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\documentclass[a4paper, 10pt, pagesize, smallheadings]{article}
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\usepackage{graphicx}
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%\usepackage[latin1]{inputenc}
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\usepackage{amsmath, amsthm, amssymb}
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\usepackage{typearea}
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\usepackage{algorithm}
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\usepackage{algorithmic}
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\usepackage{fullpage}
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\usepackage{mathtools}
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\usepackage{multirow}
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\usepackage[all]{xy}
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\addtolength{\voffset}{-20pt}
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\title{Spieltheorie \"Ubung 4}
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\author{Eugen Sawin}
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\renewcommand{\familydefault}{\sfdefault}
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\newcommand{\E}{\mathcal{E}}
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\newcommand{\R}{\mathcal{R}}
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%\include{pythonlisting}
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\pagestyle{empty}
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\begin{document}
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\maketitle
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%
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\section*{Aufgabe 4.1}
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(a) Das Picknickspiel ist das Spiel $G=\langle\{1,2\},(A,B),(u_i)\rangle$ mit $A=B=\{p_1,p_2,p_3,p_4,p_5\}$ und das $(u_i)$ definiert durch die folgende Matrix.\\\\
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\begin{tabular}{rr|c|c|c|c|c|}
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\multicolumn{1}{r}{}
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& \multicolumn{1}{r}{}
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& \multicolumn{5}{c}{\emph{Spieler 2}} \\
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\multicolumn{1}{r}{}
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& \multicolumn{1}{r}{}
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& \multicolumn{1}{c}{$p_1$}
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& \multicolumn{1}{c}{$p_2$}
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& \multicolumn{1}{c}{$p_3$}
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& \multicolumn{1}{c}{$p_4$}
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& \multicolumn{1}{c}{$p_5$} \\
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\cline{3-7}
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\multirow{5}{*}{\emph{Spieler 1}}
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& $p_1$ & $0,0$ & $1,2$ & $1,3$ & $1,4$ & $1,5$ \\\cline{3-7}
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& $p_2$ & $2,1$ & $0,0$ & $2,3$ & $2,4$ & $2,5$ \\\cline{3-7}
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& $p_3$ & $3,1$ & $3,2$ & $0,0$ & $3,4$ & $3,5$ \\\cline{3-7}
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& $p_4$ & $4,1$ & $4,2$ & $4,3$ & $0,0$ & $4,5$ \\\cline{3-7}
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& $p_5$ & $5,1$ & $5,2$ & $5,3$ & $5,4$ & $0,0$ \\\cline{3-7}
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\end{tabular}\\\\\\
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%
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Seit $(\alpha,\beta)$ ein NG mit Nutzenprofil $(u,v)$ im Spiel $G$, dann ist das LCP durch folgende Ungleichungen definiert.
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\begin{align}
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0&\leq u-\beta(p_1)\cdot 0-\beta(p_2)\cdot 1-\beta(p_3)\cdot 1-\beta(p_4)\cdot 1-\beta(p_5)\cdot 1\\
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0&\leq u-\beta(p_1)\cdot 2-\beta(p_2)\cdot 0-\beta(p_3)\cdot 2-\beta(p_4)\cdot 2-\beta(p_5)\cdot 2\\
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0&\leq u-\beta(p_1)\cdot 3-\beta(p_2)\cdot 3-\beta(p_3)\cdot 0-\beta(p_4)\cdot 3-\beta(p_5)\cdot 3\\
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0&\leq u-\beta(p_1)\cdot 4-\beta(p_2)\cdot 4-\beta(p_3)\cdot 4-\beta(p_4)\cdot 0-\beta(p_5)\cdot 4\\
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0&\leq u-\beta(p_1)\cdot 5-\beta(p_2)\cdot 5-\beta(p_3)\cdot 5-\beta(p_4)\cdot 5-\beta(p_5)\cdot 0\\
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0&\leq v-\alpha(p_1)\cdot 0-\alpha(p_2)\cdot 1-\alpha(p_3)\cdot 1-\alpha(p_4)\cdot 1-\alpha(p_5)\cdot 1\\
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0&\leq v-\alpha(p_1)\cdot 2-\alpha(p_2)\cdot 0-\alpha(p_3)\cdot 2-\alpha(p_4)\cdot 2-\alpha(p_5)\cdot 2\\
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0&\leq v-\alpha(p_1)\cdot 3-\alpha(p_2)\cdot 3-\alpha(p_3)\cdot 0-\alpha(p_4)\cdot 3-\alpha(p_5)\cdot 3\\
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0&\leq v-\alpha(p_1)\cdot 4-\alpha(p_2)\cdot 4-\alpha(p_3)\cdot 4-\alpha(p_4)\cdot 0-\alpha(p_5)\cdot 4\\
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0&\leq v-\alpha(p_1)\cdot 5-\alpha(p_2)\cdot 5-\alpha(p_3)\cdot 5-\alpha(p_4)\cdot 5-\alpha(p_5)\cdot 0\\
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0&\leq \alpha(a)&\forall a\in A\\
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0&\leq \beta(b)&\forall b\in B\\
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1&=\alpha(p_1)+\alpha(p_2)+\alpha(p_3)+\alpha(p_4)+\alpha(p_5)\\
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1&=\beta(p_1)+\beta(p_2)+\beta(p_3)+\beta(p_4)+\beta(p_5)\\
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0&=\alpha(a)\cdot(u-U_1(a,\beta))&\forall a\in A\\
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0&=\beta(b)\cdot(v-U_2(\alpha,b))&\forall b\in B
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\end{align}
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Gleichungen (15) und (16) sind zu expandieren, wie in den Gleichungen (1)-(10) gezeigt. Au\ss erdem muss f\"ur die Normalform hierbei zus\"atzliche Variablen eingef\"uhrt werden. Wir unterlassen beides um die \"Ubersichtlichkeit zu erhalten.
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\end{document}
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