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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Simulation vs. Execution in the Halting Problem
Date: Thu, 5 Jun 2025 09:14:25 +0200
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Op 04.jun.2025 om 18:31 schreef olcott:
> On 6/4/2025 4:13 AM, Fred. Zwarts wrote:
>> Op 03.jun.2025 om 21:53 schreef olcott:
>>> On 6/3/2025 7:45 AM, dbush wrote:
>>>> On 6/2/2025 10:58 PM, Mike Terry wrote:
>>>>> Even if presented with /direct observations/ contradicting his 
>>>>> position, PO can (will) just invent new magical thinking that only 
>>>>> he is smart enough to understand, in order to somehow justify his 
>>>>> busted intuitions.
>>>>
>>>> My favorite is that the directly executed D(D) doesn't halt even 
>>>> though it looks like it does:
>>>>
>>>>
>>>> On 1/24/24 19:18, olcott wrote:
>>>>  > The directly executed D(D) reaches a final state and exits normally.
>>>>  > BECAUSE ANOTHER ASPECT OF THE SAME COMPUTATION HAS BEEN ABORTED,
>>>>  > Thus meeting the correct non-halting criteria if any step of
>>>>  > a computation must be aborted to prevent its infinite execution
>>>>  > then this computation DOES NOT HALT (even if it looks like it does).
>>>
>>>
>>> If the second call of otherwise infinite recursion had
>>> to be aborted to prevent actual infinite recursion then
>>> this call always was non-halting even when it was forced
>>> to stop running.
>>>
>>
>> But since there is no infinite recursion, no abort is needed. 
> 
> *You just contradicted yourself*
> 
> void DDD()
> {
>    HHH(DDD);
>    return;
> }
> 
> HHH simulates DDD that calls HHH(DDD)
> that simulates DDD that calls HHH(DDD)
> that simulates DDD that calls HHH(DDD)
> that simulates DDD that calls HHH(DDD)...
> 
There we see that HHH aborts after a finite recursion.
It seems you do not even understand your own code and traces.
You contradict the code you provide and the traces you provide.