Overview. Regular Languages. Finite Automata. A finite automaton. Startstate : q Acceptstate : q. Transitions

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1 Overvew Regulr Lnguges Andres Krwth & Mlte Helmert Determnstc fnte utomt Regulr lnguges Nondetermnstc fnte utomt Closure oertons Regulr exressons Nonregulr lnguges The umng lemm 2 Fnte Automt A fnte utomton An ntutve exmle : suermrket door controller Fgure.4 Formlly front d door rer d To vew of n utomtc door Prolstc counterrts exst Mrkov chns, Byesn nets, etc. Not n ths course rer oth nether closed Trnston tle for the utomtc door controler: closed oen front nether nether closed closed oen front oen oen front rer oth Stte dgrm for the utomtc door controler rer closed oen oth closed oen 3 q, Sttes : q, q, q 2 3 Strtstte : q Accetstte : q Trnstons Outut : ccet or reject 2 q 3 A fnte utomton s 5-tule ( Q, Σ, δ, q, F). Qs fnte set of sttes 2. Σ s fnte set, the lhet 3. δ : Q Σ Qs the trnston functon 4. q Q s the strt stte o 5. F Qs the set of ccet sttes 4

2 q q 3 As the lnguge of mchne M we wrte LM ( ) = A A= { w wcontns t lest one nd n even numer of s follows the lst } Descre M Q= { q, q, q } 2 3 Σ= {,} δ defned y q strt stte F = { q } q q q q q q q q q 2 q Stte dgrm of the two-stte fnte utomton M 2 q Stte dgrm of the two-stte fnte utomton M Other exmles Another exmle 7,8,9 A generlston : A s the lnguge of ll strngs where the sum of the numers s multle of excet tht the sum s reset to whenever the symol reset ers Automton B =. Q = { q,..., q } 2. Σ ={,,2, reset } 3. δ ( q,) = q j δ ( q,) = q where k = ( j+ ) mod j δ ( q, 2) = q where k = ( j+ 2) mod j δ ( q, reset ) = q j k j k 4. q Q s strt nd ccet stte o 7 8

3 Forml defnton of comutton Desgnng fnte utomt Let M e fnte utomton ( Q, Σ, δ, q, F) Let w= w... w e strng over Σ M ccets wf sequence of sttes r,..., r exsts n Q such tht. r δ ( r, w ) = r for ll =,..., n 3. r F n = q n n Desgn utomton for lnguge consstng of nry strngs wth n odd numer of s Desgn frst sttes Then trnstons q even Strt stte nd ccet sttes q odd M recognzes lnguge Af A= { w M ccets w} A lnguge s regulr f some fnte utomton recognzes t. 9 Another exmle Another exmle Desgn n utomton to recognze the lnguge of nry strngs contnng the strng s sustrng We hve four ossltes:. we hven t seen ny symol of the ttern yet, or 2. we hve seen, or 3. we hve seen, or 4. we hve seen the ttern Desgn n utomton to recognze the lnguge of nry strngs contnng the strng s sustrng We hve four ossltes:. we hven t seen ny symol of the ttern yet, or 2. we hve seen, or 3. we hve seen, or 4. we hve seen the ttern q q, q q 2

4 The Regulr Oertons Regulr lnguges re closed under Let A nd B e lnguges We defne : Unon : A B= { x x A or x B} Conctenton : Ao B= { xy x A nd y B} Str : A = { xx2... xn n nd ech x A} note: lwys A A set S s closed under n oerton of lyng oon elements of S yelds elements of S. Exmle: multlcton on nturl numers Counterexmle :dvson of nturl numers Exmle A = { good, d} B = { oy, grl} A B= { good, d, oy, grl} Ao B = { goodoy, goodgrl, doy, dgrl} Theorem.2 The clss of the regulr lnguges s closed under the unon oerton. In other words, f A nd A re regulr lnguges, so s A A 2 2 A = {, good, d, goodgood, goodd, dgood, dd, goodgoodgood, goodgoodd,...} 4 5 Proof.2 (y constructon) Let M recognze A, where M = (Q, Σ, δ,q,f ), nd M recognze A, where M = (Q, Σδ,,q,F ) Construct M to recognze A A, where M = (Q, Σ,δ, q,f ). { }. Q = (r,r ) r Q nd r Q Ths set s the Crtesn roduct of sets Q nd Q (wrtten Q Q ) It s the set of ll rs of sttes, the frst from Q nd the second from Q Σ, the lhet, s the sme s n M nd M. The theorem remns true f they hve dfferent lhets, Σ nd Σ. We would then modfy the roof to let Σ = Σ Σ δ, the trnston functon, s defned s follows. For ech (r,r ) Q nd ech Σ, let δ(( r,r ), ) = ( δ ( r, ), δ ( r, )) Hence δ gets stte of M (whch ctully s r of sttes from M nd M ), together wth n nut symol, nd returns M's next stte. 4. q s the r (q,q ) F M M2 5. s the set of rs n whch ether memer s n ccet stte of nd. We cn wrte t s F = {(r,r ) r F or r F } Ths exresson s the sme s F = (F Q ) (Q F ) Note tht t s not the sme s F = F F. Wht would tht gve us? 6 7

