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\subsection{Aufgabe 3}
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\begin{frame}{Aufgabe 3}
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Zeigen Sie: Wenn der Graph eines Kreises bipartit ist, dann hat der Kreis gerade LĂ€nge.
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\end{frame}
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\pgfdeclarelayer{background}
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\pgfsetlayers{background,main}
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\begin{frame}{Aufgabe 3 - Lösung}
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Idee: Knoten abwechselnd fÀrben
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\tikzstyle{selected edge} = [draw,line width=5pt,-,black!50]
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\begin{center}
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\adjustbox{max size={\textwidth}{0.8\textheight}}{
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\begin{tikzpicture}
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\node[vertex] (a) at (0,0) {};
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\node[vertex] (b) at (2,0) {};
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\node[vertex] (c) at (2,2) {};
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\node[vertex] (d) at (0,2) {};
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\node[vertex] (e) at (1,4) {};
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\draw (a) -- (b) -- (c) -- (e) -- (d) -- (a);
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\node<2->[vertex, red] (a) at (0,0) {};
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\node<3->[vertex, blue] (b) at (2,0) {};
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\node<4->[vertex, red] (c) at (2,2) {};
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\node<5->[vertex, blue] (e) at (1,4) {};
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\node<6->[vertex, red] (d) at (0,2) {};
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\begin{pgfonlayer}{background}
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\path<3->[selected edge] (a.center) edge node {} (b.center);
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\path<4->[selected edge] (b.center) edge node {} (c.center);
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\path<5->[selected edge] (c.center) edge node {} (e.center);
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\path<6->[selected edge] (e.center) edge node {} (d.center);
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\path<7->[selected edge,lime] (d.center) edge node {} (a.center);
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\end{pgfonlayer}
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\end{tikzpicture}
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}
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\end{center}
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\end{frame}
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\subsection{Aufgabe 4}
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\begin{frame}{Aufgabe 4}
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Zeigen Sie: Ein Graph $G$ ist genau dann bipartit, wenn er nur Kreise
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gerade LĂ€nge hat.
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\end{frame}
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\begin{frame}{Aufgabe 4: Lösung, Teil 1}
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\underline{Vor.:} Sei $G = (E, K)$ ein zus. Graph. \pause
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\underline{Beh.:} $G$ ist bipartit $\Rightarrow G$ hat keine Kreis ungerader LĂ€nge \pause
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\underline{Bew.:} durch Widerspruch \pause
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\underline{Annahme:} $G$ hat Kreis ungerader LĂ€nge \pause
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$\xRightarrow[]{A.4}$ Ein Subgraph von $G$ ist nicht bipartit \pause
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$\Rightarrow$ Widerspruch zu \enquote{$G$ ist bipartit} \pause
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$\Rightarrow$ $G$ hat keinen Kreis ungerader LĂ€nge $\blacksquare$
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\end{frame}
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\begin{frame}{Aufgabe 4: Lösung, Teil 2}
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\underline{Vor.:} Sei $G = (E, K)$ ein zus. Graph. \pause
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\underline{Beh.:} $G$ hat keinen Kreis ungerader LĂ€nge $\Rightarrow G$ ist bipartit \pause
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\underline{Bew.:} Konstruktiv \pause
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FĂ€rbe Graphen mit Breitensuche $\blacksquare$
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\end{frame}
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\pgfdeclarelayer{background}
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\pgfsetlayers{background,main}
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\begin{frame}{Aufgabe 4 - Beispiel}
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\tikzstyle{selected edge} = [draw,line width=5pt,-,black!50]
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\begin{center}
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\adjustbox{max size={\textwidth}{0.8\textheight}}{
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\begin{tikzpicture}
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\node[vertex] (a) at (1,1) {};
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\node[vertex] (b) at (2,0) {};
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\node[vertex] (c) at (4,0) {};
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\node[vertex] (d) at (1,2) {};
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\node[vertex] (e) at (2,2) {};
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\node[vertex] (f) at (3,2) {};
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\node[vertex] (g) at (2,4) {};
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\node[vertex] (h) at (3,3) {};
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\node[vertex] (i) at (4,2) {};
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\node[vertex] (j) at (1,3) {};
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\draw (a) -- (b);
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\draw (a) -- (d);
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\draw (b) -- (e);
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\draw (b) -- (c);
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\draw (c) -- (f);
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\draw (d) -- (e);
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\draw (d) -- (j);
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\draw (e) -- (f);
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\draw (f) -- (i);
