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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Any honest person that knows the x86 language can see... predict
 correctly
Date: Thu, 1 Aug 2024 08:04:23 -0500
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On 8/1/2024 7:56 AM, Fred. Zwarts wrote:
> Op 01.aug.2024 om 13:51 schreef olcott:
>> On 8/1/2024 2:46 AM, Fred. Zwarts wrote:
>>> Op 01.aug.2024 om 05:51 schreef olcott:
>>>> On 7/31/2024 10:08 PM, wij wrote:
>>>>> On Tue, 2024-07-30 at 18:50 -0500, olcott wrote:
>>>>>>
>>>>>> 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 !!!
>>>>>
>>>> I have done this thousands of times and after someone
>>>> has read these thousands of times they say that I never
>>>> said it once.
>>>>
>>>> void DDD()
>>>> {
>>>>    HHH(DDD);
>>>>    return;
>>>> }
>>>>
>>>> int main()
>>>> {
>>>>    HHH(DDD);
>>>> }
>>>>
>>>> If DDD correctly emulated by HHH cannot possibly
>>>> reach its return instruction then it never halts.
>>>>
>>>>
>>>
>>> But a correct simulation is impossible.
>>
>> When HHH does what-ever-the-hell the x86 semantics specifies
>> then HHH is correct.
>>
> 
> But since HHH deviates from the semantics of the x86 language (by 
> skipping instructions of a halting  program) it is incorrect.
> 

Only a freaking moron would believe that a non
terminating input should be simulated forever.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer