ETM = { M is TM, L(M) is empty} A Decidable B Undecidable, recognizable, with recognizable complement C Undecidable, recognizable, with
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[PDF] Homework 10 Solutions
However, ETM is Turing-recognizable (HW 8, problem 4) and ATM is not Turing- recognizable (Corollary 4 23), contradicting Theorem 5 22 3 Consider the
[PDF] Homework 8 Solutions
Consider the emptiness problem for Turing machines: ETM = { 〈M〉 M is a Turing machine with L(M) = ∅} Show that ETM is co-Turing-recognizable (A
[PDF] Tutorial 5
If both L and L are recognizable then L is decidable As we know that ATM , HALTTM , ETM are recognizable, their complements cannot be rec- ognizable,
[PDF] 10 Reducibility
HALTTM is Turing-recognizable since it can be recognized by TM U ETM = {< M > M is a TM and L(M) = /0} is undecidable Proof: Reduce ATM to ETM
[PDF] and let f be the function that maps 0 1 to 1 and maps all other s
But, ¬ETM is Turing recognizable and ¬ATM is not, which contradicts Theorem 5 22 This is a contradiction Therefore, ATM is not mapping reducible to ETM 5 7 A
[PDF] 1 Reducability
ETM = {〈M〉 M is a TM such that L(M) = {}} is undecidable Let R be a TM that decides ETM If A ≤m B and B is recognizable, then A is recognizable
[PDF] CSE 105 - UCSD CSE
ETM = { M is TM, L(M) is empty} A Decidable B Undecidable, recognizable, with recognizable complement C Undecidable, recognizable, with
[PDF] ECS120 Discussion Notes
28 nov 2006 · The emptiness problem, ETM = {(M )M is a TM and L(M) = ∅}, is undecidable How to prove a language is Turing-recognizable or
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