3,438 views

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.

1. Show that $X$ is undecidable
2. Show that $X$ is not even partially decidable

### 1 comment

sir halting problem is undecidable but semi deciadable. i.e recursive enumerable. and the given problem is complement of halting problem. and we know recursive enumerable is not closed under complementation. then how can we say that the answer is not even semi decidablde

The question asks if $M$ loop forever on $w$. If $M$ loop forever on $w, M$ wouldn't halt on $w$. And if $M$ doesn't halt on $w, M$ should loop forever. So, this problem is exactly same as asking if "$M$ doesn't halt on $w$", which is the complement of halting problem and is not even partially decidable. So, $X$ is not even partially decidable.
by

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.

@arjun sir , 3 lectures on decidability and ur explanations for all  decidability questions really helped me in understanding these topic in which i was facing problem from long time thnxx for making these easy ur explanation are really awsm thankyou soo much sir :)
Undecidable means a language can be Recursively-Enumerable or may be not Recursively-Enumerable.

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.

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.

1
2
3,980 views
3
13,705 views