Re: Sequence of sequence, selection and iteration matters

Op 09.jul.2024 om 16:19 schreef olcott:

On 7/9/2024 1:22 AM, Mikko wrote:

On 2024-07-08 13:04:13 +0000, olcott said:

<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its input D
     until H correctly determines that its simulated D would never
     stop running unless aborted then
>
     H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>

Whether a partial simulation of DDD simulates the return depends
on the simulator.

>
That is false proving that you have insufficient knowledge.

>
It is true. There are partial simulators that do simulate D(I) to its termination (if it terminates) and there are simulators that don't.
That you cannot imagine something does not mean it can't exist.
>

 No pure function x86 emulator HHH can possibly emulate DDD
to its termination.

Indeed, no such HHH exists. This proves that HHH cannot possibly simulate itself correctly.

 _DDD()
[00002163] 55         push ebp      ; housekeeping
[00002164] 8bec       mov ebp,esp   ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404     add esp,+04
[00002173] 5d         pop ebp
[00002174] c3         ret
Size in bytes:(0018) [00002174]
 *When DDD is correctly emulated by any pure function*
*HHH x86 emulator that can possibly exist* which calls
an emulated HHH(DDD) to repeat this process until the
emulated DDD is aborted.

No such HHH exists. Only HHH that simulates itself incorrectly exist.
It repeats until it aborts the simulation prematurely, because the simulated HHH is programmed to also abort one cycle later, which makes the abort premature.

 At no point in this emulation does the call from DDD
correctly emulated by HHH to HHH(DDD) ever return.

Indeed, because it is aborted prematurely. A correct simulation shows that DDD does return, but HHH cannot possibly simulated itself up to the end.
The DDD is an unneeded complexity. In the following much simpler example, no DDD is needed to show the same behaviour. H cannot possibly simulate itself correctly.
        int main()
        {
          return H(main, 0);
        }
Olcott dreams of an infinite recursion, but the recursion is finite, because HHH aborts, just as in:
void Finite_Recursion (int N) {
   if (N > 0) Finite_Recursion (N - 1);
}
Olcott never told us how HHH simulates Finite_Recursion, but I would not be surprised if it thinks that it must abort it, too, for any N greater than two. Because two seems to be infinite for it.
And even although the simulation is clearly incorrect, olcott loves to remember Sipser, who only talks about correct simulations.