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l12-handout - Outline A language that is RE but not...

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Lecture 12: Undecidability, cont. David Dill Department of Computer Science 1 Outline A language that is RE but not recursive. Other undecidable problems. Rice’s theorem 2 A Language that is RE but not Recursive L u = { ( M, w ) | M accepts w } . Theorem: L u is RE. L u is accepted by the “Universal Turing Machine” M U . Idea: To check ( M, w ) , simulate M on w using one tape to hold M , use the second tape to simulate the tape of M (so the second tape starts out with w on it). 3 L u is not Recursive Theorem: L u is not recursive. Proof sketch: By reduction from L d . First, we prove that L u is not RE, by reduction. Suppose there were a TM, M , accepting L u . So, M would inputs ( M i , w ) and accepts if M i rejects w . Then we could construct a TM M 0 to accept L d , as follows: Given input string M i , M 0 first constructs the string ( M i , M i ) , then runs M on ( M i , M i ) L ( M 0 ) = L d , since M 0 accepts M i iff M i does not accept M i . But L d is not RE, so L u is not RE. Therefore, L u is not recursive, so L u must not be recursive (since the complement of a recursive language must be recursive). 4
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Other Properties of Turing Machines Problem: Is L ( M i ) empty? L e = { M i | L ( M i ) = ∅} . L ne = { M i | L ( M i ) 6 = ∅} . Theorem L ne is RE. Proof Use an NTM. Use nondeterministic choice to guess a string w . Then simulate M i on w , and accept if M i accepts w .
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