GATE CS 2014 Set 2 — Question 26
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Theory of Computation → Turing Machines & Computability → Recursive & RE Languages
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Question
Let denotes that language A is mapping reducible (also known as many-to-one reducible) to language B. Which one of the following is FALSE?
Correct answer
(D) If A ₘ B and B is not recursively enumerable then A is not recursively enumerable.
Solution
The property of mapping reducibility implies:
1.If is recursive, then is recursive. (Option A is True)
2.If is recursively enumerable (RE), then is RE. (Option C is True)
3.The contrapositive of (1) is: If is not recursive (undecidable), then is not recursive. However, Option B states "If is undecidable then is undecidable". Note that if is undecidable, it does not strictly imply is undecidable in all contexts unless we are talking about specific hardness, but generally, if a hard problem reduces to , must be at least as hard. More formally, if is undecidable, and , then cannot be decidable (recursive), because if were recursive, would be recursive. Thus, is undecidable. (Option B is True)
4.Option D states: "If is not recursively enumerable then is not recursively enumerable." This is the converse of the contrapositive of (2). The correct contrapositive of (2) is: If is not RE, then is not RE. The statement in D is false. For example, let be a regular language (which is RE) and be a non-RE language. We can define a reduction from to (mapping all strings in to an element and strings not in to , assuming such elements exist and we can construct the map). Wait, the reduction function must be computable. A constant function is computable. So we can reduce a decidable language to a non-RE language. Thus can hold with being RE and being not RE. Therefore, not RE does not imply not RE.
Thus, (D) is FALSE.Continue learning with Success Tracker
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