Re: DDD simulated by HHH cannot possibly halt (Halting Problem)

On 4/9/2025 1:58 PM, Fred. Zwarts wrote:

Op 09.apr.2025 om 19:29 schreef olcott:

>
On 4/8/2025 10:31 AM, Fred. Zwarts wrote:

Op 08.apr.2025 om 17:13 schreef olcott:

On 4/8/2025 2:45 AM, Fred. Zwarts wrote:

Op 08.apr.2025 om 06:33 schreef olcott:

>
typedef void (*ptr)();
int HHH(ptr P);
>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
int main()
{
   HHH(DD);
}
>
*Simulating termination analyzer Principle*
It is always correct for any simulating termination
analyzer to stop simulating and reject any input that
would otherwise prevent its own termination.

>
>
In this case there is nothing to prevent, because the finite string specifies a program that halts.

>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
This stuff is simply over-your-head.
HHH(DD) meets the above: *Simulating termination analyzer Principle*
Anyone with sufficient competence with the C programming language
will understand this.
>

Everyone with a little bit of C knowledge understands that if HHH returns with a value 0, then DDD halts.

>
DDD CORRECTLY SIMULATED BY HHH
NOT ANY OTHER DAMN DDD IN THE UNIVERSE NITWIT.
>

If HHH would correctly simulate DD (and the functions called by DD) then the simulated HHH would return to DD and DD would halt.

No way, José:
By "correctly simulate" I mean a simulation or an emulation
of DD by HHH that obeys the semantics of the C or the x86utm
programming languages respectfully. This expressly includes
that HHH simulates/emulates itself simulating/emulating DD.
Even though HHH does halt this does not entail that DDD halts.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer