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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1 ---
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Date: Wed, 2 Apr 2025 20:19:43 -0500
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On 4/2/2025 1:47 PM, Fred. Zwarts wrote:
> Op 02.apr.2025 om 17:55 schreef olcott:
>> On 4/2/2025 9:14 AM, joes wrote:
>>> Am Mon, 31 Mar 2025 16:26:58 -0500 schrieb olcott:
>>>> On 3/31/2025 2:10 PM, Fred. Zwarts wrote:
>>>>> Op 31.mrt.2025 om 20:16 schreef olcott:
>>>
>>>>>> A simulating termination analyzer is always correct to abort the
>>>>>> simulation and reject the input as non-halting when-so-ever this 
>>>>>> input
>>>>>> would otherwise prevent itself from halting.
>>>>>>
>>>>> But the input is halting, as proven by direct execution.
>>>>
>>>> Something other than the input is halting.
>>>> HHH1(DDD) shows the same behavior as the direct execution.
>>>>    HHH(DDD) shows the behavior of the actual input.
>>> Why are you not passing DDD as input? Why do you not call what you're
>>> doing HHH(HHH(DDD))? What is the difference in what is passed to HHH1?
>>>
>>
>> This seems to be above your level of technical competence.
>>
>> _DDD()
>> [00002172] 55         push ebp      ; housekeeping
>> [00002173] 8bec       mov  ebp,esp  ; housekeeping
>> [00002175] 6872210000 push 00002172 ; push DDD
>> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
>> [0000217f] 83c404     add  esp,+04
>> [00002182] 5d         pop  ebp
>> [00002183] c3         ret
>> Size in bytes:(0018) [00002183]
>>
>> Anyone understanding the above code where HHH
>> emulates DDD according to the semantics of the
>> x86 language knows that this DDD (not some
>> other different DDD) cannot possibly reach its
>> own final halt state.

> Yes it fails to reach the end of the simulation of a program that 
> according to the x86 semantics has an end as proven by direct execution. 

In other words you don't hardly know the x86
language at all.

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