How does this proof, that the halting problem is u

问题:

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't know when it's looping (it can't accept or reject which is why a TM is Turing Recognizable for all strings), then how would could the decider H decide if M could possibly be in a loop? The same problem will carry through when TM D does its processing.

回答1:

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.

回答2:

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.