diff --git a/übung_1/ads_übung_1.pdf b/übung_1/ads_übung_1.pdf deleted file mode 100644 index 5cc8bf0..0000000 Binary files a/übung_1/ads_übung_1.pdf and /dev/null differ diff --git a/übung_1/ads_übung_1.synctex.gz b/übung_1/ads_übung_1.synctex.gz deleted file mode 100644 index 55871b8..0000000 Binary files a/übung_1/ads_übung_1.synctex.gz and /dev/null differ diff --git a/übung_1/ads_übung_1.tex b/übung_1/ads_übung_1.tex deleted file mode 100644 index eab69a3..0000000 --- a/übung_1/ads_übung_1.tex +++ /dev/null @@ -1,86 +0,0 @@ -\documentclass{ngexrcs} -\usepackage{hyperref} - -\title{1. Übungsblatt} -\subject{Algorithmische Graphentheorie} -\author{Jasper Gude \and Pia Rötgers} - -\begin{document} -\maketitle -\points[\qquad]{20} -\section{Spannbäume \& Breitensuche} -Sei $G = \tup{V, E}$ ein zusammenhängender Graph mit Kantengewichten $w: E \to \NN$ -und $s \in V \deg$ ein ausgezeichneter Knoten. -\begin{enumerate} - \item - \begin{quote} - Wenn $w(e) = 1$ für alle $e \in E$, dann ist der Breitensuchbaum mit - Quelle $s$ ein minimaler Spannbaum. - \end{quote} - Die Breitensuche berechnet in diesem - Fall den kürzesten Weg von jedem Knoten zum Knoten $s$, also den - Breitensuchbaum. Dieser spannt also einen minimalen Spannbaum auf. - \points{2} - \item - \begin{quote} - Wenn $w(e) = 1$ für alle $e \in E$, dann ist jeder minimale Spannbaum - von $G$ ein Breitensuchbaum mit Quelle $s$. - \end{quote} - Falsch, siehe \autoref{fig:msb}. - \points{2} - \begin{figure}[h] - \centering - \includegraphics[width=0.2\textwidth]{msb.eps} - \caption{$\pi$-Zeiger des Breitensuchbaums und MSB blau hinterlegt.} - \label{fig:msb} - \end{figure} - \item - \begin{quote} - Wenn $w(e) \in \set{1, 2, 3}$ für alle $e \in E$, dann ist jeder minimale - Spannbaum von $G$ ein Tiefensuchbaum mit Quelle $s$. - \end{quote} - Sei $w(e) = 1$ für - alle $e \in E$ so gilt das Gegenbeispiel von oben. Also ist die Aussage - falsch. - \points{2} -\end{enumerate} - -\section{Kreissuche} -\begin{enumerate} - \item - Wähle Startknoten $s$ und füge ihn in eine neue Queue $Q$ ein. - - Nimm den vordersten Knoten aus $Q$ und füge seine noch nicht verbrauchten - Nachbarn ein. Markiere diesen Knoten anschließend als verbraucht ($black$). - Wiederhole den Schritt solange bis $Q$ leer ist. Merke dir dabei die zwei - letzten entnommenen Knoten. Sind sie gleich, hat der Graph einen - einfachen Kreis. - - \begin{algorithmic} - \alg{EinfacherKreis}(Graph $G$, Vertex $s$) \+ \\ - \alg{Initialize}($G$, $s$) \\ - $Q \gets$ new \alg{Queue}() \\ - $Q.$\alg{Enqueue}($s$) \\ - $t_1 \gets nil$ \\ - $t_2 \gets s$ \\ - while $Q \neq \emptyset$ do \+ \\ - $u \gets Q.$\alg{Dequeue}() \\ - $t_1 \gets t_2$ \\ - $t_2 \gets u$ \\ - foreach $v \in Adj[u]$ do \+ \\ - if $v.color = white$ then \+ \\ - $Q.