Mercurial > 13ss.theoinf
comparison notes/tex/automatons.tex @ 41:5d10471f5585
move frame-definitions out of presentations
author | Markus Kaiser <markus.kaiser@in.tum.de> |
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date | Thu, 11 Jul 2013 20:42:36 +0200 |
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children | 35e8bb96da7b |
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1 \defineUnit{dfa}{% | |
2 \begin{frame} | |
3 \frametitle{DFA} | |
4 | |
5 \begin{definition}[Deterministischer endlicher Automat] | |
6 Ein \alert{DFA} ist ein Tupel $M = (Q, \Sigma, \delta, q_0, F)$ aus einer/einem | |
7 \begin{itemize} | |
8 \item endlichen Menge von \alert{Zuständen} $Q$ | |
9 \item endlichen \alert{Eingabealphabet} $\Sigma$ | |
10 \item totalen \alert{Übergangsfunktion} $\delta : Q \times \Sigma \to Q$ | |
11 \item \alert{Startzustand} $q_0 \in Q$ | |
12 \item Menge von \alert{Endzuständen} $F \subseteq Q$ | |
13 \end{itemize} | |
14 \end{definition} | |
15 | |
16 \vfill | |
17 \pause | |
18 | |
19 \begin{center} | |
20 \begin{tikzpicture}[shorten >=1pt, node distance = 3cm, auto, bend angle=20, initial text=] | |
21 \node[state, initial] (q0) {$q_0$}; | |
22 \node[state, accepting] (q1) [right of = q0] {$q_1$}; | |
23 \node[state] (q2) [right of = q1] {$q_2$}; | |
24 | |
25 \draw[->] (q0) edge [loop above] node {0} (q0); | |
26 \draw[->] (q2) edge [loop above] node {1} (q2); | |
27 \draw[->] (q0) edge [bend left] node {1} (q1); | |
28 \draw[->] (q1) edge [bend left] node {1} (q0); | |
29 \draw[->] (q1) edge [bend left] node {0} (q2); | |
30 \draw[->] (q2) edge [bend left] node {0} (q1); | |
31 \end{tikzpicture} | |
32 \end{center} | |
33 \end{frame} | |
34 } | |
35 | |
36 \defineUnit{nfa}{% | |
37 \begin{frame} | |
38 \frametitle{NFA} | |
39 \begin{definition}[Nicht-Deterministischer endlicher Automat] | |
40 Ein \alert{NFA} ist ein Tupel $N = (Q, \Sigma, \delta, q_0, F)$ mit | |
41 \begin{itemize} | |
42 \item $Q, \Sigma, q_0, F$ wie ein DFA | |
43 \item \alert{Übergangsfunktion} $\delta : Q \times \Sigma \to P(Q)$ | |
44 \end{itemize} | |
45 \end{definition} | |
46 | |
47 \vfill | |
48 \pause | |
49 | |
50 \begin{center} | |
51 \begin{tikzpicture}[shorten >=1pt, node distance = 3cm, auto, bend angle=20, initial text=] | |
52 \node[state, initial] (q0) {$q_0$}; | |
53 \node[state, accepting] (q1) [right of = q0] {$q_1$}; | |
54 \draw[->] (q0) edge [loop above] node {0,1} (q0); \draw[->] (q0) edge node {1} (q1); \end{tikzpicture} \end{center} \end{frame} | |
55 } | |
56 | |
57 \defineUnit{enfa}{% | |
58 \begin{frame} | |
59 \frametitle{$\epsilon$-NFA} | |
60 \begin{definition}[NFA mit $\epsilon$-Übergängen] | |
61 Ein \alert{$\epsilon$-NFA} ist ein Tupel $N = (Q, \Sigma, \delta, q_0, F)$ mit | |
62 \begin{itemize} | |
63 \item $Q, \Sigma, q_0, F$ wie ein DFA | |
