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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD simulated by HHH cannot possibly halt (Halting Problem)
Date: Wed, 9 Apr 2025 16:56:30 -0400
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On 4/9/2025 4:35 PM, olcott wrote:
> 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:
>>>>>>>
>>>>>>> typedef void (*ptr)();
>>>>>>> int HHH(ptr P);
>>>>>>>
>>>>>>> int DD()
>>>>>>> {
>>>>>>>    int Halt_Status = HHH(DD);
>>>>>>>    if (Halt_Status)
>>>>>>>      HERE: goto HERE;
>>>>>>>    return Halt_Status;
>>>>>>> }
>>>>>>>
>>>>>>> int main()
>>>>>>> {
>>>>>>>    HHH(DD);
>>>>>>> }
>>>>>>>
>>>>>>> *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;
>>>>> }
>>>>>
>>>>> This stuff is simply over-your-head.
>>>>> HHH(DD) meets the above: *Simulating termination analyzer Principle*
>>>>> Anyone with sufficient competence with the C programming language
>>>>> will understand this.
>>>>>
>>>> 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.
> 
> Simply over your level of technical competence.
> 
>> 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.

Which means HHH has conflicting requirements, because to perform a 
correct simulation of its input it cannot halt itself, and therefore 
can't report that.