CS 3240: Languages and Computation

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CS 3240 Languages and Computation

• Undecidable Problems and LBA

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Undecidable Languages

• We can prove a problem is undecidable by
  contradiction
• Assume the problem is decidable
• Show that this implies something impossible

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The Halting Problem HALTTM

• HALTTM ltM,wgt M is a TM and M halts on input
  w
• Theorem HALTTM is undecidable
• Proof Idea (by contradiction)
• Show that if HALTTM is decidable then ATM is also
  decidable

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Proof

• Assume R decides HALTTM
• Let S be the following TM
• S on input ltM,wgt
• Run R on ltM,wgt
• If R rejects, reject
• If R accepts, simulate M on w until it halts
• If M accepts, accept if M rejects, reject

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Proof (cont.)

• If HALTTM is decidable then S decides ATM
• Since ATM is not decidable, HALTTM cannot be
  decidable

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Reductions and Decidability

• To prove a language is decidable, we have
  converted it to another language and used the
  decidability of that language
• Example use decidability of EDFA to determine
  decidability of EQDFA
• Thus, we reduce the problem of determining if
  EQDFA is decidable to the problem of determining
  if EDFA is decidable

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Reductions and Undecidability

• To prove a language is undecidable, we have
  assumed its decidable and found a contradiction
• Example assume decidability of HALTTM and show
  ATM is decidable which is a contradiction
• In each case, we have to do a computation to
  convert one problem to another problem
• What kind of computations can we do?

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Template for Reduction

• Assume L is decidable
• Let N be a TM that decides L
• Show that a known undecidable language L will be
  decidable if it can use N to make decisions
• This is called reducing problem L to problem L
• Conclude N cannot exist
• I.e., the language L is not decidable
• Key Figuring out which undecidable language to
  reduce from

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Undecidability of ETM

• ETM ltMgt M is a TM and L(M) ?
• Theorem ETM is undecidable
• Proof idea Assume ETM is decidable with decider
  TM R. Use R to decide ATM.
• Recall ATM ltM,wgt M is a TM that accepts
  w.
• How can we use R (which takes ltMgt as input) to
  determine if M accepts w?
• Make new TM, M1, with L(M1) ? iff M accepts w

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Proof

• New TM M1 Reject any string except for w, and
  run M on input w.
• M1 On input x
• If x ? w, reject
• If x w, run M on input w
• Accept if M accepts
• L(M1) ? ? if and only if M accepts w

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Use R and M1 to Decide ATM

• Consider the following TM
• S On input ltM,wgt
• Construct M1 that rejects all but w and simulates
  M on w
• Run R on ltM1gt
• If R accepts, reject if R rejects, accept
• S decides ATM a contradiction
• Therefore, ETM is not decidable

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Another Undecidable Language

• Let REGULARTM ltMgt M is a
• TM and L(M) is a regular language
• Theorem REGULARTM is undecidable
• Proof idea Assume R decides REGULARTM and use R
  to decide ATM (reduce the ATM problem to the
  REGULARTM problem).
• As before, make a new TM, M2, that accepts a
  regular language iff M accepts w.

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Proof

• M2 On input x
• If x 0n1n for some n, accept
• Otherwise, run M on w. If M accepts w, accept
• If M accepts w, then L(M2) S
• A regular language
• Otherwise, L(M2) 0n1n
• Not a regular language

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Proof (cont.)

• Assuming R decides REGULARTM consider the
  following TM
• S On input ltM,wgt
• Construct M2 s.t. L(M2) is regular iff M accepts
  w
• Run R on M2
• If R accepts, accept if R rejects, reject
• S decides ATM iff R decides REGULARTM

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Rices Theorem

• Determining whether a TM satisfies any
  non-trivial property is undecidable
• A property is non-trivial if
• It depends only on the language of M, and
• Some, but not all, Turing machines have the
  property
• Examples Is L(M) regular? A CFG? Finite?

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Proof of Rices Theorem

• Assume there is some decidable non-trivial
  property P for Turing machines
• Without loss of generality, assume TMs that
  accept ? do not satisfy P
• If they do, just consider ?P

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Proof of Rices Theorem

• Suppose there is a TM RP that decides P
• On input ltMgt, RP accepts iff TM M has property P
• Let MP be a TM that satisfies P
• Since P is non-trivial, there is some MP
  satisfying P
• Use RP and MP to decide ATM
• Create a new TM S that decides ATM using RP and MP

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ATM Decider Using MP

• S On input ltM,wgt
• Create the following TM N
• N On input x
• Run M on w until it accepts
• If M accepts w, run MP on x
• If MP accepts x, accept if MP rejects x, reject
• Run RP on ltN(ltM,wgt)gt
• If RP accepts, accept if RP rejects, reject
• L(N) L(MP) if M accepts w otherwise L(N) ?
• S decides ATM if RP decides TMs satisfying P
• Therefore, RP cannot exist

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Computation Histories

• Recall configuration of a TM string uqv with u,v
  ? ?, q ? Q
• The sequence of configurations M goes through on
  input w is a computation history of M on input w
• may be accepting, or rejecting
• reserve the term for halting computations
• nondeterministic machines may have several
  computation histories for a given input.

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Linear Bounded Automata

• LBA definition TM that is prohibited from moving
  head off right side of input.
• machine prevents such a move, just like a TM
  prevents a move off left of tape
• How many possible configurations for a LBA M on
  input w with w n, m states, and p ? ?
• counting gives mnpn

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LBA Problems

• Two problems we have seen with respect to TMs,
  now regarding LBAs
• LBA acceptance
• ALBA ltM, wgt LBA M accepts input w
• LBA emptiness
• ELBA ltMgt LBA M has L(M) Ø
• Both decidable? both undecidable? one decidable?

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ALBA Problem

• Theorem ALBA is decidable.
• Proof
• input ltM, wgt where M is a LBA
• key only mnpn configurations
• if M hasnt halted after this many steps, it must
  be looping forever.
• simulate M for mnpn steps
• if it halts, accept or reject accordingly,
• else reject since it must be looping

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ELBA Problem

• Theorem ELBA is undecidable.
• Proof
• reduce from co-ATM (i.e. show co-ATM m ELBA)
• Idea
• produce LBA B that accepts exactly the accepting
  computation histories of M on input w

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ELBA Problem (cont)

• Proof
• f(ltM, wgt) ltBgt described below

• is B an LBA?
• is f computable?
• YES maps to YES?
• ltM, wgt ? co-ATM ? f(M, w) ? ELBA
• NO maps to NO?
• ltM, wgt ? co-ATM ? f(M, w) ? ELBA

• on input x, check if x has form
• C1C2C3...Ck
• check whether C1 is the start configuration for
  M on input w
• check whether Ci?1 Ci1
• check whether Ck is an accepting configuration
  for M