NTM Design: Recognizing (ww^c)^n Language

Let's dive into designing a Non-Deterministic Turing Machine (NTM) that recognizes the language ${ L = \{(ww^c)^n \mid n > 1 \text{ and } w, w^c \in \{a, b\}^*\} }$, where ${ w^c }$ represents the complement of string ${ w }$. That means if ${ w = w_1w_2...w_k }$, then ${ w^c = w_1^c w_2^c ... w_k^c }$, with ${ w_i^c = b }$ if ${ w_i = a }$ and ${ w_i^c = a }$ if ${ w_i = b }$. Basically, we're looking for strings that consist of two parts repeated at least twice: a string ${ w }$ followed by its complement ${ w^c }$.

Understanding the Language

Before we jump into the NTM, let's make sure we understand the language ${ L }$ properly. A string belongs to ${ L }$ if it can be expressed as ${ (ww^c)^n }$ for some ${ n > 1 }$. Here's a breakdown:

• ${ w }$ is a string made up of ${ a }$'s and ${ b }$'s.
• ${ w^c }$ is the complement of ${ w }$, where each ${ a }$ is replaced by a ${ b }$ and each ${ b }$ is replaced by an ${ a }$.
• ${ (ww^c)^n }$ means the string ${ ww^c }$ is repeated ${ n }$ times, and ${ n }$ must be greater than 1.

For example, let's take ${ w = ab }$. Then ${ w^c = ba }$, and ${ ww^c = abba }$. A string in ${ L }$ could be ${ (abba)^2 = abbaabba }$ or ${ (abba)^3 = abbaabbaabba }$, and so on. The key is that the entire string must be made up of at least two repetitions of the ${ ww^c }$ pattern. This forms the basis of how our NTM will operate, ensuring we capture all strings that meet these criteria.

High-Level NTM Design

The NTM will work as follows:

1. Non-Deterministically Choose ${ w }$: The machine will non-deterministically guess the length of ${ w }$ and read it from the input.
2. Compute ${ w^c }$: It will then compute the complement ${ w^c }$ and verify that the next portion of the input matches this complement.
3. Repeat: It will repeat steps 1 and 2 at least once to ensure ${ n > 1 }$.

States and Transitions

Here’s a more detailed, but still informal, description of the NTM’s states and transitions:

1. Start State (q0):

   • The machine starts in state ${ q_0 }$.
   • Non-deterministically choose to either read an ${ a }$ or a ${ b }$.
   • Move to state ${ q_1 }$ (reading ${ w }$).

2. Reading w (q1):

   • In state ${ q_1 }$, the machine reads the input symbol (either ${ a }$ or ${ b }$).

   • Store the symbol in a temporary memory (conceptually).

   • Move the tape head to the right.

   • Non-deterministically decide whether to continue reading ${ w }$ or move to the complement-checking state ${ q_2 }$. This non-deterministic choice is crucial; the machine is guessing when it has reached the end of ${ w }$.

3. Complement-Checking (q2):

   • In state ${ q_2 }$, the machine checks if the next input symbol is the complement of the symbol it stored in the "memory".

   • If the current symbol is ${ a }$, it expects a ${ b }$; if it's ${ b }$, it expects an ${ a }$.

   • If the complement matches, move the tape head to the right and go to state ${ q_3 }$.

   • If the complement does not match, reject the input.

4. Checking w^c (q3):

   • In state ${ q_3 }$, the machine continues to read the complement ${ w^c }$, comparing each symbol with what's stored in its conceptual memory.

   • For each symbol, it verifies that it is the complement of the corresponding symbol in ${ w }$.

   • Non-deterministically decide whether to complete checking ${ w^c }$ or to move to state ${ q_4 }$.

5. Repeat or Accept (q4):

   • After reading ${ ww^c }$, the machine needs to ensure that the entire string consists of at least two repetitions of this pattern (i.e., ${ n > 1 }$).

   • The machine non-deterministically chooses to either go back to state ${ q_1 }$ to start reading ${ w }$ again (thereby checking for another ${ ww^c }$ sequence) or move to the accept state ${ q_5 }$.

   • This non-deterministic choice is how the NTM handles the ${ n > 1 }$ condition. It guesses whether it needs to read another ${ ww^c }$ or if it has reached the end of the input.

