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GATE | GATE CS 2013 | Question 17

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Which of the following statements is/are FALSE?

1. For every non-deterministic Turing machine, 
   there exists an equivalent deterministic Turing machine.
2. Turing recognizable languages are closed under union 
   and complementation.
3. Turing decidable languages are closed under intersection 
   and complementation.
4. Turing recognizable languages are closed under union 
   and intersection. 


1 and 4 only


1 and 3 only


2 only


3 only

Answer: (C)


A recognizer of a language is a machine that recognizes that language. A decider of a language is a machine that decides that language. Both types of machine halt in the Accept state on strings that are in the language A Decider also halts if the string is not in the language A Recognizer MAY or MAY NOT halt on strings that are not in the language On all input: A Decider MUST halt (in Accept or Reject state) A Recognizer MAY or MAY NOT halt on some strings (Q: Which ones?) A language is Turing-decidable (or decidable) if some Turing machine decides it. Aka Recursive Language. A language is Turing-recognizable if some Turing machine recognizes it. Aka Recursively Enumerable Language. Source: Recursive (Turing Decidable) languages are closed under following Kleene star, concatenation, union, intersection, complement and set difference. Recursively enumerable language are closed under Kleene star, concatenation, union, intersection. They are NOT closed under complement or set difference.

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Last Updated : 28 Jun, 2021
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