The Halting Problem is a famous problem in computation theory, which states that we cannot build a Turing Machine that can predict whether a Turing Machine will accept on an input.

Suppose we have a machine called Oracle. The machine takes as input a machine M and a word and can predict whether M will accept or not the given word. This is how Oracle works:

• accepts, if M accepts the word
• rejects, if M doesn’t accept the word

Let’s say we have another machine, D, which takes as input a machine M and utilizes the Oracle to predict whether M will accept given its own encoding (usually denoted by <M>) as input. There is a little twist though. D outputs the opposite of what the Oracle predicts! This is how it works:

• rejects, if Oracle accepts on M given word <M>
• accepts, if Oracle rejects on M given <M>

So, if we run D on a machine M, it outputs the opposite of what the machine does.

What if we input D to itself? If the Oracle predicts D will accept <D>, D will output the opposite, which means it will reject. If the Oracle predicts D will not accept its encoding, D will accept.

You can see that there is a paradox, because D cannot reject and accept at the same time. That means no such machines as D and the Oracle can exist, therefore we cannot build a machine that predicts how a machine will behave at all times.

#### Diagonalization

An interesting way to map this paradoxical behavior out is by using Cantor’s Diagonalization Technique. A node at index (i, j) in the table denotes whether Mi accepts or not Mj. Therefore, the nodes at (i, i) denote whether the machine will accept its own encoding (essentially Oracle’s work).

Note that since D is also a Turing Machine, it will also appear in the table at some point. What will that row look like? Since D outputs the opposite of what Oracle outputs, it means the row will hold values opposite to what the (i, i) nodes hold. For example, if the node (e, e) holds the value accept, the e-th node in D’s row will hold reject.

If D’s row is at index d, what then happens at node (d, d)? Will we see accept or reject? The answer is, we cannot know. The node doesn’t have a value, so we cannot write down its opposite. It’s a paradox, and thus the machine cannot exist.