Virtual Labs

Identify Non Context-free Language using Pumping Lemma

What is the primary use of the Pumping Lemma for context-free languages?

a: To prove that a language is context-free Explanation

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b: To prove that a language is not context-free Explanation

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c: To minimize a context-free grammar Explanation

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d: To remove ambiguity in a grammar Explanation

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Which of the following best describes the 'pumping' property in the Pumping Lemma for CFLs?

a: Any sufficiently long string in the language can be divided into parts that can be repeated while still remaining in the language. Explanation

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b: Every string in the language is finite. Explanation

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c: The language can be recognized by a deterministic PDA. Explanation

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d: The grammar has no useless symbols. Explanation

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What happens if a language fails to satisfy the Pumping Lemma for context-free languages?

a: The language is definitely context-free. Explanation

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b: The language is definitely not context-free. Explanation

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c: The language becomes regular. Explanation

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d: The grammar becomes ambiguous. Explanation

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Which of the following languages can be shown as not context-free using the Pumping Lemma?

a: { a^n b^n | n ≥ 0 } Explanation

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b: { a^n b^n c^n | n ≥ 0 } Explanation

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c: { a^i b^j | i, j ≥ 0 } Explanation

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d: { a^n b^m | n, m ≥ 0 } Explanation

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What are the five parts (u, v, x, y, z) in the Pumping Lemma for CFLs used for?

a: To divide the string into sections where v and y can be 'pumped' (repeated or removed). Explanation

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b: To identify the productions of the grammar. Explanation

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c: To simplify the PDA states. Explanation

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d: To eliminate useless variables. Explanation

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Why is the Pumping Lemma not a sufficient condition for context-freeness?

a: Because some non-CFLs may accidentally satisfy the lemma. Explanation

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b: Because it only applies to deterministic grammars. Explanation

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c: Because it applies only to ambiguous grammars. Explanation

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d: Because it works only for finite languages. Explanation

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