5 Exmle M wth L(M ) = {w w contns }, q M = ( Q, Σ, δ, q, F) constructed from M = ( Q, Σ, δ, q, F) nd M = ( Q, Σ, δ, q, F ) Defne. Q= {( r, r ) r Q nd r Q } 2. Σ=Σ Σ δ(( r, r ), ) = ( δ ( r, ), δ ( r, )) q = ( q, q ) 2 5. F = {( r, r ) r F or r F } M 2 wth L(M 2 ) = {w w contns t lest two s}, 2 3 Theorem.3 The clss of the regulr lnguges s closed under the conctenton oerton. In other words, f A nd A re regulr lnguges, so s A o A Non determnstc fnte utomt Determnstc versus non determnstc comutton Determnstc One successor stte trnstons not llowed Non determnstc Severl successor sttes ossle trnstons ossle Fgure 5,,, q q 3 q 4 2 2

6 Another NFA,, q, q 3 q 4 q q Inut: w = q q 3 q q 3 q q 3 q 4 q q 3 q 4 q 4 q q 3 q 4 q Nondetermnstc fnte utomton Exmle A nondetermnstc fnte utomton s 5-tule ( Q, Σ, δ, q, F). Qs fnte set of sttes 2. Σ s fnte set, the lhet 3. δ : Q Σ P( Q) s the trnston functon 4. q Q s the strt stte o 5. F Qs the set of ccet sttes Σ ncludes P( Q) the owerset of Q,,, q q 3 q 4. Q= { q, q, q, q } Σ = {,} 3. δ s gven s: q { q} { q, q } {} 2 q { q } {} { q } q {} { q } {} 3 4 q { q } { q } {} q s the strt stte 5. F = { q }

7 Forml defnton of comutton Every NFA hs n equvlent DFA Let M e nondetermnstc fnte utomton ( Q, Σ, δ, q, F) Let w= w... w e strng over Σ n M ccets wf sequence of sttes r,..., r exsts n Q such tht. r = q 2. r δ ( r, w ) for ll =,..., n r n F n Equvlence NFA nd DFA Proof: Theorem.9 Two mchnes re equvlent f they recognze the sme lnguge Theorem.9 Every nondetermnstc fnte utomton hs n equvlent fnte utomton Let N = (Q, Σ, δ,q,f ) e the NFA recognzng some lnguge A. Construct DFA M recognzng A. Frst we consder the eser cse wheren N hs no rrows. The rrows re tken nto ccount lter. Corollry.2 A lnguge s regulr f nd onl y f some nondetermnstc fnte utomton recognzes t

8 Proof: Theorem.9 (cont.) Proof: Theorem.9 (cont.) ' ' Construct M = (Q', Σ, δ,q,f').. Q' = P(Q ). Every stte of M s set of sttes of N. (Recll tht P(Q) s the ower set of Q). 2. For R Q' nd Σ let δ '(R,) = {q Q q δ(r,) for some r R}. If R s stte of M, t s lso set of sttes of N. When M reds symol n stte R, t shows where tkes ech stte n R. Becuse ech stte leds to to set of sttes, we tke the unon of ll these sets. Alterntvly we wrte: 3. q = {q }. ' U δ '( R, ) = δ( r, ). r R M strts n the stte corresondng to the collecton contnng just the strt stte of N. 4. F' = {R Q' R contns n ccet stte of N }. The mchne M ccets f one of the ossle sttes tht N could e n t ths ont s n ccet stte. Now for the rrows one needs to set u n extr t of notton. For ny stte R of M we defne E(R) to e the collecton of sttes tht cn e reched from R y gong only long rrows, ncludng the memers of R themselves. Formlly, for R Q let E( R ) = { q q cn e reched from R y trvelng long or more rrows }. The trnston functon of M s then modfed to tke nt o ccount ll sttes tht cn e reched y gong long rrows fter every ste. Relcng δ(r,) y E( δ(r, )) cheves ths. Thus δ '( R, ) = { q Q q E( δ( r, )) for some r R }. Addtonlly the strt stte of M hs to e modfed to cter for ll ossle sttes tht cn e reched from the strt stte of N long the rrows. Chngng q to e E({q }) cheves ths effect. ' We hve now comleted the constructon of the DFA M tht smultes the NFA N. 3 3 An exmle An exmle The resultng DFA The resultng DFA fter removng redundnt sttes 32 33

9 Closure under the regulr oertons Proof de Theorem.2/.22 The clss of the regulr lnguges s closed under the unon oerton. In other words, f A nd A re regulr lnguges, so s A A 2 2 Theorem.23 The clss of the regulr lnguges s closed under the conctenton oerton. Theorem.2/.22 The clss of the regulr lnguges s closed under the unon oerton. In other words, f A nd A re regulr lnguges, so s A A 2 2 Theorem.24 The clss of the regulr lnguges s closed under the str oerton Proof.2/.22 Proof de Let N = (Q, Σ, δ,q,f ) recognze A, nd N = (Q, Σ, δ,q,f ) recognze A Construct N = (Q, Σ, δ,q,f ) to recognze A A. 2 Theorem.23 The clss of the regulr lnguges s closed under the conctenton oerton.. Q = {q } Q Q. 2 The sttes of N re ll the sttes of N nd N, wth the ddton of new strt stte q. 2. The stte q s the strt stte of N. 3. The ccet sttes F = F F. 2 2 The ccet sttes of N re ll the ccet sttes of N nd N. Tht wy N ccets f ether N ccets or N ccets Defne δ so tht for ny q Q nd ny Σ, δ(q,) q Q δ2(q,) q Q2 δ (q,) = {q,} q= q nd = q = q nd

10 Proof.23 Proof de Let N = (Q, Σ, δ,q,f ) recognze A, nd N = (Q, Σ, δ,q,f ) recognze A Construct N = (Q, Σ, δ,q,f ) to recognze A o A. 2 2 Theorem.24 The clss of the regulr lnguges s closed under the str oerton.. Q = Q Q. 2 The sttes of N re ll the sttes of N nd N The stte q s the sme s the strt stte of N. 3. The ccet sttes F re the sme s the ccet sttes of N Defne δ so tht for ny q Q nd ny Σ, δ(q,) q Q nd q F δ (q,) q F nd δ (q,) = δ (q,) {q 2} q = F nd = δ (q,) q Q Proof.24 Regulr exressons Let N = (Q, Σ, δ,q,f ) recognze A. Construct N = (Q, Σ, δ,q,f ) recognze A.. Q = {q } Q. The sttes of N re the sttes of N lus new strt stte. 2. The stte q s the new strt stte. 3. F = {q } F The ccet sttes re the old ccet sttes lus the new strt stte. 4. Defne δ so tht for ny q Q nd ny Σ, δ(q,) q Q nd q F δ (q,) q F nd δ (q,) = δ (q,) {q } q F nd = {q } q= q nd = q = q nd Defnton Sy tht R s regulr exresson f R s. for some n the lhet Σ, 2., 3., 4. (R R ), where R nd R re regulr exressons, (R o R ), where R nd R re regulr exressons, or R, where Rs regulr exresson. 4 4

