Question

1. Construct a Turing machine for L-a"b", n>0 o" er w= (a,, and simulate

Answer 1

To construct a Turing machine (TM) for the language L = { a^n b^n | n > 0 }, we can outline the components of the TM, such as the states, the tape symbols, and the transition functions. This language consists of strings with an equal number of 'a's followed by 'b's, where n is a positive integer.

### Components of the Turing Machine

1. **Tape Alphabet**: We will use the following tape symbols:
- Input symbols: `a`, `b`
- Blank symbol: `#`
- Marker symbols: `X`, `Y` (used to mark visited `a`s and `b`s)

2. **States**:
- `q0`: Start state
- `q1`: Mark the first `a`
- `q2`: Move to the right to find a `b`
- `q3`: Mark the first `b`
- `q4`: Return to the left to find the next `a`
- `q5`: Check for unmarked `a`s and `b`s
- `q_accept`: Accept state
- `q_reject`: Reject state

### Transition Function

We'll describe the transitions using `(current state, read symbol) -> (next state, write symbol, move direction)`:

1. **Initialization**:
- `(q0, a) -> (q1, X, R)` // Mark `a` as `X` and move right
- `(q0, b) -> (q_reject, b, R)` // If the first symbol is `b`, it's invalid
- `(q0, #) -> (q_accept, #, R)` // If only `#` is found (empty input after marking), accept

2. **Marking `b` and Moving Left**:
- `(q1, a) -> (q1, a, R)` // Skip over unmarked `a`s
- `(q1, b) -> (q2, Y, R)` // Mark `b` as `Y` and move right
- `(q1, #) -> (q_reject, #, R)` // If reached end of input and still `a`s left, reject

3. **Finding Unmarked `a`**:
- `(q2, a) -> (q2, a, R)` // Move right past `a`s
- `(q2, b) -> (q2, b, R)` // Move right past `b`s
- `(q2, #) -> (q3, #, L)` // If reached the end after marking, move to check for more `a`s

4. **Returning to Mark `a`**:
- `(q3, Y) -> (q3, Y, L)` // Move back over marked `b`s
- `(q3, X) -> (q3, X, L)` // Move back over marked `a`s
- `(q3, a) -> (q1, X, R)` // If find `a`, mark it and go to find `b`
- `(q3, #) -> (q_accept, #, R)` // If find no more unmarked `a`, accept

5. **Completing the Cycle**:
- Continue cycling between `q1` and `q2` until you either accept or reject the input.
- If at any point, you reach a configuration that cannot continue (e.g., if there's an imbalance or remaining unmarked symbols), lead to the `q_reject` state.

6. **Final Check**:
- Ensure no unmarked `a`s or `b`s remain before transitioning to `q_accept`.

### Summary

This Turing machine effectively checks for an equal count of `a`s and `b`s, utilizing marking for both types of symbols. Once there are no unmarked `a`s or `b`s left, and the steps are correctly aligned, the TM will accept the string.


### Example Trace

For the input `aabb`, the following sequence of transitions occurs:
1. `q0` reads `a`, marks it as `X`, and moves to `q1`.
2. `q1` then reads `a` (skips), finds `b`, marks it (changes to `Y`), and then transitions to `q2`.
3. Continues marking `a` and `b` until all are marked.

If the input has a perfect matching count, the TM will end in the `q_accept` state.

You can adjust the specific transitions and states as you implement the TM based on your formal TM specifications.