$\alg{Enqueue}($v$) \-\- \\ - $u.color \gets black$ \- \\ - if $t_1 = t_2$ then \+ \\ - return $true$ \- \\ - else \+ \\ - return $false$ - - \end{algorithmic} -\end{enumerate} - -\section{Eulerwege} - -\section{Graphmodellierung} - -\end{document} diff --git a/übung_1/ads_übung_1.aux b/übung_1/agt_übung_1.aux similarity index 58% rename from übung_1/ads_übung_1.aux rename to übung_1/agt_übung_1.aux index 891efca..1c80dd6 100644 --- a/übung_1/ads_übung_1.aux +++ b/übung_1/agt_übung_1.aux @@ -7,15 +7,15 @@ 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listings.sty no longer applied! + Expected: + 2024/09/23 1.10c (Carsten Heinz) + but found: + 2025/11/14 1.11b (Carsten Heinz) + so I'm assuming it got fixed. +\c@algorithm=\count296 +\theoremskip=\skip67 +) (/usr/local/texlive/2025/texmf-dist/tex/latex/enumitem/enumitem.sty Package: enumitem 2025/02/06 v3.11 Customized lists \labelindent=\skip68 -\enit@outerparindent=\dimen171 -\enit@toks=\toks20 -\enit@inbox=\box57 -\enit@count@id=\count288 -\enitdp@description=\count289 +\enit@outerparindent=\dimen175 +\enit@toks=\toks21 +\enit@inbox=\box59 +\enit@count@id=\count297 +\enitdp@description=\count298 ) (/usr/local/texlive/2025/texmf-dist/tex/latex/fontspec/fontspec.sty (/usr/local/texlive/2025/texmf-dist/tex/latex/l3packages/xparse/xparse.sty @@ -408,7 +442,7 @@ Package: expl3 2026-01-19 L3 programming layer (loader) (/usr/local/texlive/2025/texmf-dist/tex/latex/l3backend/l3backend-luatex.def File: l3backend-luatex.def 2025-10-09 L3 backend support: PDF output (LuaTeX) 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Übungsblatt} +\subject{Algorithmische Graphentheorie} +\author{Jasper Gude \and Pia Röttgers} + +\begin{document} +\maketitle +\points[\qquad]{20} +\section{Spannbäume \& Breitensuche} +Sei $G = \tup{V, E}$ ein zusammenhängender Graph mit Kantengewichten $w: E \to \NN$ +und $s \in V \deg$ ein ausgezeichneter Knoten. + +\begin{tasks} + +\item +\begin{quote} +Wenn $w(e) = 1$ für alle $e \in E$, dann ist der Breitensuchbaum mit +Quelle $s$ ein minimaler Spannbaum. +\end{quote} +Die Breitensuche berechnet in diesem +Fall den kürzesten Weg von jedem Knoten zum Knoten $s$, also den +Breitensuchbaum. Dieser spannt also einen minimalen Spannbaum auf. +\points{2} + +\item +\begin{quote} +Wenn $w(e) = 1$ für alle $e \in E$, dann ist jeder minimale Spannbaum +von $G$ ein Breitensuchbaum mit Quelle $s$. +\end{quote} +Falsch, siehe \autoref{fig:msb}. +\points{2} +\begin{figure}[h] + \centering + \includegraphics[width=0.2\textwidth]{msb.eps} + \caption{$\pi$-Zeiger des Breitensuchbaums und MSB blau hinterlegt.} + \label{fig:msb} +\end{figure} + +\item +\begin{quote} +Wenn $w(e) \in \set{1, 2, 3}$ für alle $e \in E$, dann ist jeder minimale +Spannbaum von $G$ ein Tiefensuchbaum mit Quelle $s$. +\end{quote} +Sei $w(e) = 1$ für +alle $e \in E$ so gilt das Gegenbeispiel von oben. Also ist die Aussage +falsch. +\points{2} + +\end{tasks} + +\section{Kreissuche} +\begin{tasks} +\item +Wähle Startknoten $s$ und füge ihn in eine neue Queue $Q$ ein. + +Nimm den vordersten Knoten $u$ aus $Q$ und füge seine noch nicht entdeckten (weißen) +Nachbarn ein und färbe sie grau. Wenn einer der Nachbarn schon entdeckt wurde, +also grau ist, gibt es einen Kreis in $G$. +Markiere den Knoten $u$ als abgeschlossen ($black$) und entferne ihn aus $Q$ + +Wiederhole den Schritt solange bis $Q$ leer ist. + +% \begin{pseudocode}[caption=Hello World Algorithmische] +% EinfacherKreis(|Graph| $G$, |Vertex| $s$) +% Initialize($G$, $s$) +% $Q \gets$ new Queue +% $Q.