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: Analyze stack behavior for input symbols.
Each input symbol \( a \) causes the PDA to push one symbol \( A \) onto the stack.
Each input symbol \( b \) causes the PDA to pop one symbol \( A \) from the stack.

Step 2: Acceptance condition.
The PDA reaches the accepting state only if there is at least one symbol \( A \) remaining on the stack.
Therefore, the total number of \( a \)'s must be strictly greater than the total number of \( b \)'s.

Step 3: Describe the accepted language.
All accepted strings are of the form:
\[ a^n b^m \] subject to the condition:
\[ n > m \]

Step 4: Apply the length constraint.
Given that the length of the string is 100:
\[ n + m = 100 \] with the condition \( n > m \).

Step 5: Count valid strings.
The inequality \( n > m \) implies:
\[ n > 50 \] So \( n \) can take integer values from 51 to 100.
Number of valid values of \( n \):
\[ 100 - 50 = 50 \] % Final Answer

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