6. Accept State (q5):

   • If the machine reaches the end of the input tape in state ${ q_5 }$, it accepts the input string. This implies that the input string is of the form ${ (ww^c)^n }$ for some ${ n > 1 }$.

7. Reject State (q_reject):

   • If at any point the machine encounters a mismatch (e.g., the complement does not match) or reaches the end of the input unexpectedly, it transitions to a reject state.

Transition Function (Informal)

Here’s a more formal, but still relatively informal, representation of the transition function:

• ${ \delta(q_0, a) = \{(q_1, a, R)\} }$
• ${ \delta(q_0, b) = \{(q_1, b, R)\} }$
• ${ \delta(q_1, a) = \{(q_1, a, R), (q_2, a, R)\} }$ (Non-deterministic choice: continue reading ${ w }$ or start checking ${ w^c }$)
• ${ \delta(q_1, b) = \{(q_1, b, R), (q_2, b, R)\} }$ (Non-deterministic choice: continue reading ${ w }$ or start checking ${ w^c }$)
• ${ \delta(q_2, a) = \{(q_3, b, R)\} }$ (If the symbol in ${ w }$ was ${ a }$, expect ${ b }$)
• ${ \delta(q_2, b) = \{(q_3, a, R)\} }$ (If the symbol in ${ w }$ was ${ b }$, expect ${ a }$)
• ${ \delta(q_3, a) = \{(q_4, a, R)\} }$ (Similar to ${ q_2 }$, but for the remaining symbols of ${ w^c }$)
• ${ \delta(q_3, b) = \{(q_4, b, R)\} }$
• ${ \delta(q_4, \text{end of input}) = \{(q_5, \text{end of input}, R)\} }$ (Accept if end of input is reached)
• ${ \delta(q_4, a) = \{(q_1, a, R)\} }$ (Non-deterministically choose to read another ${ ww^c }$)
• ${ \delta(q_4, b) = \{(q_1, b, R)\} }$ (Non-deterministically choose to read another ${ ww^c }$)

Example Execution

Let's trace how the NTM might process the input string abbaabba (which is ${ (abba)^2 }$, where ${ w = ab }$).

1. Start (q0): The machine starts in ${ q_0 }$.
2. First w (q1): It non-deterministically chooses to read ${ w = ab }$.
3. Complement Check (q2, q3): It reads ${ ba }$ and verifies it's the complement of ${ ab }$.
4. First ww^c (q4): Now it has read ${ abba }$.
5. Repeat or Accept (q4): Here's the crucial non-deterministic choice. The machine guesses that it needs to read another ${ ww^c }$. It goes back to ${ q_1 }$.
6. Second w (q1): It reads ${ ab }$ again.
7. Complement Check (q2, q3): It reads ${ ba }$ and verifies it's the complement of ${ ab }$.
8. Second ww^c (q4): Now it has read ${ abbaabba }$.
9. Repeat or Accept (q4): Again, the machine makes a non-deterministic choice. This time, it guesses that it has reached the end of the input (correctly!). It transitions to ${ q_5 }$.
10. Accept (q5): The machine accepts the input because it's in ${ q_5 }$ at the end of the input.

If, at step 5, the machine had incorrectly guessed it was at the end of the input, it would have rejected because it wouldn't have consumed the entire input.

Key Points about Non-Determinism

• Guessing: The NTM guesses the length of ${ w }$ and when to transition between states. This "guessing" is what makes it non-deterministic.
• Multiple Paths: For a given input, the NTM can explore multiple possible execution paths simultaneously. If any of these paths leads to an accept state, the NTM accepts the input.
• Rejection: If all possible execution paths lead to a reject state, the NTM rejects the input.

Conclusion

This informal description provides a clear overview of how an NTM can recognize the language ${ L = \{(ww^c)^n \mid n > 1 \text{ and } w, w^c \in \{a, b\}^*\} }$. The key to its operation is the non-deterministic choices, which allow the machine to explore multiple possibilities and accept the input if any of those possibilities conform to the language's requirements. Designing NTMs often involves cleverly using non-determinism to "guess" the correct path, making them a powerful tool for recognizing complex languages. Remember, guys, the beauty of NTMs lies in their ability to explore all possibilities at once, making them a theoretical powerhouse in computer science!