Let a decision problem $X$ be defined as follows:

$X$: Given a Turing machine $M$ over $\Sigma$ and any word $w \in \Sigma$, does $M$ loop forever on $w$?

You may assume that the halting problem of Turing machine is undecidable but partially decidable.

- Show that $X$ is undecidable
- Show that $X$ is not even partially decidable

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Consider a decision problem. A decision can be either "yes" or "no". So. we can divide the problem cases to 2 parts- one where the answer is "yes", and one where the answer is "no".

Now, if the "yes" instance of the problem is given, suppose we (our TM) can always say "yes", then the problem is partially decidable (its language is recursively enumerable).

If we can also say "no" for all "no" instances of the problem, then the problem is decidable (its language is recursive).

NB: TM for a partially decidable but not decidable problem, will go to an infinite loop for **SOME** "no" instances. For some "no" instances it may say "no". (Obviously, it won't say "yes" for any no" instance)

Given a Turing machine MM over ΣΣ and any word w∈Σw∈Σ, does MM loop forever on w

If you answer "yes" or "no" for looping question indirectly you are answering looping problem of TM which is undecidable.

Summarizing it: Decidable - we can say yes or no for both instance- recursive

partially decidable - we can say yes only for "yes" instance. REL.

Totally undecidable : cannot say yes for **every** yes case. Not even REL.

Partially decidable is same as semi-decidable language which means that there exists a Turing Machine for such languages(given an input, the machine can accept/reject/loop) but we are not sure whether there is any Total Turing Machine(given an input, the machine can accept or reject but does not loop) or not.

Ref: https://www.youtube.com/watch?v=v2VuM0AIZVU&t=13s **<2:47 to 4:30>**

Here the question is whether M loops on a given string 'w' which is same as saying whether M does not halt on w. As Arjun Sir said, this is just the complement of Halting problem.

Halting problem is an undecidable problem. Given an input x to a machine M, there is no way to decide whether M will halt(accept or reject) or not(loop). The language L={M#x | M halts on x} is **R.E but not Recursive. **

So** complement of Halting problem** would be **complement of language which is (R.E. but not Recursive)** which is **Non-R.E.**

Why is it so? Ref: https://www.youtube.com/watch?v=VqDMudEzZCo **<0:00-7:00>**

Non R.E. means unrecognizable meaning there does not exist any Turing machine also for such languages hence they are not even partially decidable.