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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: HHH(DDD) is correct to reject its input as non-halting --- EVIDENCE THAT I AM CORRECT
Date: Wed, 25 Jun 2025 09:27:08 +0300
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On 2025-06-24 14:06:12 +0000, olcott said:

> 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?

Why would anyone ask that question or care about the answer?

-- 
Mikko