Let < M > be the encoding of a Turing machine as a string over {0, 1}. Let L = { < M > |M is a Turing machine that accepts a string of length 2014 }. Then, L isa)decidable and recursively enumerableb)undecidable but recursively enumerablec)undecidable and not recursively enumerabled)decidable but not recursively enumerableCorrect answer is option 'B'. Can you explain this answer?

Question

Let &#60; M &#62; be the encoding of a Turing machine as a string over {0, 1}. Let L = { &#60; M &#62; |M is a Turing machine that accepts a string of length 2014 }. Then, L isa)decidable and recursively enumerableb)undecidable but recursively enumerablec)undecidable and not recursively enumerabled)decidable but not recursively enumerableCorrect answer is option 'B'. Can you explain this answer?

Answer

There are finite number of strings of length ‘2014’. So, a turing machine will take the input string of length ‘2014’ and test it.
If, input string is present in the language then turing machine will halt in final state . But, if turing machine is unable to accept the input string then it will halt in non-final state or go in an infinite loop and never halt.

Thus, ‘L’ is undecidable and recursively enumerable .

Please comment below if you find anything wrong in the above post.

Answer

There are finite number of strings of length ‘2014’. So, a turing machine will take the input string of length ‘2014’ and test it.
If, input string is present in the language then turing machine will halt in final state . But, if turing machine is unable to accept the input string then it will halt in non-final state or go in an infinite loop and never halt.

Thus, ‘L’ is undecidable and recursively enumerable .

Please comment below if you find anything wrong in the above post.

Answer

Explanation:

Undecidability of L:
- The language L is undecidable because it is not possible to determine whether a given Turing machine M accepts a string of length 2014. This is a variation of the Halting Problem, which is known to be undecidable.
- To prove this, we can reduce the Halting Problem to L. Given an input (M, w) for the Halting Problem, we can construct a Turing machine M' that simulates M on input w and then accepts if M halts within 2014 steps. Thus, if we could decide L, we could also decide the Halting Problem, which is a contradiction.

Recursively Enumerable nature of L:
- Despite being undecidable, L is recursively enumerable because we can construct a Turing machine that enumerates all possible Turing machines and checks if each one accepts a string of length 2014. If it does, the machine can output that Turing machine. This way, we can list all Turing machines in L, making it recursively enumerable.
Therefore, the correct answer is option 'b' - L is undecidable but recursively enumerable.

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