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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: How do simulating termination analyzers work?
Date: Mon, 30 Jun 2025 12:58:08 +0300
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On 2025-06-29 14:21:55 +0000, olcott said:

> On 6/29/2025 3:44 AM, Mikko wrote:
>> On 2025-06-28 13:17:17 +0000, olcott said:
>> 
>>> On 6/28/2025 6:41 AM, Mikko wrote:
>>>> On 2025-06-21 17:34:55 +0000, olcott said:
>>>> 
>>>>> On 6/21/2025 4:52 AM, Mikko wrote:
>>>>>> On 2025-06-20 13:59:02 +0000, olcott said:
>>>>>> 
>>>>>>> On 6/20/2025 4:20 AM, Fred. Zwarts wrote:
>>>>>>>> Op 19.jun.2025 om 17:17 schreef olcott:
>>>>>>>>> On 6/19/2025 4:21 AM, Fred. Zwarts wrote:
>>>>>>>>>> Op 18.jun.2025 om 15:46 schreef olcott:
>>>>>>>>>>> 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*
>>>>>>>>>>> 
>>>>>>>>>> 
>>>>>>>>>> At least it is true for all aborting ones, such as the one you 
>>>>>>>>>> presented in Halt7.c.
>>>>>>>>> 
>>>>>>>>> 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?
>>>>>>> 
>>>>>>>> Yes, I confirmed many times that we can confirm this vacuous claim, 
>>>>>>>> because no such HHH exists. All of them fail to do a correct simulation 
>>>>>>>> up to the point where they can see whether the input specifies a 
>>>>>>>> halting program.
>>>>>>> 
>>>>>>> if DDD correctly simulated by any simulating termination
>>>>>>> analyzer HHH never aborts its simulation of DDD then
>>>>>> 
>>>>>> that HHH is not interesting.
>>>>> 
>>>>> *then the HP proofs are proved to be wrong*
>>>> 
>>>> Does not follow. HHH and DDD are irrelevant to those proofs.
>>> 
>>> void DDD()
>>> {
>>>    HHH(DDD);
>>>    return;
>>> }
>>> 
>>> When I dumbed the original self-referential proof down
>>> to HHH(DDD) everyone here proved that they did not even
>>> understand what ordinary recursion is.
>> 
>> That you are dumb does not mean that others don't understand
>> ordinary recursion.
>> 
> 
> Mensa scored me on the top 3% of the population.

Your intelligence, not wisdom.

> This is a little more difficult than ordinary recursion.

Perhaps to your little mind.

> void DDD()
> {
>    HHH(DDD);
>    return;
> }
> 
> _DDD()
> [00002192] 55             push ebp
> [00002193] 8bec           mov ebp,esp
> [00002195] 6892210000     push 00002192  // push DDD
> [0000219a] e833f4ffff     call 000015d2  // call HHH
> [0000219f] 83c404         add esp,+04
> [000021a2] 5d             pop ebp
> [000021a3] c3             ret
> Size in bytes:(0018) [000021a3]
> 
> The x86 source code of DDD specifies that this emulated
> DDD cannot possibly reach its own emulated "ret" instruction
> final halt state when emulated by HHH according to the
> semantics of the x86 language.

That defect in HHH is already known and a possible fix has been proposed.

-- 
Mikko