Step 1: Understanding NDFA to DFA conversion.
In the conversion from a Non-deterministic Finite Automaton (NDFA) to a Deterministic Finite Automaton (DFA), the number of states in the DFA is determined by the power set of the states of the NDFA. For an NDFA with \( N \) states, the DFA may have up to \( 2^N \) states because each state of the DFA corresponds to a subset of states of the NDFA.
Step 2: Conclusion.
Therefore, the possible number of states in the equivalent DFA is \( 2^N \), making the correct answer (3).
Find the least upper bound and greatest lower bound of \( S = \{X, Y, Z\} \) if they exist, of the poset whose Hasse diagram is shown below:
Suppose \( D_1 = (S_1, \Sigma, q_1, F_1, \delta_1) \) and \( D_2 = (S_2, \Sigma, q_2, F_2, \delta_2) \) are finite automata accepting languages \( L_1 \) and \( L_2 \), respectively. Then, which of the following languages will also be accepted by the finite automata:
(A) \( L_1 \cup L_2 \)
(B) \( L_1 \cap L_2 \)
(C) \( L_1 - L_2 \)
(D) \( L_2 - L_1 \)
Choose the correct answer from the options given below:
Match LIST-I with LIST-II \[\begin{array}{|c|c|c|}\hline \text{ } & \text{LIST-I} & \text{LIST-II} \\ \hline \text{A.} & \text{A Language L can be accepted by a Finite Automata, if and only if, the set of equivalence classes of $L$ is finite.} & \text{III. Myhill-Nerode Theorem} \\ \hline \text{B.} & \text{For every finite automaton M = $(Q, \Sigma, q_0, A, \delta)$, the language L(M) is regular.} & \text{II. Regular Expression Equivalence} \\ \hline \text{C.} & \text{Let, X and Y be two regular expressions over $\Sigma$. If X does not contain null, then the equation $R = Y + RX$ in R, has a unique solution (i.e. one and only one solution) given by $R = YX^*$.} & \text{I. Arden's Theorem} \\ \hline \text{D.} & \text{The regular expressions X and Y are equivalent if the corresponding finite automata are equivalent.} & \text{IV. Kleen's Theorem} \\ \hline \end{array}\]
\[\text{Matching List-I with List-II}\]
Choose the correct answer from the options given below: