Re: 197 page execution trace of DDD correctly simulated by HHH

On 7/1/24 10:34 PM, olcott wrote:

On 7/1/2024 9:24 PM, Richard Damon wrote:

On 7/1/24 9:36 PM, olcott wrote:

On 7/1/2024 7:38 PM, Richard Damon wrote:

On 7/1/24 8:59 AM, olcott wrote:

On 7/1/2024 3:23 AM, Fred. Zwarts wrote:

Op 30.jun.2024 om 19:20 schreef olcott:

>
_DDD()
[00002172] 55               push ebp      ; housekeeping
[00002173] 8bec             mov ebp,esp   ; housekeeping
[00002175] 6872210000       push 00002172 ; push DDD
[0000217a] e853f4ffff       call 000015d2 ; call HHH(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]
>

>
It cannot possibly return, because HHH aborts itself one cycle too early, showing that the emulation is incorrect. If that is over your head, try to learn how x86 instructions work.

>
_DDD()
[00002172] 55               push ebp      ; housekeeping
[00002173] 8bec             mov ebp,esp   ; housekeeping
[00002175] 6872210000       push 00002172 ; push DDD
[0000217a] e853f4ffff       call 000015d2 ; call HHH(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]
>
DDD is correctly emulated by HHH which calls an
emulated HHH(DDD) to repeat the process until aborted.
>
>

>
CAN'T BE.
>
A "Correct Emulation" is one that produces the same result as the program at the input.
>

>
Which can only possibly occur be disregarding the semantics
of the x86 language. Liars would do that ignoramuses would do
that. Everyone with the equivalent of a BSCS would know that
what I said is true.
>
>

>
Why do you say that? That is EXACTLY the definition of Correct Emulation.

  WELL INDOCTRINATED FALSE ASSUMPTIONS ARE NOT TRUTH.
WELL INDOCTRINATED FALSE ASSUMPTIONS ARE NOT TRUTH.
WELL INDOCTRINATED FALSE ASSUMPTIONS ARE NOT TRUTH.

And denying definitions is just lying.

 void Infinite_Loop()
{
   HERE: goto HERE;
}
 void Infinite_Recursion()
{
   Infinite_Recursion();
}
 void DDD()
{
   HHH(DDD);
}
 Every C programmer that knows what an x86 emulator is knows
that when HHH emulates the machine language of Infinite_Loop, Infinite_Recursion, and DDD that it must abort these emulations
so that itself can terminate normally.
 SO THESE THREE INPUTS DO NOT FREAKING HALT
SO THESE THREE INPUTS DO NOT FREAKING HALT
SO THESE THREE INPUTS DO NOT FREAKING HALT
 

No, DDD does halt if HHH is a decider and HHH(DDD) returns.
The fact that HHH need to abort its emulation to be a decider doesn't mean it gets to be wrong about the question put to it as a halt decider.
It just shows that Halting is not Computable.