Input − A Turing machine and an input string w.

Problem − Does the Turing machine finish computing of the string w in a finite number of steps? The answer must be either yes or no.

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 −

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’ −

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.