64 \item \alert{Übergangsfunktion} $\delta : Q \times \left( \Sigma \cup \{\epsilon\} \right) \to P(Q)$ | |
65 \end{itemize} | |
66 \end{definition} | |
67 | |
68 \vfill | |
69 \pause | |
70 | |
71 \begin{center} | |
72 \begin{tikzpicture}[shorten >=1pt, node distance = 3cm, auto, bend angle=30, initial text=] | |
73 \node[state] (q1) {$q_1$}; | |
74 \node[state, initial] (q0) [left of = q1] {$q_0$}; | |
75 \node[state, accepting] (q2) [right of = q1] {$q_2$}; | |
76 \draw[->] (q0) edge [red] node {$\epsilon$} (q1); \draw[->] (q1) edge [loop above] node {0,1} (q1); \draw[->] (q1) edge node {1} (q2); \draw[->] (q0) edge [bend right, red] node {$\epsilon$} (q2); \end{tikzpicture} \end{center} \end{frame} | |
77 } | |
78 | |
79 \defineUnit{endlicheautomaten}{% | |
80 \begin{frame} | |
81 \frametitle{Endliche Automaten} | |
82 \begin{block}{Übergangsfunktionen} | |
83 Die Automaten $A = (Q, \Sigma, \delta, q_0, F)$ unterscheiden sich nur durch ihre Übergangsfunktionen. | |
84 | |
85 \begin{description} | |
86 \item[DFA] $\delta : Q \times \Sigma \to Q$ | |
87 \item[NFA] $\delta : Q \times \Sigma \to \alert{P(Q)}$ | |
88 \item[$\epsilon$-NFA] $\delta : Q \times \alert{\left( \Sigma \cup \{\epsilon\} \right)} \to \alert{P(Q)}$ | |
89 \end{description} | |
90 \end{block} | |
91 | |
92 \vfill | |
93 | |
94 \begin{theorem} | |
95 \alert{DFA}, \alert{NFA} und \alert{$\epsilon$-NFA} sind gleich mächtig und lassen sich ineinander umwandeln. | |
96 \end{theorem} | |
97 \end{frame} | |
98 } | |
99 | |
100 \defineUnit{regex}{% | |
101 \begin{frame} | |
102 \frametitle{Reguläre Ausdrücke} | |
103 \setbeamercovered{dynamic} | |
104 | |
105 \begin{definition}[Regulärer Ausdruck] | |
106 \alert{Reguläre Ausdrücke} sind induktiv definiert | |
107 \begin{itemize} | |
108 \item \alert{$\emptyset$} ist ein regulärer Ausdruck | |
109 \item \alert{$\epsilon$} ist ein regulärer Ausdruck | |
110 \item Für alle $a \in \Sigma$ ist \alert{$a$} ein regulärer Ausdruck | |
111 \item Sind $\alpha$ und $\beta$ reguläre Ausdrücke, dann auch | |
112 \begin{description} | |
113 \item[Konkatenation] \alert{$\alpha\beta$} | |
114 \item[Veroderung] \alert{$\alpha \mid \beta$} | |
115 \item[Wiederholung] \alert{$\alpha^*$} | |
116 \end{description} | |
117 \end{itemize} | |
118 Analoge Sprachdefinition, z.b. $L(\alpha\beta) = L(\alpha)L(\beta)$ | |
119 \end{definition} | |
120 | |
121 \begin{example} | |
122 $\alpha = (0|1)^*00$ \hfill $L(\alpha) = \left\{x \mid x \text{ Binärzahl}, x \mod 4 = 0 \right\}$ | |
123 \end{example} | |
124 \end{frame} | |
125 } | |
126 | |
127 \defineUnit{automatenkonversionen}{% | |
128 \begin{frame}[c] | |
129 \frametitle{Konversionen} | |
130 \setbeamercovered{dynamic} | |
131 | |
132 \begin{center} | |
133 \begin{tikzpicture}[node distance=2cm] | |
