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$L = \{ \langle M \rangle \mid \text{there is at least one input string on which the \(M\) does not halt} \}$

Here, for a Turing machine $M$, the notation $\langle M \rangle$ denotes an encoding, over some alphabet, of the code of the Turing machine. To which of the following language classes does $L$ belong?

  1. Regular.
  2. Context-free but not Regular.
  3. Recursive but not Context-free.
  4. Recursively enumerable but not recursive.
  5. Not recursively enumerable.
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  • $\begingroup$ since it is given that M doesn't halt(i'm stuck here) if it is just that M "halts on some input x",then it is easy to say that it is regular,non-recusive..but it is given that M doesn't halt. $\endgroup$ – Harshil Sep 26 '13 at 2:22
  • $\begingroup$ You're missing the point. The language is not the one accepted by $M$, but rather the one accepted by $L$. The question is, how hard is it to tell whether a given Turing machine halts on all inputs? $\endgroup$ – Yuval Filmus Sep 26 '13 at 4:52
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The language $L$ consists of (descriptions of) Turing machines $M$ such that $M$ does not halt on all inputs. In other words, $\langle M \rangle \notin L$ if $M$ halts on all inputs. Can you think of any connection to the halting problem?

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  • $\begingroup$ i don't think any connection to halting problem $\endgroup$ – Harshil Sep 26 '13 at 2:58
  • $\begingroup$ can u please elabroate ur answer i didn't get it!! $\endgroup$ – Harshil Sep 26 '13 at 2:59
  • $\begingroup$ This is not an answer, but rather a hint. You will have to take it from here on your own. $\endgroup$ – Yuval Filmus Sep 26 '13 at 4:01
  • $\begingroup$ for ur inteligency..._/\ _.... $\endgroup$ – Harshil Sep 26 '13 at 4:41
  • $\begingroup$ @user561298 The language is defined in terms of whether Turing machines halt or not. Are you sure there's no connection with the halting problem? $\endgroup$ – David Richerby Sep 26 '13 at 12:47

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