Re: The input to HHH(DDD) specifies a non-halting sequence of configurations +++

On 6/16/2025 6:40 AM, Mikko wrote:

On 2025-06-15 13:57:01 +0000, olcott said:
 

On 6/15/2025 3:44 AM, Fred. Zwarts wrote:

Op 14.jun.2025 om 16:07 schreef olcott:

On 6/13/2025 6:02 AM, Mikko wrote:

On 2025-06-11 14:03:41 +0000, olcott said:
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On 6/11/2025 3:20 AM, Mikko wrote:

On 2025-06-10 15:41:33 +0000, olcott said:
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On 6/10/2025 6:41 AM, 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,

>
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.
>

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.
>

>
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?

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If HHH is not a decider the question is not interesting.

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I switched to the term: "termination analyzer" because halt deciders
have the impossible task of being all knowing.

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The termination problem is in certain sense harder than the halting
problem.

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Not at all

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That's in another sense in which nothing is harder than impossible.
>

void DDD()
{
   HHH(DDD);
   return;
}
>
If HHH only determines non-halting correctly for the
above input and gets the wrong answer on everything
else then HHH *is* a correct termination analyzer.

>
It is not a correct termination analyzer if if gives the wrong answer.

>
*Key verified facts such that disagreement is inherently incorrect*
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(a) HHH(DDD) does not correctly report on the behavior of its caller.

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Irrelevant. HHH should decide about the program specified in the input, whether or not it is the same code used by the caller.

>
In other words you do not understand that a partial
halt decider is not allowed to report on the behavior
of its caller and only allowed to report on the behavior
specified by the sequence of state transitions specified
by its input.

 It is not allowed to report incorrectly. There are no prohibitions
against correct reporting.
 

And you do not understand which is which.
int main()
{
   DDD(); // Calls HHH(DDD) that cannot report on the
}        // behavior of its caller because it cannot
          // see its caller. Its caller could have been main()
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer