Aufgabe 2 und 3 fertig
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\documentclass{ngexrcs}
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\usepackage{hyperref}
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\title{1. Übungsblatt}
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\subject{Algorithmische Graphentheorie}
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\author{Jasper Gude \and Pia Rötgers}
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\begin{document}
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\maketitle
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\points[\qquad]{20}
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\section{Spannbäume \& Breitensuche}
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Sei $G = \tup{V, E}$ ein zusammenhängender Graph mit Kantengewichten $w: E \to \NN$
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und $s \in V \deg$ ein ausgezeichneter Knoten.
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\begin{enumerate}
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\item
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\begin{quote}
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Wenn $w(e) = 1$ für alle $e \in E$, dann ist der Breitensuchbaum mit
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Quelle $s$ ein minimaler Spannbaum.
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\end{quote}
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Die Breitensuche berechnet in diesem
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Fall den kürzesten Weg von jedem Knoten zum Knoten $s$, also den
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Breitensuchbaum. Dieser spannt also einen minimalen Spannbaum auf.
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\points{2}
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\item
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\begin{quote}
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Wenn $w(e) = 1$ für alle $e \in E$, dann ist jeder minimale Spannbaum
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von $G$ ein Breitensuchbaum mit Quelle $s$.
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\end{quote}
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Falsch, siehe \autoref{fig:msb}.
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\points{2}
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.2\textwidth]{msb.eps}
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\caption{$\pi$-Zeiger des Breitensuchbaums und MSB blau hinterlegt.}
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\label{fig:msb}
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\end{figure}
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\item
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\begin{quote}
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Wenn $w(e) \in \set{1, 2, 3}$ für alle $e \in E$, dann ist jeder minimale
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Spannbaum von $G$ ein Tiefensuchbaum mit Quelle $s$.
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\end{quote}
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Sei $w(e) = 1$ für
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alle $e \in E$ so gilt das Gegenbeispiel von oben. Also ist die Aussage
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falsch.
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\points{2}
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\end{enumerate}
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\section{Kreissuche}
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\begin{enumerate}
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\item
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Wähle Startknoten $s$ und füge ihn in eine neue Queue $Q$ ein.
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Nimm den vordersten Knoten aus $Q$ und füge seine noch nicht verbrauchten
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Nachbarn ein. Markiere diesen Knoten anschließend als verbraucht ($black$).
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Wiederhole den Schritt solange bis $Q$ leer ist. Merke dir dabei die zwei
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letzten entnommenen Knoten. Sind sie gleich, hat der Graph einen
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einfachen Kreis.
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\begin{algorithmic}
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\alg{EinfacherKreis}(Graph $G$, Vertex $s$) \+ \\
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\alg{Initialize}($G$, $s$) \\
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$Q \gets$ new \alg{Queue}() \\
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$Q.$\alg{Enqueue}($s$) \\
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$t_1 \gets nil$ \\
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$t_2 \gets s$ \\
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while $Q \neq \emptyset$ do \+ \\
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$u \gets Q.$\alg{Dequeue}() \\
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$t_1 \gets t_2$ \\
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$t_2 \gets u$ \\
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foreach $v \in Adj[u]$ do \+ \\
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if $v.color = white$ then \+ \\
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$Q.$\alg{Enqueue}($v$) \-\- \\
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$u.color \gets black$ \- \\
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if $t_1 = t_2$ then \+ \\
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return $true$ \- \\
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else \+ \\
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return $false$
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\end{algorithmic}
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\end{enumerate}
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\section{Eulerwege}
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\section{Graphmodellierung}
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\end{document}
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OUTPUT agt_übung_1.pdf
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INPUT ./ads_übung_1.aux
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INPUT ./agt_übung_1.aux
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|||
This is LuaHBTeX, Version 1.22.0 (TeX Live 2025) (format=lualatex 2026.2.8) 17 APR 2026 18:49
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This is LuaHBTeX, Version 1.22.0 (TeX Live 2025) (format=lualatex 2026.2.8) 19 APR 2026 18:21
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restricted system commands enabled.
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**/home/never/Documents/InNa/26_SoSe/AGT/agt_exercise/übung_1/ads_übung_1.tex
|
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**/home/never/Documents/InNa/26_SoSe/AGT/agt_exercise/übung_1/agt_übung_1.tex
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||||
(/home/never/Documents/InNa/26_SoSe/AGT/agt_exercise/übung_1/ads_übung_1.tex
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(/home/never/Documents/InNa/26_SoSe/AGT/agt_exercise/übung_1/agt_übung_1.tex
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but found:
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so I'm assuming it got fixed.
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LaTeX2e <2025-11-01>
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\documentclass{ngexrcs}
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\usepackage{hyperref}
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\title{1. Übungsblatt}
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\subject{Algorithmische Graphentheorie}
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\author{Jasper Gude \and Pia Röttgers}
|
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\begin{document}
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\maketitle
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\points[\qquad]{20}
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\section{Spannbäume \& Breitensuche}
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Sei $G = \tup{V, E}$ ein zusammenhängender Graph mit Kantengewichten $w: E \to \NN$
|
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und $s \in V \deg$ ein ausgezeichneter Knoten.
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\begin{tasks}
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\item
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\begin{quote}
|
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Wenn $w(e) = 1$ für alle $e \in E$, dann ist der Breitensuchbaum mit
|
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Quelle $s$ ein minimaler Spannbaum.
|
||||
\end{quote}
|
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|
||||
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|
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||||
\item
|
||||
\begin{quote}
|
||||
Wenn $w(e) = 1$ für alle $e \in E$, dann ist jeder minimale Spannbaum
|
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von $G$ ein Breitensuchbaum mit Quelle $s$.
|
||||
\end{quote}
|
||||
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|
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|
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|
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\centering
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\includegraphics[width=0.2\textwidth]{msb.eps}
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\caption{$\pi$-Zeiger des Breitensuchbaums und MSB blau hinterlegt.}
|
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\label{fig:msb}
|
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\end{figure}
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||||
\item
|
||||
\begin{quote}
|
||||
Wenn $w(e) \in \set{1, 2, 3}$ für alle $e \in E$, dann ist jeder minimale
|
||||
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|
||||
\end{quote}
|
||||
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|
||||
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|
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falsch.
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\section{Kreissuche}
|
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\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)
|
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Nachbarn ein und färbe sie grau. Wenn einer der Nachbarn schon entdeckt wurde,
|
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also grau ist, gibt es einen Kreis in $G$.
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Markiere den Knoten $u$ als abgeschlossen ($black$) und entferne ihn aus $Q$
|
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|
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% \begin{pseudocode}[caption=Hello World Algorithmische]
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|
||||
% $Q.$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}
|
||||
Loading…
Add table
Add a link
Reference in a new issue