CS 3240: Languages and Computation

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CS 3240 Languages and Computation

• Decidable Languages

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Decidability

• A language is decidable if some Turing machine
  decides it
• Not all languages are decidable
• How to show a language is decidable?
• Write a decider that decides it
• Accepts w iff w is in the language
• Must halt on all inputs

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DFA Acceptance Problem

• Consider the language ADFA ltB,wgt B is a DFA
  that accepts the string w
• Theorem ADFA is a decidable language
• Proof Consider the following TM, M
• On input string ltB,wgt, where B is a DFA and w is
  an input to B
• Simulate B on input w
• If simulation ends in accept state, accept.
  Otherwise, reject.

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NFA Acceptance Problem

• Consider the language
• ANFA ltB,wgt B is a NFA that accepts the
  string w
• Theorem ANFA is a decidable language
• Proof Consider the following TM, NOn input
  string ltB,wgt
• Convert B to a DFA C
• Run TM M from previous slide on ltC,wgt
• If M accepts, accept. Otherwise, reject.

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RE Acceptance Problem

• Consider the language
• AREX ltR,wgt R is an RE that generates the
  string w
• Theorem AREX is a decidable language
• Proof Consider the following TM, P
• On input string ltR,wgt
• Convert R to a DFA C using algorithm discussed in
  class and in text
• Run TM M from earlier slide on ltC,wgt
• If M accepts, accept. Otherwise, reject.

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Emptiness Testing Problem

• Consider the language
• EDFA ltAgt A is a DFA and L(A) ?
• Theorem EDFA is a decidable language
• Proof Consider the following TM, T
• On input string ltAgt, where A is a DFA
• Mark the start state
• Repeat until no new states get marked
• Mark any state that has a transition coming into
  it from any state already marked
• If no accept states are marked, accept.
  Otherwise, reject.

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DFA Equivalence Problem

• EQDFA ltA,Bgt A and B are DFAs and L(A)
  L(B)
• Theorem EQDFA is a decidable language
• Proof Consider the following language
• (L(A) ? L(B)) ? (L(A) ? L(B))

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DFA Equivalence Problem

• (L(A) ? L(B)) ? (L(A) ? L(B))

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DFA Equivalence Problem

• EQDFA ltA,Bgt A and B are DFAs and L(A)
  L(B)
• Theorem EQDFA is a decidable language
• Proof idea Consider DFA C that accepts L(C)
  (L(A) ? L(B)) ? (L(A) ? L(B))
• How do we know such a DFA exists?
• If L(C) ?, then L(A) L(B)

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TM That Decides EQDFA

• On input string ltA,Bgt, where A and B are DFAs
• Create DFA C such that
• L(C) (L(A) ? L(B)) ? (L(A) ? L(B))
• Submit C to Turing machine T that decides EDFA
• If T accepts C, accept. Otherwise, reject.

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Deciders and CFGs

• Consider the following language
• ACFG ltG,wgt G is a CFG that generates string
  w
• Is ACFG decidable?
• How can we get a TM to simulate a CFG?
• Must be certain CFG tries a finite number of
  steps!
• Solution Use Chomsky normal form

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Chomsky Normal Form Review

• All rules are of the form
• A ?BC
• A ?a
• Where A, B, and C are any variables (B and C
  cannot be the start variable)
• S ?eis the only erule, where S is the start
  variable

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How Many Steps to Generate w?

• If w 0
• 1 step
• If w n gt 0?
• 2n 1 steps

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TM Simulating ACFG

• On input ltGgt, where G is a CFG
• Convert G into Chomsky normal form
• If w 0
• If there is an S ? e rule, accept
• Otherwise, reject
• List all derivations with 2w-1 steps
• If any generate w, accept
• Otherwise, reject

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Empty CFGs

• Consider the following language
• ECFG ltGgt G is a CFG and L(G) ?
• Theorem ECFG is decidable
• Can we use the TM in ACFG to prove this?
• No. There are infinitely many possible strings
  in S
• We need to check if there is any way to get from
  the start variable to some string of terminals

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Work Backwards

• On input ltGgt, where G is a CFG
• Mark all terminals
• Repeat until no new variables are marked
• Mark any variable A if G has a rule A?U1U2Uk
  where U1, U2, , Uk are all marked
• If S is marked, reject Otherwise, accept

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What About EQCFG?

• Recall for EQDFA, we considered
• (L(A) ? L(B)) ? (L(A) ? L(B))
• Will this work for CFGs?
• No. CFGs are not closed under complementation or
  intersection
• EQCFG is not a decidable language!
• We will show this later

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Decidability of CFLs

• Theorem Every context-free language L is
  decidable
• Can we make a TM that simulates a PDA?
• No. May run into infinite loop.
• Proof idea For each w, we need to decide
  whether or not w is in L. Let G be a CFG for L.
  This problem boils down to ACFG, which we showed
  is decidable.

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Summary of Decidable Languages