{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

UTF-8__http-__www.jmchilton.net_files_school_csci4011-SU07_rec_6

# UTF-8__http-__www.jmchilton.net_files_school_csci4011-SU07_rec_6

This preview shows pages 1–8. Sign up to view the full content.

Recitation 6 John Chilton July 18, 2007 John Chilton Recitation 6

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
LEFT TM = { < M , w > | M moves its head left while on leftmost tape position during computation of w } John Chilton Recitation 6

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 based on M but a) M adds @ at the beginning of input and moves all input to the right b) Adjust transitions so that on @ M 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
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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 . Method 2: Mapping Reduction. The following proves A TM M J T = On input < M , w > 1. Output 1 < M , w > .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 29

UTF-8__http-__www.jmchilton.net_files_school_csci4011-SU07_rec_6

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online