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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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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
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20.2 +++ /dev/null Thu Jan 01 00:00:00 1970 +0000
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20.18 -
20.19 -\bibstyle{alphadin}
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21.1 --- a/exzerpt.bbl Tue Feb 22 18:58:21 2011 +0100
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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
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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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25.93 -/Action/GoTo/Dest(subsection.4.2)cvn
25.94 -/OUT pdfmark
25.95 -[
25.96 -/Title(Union Find)
25.97 -/Action/GoTo/Dest(subsection.4.3)cvn
25.98 -/OUT pdfmark
25.99 -[
25.100 -/Title(Greedy)
25.101 -/Count -1
25.102 -/Action/GoTo/Dest(section.5)cvn
25.103 -/OUT pdfmark
25.104 -[
25.105 -/Title(K\374rzeste \(billigste\) Wege)
25.106 -/Action/GoTo/Dest(subsection.5.1)cvn
25.107 -/OUT pdfmark
25.108 -[
25.109 -/Title(Bin Packing)
25.110 -/Count -2
25.111 -/Action/GoTo/Dest(section.6)cvn
25.112 -/OUT pdfmark
25.113 -[
25.114 -/Title(Online Verfahren)
25.115 -/Count -3
25.116 -/Action/GoTo/Dest(subsection.6.1)cvn
25.117 -/OUT pdfmark
25.118 -[
25.119 -/Title(Next Fit \(NF\))
25.120 -/Action/GoTo/Dest(subsubsection.6.1.1)cvn
25.121 -/OUT pdfmark
25.122 -[
25.123 -/Title(First-Fit \(FF\))
25.124 -/Action/GoTo/Dest(subsubsection.6.1.2)cvn
25.125 -/OUT pdfmark
25.126 -[
25.127 -/Title(Best-Fit \(BF\))
25.128 -/Action/GoTo/Dest(subsubsection.6.1.3)cvn
25.129 -/OUT pdfmark
25.130 -[
25.131 -/Title(Offline Verfahren)
25.132 -/Count -2
25.133 -/Action/GoTo/Dest(subsection.6.2)cvn
25.134 -/OUT pdfmark
25.135 -[
25.136 -/Title(First Fit Decreasing \(FFD oder FFNI\))
25.137 -/Action/GoTo/Dest(subsubsection.6.2.1)cvn
25.138 -/OUT pdfmark
25.139 -[
25.140 -/Title(Best Fit Decreasing \(BFD\))
25.141 -/Action/GoTo/Dest(subsubsection.6.2.2)cvn
25.142 -/OUT pdfmark
25.143 -[
25.144 -/Title(Dynamische Programmierung)
25.145 -/Action/GoTo/Dest(section.7)cvn
25.146 -/OUT pdfmark
25.147 -[
25.148 -/Title(Landau-Symbole)
25.149 -/Action/GoTo/Dest(appendix.A)cvn
25.150 -/OUT pdfmark
26.1 --- a/exzerpt.tex Tue Feb 22 18:58:21 2011 +0100
26.2 +++ /dev/null Thu Jan 01 00:00:00 1970 +0000
26.3 @@ -1,250 +0,0 @@
26.4 -\documentclass[a4paper,11pt,twoside]{article}
26.5 -
26.6 -\usepackage[ngerman]{babel}
26.7 -\usepackage[T1]{fontenc}
26.8 -\usepackage[latin1]{inputenc}
26.9 -\usepackage[colorlinks=true,linkcolor=black]{hyperref}
26.10 -
26.11 -\usepackage{vaucanson-g}
26.12 -\usepackage{hyperref}
26.13 -\usepackage{hypcap}
26.14 -\usepackage{fancyhdr}
26.15 -\usepackage{graphicx}
26.16 -\usepackage{lastpage}
26.17 -\usepackage[cm]{fullpage}
26.18 -\usepackage{amsthm, amsmath, amsfonts}
26.19 -\usepackage{float}
26.20 -\usepackage{paralist}
26.21 -\usepackage{listings}
26.22 -\usepackage{color}
26.23 -\usepackage{bookmark}
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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