Which of the following is true for the language: { ap| p is a prime}
2025
Which of the following is true for the language: { ap| p is a prime}
Answer: B. It is neither regular nor context-free, but accepted by a Turing machine — ConceptA language over a single-letter alphabet (a unary language) can only be regular, or context-free, or neither, based on whether its set of string…
- A.
It is regular but not context-free
- B.
It is neither regular nor context-free, but accepted by a Turing machine
- C.
It is not accepted by a Turing Machine
- D.
It is context-free but not regular
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Show answer & explanation
Correct answer: B
Concept
A language over a single-letter alphabet (a unary language) can only be regular, or context-free, or neither, based on whether its set of string lengths is "ultimately periodic" — i.e. eventually repeats with some fixed period. The pumping lemma for regular languages, and Parikh's theorem for context-free languages, both force this same periodicity requirement on any unary regular or context-free language. Separately, a language is accepted (decided) by a Turing machine whenever SOME algorithm — however slow — can determine membership in finite time; this is a much weaker requirement than periodicity.
Application to This Language
Here the string lengths are exactly the prime numbers. Primes are famously NOT ultimately periodic: the gaps between consecutive primes grow without bound — for any n, the n consecutive integers n!+2, n!+3, ..., n!+n are all composite, so arbitrarily long gaps exist — so no fixed period eventually repeats. Applying the pumping lemma for regular languages: for a large enough prime p, pumping any nonempty substring y within the first n characters produces lengths p + k*(|y|) for every k >= 0 — an arithmetic progression that must eventually hit a composite value, so ap cannot be regular. Parikh's theorem then rules out context-freeness too, since it forces every unary CFL to be ultimately periodic, and primes are not. On the other hand, checking primality IS decidable — trial division (or the polynomial-time AKS algorithm) lets a Turing machine test any given length and halt with a definite yes/no — so the language IS accepted by a Turing machine.
Cross-Check
As a cross-check: decidable languages are a strict superset of context-free languages, which are themselves a strict superset of regular languages. A language sitting outside both the regular and context-free classes but still decidable is exactly the middle ground the Chomsky hierarchy predicts for a language whose lengths are governed by an irregular arithmetic property like primality — consistent with ruling out options 0, 2, and 3, and confirming that it is neither regular nor context-free, but is accepted by a Turing machine.
Conclusion: The language { ap | p is prime } is neither regular nor context-free, but it is accepted by a Turing machine.