Alright, let's dive into designing a Pushdown Automaton (PDA) for the language L = {a^(2n) s b^n s e | n >= 0}. This is a classic computer science problem, and walking through the design will help solidify your understanding of PDAs and context-free languages. So, buckle up, and let's get started!

Understanding the Language

Before we even think about drawing states and transitions, we need to truly understand what this language is all about. L consists of strings that follow a very specific pattern: some number of 'a's, followed by a single 's', then some number of 'b's, another 's', and finally an 'e'. The number of 'a's must be twice the number of 'b's. Let's break it down with some examples:

• If n = 0: The string is "s s e".
• If n = 1: The string is "aa s b s e".
• If n = 2: The string is "aaaa s bb s e".
• If n = 3: The string is "aaaaaa s bbb s e".

Notice how the number of 'a's is always double the number of 'b's. This is crucial for our PDA design because we'll need a way to keep track of this relationship. This tracking is where the stack comes into play. A PDA (Pushdown Automaton) is a finite automaton with an extra stack. The stack is a last-in, first-out (LIFO) data structure, which is extremely useful for keeping track of the relationship between different parts of the input string, especially for context-free grammars.

High-Level Strategy

Here’s the general idea of how we'll design the PDA:

1. Read 'a's: For every two 'a's we read, push a symbol (let’s use 'X') onto the stack.
2. Encounter the first 's': When we see the first 's', we know we're done reading the 'a's. We'll transition to a new state without modifying the stack. This marks the separation between the 'a's and 'b's.
3. Read 'b's: For every 'b' we read, pop an 'X' from the stack. This ensures that the number of 'b's is half the number of 'a's.
4. Encounter the second 's': When we see the second 's', we transition to another state. The stack should be empty at this point if the number of 'a's and 'b's were correct.
5. Read 'e': Finally, we read the 'e'. If the stack is empty and we read 'e', we accept the string. If the stack is not empty, the string is rejected.

This strategy uses the stack to enforce the a^(2n) and b^n relationship. We are using a stack to store the ‘a’ and comparing it with ‘b’. If the number of ‘b’ matches with half the number of ‘a’ then the machine will accept it otherwise it will reject it.

Formal Definition of the PDA

Now, let's formalize this into a PDA. A PDA is formally defined as a 7-tuple: (Q, Σ, Γ, δ, q0, Z0, F), where:

• Q is a finite set of states.
• Σ is a finite input alphabet.
• Γ is a finite stack alphabet.
• δ is the transition function: Q x (Σ ∪ {ε}) x (Γ ∪ {ε}) -> P(Q x Γ*).
• q0 is the start state.
• Z0 is the initial stack symbol.
• F is the set of accept states.

For our language L, we can define the PDA as follows:

• Q = {q0, q1, q2, q3, q4} (States)
• Σ = {a, b, s, e} (Input Alphabet)
• **Γ = {X, }** (*Stack Alphabet*. '' is the initial stack symbol.)
• q0 (Start State)
• Z0 = $ (Initial Stack Symbol)
• F = {q4} (Accept State)

Now, the trickiest part: the transition function δ. Let's define it step-by-step. I am using the ‘$’ to determine whether the stack is empty or not. If the stack is empty then there will be ‘$’ otherwise there will be other values.

Transition Function δ

Here's how the transition function works:

1. (q0, a, $) -> (q0, X$)
   • In state q0, if we read an 'a' and the top of the stack is '$', we push 'X' onto the stack.
2. (q0, a, X) -> (q0, XX)
   • In state q0, if we read an 'a' and the top of the stack is 'X', we push another 'X' onto the stack. This ensures that for every two 'a's, we push one 'X'.
3. (q0, s, $) -> (q1, $)
   • In state q0, if we read an 's' and the top of the stack is '$', we move to state q1 without changing the stack. (This handles the case when n=0)
4. (q0, s, X) -> (q1, X)
   • In state q0, if we read an 's' and the top of the stack is 'X', we move to state q1 without changing the stack.
5. (q1, b, X) -> (q1, ε)
   • In state q1, if we read a 'b' and the top of the stack is 'X', we pop 'X' from the stack.
6. (q1, s, $) -> (q2, $)
   • In state q1, if we read an 's' and the top of the stack is '$', we move to state q2 without changing the stack. This is used to verify that all ‘b’ are processed and the stack is empty.
7. (q1, s, X) -> (q_reject, X)
   • In state q1, if we read an 's' and the top of the stack is 'X', we go to a reject state since not all the ‘b’ are processed.
8. (q2, e, $) -> (q4, $)
   • In state q2, if we read an 'e' and the top of the stack is '$', we move to the accept state q4 without changing the stack.
9. (q_reject, a, X) -> (q_reject, X)
   • This is a trap state for all invalid sequence of input.
10. (q_reject, b, X) -> (q_reject, X)
    • This is a trap state for all invalid sequence of input.
11. (q_reject, s, X) -> (q_reject, X)
    • This is a trap state for all invalid sequence of input.
12. (q_reject, e, X) -> (q_reject, X)
    • This is a trap state for all invalid sequence of input.

Explanation of the Transitions

• State q0: This is our initial state. We read the 'a's here, pushing 'X' onto the stack for every two 'a's. When we encounter the first 's', we move to state q1.
• State q1: In this state, we read the 'b's, popping an 'X' from the stack for each 'b'. This ensures that the number of 'b's is half the number of 'a's. When we encounter the second 's', we move to state q2.
• State q2: In this state, we check if the stack is empty and we read ‘e’ from the input. If the stack is empty it means that the input follows the language and we accept the input.
• State q4: This is our accept state. If we reach this state, the string is in the language L.

Example Trace

Let's trace the string "aaaa s bb s e" to see how this PDA works:

1. Start: (q0, aaaa s bb s e, $)
2. **(q0, a, $)** -> (q0, aaa s bb s e, X$)
3. (q0, a, X) -> (q0, aa s bb s e, XX$)
4. (q0, a, X) -> (q0, a s bb s e, XXX$)
5. (q0, a, X) -> (q0, s bb s e, XXXX$)
6. (q0, s, X) -> (q1, bb s e, XXXX$)
7. (q1, b, X) -> (q1, b s e, XXX$)
8. (q1, b, X) -> (q1, s e, XX$)
9. (q1, s, X) -> (q_reject, e, XX$)

In this case, the string is rejected.

Let’s try “aa s b s e”

1. Start: (q0, aa s b s e, $)
2. **(q0, a, $)** -> (q0, a s b s e, X$)
3. (q0, a, X) -> (q0, s b s e, XX$)
4. (q0, s, X) -> (q1, b s e, X$)
5. (q1, b, X) -> (q1, s e, $)
6. (q1, s, $) -> (q2, e, $)
7. (q2, e, $) -> (q4, $, $)

The string "aa s b s e" is accepted.

Conclusion

So, there you have it! A complete PDA design for the language L = {a^(2n) s b^n s e | n >= 0}. The key to designing PDAs is to understand the language's structure and use the stack to keep track of the relationships between different parts of the input. Remember to break down the problem into smaller, manageable steps, and always test your PDA with various inputs to ensure it works correctly. Now, go forth and conquer more PDA challenges!