Re: Any honest person that knows the x86 language can see... predict correctly

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Sujet : Re: Any honest person that knows the x86 language can see... predict correctly
De : wyniijj5 (at) *nospam* gmail.com (wij)
Groupes : comp.theory
Date : 01. Aug 2024, 05:08:35
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <ddb7a467da20b6a6bd90aee9735a62ae68cac50e.camel@gmail.com>
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On Tue, 2024-07-30 at 18:50 -0500, olcott wrote:
On 7/30/2024 6:45 PM, Mad Hamish wrote:
On Tue, 30 Jul 2024 16:13:55 -0500, olcott <polcott333@gmail.com>
wrote:
 
On 7/30/2024 4:07 PM, joes wrote:
Am Tue, 30 Jul 2024 15:05:54 -0500 schrieb olcott:
On 7/30/2024 1:48 PM, Fred. Zwarts wrote:
Op 30.jul.2024 om 17:14 schreef olcott:
On 7/30/2024 9:51 AM, Fred. Zwarts wrote:
Op 30.jul.2024 om 16:21 schreef olcott:
On 7/30/2024 1:52 AM, Mikko wrote:
On 2024-07-29 14:07:53 +0000, olcott said:
 
HHH(Infinite_Recursion) and HHH(DDD) show the same non-halting
behavior pattern in their derived execution traces of their
inputs.
Hard to believe as their behaviour is so different and you don't
say what pattern the see.
 
*Its all in the part that you erased*
 
We all see the differences between these two.
 
They both correctly predict behavior that must be aborted to prevent
the infinite execution of the simulating halt decider.
 
Except that the prediction for the second one is wrong. The simulation
of an aborting and halting function, like HHH, does not need to be
aborted.
I proved otherwise. When the abort code is commented out then it keeps
repeating again and again, thus conclusively proving that is must be
aborted or HHH never halts.
But the abort is not commented out in the running code!
 
 
I modified the original code by commenting out
the abort and it does endlessly repeat just like
HHH correctly predicted.
 
Which means that it works in some cases, not all cases
Which means it doesn't work universally
 
So it's not a general solution to the halting problem
 
It is not supposed to be a general solution to the halting problem.
it only shows how the "impossible" input is correctly determined
to be non halting.
 

But how do you determine it is non-halting?

As I know you are even unable to define what 'halt' mean !!!

int main()
{
  H(D,D);
}

Above is the last 'explain' of POOH. It does not reply anything!!! How is it a decider?
The HP asks a TM to reply the y/n answer, not from your mouth.



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