Re: How do simulating termination analyzers work? (V2)

On 6/20/2025 7:01 PM, Richard Damon wrote:

On 6/20/25 3:08 PM, olcott wrote:

void Infinite_Recursion()
{
   Infinite_Recursion();
   return;
}
>
void Infinite_Loop()
{
   HERE: goto HERE;
   return;
}
>
void DDD()
{
   HHH(DDD);
   return;
}
>
int Sipser_D()
{
   if (HHH(Sipser_D) == 1)
     return 0;
   return 1;
}
>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
My claim is that each of the above functions correctly
simulated by any termination analyzer HHH that can possibly
exist will never stop running unless aborted by HHH.

 The problem is that NO "Simulating Halt Decider" HHH, can correctly simulte ANY of those inputs.
 For the first two, it is possible for a smart enough Simulation Halt Decider to determine that the correct simulation of the input will not halt, no matter what HHH actually does, since it doesn't depend on the decider.
 For the last 3, it can not prove that they will not halt, as, in fact,

the correct simulation of all those inputs *WILL* halt

_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]
Exactly how would N instructions of DDD emulated by HHH
according to the semantics of the x86 language reach
machine address 000021a3 ?
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer