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missing accepting state
author | Markus Kaiser <markus.kaiser@in.tum.de> |
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date | Tue, 30 Apr 2013 01:59:25 +0200 |
parents | 2cabb8d02269 |
children | 20c64dea84fa |
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\documentclass[compress, german, t]{beamer} \usepackage[ngerman,english]{babel} \uselanguage{German} \languagepath{German} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{helvet} \usepackage{url} \usepackage{listings} \usepackage{xcolor} \usepackage{tikz} \usepackage{pgfplots} \usetikzlibrary{automata} \usetikzlibrary{calc} \usetikzlibrary{shapes.geometric} \usetikzlibrary{positioning} \usepackage{tabu} \usepackage{beamerthemeLEA2} \newcommand{\N} {\mathbb{N}} % natürliche Zahlen \newcommand{\Z} {\mathbb{Z}} % ganze Zahlen \newcommand{\R} {\mathbb{R}} % reelle Zahlen \newcommand{\Prob} {\mathrm{P}} % Wahrscheinlichkeit \newcommand{\Oh} {\mathcal{O}} % O-Notation (Landau-Symbole) \newcommand{\mycite}[1]{\textcolor{tumgreen}{[#1]}} \tikzstyle{every edge} = [draw,very thick,->,>=latex] \tikzstyle{every state} = [circle,thick,draw,fill=tumblue!10] \tikzstyle{automaton} = [shorten >=1pt, node distance = 3cm, auto, bend angle=20, initial text=] \tikzstyle{small} = [every node/.style={scale=0.5}, baseline=(current bounding box.north), font=\LARGE] \title{Übung 2: Konversion RE $\rightarrow$ DFA} \subtitle{Theoretische Informatik Sommersemester 2013} \author{\href{mailto:markus.kaiser@in.tum.de}{Markus Kaiser}} \begin{document} \begin{frame} \titlepage \end{frame} \begin{frame} \setbeamercovered{dynamic} \frametitle{Reguläre Ausdrücke} \begin{definition}[Regulärer Ausdruck] \alert{Reguläre Ausdrücke} sind induktiv definiert \begin{itemize} \item \alert{$\emptyset$} ist ein regulärer Ausdruck \item \alert{$\epsilon$} ist ein regulärer Ausdruck \item Für alle $a \in \Sigma$ ist \alert{$a$} ein regulärer Ausdruck \item Sind $\alpha$ und $\beta$ reguläre Ausdrücke, dann auch \begin{description} \item[Konkatenation] \alert{$\alpha\beta$} \item[Veroderung] \alert{$\alpha \mid \beta$} \item[Wiederholung] \alert{$\alpha^*$} \end{description} \end{itemize} Analoge Sprachdefinition, z.b. $L(\alpha\beta) = L(\alpha)L(\beta)$ \end{definition} \begin{example} $\alpha = (0|1)^*00$ \hfill $L(\alpha) = \left\{x \mid x \text{ Binärzahl}, x \mod 4 = 0 \right\}$ \end{example} \end{frame} \begin{frame}[c] \setbeamercovered{dynamic} \frametitle{Konversionen} \begin{center} \begin{tikzpicture}[node distance=2cm] \node (nfa) {NFA}; \node (dfa) [left of=nfa] {DFA}; \node (enfa) [right of=nfa] {$\epsilon$-NFA}; \node (re) [below of=nfa] {RE}; \draw [every edge, tumred] (nfa) -- (dfa); \draw [every edge, tumred] (enfa) -- (nfa); \draw [every edge] (dfa) -- (re); \draw [every edge] (nfa) -- (re); \draw [every edge, tumred] (re) -- (enfa); \end{tikzpicture} \end{center} \end{frame} \begin{frame} \setbeamercovered{dynamic} \frametitle{RE $\rightarrow$ $\epsilon$-NFA} \begin{block}{Idee (Kleene)} Für einen Ausdruck \alert{$\gamma$} wird rekursiv mit struktureller Induktion ein $\epsilon$-NFA konstruiert. \end{block} \begin{tabu} to \linewidth {XXX} \alert{$\gamma = \emptyset$} & \alert{$\gamma = \epsilon$} & \alert{$\gamma = a \in \Sigma$} \\ \begin{tikzpicture}[automaton, small, baseline=(current bounding box.north)] \node[state, initial] () {}; \end{tikzpicture} & \begin{tikzpicture}[automaton, small, baseline=(current bounding box.north)] \node[state, initial, accepting] () {}; \end{tikzpicture} & \begin{tikzpicture}[automaton, small, baseline=(current bounding box.north)] \node[state, initial] (i) {}; \node[state, accepting] (j) [right of=i] {}; \draw[->] (i) edge node {$a$} (j); \end{tikzpicture} \\ \vspace{2em} \alert{$\gamma = \alpha\beta$} \\ \multicolumn3{c}{ \begin{tikzpicture}[automaton, small] \draw[tumgreen, fill=tumgreen!20] (-0.3, 1) rectangle (1.8, -1); \node[tumgreen] () at (0.75, -1.2) {$N_\alpha$}; \draw[tumgreen, fill=tumgreen!20] (3.7, 1) rectangle (5.8, -1); \node[tumgreen] () at (4.75, -1.2) {$N_\beta$}; \node[state, initial] (i) at (0, 0) {}; \node[state] (j) at (1.5, 0.5) {}; \node[state] (k) at (1.5, -0.5) {}; \node[state] (l) at (4, 0) {}; \node[state, accepting] (m) at (5.5, 0) {}; \draw[->] (j) edge node {$\epsilon$} (l); \draw[->] (k) edge node {$\epsilon$} (l); \end{tikzpicture} }\\ \end{tabu} \end{frame} \begin{frame} \setbeamercovered{dynamic} \frametitle{RE $\rightarrow$ $\epsilon$-NFA} \begin{tabu} to \linewidth {X} \alert{$\gamma = \alpha \mid \beta$} \\ \centering \begin{tikzpicture}[automaton, small] \draw[tumgreen, fill=tumgreen!20] (2, 1.5) rectangle (4.5, 0.5); \node[tumgreen] () at (3.25, 0.3) {$N_\alpha$}; \draw[tumgreen, fill=tumgreen!20] (2, -0.5) rectangle (4.5, -1.5); \node[tumgreen] () at (3.25, -1.7) {$N_\beta$}; \node[state, initial] (i) at (0, 0) {}; \node[state] (j) at (2.5, 1) {}; \node[state, accepting] (k) at (4, 1) {}; \node[state] (l) at (2.5, -1) {}; \node[state, accepting] (m) at (4, -1) {}; \draw[->] (i) edge node {$\epsilon$} (j); \draw[->] (i) edge node {$\epsilon$} (l); \end{tikzpicture} \\ \vfill \alert{$\gamma = \alpha^*$} \\ \centering \begin{tikzpicture}[automaton, small, bend angle=70] \draw[tumgreen, fill=tumgreen!20] (2, 1) rectangle (4.5, -1); \node[tumgreen] () at (3.25, -1.2) {$N_\alpha$}; \node[state, initial, accepting] (i) at (0, 0) {}; \node[state] (j) at (2.5, 0) {}; \node[state, accepting] (k) at (4, 0.5) {}; \node[state, accepting] (m) at (4, -0.5) {}; \draw[->] (i) edge node {$\epsilon$} (j); \draw[->] (k) edge [bend right] node {$\epsilon$} (j); \draw[->] (m) edge [bend left] node[above] {$\epsilon$} (j); \end{tikzpicture} \end{tabu} \end{frame} \begin{frame} \setbeamercovered{dynamic} \frametitle{$\epsilon$-NFA $\rightarrow$ NFA} \begin{block}{Idee} Entferne $\epsilon$-Kanten durch das Bilden von $\epsilon$-Hüllen. \begin{enumerate} \item<1-> Entferne \alert{unnötige Knoten}. \item<1,3-> Für jeden \alert{Pfad} der Form $\epsilon\ldots\epsilon \alert{a} \epsilon\ldots\epsilon$ verbinde Anfangs- und Endknoten mit einer \alert{$a$}-Kante. \item<1,4-> Entferne alle \alert{$\epsilon$-Kanten} und unerreichbare Knoten. \item<1,5-> Wurde das leere Wort akzeptiert mache den \alert{Anfangszustand} zum Endzustand. \end{enumerate} \end{block} \vfill \begin{tikzpicture}[automaton, bend angle=40, node distance=2.1cm] \useasboundingbox (-1.4,2) rectangle (9, -2); \node<-4>[state, initial] (q0) {$q_0$}; \node[state] (q2) [right = 3.2cm of q0] {$q_2$}; \node[state] (q3) [right of = q2] {$q_3$}; \node[state, accepting] (q4) [right of = q3] {$q_4$}; \draw[->] (q2) edge node {$0$} (q3); \draw[->] (q3) edge node {$1$} (q4); \draw<1-4>[->] (q3) edge [bend right] node [above] {$\epsilon$} (q2); \draw[->] (q4) edge [bend right] node [above] {$1$} (q3); \draw<1-4>[->] (q0) edge [bend right=20] node [below] {$\epsilon$} (q4); \node<1>[state] (q1) [right of = q0] {$q_1$}; \draw<1>[->] (q0) edge node {$\epsilon$} (q1); \draw<1>[->] (q1) edge node {$1$} (q2); \node<2>[state, fill=tumred!20] (q1) [right of = q0] {$q_1$}; \draw<2>[->, tumred] (q0) edge node {$\epsilon$} (q1); \draw<2>[->, tumred] (q1) edge node {$0$} (q2); \draw<2->[->, tumblue] (q0) edge [bend left] node {$0$} (q2); \draw<3,4,5>[->, tumred] (q0) edge [bend right=20] node [below] {$\epsilon$} (q4); \draw<3>[->, tumred] (q4) edge [bend right] node [above] {$1$} (q3); \draw<3,4>[->, tumred] (q3) edge [bend right] node [above] {$\epsilon$} (q2); \draw<3->[->, tumgreen] (q0) edge node {$1$} (q2); \draw<4->[->, tumgreen] (q2) edge [loop above] node [above] {$0$} (q2); \draw<4->[->, tumgreen] (q3) edge [loop above] node [above] {$0$} (q3); \draw<4->[->, tumgreen] (q4) edge [bend right=70] node [above] {$1$} (q2); \node<5>[state, initial, accepting, fill=tumgreen!20] (q0) {$q_0$}; \node<6->[state, initial, accepting] (q0) {$q_0$}; \end{tikzpicture} \end{frame} \begin{frame} \setbeamercovered{dynamic} \frametitle{NFA $\rightarrow$ DFA} \begin{block}{Idee (Potenzmengenkonstruktion)} Konstruiere aus einem NFA $N = (Q, \Sigma, \delta, q_0, F)$ einen DFA $D = (P(Q), \Sigma, \overline{\delta}, \{q_0\}, F_M)$ mit Zuständen aus \alert{$P(Q)$}. \begin{itemize} \item $\overline{\delta}: \alert{P(Q)} \times \Sigma \mapsto P(Q)$ \\ \[\overline{\delta}(S, a) := \bigcup_{q \in S} \delta(q, a)\] \item $F_M := \left\{S \subseteq Q \mid \alert{S \cap F} \neq \emptyset\right\}$ \end{itemize} \end{block} \begin{tikzpicture}[automaton, bend angle=20, node distance=2.1cm] \useasboundingbox (-1.4,2) rectangle (9, -2); \node[state, initial] (q0) {$q_0$}; \node[state, accepting] (q1) [right of = q0] {$q_1$}; \draw[->] (q0) edge [loop above] node {$0,1$} (q0); \draw[->] (q0) edge node {$1$} (q1); \node<2->(sep) [right of = q1] {$\rightarrow$}; \node<2->[state, initial, inner sep=1pt] (pq0) [right of = sep] {$q_{\{0\}}$}; \node<3->[state, accepting, inner sep=0pt] (pq01) [right of = pq0] {$q_{\{0,1\}}$}; \draw<3->[->] (pq0) edge [loop above] node {$0$} (pq0); \draw<3->[->] (pq0) edge [bend left] node {$1$} (pq01); \draw<4->[->] (pq01) edge [loop above] node {$1$} (pq01); \draw<4->[->] (pq01) edge [bend left] node {$0$} (pq0); \end{tikzpicture} \end{frame} \begin{frame} \setbeamercovered{dynamic} \frametitle{Produktautomat} \begin{theorem} Sind $M_1 = (Q_1, \Sigma, \delta_1, s_1, F_1)$ und $M_2 = (Q_2, \Sigma, \delta_2, s_2, F_2)$ DFAs, dann ist der \alert{Produkt-Automat} \begin{align*} M &:= (\alert{Q_1 \times Q_2}, \Sigma, \delta, (s_1, s_2), F_1 \times F_2) \\ \delta\left( (q_1, q_2), a \right) &:= \left( \alert{\delta_1}(q_1, a), \alert{\delta_2}(q_2, a) \right) \end{align*} ein DFA, der $L(M_1) \cap L(M_2)$ akzeptiert. \end{theorem} \end{frame} \end{document}