Regular Languages and Finite- State Automata. Formal Languages. Regular Expressions. 22c:19

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1 Regulr Lnguges nd Finite- Stte Automt 22c:19 Forml Lnguges Σ* for the set of ll words over Σ Lnguges = susets of Σ* Lnguge-forming opertions L 1 L 2, L 1 L 2 L 1* (Kleene closure) Regulr Expressions Inductive definition: Bse cses: is regulr expression (over Σ) nd denotes lnguge is regulr expression (over Σ) nd denotes lnguge {} inσis regulr expression (over Σ) nd denotes lnguge {} 1

2 Regulr Expressions Inductive cses: Assumer 1 nd r 2 re regulr expressions nd denote L 1 nd L 2,resp (r 1 r 2 ) is regulr expression nd denotes L 1 L 2 (r 1 r 2 ) is regulr expression nd denotes L 1 L 2 (r 1 *) is regulr expression nd denotes L 1 * Regulr Expressions Denote Lnguges * denotes lnguge { n n 0} ()() or just denotes unit lnguge {} ** denotes { n m n, m 0} 2 3 denotes {} n not regulr expression + denotes { n n 1}? denotes {, } Regulr expressions with prcticl convention Chrcter Mening Exmples [ ] lterntives /[eiou]/, /m[e]n/ - rnge /[-z]/ [^ ] not /[^pm]/, /[^ox]s/? optionlity /Kth?mndu/ * zero or more /*!/ + one or more /+!/ ny chrcter /ct[eiou]/ ^, $ strt, end of line \ not specil chrcter \\?\^ lternte strings /ct dog/ ( ) sustring /cit(y ies)/ etc 2

3 Regulr Lnguges L e lnguge over lphet Σ, ie, L Σ* Then L sid to e regulr lnguge if L denoted y some regulr expression over Σ Σ finite lphet nd L 1 nd L 2 regulr lnguges over Σ Then L 1 L 2, L 1 L 2,nd L 1* re lso regulr Remrks Σ finite lphet nd w ny word over Σ Then unit lnguge {w} regulr Any finite lnguge over Σ regulr Deterministic Finite-Stte Automt q q 0 1 q 2 Figure 921() 3

4 Another Exmple strt Figure 921() Determinism Determinism mens tht, within ny stte digrm for FSA, pth leled y given word w unique: for word w Σ*,, there is exctly one pth strting t q 0 nd leled y w Trnsition Functions Figure 921(c) 4

5 Forml Definition FSA is quintuple Σ, Q, q init, F, δ Μ Σ is input lphet Q is finite, nonempty set of sttes q init Q initil stte or strt stte F Q is (possily empty) set of ccepting or terminl sttes δ Μ : Q Σ Qtrnsition function (totl) Word Acceptnce Deterministic finite-stte utomton M ccepts word w Σ* if unique pth strting t q init nd leled y w leds to some memer of F, ie, to some ccepting stte of M Lnguge Acceptnce The lnguge ccepted y M is the set of ll nd only those words over Σ tht re ccepted y M L(M) for the lnguge ccepted y M FSAs re lnguge cceptors only 5

6 A Nondeterministic Mchine q q 1 2 q 0 q 4 q 3 Figure 933 Nondeterminism δ Μ : Q Σ Q is trnsition mpping Assumed to e totl ( fully defined ) ut permitted to e multivlued Word Acceptnce Word w Σ* ccepted y M provided there exists pth, leled y w, in the stte digrm of M leding from q init to terminl stte Compre deterministic cse 6

7 Word Acceptnce Word w Σ* ccepted y M provided there exists pth, leled y w, in the stte digrm of M leding from q init to terminl stte Compre deterministic cse Lnguge Acceptnce The lnguge ccepted y nondeterministic is set of words ccepted y M Nondeterministic Design Often Esier q 1 q 0 q 3 Figure 934 This nondeterministic finitestte utomton ccepts he t lnguge denoted y regulr expression * ((*) + ) q 2 7

8 Gol Suppose given nondeterministic M tht ccepts L We seek lgorithm for constructing, on sis of M, new deterministic M tht ccepts L Suset Construction Sttes of nondeterministic M will correspond to nonempty sets of sttes of deterministic M Whereq 0 is strt stte of M, use{q 0 }sstrt stte of M Accepting sttes of M will e those sttesets contining t lest one ccepting stte of M Suset Construction (cont) For ech stte-set S nd for ech s in M s lphet, we drw n rc leled s from stte S to tht stte-set (cll it S s-succ) ) consisting of ll nd only the s-successors of memers of S Eliminte ny stte-set, s well s ll rcs incident upon it, such tht there is no pth leding to it from {q 0 } 8

9 Exmple M hs 4 sttes So M will hve stte(-sets) Terminl stte(-sets) sets) mrked y ullet -rc from stte {q 0, q 1 } will led to stte {q 1, q 2, q 3 } in new stte digrm for M Theorem (Kleene) LetM e nondeterministic FSA ccepting L Then there exists deterministic finite-stte utomton M tht ccepts L s well Theorem Any finite lnguge is FSA-cceptle Exmple L = {,, } 9

10 Finite-Stte Automt with Executing rcs leled do not dvnce input -rcs my or my not introduce nondeterminism Exmple c c c 3 c This FSA ccepts the lnguge L(**c*) Figure 971 Equivlence Result LetM e FSA with -moves Then there exists FSA M with no -moves such tht L(M) ( ) = L(M ) ( ) 10

11 Equivlence of FA nd RE Finite Automt nd Regulr Expressions re equivlent To show this: Show we cn express DFA s n equivlent RE Show we cn express RE s n -NFA Since the -NFA cn e converted to DFA nd the DFA to n NFA, then RE will e equivlent to ll the utomt we hve descried Turning DFA into RE Theorem: If L=L(A) for some DFA A, then there is regulr expression R such tht L=L(R) Proof Construct GNFA, Generlized NFA We ll skip this in clss, ut see the textook for detils Stte Elimintion We ll see how to do this next, esier thn inductive construction, there is no exponentil numer of expressions DFA to RE: Stte Elimintion Elimintes sttes of the utomton nd replces the edges with regulr expressions tht includes the ehvior of the eliminted sttes Eventully we get down to the sitution with just strt nd finl node, nd this is esy to express s RE 11

12 Stte Elimintion Consider the figure elow, which shows generic stte s out to e eliminted The lels on ll edges re regulr expressions To remove s, we must mke lels from ech q i to p 1 up to p m tht include the pths we could hve mde through s R 1m R q 1 p 11 +Q 1 S*P 1 R 1 11 q 1 p 1 Q 1 S P 1 R 1m +Q 1 S*P m s Q K P m R k1 +Q k S*P 1 R km q k p m q k R km +Q k S*P m p m R k1 Note: q nd p my e the sme stte! DFA to RE vi Stte Elimintion (1) 1 Strting with intermedite sttes nd then moving to ccepting sttes, pply the stte elimintion process to produce n equivlent utomton with regulr expression lels on the edges The result will e one or two stte utomton with strt stte nd ccepting stte DFA to RE Stte Elimintion (2) 2 If the two sttes re different, we will hve n utomton tht looks like the following: R U Strt S T We cn descrie this utomton s: (R+SU*T)*SU* 12

13 DFA to RE Stte Elimintion (3) 3 If the strt stte is lso n ccepting stte, then we must lso perform stte elimintion from the originl utomton tht gets rid of every stte ut the strt stte This leves the following: R Strt We cn descrie this utomton s simply R* DFA to RE Stte Elimintion (4) 4 If there re n ccepting sttes, we must repet the ove steps for ech ccepting sttes to get n different regulr expressions, R 1, R 2, R n For ech repet we turn ny other ccepting stte to non-ccepting The desired regulr expression for the utomton is then the union of ech of the n regulr expressions: R 1 R 2 R N DFA RE Exmple Convert the following to RE 0 0,1 1 1 Strt First convert the edges to RE s: Strt

14 DFA RE Exmple (2) Eliminte Stte 1: Strt To: Note edge from Answer: (0+10)*11(0+1)* 11 Strt 3 2 Second Exmple Automt tht ccepts even numer of 1 s Strt Eliminte stte 2: *1 10*1 Strt 3 1 Second Exmple (2) *1 10*1 Strt 3 1 Two ccepting sttes, turn off stte 3 first *1 Strt 1 10*1 3 This is just 0*; cn ignore going to stte 3 since we would die 14

15 Second Exmple (3) *1 10*1 Strt 3 Turn off stte 1 second: *1 10*1 Strt 3 1 This is just 0*10*1(0+10*1)* Comine from previous slide to get 0* + 0*10*1(0+10*1)* Converting RE to n Automt We hve shown we cn convert n utomt to RE To show equivlence we must lso go the other direction, convert RE to n utomton We cn do this esiest y converting RE to n - NFA Inductive construction Strt with simple sis, use tht to uild more complex prts of the NFA Bsis: RE to -NFA R= R= R=Ø Next slide: More complex RE s 15

16 R=S+T S T R=ST S T R=S * S RE to -NFA Exmple Convert R= (+)* to n NFA We proceed in stges, strting from simple elements nd working our wy up RE to -NFA Exmple (2) + (+)* 16

17 Wht hve we shown? Regulr expressions nd finite stte utomt re relly two different wys of expressing the sme thing In some cses you my find it esier to strt with one nd move to the other Eg, the lnguge of n even numer of one s is typiclly esier to design s NFA or DFA nd then convert it to RE 17

Homework 3 Solutions

CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.