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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: How do computations actually work?
Date: Fri, 30 May 2025 10:56:35 +0300
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On 2025-05-28 14:33:56 +0000, olcott said:

> On 5/28/2025 2:43 AM, Mikko wrote:
>> On 2025-05-27 15:48:14 +0000, olcott said:
>> 
>>> On 5/27/2025 3:37 AM, Mikko wrote:
>>>> On 2025-05-26 15:57:11 +0000, olcott said:
>>>> 
>>>>> On 5/26/2025 3:46 AM, Mikko wrote:
>>>>>> On 2025-05-25 14:53:06 +0000, olcott said:
>>>>>> 
>>>>>>> On 5/25/2025 4:14 AM, Mikko wrote:
>>>>>>>> On 2025-05-24 15:18:57 +0000, olcott said:
>>>>>>>> 
>>>>>>>>> On 5/24/2025 2:47 AM, Mikko wrote:
>>>>>>>>>> On 2025-05-23 02:47:40 +0000, olcott changed the subject to
>>>>>>>>>>> How do computations actually work?
>>>>>>>>>> 
>>>>>>>>>> Each computation works differently. It does not matter how it works as
>>>>>>>>>> long as there are instructions that fully specify how that computation
>>>>>>>>>> shall be performed.
>>>>>>>>> 
>>>>>>>>> All termination analyzers are required to report on the
>>>>>>>>> behavior that their input finite string specifies.
>>>>>>>> 
>>>>>>>> To report correctly. Though the input string to a termination analyzer
>>>>>>>> usially is incomlete: the input string usually specifies different
>>>>>>>> behavours depending on the input that is not shown to the termination
>>>>>>>> analyzer, and the analyzer's report must cover all of them.
>>>>>>>> 
>>>>>>>> A partial termination analyzer may fail to report but is not allowed
>>>>>>>> to report incorrectly.
>>>>>>> 
>>>>>>> void DDD()
>>>>>>> {
>>>>>>>    HHH(DDD);
>>>>>>>    return;
>>>>>>> }
>>>>>>> 
>>>>>>> DDD simulated by HHH cannot possibly reach its
>>>>>>> "return" statement final halt state, only liars
>>>>>>> will disagree.
>>>>>> 
>>>>>> Again a straw man deception. Where reasonable people disgraee is the
>>>>>> relevance of that claim. The behavour specified by DDD does reach the
>>>>>> final return from DDD. Whether HHH can simulate that part of the
>>>>>> behaviour is irrelevant. Even without the simjlation HHH decides
>>>>>> correctly if and only if it determines and reports that DDD halts.
>>>>> 
>>>>> _DDD()
>>>>> [00002192] 55             push ebp
>>>>> [00002193] 8bec           mov ebp,esp
>>>>> [00002195] 6892210000     push 00002192
>>>>> [0000219a] e833f4ffff     call 000015d2  // call HHH
>>>>> [0000219f] 83c404         add esp,+04
>>>>> [000021a2] 5d             pop ebp
>>>>> [000021a3] c3             ret
>>>>> Size in bytes:(0018) [000021a3]
>>>>> 
>>>>> How many recursive emulations does HHH have to
>>>>> wait before its emulated DDD magically halts
>>>>> on its own without ever needing to be aborted?
>>>> 
>>>> In order to determine that you need a program that simulates DDD and all
>>>> functios DDD calls and detects and counts and reports emulation levels.
>>> 
>>> No you don't. All that you need to do is simply imagine
>>> that HHH is an x86 emulator capable of emulating itself
>>> emulating DDD. // *This is fully operational code*
>> 
>> Well, you needn't if you have a good substitution. But imagination is
>> not a good substitution of real work with a real tool.
> 
> https://github.com/plolcott/x86utm/blob/master/Halt7.c
> has an HHH that does simulate itself simulating DDD.

Nothing behind that pointer refutes my claim that it is not sufficent
to imagine.

-- 
Mikko