Consider L₁ = {a^(n+m) bⁿ aᵐ | n, m ≥ 0} and L₂ = {a^(n+m) b^(n+m) a^(n+m) | n, m ≥ 0}. Which one is correct?
Pick an option to check yourself, or open the answer below.
Show answer & explanation
Correct answer: A. Only L₁ is context free
Explanation
L₁ = aᵐ aⁿ bⁿ aᵐ has nested matching, so it is context free. L₂ = aᵏ bᵏ aᵏ needs three equal counts, which no PDA can check.
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