11 RE Exmles RE Exmles (cont.) In the followng exmles we ssume tht the lhet Σ s {,}.. = {w whs exctly sngle }. 2. Σ Σ = {w w hs t lest one }. 3. Σ Σ = { w w contns the strng s sustrng }. 4. ( ΣΣ) = {w w s strng of even length }. 5. ( ΣΣΣ ) = {w the length of w s multle of three }. 6. = {, }. 7. Σ Σ = { w w strts nd ends wth the sme symol }. 8. ( )( ) =. The exresson descres the lnguge {, }, so the conctenton oerton dds ether or efore every strng n. 9. ( )( ) = {,,,}.. =. Conctentng the emty set to ny set yelds the emty set.. = { }. The str oerton uts together ny numer of strngs from the lnguge to get strng n the result. If the lnguge s emty, the str oerton cn ut together strng, gvng only the emty strng. R = R Ro = R R Ro R R Alctons Equvlence RE nd NFA Desgn of comlers { +,, }( DD DD. D D. DD ) where D = {,...,9} Theorem.28 A lnguge s regulr f nd only f some regulr exresson descres t Proof through : wk, gre, v n unx (serch for strngs) Perl, Python, or Jv rogrmmng lnguges Bonformtcs So clled motfs (tterns occurrng n sequences, e.g. rotens) Lemm.29 If lnguge s descred y some regulr exresson, then t s regulr Lemm.32 If lnguge s regulr, then t s descred y some regulr exresson 44 45

12 Proof for Lemm.29 (cont.) Proof for Lemm.29 (cont.) 2. R =. Then L(R) = { }, nd the followng NFA recognzes L(R). 3. R =. Then L(R) =, nd the followng NFA recognzes L(R). Formlly, N = ({q }, Σ, δ,q,{q }), where δ (r,) = for ny r nd. Formlly, N = ({q}, Σ, δ,q, ), where δ (r,) = for ny r nd Proof for Lemm.29 (cont.) Exmle.3 4. R= R R R= Ro R R = R. For the lst three cses we use the constructons gven n the roofs tht the clss of regulr lnguges s closed under the regulr oertons. In other words, we construct the NFA for R from the NFAs for R nd R (or just R n cse 6) nd the rorte closure constructon. 2 We convert the regulr exresson ( ) to n NFA n sequence of stges. We uld u from the smllest suexressons to lrger suexressons untl we hve n NFA for the orgnl exresson, s shown n the followng dgrm. Note tht ths rocedure generlly doesn't gve the NFA wth the fewest sttes! 49 5

13 Exmle: NFA for: ( ) Exercse: NFA for: ( ) : : : : : ( ) ( ) 5 52 Exmle: NFA for: ( ) (cont.) : Lemm.32 If lnguge s regulr, then t s descred y some regulr exresson ( ): Two stes DFA nto GNFA (generlzed nondetermnstc fnte utomton) Convert GNFA nto regulr exresson 53 54

14 GNFAs Formlly Lels re regulr exressons Two sttes q nd r re connected n oth drectons (fully connected) Exceton : One drecton only Strt stte (extng trnston rrows) Accet stte (only one!) (only ncomng trnston rrows) q strt Ø () q ccet 55 A generlzed nondetermnstc fnte utomton s 5-tule ( Q, Σ, δ, q, q ). Qs fnte set of sttes 2. Σ s fnte set, the lhet 3. δ : ( Q { q }) ( Q { q }) R s the trnston functon ccet 4. q Q s the strt stte strt strt 56 strt ccet 5. qccet Q the ccet stte A GNFA ccets w= w... wk where ech w Σ f sequence of sttes r,..., r exsts n Q such tht. r 2. r k = q = q strt ccet 3. for ll =,..., n, we h ve tht w L( R) where R = δ ( r, r) n Convert DFA nto GNFA Convert GNFA nto regulr exresson Add new strt stte, wth rrow to old strt stte Add new ccet stte, wth rrows from old ccet sttes If ny rrows hve multle lels nd, relce y Add rrows wth lel etween sttes where necessry (:etween sttes tht hd no rrows efore) 3 stte DFA 5 stte GNFA 4 stte GNFA Regulr Exresson 2 stte GNFA 3 stte GNFA 57 58

15 Rng of sttes Convert(G) Relce one stte y the corresondng RE R 4 q R q R r 3 R 2 (R )(R 2 ) ) (R 3 ) R 4 q Convert( G ) :. Let k e the numer of sttes of G. 2. If k = 2, then G must consst of strt stte, n ccet stte, nd sngle rrow connectong them nd leled wth regulr exresson R. Return the exresson R. 3. If k > 2, we select ny stte q Q dfferent from q nd q nd let G' e the GNFA (Q', Σδ, ',q,q ), where strt r strt ccet ccet Q' = Q { q }, nd for ny q Q' {q } nd ny q Q' {q } let r ccet j strt δ '(q,q ) = (R )(R )(R ) (R ), j 3 4 for R =δ (q,q ),R =δ (q,q ),R =δ (q,q ), nd R =δ(q,q ). r 2 r r 3 r j 4 j 4. Comute Convert( G') nd return ths vlue Exmle Another Exmle R q R 4 q r R 2 R 3 q (R )(R 2 ) (R 3 ) R 4 DFA: 2 3 GNFA: s 2 3 R 2: R : R : R 2: 2 s ( ) s ( ) 3 3 ( ) ( ) ( ) ( ) R 3: s (( ) )(( ) ( ) )(( ) ( ) ) ( ) 6 63

16 Inducton Proof Inducton Proof (cont.) R 4 q (R )(R 2 ) (R 3 ) R 4 q R q r R 2 R 2 Clm For ny GNFA G, Convert( G ) s equvlent to G. We rove ths clm y nducton on k, the numer of sttes of the GNFA. Bss: Prove the clm true for k = 2 sttes. If G hs only two sttes, t cn hve only sngle rrow, whch goes from the strt stte to the ccet stte. The regulr exresson lel on ths rrow descres ll the strngs tht llow G to get to the ccet stte. Hence ths exresson s equvlent to G. Inducton ste: Assume tht the clm s true for k sttes nd use ths ssumton to rove tht the clm s true for k sttes. Frst we show tht G nd G' recognze the sme lnguge. Suose tht G ccets n nut w. Then n n ccetng rnch of the comutton G enters sequence of sttes q,q,q,q,...,q. strt 2 3 ccet If none of them s the removed stte q, clerly G' lso ccets w. The reson s tht ech of the new regulr exressons lelng the rrows of G' contns the old regulr exresson s rt of unon. r 64 R 4 q (R )(R 2 ) (R 3 ) R 4 q R q r R 2 R 2 If q does er, removng ech run of consecutve q sttes r forms n ccetng comutton for G'. The sttes q nd q rcketng run hve new regulr exresson on the rrow etween them tht descres ll strngs tkng q to q v q on G. So G' ccets w. j r For the other drecton, suose tht G' ccets n nut w. As ech rrow etween ny two sttes q nd q n G' descres the collecton of strngs tkng q to q n G, ether drectly or v q,g must lso j r ccet wthus G nd G' re equvlent. The nducton hyothess sttes tht when the lorthm clls tself recursvely on nut G', the result s regulr exresson tht s equvlent to G' ecuse G' hs k sttes. Hence the regulr exresson lso s equvlent to G, nd the lgorthm s roved correct. Ths concludes the roof of Clm.34, Lemm.32, nd theorem j r j Nonregulr Lnguges The umng lemm Fnte Automt hve fnte memory Are the followng lnguges regulr? n n B= { n } C = { w whs n equl numer of s nd s} D= { w whs n equl numer of occurences of nd } Mthemtcl roof necessry If A s regulr lnguge, then there s numer (the umng length), where, f s s ny strng n A of length t lest then s my e dvded nto three eces s = xyz such tht. for ech, xy z A 2. y > 3. xy Note from 2: y 66 67

17 Proof Ide Proof: Pumng Lemm Let M e DFA recognzng A. Assgn to e the numer of sttes n M. Show tht strng s, wth length t lest, cnerokenntoxyz. Pgeonhole rncle Now rove tht ll three condtons re met Let M = (Q, Σ, δ, q, F) e DFA recognzng A nd Q =. Let s = s s 2...s n e strng n A, wth s = n, nd n Let r =r,...,r n+ e the sequence of sttes tht M enters for s, so r + = δ(r,s ) wth n. r,...,r n+ = n+, n+ +. Amoung the frst + elements n r, there must e r j nd r l eng the sme stte q m, wth j l. As r l occurs n the frst + sttes: l +. Let x = s...s j-, y = s j...s l- nd z = s l...s n : s x tkes M from r to r j, y from r j to r l, nd z from r l to r n+, eng n ccet stte, M must ccet xy z for wth j l, y > wth l +, xy Pumng Lemm (cont.) Nonregulr lnguges exmles Use umng lemm to rove tht lnguge A s not regulr:. Assume tht A s regulr (Proof y contrdcton) 2. use the lemm to gurntee the exstence of, such tht strngs of length or greter cn e umed 3. fnd strng s of A, wth s tht cnnot e umed 4. demonstrte tht s cnnot e umed usng ll dfferent wys of dvdng s nto x,y, nd z (usng condton 3. s here very useful ) 5. the exstence of s contrdcts the ssumton, therefore A s not regulr lnguge B n n = { n } Choose s = for s :. for ech, xy z A 2. y > 3. xy If would now only consder condton2, then we would hve tht:. strng y conssts only of s 2. strng y conssts only of s 3. strng y conssts of oth s nd s 7 7

