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28 votes

Let $L$ be a language and $\bar{L}$ be its complement. Which one of the following is NOT a viable possibility?

  1. Neither $L$ nor  $\bar{L}$ is recursively enumerable $(r.e.)$. 
  2. One of $L$ and $\bar{L}$ is r.e. but not recursive; the other is not r.e.
  3. Both $L$ and $\bar{L}$ are r.e. but not recursive.
  4. Both $L$ and $\bar{L}$ are recursive.
in Theory of Computation
edited by
plz explain a and b in brief ?

@Shubham Option A: In figure if your language L lies in outer space(not R E) then its compliment will also lie in outer space(not RE HERE it can idientified using ALAN) so This statement is true.

Option B:: if langugae L is Re but not rec(meaning it is not recursive so Total turing machine not possible but RE so Univarsal turing machine possible) . L compliment  will not be RE (as only recursive lang is closed under compliment).

 Image result for automata languages graph like regular context free


It can be a MSQ qus

2 Answers

33 votes
Best answer

(C) is not possible. If $L$ is re we have a TM that accepts string in $L$. If $L$' is re, we have a TM that accepts strings in $L$'. So, using both these TMs we can make a new TM M which accepts strings in $L$ and rejects strings in $L$' - that is $M$ decides $L$, making $L$ recursive.

edited by
Can you give an example for option A?
@Shalini If $L$ itself is not RE then $L'$ is also not RE
IF L is recursive then L' is also Recursive
IF L is Recursively enumerable then L' is NOT recursively enumerable .

Correct me if wrong...
That seems correct
The number of turing machines are countable but total number of languages are uncountable. There are many languages for which turing machines do not exist i.e. they are not recursively enumerable.
Please explain option A.
If L is rec enum then Lc can be rec enum or not rec enum.

But if L is rec enum but not recursive Lc is definitely not Rec enum
1 vote
A) It is possible if L itself is NOT RE. Then L' will also not be RE. B) Suppose there is a language such that turing machine halts on the input. The given language is RE but not recursive and its complement is NOT RE. C) This is not possible because if we can write enumeration procedure for both languages and it's complement, then the language becomes recursive. D) It is possible because L is closed under complement if it is recursive.   Thus, C is the correct choice.

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