Consider the following languages. \(L_1 = \{a^p \mid p \text{ is a prime…

Concept

Language-class questions are decided by closure properties and machine models. Every unary context-free language is regular, a context-free grammar may coordinate two dependent blocks, and a deterministic pushdown automaton can compare two consecutive blocks.

A language can be recursive without being context-free: a terminating algorithm may verify several length constraints even when one stack cannot enforce all of them.

Application

1. L1 = {a^p | p is prime}: If L1 were context-free, then as a unary context-free language it would be regular. Prime lengths are not ultimately periodic, so L1 is not regular and therefore not context-free. Statement I is false.

2. L2 = {a^n b^m c^{2m} | n ≥ 0, m ≥ 0}: The grammar S → aS | T and T → bTcc | ε generates any number of a symbols, followed by m b symbols and 2m c symbols. Hence L2 is context-free, so statement II is false.

3. L3 = {a^n b^n c^{2n} | n ≥ 0}: It requires the same n across three blocks, with the c block twice as long. It is not context-free. It is recursive because an algorithm can verify the block order, count a and b symbols, and check that the number of c symbols is twice that count. Statement III is true.

4. L4 = {a^n b^n | n ≥ 1}: A deterministic pushdown automaton pushes one marker for each a, switches when the first b appears, and pops one marker per b. It accepts exactly when the input ends with the stack balanced. Statement IV is true.

Cross-check

• For L3, let h(a)=a, h(b)=b and h(c)=cc. If L3 were context-free, closure under inverse homomorphism and intersection with the regular language a*b*c* would make {a^n b^n c^n | n ≥ 0} context-free, a contradiction.

• The proposed grammar S → aSbcc | ε does not generate a^n b^n c^{2n}; for n=2 it yields aabbccbcc, so it does not preserve one contiguous b block followed by one contiguous c block.

Result

Only statements III and IV are correct. Therefore the answer is III and IV only.