Let Ξ£ = {π‘Ž, 𝑏, 𝑐, 𝑑} and let 𝐿 = {π‘Ž 𝑖𝑏 𝑗 𝑐 π‘˜π‘‘ β„“ ∣ 𝑖,𝑗, π‘˜, β„“ β‰₯…

2026

Let Ξ£ = {π‘Ž, 𝑏, 𝑐, 𝑑} and let 𝐿 = {π‘Ž 𝑖𝑏 𝑗 𝑐 π‘˜π‘‘ β„“ ∣ 𝑖,𝑗, π‘˜, β„“ β‰₯ 0}. Which of the following constraints ensure(s) that the language 𝐿 is context-free?

  1. A.

    i + π‘˜ = 𝑗 + β„“

  2. B.

    𝑖 = π‘˜ and 𝑗 = β„“

  3. C.

    i= β„“ and 𝑗 = k

  4. D.

    𝑖 + 𝑗 = π‘˜ + β„“

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Correct answer: A, C, D

Analysis of Constraints

The base language L = {a^i b^j c^k d^β„“ | i, j, k, β„“ β‰₯ 0} is regular. The context-free status depends on the constraints linking the counts.

Option A: i + k = j + β„“

This constraint implies i - j = β„“ - k. A PDA can push for 'a' and 'c', and pop for 'b' and 'd'. State transitions handle stack underflow, making this language context-free.

Option B: i = k and j = β„“

This requires matching 'a' with 'c' and 'b' with 'd' independently. The structure a^i b^j c^i d^j is not context-free because the stack cannot track two independent counts separated by other symbols.

Option C: i = β„“ and j = k

This forms a^i b^j c^j d^i. A PDA pushes 'a' and 'b', pops 'b' for 'c', and pops 'a' for 'd'. The nesting is valid, so this language is context-free.

Option D: i + j = k + β„“

This balances the first half against the second half. A PDA pushes for 'a' and 'b', then pops for 'c' and 'd'. The linear relationship allows verification with a single stack.

Conclusion

Options A, C, and D ensure the language is context-free. Option B does not.

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