Re: Analysis of Flibble’s Latest: Detecting vs. Simulating Infinite Recursion ZFC

Op 26.mei.2025 om 17:42 schreef olcott:

On 5/25/2025 12:48 PM, Richard Heathfield wrote:

On 25/05/2025 16:55, olcott wrote:

On 5/25/2025 5:19 AM, Richard Heathfield wrote:

On 24/05/2025 17:13, olcott wrote:

No HHH can report on the behavior of its caller

>
 From Halt7.c:
>
void DDD()
{
   HHH(DDD);
   return;
}
>
Since (as you say) no HHH can report on the behaviour of its caller, and since (as your code shows) DDD is HHH's caller, we deduce that HHH cannot report on DDD.
>
So HHH is not (according to you) a halt analyser for DDD.
>
I'm not sure you've left anything to discuss, have you?
>

>
HHH(DDD) does correctly reject
*ITS INPUT THUS NOT ITS CALLER*
as non-halting.

>
Its input, as you show by the notation HHH(DDD), is DDD. So it's reporting on DDD.
>
 > HHH(DDD) does correctly reject
 > *ITS INPUT THUS NOT ITS CALLER*
 > as non-halting.
>
But as this code shows:
>
 >> void DDD()
 >> {
 >>    HHH(DDD);
 >>    return;
 >> }
>
DDD calls HHH. Therefore DDD is HHH's caller.
>
DDD is HHH's caller AND its input.

 The requirement for HHH to report on the direct
execution of its input is WRONG because this require
HHH to report on the behavior of its caller and no
C function can see its own caller.

HHH needs to report on the input. It is irrelevant whether it is its caller or not. The input is a pointer to a function in memory. This memory also includes all functions called by DDD. In particular it includes the code of Halt7.c, which specifies the abort. Therefore the input specifies a halting program. That specification is independent of whether this function is also the caller of HHH. If HHH is programmed in such a way that it cannot see this specification in the input, then that is clearly a bug.
Even if DDD is not the caller, such as in:
int main()
  {
     HHH(DDD);
     return;
  }
the same input is specified to HHH and it should return the same result: not halting.