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From: olcott <polcott333@gmail.com>
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 09:14:07 -0500
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On 6/25/2025 1:27 AM, Mikko wrote:
> 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?
>
In computer science the only measure of halting
is reaching a final halt state. Stopping running
for any other reason does not count as halting.
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
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