which one is R.E??

Question

Consider the following languages:
L1={0^(2^k)│k≥0}
L2={b∈{0,1}*│b∈L(M_b] ) }
Which of the above languages is TM recognizable?

please explain second one

Comments

Now, L(M) = { $\epsilon$ } or L(M) = {0}  or L(M) = {1} and so on(i.e M will be turing machine for each of the strings of (0+1)* ).

Now, this is non-trivial and hence, undecidable.

Considering extended Rice theorem also, you dont get any voilation. So, I think this is RE.

 

But, I am not sure of this.

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is first one RE?

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first one is given as R.E ND SECOND ONE IS GIVEN AS NON-r.e

WHAT IS THE MENAING OF SECOND ONE??

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First one is CSL so it should be recursive.

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@rahul,can you pls explain second language?

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@akriti, i guess 2nd one is in set form b is set of strings over all possible strings such that string belong to

language accepted by T.M  Mb. i may be wrong on the interpretation, but 2nd one is quite confusing

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Actually I am not able to understand, but I think they want to say that if a string is accpetable by TM in {0,1}*  then it belongs to L2. L2 may be CSL it may be CFL or regular