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GATE | GATE-CS-2009 | Question 60

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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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Last Updated : 11 Oct, 2021
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