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  • Re: Does the call from P() to H() specify infinite recursion?

    From olcott@21:1/5 to Richard Damon on Thu Nov 11 10:31:31 2021
    XPost: comp.theory, sci.logic, sci.math

    On 11/11/2021 10:14 AM, Richard Damon wrote:
    On 11/11/21 11:00 AM, olcott wrote:
    #define ptr uintptr_t

    void P(ptr x)
    {
       H(x, x);
    }

    int H(ptr x, ptr y)
    {
       ((void(*)(ptr))x)(y);
       return 1;
    }

    int main()
    {
       H((ptr)P, (ptr)P);
       return 0;
    }


    Yes.

    So all you have proven is that if H unconditionally executes its input
    you get infinite recursion.

    And, H will never return.

    If H isn't that function, then the computation of P changes so you have
    no proof for what the behavior of that machine is.


    _P()
    [00001a5e](01) 55 push ebp
    [00001a5f](02) 8bec mov ebp,esp
    [00001a61](03) 8b4508 mov eax,[ebp+08]
    [00001a64](01) 50 push eax // push P
    [00001a65](03) 8b4d08 mov ecx,[ebp+08]
    [00001a68](01) 51 push ecx // push P
    [00001a69](05) e810000000 call 00001a7e // call H
    [00001a6e](03) 83c408 add esp,+08
    [00001a71](01) 5d pop ebp
    [00001a72](01) c3 ret
    Size in bytes:(0021) [00001a72]


    If H simulates the x86 machine language of its input and sees that its simulated P is calling H with the same parameters that H was called with
    H can abort its simulation of P and correctly report that P would never
    reach its final state at 1a72.


    --
    Copyright 2021 Pete Olcott

    "Great spirits have always encountered violent opposition from mediocre
    minds." Einstein

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From olcott@21:1/5 to Richard Damon on Thu Nov 11 11:14:12 2021
    XPost: comp.theory, sci.logic, sci.math

    On 11/11/2021 10:46 AM, Richard Damon wrote:
    On 11/11/21 11:31 AM, olcott wrote:
    On 11/11/2021 10:14 AM, Richard Damon wrote:
    On 11/11/21 11:00 AM, olcott wrote:
    #define ptr uintptr_t

    void P(ptr x)
    {
       H(x, x);
    }

    int H(ptr x, ptr y)
    {
       ((void(*)(ptr))x)(y);
       return 1;
    }

    int main()
    {
       H((ptr)P, (ptr)P);
       return 0;
    }


    Yes.

    So all you have proven is that if H unconditionally executes its
    input you get infinite recursion.

    And, H will never return.

    If H isn't that function, then the computation of P changes so you
    have no proof for what the behavior of that machine is.


    _P()
    [00001a5e](01)  55              push ebp
    [00001a5f](02)  8bec            mov ebp,esp
    [00001a61](03)  8b4508          mov eax,[ebp+08]
    [00001a64](01)  50              push eax        // push P
    [00001a65](03)  8b4d08          mov ecx,[ebp+08]
    [00001a68](01)  51              push ecx        // push P
    [00001a69](05)  e810000000      call 00001a7e   // call H
    [00001a6e](03)  83c408          add esp,+08
    [00001a71](01)  5d              pop ebp
    [00001a72](01)  c3              ret
    Size in bytes:(0021) [00001a72]


    If H simulates the x86 machine language of its input and sees that its
    simulated P is calling H with the same parameters that H was called
    with H can abort its simulation of P and correctly report that P would
    never reach its final state at 1a72.



    No it can't.

    If H can abort its simulation, then it needs to take into account that H
    can abort its simulation. PERIOD.

    H, in making that conclusion, is PRESUMING that the called H will not
    abort its simulation, which it is wrong about.

    Because I keep repeating these same words dozens of times and you never acknowledge that you have ever seen them I really believe that you
    actually have attention deficit disorder (ADD):

    (a) P only halts if it reaches its final state at 1a72.

    (b) If H does not abort its simulation of P then P never reaches its
    final state at 1a72.

    (c) If H aborts its simulation of P then P never reaches its final state
    as 1a72.

    Because P never halts in all possible cases H(P,P)==0 is always correct.




    --
    Copyright 2021 Pete Olcott

    "Great spirits have always encountered violent opposition from mediocre
    minds." Einstein

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)