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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Mike can't even count to eight --- HHH(DDD)
Date: Tue, 10 Jun 2025 12:36:24 -0500
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On 6/10/2025 11:42 AM, Mike Terry wrote:
> On 10/06/2025 17:17, olcott wrote:
>> On 6/10/2025 11:09 AM, Mike Terry wrote:
>>> On 10/06/2025 12:41, Mikko wrote:
>>>> On 2025-06-10 00:47:12 +0000, olcott said:
>>>>
>>>>> On 6/9/2025 7:26 PM, Richard Damon wrote:
>>>>>> On 6/9/25 10:43 AM, olcott wrote:
>>>>>>> On 6/9/2025 5:31 AM, Fred. Zwarts wrote:
>>>>>>>> Op 09.jun.2025 om 06:15 schreef olcott:
>>>>>>>>> On 6/8/2025 10:42 PM, dbush wrote:
>>>>>>>>>> On 6/8/2025 11:39 PM, olcott wrote:
>>>>>>>>>>> On 6/8/2025 10:32 PM, dbush wrote:
>>>>>>>>>>>> On 6/8/2025 11:16 PM, olcott wrote:
>>>>>>>>>>>>> On 6/8/2025 10:08 PM, dbush wrote:
>>>>>>>>>>>>>> On 6/8/2025 10:50 PM, olcott wrote:
>>>>>>>>>>>>>>> void DDD()
>>>>>>>>>>>>>>> {
>>>>>>>>>>>>>>>    HHH(DDD);
>>>>>>>>>>>>>>>    return;
>>>>>>>>>>>>>>> }
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> The *input* to simulating termination analyzer HHH(DDD)
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> No it's not, as halt deciders / termination analyzers work 
>>>>>>>>>>>>>> with algorithms,
>>>>>>>>>>>>>
>>>>>>>>>>>>> That is stupidly counter-factual.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> That you think that shows that
>>>>>>>>>>>
>>>>>>>>>>> My understanding is deeper than yours.
>>>>>>>>>>> No decider ever takes any algorithm as its input.
>>>>>>>>>>
>>>>>>>>>> But they take a description/specification of an algorithm,
>>>>>>>>>
>>>>>>>>> There you go.
>>>>>>>>>
>>>>>>>>>> which is what is meant in this context.
>>>>>>>>>
>>>>>>>>> It turns out that this detail makes a big difference.
>>>>>>>>>
>>>>>>>>>> And because your HHH does not work with the description/ 
>>>>>>>>>> specification of an algorithm, by your own admission, you're 
>>>>>>>>>> not working on the halting problem.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> HHH(DDD) takes a finite string of x86 instructions
>>>>>>>>> that specify that HHH simulates itself simulating DDD.
>>>>>>>>
>>>>>>>> And HHH fails to see the specification of the x86 instructions. 
>>>>>>>> It aborts before it can see how the program ends.
>>>>>>>>
>>>>>>>
>>>>>>> This is merely a lack of sufficient technical competence
>>>>>>> on your part. It is a verified fact that unless the outer
>>>>>>> HHH aborts its simulation of DDD that DDD simulated by HHH
>>>>>>> the directly executed DDD() and the directly executed HHH()
>>>>>>> would never stop running. That you cannot directly see this
>>>>>>> is merely your own lack of sufficient technical competence.
>>>>>>
>>>>>> And it is a verified fact that you just ignore that if HHH does in 
>>>>>> fact abort its simulation of DDD and return 0, then the behavior 
>>>>>> of the input, PER THE ACTUAL DEFINITIONS, is to Halt, and thus HHH 
>>>>>> is just incorrect.
>>>>>>
>>>>>
>>>>> void DDD()
>>>>> {
>>>>>    HHH(DDD);
>>>>>    return;
>>>>> }
>>>>>
>>>>> How the f-ck does DDD correctly simulated by HHH
>>>>> reach its own "return" statement final halt state?
>>>>
>>>> If HHH is not a decider the question is not interesting.
>>>> If HHH is a deider it returns. The first insturunction
>>>> after the return terminates the execution of DDD.
>>>
>>> But then DDD *simulated by HHH* does is not simulated as far as its 
>>> own return statement.  In fact zero steps of DDD are simulated.
>>>
>>
>> Since I have shown you the execution trace of DDD simulated
>> by HHH many times it is gross negligence that you say that
>> "zero steps of DDD are simulated."
> 
> Upon closer inspection it seems I misread what Mikko was saying.  I 
> misread that he was suggesting an HHH which returned straight away.  So 
> just ignore my "zero steps" remark.
> 

That is great. I continue to trust that you are honest
because of all of the evidence that you provided showing
a very deep and correct understanding of my work.

> My other remarks look right:
> 
> -   But then DDD *simulated by HHH* is not simulated as far as its own 
> return statement.
>      HINT:  I'm agreeing with you.
>      HINT:  Mikko, Fred and Richard have all also agreed with that 
> statement,
>             Nobody has agreed that that implies that DDD halts. <=== 
> THAT is your disagreement.
>             You're welcome.
> -   Of course, DDD directly executed will reach its own return statement,
>      so it is true that this DDD halts...
> 
> 
> Mike.
> 

The input to HHH(DDD) specifies non-halting behavior.

int main()
{
   DDD(); // calls HHH(DDD)
}

The HHH(DDD) that DDD() calls cannot report on the
behavior of its caller because it can't even see its
caller. Its caller could have been main() and HHH(DDD)
has no way to see this.

We could define the requirements for an algorithm
to correctly predict the name of president of the
United States 4 years from now and its only input
is a single integer value.

This could not be computable yet does not show any
actual limitation to computer science. Likewise for
a termination analyzer that is required to report
on the behavior of its caller.



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