134 \node (nfa) {NFA}; | |
135 \node (dfa) [left of=nfa] {DFA}; | |
136 \node (enfa) [right of=nfa] {$\epsilon$-NFA}; | |
137 \node (re) [below of=nfa] {RE}; | |
138 | |
139 \draw [every edge, tumred] (nfa) -- (dfa); | |
140 \draw [every edge, tumred] (enfa) -- (nfa); | |
141 \draw [every edge] (dfa) -- (re); | |
142 \draw [every edge] (nfa) -- (re); | |
143 \draw [every edge, tumred] (re) -- (enfa); | |
144 \end{tikzpicture} | |
145 \end{center} | |
146 \end{frame} | |
147 } | |
148 | |
149 \defineUnit{rezunfa}{% | |
150 \begin{frame} | |
151 \frametitle{RE $\rightarrow$ $\epsilon$-NFA} | |
152 \setbeamercovered{dynamic} | |
153 | |
154 \begin{block}{Idee (Kleene)} | |
155 Für einen Ausdruck \alert{$\gamma$} wird rekursiv mit struktureller Induktion ein $\epsilon$-NFA konstruiert. | |
156 \end{block} | |
157 | |
158 \begin{tabu} to \linewidth {XXX} | |
159 \alert{$\gamma = \emptyset$} & \alert{$\gamma = \epsilon$} & \alert{$\gamma = a \in \Sigma$} \\ | |
160 \begin{tikzpicture}[automaton, small, baseline=(current bounding box.north)] | |
161 \node[state, initial] () {}; | |
162 \end{tikzpicture} & | |
163 | |
164 \begin{tikzpicture}[automaton, small, baseline=(current bounding box.north)] | |
165 \node[state, initial, accepting] () {}; | |
166 \end{tikzpicture} & | |
167 | |
168 \begin{tikzpicture}[automaton, small, baseline=(current bounding box.north)] | |
169 \node[state, initial] (i) {}; | |
170 \node[state, accepting] (j) [right of=i] {}; | |
171 | |
172 \draw[->] (i) edge node {$a$} (j); | |
173 \end{tikzpicture} \\ | |
174 \vspace{2em} | |
175 \alert{$\gamma = \alpha\beta$} \\ | |
176 \multicolumn3{c}{ | |
177 \begin{tikzpicture}[automaton, small] | |
178 \draw[tumgreen, fill=tumgreen!20] (-0.3, 1) rectangle (1.8, -1); | |
179 \node[tumgreen] () at (0.75, -1.2) {$N_\alpha$}; | |
180 | |
181 \draw[tumgreen, fill=tumgreen!20] (3.7, 1) rectangle (5.8, -1); | |
182 \node[tumgreen] () at (4.75, -1.2) {$N_\beta$}; | |
183 | |
184 \node[state, initial] (i) at (0, 0) {}; | |
185 \node[state] (j) at (1.5, 0.5) {}; | |
186 \node[state] (k) at (1.5, -0.5) {}; | |
187 \node[state] (l) at (4, 0) {}; | |
188 \node[state, accepting] (m) at (5.5, 0) {}; | |
189 | |
190 \draw[->] (j) edge node {$\epsilon$} (l); | |
191 \draw[->] (k) edge node {$\epsilon$} (l); | |
192 \end{tikzpicture} | |
193 }\\ | |
194 \end{tabu} | |
195 \end{frame} | |
196 } | |
197 | |
198 \defineUnit{rezuenfabeispiel}{% | |
199 \begin{frame} | |
200 \frametitle{RE $\rightarrow$ $\epsilon$-NFA} | |
201 \setbeamercovered{dynamic} | |
202 | |
203 \begin{tabu} to \linewidth {X} | |
204 \alert{$\gamma = \alpha \mid \beta$} \\ | |
205 \centering | |
206 \begin{tikzpicture}[automaton, small] | |
207 \draw[tumgreen, fill=tumgreen!20] (2, 1.5) rectangle (4.5, 0.5); | |
208 \node[tumgreen] () at (3.25, 0.3) {$N_\alpha$}; | |
