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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic,comp.ai.philosophy
Subject: Re: HHH(DDD) is correct to reject its input as non-halting ---
 EVIDENCE THAT I AM CORRECT
Date: Tue, 24 Jun 2025 09:06:12 -0500
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On 6/24/2025 2:54 AM, Fred. Zwarts wrote:
> Op 23.jun.2025 om 16:50 schreef olcott:
>> On 6/23/2025 3:24 AM, Fred. Zwarts wrote:
>>> Op 22.jun.2025 om 21:27 schreef olcott:
>>>> On 6/22/2025 11:01 AM, Fred. Zwarts wrote:
>>>>> Op 20.jun.2025 om 16:53 schreef olcott:
>>>>>> On 6/20/2025 4:42 AM, Fred. Zwarts wrote:
>>>>>>> Op 19.jun.2025 om 17:23 schreef olcott:
>>>>>>>> On 6/19/2025 3:55 AM, Fred. Zwarts wrote:
>>>>>>>>> Op 18.jun.2025 om 17:41 schreef olcott:
>>>>>>>>>> On 6/18/2025 4:36 AM, Fred. Zwarts wrote:
>>>>>>>>>>> Op 17.jun.2025 om 16:36 schreef olcott:
>>>>>>>>>>>> On 6/17/2025 4:28 AM, Fred. Zwarts wrote:
>>>>>>>>>>>>> Op 17.jun.2025 om 00:26 schreef olcott:
>>>>>>>>>>>>>> On 6/16/2025 3:53 AM, Fred. Zwarts wrote:
>>>>>>>>>>>>>>> Op 15.jun.2025 om 22:10 schreef olcott:
>>>>>>>>>>>>>>>> void DDD()
>>>>>>>>>>>>>>>> {
>>>>>>>>>>>>>>>>    HHH(DDD);
>>>>>>>>>>>>>>>>    return;
>>>>>>>>>>>>>>>> }
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> When I challenge anyone to show the details of exactly
>>>>>>>>>>>>>>>> how DDD correctly simulated by ANY simulating termination
>>>>>>>>>>>>>>>> analyzer HHH can possibly reach its own simulated "return"
>>>>>>>>>>>>>>>> statement final halt state they ignore this challenge.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> It seems very difficult for you to read.
>>>>>>>>>>>>>>> We clearly stated that the challenge is improper.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Are you too stupid to understand that dogmatic
>>>>>>>>>>>>>> assertions that are utterly bereft of any supporting
>>>>>>>>>>>>>> reasoning DO NOT COUNT AS REBUTTALS ???
>>>>>>>>>>>>>
>>>>>>>>>>>>> No, you are too stupid to realise that challenging for a 
>>>>>>>>>>>>> recipe to draw a square circle does not count as a proof 
>>>>>>>>>>>>> that square circles exist.
>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Claiming that I made a mistake with no ability to
>>>>>>>>>>>>>> show this mistake is DISHONEST.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Indeed, but irrelevant,
>>>>>>>>>>>>
>>>>>>>>>>>> That alternative is that you are dishonest.
>>>>>>>>>>>> When you claim that I am wrong and have
>>>>>>>>>>>> no ability to show how and where I am wrong
>>>>>>>>>>>> this would seem to make you a liar.
>>>>>>>>>>>>
>>>>>>>>>>>> No one has ever even attempted to show the details
>>>>>>>>>>>> of how this is not correct:
>>>>>>>>>>>>
>>>>>>>>>>>> void DDD()
>>>>>>>>>>>> {
>>>>>>>>>>>>    HHH(DDD);
>>>>>>>>>>>>    return;
>>>>>>>>>>>> }
>>>>>>>>>>>>
>>>>>>>>>>>> When one or more instructions of DDD are correctly
>>>>>>>>>>>> simulated by ANY simulating termination analyzer HHH
>>>>>>>>>>>> then this correctly simulated DDD never reaches its
>>>>>>>>>>>> simulated "return" statement final halt state.
>>>>>>>>>>>
>>>>>>>>>>> Indeed, HHH fails to reach the end of the simulation, even 
>>>>>>>>>>> though the end is only one cycle further from the point where 
>>>>>>>>>>> it gave up the simulation.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> That is counter-factual and over-your-head.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> No evidence presented for this claim. Dreaming again?
>>>>>>>>> Even a beginner understands that when HHH has code to abort and 
>>>>>>>>> halt, the simulated HHH runs one cycle behind the simulating 
>>>>>>>>> HHH, so that when the simulating HHH aborts, the simulated HHH 
>>>>>>>>> is only one cycle away from the same point.
>>>>>>>>
>>>>>>>> Proving that you do not understand what unreachable code is.
>>>>>>>> First year CS students and EE majors may not understand this.
>>>>>>>> All CS graduates would understand this.
>>>>>>>
>>>>>>> That you do not understand what I write makes it difficult for 
>>>>>>> you to learn from your errors.
>>>>>>> It is not that difficult. Try again and pay full attention to it.
>>>>>>> Even a beginner understands that when HHH has code to abort and 
>>>>>>> halt,
>>>>>>> the simulated HHH runs one cycle behind the simulating HHH, so 
>>>>>>> that when the simulating HHH aborts, the simulated HHH is only 
>>>>>>> one cycle away from the same point.
>>>>>>
>>>>>> Yes this is factual.
>>>>>>
>>>>>> *This is only ordinary computer programming with*
>>>>>> *no theory of computation computer science required*
>>>>>>
>>>>>> Every simulated HHH remains one cycle behind its simulator
>>>>>> no matter how deep the recursive simulations go. This means
>>>>>> that the outermost directly executed HHH reaches its abort
>>>>>> criteria first.
>>>>>
>>>>> And it fails to see that the simulated HHH would reach exactly the 
>>>>> same abort criteria one cycle later.
>>>>> In this way, it misses the fact that it is simulating an HHH that 
>>>>> would abort and halt.
>>>>>
>>>>
>>>> void Infinite_Loop()
>>>> {
>>>>    HERE: goto HERE;
>>>>    printf("Fred Zwarts can't understand this is never reached\n");
>>>> }
>>>>
>>> Another claim without any evidence.
>>>
>>> Olcott does not understand that his HHH does not see an infinite loop.
>>> It aborts and halt, so the recursion is finite. 
>>
>> You didn't even use the term recursion correctly.
>> Infinite loops have nothing to do with recursion.
> 
> And infinite loops have nothing to do with a simulator simulating 
> itself. Therefore, talking about infinite loops is changing the subject.
> 
>> Mike understands that HHH could recognize an infinite
>> loop correctly.
>>
>>     The process in which a function calls itself directly
>>     or indirectly is called recursion and the corresponding
>>     function is called a recursive function.
>> https://www.geeksforgeeks.org/introduction-to-recursion-2/
>>
>> Lines 987 to 992 is where infinite loops are recognized
>> Lines 996 to 1005 is where infinite recursion is recognized
>> https://github.com/plolcott/x86utm/blob/master/Halt7.c
>>
>> HHH correctly emulates the x86 machine code of its
>> input until one of those two patterns is matched.
> 
> But there is a bug in the code that tries to recognise an infinite 
> recursion. 

There is no bug. Quit your defamation.

> It forgets to count the conditional branch instructions when 
> simulating the simulator. 

*It does not forget them. They are irrelevant*

The question being asked is this:
Can DDD correctly simulated by any termination analyzer
HHH that can possibly exist reach its own "return" statement
final halt state?

> In this way it fails to do a correct analysis 
> of the code that specifies the abort, which makes the recursion finite. 
> Because of this bug it 'recognises' an infinite recursion, when there is 
> only a finite recursion.
> The x86 machine code also specifies the abort code and the halting 
> behaviour of the input, but HHH is blind for that part of the 
> specification.
> 
> 


-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer