Objekt jeder Größe in eine Kiste und benötigt somit nur $6m$ Kisten.\\
  26.213 -                Laufzeit für Inputfolgen der Länge $n$: FF $O(n^2) \rightarrow O(n \log n)$
  26.214 -            \subsubsection{Best-Fit (BF)}
  26.215 -                Verpacke das nächste Objekt in diejenige Kiste, in die es am besten passt (d.h. den geringsten Platz ungenutzt lässt).
  26.216 -                Verhalten von BF ähnlich zu FF.
  26.217 -                Laufzeit für Inputfolgen der Länge $n$: BF $O(n^2) \rightarrow O(n \log n)$
  26.218 -        \subsection{Offline Verfahren}
  26.219 -            \underline{NP-schwieriges Problem! Entscheidungsproblem ist NP-vollständig!}\\
  26.220 -            Zunächst wird die Anzahl $n$ und alle $n$ Objekte vorgegeben. Dann beginnt die Verpackung.\\
  26.221 -            $n$ und $s_1, ..., s_n$ sind gegeben, bevor die Verpackung beginnt!\\
  26.222 -            $\rightarrow$ \underline{Optimale Verpackung} kann durch erschöpfende Suche bestimmt werden. ABER: Benötigt exponentielle Zeit! Daher Approximimationsalgorithmen, die schneller sind aber dafür unter Umständen nicht optimal!\\
  26.223 -            Idee für Offline Approximationsalgorithmen: Sortiere die Objekte zunächst nach abnhemender Größe und verpacke größere Objekte zuerst!
  26.224 -            \subsubsection{First Fit Decreasing (FFD oder FFNI)}
  26.225 -                \textbf{Lemma 1}: Sei $I$ eine Folge von $n$ Objekten mit Größen $s_1\geq s_2\geq\ldots\geq s_n$ und sei $m=OPT(I)$. Dann haben alle von FFD in den Bins $B_{m+1},B_{m+2},\ldots,B_{FFD(I)}$ verpackten Objekte eine Größe von höchstens $\frac{1}{3}$.\\
  26.226 -                \textbf{Beweis}: \$\$\$TODO:Elsässer hat das in seiner VL gemacht - aber hässlich lang\$\$\$\\
  26.227 -                \textbf{Lemma 2}: Sei $I$ eine Folge von $n$ Objekten mit Größen $s_1\geq s_2\geq\ldots\geq s_n$ und sei $m=OPT(I)$. Dann ist die Anzahl der Objekte, die FFD in die Kisten $B_{m+1},B_{m+2},\ldots,B_{FFD(I)}$ verpackt, höchstens $m-1$.\\
  26.228 -                \textbf{Beweis}: Annahme: Es gibt mehr als $m-1$ Objekte. $x_1,\ldots,x_m$ in $I$, die FFD in extra Kisten verpacken. Dies führt zu einem Widerspruch. \$\$\$TODO:Warum?\$\$\$\\
  26.229 -                \textbf{Satz}: Für alle Inputfolgen I gilt $FFD(I)\leq(4 OPT(I)+\frac{1}{3})$.\\
  26.230 -                \textbf{Satz}:
  26.231 -                \begin{itemize}
  26.232 -                    \item[1.] Für alle Inputfolgen $I$ gilt: $FFD(I)\leq \frac{11}{9} OPT(I)+4$
  26.233 -                    \item[2.] Es gibt Inputfolgen $I$ mit: $FFD(I)=\frac{11}{9}OPT(I)$.
  26.234 -                \end{itemize}
  26.235 -                \textbf{Beweis}: \$\$\$TODO\$\$\$
  26.236 -            \subsubsection{Best Fit Decreasing (BFD)}
  26.237 -                Nicht im Script???
  26.238 -    
  26.239 -    \newpage
  26.240 -    \section{Dynamische Programmierung}
  26.241 -        
  26.242 -        
  26.243 -    \newpage
  26.244 -    \begin{appendix}
  26.245 -        \section{Landau-Symbole}
  26.246 -            \begin{figure}[H]
  26.247 -                \centering
  26.248 -                \includegraphics[scale=0.45]{img/landau_sym}
  26.249 -                \caption{Definition der Landau Symbole}
  26.250 -                \label{fig:landau_sym}
  26.251 -            \end{figure}
  26.252 -    \end{appendix}
  26.253 -\end{document}
    27.1 --- a/exzerpt.toc	Tue Feb 22 18:58:21 2011 +0100
    27.2 +++ /dev/null	Thu Jan 01 00:00:00 1970 +0000
    27.3 @@ -1,34 +0,0 @@
    27.4 -\select@language {ngerman}
    27.5 -\contentsline {section}{\numberline {1}Einleitung}{2}{section.1}
    27.6 -\contentsline {subsection}{\numberline {1.1}Themen:}{2}{subsection.1.1}
    27.7 -\contentsline {subsection}{\numberline {1.2}Problem-/Anwendungsbereiche:}{2}{subsection.1.2}
    27.8 -\contentsline {subsection}{\numberline {1.3}Literatur:}{2}{subsection.1.3}
    27.9 -\contentsline {section}{\numberline {2}Divide and Conquer}{3}{section.2}
   27.10 -\contentsline {subsection}{\numberline {2.1}Formulierung des Divide and Conquer Prinzips}{3}{subsection.2.1}
   27.11 -\contentsline {subsection}{\numberline {2.2}Quicksort}{3}{subsection.2.2}
   27.12 -\contentsline {subsubsection}{\numberline {2.2.1}Analyse}{3}{subsubsection.2.2.1}
   27.13 -\contentsline {subsection}{\numberline {2.3}N\"achste Paare}{3}{subsection.2.3}
   27.14 -\contentsline {subsubsection}{\numberline {2.3.1}Algorithmus}{3}{subsubsection.2.3.1}
   27.15 -\contentsline {subsubsection}{\numberline {2.3.2}Analyse}{4}{subsubsection.2.3.2}
   27.16 -\contentsline {subsection}{\numberline {2.4}Segmentschnitt}{4}{subsection.2.4}
   27.17 -\contentsline {subsubsection}{\numberline {2.4.1}Algorithmus}{4}{subsubsection.2.4.1}
   27.18 -\contentsline {subsubsection}{\numberline {2.4.2}Analyse}{5}{subsubsection.2.4.2}
   27.19 -\contentsline {subsection}{\numberline {2.5}Polynomprodukt und Fast Fourier-Transformation}{5}{subsection.2.5}
   27.20 -\contentsline {section}{\numberline {3}Randomisierung}{6}{section.3}
   27.21 -\contentsline {subsection}{\numberline {3.1}Hashing}{6}{subsection.3.1}
   27.22 -\contentsline {section}{\numberline {4}Amortisierte Analyse}{7}{section.4}
   27.23 -\contentsline {subsection}{\numberline {4.1}Binomial Heaps}{7}{subsection.4.1}
   27.24 -\contentsline {subsection}{\numberline {4.2}Fibonacci Heaps}{7}{subsection.4.2}
   27.25 -\contentsline {subsection}{\numberline {4.3}Union Find}{7}{subsection.4.3}
   27.26 -\contentsline {section}{\numberline {5}Greedy}{8}{section.5}
   27.27 -\contentsline {subsection}{\numberline {5.1}K\"urzeste (billigste) Wege}{8}{subsection.5.1}
   27.28 -\contentsline {section}{\numberline {6}Bin Packing}{9}{section.6}
   27.29 -\contentsline {subsection}{\numberline {6.1}Online Verfahren}{9}{subsection.6.1}
   27.30 -\contentsline {subsubsection}{\numberline {6.1.1}Next Fit (NF)}{9}{subsubsection.6.1.1}
   27.31 -\contentsline {subsubsection}{\numberline {6.1.2}First-Fit (FF)}{9}{subsubsection.6.1.2}
   27.32 -\contentsline {subsubsection}{\numberline {6.1.3}Best-Fit (BF)}{10}{subsubsection.6.1.3}
   27.33 -\contentsline {subsection}{\numberline {6.2}Offline Verfahren}{10}{subsection.6.2}
   27.34 -\contentsline {subsubsection}{\numberline {6.2.1}First Fit Decreasing (FFD oder FFNI)}{10}{subsubsection.6.2.1}
   27.35 -\contentsline {subsubsection}{\numberline {6.2.2}Best Fit Decreasing (BFD)}{10}{subsubsection.6.2.2}
   27.36 -\contentsline {section}{\numberline {7}Dynamische Programmierung}{11}{section.7}
   27.37 -\contentsline {section}{\numberline {A}Landau-Symbole}{12}{appendix.A}
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