Re: Defining a correct simulating halt decider

Op 03.sep.2024 om 00:22 schreef olcott:

On 9/2/2024 12:52 PM, Fred. Zwarts wrote:

Op 02.sep.2024 om 18:38 schreef olcott:

A halt decider is a Turing machine that computes
the mapping from its finite string input to the
behavior that this finite string specifies.
>
If the finite string machine string machine
description specifies that it cannot possibly
reach its own final halt state then this machine
description specifies non-halting behavior.
>
A halt decider never ever computes the mapping
for the computation that itself is contained within.
>
Unless there is a pathological relationship between
the halt decider H and its input D the direct execution
of this input D will always have identical behavior to
D correctly simulated by simulating halt decider H.
>
*Simulating Termination Analyzer H Not Fooled by Pathological Input D*
https://www.researchgate.net/ publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D
>
A correct emulation of DDD by HHH only requires that HHH
emulate the instructions of DDD** including when DDD calls
HHH in recursive emulation such that HHH emulates itself
emulating DDD.

>
Indeed, it should simulate *itself* and not a hypothetical other HHH with different behaviour.
If HHH includes code to see a 'special condition' and aborts and halts, then it should also simulate the HHH that includes this same code and

  DDD has itself and the emulated HHH stuck in recursive emulation.
 void DDD()
{
   HHH(DDD);
   return;
}

It is not DDD. It is HHH that has the problem when trying to simulate itself.
        int main() {
          return HHH(main);
        }
Only when HHH would not abort (as in olcott's dream) there would be an infinite recursion. But for the HHH that aborts, it is stuck in the recursion for only a few recursions. Then it aborts and halts.

 When HHH emulates itself emulating DDD the emulated
HHH cannot possibly return because each DDD keeps
calling HHH to emulate itself again until the outer
executed HHH kills the whole emulated process at
the very first emulated DDD before it ever reaches
its own second line.

And this outer HHH does it one cycle before the inner HHH would do the same and halt. This means that the outer HHH misses the halting behaviour of the inner HHH by aborting too soon.
In this way the outer HHH prevents the inner HHH to reach the end of the simulation. The outer HHH, therefore, violates the semantics of the x86 language, by skipping the last few instructions of the halting program.
But olcott is still dreaming of the inner HHH that does not abort and thinks that the abort code only affects the outer HHH, not the inner HHH. A beginners error to think that a code change will not affect the behaviour of a program.