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Yesterday I was reading about Computability and they mention the Halting Problem. It got stuck in mind all day until I remember that some weeks ago, when learning Java, the IDE (Netbeans) show me a warning saying that my code was going to result in an infinite loop. Code Warning

My question is, why can't this problem, be solved by using an "Intelligent" program to determine if another program will halt or end in an infinite loop? This hypothetical program could look for repetitive structures, conditions, etc and even predict under which circumstances (possible inputs) will halt, loop infinitely or end. Maybe like looking for the "acceptable range", like the domain of a mathematical function.

Raphael
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nmomn
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  • Why is this not answered by the fact that the halting problem is incomputable, which you seem to be aware of (see also our reference question)? Note that "intelligent" has no real meaning; we build algorithms, period. If they are randomized, they do not contradict the halting problem (because its model does not allow for randomness) but also can't solve it, always. – Raphael Mar 10 '15 at 14:52
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    See also here, here, here and, most importantly, here. Oh, and here -- yes, the Halting problem is decidable on some classes of programs, which is why IDEs can detect some forms of looping. (Note how they never tell you "terminates always".) – Raphael Mar 10 '15 at 14:57
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    I know this problem is not computable. I am looking for someone to prove why my idea is wrong. I am a beginner trying to understand. Sorry if my question is insulting to you. – nmomn Mar 10 '15 at 14:58
  • Well, the question I link to provides the proof you are looking for. If you don't understand the proofs there, you will have to do some reading because it's unlikely you'll understand (and accept) any proof. Sorry. – Raphael Mar 10 '15 at 15:00
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    @Raphael: In OP's defense, he's not really asking for a computability proof, he's asking about the common logical error (made, for example, by the famous mathematical physicist Roger Penrose) that "intelligence" is a magic property stronger than an algorithm. (Penrose goes so far as to propose that consciousness must be a result of quantum gravitational effects in cellular microtubules. I kid you not.) – Wandering Logic Mar 10 '15 at 15:38
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    @NMO Your question certainly isn't insulting: don't worry. The point is that any program that uses artificial intelligence is still a program, so it still can't solve the halting problem. It's certainly possible to write a program that says either "This halts", "This doesn't halt" or "I don't know", but any such program will always say "I don't know" for infinitely many inputs. – David Richerby Mar 10 '15 at 15:59
  • @WanderingLogic but given really smart person with a really big memory and sufficient time, isn't he able to completely determine if a program is going to halt or loop infinitely? I mean by looking for repetitive structures, counters, flags, etc... If I am given a certain program and sufficient time I can decide if it is going to halt o loop. Could I make a program to check the same things I do? – nmomn Mar 10 '15 at 17:00
  • @WanderingLogic and I also notice some people rise the question of what happens if I pass as a parameter my own program. But what If I could make my program notice If it is itself and then halt. To make it self-concious that is analysing himself. – nmomn Mar 10 '15 at 17:02
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    @NMO: No, the halting program shows that there are programs that a really smart person with a really big memory and sufficient (but finite) time can't determine if it is going to halt or loop infinitely. There is no mathematical way to do it correctly in all cases, it has nothing to do with how smart you are. The way to see it is by working through the proof by yourself. – Wandering Logic Mar 10 '15 at 17:04
  • @WanderingLogic okay, thank you very much for your time. That means I can approach an answer in some cases through induction, which is better than none, but there is no axiom i could use to deduce a completely true answer? – nmomn Mar 10 '15 at 17:06
  • Exactly! You might find this interesting: http://cs.stackexchange.com/a/33910/7459. – Wandering Logic Mar 10 '15 at 17:34
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    @NMO Your program does not stand a chance with that. Of course, you can check "Is the input my program up to whitespace?" but that does not help you, ultimately: there are infinitely many programs that compute the same function as yours, and any serves to prove the contradiction. You can't hope to detect all of and only these (proof by Rice's theorem), so no. Not a chance. – Raphael Mar 11 '15 at 11:45
  • think this is excellent question; there is some research into this eg via Busy Beaver problem & other related areas, will post some answer/ links if its reopened. agreed other questions cited by R are relevant but dont think the current dup marked is actually a dup. – vzn Mar 12 '15 at 15:08

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The halting problem is not a statement about intelligence (human or artificial) it is a statement about the limits of mathematics. It is an historically important example of an undecidable problem.

An artificial (or human) intelligence can certainly look for and find many sorts of different infinite loops in real programs. And the halting problem doesn't say that such a thing is impossible or even very hard.

What the halting problem says is that given any program $I$ ("I" for "intelligent"), (say a really complicated AI program written in Netbeans) that looks for infinite loops we can produce a program that $I$ can't analyze. The way we do this is a clever technique called diagonalization, which is described in the question @Raphael linked: How to show that a function is not computable?, and also in Why, really, is the Halting Problem so important?

The halting problem is similar to the Liar's Paradox in that it demonstrates the limits of mathematical and logical definitions of "truth", and "provability". The Liar's Paradox is: suppose I say "this statement is a lie." Is that statement true or false? If it is true then it must be a lie, so must be false. If it is false, then it is not a lie, so must be true. Writing a complicated artificial intelligence (or relying on a really smart person) won't make the question about whether I lied or not any more meaningful.

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Wandering Logic
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  • Thank you very much for your answer! I am finally starting to understand this problem. When you said we can always make a program that "I" is not able to analyse, why is this? I mean, aren't finite repetition structures we can find in a program? Because the problem seems to arise here. – nmomn Mar 10 '15 at 15:21
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    The answers @Raphael linked will show you how to construct a program that will take $I$ and (essentially) run it on itself in such a way that it itself must enter an infinite loop. The proof sketch of the halting problem on Wikipedia is also very clearly written. – Wandering Logic Mar 10 '15 at 15:47