Match the correct automation given in Y to its transition function in $X$:
$\begin{array}{|l|l|} \hline {} & X & {} & Y \\ \hline I. & Q^* \Sigma \rightarrow Q & A. & \text{NFA with null moves} \\ \hline II. & Q^* \Sigma \rightarrow 2Q & B. & \text{DFA} \\ \hline III. & Q^* \Sigma \rightarrow Q \times \{L,R\} & C. & \text{NFA} \\ IV. & Q^*\{ \Sigma \cup \epsilon \} \rightarrow 2Q & D. & \text{Turing Machine} \\ \hline {} & {} & E. & \text{2 way DFA} \\ \hline {} & {} & F. & \text{PDA} \\ \hline \end{array}$
- I-B, II-C, III-D, IV-F
- I-B, II-C, III-E, IV-A
- I-C, II-B, III-E, IV-F
- I-B, II-C, III-F, IV-D