Let L = L1∩L2, where L1 and L2 are languages as defined below:
L1 = {am bm can bn | m, n >= 0}
L2 = {ai bj ck | 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.
Answer: (C)
Explanation:
The language L1 accept strings {c, abc, abcab, aabbcab, aabbcaabb, …} and L2 accept strings {a, b, c, ab, abc, aabc, aabbc, … }. Intersection of these two languages is L1 ∩ L2 = ak bk c ∣ k >= 0 which is context free, but not regular.
Hence, Option C is the correct answer
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