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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: How do simulating termination analyzers work?
Date: Sat, 21 Jun 2025 12:33:10 -0500
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On 6/21/2025 4:46 AM, Mikko wrote:
> On 2025-06-20 17:12:30 +0000, olcott said:
> 
>> On 6/20/2025 3:45 AM, Mikko wrote:
>>> On 2025-06-19 09:09:34 +0000, Fred. Zwarts said:
>>>
>>>> Op 19.jun.2025 om 08:59 schreef olcott:
>>>>> On 6/19/2025 1:17 AM, Mikko wrote:
>>>>>> On 2025-06-18 13:46:16 +0000, olcott said:
>>>>>>
>>>>>>> On 6/18/2025 5:12 AM, Fred. Zwarts wrote:
>>>>>>>> Op 18.jun.2025 om 03:54 schreef olcott:
>>>>>>>>> On 6/17/2025 8:19 PM, Richard Damon wrote:
>>>>>>>>>> On 6/17/25 4:34 PM, olcott wrote:
>>>>>>>>>>> void Infinite_Recursion()
>>>>>>>>>>> {
>>>>>>>>>>>    Infinite_Recursion();
>>>>>>>>>>>    return;
>>>>>>>>>>> }
>>>>>>>>>>>
>>>>>>>>>>> void Infinite_Loop()
>>>>>>>>>>> {
>>>>>>>>>>>    HERE: goto HERE;
>>>>>>>>>>>    return;
>>>>>>>>>>> }
>>>>>>>>>>>
>>>>>>>>>>> void DDD()
>>>>>>>>>>> {
>>>>>>>>>>>    HHH(DDD);
>>>>>>>>>>>    return;
>>>>>>>>>>> }
>>>>>>>>>>>
>>>>>>>>>>> When it is understood that HHH does simulate itself
>>>>>>>>>>> simulating DDD then any first year CS student knows
>>>>>>>>>>> that when each of the above are correctly simulated
>>>>>>>>>>> by HHH that none of them ever stop running unless aborted.
>>>>>>>>>>
>>>>>>>>>> WHich means that the code for HHH is part of the input, and 
>>>>>>>>>> thus there is just ONE HHH in existance at this time.
>>>>>>>>>>
>>>>>>>>>> Since that code aborts its simulation to return the answer 
>>>>>>>>>> that you claim, you are just lying that it did a correct 
>>>>>>>>>> simulation (which in this context means complete)
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> *none of them ever stop running unless aborted*
>>>>>>>>
>>>>>>>> All of them do abort and their simulation does not need an abort.
>>>>>>>>
>>>>>>>
>>>>>>> *It is not given that any of them abort*
>>>>>>
>>>>>> It is known a priori that HHH either does or does not abort.
>>>>>
>>>>> Very good.
>>>>>
>>>>>> If HHH does
>>>>>> not abort it does not terminate the simulation of DDD and therefore
>>>>>
>>>>> DDD never stops running.
>>>>
>>>> because HHH never stops running and therefore this HHH
>>>>
>>>>>
>>>>>> does
>>>>>> not report correctly. If HHH does abort it reports that DDD does not
>>>>>> halt, which is incorrect as in that case DDD does halt. HHH is 
>>>>>> correct
>>>>>> about DDD only if it does abort its simulation and reports "halts".
>>>>>> But you HHH does not do that.
>>>>>>
>>>>>
>>>>>
>>>>
>>>> So, both the aborting and the non-aborting HHH do not provide a 
>>>> correct report.
>>>
>>> My HHH, if given DDD for input, does abort and does give the correct 
>>> report
>>> but gives the worng report if given DD.
>>
>> My claim is that each of the above functions correctly
>> simulated by any termination analyzer HHH that can possibly
>> exist will never stop running unless aborted by HHH.
>> Can you affirm or correctly refute this?
> 
> No, because that is not well claimed. You have used "HHH" in at least
> two different meanings and it is not clear which meaning is intended.
> 

*clearer words*
My claim is that each of the above functions correctly
simulated by any termination analyzer HHH that can possibly
exist will never stop running unless aborted by HHH.
Can you affirm or correctly refute this?


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