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10-reducibility - Reducibility We say that a problem Q...

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Reducibility We say that a problem Q reduces to problem P if we can use P to solve Q . In the context of decidability we have the following “templates”: If A reduces to B and B is decidable, then so is A . (1) and If A reduces to B and A is undecidable, then so is B . (2) The template (1) allows us to set up proofs by contradiction to prove undecidability. – p. 1/1
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Reducibility Theorem: The language HALT T M = { M, w | M is a TM and M halts on input w } is undecidable. Proof: Proof by contradiction. We assume that HALT T M is decidable. We show that A T M is reducible to HALT T M by constructing a machine based on HALT T M that will decide A T M . Let Q be a TM that decides HALT T M . The we can construct a decider S that decides A T M as follows, S = "On input M, w , where M is a TM and w a string: 1. Run Q on M, w . 2. If Q rejects, reject . 3. If Q accepts, simulate M on w until it halts. 4. If M has accepted, accept ; if M has rejected, reject ." We have shown that A T M is undecidable, therefore this is a contradiction and our assumption that HALT T M is decidable must be incorrect. – p. 2/1
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Properties of L ( M ) Theorem: The language E T M = { M | M is a TM and L ( M ) = ∅} is undecidable. Proof: By contradiction. Assume E T M is decidable and Q is the decider. We show that A T M reduces to E T M by constructed the following decider S for A T M , S = "On input M, w , where M is a TM and w a string: 1. Build the machine M 1 as follows, M 1 = "On input x : 1. If x = w , reject . 2. If x = w , run M on input w and accept if M does." 2. Run Q in M 1 . 3. If Q accepts, reject ; if Q rejects, accept ." But this machine cannot exist, therefore our assumption must be wrong.
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