Re: Any honest person that knows the x86 language can see...

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Sujet : Re: Any honest person that knows the x86 language can see...
De : F.Zwarts (at) *nospam* HetNet.nl (Fred. Zwarts)
Groupes : comp.theory comp.ai.philosophy
Date : 30. Jul 2024, 09:59:49
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v8a6hn$uejg$1@dont-email.me>
References : 1 2 3 4 5
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Op 29.jul.2024 om 21:48 schreef olcott:
On 7/29/2024 2:38 PM, Fred. Zwarts wrote:
Op 29.jul.2024 om 21:35 schreef olcott:
On 7/29/2024 2:18 PM, Fred. Zwarts wrote:
Op 29.jul.2024 om 16:07 schreef olcott:
HHH(Infinite_Recursion) and HHH(DDD) show the same non-halting
behavior pattern in their derived execution traces of their
inputs.
>
Correct emulation is defined as emulating the machine language
input according to the x86 semantics specified by this input.
>
For DDD correctly emulated by HHH this includes HHH emulating
itself emulating DDD according to the x86 semantics of itself.
>
HHH(DDD) shows the exact same execution trace behavior pattern
as HHH(Infinite_Recursion) where 3-4 instructions are repeated
with no conditional branch instructions in this trace that could
prevent them from endlessly repeating.
>
void Infinite_Recursion()
{
   Infinite_Recursion();
}
>
No, the HHH that aborts after N cycles has a similar behaviour as
>
>
So you don't even know that infinite recursion is non-halting behavior.
You can go back and try again on this same post I am not looking at
anything else that you say.
>
Non halting is only in your dreams. HHH that aborts halts. Dreams are no substitute for logic.
>
 void Infinite_Recursion()
{
   Infinite_Recursion();
}
 In other words you are confirming that you
honestly believe the above function halts?
No, as usual you are twisting my words. Further Infinite_Recursion is irrelevant, because we are talking about a HHH that aborts, so its simulation looks more like:
void Finite_Recursion (int N) {
   if (N > 0) Finite_Recursion (N - 1);
}

 Do you understand that Halts means terminates normally
on its own without being forced to stop running?
 void Infinite_Loop()
{
   HERE: goto HERE;
}
Irrelevant text ignored. HHH, when simulating itself, is simulating a program that aborts after two cycles of recursion. So, when HHH is simulated, no abort is needed.
Do you understand that Finite_Recursion halts after N recursions, without a need to abort it?

 Do you understand that yanking the power cord out of
the wall does not cause Infinite_Loop() to halt?
 
Do you understand that Finite_Recursion halts, even when the power cord is not yanked out?

Date Sujet#  Auteur
29 Jul 24 * Any honest person that knows the x86 language can see...7olcott
29 Jul 24 +* Re: Any honest person that knows the x86 language can see...5Fred. Zwarts
29 Jul 24 i`* Re: Any honest person that knows the x86 language can see...4olcott
29 Jul 24 i `* Re: Any honest person that knows the x86 language can see...3Fred. Zwarts
29 Jul 24 i  `* Re: Any honest person that knows the x86 language can see...2olcott
30 Jul 24 i   `- Re: Any honest person that knows the x86 language can see...1Fred. Zwarts
30 Jul 24 `- Re: Any honest person that knows the x86 language can see... DDD will HALT if HHH answers1Richard Damon

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