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GATE | GATE-CS-2006 | Question 33

  • Last Updated : 28 Jun, 2021

Let L1 be a regular language, L2 be a deterministic context-free language and L3 a recursively enumerable, but not recursive, language. Which one of the following statements is false?
(A) L1 ∩ L2 is a deterministic CFL
(B) L3 ∩ L1 is recursive
(C) L1 ∪ L2 is context free
(D) L1 ∩ L2 ∩ L3 is recursively enumerable


Answer: (B)

Explanation:
(A) This statement is true because deterministic context free languages are closed under intersection with regular languages.

(B) This statement is false because L1 is recursive and every recursive language is decidable. L3 is recursively enumerable but not recursive. So, L3 is undecidable. Intersection of recursive language and recursive enumerable language is recursively enumerable .

(C) This statement is true because L1 is regular. Every regular language is also a context free languages. L2 is a deterministic context free language and every DCFL is also a context free languages. Every context free language is closed under Union.


(D) This statement is true beacuse L1 is regular hence it is also recursively enumerable. L2 is deterministic context free language so, it is also recursively enumerable . Recursively enumerable languages are closed under intersection.

 
Thus, problem mentioned in option (A) is undecidable.

 
Please comment below if you find anything wrong in the above post.


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