• Re: Recursive simulation

    From Fred. Zwarts@21:1/5 to All on Sun Jul 21 18:45:10 2024
    Op 21.jul.2024 om 16:53 schreef olcott:
    On 7/21/2024 9:37 AM, joes wrote:
    Am Sun, 21 Jul 2024 09:25:51 -0500 schrieb olcott:
    On 7/21/2024 5:25 AM, Fred. Zwarts wrote:
    Op 20.jul.2024 om 22:08 schreef olcott:

    It *is* a fact that no DDD correctly simulated by any pure function
    HHH ever reaches its own return instruction.
    Which proves that these simulations are incorrect.
    *You don't get to be the judge of this*
    As long as the x86 machine language instructions of DDD are emulated by
    HHH according to the semantic meaning of these instructions then the
    emulation is correct and anyone that disagrees is disagreeing with a
    tautology.

    They are not simulated correctly, they are aborted.


    N instructions of DDD[N] are correctly emulated by some HHH[N]
    where N is each element of the ordered set of positive integers.

    This correct emulation must take into account the fact that DDD is
    calling its own emulator: HHH(DDD) in recursive emulation.

    Only in so far that it also simulates all recursive calls.
    Why is the return of HHH not simulated?


    The return of HHH is not simulated for the same reason
    that the following OutputString("Hello World!"); would
    never be simulated.

    void Infinite_Recursion()
    {
      Infinite_Recursion();
      OutputString("Hello World!");
    }



    But why is the return if Finite_Recursion not simulated?

    void Finite_Recursion (int N) {
    if (N > 0) Finite_Recursion (N - 1);
    }

    This looks more like HHH, which also halts after a finite number of
    recursions.

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