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\subsection{Aufgabe 3}1\begin{frame}{Aufgabe 3}2Zeigen Sie: Wenn der Graph eines Kreises bipartit ist, dann hat der Kreis gerade LĂ€nge.3\end{frame}45\pgfdeclarelayer{background}6\pgfsetlayers{background,main}7\begin{frame}{Aufgabe 3 - Lösung}8Idee: Knoten abwechselnd fĂ€rben910\tikzstyle{selected edge} = [draw,line width=5pt,-,black!50]11\begin{center}12\adjustbox{max size={\textwidth}{0.8\textheight}}{13\begin{tikzpicture}14\node[vertex] (a) at (0,0) {};15\node[vertex] (b) at (2,0) {};16\node[vertex] (c) at (2,2) {};17\node[vertex] (d) at (0,2) {};18\node[vertex] (e) at (1,4) {};1920\draw (a) -- (b) -- (c) -- (e) -- (d) -- (a);2122\node<2->[vertex, red] (a) at (0,0) {};23\node<3->[vertex, blue] (b) at (2,0) {};24\node<4->[vertex, red] (c) at (2,2) {};25\node<5->[vertex, blue] (e) at (1,4) {};26\node<6->[vertex, red] (d) at (0,2) {};2728\begin{pgfonlayer}{background}29\path<3->[selected edge] (a.center) edge node {} (b.center);30\path<4->[selected edge] (b.center) edge node {} (c.center);31\path<5->[selected edge] (c.center) edge node {} (e.center);32\path<6->[selected edge] (e.center) edge node {} (d.center);33\path<7->[selected edge,lime] (d.center) edge node {} (a.center);34\end{pgfonlayer}35\end{tikzpicture}36}37\end{center}38\end{frame}3940\subsection{Aufgabe 4}41\begin{frame}{Aufgabe 4}42Zeigen Sie: Ein Graph $G$ ist genau dann bipartit, wenn er nur Kreise43gerade LĂ€nge hat.44\end{frame}4546\begin{frame}{Aufgabe 4: Lösung, Teil 1}47\underline{Vor.:} Sei $G = (E, K)$ ein zus. Graph. \pause4849\underline{Beh.:} $G$ ist bipartit $\Rightarrow G$ hat keine Kreis ungerader LĂ€nge \pause5051\underline{Bew.:} durch Widerspruch \pause5253\underline{Annahme:} $G$ hat Kreis ungerader LĂ€nge \pause5455$\xRightarrow[]{A.4}$ Ein Subgraph von $G$ ist nicht bipartit \pause5657$\Rightarrow$ Widerspruch zu \enquote{$G$ ist bipartit} \pause5859$\Rightarrow$ $G$ hat keinen Kreis ungerader LĂ€nge $\blacksquare$60\end{frame}6162\begin{frame}{Aufgabe 4: Lösung, Teil 2}63\underline{Vor.:} Sei $G = (E, K)$ ein zus. Graph. \pause6465\underline{Beh.:} $G$ hat keinen Kreis ungerader LĂ€nge $\Rightarrow G$ ist bipartit \pause6667\underline{Bew.:} Konstruktiv \pause6869FĂ€rbe Graphen mit Breitensuche $\blacksquare$70\end{frame}7172\pgfdeclarelayer{background}73\pgfsetlayers{background,main}74\begin{frame}{Aufgabe 4 - Beispiel}75\tikzstyle{selected edge} = [draw,line width=5pt,-,black!50]76\begin{center}77\adjustbox{max size={\textwidth}{0.8\textheight}}{78\begin{tikzpicture}79\node[vertex] (a) at (1,1) {};80\node[vertex] (b) at (2,0) {};81\node[vertex] (c) at (4,0) {};82\node[vertex] (d) at (1,2) {};83\node[vertex] (e) at (2,2) {};84\node[vertex] (f) at (3,2) {};85\node[vertex] (g) at (2,4) {};86\node[vertex] (h) at (3,3) {};87\node[vertex] (i) at (4,2) {};88\node[vertex] (j) at (1,3) {};8990\draw (a) -- (b);91\draw (a) -- (d);92\draw (b) -- (e);93\draw (b) -- (c);94\draw (c) -- (f);95\draw (d) -- (e);96\draw (d) -- (j);97\draw (e) -- (f);98\draw (f) -- (i);99\draw (g) -- (j);100\draw (g) -- (h);101102\node<2->[vertex, red] (a) at (1,1) {};103\node<3->[vertex, blue] (b) at (2,0) {};104\node<3->[vertex, blue] (d) at (1,2) {};105\node<4->[vertex, red] (c) at (4,0) {};106\node<4->[vertex, red] (e) at (2,2) {};107\node<4->[vertex, red] (j) at (1,3) {};108\node<5->[vertex, blue] (f) at (3,2) {};109\node<5->[vertex, blue] (g) at (2,4) {};110\node<6->[vertex, red] (h) at (3,3) {};111\node<6->[vertex, red] (i) at (4,2) {};112113\begin{pgfonlayer}{background}114\path<3->[selected edge] (a.center) edge node {} (b.center);115\path<3->[selected edge] (a.center) edge node {} (d.center);116\path<4->[selected edge] (b.center) edge node {} (c.center);117\path<4->[selected edge] (b.center) edge node {} (e.center);118\path<4->[selected edge] (d.center) edge node {} (j.center);119\path<4->[selected edge] (d.center) edge node {} (e.center);120\path<5->[selected edge] (j.center) edge node {} (g.center);121\path<5->[selected edge] (e.center) edge node {} (f.center);122\path<5->[selected edge] (c.center) edge node {} (f.center);123\path<6->[selected edge] (g.center) edge node {} (h.center);124\path<6->[selected edge] (f.center) edge node {} (i.center);125\end{pgfonlayer}126\end{tikzpicture}127}128\end{center}129\end{frame}130131\subsection{Aufgabe 9}132\begin{frame}{Aufgabe 9, Teil 1}133Im folgenden sind die ersten drei Graphen $G_1, G_2, G_3$ einer134Folge $(G_n)$ aus Graphen abgebildet. Wie sieht $G_4$ aus?135136\begin{gallery}137\galleryimage{graphs/triangular-1}138\galleryimage{graphs/triangular-2}139\galleryimage{graphs/triangular-3}140\end{gallery}141\end{frame}142143\begin{frame}{Aufgabe 9, Teil 1 (Lösung)}144\begin{center}145\input{graphs/triangular-4}146\end{center}147\end{frame}148149\begin{frame}{Aufgabe 9, Teil 1 (Lösung)}150\begin{center}151\input{graphs/triangular-5}152\end{center}153\end{frame}154155\begin{frame}{Aufgabe 9, Teil 1 (Lösung)}156\begin{center}157\input{graphs/triangular-6}158\end{center}159\end{frame}160161\begin{frame}{Aufgabe 9, Teil 2}162Wie viele Ecken und wie viele Kanten hat $G_i$?163164\begin{gallery}165\galleryimage{graphs/triangular-1}166\galleryimage{graphs/triangular-2}167\galleryimage{graphs/triangular-3}168\end{gallery}169\end{frame}170171\begin{frame}{Aufgabe 9, Teil 2: Antwort}172Ecken:173174\[|E_n| = |E_{n-1}| + (n+1) = \sum_{i=1}^{n+1} i = \frac{n^2 + 2n+2}{2}\]175176Kanten:177178\begin{align}179|K_n| &= |K_{n-1}| + \underbrace{((n+1)-1)+2}_{\text{auĂen}} + (n-1) \cdot 2\\180&= |K_{n-1}| + n+2+2n-2\\181&= |K_{n-1}| + 3n\\182&= \sum_{i=1}^{n} 3i = 3 \sum_{i=1}^{n} i \\183&= 3 \frac{n^2 + n}{2}184\end{align}185\end{frame}186187\begin{frame}{Aufgabe 9, Teil 3}188Gebe $G_i$ formal an.189190\begin{gallery}191\galleryimage{graphs/triangular-1}192\galleryimage{graphs/triangular-2}193\galleryimage{graphs/triangular-3}194\end{gallery}195\end{frame}196197\begin{frame}{Aufgabe 9, Teil 3 (Lösung)}198Gebe $G_n$ formal an.199200\begin{gallery}201\galleryimage{graphs/triangular-1}202\galleryimage{graphs/triangular-2}203\galleryimage{graphs/triangular-3}204\end{gallery}205206\begin{align*}207E_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}\\208K_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)}\\209G_n &= (E_n, K_n)210\end{align*}211212\end{frame}213214\begin{frame}{{\sc RectangleFreeColoring}}215\begin{block}{{\sc RectangleFreeColoring}}216Gegeben ist $n, m \in \mathbb{N}_{\geq 1}$ und ein217ungerichteter Graph $G = (E, K)$ mit218\[E = \Set{e_{x,y} | 1 \leq x \leq n \land 1 \leq y \leq m}\]219und220\[K = \Set{k=\Set{e_{x,y}, e_{x',y'}} \in E \times E : |x-x'| + |y-y'| = 1} \]221222FĂ€rbe die Ecken von $G$ mit einer minimalen Anzahl von Farben so, dass gilt:223\begin{align*}224\forall e_{x,y}, e_{x',y'} \in E: (x \neq x' \land y \neq y') \Rightarrow\\225\neg \left (c(e_{x,y}) = c(e_{x',y'}) = c(e_{x',y}) = c(e_{x,y'}) \right )226\end{align*}227\end{block}228\end{frame}229230\begin{frame}{{\sc RectangleFreeColoring}}231$4 \times 4$ - Instanz:\\232233\vspace{1cm}234\begin{tikzpicture}235\newcommand{\n}{4}236\newcommand{\m}{4}237\foreach \x in {1, ..., \n}{238\foreach \y in {1, ..., \m}{239\node[vertex] (n-\x-\y) at (\x,\y) {};240}241}242243\foreach \x in {1, ..., \n}{244\foreach \y in {1, ..., \m}{245\ifthenelse{\x<\n}{\draw (\x,\y) -- (\x+1,\y);}{}246}247}248\foreach \y in {1, ..., \m}{249\foreach \x in {1, ..., \n}{250\ifthenelse{\y<\m}{\draw (\x,\y) -- (\x,\y+1);}{}251}252}253254\node[vertex,blue] (n-1-1) at (1,1) {};255\node[vertex,blue] (n-2-1) at (2,1) {};256\node[vertex,blue] (n-3-1) at (3,1) {};257\node[vertex,red] (n-4-1) at (4,1) {};258259\node[vertex,blue] (n-1-1) at (1,2) {};260\node[vertex,red] (n-2-1) at (2,2) {};261\node[vertex,red] (n-3-1) at (3,2) {};262\node[vertex,blue] (n-4-1) at (4,2) {};263264\node[vertex,red] (n-1-1) at (1,3) {};265\node[vertex,blue] (n-2-1) at (2,3) {};266\node[vertex,red] (n-3-1) at (3,3) {};267\node[vertex,blue] (n-4-1) at (4,3) {};268269270\node[vertex,red] (n-1-1) at (1,4) {};271\node[vertex,red] (n-2-1) at (2,4) {};272\node[vertex,blue] (n-3-1) at (3,4) {};273\node[vertex,blue] (n-4-1) at (4,4) {};274\end{tikzpicture}275\end{frame}276277\subsection{Bildquellen}278\begin{frame}{Bildquellen}279\begin{itemize}280\item \href{http://commons.wikimedia.org/wiki/File:Hypercube.svg}{http://commons.wikimedia.org/wiki/File:Hypercube.svg}281\item \href{http://commons.wikimedia.org/wiki/File:Konigsberg\_bridges.png}{http://commons.wikimedia.org/wiki/File:Konigsberg\_bridges.png}282\item \href{http://commons.wikimedia.org/wiki/File:Unit\_disk\_graph.svg}{http://commons.wikimedia.org/wiki/File:Unit\_disk\_graph.svg}283\item \href{http://goo.gl/maps/WnXRh}{Google Maps} (Grafiken \TCop 2013 Cnes/Spot Image, DigitalGlobe)284\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}285\end{itemize}286\end{frame}287288\subsection{Literatur}289\begin{frame}{Literatur}290\begin{itemize}291\item A. Beutelspacher: \textit{Diskrete Mathematik fĂŒr Einsteiger}, 4. Auflage, ISBN 978-3-8348-1248-3292\end{itemize}293\end{frame}294295\subsection{Folien, \LaTeX und Material}296\begin{frame}{Folien, \LaTeX und Material}297Der Foliensatz und die \LaTeX und Ti\textit{k}Z-Quellen sind unter298299\href{https://github.com/MartinThoma/LaTeX-examples/tree/master/presentations/Diskrete-Mathematik}{github.com/MartinThoma/LaTeX-examples/tree/master/presentations/Diskrete-Mathematik}300\\301302Kurz-URL:303\href{http://goo.gl/uTgam}{goo.gl/uTgam}304\end{frame}305306307