Designing a PDA for Language L = { a^nb^m }

What is the PDA for the language L = { a^nb^m }?

How can we create a Pushdown Automaton (PDA) for the language L = { a^nb^m } given the conditions m <= n <= 2m? Is the PDA deterministic or not?

The PDA for the language L = { a^nb^m }

The PDA for the language L = { a^nb^m }, where m <= n <= 2m, is deterministic. To construct a Pushdown Automaton (PDA) for the language L = { a^nb^m }, we can follow these steps:

1. Design the PDA's states:

  • Start state (q0)
  • Accept state (qf)
  • Additional states as needed

2. Define the PDA's alphabet:

  • Input alphabet: {a, b}
  • Stack alphabet: {Z0} (includes the initial stack symbol Z0)

3. Set the transition rules:

  • From state q0:
    • On input symbol 'a' and stack symbol Z0, push 'A' onto the stack and stay in q0.
    • On input symbol 'a' and stack symbol 'A', push 'A' onto the stack and stay in q0.
    • On input symbol 'b' and stack symbol 'A', pop 'A' from the stack and stay in q0.
    • On input symbol 'b' and stack symbol Z0, move to state qf only if the stack is empty.

4. Determine if the PDA is deterministic or not:

  • The PDA is deterministic if, for every input symbol and stack symbol, there is at most one transition defined. In this case, the PDA is deterministic.

5. Draw the transition diagram:

Here's a simplified diagram to help visualize the PDA:

        a, Z0 -> A
        a, A  -> A
        b, A  -> ε
        b, Z0 -> ε (to qf)

Please note that the diagram above is a simplified representation of the PDA's transition rules.

Remember to follow the conventions used in your specific compiler construction lecture or course material while constructing the PDA.

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