Re: This function proves that only the outermost HHH examines the execution trace

Op 27.jul.2024 om 16:34 schreef olcott:

On 7/27/2024 9:23 AM, Fred. Zwarts wrote:

Op 27.jul.2024 om 15:48 schreef olcott:

On 7/27/2024 3:36 AM, Fred. Zwarts wrote:

Op 26.jul.2024 om 17:56 schreef olcott:

This is meant for Mike, Joes and Fred don't have the technical competence to understand it.

>
I have pity with you,

>
I am not the one that stupidly believes that a non-terminating
input must be emulated to non-existent completion or the emulation
is wrong.

>
(That is something I never said, but it seems too difficult for you.)
You are the one that believes that the simulation of a halting program must be aborted to prevent non-halting. Ha ha.
>

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That the first HHH to see the non-halting behavior pattern must
abort or none of them abort is simply too difficult for you.

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That two recursions is not equal to an infinite recursion is already too difficult for you.

 Two recursions with no conditional branch instructions
inbetween *IS* the correct non-halt status criteria.

Are you really so stupid to think that HHH has no branch instructions?
HHH is a halting program. So, when DDD calls HHH, we know that DDD also halts. It is clear that DDD is a misleading and unneeded complication. It is easy to eliminate DDD:
        int main() {
          return HHH(main);
        }
This has the same problem. This proves that the problem is not in DDD, but in HHH, which halts when it aborts the simulation, but it decides that the simulation of itself does not halt.
It shows that HHH cannot possibly simulate itself correctly.
HHH is simply unable to decide about comparable finite recursions.
void Finite_Recursion (int N) {
   if (N > 0) Finite_Recursion (N - 1);
}
It decides after two recursions that there is an infinite recursion, which is incorrect.
Two differs from infinite!