For alphabet {0, 1], the behavior of a Turing machine M can be described as follows: 1. Scan the tape and mark the first O that has not been marked. If no unmarked 0 is found, go to stage 4. Otherwise, move the head back to the front of the tape. 2. Scan the tape and mark the first 1 that has not been marked. If no unmarked 1 is found, reject. 3. Move the head back to the front of the tape and go to stage 1. 4. Move the head back to the front of the tape. Scan the tape to see if any unmarked 1s remain. If none are found, accept; otherwise, reject. The correct statements are: OL(M)= {w|w contains more Os than 1s} L(M) = {w|w contains less Os than 1s} OL(M) = {w|w contains an equal number of Os and 1s}

Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
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Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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For alphabet {0, 1}, the behavior of a Turing machine M can be described as follows:
1. Scan the tape and mark the first O that has not been marked. If no unmarked 0 is found, go to
stage 4. Otherwise, move the head back to the front of the tape.
2. Scan the tape and mark the first 1 that has not been marked. If no unmarked 1 is found, reject.
3. Move the head back to the front of the tape and go to stage 1.
4. Move the head back to the front of the tape. Scan the tape to see if any unmarked 1s remain. If
none are found, accept; otherwise, reject.
The correct statements are:
L(M) = {w|w contains more Os than 1s}
OL(M) = {w|w contains less Os than 1s}
OL(M) = {w|w contains an equal number of Os and 1s}
Transcribed Image Text:For alphabet {0, 1}, the behavior of a Turing machine M can be described as follows: 1. Scan the tape and mark the first O that has not been marked. If no unmarked 0 is found, go to stage 4. Otherwise, move the head back to the front of the tape. 2. Scan the tape and mark the first 1 that has not been marked. If no unmarked 1 is found, reject. 3. Move the head back to the front of the tape and go to stage 1. 4. Move the head back to the front of the tape. Scan the tape to see if any unmarked 1s remain. If none are found, accept; otherwise, reject. The correct statements are: L(M) = {w|w contains more Os than 1s} OL(M) = {w|w contains less Os than 1s} OL(M) = {w|w contains an equal number of Os and 1s}
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