Turing machine

Question

Though this question has already been discussed but still it's not getting clear to me. Can somebody please explain?

Define languages L0 and L1 as follows : 

L0={⟨M,w,0⟩∣M halts on w}

L1={⟨M,w,1⟩∣M does not halts on w}

Here ⟨M,w,i⟩ is a triplet, whose first component M is an encoding of a Turing 
Machine, second component w is a string, and third component i is a bit. 

Let L=L0∪L1. Which of the following is true?

1. L is recursively enumerable, but L′ is not
2. L′ is recursively enumerable, but L is not
3. Both L and L′ are recursive  
4. Neither L nor L′ is recursively enumerable

Comments

mention your doubt

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hey srestha and sourav my doubt is on understanding the basics of the question.1) How are the strings stored in the language and how to solve it.But before that if u can please look into this question : https://gateoverflow.in/1994/gate2014-2-35 and i have given my explaination in that link as well but got no reply(as far as i have understood) and correct me there it would be nice!Explaination : <M> is encoding of Turing machine(bits of 0 and 1) and the Language is like : L ={ <011010>,<0110101011>...finite no. of such Turing machines which can accept a string of length 2014} .Here <011010> represents the initial state,transition,final states all inputs and other usual things. Now we will give this Language (L) to a SINGLE Turing machine(let me call this T1) which will process the encodings of all other turing machines.Now we can say "yes"  if the T1 can halt after processing the <011010> etc encodings after processing an arbitrary string based on the input symbols of the encoding.If the TM halts after processing 2014 length and ends in final state then it is accepted and says "yes".Coming to "no" part it can say no or can go to loop.So to confirm for loop we use Rice's Theorem which states that as there exists a turing machine which will take a string(based on input symbols of the TM) and halt on length 2014 and there exists another TM which wont halt after processing a string of length 2014 hence we can confirm its undecidable. @Arjun Sir Please check ( These question's basics are quite uneasy to me hence i need you to clarify it by word and see if i made mistake(s) ) .Thank you!

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@ Sayan Ghosh
u mean for no part , u r taking another turing machine?
But break the loop in such a way is not required , because before that we got the result.