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

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,

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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.
>

>
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.
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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*
 (a) HHH(DDD) does not correctly report on the behavior of its caller.

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

 (b) Within the theory of computation HHH is not allowed to report
     on the behavior of its caller.

No, it should report on the behaviour of the program specified in the input, even if the caller uses the same code.

 (c) HHH(DDD) does correctly report on the behavior that its
     input specifies.

It does not. It seems you still does not understand the verified fact that the input is a pointer to a program including DDD and all the functions it uses, including the code that aborts and halts.
That HHH has a bug that makes that it does not see that part of the specification does not change the specification of a halting program.
It is not the HHH in your dream that does not abort.
Sum(2,3) should report the sum of 2 and 3, not the sum of 7 and 9, even if you dream about 7 and 9.