Recursively Enumerable and Recursive Languages

Transcription

1 CS 0 - ecture Recursve nd Recursvely Enumerble nguges Fll 008 Revew nguges nd Grmmrs Alphbets, strngs, lnguges Regulr nguges Determnstc Fnte nd Nondetermnstc Automt Equvlence of NFA nd DFA Regulr Expressons Regulr Grmmrs Propertes of Regulr nguges nguges tht re not regulr nd the pumpng lemm Context Free nguges Context Free Grmmrs Dervtons: leftmost, rghtmost nd dervton trees Prsng nd mbguty Smplfctons nd Norml Forms Nondetermnstc Pushdown Automt Pushdown Automt nd Context Free Grmmrs Determnstc Pushdown Automt Pumpng emm for context free grmmrs Propertes of Context Free Grmmrs Turng chnes Defnton, Acceptng nguges, nd Computng Functons Combnng Turng chnes nd Turng s Thess Turng chne Vrtons Unversl Turng chne nd ner Bounded Automt Tody: Recursve nd Recursvely Enumerble nguges Recursvely Enumerble nd Recursve nguges Defnton: A lnguge s recursvely enumerble f some Turng mchne ccepts t

2 et nd For strng : w be recursvely enumerble lnguge the Turng chne tht ccepts t w f then hlts n fnl stte f w then hlts n non-fnl stte or loops forever Defnton: A lnguge s recursve f some Turng mchne ccepts t nd hlts on ny nput strng In other words: A lnguge s recursve f there s membershp lgorthm for t et nd For strng : w be recursve lnguge the Turng chne tht ccepts t w f then hlts n fnl stte Turng cceptble lnguges nd Enumerton Procedures f w then hlts n non-fnl stte

3 We wll prove: (wek result) If lnguge s recursve then there s n enumerton procedure for t Theorem: f lnguge s recursve then there s n enumerton procedure for t (strong result) A lnguge s recursvely enumerble f nd only f there s n enumerton procedure for t Proof: ~ Enumerton chne Enumertes ll strngs of nput lphbet Accepts {, b} If the lphbet s then ~ cn enumerte strngs s follows: b b b bb b...

4 Enumerton procedure Exmple: = { b, b, bb,,...} Repet: ~ genertes strng checks f w YES: prnt NO: End of Proof w gnore w w to output ~ b b b bb b... ( ) b b bb... Enumerton Output b b bb... Theorem: Proof: f lnguge s recursvely enumerble then there s n enumerton procedure for t Enumerton chne ~ Enumertes ll strngs of nput lphbet Accepts

5 {, b} If the lphbet s then ~ cn enumerte strngs s follows: b b b bb b Enumerton procedure Repet: ~ genertes strng NAIVE APPROACH checks f w YES: prnt NO: w Problem: If mchne w w gnore w to output my loop forever ~ BETTER APPROACH Genertes frst strng ~ Genertes thrd strng w w executes frst step on ~ Genertes second strng w w executes frst step on w second step on w executes frst step on w And so on... second step on w thrd step on w 5

6 Step n strng w w w w If for ny strng w mchne hlts n fnl stte then t prnts w on the output End of Proof Theorem: If for lnguge there s n enumerton procedure then s recursvely enumerble Proof: chne tht ccepts Input Tpe w Enumertor for Compre 6

7 Turng mchne tht ccepts For nput strng w Repet: Usng the enumertor, generte the next strng of We hve proven: A lnguge s recursvely enumerble f nd only f there s n enumerton procedure for t Compre generted strng wth w If sme, ccept nd ext loop End of Proof We wll prove:. There s specfc lnguge whch s not recursvely enumerble (not ccepted by ny Turng chne) Non Recursvely Enumerble Recursvely Enumerble. There s specfc lnguge whch s recursvely enumerble but not recursve Recursve 7

8 Some nguges Are Not Recursvely Enumerble Defnton: A set s uncountble f t s not countble Proof Usng Uncountble Sets Theorem: et S be n nfnte countble set S The powerset of S s uncountble Proof: Snce S s countble, we cn wrte S = { s, s, s, } Elements of S 8

9 Elements of the powerset hve the form: We encode ech element of the power set wth bnry strng of 0 s nd s { s, s} { s5, s7, s9, s0} Powerset element { s } Encodng s s s s { s, s} 0 0 { s, s, s} 0 et s ssume (for contrdcton) tht the powerset s countble. Powerset element Encodng t Then: we cn enumerte the elements of the powerset t t t 0 0 9

10 t Tke the powerset element whose bts re the complements n the dgonl t t t 0 0 New element: 00 (brry complement of dgonl) The new element must be some of the powerset However, tht s mpossble: from defnton of t t Snce we hve contrdcton: The powerset of s uncountble S S the -th bt of t must be the complement of tself Contrdcton!!! 0

11 An Applcton: nguges Exmple Alphbet : {, b} The set of ll Strngs: * S = {, b} = { λ,, b,, b, b, bb,, b, } nfnte nd countble Exmple Alphbet : {, b} The set of ll Strngs: * S = {, b} = { λ,, b,, b, b, bb,, b, } nfnte nd countble A lnguge s subset of : S = {, b, b} Exmple Alphbet : {, b} The set of ll Strngs: * S = {, b} = { λ,, b,, b, b, bb,, b, } nfnte nd countble S The powerset of contns ll lnguges: S = {{ λ},{ },{, b}{, b, b}, } uncountble nguges: uncountble Turng mchnes: countble There re more lnguges thn Turng chnes k?

12 Concluson: There re some lnguges not ccepted by Turng chnes (These lnguges cnnot be descrbed by lgorthms) nguges not ccepted by Turng chnes k nguges Accepted by Turng chnes A nguge whch s not Recursvely Enumerble We wnt to fnd lnguge tht s not Recursvely Enumerble Ths lnguge s not ccepted by ny Turng chne

13 Consder lphbet {} Strngs:,,,, Consder Turng chnes tht ccept lnguges over lphbet {} They re countble:,,,, ( Exmple lnguge ccepted by ( ) = {,, } ( ) = {,, Alterntve representton ) 6 5 } ( ) ( ) ( ) ( ) 0 0 0

14 Consder the lnguge = { : ( )} ( ) ( ) ( ) conssts from the s n the dgonl ( ) = {,, } Consder the lnguge ( ) 0 0 = { : ( )} ( ) 0 0 = { : ( )} conssts of the 0 s n the dgonl ( ) 0 ( ) = {,, }

15 Theorem: Proof: nguge s not recursvely enumerble Assume for contrdcton tht s recursvely enumerble There must exst some mchne tht ccepts k ( k ) = ( ) 0 0 ( ) 0 0 ( ) 0 0 ( ) 0 0 ( ) 0 ( ) 0 ( ) Queston: k =? ( ) Answer: k ( k ) ( ) 5

16 ( ) 0 0 ( ) 0 0 ( ) 0 0 ( ) 0 0 ( ) 0 ( ) 0 ( ) Queston: k =? ( ) Answer: k ( k ) ( ) ( ) 0 0 ( ) 0 0 ( ) 0 0 ( ) 0 0 ( ) 0 ( ) 0 ( ) Queston: k =? ( ) Answer: k ( k ) ( ) 6

17 Smlrly: k for ny Therefore, the mchne k cnnot exst Becuse ether: ( k ) ( ) or ( k ) ( ) Therefore, the lnguge s not recursvely enumerble End of Proof Observton: There s no lgorthm tht descrbes Non Recursvely Enumerble Recursvely Enumerble (otherwse would be ccepted by some Turng chne) Recursve 7

18 Red Wht s Next nz Chpter,.,.,., (skp.),,, 5, 6., 6., (skp 6.), 7., 7., 7., (skp 7.), 8, 9, 0, JFAP Chpter,., (skp.),,, 5, 6, 7, (skp 8), 9, (skp 0), Next ecture Topcs From.,.,., nd. ore Recursve nguges, Unrestrcted Grmmrs, Context Senstve Grmmrs, nd the Chomsky Herrchy Quz n Rectton on Wednesdy / Covers nz 7., 7., 7., (skp 7.), 8, nd JFAP 5,6,7 Closed book, but you my brng one sheet of 8.5 x nch pper wth ny notes you lke. Quz wll tke the full hour Homework Homework Due Thursdy 8