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C SC 473 Automata, Grammars & Languages
Automata, Grammars and Languages
Lecture 06
Decidability and Undecidability
C SC 473 Automata, Grammars & Languages 2
Decidable Problems for Regular Languages
- Theorem 4.1: (Membership/Acceptance Prob. for DFAs) is a decidable language. Pf: A decider for the language is:
A DFA = { A w , | A is a DFA and w ∈ L A ( ) }
- M = “On input :
- Simulate A on w.
- If the simulation reaches an accept state, halt and accept. If it ends (jams) in a non-accept state, halt and reject .” •
C SC 473 Automata, Grammars & Languages 3
Decidable—Regular Lang.s (cont’d)
- Theorem 4.2: (Membership/Acceptance Prob. for NFAs) is a decidable language. Pf: A decider for the language is:
A NFA = { N w , | N is an NFA and w ∈ L N ( ) }
- M = “On input :
- Use the Rabin-Scott algorithm to convert N to DFA A
- Run the Theorem 4.1 algorithm with input < A,w>.”
- If that algorithm accepts, then accept; otherwise reject.” •
C SC 473 Automata, Grammars & Languages 4
Decidable—Regular Lang.s (cont’d)
- Theorem 4.3: (Membership Prob. for RegEx’s) is a decidable language. Pf: Use the algorithm to convert R to an equivalent NFA N and use the algorithm of Theorem 4.2 with input __ •
A (^) REX= { R w , | R is a regular expression and w ∈ L R ( ) }
C SC 473 Automata, Grammars & Languages 5
Decidable—Regular Lang.s (cont’d)
- Theorem 4.4: ( Emptiness Problem for DFAs) is a decidable language. Pf: A decider for the language is:
E (^) DFA= { A | A is a DFA and L A ( ) = ∅}
T = “On input :
- repeat {
- } until ( )
- if accept; otherwise reject.”
{ | ( )( ) δ( , ) }
M S
S M q a p M p a q
S = M
S ←{ q 0 }
F ∩ M = ∅
C SC 473 Automata, Grammars & Languages 6
Decidable—Regular Lang.s (cont’d)
- Theorem 4.5: ( Equivalence Problem for DFAs) is a decidable language. Pf: Observe that
Because of closure properties, there is an algorithm to construct a DFA C from A , B that accepts Use Theorem 4.4 with __ to test whether If that algorithm accepts, then accept; otherwise, reject.
EQ (^) DFA= { A B , | A B , are DFA and L A ( ) = ( L B )}
( ) ( ) ( ) ( ) where ( ) ( ) ( ( ) ( )) ( ( ) ( ))
L A L B L A L B
L A L B L A L B L A L B
L A ( ) ⊕ L B ( )
L C ( ) = ∅.
C SC 473 Automata, Grammars & Languages 10
The Halting Problem
- Although the following language is TM-recognizable, we will show it is not decidable.
- This is called the Membership Problem for TMs, and by some authors the Halting Problem for TMs: given a TM M and string w, does M accept w?
- Why called “Halting Problem”? Given a TM M , can always alter it to an equivalent TM M ′ such that: M ′ halts on w iff M ′ accepts w (iff M accepts w ). Pf: For each undefined transition δ( q,a ) in M, M ′ will transition to a state and loop forever; also goes to the same loop state •
- Thus acceptance can be made synonymous with halting
A (^) TM= { M w , | M is a TM and w ∈ L M ( )}
q (^) lo o p q re je c t
C SC 473 Automata, Grammars & Languages 11
The Halting Problem (cont’d)
- Thm: is Turing-recognizable. Pf: Let U be a UTM. A recognizer for is:
A (^) TM= { M w , | M is a TM and w ∈ L M ( )}
U
yes
no
M w , ∈ L R ( ) ⇔ M w , ∈ L U ( ) ⇔ w ∈ L M ( ) ⇔ M w , ∈ A
TM
yes M w , no
A TM
R
C SC 473 Automata, Grammars & Languages 12
The Halting Problem (cont’d)
- Thm: is undecidable. Pf: Proof by contradiction. Assume that is decidable. Then it has a decider; call it H. H behaves as follows:
Construct a TM D that calls H as a subroutine. On input 〈 M 〉, D runs H on 〈 M , 〈 M 〉〉. That is, D determines if M accepts or rejects its own description as input. If M accepts 〈 M 〉, then D rejects; If M rejects 〈 M 〉, then D accepts. Here is the picture of how D behaves:
A (^) TM= { M w , | M is a TM and w ∈ L M ( )}
H
{ if M accepts w }
accept M w , reject
A TM
{ if M does not accept w }
C SC 473 Automata, Grammars & Languages 13
The Halting Problem (cont’d)
- What happens if we run D on its own description 〈 D 〉? Set 〈 M 〉 = 〈 D 〉 in the above. Then This contradiction shows that decider H cannot exist. •
H
{ if M accepts 〈 M 〉 } accept M reject { if M does not accept 〈 M 〉 }
M , M
D accept
reject
{ if M does not accept 〈 M 〉 }
{ if M accepts 〈 M 〉 } M ∈ L D ( ) ⇔ M ∉ L M ( )
D ∈ L D ( ) ⇔ D ∉ L D ( ).
C SC 473 Automata, Grammars & Languages 14
“Diagonalization”—Why called?
1 2 3 1 1 1 2 1 3 1 2 1 2 2 2 3 2 3 1 3 2 3 3 3
M M M
M M M M M M M
M M M M M M M
M M M M M M M
D D ( D )
is D ( D ) = accept or reject?
D
D ( Mi ) = accept ⇔ Mi ( M i )= reject
D ( D ) = accept ⇔ D ( D )= reject
D computes the opposite of the diagonal entries
TMs →
Encodings
(^) →
C SC 473 Automata, Grammars & Languages 15
Decidable vs Recognizable Sets: Basics
- Theorem: decidable decidable Proof: If L is decidable, it has a decider M. The decider halts for every input in either the accepting halt state or in the rejecting halt state. Construct from M as follows: make the accepting halt state the rejecting state and the rejecting halt state the accepting state. Then that is, •
⇒ L
q accept
M
w ∈ L ( M )⇔ w ∉ L ( M ), L ( M )= L ( M ).
L
q reject