$Enqueue($s$) +% $t_1 \gets nil$ +% $t_2 \gets s$ +% // Comment +% while $Q \neq \emptyset$ do +% $u \gets Q.$Dequeue +% $t_1 \gets t_2$ +% $t_2 \gets u$ +% foreach $v \in Adj[u]$ do +% if $v.color = white$ then +% $Q.$Enqueue($v$) +% $u.color \gets black$ +% if $t_1 = t_2$ then +% return $true$ +% else +% return $false$ +% \end{pseudocode} + +\begin{algorithm}[h] +\centering +\begin{algorithmic} +\alg{EinfacherKreis}{Graph $G$, Vertex $s$} \+ \\ + \alg{Initialize}{$G$, $s$} \com{So wie in der Breitensuche} \\ + $Q \gets$ new \alg{Queue}{} \\ + $Q.$\alg{Enqueue}{$s$} \\ + while $Q \neq \emptyset$ do \+ \\ + $u \gets Q.$\alg{Dequeue}{} \\ + $u.color \gets gray$ \\ + foreach $v \in Adj[u]$ do \+ \\ +\com{Füge alle noch nicht entdeckten Knoten ein} \\ + if $v.color = white$ then \+ \\ + $v.color \gets gray$ \\ + $Q.$\alg{Enqueue}{$v$} \- \\ +\com{Wenn ein Knoten schon entdeckt wurde, gibt es einen Kreis} \\ + else return $true$ \- \\ + $u.color \gets black$ \- \\ + return $false$ +\end{algorithmic} +\end{algorithm} + +\item + Dadurch, dass wir nur Knoten + einfügen, die noch nicht entdeckt wurden, können wir nie auf dem selben Pfad + zu einem Knoten kommen. Das heißt, wenn wir einen schon entdeckten Knoten + finden, haben wir einen Kreis im Graphen gefunden. + + Der Algorithmus kann aufgrund der Struktur des Graphens (kein Multigraph, + keine Selbstkanten) nur Kreise der Länge mindestens 3 finden. + + Jeder Knoten wird nur einmal in die Queue eingefügt und nur einmal herasgenommen. + Somit läuft der Algorithmus in $\Oh(\abs{V})$. + +\item + Solange es weiße Knoten im Graphen gibt, wählen wir einen neuen Startknoten + für diese Zusammenhangskomponente. Als Ausgabe geben wir ein Array von + Tupeln $\tup{s_i, c_i}$ zurück, wobei $s_i$ der Startknoten einer + Zusammenhangskomponente und $c_i \in \set{true, false}$ der Wahrheitswert, ob + ein Kreis in der Komponente existiert. +\end{tasks} + +\section{Eulerwege} +\begin{quote} +Sei $G = \tup{V, E}$ ein ungerichteter, zusammenhängender Graph. Dann gilt: $G$ hat genau +dann einen Eulerweg, wenn die Anzahl an Knoten $v \in V$, für die gilt, dass $deg(v)$ +ungerade ist, genau $0$ oder $2$ ist. +\end{quote} + +\begin{itemize} + \item[$\seilpmi$] + 1. Fall: $0$ Knoten mit ungeradem Grad. Nach dem Satz in der Vorlesung + gibt es einen Eulerkreis. Im Eulerweg sind also Start- und Endknoten + identisch. + + 2. Fall: $2$ Knoten mit ungeradem Grad. Die beiden Knoten bilden den Start- + und Endknoten des Eulerwegs. Die Kante die den Eulerkreis schließen würde + braucht genau zwei Knoten, zu denen sie inzident ist. Nehmen wir diese Kante + weg, ergibt sich eine ungerader Grad an diesen beiden Knoten. + \item[$\implies$] + Ein Graph mit ungerader Anzahl an Knoten mit ungeradem Grad kann nicht + existieren, da die Summe aller Knoten mit ungeradem Grad gerade ist. + + Für alle anderen Fälle gilt, wenn ein Knoten ungeraden Grad hat, dann gibt + es keinen Weg aus dem Knoten heraus, wenn man hineingelaufen ist. + +\end{itemize} + + +\section{Graphmodellierung} + +\end{document}