209 | |
210 \draw[tumgreen, fill=tumgreen!20] (2, -0.5) rectangle (4.5, -1.5); | |
211 \node[tumgreen] () at (3.25, -1.7) {$N_\beta$}; | |
212 | |
213 \node[state, initial] (i) at (0, 0) {}; | |
214 | |
215 \node[state] (j) at (2.5, 1) {}; | |
216 \node[state, accepting] (k) at (4, 1) {}; | |
217 \node[state] (l) at (2.5, -1) {}; | |
218 \node[state, accepting] (m) at (4, -1) {}; | |
219 | |
220 \draw[->] (i) edge node {$\epsilon$} (j); | |
221 \draw[->] (i) edge node {$\epsilon$} (l); | |
222 \end{tikzpicture} \\ | |
223 \vfill | |
224 | |
225 \alert{$\gamma = \alpha^*$} \\ | |
226 \centering | |
227 \begin{tikzpicture}[automaton, small, bend angle=70] | |
228 \draw[tumgreen, fill=tumgreen!20] (2, 1) rectangle (4.5, -1); | |
229 \node[tumgreen] () at (3.25, -1.2) {$N_\alpha$}; | |
230 | |
231 \node[state, initial, accepting] (i) at (0, 0) {}; | |
232 | |
233 \node[state] (j) at (2.5, 0) {}; | |
234 \node[state, accepting] (k) at (4, 0.5) {}; | |
235 \node[state, accepting] (m) at (4, -0.5) {}; | |
236 | |
237 \draw[->] (i) edge node {$\epsilon$} (j); | |
238 \draw[->] (k) edge [bend right] node {$\epsilon$} (j); | |
239 \draw[->] (m) edge [bend left] node[above] {$\epsilon$} (j); | |
240 \end{tikzpicture} | |
241 \end{tabu} | |
242 \end{frame} | |
243 } | |
244 | |
245 \defineUnit{enfazunfa}{% | |
246 \begin{frame} | |
247 \frametitle{$\epsilon$-NFA $\rightarrow$ NFA} | |
248 \setbeamercovered{dynamic} | |
249 | |
250 \begin{block}{Idee} | |
251 Entferne $\epsilon$-Kanten durch das Bilden von $\epsilon$-Hüllen. | |
252 \begin{enumerate} | |
253 \item<1-> Entferne \alert{unnötige Knoten}. | |
254 \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. | |
255 \item<1,4-> Entferne alle \alert{$\epsilon$-Kanten} und unerreichbare Knoten. | |
256 \item<1,5-> Wurde das leere Wort akzeptiert mache den \alert{Anfangszustand} zum Endzustand. | |
257 \end{enumerate} | |
258 \end{block} | |
259 | |
260 \vfill | |
261 | |
262 \begin{tikzpicture}[automaton, bend angle=40, node distance=2.1cm] | |
263 \useasboundingbox (-1.4,2) rectangle (9, -2); | |
264 | |
265 \node<-4>[state, initial] (q0) {$q_0$}; | |
266 \node[state] (q2) [right = 3.2cm of q0] {$q_2$}; | |
267 \node[state] (q3) [right of = q2] {$q_3$}; | |
268 \node[state, accepting] (q4) [right of = q3] {$q_4$}; | |
269 | |
270 \draw[->] (q2) edge node {$0$} (q3); | |
271 \draw[->] (q3) edge node {$1$} (q4); | |
272 | |
273 \draw<1-4>[->] (q3) edge [bend right] node [above] {$\epsilon$} (q2); | |
274 \draw[->] (q4) edge [bend right] node [above] {$1$} (q3); | |
275 \draw<1-4>[->] (q0) edge [bend right=20] node [below] {$\epsilon$} (q4); | |
276 | |
277 \node<1>[state] (q1) [right of = q0] {$q_1$}; | |
278 \draw<1>[->] (q0) edge node {$\epsilon$} (q1); | |
279 \draw<1>[->] (q1) edge node {$1$} (q2); | |
