Question

Consider the pushdown automaton (PDA) \(P\) over input alphabet \(\{a,b\}\) and stack alphabet \(\{\perp, A\}\) with start state \(s\) and acceptance by empty stack.
The transition sketch shows:
- In \(s\) on each \(a\) it pushes one \(A\) (i.e., \(a/\perp \to A\perp\) and \(a/A \to AA\)).
- On the first \(b\) it goes to \(p\) with \(b/A \to \epsilon\) and stays in \(p\) popping one \(A\) per \(b\) (\(b/A \to \epsilon\)).
- From \(p\) it may go to \(q\) by \(\epsilon/A \to \epsilon\), and in \(q\) it can keep popping \(A\)’s and finally pop \(\perp\) by \(\epsilon/\perp \to \epsilon\).
Question: Which option describes \(L(P)\)? IMAGE8

• A. $\{a^m b^n \mid 1 \le m \text{ and } n<m\}$
• B. $\{a^m b^n \mid 0 \le n \le m\}$
• C. $\{a^m b^n \mid 0 \le m \text{ and } 0 \le n\}$
• D. $\{a^m \mid 0 \le m\} \;\cup\; \{b^n \mid 0 \le n\}$

Hint:

For PDAs that accept by empty stack, check where} the $\epsilon/\perp$ pop is available. If reaching that state requires a nonempty stack (e.g., an $\epsilon/A$ edge), equal counts (here $n=m$) often get blocked from emptying the stack.

Answer 1

The Correct Option is A

To determine the language \(L(P)\) described by the given pushdown automaton (PDA) \(P\), let's examine the transition operations carefully:

1. The PDA starts in state \(s\). On reading each input \(a\), it pushes one \(A\) onto the stack. This operation can be represented by:
   • \(a/\perp \to A\perp\)
   • \(a/A \to AA\)
2. Upon encountering the first \(b\), the PDA transitions to state \(p\). Here, the PDA pops an \(A\) for each \(b\) it reads:
   • \(b/A \to \epsilon\)
3. In state \(p\), the PDA can transition to state \(q\) via an epsilon (\(\epsilon\)) transition, popping an \(A\) if there is any at the top of the stack:
   • \(\epsilon/A \to \epsilon\)
4.  Finally, in state \(q\), the PDA continues popping \(A\)'s and eventually pops \(\perp\) using:
   • \(\epsilon/\perp \to \epsilon\)

Since the PDA accepts by empty stack, acceptance occurs only when the stack is fully emptied. Now, let's derive what this means for the language:

• The PDA reads \(a^m\) where each \(a\) pushes an \(A\) onto the stack. Thus we have \(m\) occurrences of \(A\).
• It then reads \(b^n\), where each \(b\) pops an \(A\). Therefore, \(n\) must be less than \(m\) to have some \(A\)'s remaining after processing all \(b\)'s.
• After processing all \(b\)'s, it uses \(\epsilon\) transitions to pop the remaining \(A\)'s and eventually \(\perp\) to accept the string with an empty stack.
• This implies that the PDA accepts strings of the form \(a^m b^n\) where \(m > n\) and \(m \geq 1\).

Therefore, the language \(L(P)\) that the PDA accepts is correctly described by:

$\{a^m b^n \mid 1 \le m \text{ and } n < m\}$

This option ensures that there are always more \(a\)'s than \(b\)'s, which is necessary to leave an empty stack at the end.