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GATE | GATE-CS-2009 | Question 60
  • Difficulty Level : Hard
  • Last Updated : 12 Nov, 2018

Let L = L1∩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.


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 \cap L2 = \{a^{k}b^{k}c | k >= 0\} which is context free, but not regular.

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