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\draw (g) -- (j);
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\draw (g) -- (h);
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\node<2->[vertex, red] (a) at (1,1) {};
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\node<3->[vertex, blue] (b) at (2,0) {};
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\node<3->[vertex, blue] (d) at (1,2) {};
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\node<4->[vertex, red] (c) at (4,0) {};
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\node<4->[vertex, red] (e) at (2,2) {};
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\node<4->[vertex, red] (j) at (1,3) {};
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\node<5->[vertex, blue] (f) at (3,2) {};
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\node<5->[vertex, blue] (g) at (2,4) {};
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\node<6->[vertex, red] (h) at (3,3) {};
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\node<6->[vertex, red] (i) at (4,2) {};
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\begin{pgfonlayer}{background}
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\path<3->[selected edge] (a.center) edge node {} (b.center);
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\path<3->[selected edge] (a.center) edge node {} (d.center);
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\path<4->[selected edge] (b.center) edge node {} (c.center);
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\path<4->[selected edge] (b.center) edge node {} (e.center);
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\path<4->[selected edge] (d.center) edge node {} (j.center);
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\path<4->[selected edge] (d.center) edge node {} (e.center);
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\path<5->[selected edge] (j.center) edge node {} (g.center);
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\path<5->[selected edge] (e.center) edge node {} (f.center);
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\path<5->[selected edge] (c.center) edge node {} (f.center);
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\path<6->[selected edge] (g.center) edge node {} (h.center);
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\path<6->[selected edge] (f.center) edge node {} (i.center);
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\end{pgfonlayer}
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\end{tikzpicture}
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}
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\end{center}
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\end{frame}
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\subsection{Aufgabe 9}
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\begin{frame}{Aufgabe 9, Teil 1}
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Im folgenden sind die ersten drei Graphen $G_1, G_2, G_3$ einer
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Folge $(G_n)$ aus Graphen abgebildet. Wie sieht $G_4$ aus?
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\begin{gallery}
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\galleryimage{graphs/triangular-1}
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\galleryimage{graphs/triangular-2}
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\galleryimage{graphs/triangular-3}
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\end{gallery}
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\end{frame}
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\begin{frame}{Aufgabe 9, Teil 1 (Lösung)}
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\begin{center}
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\input{graphs/triangular-4}
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\end{center}
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\end{frame}
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\begin{frame}{Aufgabe 9, Teil 1 (Lösung)}
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\begin{center}
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\input{graphs/triangular-5}
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\end{center}
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\end{frame}
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\begin{frame}{Aufgabe 9, Teil 1 (Lösung)}
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\begin{center}
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\input{graphs/triangular-6}
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\end{center}
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\end{frame}
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\begin{frame}{Aufgabe 9, Teil 2}
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Wie viele Ecken und wie viele Kanten hat $G_i$?
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\begin{gallery}
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\galleryimage{graphs/triangular-1}
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\galleryimage{graphs/triangular-2}
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\galleryimage{graphs/triangular-3}
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\end{gallery}
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\end{frame}
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\begin{frame}{Aufgabe 9, Teil 2: Antwort}
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Ecken:
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\[|E_n| = |E_{n-1}| + (n+1) = \sum_{i=1}^{n+1} i = \frac{n^2 + 2n+2}{2}\]
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Kanten:
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\begin{align}
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|K_n| &= |K_{n-1}| + \underbrace{((n+1)-1)+2}_{\text{außen}} + (n-1) \cdot 2\\
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&= |K_{n-1}| + n+2+2n-2\\
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&= |K_{n-1}| + 3n\\
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&= \sum_{i=1}^{n} 3i = 3 \sum_{i=1}^{n} i \\
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&= 3 \frac{n^2 + n}{2}
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\end{align}
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\end{frame}
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\begin{frame}{Aufgabe 9, Teil 3}
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Gebe $G_i$ formal an.
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\begin{gallery}
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\galleryimage{graphs/triangular-1}
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\galleryimage{graphs/triangular-2}
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\galleryimage{graphs/triangular-3}
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\end{gallery}
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\end{frame}
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\begin{frame}{Aufgabe 9, Teil 3 (Lösung)}
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Gebe $G_n$ formal an.
