Question

Q1. If you cannot encode the input in a _________ language, a TM/program is not guaranteed to halt.

     

Q2. The Universal TM takes ______ and ___________ as its input.

Q3:

       Is this question Decidable or Undecidable?

               "Does a given Turing Machine M answer yes to a given problem?" __________

Answer 1

1)Turing Recognizable language

2)Takes as input a description of a Turing Machine, and the input string for the Turing Machine .And Simulates running the machine on the input string.

The universal language U over the alphabet { 0, 1 } is

U = { 〈M, w〉 | w ∈ L(M) }.

• The language U contains information on all Turing-recognizable languager over { 0, 1 }:

• Let A ⊆ { 0, 1 }* be some Turing-recognizable language and M a standard TM recognizing A. Then

A = { w ∈ { 0, 1 }* | 〈M, w〉 ∈ U }.

• Also U is Turing-recognizable.

• Turing machines recognizing U are called universal Turing machines.

3) Undecidable

Proof − At first, we will assume that such a Turing machine exists to solve this problem and then we will show it is contradicting itself. We will call this Turing machine as a Halting machine that produces a ‘yes’ or ‘no’ in a finite amount of time. If the halting machine finishes in a finite amount of time, the output comes as ‘yes’, otherwise as ‘no’. The following is the block diagram of a Halting machine −

Yes (HM halts on input w) Input string Halting Machine No (HM does not halt on input w)

Now we will design an inverted halting machine (HM)’ as −

• If H returns YES, then loop forever.

• If H returns NO, then halt.

The following is the block diagram of an ‘Inverted halting machine’ −

Infinite loop Yes Halting Input string Machine No

Further, a machine (HM)2 which input itself is constructed as follows −

• If (HM)2 halts on input, loop forever.
• Else, halt.

Here, we have got a contradiction. Hence, the halting problem is undecidable.