Re: Hypothetical possibilities --- Correct emulation has been proven for three years

Op 31.jul.2024 om 23:23 schreef olcott:

On 7/31/2024 3:01 PM, Fred. Zwarts wrote:

Op 31.jul.2024 om 17:14 schreef olcott:

On 7/31/2024 3:44 AM, Fred. Zwarts wrote:

Op 31.jul.2024 om 06:09 schreef olcott:

>
  machine   stack     stack     machine    assembly
  address   address   data      code       language
  ========  ========  ========  =========  =============
[00002192][00103820][00000000] 55         push ebp
[00002193][00103820][00000000] 8bec       mov ebp,esp
[00002195][0010381c][00002172] 6872210000 push 00002172 ; push DDD
[0000219a][00103818][0000219f] e833f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:1038c4
>
We don't show any of HHH and show the execution trace of
of just DDD assuming that HHH is an x86 emulator.

>
This assumption is incorrect if it means that HHH is an unconditional simulator that does not abort.

This algorithm is used by all the simulating termination analyzers:
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     *If simulating halt decider H correctly simulates its input D*
     *until H correctly determines that its simulated D would never*
     *stop running unless aborted* then
>
     H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>

>
So, Sipser only agreed to a correct simulation, not with an incorrect simulation that violates the semantics of the x86 language by skipping the last few instructions of a halting program.
>

 int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
 int main()
{
   HHH(DD);
}
 DD correctly emulated by HHH cannot possibly reach its own
second line. I switched to DDD correctly emulated by HHH

But it has been proven that no such HHH exists that simulates itself correctly. So, talking about a correct simulation by HHH is vacuous word salad.

because only C experts understood the above example and we
never had any of those here.

There are many C experts that looked at it, but you only got critic, because you keep hiding important properties of HHH, which made the conclusion impossible. Many C experts are in this newsgroup.
All C experts say: "if HHH halts, then there is no reason why the second line cannot be reached".
Of course you ignored it, because you you dream of a correct simulation.
But dreams are no substitute for facts.
No matter how much olcott wants it to be correct, or how many times olcott repeats that it is correct, it does not change the fact that such a simulation is incorrect, because it is unable to reach the end of a halting program.
The difference between DDD and DD is irrelevant, because HHH is unable to reach the code where Halt_Status is used.
DDD is a misleading and unneeded complication. It is easy to eliminate DDD:
        int main() {
          return HHH(main);
        }
This has the same problem. This proves that the problem is not in DDD, but in HHH, which halts when it aborts the simulation, but it decides that the simulation of itself does not halt.
It shows that HHH cannot possibly simulate itself correctly.
When HHH cannot possibly simulate itself correctly, talking about a correct simulation by HHH is vacuous word salad.
Sipser would agree that this incorrect simulation cannot be used to detect a non-halting behaviour.