Question

In a pushdown automaton \( P = (Q, \Sigma, \Gamma, \delta, q_0, F) \), a transition of the form 

where \( p, q \in Q \), \( a \in \Sigma \cup \{\epsilon\} \), and \( X, Y \in \Gamma \cup \{\epsilon\} \), represents \[ (q, Y) \in \delta(p, a, X). \] Consider the following pushdown automaton over the input alphabet \( \Sigma = \{a, b\} \) and stack alphabet \( \Gamma = \{\#, A\} \):

The number of strings of length 100 accepted by the above pushdown automaton is \(\underline{\hspace{2cm}}\).

Hint:

In PDAs, push operations for one symbol and pop operations for another typically enforce counting constraints like \( n > m \) or \( n = m \).

Answer 1

Correct Answer: 50

Step 1: Stack behavior of the PDA.
For each input symbol: 
- Every \( a \) causes the PDA to push one symbol \( A \) onto the stack.  
- Every \( b \) causes the PDA to pop one symbol \( A \) from the stack.

Step 2: Acceptance condition.
The PDA accepts an input string only if, after processing the entire string, there is at least one \( A \) remaining on the stack. 
This means the total number of \( a \)'s must be strictly greater than the total number of \( b \)'s.

Step 3: Form of accepted strings.
All accepted strings are of the form: \[ a^n b^m \] with the constraint: \[ n > m \]

Step 4: Apply the length constraint.
The total length of the string is given as 100: \[ n + m = 100 \] along with: \[ n > m \]

Step 5: Count the valid strings.
From \( n + m = 100 \) and \( n > m \), we get: \[ n > 50 \] Hence, \( n \) can take integer values from 51 to 100.

The number of valid values of \( n \) is: \[ 100 - 50 = 50 \]

Final Answer:
\[ \boxed{50} \]