Toc-Recursively Ennumerable

Question

A is a decision problem. If A is recursively ennumerable then there exists an algorithm which halts on every string in A.

True or false.Cite with reasons.

Comments

Yes surely answer is false.

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For RE problem A, we can say YES to one half of the problem. By one half means : we can say yes if answer is yes. On the other hand we might end up in a inf loop if answer is NO. Example : Assume we have TM_A for A , if w is a string which is a member of A, then if we give TM_A and w to a universal Turing machine TM_U , and give rum command. Since we have given a member, TM_U will stop and say YES. For non-member string TM_U may run into a inf loop. 

What I can not figure out here is that, If $\sum$  is the alphabet associated with A, then what is

EVERY STRING IN A ? is it every string which belongs to A or simply  $\sum$^* ?

If it means $\sum$^*  then answer would be surely False.

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@Debashish, I edit the answer with an algorithm (found in Peter Linz) that will enumerate every string that is present in A.  This algorithm is really beautiful and simple. It passed the hurdle for those strings that will fall into an infinite loop.

Answer 1

True, There exists an algorithm which will find every string in A.
A is recursively enumerable. Means there exists a turning machine for A which will accept every string inside A and produce output as 'Yes'.
But the problem arises when the strings are not in A. For such string (that are not in A) that Turing machine either reject these string and produce output as 'No' or fall into an infinite loop. 

The algorithm is given in Peter Linz. Actually, this algorithm enumerates all the strings that are present in A. Hence it is showing that recursively enumerable language is countable. First, it is finding the algorithm for recursive language. Means if A is a recursive language, then it is enumerating every string that is present in A. Hence showing that Recursive language is countable. 

Answer 2

Yes! A Non Deterministic Turing Machine will always halt on the Recursively Enumerable. But please note that there are some languages that are out side the Type-0 grammars in chomsky hierarchy for which we can't construct a Turing Machine! These are generally NP-Hard problems.

Actually we can't prove the Non Deterministic Turing Machine!

Like Alan Turing said "We can't construct a Universal Turing Machine"

So I think practically it is not possible but theoretically it may possible! That is what the definition of RE language "A language for which a turing machine may or may not halt"!

Please tell me if any thing is wrong!