
Computer Networking: A Top-Down Approach (7th Edition)
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ISBN: 9780133594140
Author: James Kurose, Keith Ross
Publisher: PEARSON
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Transcribed Image Text:**Turing Machine Design with Input Alphabet {a, b}**
*Objective*:
Create a Turing machine with the input alphabet {a, b} that accepts a string if and only if the string has the specified property.
*Property*:
The input string contains more a's than b's.
### Detailed Explanation
In order to design a Turing machine that accepts any string with more `a`s than `b`s, consider the following high-level steps:
1. **Initialization**:
- Start from the beginning of the tape and leave an indicator or marker (such as `*`) to know where you began.
2. **Scan for Characters**:
- Traverse the tape to find the first unmarked `a` and replace it with a special symbol (like `X` or `Y`) to indicate it has been processed.
- Similarly, traverse the tape to find the first unmarked `b` and mark it similarly.
3. **Balancing a's and b's**:
- For each `a` marked, find a matching `b` and vice versa. If you manage to pair all `b`s but still have `a`s left, it means there are more `a`s than `b`s.
- Conversely, if you pair all `a`s but have `b`s left, the string doesn't have more `a`s.
4. **Final Check**:
- If you reach the end of the tape with more `a`s marked than `b`s, stop and accept the string.
- If `b`s exceed, the machine enters a reject state.
### Example Algorithm:
1. **Start State**: Traverse right looking for an `a` or `b`.
2. **Marking an `a`**:
- Upon finding an unmarked `a`, change it to `X` and move to a state to find a `b`.
3. **Marking a `b`**: On finding an unmarked `b`, change it to `Y` and revert back to the state searching for an `a`.
- Repeat the above two steps until all characters are marked.
4. **Decision**:
- Transition to the accept state if excess `a`s are found, else reject.
### Note:
The machine relies on marking `a`s and `b`s to balance them. By the end of the traversal, the excess of
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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}arrow_forwardUse JFLAP software only. for turing machinearrow_forwardGive the state diagram for a Turing machine that decides the following language over E = {0, 1}: LF {w :w contains at least two 0's and at most two 1's}arrow_forward
- Create a Turing machine for the following language: On Σ = {0,1} Language 1 (L1) every string in the language begin and end with 1 Language 2 (L2) all strings in the language begin with 001. The union of L1 and L2arrow_forwardDesign a Turing Machine that accepts the language {a^cb", n >0}arrow_forwardT/F Any set of strings that can be described by a regular expression can be recognized by a Turing machine. T/F Any set of strings that can be recognized by a Turing Machine can be described by a regular expressionarrow_forward
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