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Recitation 6 John Chilton July 18, 2007 John Chilton Recitation 6
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At the bottom of the proof for 5.13, write down one specific thing that you were the most unclear about while doing homework 5. Try to be as specific as possible. John Chilton Recitation 6
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LEFT TM = { < M , w > | M moves its head left while on leftmost tape position during computation of w } John Chilton Recitation 6
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LEFT TM = { < M , w > | M moves its head left while on leftmost tape position during computation of w } Assume LEFT TM is decidable. Let L be some decider of it. Now consider the following decider of A TM . S = On input < M , w > 1. Create a new TM, M 0 based on M but a) M 0 adds @ at the beginning of input and moves all input to the right b) Adjust transitions so that on @ M 0 stays in the same state and moves to the right c) Replace q accept with a state that moves left indefinitely 2. Run decider L on this new TM 3. Accept if L accepts, else reject . John Chilton Recitation 6
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J = { w | w = 0 x for some x A TM or w = 1 y for some y A TM } Show J and J are not Turing- recognizable . John Chilton Recitation 6
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J = { w | w = 0 x for some x A TM or w = 1 y for some y A TM } Method 1: Normal Reduction. Assume J is recognizable by some TM S . Then the following would recognize A TM , a contradiction. T = On input < M , w > 1. Run S on input 1 < M , w > . 2. If at some point S accepts, accept . John Chilton Recitation 6
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J = { w | w = 0 x for some x A TM or w = 1 y for some y A TM } Method 1: Normal Reduction. Assume J is recognizable by some TM S . Then the following would recognize A TM , a contradiction. T = On input < M , w > 1. Run S on input 1 < M , w > . 2. If at some point
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This note was uploaded on 02/22/2010 for the course CS 881 taught by Professor H.f. during the Spring '10 term at Shahid Beheshti University.

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