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\begin{gallery}
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\galleryimage{graphs/triangular-1}
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\galleryimage{graphs/triangular-2}
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\galleryimage{graphs/triangular-3}
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\end{gallery}
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\begin{align*}
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E_n &= \Set{e_{x,y} | y \in 1, \dots, n;\; x \in y, \dots, 2 \cdot n - y \text{ mit } x-y \equiv 0 \mod 2}\\
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K_n &= \Set{\Set{e_{x,y}, e_{i,j}} \in E_n^2 | (x+2=i \land y=j) \lor (x+1=i \land y\pm1=j)}\\
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G_n &= (E_n, K_n)
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\end{align*}
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\end{frame}
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\begin{frame}{{\sc RectangleFreeColoring}}
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\begin{block}{{\sc RectangleFreeColoring}}
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Gegeben ist $n, m \in \mathbb{N}_{\geq 1}$ und ein
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ungerichteter Graph $G = (E, K)$ mit
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\[E = \Set{e_{x,y} | 1 \leq x \leq n \land 1 \leq y \leq m}\]
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und
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\[K = \Set{k=\Set{e_{x,y}, e_{x',y'}} \in E \times E : |x-x'| + |y-y'| = 1} \]
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FĂ€rbe die Ecken von $G$ mit einer minimalen Anzahl von Farben so, dass gilt:
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\begin{align*}
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\forall e_{x,y}, e_{x',y'} \in E: (x \neq x' \land y \neq y') \Rightarrow\\
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\neg \left (c(e_{x,y}) = c(e_{x',y'}) = c(e_{x',y}) = c(e_{x,y'}) \right )
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\end{align*}
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\end{block}
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\end{frame}
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\begin{frame}{{\sc RectangleFreeColoring}}
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$4 \times 4$ - Instanz:\\
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\vspace{1cm}
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\begin{tikzpicture}
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\newcommand{\n}{4}
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\newcommand{\m}{4}
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\foreach \x in {1, ..., \n}{
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\foreach \y in {1, ..., \m}{
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\node[vertex] (n-\x-\y) at (\x,\y) {};
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}
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}
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\foreach \x in {1, ..., \n}{
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\foreach \y in {1, ..., \m}{
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\ifthenelse{\x<\n}{\draw (\x,\y) -- (\x+1,\y);}{}
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}
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}
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\foreach \y in {1, ..., \m}{
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\foreach \x in {1, ..., \n}{
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\ifthenelse{\y<\m}{\draw (\x,\y) -- (\x,\y+1);}{}
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}
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}
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\node[vertex,blue] (n-1-1) at (1,1) {};
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\node[vertex,blue] (n-2-1) at (2,1) {};
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\node[vertex,blue] (n-3-1) at (3,1) {};
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\node[vertex,red] (n-4-1) at (4,1) {};
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\node[vertex,blue] (n-1-1) at (1,2) {};
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\node[vertex,red] (n-2-1) at (2,2) {};
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\node[vertex,red] (n-3-1) at (3,2) {};
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\node[vertex,blue] (n-4-1) at (4,2) {};
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\node[vertex,red] (n-1-1) at (1,3) {};
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\node[vertex,blue] (n-2-1) at (2,3) {};
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\node[vertex,red] (n-3-1) at (3,3) {};
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\node[vertex,blue] (n-4-1) at (4,3) {};
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\node[vertex,red] (n-1-1) at (1,4) {};
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\node[vertex,red] (n-2-1) at (2,4) {};
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\node[vertex,blue] (n-3-1) at (3,4) {};
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\node[vertex,blue] (n-4-1) at (4,4) {};
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\end{tikzpicture}
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\end{frame}
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\subsection{Bildquellen}
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\begin{frame}{Bildquellen}
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\begin{itemize}
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\item \href{http://commons.wikimedia.org/wiki/File:Hypercube.svg}{http://commons.wikimedia.org/wiki/File:Hypercube.svg}
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\item \href{http://commons.wikimedia.org/wiki/File:Konigsberg\_bridges.png}{http://commons.wikimedia.org/wiki/File:Konigsberg\_bridges.png}
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\item \href{http://commons.wikimedia.org/wiki/File:Unit\_disk\_graph.svg}{http://commons.wikimedia.org/wiki/File:Unit\_disk\_graph.svg}
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\item \href{http://goo.gl/maps/WnXRh}{Google Maps} (Grafiken \TCop 2013 Cnes/Spot Image, DigitalGlobe)
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\item \href{http://cf.drafthouse.com/\_uploads/galleries/29140/good_will\_hunting\_3.jpg}{cf.drafthouse.com/\_uploads/galleries/29140/good\_will\_hunting\_3.jpg}
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\end{itemize}
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\end{frame}
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\subsection{Literatur}
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\begin{frame}{Literatur}
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\begin{itemize}
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\item A. Beutelspacher: \textit{Diskrete Mathematik fĂŒr Einsteiger}, 4. Auflage, ISBN 978-3-8348-1248-3
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\end{itemize}
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\end{frame}
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\subsection{Folien, \LaTeX und Material}
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\begin{frame}{Folien, \LaTeX und Material}
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Der Foliensatz und die \LaTeX und Ti\textit{k}Z-Quellen sind unter
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\href{https://github.com/MartinThoma/LaTeX-examples/tree/master/presentations/Diskrete-Mathematik}{github.com/MartinThoma/LaTeX-examples/tree/master/presentations/Diskrete-Mathematik}
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\\
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Kurz-URL:
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\href{http://goo.gl/uTgam}{goo.gl/uTgam}
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\end{frame}
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