Re: How do computations actually work?

On 2025-05-28 14:31:01 +0000, olcott said:

On 5/28/2025 2:36 AM, Mikko wrote:

On 2025-05-27 15:40:33 +0000, olcott said:
 

On 5/27/2025 3:29 AM, Mikko wrote:

On 2025-05-26 16:40:25 +0000, olcott said:
 

On 5/25/2025 10:46 AM, Fred. Zwarts wrote:

Op 25.mei.2025 om 16:50 schreef olcott:

On 5/25/2025 4:09 AM, Mikko wrote:

On 2025-05-24 15:25:21 +0000, olcott said:
 

On 5/24/2025 2:54 AM, Mikko wrote:

On 2025-05-23 16:04:49 +0000, olcott said:
 

On 5/23/2025 2:09 AM, Mikko wrote:

On 2025-05-23 02:47:40 +0000, olcott said:
 

On 5/22/2025 8:24 PM, Mike Terry wrote:

On 22/05/2025 06:41, Richard Heathfield wrote:

On 22/05/2025 06:23, Keith Thompson wrote:

Richard Heathfield <rjh@cpax.org.uk> writes:

On 22/05/2025 00:14, olcott wrote:

On 5/21/2025 6:11 PM, Richard Heathfield wrote:

[...]

Turing proved that what you're asking is impossible.
 

That is not what he proved.

 Then you'll be able to write a universal termination analyser that can
correctly report for any program and any input whether it halts. Good
luck with that.

 Not necessarily.

 Of course not. But I'm just reflecting. He seemed to think that my inability to write the kind of program Turing envisaged (an inability that I readily concede) is evidence for his argument. Well, what's sauce for the goose is sauce for the gander.
 

Even if olcott had refuted the proofs of the
insolvability of the Halting Problem -- or even if he had proved
that a universal halt decider is possible

 And we both know what we both think of that idea.
 

-- that doesn't imply
that he or anyone else would be able to write one.

 Indeed.
 

I've never been entirely clear on what olcott is claiming.

 Nor I. Mike Terry seems to have a pretty good handle on it, but no matter how clearly he explains it to me my eyes glaze over and I start to snore.

 Hey, it's the way I tell 'em!
 Here's what the tabloids might have said about it, if it had made the front pages when the story broke:
   COMPUTER BOFFIN IS TURING IN HIS GRAVE!
   An Internet crank claims to have refuted Linz HP proof by creating a
  Halt Decider that CORRECTLY decides its own "impossible input"!
  The computing world is underwhelmed.
 Better?  (Appologies for the headline, it's the best I could come up with.)
 Mike.
 

 There is a key detail about ALL of these proofs
that no one has paid attention to for 90 years.
 It is impossible to define *AN INPUT* to HHH that
does the opposite of whatever value that HHH returns.

 That is a key detail about HHH. Your HHH is not a part of those proofs.

 All of the proofs work this same way.

 No, they don't. Some proofs derive the same conclusion with an essentially
different approach.
 However, in spite of the differences, they do share a common fieature:
your HHH is not a part of any of the proofs.

 All of the conventional proofs of the HP assume that
there is an *input D* that can actually do the opposite
of whatever value that HHH returns.

 Depends on what you mean by "conventional". If you merely mean proofs
that apply ordinary logic then there are proofs with a different
strategy. If you mean only proofs that use the same strategy that
Turing used then you are closer to the truth. But there is no assumption
about the exstence of such D. Its existence is proven.
 

 In seems that way until you pay much closer attention.
 int main()
{
   DDD(); // The HHH that DDD calls cannot report on the
}        // behavior of its caller because it cannot see
          // is caller.
 Even if HHH could see and report on the behavior of
its caller because its caller is not its input this
too is no good.

 It seems that way to you, until you pay somewhat closer attention.

 The HHH(DDD) must report on the behavior that its actual input
actually specified CANNOT BE VIOLATED.

 Of course it can. In fact HHH does violate that. DDD specifies a halting
behaviour but HHH reports that DDD specifies a non-halting behaviour.
That is a violation of that rquirement.

 If DDD simulated by HHH stops running for any
reason besides reaching its own "ret" instruction
final halt state THEN DDD HAS NOT HALTED.

 Irrelevant. The requirement is that a halt decider predicts whether the
complete execution of the computation described by the input will halt.

 Halting is defined as reaching a final state and
terminating normally.

The usual meaning of "final state" is a state where the computation cannot
be continued. The usual meaning of halting is reaching a state where the
computation cannot be computed. By the usual meaning of halting it does
not matter whether the termination is "normal" and it is usually not even
defined what would constiture a "normal" termination.
If different meanngs are intended for some terms or different terms for
some meanings then those terms must be defined in the opus where they are
used. Otherwise readers apply the usual meanings and the intent of the
author will be misunderstood.
It is also possible to define variants of the halting problem where the
question is about halting in a particular way, for example whether the
computation accepts or the computation rejects. The question whether the
computation halts normally, if the meaning of "normally" is defined, is
of the same type. Another kind of variation is the question whether the
computation reaches a particular kind of non-termination state at some
point.
Typical variants of the halting problem are as hard but not harder than
the standard halting problem. They are not Turing-decidable but are
decidable with an oracle machine that uses an oracle for the standard
halting problem.
--
Mikko