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

Op 15.jun.2025 om 15:57 schreef olcott:

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.
>

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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.

>
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.
 

So, you do not understand that the behaviour of the caller is irrelevant. The decider must report on the behaviour of its input, even if this input has exactly the same behaviour as the caller.

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

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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.

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It does not.

 You cannot show the detailed steps that "it does not" because
you are wrong. You simply don't understand these things.

I did that several times, but apparently you do not understand it.

 When you take a guess and provide zero supporting reasoning
for this guess it does not count as any actual rebuttal at all.

My rebuttal is the verified fact that the input for HHH is a pointer to code that includes the code to abort and halt.
You ignore that and keep dreaming of an infinite recursion
Your HHH aborts a finite recursion, makes a wild guess and you keep dreaming that it is correct.
A dream is not a rebuttal.