Let L = L₁ \cap L₂, where L₁ and L₂ are languages as defined below: L₁ = {a^{m}b^{m}ca^{n}b^{n} | m, n = 0 } L₂ = {a^{i}b^{j}c^{k} | i, j, k = 0 } Then L is A)Not recursive B)Regular C)Context free but not regular D)Recursively enumerable but not context free.

Question

Let L = L₁ \cap L₂, where L₁ and L₂ are languages as defined below: L₁ = {a^{m}b^{m}ca^{n}b^{n} | m, n >= 0 } L₂ = {a^{i}b^{j}c^{k} | i, j, k >= 0 } Then L is A)Not recursive B)Regular C)Context free but not regular D)Recursively enumerable but not context free.

Quizplus · Accepted Answer

The language L = L₁ ∩ L₂ is context-free but not regular. This is because L is the intersection of two languages, where L₁ = {a^m b^m ca^n b^n | m, n >= 0} is a context-free language that requires a matching number of a's and b's both before and after a 'c', which cannot be recognized by a finite automaton, making it non-regular. L₂ = {a^i b^j c^k | i, j, k >= 0} is a regular language as it can be recognized by a finite automaton (it does not require matching numbers of symbols). The intersection of a context-free language and a regular language is context-free, thus making L context-free. However, since L requires a specific pattern that cannot be recognized by a finite automaton (due to the need to match numbers of a's and b's), it is not regular.

Let L = L1 \Cap L2, Where L1 and L2

Let L = L₁ \Cap L₂, Where L₁ and L₂