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Aufgabe 1 korrektur und Overfull \hbox in 2 beheben

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Never Gude 2026-05-18 19:20:02 +02:00
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übung_4/aufgabe_1.tex
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@ -11,23 +11,33 @@
\end{figure}
\item
Das Problem kann als Maximalflussproblem modelliert werden. Dafür haben wir
eine Quelle $s$, eine Senke $t$ und Knoten $v_i$ mit $i \in \NN_{\leq n}$.
Die Eingangskapazität eines Knoten $v_i$ entspricht dem Eingangsgrad $e_i$.
Die Ausgangskapazität entspricht dem Ausgangsgrad $a_i$. Siehe \autoref{fig:task1b}.
eine Quelle $s$, eine Senke $t$ und Knoten $v_{i_{e/a}}$ mit $i \in \NN_{\leq n}$.
Wenn es einen vollständigen Fluss gibt, gibt es eine Lösung für das Problem,
Die Eingangskapazität eines Knoten $v_{i_a}$ entspricht dem Ausgangsgrad $a_i$.
Die Ausgangskapazität eines Knoten $v_{i_e}$ entspricht dem Eingangsgrad $e_i$.
Für die Kantenkapazität $c \in \NN$ einer Kante $\tup{v_{i_a}, v_{j_e}}$ mit $i, j \leq n$ gilt,
wenn $i = j$, sind Selbstkanten erlaubt, wenn $c > 0$.
Ausserdem sind Mehrfachkanten erlaubt, wenn $c > 1$.
Siehe \autoref{fig:task1b}.
Wenn es einen vollständigen Fluss gibt, also alle in $t$ eingehenden Kanten vollständig benutzt werden, gibt es eine Lösung für das Problem,
sonst nicht.
\points{4}
\begin{figure}
\centering
\includegraphics[page=2, width=0.5\textwidth]{figures.pdf}
\includegraphics[page=2, width=0.8\textwidth]{figures.pdf}
\caption{Das Problem als Maximalflussproblem.}
\label{fig:task1b}
\end{figure}
\item
Die Modellierung ist korrekt, da der Eingangs- und Ausgangsgrad jedes Knotens $v_i$
von oben beschränkt wird von der Eingangskapazität $e_i$ bzw. Ausgangskapazität $a_i$.
Die Modellierung ist korrekt, da die Eingangs- und Ausgangsgrade der Knoten
$v_i$ im Graphen durch die Kantenkapazitäten $a_i$ und $e_i$ Fluss beschränkt werden.
Zudem erfüllt die Flusserhaltung den Zweck, dass die Summe der Eingangsgrade gleich der Summe der Ausgangsgrade sind, was in einem Graphen erfüllt sein muss.
Durch wählen der Kantenkapazitäten der Kanten $(v_{i_a}, v_{j_e})$ im Fluss
können Selbst- und Mehrfachkanten ein- oder ausgeschlossen werden.
Damit entsprechen diese Kanten im Fluss den Kanten im Graphen.
\points{2}
\end{tasks}

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übung_4/aufgabe_2.tex
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@ -23,7 +23,11 @@ entspricht, da dann alle neuen von $s$ ausgehenden Kanten vollständig ausgelast
Also ist $f'$ ein zulässiger $s$-$t$-Fluss mit Wert $|f'|=\sum_{v:b(v)>0}b(v)$
\item Sei $f'$ ein $s$-$t$-Fluss in $G'$. Da sein Wert maximal ist, sind alle Kanten zu $s$ und $t$ voll ausgelastet.\\
Für die Einschränkung $f:=f'\vert_E$ betrachtet man die Flusserhaltung jedes Knotens $v\in V$ in $G'$:
$$\textnormal{Nettozufluss}_f(v)=\textnormal{Nettozufluss}_{f'}(v)-[\textnormal{Beitrag der neuen Kante]}=0-(-b(v))=b(v)$$
\begin{align*}
\textnormal{Nettozufluss}_f(v) &= \textnormal{Nettozufluss}_{f'}(v)-[\textnormal{Beitrag der neuen Kante]} \\
&= 0-(-b(v)) \\
&= b(v)
\end{align*}
Also ist $f$ ein zulässiger b-Fluss in $G$.
\end{enumerate}
\points{2}

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416 256 m
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416 256 m
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416 256 m
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416 256 m
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416 256 m
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416 256 m
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416 64 m
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416 64 m
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416 64 m
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416 64 m
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416 64 m
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560 112 m
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560 112 m
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560 112 m
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560 112 m
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560 112 m
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608 256 m
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608 256 m
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608 256 m
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512 416 m
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320 320 m
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320 192 m
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320 192 m
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320 256 m
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320 256 m
448 192 l
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320 256 m
448 256 l
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320 320 m
448 320 l
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320 320 m
448 256 l
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320 256 m
448 256 l
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320 192 m
448 256 l
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192 256 m
320 320 l
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192 256 m
320 256 l
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192 256 m
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\item[\color{dark purple}{Lila Kanten}] Kantenkapazität $1$ wenn keine Mehrfachkanten erlaubt sind.
\item[\color{dark gray}{Graue Kanten}] Kantenkapazität $0$ wenn keine Selbstkanten erlaubt sind.
\end{description}</text>
</page>
</ipe>