self doubt

Question

L_B = { < M > | M is a valid Turing Machine } is decidable or undecidable.

Comments

A Turing machine that halts-->means given an input it will either say yes or no-->therefore we can say it is decidable for a given input to say yes or no.

A Turing machine that halts means there exists an algorithm --->and as known algorithm is step by step guide for solving the problem,so we can definitely say that this problem can be solved in these many steps.

But then there are other Turing machine that doesn't halt/stop for given input-->if they accept the input they will definitely say yes but if don't accept it then the Turing machine may say no or can go into infinite loop-->here we cant say anything as ambiguity is there.

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I did 'nt get it . Can you please explain in simple terms . How you get to know this Turing machine will halt ?

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those who are saying it is not decidable or it will loop, can you tell why do you think that L is undecidable??

Answer 1

As we can encode $every \,\, Turing\,\, machine$ using $0's$ and $1's$. So weather

 $L_{B} = \left \{ < M > | M\, is\, a\, valid \,Turing\, Machine \right \}$

is $decidable$ or not will depend upon the type of encoding we will use for a Turing machine.

Answer 2

It is trivial so it is Decidable we cannot use Rice theorem here.

Comments

"It is trivial" so it is decidable...

Means??

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Any “trivial” property is always decidable – because the decision is known ahead and that is why the property is trivial.

and trivial means if it contains all the TM or it is Empty. here if M is valid TM will contains all the Turing machines so it is trivial