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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: HHH(DDD) computes the mapping from its input to HHH emulating
 itself emulating DDD --- anyone that says otherwise is a liar
Date: Fri, 22 Nov 2024 08:50:33 -0600
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On 11/22/2024 6:20 AM, joes wrote:
> Am Thu, 21 Nov 2024 15:19:43 -0600 schrieb olcott:
>> On 11/21/2024 3:11 PM, joes wrote:
>>> Am Thu, 21 Nov 2024 09:19:03 -0600 schrieb olcott:
>>>> On 11/20/2024 10:00 PM, Richard Damon wrote:
>>>>> On 11/20/24 9:57 PM, olcott wrote:
>>>>>> On 11/20/2024 5:51 PM, Richard Damon wrote:
>>>>>>> On 11/20/24 5:03 PM, olcott wrote:
>>>>>>>> On 11/20/2024 3:53 AM, Mikko wrote:
>>>>>>>>> On 2024-11-20 03:23:12 +0000, olcott said:
>>>>>>>>>> On 11/19/2024 4:12 AM, Mikko wrote:
>>>>>>>>>>> On 2024-11-18 20:42:02 +0000, olcott said:
>>>>>>>>>>>> On 11/18/2024 3:41 AM, Mikko wrote:
>>>>>>>>>>>>> The "the mapping" on the subject line is not correct. The
>>>>>>>>>>>>> subject line does not specify which mapping and there is no
>>>>>>>>>>>>> larger context that could specify that. Therefore it should
>>>>>>>>>>>>> be "a mapping".
>>>>>>>>>>>>> On 2024-11-17 18:36:17 +0000, olcott said:
>>>
>>>>>> My code is one example of the infinite set of every possible HHH
>>>>>> that emulates DDD according to the semantics of the x86 language.
>>> Like all of them, it is unable to simulate DDD to its undeniable
>>> halting state.
>> In your case you may simply not even understand what infinite recursion
>> is, thus cannot see the isomorphism.
> 
> 
>>>>> But it gets the wrong answer for the halting problem, as DDD dpes
>>>>> halt.
>>>> DDD emulated by HHH does not halt.
>>> Whatever. DDD halts and HHH should return that.
>> IT IS NOT THE SAME INSTANCE OF DDD.
> All instances of DDD behave the same (if it is a pure function and
> the HHH called from it doesn't switch behaviour by a static variable).
> 

Only HHH is required to be a pure function, DDD is expressly
allowed to be any damn thing. The actual behavior of DDD
emulated by HHH is different than the actual behavior of DDD
emulated by HHH1.

*ONLY BECAUSE DDD CALLS HHH AND DDD DOES NOT CALL HHH1*
Even Mike does not seem to be able to understand this.

Note we have been on this one point about the behavior of
DDD emulated by HHH for many months and have not yet even
begun to talk about how HHH would report this behavior.

The question does DDD halt?
Is answered by Can DDD emulated by any HHH reach its own
"return" instruction final state?

The question: How could HHH determine whether or not DDD
emulated by HHH can possibly reach its own final state?
is another entirely different question.


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