280 | |
281 \node<2>[state, fill=tumred!20] (q1) [right of = q0] {$q_1$}; | |
282 \draw<2>[->, tumred] (q0) edge node {$\epsilon$} (q1); | |
283 \draw<2>[->, tumred] (q1) edge node {$0$} (q2); | |
284 \draw<2->[->, tumblue] (q0) edge [bend left] node {$0$} (q2); | |
285 | |
286 \draw<3,4,5>[->, tumred] (q0) edge [bend right=20] node [below] {$\epsilon$} (q4); | |
287 \draw<3>[->, tumred] (q4) edge [bend right] node [above] {$1$} (q3); | |
288 \draw<3,4>[->, tumred] (q3) edge [bend right] node [above] {$\epsilon$} (q2); | |
289 \draw<3->[->, tumgreen] (q0) edge node {$1$} (q2); | |
290 | |
291 \draw<4->[->, tumgreen] (q2) edge [loop above] node [above] {$0$} (q2); | |
292 \draw<4->[->, tumgreen] (q3) edge [loop above] node [above] {$0$} (q3); | |
293 \draw<4->[->, tumgreen] (q0) edge [bend right=20] node [above] {$1$} (q3); | |
294 \draw<4->[->, tumgreen] (q4) edge [bend right=70] node [above] {$1$} (q2); | |
295 | |
296 \node<5>[state, initial, accepting, fill=tumgreen!20] (q0) {$q_0$}; | |
297 | |
298 \node<6->[state, initial, accepting] (q0) {$q_0$}; | |
299 \end{tikzpicture} | |
300 \end{frame} | |
301 } | |
302 | |
303 \defineUnit{nfazudfa}{% | |
304 \begin{frame} | |
305 \frametitle{NFA $\rightarrow$ DFA} | |
306 \setbeamercovered{dynamic} | |
307 | |
308 \begin{block}{Idee (Potenzmengenkonstruktion)} | |
309 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{$P(Q)$}. | |
310 | |
311 \begin{itemize} | |
312 \item $\overline{\delta}: \alert{P(Q)} \times \Sigma \to P(Q)$ \\ | |
313 \[\overline{\delta}(S, a) := \bigcup_{q \in S} \delta(q, a)\] | |
314 \item $F_M := \left\{S \subseteq Q \mid \alert{S \cap F} \neq \emptyset\right\}$ | |
315 \end{itemize} | |
316 \end{block} | |
317 | |
318 \begin{tikzpicture}[automaton, bend angle=20, node distance=2.1cm] | |
319 \useasboundingbox (-1.4,2) rectangle (9, -2); | |
320 | |
321 \node[state, initial] (q0) {$q_0$}; | |
322 \node[state, accepting] (q1) [right of = q0] {$q_1$}; | |
323 | |
324 \draw[->] (q0) edge [loop above] node {$0,1$} (q0); | |
325 \draw[->] (q0) edge node {$1$} (q1); | |
326 | |
327 \node<2->(sep) [right of = q1] {$\rightarrow$}; | |
328 | |
329 \node<2->[state, initial, inner sep=1pt] (pq0) [right of = sep] {$q_{\{0\}}$}; | |
330 | |
331 \node<3->[state, accepting, inner sep=0pt] (pq01) [right of = pq0] {$q_{\{0,1\}}$}; | |
332 \draw<3->[->] (pq0) edge [loop above] node {$0$} (pq0); | |
333 \draw<3->[->] (pq0) edge [bend left] node {$1$} (pq01); | |
334 | |
335 \draw<4->[->] (pq01) edge [loop above] node {$1$} (pq01); | |
336 \draw<4->[->] (pq01) edge [bend left] node {$0$} (pq0); | |
337 | |
338 \end{tikzpicture} | |
339 \end{frame} | |
340 } | |
341 | |
342 \defineUnit{produktautomat}{% | |
343 \begin{frame} | |
344 \frametitle{produktautomat} | |
345 \setbeamercovered{dynamic} | |
346 \begin{theorem} | |
347 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} | |
348 | |
349 \begin{align*} | |
350 m &:= (\alert{q_1 \times q_2}, \sigma, \delta, (s_1, s_2), f_1 \times f_2) \\ | |
351 \delta\left( (q_1, q_2), a \right) &:= \left( \alert{\delta_1}(q_1, a), \alert{\delta_2}(q_2, a) \right) | |
352 \end{align*} | |
353 | |
354 ein dfa, der $l(m_1) \cap l(m_2)$ akzeptiert. | |
355 \end{theorem} | |
356 \end{frame} | |
357 } | |
358 | |
359 \defineUnit{regexrechnen}{% | |
360 \begin{frame} | |
361 \frametitle{Nochmal Reguläre Ausdrücke} | |
362 \setbeamercovered{dynamic} | |
363 | |
364 \begin{theorem} | |
365 Die regulären Ausdrücke $\mathfrak{R}$ über einem Alphabet $\Sigma$ bilden mit Konkatenation $\circ$ und Veroderung $\mid$ einen \alert{Halbring} $\langle \mathfrak{R}, \mid, \circ, \emptyset, \epsilon \rangle$. | |
366 | |
367 \begin{itemize} | |
368 \item \alert{Assoziative} Operationen | |
369 \item Veroderung \alert{kommutativ} | |
370 \item \alert{Distributivität}: $\alpha (\beta \mid \gamma) \equiv \alpha\beta \mid \alpha\gamma$ | |
371 \item $\emptyset$ \alert{neutral} bezüglich Oder | |
372 \item $\epsilon$ \alert{neutral} bezüglich Konkatenation | |
373 \end{itemize} | |
374 \end{theorem} | |
375 | |
376 \begin{example} | |
377 \[ | |
378 1\psi \mid 0\phi \mid \psi \equiv 0 \phi \mid (1 \mid \epsilon) \psi | |
379 \] | |
380 \end{example} | |
381 \end{frame} | |
382 } | |
383 | |
384 \defineUnit{arden}{% | |
385 \begin{frame} | |
386 \frametitle{Ardens Lemma} | |
387 \setbeamercovered{dynamic} | |
388 | |
389 \begin{theorem}[Ardens Lemma] | |
390 Sind $A$, $B$ und $X$ Sprachen mit $\epsilon \not \in A$, dann gilt | |
391 \[ | |
392 X = AX \cup B \Longrightarrow X = A^* B | |
393 \] | |
394 Speziell gilt für reguläre Ausdrücke | |
395 \[ | |
396 X \equiv \alpha X \mid \beta \Longrightarrow X \equiv \alpha^* \beta | |
397 \] | |
398 \end{theorem} | |
399 | |
400 \begin{example} | |
401 \[ | |
402 \psi \equiv 0 \psi \mid (1 \mid \epsilon) \phi \Longrightarrow \psi \equiv 0^*(1\mid \epsilon) \phi | |
403 \] | |
404 \end{example} | |
405 \end{frame} | |
406 } | |
407 | |
408 \defineUnit{nfazure}{% | |
409 \begin{frame} | |
410 \frametitle{NFA $\rightarrow$ RE} | |
411 \setbeamercovered{dynamic} | |
412 | |
413 \begin{block}{Idee} | |
414 Erzeuge ein Gleichungssystem aus allen Zuständen. | |
415 \begin{enumerate} | |
416 \item<1,2-> Ausdruck für jeden Zustand | |
417 \item<1,3-> Auflösen nach $X_0$ mit Algebra und Ardens Lemma | |
418 \end{enumerate} | |
419 \end{block} | |
420 \begin{columns}<2-> | |
421 \begin{column}[b]{.65\textwidth} | |
422 \begin{align*} | |
423 X_0 &\equiv 1X_0 \mid 0X_1 \\ | |
424 &\equiv \uncover<4->{1X_0 \mid 00^*(\epsilon \mid 1X_0)} \\ | |
425 &\equiv \uncover<4->{(1 \mid 00^*1) X_0 \mid 00^*} \\ | |
426 &\equiv \uncover<4->{(1 \mid 00^*1)^*(00^*)} \\ | |
427 \\ | |
428 X_1 &\equiv 1X_0 \mid 0X_1 \alt<3->{\mid \epsilon}{\alert{\mid \epsilon}} \\ | |
429 &\equiv \uncover<3-> {0X_1 \mid (\epsilon \mid 1 X_0)}\\ | |
430 &\equiv \uncover<3-> {\alt<-2,4->{0^*(\epsilon \mid 1X_0)}{\alert{0^*(\epsilon \mid 1X_0)}}} | |
431 \end{align*} | |
432 \end{column} | |
433 \begin{column}[t]{.35\textwidth} | |
434 \begin{tikzpicture}[automaton] | |
435 \node[state, initial] (q0) {$q_0$}; | |
436 \node[state, accepting] (q1) [below of=q0] {$q_1$}; | |
437 | |
438 \draw[->] (q0) edge [bend right] node [left] {$0$} (q1); | |
439 \draw[->] (q1) edge [bend right] node [right] {$1$} (q0); | |
440 \draw[->] (q0) edge [loop right] node {$1$} (q0); | |
441 \draw[->] (q1) edge [loop right] node {$0$} (q1); | |
442 \end{tikzpicture} | |
443 \end{column} | |
444 \end{columns} | |
445 \end{frame} | |
446 } | |
447 | |
448 \defineUnit{rpl}{% | |
449 \begin{frame} | |
450 \frametitle{Pumping Lemma} | |
451 \setbeamercovered{dynamic} | |
452 | |
453 \begin{theorem}[Pumping Lemma für reguläre Sprachen] | |
454 Sei $R \subseteq \Sigma^*$ regulär. Dann gibt es ein $n > 0$, so dass sich \alert{jedes} $z \in R$ mit $|z| \geq n$ so in $z = uvw$ zerlegen lässt, dass | |
455 \begin{itemize} | |
456 \item $v \neq \epsilon$ | |
457 \item $|uv| \alert{\leq n}$ | |
458 \item $\forall i \alert{\geq 0}. uv^iw \in R$ | |
459 \end{itemize} | |
460 \end{theorem} | |
461 | |
462 \vfill | |
463 | |
464 \begin{center} | |
465 \begin{tikzpicture}[automaton] | |
466 \node[state, initial] (q0) {}; | |
467 \node[state, fill=tumred!20] (q1) [right of=q0] {}; | |
468 \node[state, accepting] (q2) [right of=q1] {}; | |
469 | |
470 \draw[->, densely dashed] (q0) edge node {$u$} (q1); | |
471 \draw[->, tumred] (q1) edge [loop above] node {$v$} (q1); | |
472 \draw[->, densely dashed] (q1) edge node {$w$} (q2); | |
473 \end{tikzpicture} | |
474 \end{center} | |
475 \end{frame} | |
476 } | |
477 | |
478 \defineUnit{rplanwenden}{% | |
479 \begin{frame} | |
480 \frametitle{Nichtregularität beweisen} | |
481 \setbeamercovered{dynamic} | |
482 | |
483 \begin{block}{Idee} | |
484 Gegenbeispiel fürs Pumpinglemma suchen. | |
485 \[ | |
486 \alert{\forall} n \in \N_0 \alert{\exists} z \in L. |z| \geq n \ \alert{\forall} u,v,w. \ z = uvw \ \text{\alert{nicht} pumpbar} | |
487 \] | |
488 \end{block} | |
489 | |
490 \vfill | |
491 | |
492 \begin{example}<2-> | |
493 Ist $L = \left\{ a^ib^i \mid i \in \N_0 \right\}$ regulär? | |
494 \begin{enumerate} | |
495 \item \alert{Sei $n$} PL-Zahl | |
496 \item \alert{Wähle} $\alert{z} = a^nb^n$ | |
497 \item Dann ist \alert{$z = uvw$} mit \alert{$|uv| \leq n$}, hier: $v=a^k$ mit $k > 0$ | |
498 \item Dann ist $uv^0w \not \in L$ | |
499 \item Damit ist L \alert{nicht} regulär. | |
