CS 3240: Languages and Computation

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CS 3240 Languages and Computation

• Mapping Reducibility

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A Simple Undecidable Problem

• Is there an example of undecidable problem that
  is not related to automata?
• Post correspondence problem (PCP) is such an
  example
• A domino looks like , where u and v are
  two strings
• Given a collection of dominos, can you make a
  list of dominos (repetition allowed) so that the
  concatenated string on the top is the same as
  that on the bottom?
• Example

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PCP is Undecidable

• Proof idea
• Reduction from ATM via computation histories
• Define collection of dominos so that a match
  corresponds to an accepting computational history
• Internally, define sets of dominos for each
  transition function
• Many technical details for treatments of
  boundaries, which depend on input string to ATM

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TMs and Computation

• TMs can do more than just accept and reject
  strings
• They can perform functions
• Definition A function f? ? ? is a computable
  function if there is some TM M that, on every
  input w, halts with f(w) on the tape

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Examples

• The copying TM
• Start with w on the tape, halt with ww on the
  tape
• Summation
• Takes input and returns value of mn
• Finding intersection of two DFAs
• Start with on the tape, where A and B are
  DFAs, halt with on the tape, where L(C)
  L(A) ? L(B)

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Mapping Reducibility

• Definition Language A is mapping reducible to
  language B, written A?mB, if there is a
  computable function f? ? ?, where for every w,
• w ? A iff f(w) ? B
• The function f is called the reduction of A to B.

B
A
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Mapping Reductions Decidability

• Theorem If A ?m B and B is decidable, then A is
  decidable.
• Proof Let M be a decider for B and let f be a
  reduction from A to B.
• Consider the following TM, N
• N On input w
• Compute f(w)
• Run M on f(w) and report Ms output
• Then N decides A

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Mapping Reductions Undecidability

• Corollary If A ?m B and A is undecidable, then
  B is undecidable.
• We have been using this corollary implicitly
  already

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Example

• We showed that HALTTM is undecidable by
  contradiction
• Now lets show its undecidable using mapping
  reduction
• Need a function that takes input
• Halts if M accepts w, and loops if not

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Mapping from ATM to HALTTM

• F On input x
• If x ? for some TM M, output x
• Otherwise, construct the following TM
• M On input x
• 1. Run M on x
• 2. If M accepts, accept
• 3. If M rejects, enter a loop
• 2. Output
• If M accepts w M halts on w, otherwise M loops
  or generates a string not in HALTTM
• I.e., ?ATM iff ?HALTTM

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HALTTM is Undecidable

• We just showed that ATM m HALTTM
• Since we know ATM is undecidable, we can conclude
  that HALTTM is undecidable

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Reductions TM-recognizability

• Theorem If A ?m B and B is Turing-recognizable,
  then A is Turing-recognizable.
• Proof (same as decidable proof) Let M be a
  recognizer for B and let f be a reduction from A
  to B.
• Consider the following TM, N
• N On input w
• Compute f(w)
• Run M on f(w) and report Ms output
• Then N recognizes A

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Appls. of Mapping Reductions

• If A ?m B and B is decidable, then A is decidable
• If A ?m B and A is undecidable, then B is
  undecidable
• If A ?m B and B is Turing-recognizable, then A is
  Turing-recognizable
• Equivalently, A ?m B
• If A ?m B and A is not Turing-recognizable, then
  B is not Turing-recognizable

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Reductions non-TM-recognizability

• Corollary If A ?m B and A is not
  Turing-recognizable, then B is not
  Turing-recognizable.
• Question Which language that have we seen is
  not Turing-recognizable?
• Answer ATM

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EQTM

• Theorem EQTM is neither Turing-recognizable nor
  co-Turing-recognizable.
• Proof
• Reduce ATM to EQTM.
• Given input
• Construct M1 that accepts nothing
• Construct M2 that accepts nothing if M accepts w
• Reduce ATM to EQTM.