18 C = { w whs n equl numer of s nd s} Exmle lnguge B gn Choose s = Would seem ossle wthout condton 3! However, condton 3 of lemm sttes xy Thus y conssts of s only Then xyyz C Choc e of s crucl. Consder s = () Alterntve roof : B s nonregulr If C were C = B regulr, then regulr Regulr lnguges closed under ntersecton for s :. for ech, xy z A 2. y > 3. xy n n B= { n } Choose s = condton 3 of lemm sttes xy Thus y conssts of s only Then xyyz B for s :. for ech, xy z A 2. y > 3. xy F = { ww w {,}} D 2 n = { n> } Choose s = Condton 3 of lemm sttes xy Thus y conssts of s only Then xyyz F would not work, s t cn e umed! for s :. for ech, xy z A 2. y > 3. xy Choose s = 2 Perfect squres :,,4,9,6,25,36,49,... Consder + xy z xy z nd xyz xy z + Prove tht for lrge : nd cnnot oth e erfect squres, whch should e true ccordng to umng lemm. Therefore, D s not regulr lnguge Proof: 2 Let m= n = xy z 2 2 Then: ( ) 2 2 n+ n = n+ = m+ Choose y < 2 m+ = 2 xy z + Indeed, oserve 2 4 ; let then: y s = = <2 y = + 2 xy z + for s :. for ech, xy z A 2. y > 3. xy 74 75

19 j E = { w where > j} Exmle Exm Queston for s :. for ech, xy z A 2. y > 3. xy + Choose s = Condton 3 of lemm sttes xy Thus y conssts of s only Then xy z F for s :. for ech, xy z A 2. y > 3. xy Q: Use the umng lemm to rove tht L = { k j : k,j nd k 2j} s not regulr. A: Assume tht L = { k j : k,j nd k 2j} s regulr. Let e the umng length of L. The umng lemm sttes tht for ny strng s L of t lest length, there exst strng x,y, nd z such tht s = xyz, xy, y >, nd for ll : xy z L. Choose s = 2. Becuse s L nd s =3, we otn from the umng lemm the strngs x,y, nd z wth the ove roertes. As s = xyz, xy, nd s egns wth 2 zeros, one cn see tht xy cn only consst of zeros. If we um s down,.e. select =, the strng xy z = xz = 2- y. As xz hs ones, nd y >, xz hs fewer thn 2 zeros. Hence xz L CONTRADICTION. Therfore L s not regulr! Exmle Exm Queston for s :. for ech, xy z A 2. y > 3. xy Summry Q: Use the umng lemm to rove tht L = { k j : k,j nd k 2j} s not regulr. A: Assume tht L = { k j : k,j nd k 2j} s regulr. Let e the umng length of L. The umng lemm sttes tht for ny strng s L of t lest length, there exst strng x,y, nd z such tht s = xyz, xy, y >, nd for ll : xy z L. Choose s = 2. Becuse s L nd s =3, we otn from the umng lemm the strngs x,y, nd z wth the ove roertes. As s = xyz, xy, nd s egns wth 2 zeros, one cn see tht xy cn only consst of zeros. If we um s down,.e. select =, the strng xy z = xz = 2- y. Determnstc fnte utomt Regulr lnguges Nondetermnstc fnte utomt Closure oertons Regulr exressons Nonregulr lnguges The umng lemm As xz hs ones, nd y >, xz hs fewer thn 2 zeros. Hence xz L CONTRADICTION. Therfore L s not regulr! 78 79