Undecidable Problems for CFGs CSCI 3130 Formal Languages and - PowerPoint PPT Presentation

1/21 Undecidable Problems for CFGs CSCI 3130 Formal Languages and Automata Theory Siu On CHAN Chinese University of Hong Kong Fall 2016

2/21 Decidable vs undecidable Decidable Undecidable DFA D accepts w TM M accepts w CFG G generates w TM M halts on w inputs TM M accepts some input same inputs CFG G generates all inputs? CFG G is ambiguous? DFAs D and D ′ accept same TM M and M ′ accept the

3/21 . . . q a . Representing computations . . L 1 = { w % w | w ∈ { a , b } ∗ } a / a R b / b R x / x R q 0 abb%abb q 2 abb%abb % / % R q 2 q 3 a / a L a / x L b / b L a / a L a / x R q 2 abb%abb x / x R x / x L b / b L q 3 abb%abb % / % R � / � L % / % L q 0 q 1 q 6 q 7 q 6 abb%xbb b / x R b / x L a / a R b / b R x / x R q 1 xxx%xxx � % / % R q a xxx%xxx q 4 q 5 x / x R

4/21 q acc A configuration consists of current state, head position, and tape contents Configuration (abbreviation) … Configurations … q acc a b a � q 1 ab q 1 a a / b R q 1 a b b � abb q acc

5/21 Computation histories computation . . . . . . history q a q 0 abb%abb a / a R b / b R x / x R a q 2 bb%abb % / % R q 2 q 3 a / a L abb q 2 %abb a / x L b / b L a / a L a / x R x / x R x / x L b / b L abb% q 3 abb abb q 2 %xbb % / % R � / � L % / % L q 0 q 1 q 6 q 7 b / x R b / x L a / a R xxx%xxx q 1 b / b R x / x R xxx%xx q a x % / % R q 4 q 5 x / x R

6/21 . The computation history can be written as a string h accepting history: Computation histories as strings M accepts w M rejects w rejecting history: . . C k If M halts on w , the computation history of ( M , w ) is the sequence of configurations C 1 , . . . , C k that M goes through on input w q 0 ab%ab # q 0 ab%ab #x q 1 b%ab # … #xx%x q a x # a q 2 b%ab � �� � � �� � � �� � C 1 C 2 over alphabet Γ ∪ Q ∪ { # } xx%xx q 1 xx%x q a x ⇔ q acc appears in h ⇔ q rej appears in h

7/21 Undecidable problems for CFGs We will argue that ALL CFG = {� G � | G is a CFG that generates all strings } The language ALL CFG is undecidable If ALL CFG can be decided, so can A TM A TM = {� M , w � | M is a TM that rejects or loops on w }

8/21 Undecidable problems for CFGs G generates all strings if M rejects or loops on w S reject if M accepts w loops on w accept if M rejects or A to G Convert We want to construct a Turing machine S that decides A TM reject otherwise accept if G generates all strings A Suppose some Turing machine A decides ALL CFG Proof by contradiction � G � � G � � M , w � G fails to generate some string if M accepts w

9/21 Undecidable problems for CFGs Convert to G M accepts w First we construct a PDA P , then convert it to CFG G � G � G fails to generate some string � � M , w � The alphabet of G will be Γ ∪ Q ∪ { # } G will generate all strings except accepting computation histories of ( M , w )

10/21 except accepting h of M , accept Reject Undecidablility via computation histories Otherwise, h must be an accepting history, reject accept everything candidate computation P history h of ( M , w ) ⇒ # q 0 ab%ab#x q 1 b%ab#…#xx%x q a x# P = on input h (try to spot a mistake in h ) ◮ If h is not of the form # w 1 # w 2 # … # w k # , accept ◮ If w 1 � = q 0 w or w k does not contain q a , accept ◮ If two consecutive blocks w i # w i +1 do not follow from the transitions

11/21 Computation is local . . . Changes between configurations always occur around the head q a a / a R b / b R x / x R q 0 ab%ab % / % R a q 2 b%ab q 2 q 3 a / a L ab q 2 %ab a / x L b / b L a / a L a / x R ab% q 3 ab x / x R x / x L b / b L ab q 2 %xb % / % R � / � L % / % L q 0 q 1 q 6 q 7 b / x R b / x L a / a R xx%xx q 1 b / b R x / x R xx%x q a x % / % R q 4 q 5 x / x R

12/21 … … Legal and illegal transitions windows … … … … aba … … … a/xL … … … … … … … … … … … … abx … abx … … … illegal windows … … legal windows … … … … … q 3 ab ab q 3 a q 3 a q 3 q 3 a q 3 q 6 ax q 3 q 3 x a q 3 a q 6 ab q 6 q 6 ab aa � a q 3 a xa � a q 6 x

13/21 Implementing P For every position of w i : Remember first row of window in state Remember second row of window in state #xb% q 3 ab If two consecutive blocks w i # w i +1 do not follow #xb q 5 %xb from the transitions of M , accept Remember ofgset from # in w i on stack Afuer reaching the next # : Pop ofgset from # from stack as you consume input If window is illegal, accept; Otherwise reject

14/21 The computation history method Convert to G G accepts all strings except accepting We first construct a PDA P , then convert it to CFG G ALL CFG = {� G � | G is a CFG that generates all strings } If ALL CFG can be decided, so can A TM computation histories of ( M , w ) � G � � M , w �

15/21 Post Correspondence Problem baba a cc bab ab c ab c bab a baa ab a bottom? Given an infinite supply of tiles from a particular set, can you match top and bab baba a a baa ab a ab c cc bab Input: A fixed set of tiles, each containing a pair of strings Top and bottom are both abaababccbaba ε ε

16/21 Undecidability of PCP The language PCP is undecidable PCP = {� T � | T is a collection of tiles that contains a top-bottom match }

17/21 Ambiguity of CFGs The language AMB is undecidable We will argue that If AMB can be decided, then so can PCP AMB = {� G � | G is an ambiguous CFG }

18/21 cc 2 1 ab a ab c bab Ambiguity of CFGs First, let’s number the tiles If T cannot be matched, then G is unambiguous If T can be matched, then G is ambiguous G (CFG) T (collection of tiles) 3 �− →

19/21 a Productions: Variables: S , T , B Ambiguity of CFGs 3 2 1 ab Terminals: a, b, c, 1, 2, 3 ab c cc bab G (CFG) T (collection of tiles) �− → S → T | B T → bab T 1 T → c T 2 T → a T 3 B → cc B 1 B → ab B 2 B → ab B 3 T → bab1 T → c2 T → a3 B → cc1 B → ab2 B → ab3

19/21 a Productions: Variables: S , T , B Ambiguity of CFGs 3 2 1 ab Terminals: a, b, c, 1, 2, 3 ab c cc bab G (CFG) T (collection of tiles) �− → S → T | B T → bab T 1 T → c T 2 T → a T 3 B → cc B 1 B → ab B 2 B → ab B 3 T → bab1 T → c2 T → a3 B → cc1 B → ab2 B → ab3

20/21 Ambiguity of CFGs Each sequence of tiles gives a pair of derivations bab cc c ab c ab 1 2 2 (with difgerent parse trees) S ⇒ T ⇒ bab T 1 ⇒ babc T 21 ⇒ babcc221 S ⇒ B ⇒ cc B 1 ⇒ ccab B 21 ⇒ ccabab221 If the tiles match, these two derive the same string

21/21 . n i Ambiguity of CFGs B B . . B T b j m j a i b i n i … a i S . If T cannot be matched, then G is unambiguous . T (collection of tiles) T G (CFG) If T can be matched, then G is ambiguous T . S If G is ambiguous, then the two parse trees will look like �− → ✓ ✓ n 1 n 2 m 1 a 1 n 1 b 1 a 1 a 2 m 2 a 2 n 2 b 2 b 1 b 2 Therefore n 1 n 2 . . . n i = m 1 m 2 . . . m j , and there is a match