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Re:Weakening Cycles So That Turing Can Halt by cortesoft on Thursday May 14, 2020 @09:34PM (#23187862)
The best way the halting problem was explained to me was to think of it like this: imagine you write your perfect halting detection program. It reads in the contents of some program, and returns true if it halts or false if it doesn't.

Now, all I have to do is write a program that includes an embedded version of your program. My program reads in the contents of itself and passes it to the embedded program. Now, all my program has to do is take the output of the embedded halt checker, and do the opposite; if the halt checker says the program halts, just keep looping. If it says I don't halt, exit immediately.

Now, you might think this is cheating, because it is using your own program against you. However, this is at the core of the problem; it has to work for ALL programs, even ones written with knowledge of your program.

Re:Weakening Cycles So That Turing Can Halt by infogulch on Friday May 15, 2020 @08:47PM (#23199338)
But why do I have to accept your arbitrary program? Why couldn't I only consider programs that come with a proof that they terminate? I don't care how you made the proof, maybe you used some restrictive programming language that can only express concepts that terminate, or your program implements paxos and you prove it by some embedding into Coq. Some restrictions though: the proof has to be formulated in a way that's compact and takes linear-time to check [1], and the proof goes down to some formalization of first-order logic that we can agree on.

[1]: Yes, I think this is possible. [2]

[2]: https://github.com/digama0/mm0#introduction

Re:Weakening Cycles So That Turing Can Halt by infogulch on Monday May 18, 2020 @09:55PM (#23230556)
After a bit of searching, proof carrying code seems to be related to what I'm describing [3].

So I guess my point is: Does it really matter that you can come up with arbitrary programs that I will be unable to determine if they halt, if I can build up an entire OS platform of provably correct machine code with correctness proofs that I can scan out of a file to verify?

Ultimately, here's the point of re-framing it like this in the first place:

Now it's just a tooling problem.

I recognize that this is a very big 'just'.

[3] https://people.eecs.berkeley.edu/~necula/pcc.html

Re:Weakening Cycles So That Turing Can Halt by galaxyLogic on Friday May 15, 2020 @01:06AM (#23189065)
Doesn't this assume that your program can reliably access it's own source-code? Can it? It can of course read some file containing some source-code. But assume it accesses some file which contains the source-code of some other program, then wouldn't your logic still hold the same, since from the point of your program it is simply reading some file?
Re:Weakening Cycles So That Turing Can Halt by pjc50 on Friday May 15, 2020 @05:07AM (#23190262)
Doesn't matter. Doesn't have to be the executing source, can work with compiled representations; the original Turing paper "compiles" every program to a single giant integer.
Re:Weakening Cycles So That Turing Can Halt by kazinator on Friday May 15, 2020 @11:37AM (#23193407)
No, it has nothing to do with source code.

The halting verifier program V never accesses itself. It takes two parameters: a program to check for halting (P) and an input to the program (I). It's job is to test whether P(I) (the program applied to the input) will halt.

To trick the halting verifier V we construct a diabolic P like this:

  program P(I) {
    if (V(I, I) == YES) {    // if Verifier says I(I) halts
      inifinite_loop();      // ... then do *not* halt
    } else {
      halt();                // ... otherwise halt
    }
  }
Like V, P is a special program that takes a program as input. It then asks the verifier: if the input program is given itself as input, does it halt? And then it behaves in the opposite manner: if the verifier says that I(I) halts, then P deliberately infinite-loops. Otherwise P halts.

Given that P, we ask the verifier the devastating question: V(P, P).

Does P halt if given itself as input; does P(P) halt?

V cannot decide. Note that if we consider P(P), then it means that in the above program, the input parameter I becomes P itself: I = P. And so the V(I, I) == YES expression is actually testing V(P, P) == YES.

But V(P, P) is the devastating question itself!

If V(P, P) yields YES (P(P) halts) P goes into an infinite loop (P(P) in fact fails to halt), and so V has calculated the wrong answer: the actual behavior of P(P) contradicts the YES answer from V(P, P), rendering V incorrect. The same problem occurs if V(P, P) yields NO.

Note that P contains no self refererence. The fact that P is applied to itself as P(P) doesn't come from P; it comes from the way P is used. P is used that way when we ask the verifier V(P, P). The verifier itself is checking what happens if P is given itself as input. P can happily accept other programs as the input I, programs which are not itself.

What P needs is a way to embed V, because it relies on it. The assumption is that whatever halting verifier V someone tries to develop, it has to be public. So that is to say, some CS researcher claims he or she has developed a universal halting verifier algorithm V and they publish it. Then, promptly, their adversary takes the algorithm V and sticks it into their program P(I) as a subroutine, and has an instant test case which defeats V.

The tricky thing is not self-reference, but the fact that P contains the verifier and turns it against itself. The self-reference is instigated from the outside by choice of inputs. In other words, the devastating question V(P, P) is what perpetrates the self-reference which breaks V.

Re:Weakening Cycles So That Turing Can Halt by cortesoft on Friday May 15, 2020 @11:52AM (#23193615)
Even if the code isn't public... it still exists. And since we are talking about "does their exist any possible program it can't determine", we know that some sequence of bytes will be that program, and therefore it is possible to exist.
Re:Weakening Cycles So That Turing Can Halt by cortesoft on Friday May 15, 2020 @03:50AM (#23189900)
Even if it can't access its source code, you can... you know what the code is