Consider the NPDA $(Q = \{q_0, q_1, q_2\}, \Sigma = \{0, 1\}, \Gamma = \{0, 1, \bot\}, \delta, q_0, \bot, F = \{q_2\})$, where (as per usual convention) $Q$ is the set of states, $\Sigma$ is the input alphabet, $\Gamma$ is the stack alphabet, $\delta$ is the state transition function, $q_0$ is the initial state, $\bot$ is the initial stack symbol, and $F$ is the set of accepting states. The state transition is as follows: Which one of the following sequences must follow the string 101100 so that the overall string is accepted by the automaton?

Let us trace the operation of the given Non-deterministic Pushdown Automaton (NPDA):

1. In state $q_0$, every input bit ($0$ or $1$) is pushed onto the stack on top of whatever symbol is present.

2. The transition from $q_0$ to $q_1$ can be taken non-deterministically on input $0$, $1$, or $\epsilon$, without changing the stack symbol ($Z \to Z$).

3. In state $q_1$, reading a $0$ pops a $1$ from the stack ($0, 1Z \to Z$), and reading a $1$ pops a $0$ from the stack ($1, 0Z \to Z$).

4. Once the stack is cleared down to $\bot$, the automaton moves to accepting state $q_2$ via the $\epsilon$-transition $\epsilon, \bot \to \epsilon$.

Given the input prefix 101100:

• If the first $5$ characters 10110 are processed in state $q_0$, the stack (from bottom to top above $\bot$) contains: 1, 0, 1, 1, 0 (with top of the stack being 0).

• The $6^{\text{th}}$ character 0 is consumed during the transition $q_0 \to q_1$ via the transition $0, Z \to Z$.

• In state $q_1$, to pop the remaining symbols from top to bottom (0, 1, 1, 0, 1), the automaton requires the complement of each symbol in order:

  • Top 0 requires input 1 (pops 0)

  • Next 1 requires input 0 (pops 1)

  • Next 1 requires input 0 (pops 1)

  • Next 0 requires input 1 (pops 0)

  • Next 1 requires input 0 (pops 1)

Thus, the string that must follow is 10010, corresponding to option (B).