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

On 2025-05-26 15:59:50 +0000, olcott said:

On 5/26/2025 4:43 AM, Mikko wrote:

On 2025-05-25 15:55:42 +0000, olcott said:
 

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.

 If the caller is DDD then the input specifies a halting behaviour because
DDD calls HHH with an input that specifies a halting behavour. But HHH
cannot correctly reject a halting input as non-halting.
 The requirements of a halting decider cannot be met if the decider reports
differently dependig on who calls it. Consequently, HHH is required to
return true also when the following FFF calls it:
 void FFF(void) {
 HHH(DDD);
 return;
}
 

 _DDD()
[00002192] 55             push ebp
[00002193] 8bec           mov ebp,esp
[00002195] 6892210000     push 00002192
[0000219a] e833f4ffff     call 000015d2  // call HHH
[0000219f] 83c404         add esp,+04
[000021a2] 5d             pop ebp
[000021a3] c3             ret
Size in bytes:(0018) [000021a3]
 How many recursive emulations does HHH have to
wait before its emulated DDD magically halts
on its own without ever needing to be aborted?

HHH only needs to simulate eoungh to see that it is called. You know
what HHH is and what it does so you needn't simulate HHH in order to
find out. But you need to find out what happens after the return from
HHH, either by simulating the rest of the ode of DDD or otherwise.
--
Mikko