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ad/AD-Gruppe_8_Koehler_Krabbe_Martens_Blatt6.tex
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ad/AD-Gruppe_8_Koehler_Krabbe_Martens_Blatt6.tex
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\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
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\usepackage[T1]{fontenc}
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\usepackage[utf8]{inputenc}
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\usepackage[ngerman]{babel}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{bytefield}
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\usepackage{paralist}
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\usepackage{gauss}
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\usepackage{pgfplots}
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\usepackage{textcomp}
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\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
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\usepackage{tikz}
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\usepackage{algorithm}
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\usepackage{algorithmic}
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%\usepackage{minted}
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\usetikzlibrary{automata,matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
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\usepackage{polynom}
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\polyset{style=C, div=:,vars=x}
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\pgfplotsset{compat=1.8}
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\pagenumbering{arabic}
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\def\thesection{\arabic{section})}
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\def\thesubsection{(\alph{subsection})}
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\def\thesubsubsection{(\roman{subsubsection})}
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\makeatletter
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\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
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\hskip -\arraycolsep
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\let\@ifnextchar\new@ifnextchar
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\array{#1}}
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\makeatother
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\parskip 12pt plus 1pt minus 1pt
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\parindent 0pt
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\begin{document}
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\author{Reinhard Köhler (6425686), Tronje Krabbe (6435002), \\
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Jim Martens (6420323)}
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\title{Hausaufgaben zum 15. Januar}
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\subtitle{Gruppe 8}
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\maketitle
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\section{} %1
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\subsection{} %a
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\begin{tabular}{c|c|c|c|c|c|c|c|c}
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EXTRACT & 1 & 2 & 3 & 4 & 5 & 6 & 7 &\\
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\hline
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- & 0 & $\infty$ & $\infty$ & $\infty$ & $\infty$ & $\infty$ & $\infty$ & (v.dist) \\
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& - & - & - & - & - & - & - & (v.$\pi$) \\
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\hline
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1 & 0 & 4 & $\infty$ & $\infty$ & $\infty$ & 5 & $\infty$ & \\
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& - & 1 & - & - & - & 1 & - & \\
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\hline
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2 & 0 & 4 & 14 & $\infty$ & $\infty$ & 5 & 7 & \\
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& - & 1 & 2 & - & - & 1 & 2 & \\
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\hline
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6 & 0 & 4 & 14 & $\infty$ & 14 & 5 & 7 & \\
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& - & 1 & 2 & - & 6 & 1 & 2 & \\
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\hline
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7 & 0 & 4 & 13 & $\infty$ & 10 & 5 & 7 & \\
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& - & 1 & 7 & - & 7 & 1 & 2 & \\
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\hline
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5 & 0 & 4 & 12 & 15 & 10 & 5 & 7 & \\
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& - & 1 & 5 & 5 & 7 & 1 & 2 & \\
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\hline
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3 & 0 & 4 & 12 & 14 & 10 & 5 & 7 & \\
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& - & 1 & 5 & 3 & 7 & 1 & 2 & \\
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\hline
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4 & 0 & 4 & 12 & 14 & 10 & 5 & 7 & \\
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& - & 1 & 5 & 3 & 7 & 1 & 2 &
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\end{tabular}
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Der kürzeste Pfad von 1 nach 4 verläuft über 2, 7, 5 und 3 nach 4. Insgesamt kostet der Pfad 14.
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\subsection{} %b
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In $G_{2}$ ist 3 die Quelle.
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\begin{tabular}{c|c|c|c|c|c|c|c|c}
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EXTRACT & 3 & 1 & 2 & 4 & 5 &\\
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\hline
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- & 0 & $\infty$ & $\infty$ & $\infty$ & $\infty$ & (v.dist) \\
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& - & - & - & - & - & (v.$\pi$) \\
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\hline
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3 & 0 & $\infty$ & 9 & 4 & $\infty$ & \\
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& - & - & 3 & 3 & - & \\
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\hline
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4 & 0 & 5 & 8 & 4 & 6 & \\
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& - & 4 & 4 & 3 & 4 & \\
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\hline
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1 & 0 & 5 & 8 & 4 & 6 & \\
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& - & 4 & 4 & 3 & 4 & \\
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\hline
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5 & 0 & 1 & 8 & 4 & 6 & \\
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& - & 5 & 4 & 3 & 4 & \\
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\end{tabular}
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Durch die negative Kante von 5 nach 1, würde sich der kürzeste Pfad von 1 von 5 auf 1 ändern, was jedoch nicht geht, da 1 bereits besucht wurde. Daher liefert Dijkstra für das Single-Source-Shortest-Path Problem in $G_{2}$ ein falsches Ergebnis.
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\section{} %2
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\section{} %3
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\subsection{} %a
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\subsection{} %b
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\subsection{} %c
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\section{} %4
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\subsection{} %a
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\subsection{} %b
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\end{document}
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