500 \end{enumerate} | |
501 \end{example} | |
502 \end{frame} | |
503 } | |
504 | |
505 \defineUnit{aequivalenteZustaende}{% | |
506 \begin{frame} | |
507 \frametitle{Äquivalenzen} | |
508 \setbeamercovered{dynamic} | |
509 | |
510 \begin{definition}[Äquivalente Worte] | |
511 Jede Sprache $L \subseteq \Sigma^*$ induziert eine Äquivalenzrelation $\alert{\equiv_L \subseteq \Sigma^* \times \Sigma^*}$: | |
512 \[ | |
513 u \alert{\equiv_L} v \Longleftrightarrow \left( \forall w \in \Sigma^*. \alert{uw} \in L \Leftrightarrow \alert{vw} \in L\right) | |
514 \] | |
515 \end{definition} | |
516 | |
517 \vfill | |
518 | |
519 \pause | |
520 | |
521 \begin{definition}[Äquivalente Zustände] | |
522 Zwei Zustände im DFA $A$ sind \alert{äquivalent} wenn sie die selbe Sprache akzeptieren. | |
523 | |
524 \[ | |
525 p \alert{\equiv_A} q \Longleftrightarrow \left( \forall w \in \Sigma^*. \alert{\hat{\delta}(p, w)} \in F \Leftrightarrow \alert{\hat{\delta}(q, w)} \in F \right) | |
526 \] | |
527 \end{definition} | |
528 \end{frame} | |
529 } | |
530 | |
531 \defineUnit{unterscheidbareZustaende}{% | |
532 \begin{frame} | |
533 \frametitle{Unterscheidbare Zustände} | |
534 \setbeamercovered{dynamic} | |
535 | |
536 \begin{definition}[Unterscheidbarkeit] | |
537 Zwei Zustände sind \alert{unterscheidbar}, wenn sie unterschiedliche Sprachen akzeptieren. | |
538 \[ | |
539 p \alert{\not\equiv_A} q \Longleftrightarrow \left( \exists w \in \Sigma^*. \hat{\delta}(p, w) \alert{\in} F \wedge \hat{\delta}(q, w) \alert{\not\in} F \right) | |
540 \] | |
541 \end{definition} | |
542 | |
543 \begin{theorem} | |
544 Sind $\delta(p, a)$ und $\delta(q, a)$ unterscheidbar, dann auch $p$ und $q$. | |
545 \end{theorem} | |
546 | |
547 \pause | |
548 | |
549 \begin{tikzpicture}[automaton, bend angle=40, node distance=2.5cm] | |
550 \node[state, initial] (q0) {$q_0$}; | |
551 \node[state] (q1) [right of = q0] {$q_1$}; | |
552 \node[state] (q2) [right of = q1] {$q_2$}; | |
553 \node[state, accepting] (q3) [right of = q2] {$q_3$}; | |
554 | |
555 \draw[->] (q0) edge node {$a$} (q1); | |
556 \draw[->] (q0) edge [bend left] node {$b$} (q2); | |
557 \draw[->] (q1) edge node {$a$} (q2); | |
558 \draw[->] (q1) edge [bend right] node {$b$} (q3); | |
559 \draw[->] (q2) edge node {$a,b$} (q3); | |
560 \draw[->] (q3) edge [loop right] node {$a,b$} (q3); | |
561 | |
562 \node<3>[state, fill=tumred!35] () at (q2) {$q_2$}; | |
563 \node<3->[state, accepting, fill=tumgreen!35] () at (q3) {$q_3$}; | |
564 | |
565 \node<4>[state, fill=tumred!35] () at (q0) {$q_0$}; | |
566 \node<4>[state, fill=tumred!35] () at (q1) {$q_1$}; | |
567 \draw<4>[->, tumred] (q0) edge [bend left] node {$b$} (q2); | |
568 \draw<4>[->, tumgreen] (q1) edge [bend right] node {$b$} (q3); | |
569 \end{tikzpicture} | |
570 \end{frame} | |
571 } |