How does this proof, that the halting problem is undecidable, work?

Question

I\'m going over the proof for The Halting Problem in Intro to the Theory of Computation by Sipser and my main concern is about the proof below:

If TM M doesn\'

Answer 1

After reading this and trying to visualize the proof I came up with this code which is a simplified version of the code in this answer to a related question:

function halts(func) {
  // Insert code here that returns "true" if "func" halts and "false" otherwise.
}

function deceiver() {
  if(halts(deceiver))
    while(true) { }
}

If halts(deceiver) returns true, deceiver will run forever, and if it returns false, deceiver will halt, which contradicts the definition of halts. Hence, the function halts is impossible.

Answer 2

This is a "proof by contradiction", a reductio ad absurdum. (Latin phrases are always good in theory classes... as long as they make sense, of course.)

This program H is just a program with two inputs: a string representing a program for some machine, and an input. For purposes of the proof, you simply assume the program H is correct: it simply will halt and accept if M accepts with w. You don't need to think about how it would do that; in fact, we're about to prove it can't, that no such program H can exist, ...

BECAUSE

if such a program existed, we could immediately construct another program H' that H couldn't decide. But, by the assumption, there is no such program: H can decide everything. So, we're forced to conclude that no program defined as we defined H is possible.

By the way, the reductio method of proof is more controversial than you might expect, considering how often its used, especially in Computer Science. You shouldn't be embarrassed to find it a little odd. The magic term is "non-constructive" and if you feel really ambitious, ask one of your professors about Errett Bishop's critique of non-constructive mathematics.