Re: Simulation vs. Execution in the Halting Problem

On 6/6/25 2:42 PM, olcott wrote:

On 6/6/2025 3:29 AM, Fred. Zwarts wrote:

Op 05.jun.2025 om 17:50 schreef olcott:

On 6/5/2025 2:14 AM, Fred. Zwarts wrote:

Op 04.jun.2025 om 18:31 schreef olcott:

On 6/4/2025 4:13 AM, Fred. Zwarts wrote:

Op 03.jun.2025 om 21:53 schreef olcott:

On 6/3/2025 7:45 AM, dbush wrote:

On 6/2/2025 10:58 PM, Mike Terry wrote:

Even if presented with /direct observations/ contradicting his position, PO can (will) just invent new magical thinking that only he is smart enough to understand, in order to somehow justify his busted intuitions.

>
My favorite is that the directly executed D(D) doesn't halt even though it looks like it does:
>
>
On 1/24/24 19:18, olcott wrote:
 > The directly executed D(D) reaches a final state and exits normally.
 > BECAUSE ANOTHER ASPECT OF THE SAME COMPUTATION HAS BEEN ABORTED,
 > Thus meeting the correct non-halting criteria if any step of
 > a computation must be aborted to prevent its infinite execution
 > then this computation DOES NOT HALT (even if it looks like it does).

>
>
If the second call of otherwise infinite recursion had
to be aborted to prevent actual infinite recursion then
this call always was non-halting even when it was forced
to stop running.
>

>
But since there is no infinite recursion, no abort is needed.

>
*You just contradicted yourself*
>
void DDD()
{
   HHH(DDD);
   return;
}
>
HHH simulates DDD that calls HHH(DDD)
that simulates DDD that calls HHH(DDD)
that simulates DDD that calls HHH(DDD)
that simulates DDD that calls HHH(DDD)...
>

There we see that HHH aborts after a finite recursion.

>
HHH is supposed to abort as soon as it detects that
the simulated input cannot possibly reach its own
simulated "return" instruction final state.
>

But HHH does not *detect* that. The abort is based on an *invalid assumption*, not on what HHH sees. HHH sees only the first few recursive recursions of a finite recursion and misses the fact that the next cycle if the simulated HHH will abort and halt.

 You seem to remain unqualified to make this assessment
by continuing to fail to understand that because every
instance of HHH has the exact same machine code that
unless the outermost HHH aborts then none of them abort.
 

And if the outer one aborts, they all abort.
Since the behavior of the input to an aborted simulation is not the aborted simulation, but the behavior of the complete simulation of that exact same input, the fact that all of the HHH abort, means that all of the DDD halt, and thus all the HHH were wrong.
If you think that aborting a simulation, which means the simulation didn't see the final state, makes that actual behavior of the input to be non-halting, then perhaps you can make yourself immortal by aborting your observation of your life. Since you then will never see the point where you die, you must be non-dying and thus immortal.
Does that make logical sense to you?
Why does the simulator stopping its simulation, and thus obersvation of the behavior of the program it is simulating, affect what happens after is stops observing.
Simulation is JUST an observation of the recreation of the actual behavior, it is NOT the actual behavior itself.
UTM are sort of special machines, because they are defined that they are only a UTM if their simulation DOES exactly match the behavior of the program that its input represents, but even for them, the definiton of behavior comes from the actual execution, and the UTMness is shown by the fact that they match that.