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

Undecidable Problems for CFGs CSCI 3130 Formal Languages and Automata Theory Siu On CHAN Fall 2018 Chinese University of Hong Kong 1/21

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? 2/21 DFAs D and D ′ accept same TM M and M ′ accept the

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

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

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

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

Undecidable problems for CFGs We will argue that 7/21 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 }

Undecidable problems for CFGs to G 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 Proof by contradiction A 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 8/21 � G � � G � � M , w � G fails to generate some string if M accepts w

Undecidable problems for CFGs Convert to G M accepts w First we construct a PDA P , then convert it to CFG G 9/21 � 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 history of ( M , w )

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

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

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

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

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

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

Undecidability of PCP Next lecture we will show (using computation history method) The language PCP is undecidable 16/21 PCP = {� T � | T is a collection of tiles that contains a top-bottom match }

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

Ambiguity of CFGs c 3 2 1 ab a ab cc T (collection of tiles) bab 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) 18/21 �− →

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

Ambiguity of CFGs Each sequence of tiles gives a pair of derivations bab cc c ab c ab 1 2 2 (with different parse trees) 20/21 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

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