Re: Mike can't even count to eight --- HHH(DDD)

On 6/10/2025 1:22 PM, Mike Terry wrote:

On 10/06/2025 17:42, 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:
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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;
}
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The *input* to simulating termination analyzer HHH(DDD)

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No it's not, as halt deciders / termination analyzers work with algorithms,

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That is stupidly counter-factual.
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That you think that shows that

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My understanding is deeper than yours.
No decider ever takes any algorithm as its input.

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But they take a description/specification of an algorithm,

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There you go.
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which is what is meant in this context.

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It turns out that this detail makes a big difference.
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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.
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HHH(DDD) takes a finite string of x86 instructions
that specify that HHH simulates itself simulating DDD.

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And HHH fails to see the specification of the x86 instructions. It aborts before it can see how the program ends.
>

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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.

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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.
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void DDD()
{
   HHH(DDD);
   return;
}
>
How the f-ck does DDD correctly simulated by HHH
reach its own "return" statement final halt state?

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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.

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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."

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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.
>
My other remarks look right:
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-   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.

 LOL, oh dear, of course I meant:
                Nobody has agreed that that implies that DDD /never/ halts. <=== THAT is
               your disagreement.
 

Infinite_Loop()
{
   HERE: goto HERE;
   return;
}
Yet most people here don't know that halts means
reaching the simulated "return" statement final
halt state.

i.e. consistent with the next statement.  [I did reread my post before clicking send, but I guess I just saw what I expected to see...]  Also I should apologise in advance if I've misunderstood M/F/R's position - I'm fairly confident they all /have/ agreed that the /simulation/ never proceeds as far as DDD's return statement (due to premature abort

If that was not a ridiculously stupid statement then
exactly what fixed integer number of instructions of
DDD must be correctly simulated by HHH so that DDD
simulated by HHH reaches its own simulated "return"
instruction final halt state?

etc..)  I just get fed up reading PO's repeated claims that everybody "lies" about that statement, when PO's memory/comprehension skills are at fault on this.
 

-   Of course, DDD directly executed will reach its own return statement,
     so it is true that this DDD halts...
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>
Mike.
>

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