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From: olcott
Newsgroups: comp.theory
Subject: Re: DDD simulated by HHH cannot possibly halt (Halting Problem)
Date: Sun, 13 Apr 2025 14:52:10 -0500
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On 4/13/2025 4:18 AM, joes wrote:
> Am Wed, 09 Apr 2025 15:35:30 -0500 schrieb olcott:
>> On 4/9/2025 1:58 PM, Fred. Zwarts wrote:
>>> Op 09.apr.2025 om 19:29 schreef olcott:
>>>> On 4/8/2025 10:31 AM, Fred. Zwarts wrote:
>>>>> Op 08.apr.2025 om 17:13 schreef olcott:
>>>>>> On 4/8/2025 2:45 AM, Fred. Zwarts wrote:
>>>>>>> Op 08.apr.2025 om 06:33 schreef olcott:
>
>>>>>>>> *Simulating termination analyzer Principle*
>>>>>>>> It is always correct for any simulating termination analyzer to
>>>>>>>> stop simulating and reject any input that would otherwise prevent
>>>>>>>> its own termination.
>>>>>>>
>>>>>>> In this case there is nothing to prevent, because the finite string
>>>>>>> specifies a program that halts.
>>>>>>
>>>>>> int DD()
>>>>>> {
>>>>>> int Halt_Status = HHH(DD);
>>>>>> if (Halt_Status)
>>>>>> HERE: goto HERE;
>>>>>> return Halt_Status;
>>>>>> }
>>>>>> HHH(DD) meets the above: *Simulating termination analyzer Principle*
>>>>>>
>>>>> Everyone with a little bit of C knowledge understands that if HHH
>>>>> returns with a value 0, then DDD halts.
>>>>
>>>> DDD CORRECTLY SIMULATED BY HHH NOT ANY OTHER DAMN DDD IN THE UNIVERSE
>>>> NITWIT.
>>>>
>>> If HHH would correctly simulate DD (and the functions called by DD)
>>> then the simulated HHH would return to DD and DD would halt.
>>> But HHH failed to complete the simulation of the halting program,
>>
>> HHH is only required to report on the behavior of its own correct
>> simulation (meaning the according to the semantics of the C programming
>> language) and would be incorrect to report on any other behavior.
> Such as what actually happens when you run it?
>
Why do you insist on dishonestly ignoring the pathological
self-reference difference?
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer