DecidableProblemsand RecursiveLanguages. We examine the computation of TMs thatdoanddon thalt.

Transcription

1 DecidableProblemsand RecursiveLanguages We examine the computation of TMs thatdoanddon thalt.

2 Recursive and Recursively Enumerable A language is recursively enumerable(r.e.) if itisthesetofstringsacceptedbysometm. Alanguageisrecursiveifitisthesetofstrings acceptedbysometmthathaltsoneveryinput. For example, any regular language is recursive. Wewillshowlaterthattherearemanyr.e.languages that are not recursive, and many languagesthatarenotevenr.e. Goddard 13: 2

3 Closure Facts Fact. (a)thesetofr.e.languagesisclosedunder union and intersection. (b)thesetofrecursivelanguagesisclosedunder union and intersection. Weprove(a)forunion. Saylanguages L 1 and L 2 arer.e.,acceptedby TMs M 1 and M 2. ATMfor L 1 L 2 simplyruns M 1 and M 2 inparallel.theinputisintheunion exactlywhenatleastonemachinehaltsandaccepts. Goddard 13: 3

4 Recursive, R.e., and Complements Theorem.Alanguageisrecursiveifandonlyif bothitanditscomplementarer.e. If Lisrecursive, thensoisitscomplement (interchange h a and h r ). Assumeboth Land Larer.e.; thatis, they havetms. ThenrunthetwoTMsinparallel. Atleastonewillhalt,andthatgivesusthe answer. Goddard 13: 4

5 A Printer Turing Machine A printer-tm is TM with an added printer-tape. The printer-tm writes strings on the printertape(separatedby );oncewrittenastringis not altered. Theorem.Alanguageisr.e.ifandonlyifsome printer-tm outputs precisely those strings. Theproofisintwoconstructions... Goddard 13: 5

6 From Printer-TM to Standard TM Armedwithprinter-TM M forlanguage L, we buildstandardtm N:Oninput x,tm Nruns M andmonitors M;if Neverfinds xontheprintertape,thenitaccepts.so Nacceptsstringsin L, and does not halt otherwise. Goddard 13: 6

7 From Standard TM to Printer-TM ArmedwithstandardTM Nfor L,webuildprinter- TM M. Theideaistorun N oneverypossible string in parallel an infinite number of tasks! The printer-tm works in rounds. In round i, M starting from scratch, generates the first i strings lexicographically(in dictionary order), runs N oneachfor isteps,andoutputsanystring that is accepted. Eventually, every string in L(N)willbegeneratedand Lrunforlongenough, andwillappearintheoutput. Goddard 13: 7

8 Decidable Questions Givenaquestion,webuildalanguagebytakingalltheinstancesofthequestionwherethe answerisyesandconvertingtheinstancetoa string. A yes/no question is decidable if the associated language is recursive. This is equivalent tofindingaprogramthatalwayshaltsandanswers the question correctly. Goddard 13: 8

9 Encodings We assume a standard encoding of a machine. E.g.,theformatyouwouldusetodescribeitif you wrote a Java program that simulated that type. The actual encoding does not matter, as longasitisfixedandonecaneasilygetbetween A and its encoding A. We use A, B to denote the encoding of pair {A, B}. Goddard 13: 9

10 Questions about Regular Languages It is easy to answer questions about regular languages: Recursive are: a)theacceptanceproblem: A fa = { M, w : Mis FAthataccepts w }. b)theemptinessproblem: Empty fa = { M : M isfawithemptylanguage }. c)theequivalenceproblem: EQ fa = { A, B : A and BareFAswith L(A) = L(B) }. Weprovepart(c). Goddard 13: 10

11 Proof that Equivalence is Decidable WebuildaTMthatfirstcheckstheinputhas the correct syntax(encodes some pair of FAs). If not, it rejects. Then,oneapproachistoconstructtheFAfor: (A, B) = (L(A) L(B)) (L(B) L(A)) and test (A, B) for emptiness. Goddard 13: 11

12 Questions about Context-Free Languages Many questions about context-free grammars or languages can be readily answered: Recursive are: a)theacceptanceproblem: A cfg = { G, w : Gis acfgand wastringof L(G) } b) Any context-free language with grammar G c) The emptiness problem: Empty cfg = { G : L(G) is empty } Weprovepart(a). Goddard 13: 12

13 Proof that Acceptance is Decidable For example, convert to Chomsky Normal Form. Examineallderivationsoflength 2 w 1and conclude.(see also CYK algorithm from earlier.) Goddard 13: 13

14 Totality is Not Decidable Surprisingly, perhaps, testing whether a CFG generates every string is hard: Total cfg = { G : L(G) = Σ }isnotrecursive. This will be established later. Goddard 13: 14

15 Configurations Recallthataconfigurationofamachineisa complete record of the machine that tells one the current state and contents of memory. Specifically,fora1-tapeTM,wewritethestate wheretheheadis.thatis,theconfigurationis written as t L S t R where t L istheusedtapetotheleftofthehead, Sthecurrentstate,and t R theusedtapetothe rightofthehead. Goddard 13: 15

16 Configurations and Loops For a deterministic machine, if the same configurationrecurs,thenthemachineisstuckinan infinite loop. Hence: Fact. If there are at most Q possible configurationsofadeterministicmachineanditrunsfor longerthan Q,thenitisinaninfiniteloop. Goddard 13: 16

17 Computation Strings A computation string for machine M accepting string w is the string of configurations from start to finish(separated by some special symbol). Fact.Itisdecidablewhetheragivenstringisa computation string or not. Check that first configuration is correct and matches the input; check that each configuration follows frompreviousbytherulesof M;andcheckthat final configuration is accepting. Goddard 13: 17

18 Other Models Recall that we introduced the Chomsky Hierarchyearlier. Thoughwedonotproveit,thetop level corresponds to TMs: Theorem. There is an unrestricted grammar for alanguageifandonlyifitisr.e. Goddard 13: 18

19 Linearly Bounded Automata Alinearlyboundedautomaton(LBA)isa1- tapetmwhoseheadisnotallowedtomoveoff theinputportionofthetape(andthereisadevicethattellsitwherethetapestartsandfinishes). Thoughwedonotproveit,itturnsoutthat: Theorem. There is a nondeterministic LBA for a language if and only if there is a contextsensitive grammar for it. Goddard 13: 19

20 Practice 1.ShowthatitisdecidablewhetheranNFA M generates all strings from the alphabet. 2. Showthatthesetofrecursivelanguagesis closed under reversal. Goddard 13: 20

21 Solutions to Practice 1.Oneapproachistoconvert MtoaDFA.Itis easytoseewhetheradfaacceptseverystring: only if every state is accepting. 2. Thequestionsasksonetoshowthat,if Lis recursive,thensois L R = { x R : x L }. If L isdecidedbytm M,then L R canbedecidedby atmthatsimplyreversestheinputandthen calls M. Goddard 13: 21

22 Summary Recursive languages are accepted by TMs that always halt; r.e. languages are accepted by TMs. These two families are closed under intersection andunion.ifalanguageisrecursive,thensois itscomplement;ifbothalanguageanditscomplement are r.e., then the language is recursive. There is a connection with printer-tms. A problem is decidable if the associated language is recursive. All problems about FAs and REs are decidable; most problems about CFGs and PDAs are decidable. A computation string isarecordofthecomputationofamachine. Goddard 13: 22