Question 40

Evaluate the four languages:

1. $L_1 = \{a^p \mid p \text{ is prime}\}$:
   By Parikh's theorem, any context-free language over a single-letter alphabet is regular. Since $L_1$ is not regular (primes do not form an arithmetic progression), $L_1$ cannot be context-free. Thus, Statement I is INCORRECT.
2. $L_2 = \{a^n b^m c^{2m} \mid n \ge 0, m \ge 0\}$:
   $L_2 = a^* \{b^m c^{2m} \mid m \ge 0\}$. Since $a^*$ is regular and $\{b^m c^{2m}\}$ is a deterministic context-free language, their concatenation is a context-free language. Thus, Statement II (which asserts $L_2$ is not context-free) is INCORRECT.
3. $L_3 = \{a^n b^n c^{2n} \mid n \ge 0\}$:
   Requires matching the counts of three separate symbol blocks ($a, b, c$) simultaneously. By the pumping lemma for CFLs, $L_3$ is not context-free. However, a multi-track Turing machine can easily compare the counts, so $L_3$ is recursive (decidable). Thus, Statement III is CORRECT.
4. $L_4 = \{a^n b^n \mid n \ge 1\}$:
   A deterministic pushdown automaton (DPDA) can push each '$a$' onto the stack and pop an '$a$' for each '$b$', accepting when the input and stack are both empty. Thus, $L_4$ is a deterministic context-free language (DCFL). Thus, Statement IV is CORRECT.

Statements III and IV only are correct, which corresponds to option D.