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From: olcott <abc@def.com>
Newsgroups: comp.theory
Subject: HHH computes the mapping from its input finite sting to the actual
 behavior specified by this finite string
Date: Wed, 7 Aug 2024 08:54:41 -0500
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On 8/7/2024 2:29 AM, Mikko wrote:
> On 2024-08-05 13:49:44 +0000, olcott said:
> 
>> On 8/5/2024 2:39 AM, Mikko wrote:
>>> On 2024-08-04 18:59:03 +0000, olcott said:
>>>
>>>> On 8/4/2024 1:51 PM, Richard Damon wrote:
>>>>> On 8/4/24 9:53 AM, olcott wrote:
>>>>>> On 8/4/2024 1:22 AM, Fred. Zwarts wrote:
>>>>>>> Op 03.aug.2024 om 18:35 schreef olcott:
>>>>>>  >>>> ∞ instructions of DDD correctly emulated by HHH[∞] never
>>>>>>>> reach their own "return" instruction final state.
>>>>>>>>
>>>>>>>> So you are saying that the infinite one does?
>>>>>>>>
>>>>>>>
>>>>>>> Dreaming again of HHH that does not abort? Dreams are no 
>>>>>>> substitute for facts.
>>>>>>> The HHH that aborts and halts, halts. A tautology.
>>>>>>
>>>>>> void DDD()
>>>>>> {
>>>>>>    HHH(DDD);
>>>>>>    return;
>>>>>> }
>>>>>>
>>>>>> That is the right answer to the wrong question.
>>>>>> I am asking whether or not DDD emulated by HHH
>>>>>> reaches its "return" instruction.
>>>>>
>>>>> But the "DDD emulated by HHH" is the program DDD above,
>>>>
>>>> When I say DDD emulated by HHH I mean at any level of
>>>> emulation and not and direct execution.
>>>
>>> If you mean anything other than what the words mean you wihout
>>> a definition in the beginning of the same message then it is
>>> not reasonable to expect anyone to understand what you mean.
>>> Instead people may think that you mean what you say or that
>>> you don't know what you are saying.
>>
>> If you don't understand what the word "emulate" means look it up.
> 
> I know what it means. But the inflected form "emulated" does not
> mean what you apparently think it means. You seem to think that
> "DDD emulated by HHH" means whatever HHH thinks DDD means but it
> does not. DDD means what it means whether HHH emulates it or not.
> 

In other words when DDD is defined to have a pathological
relationship to HHH we can just close our eyes and ignore
it and pretend that it doesn't exist?

DDD does specify non-halting behavior to HHH and HHH must
report on this non-halting behavior that DDD specifies.

All halt deciders compute the mapping from their input
finite string to the behavior that this finite string
specifies.

No halt decider is ever allowed to report on the behavior
of any computation that itself is contained within unless
this is the same behavior that its finite string input
specifies.

void DDD()
{
   HHH(DDD);
   return;
}

Any expert in the C language that knows what x86 emulators
are knows that DDD correctly emulated by HHH specifies what
is essentially equivalent to infinite recursion that cannot
possibly reach its "return" instruction halt state.

It seems that no one here has that degree of expertise.
That they know that they don't understand these things
and still say that I am wrong is dishonest.

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