Re: DDD correctly emulated by HHH --- Correct Emulation Defined

On 3/21/25 9:07 AM, olcott wrote:

On 3/21/2025 2:59 AM, Mikko wrote:

On 2025-03-20 22:43:34 +0000, olcott said:
>

On 3/20/2025 4:16 AM, Mikko wrote:

On 2025-03-20 02:32:43 +0000, olcott said:
>

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]
>
When N steps of DDD are emulated by HHH according
to the semantics of the x86 language then these
N steps are emulated correctly.

>
That does not make much sense to define the correct emulation of DDD as
it should mean whatever "correct emulation" means when applied to DDD.
>
Althouth promised otherwise on the subject line the meaning of "DDD
correctly emulated by HHH" when N is not specified is not defined.

>
N in this context always means any element of the
set of natural numbers.

>
No, it does not. It means a number that makes sense in the context and
will be specified when the definition shall be applied.
>

1,2,3...4,294,967,296 steps of DDD are correctly emulated
by HHH and DDD never reaches its "ret" instruction and
terminates normally.

>
But your HHH does not simulate correctly more steps of DDD than your
HHH1 does.
>

 This HHH is the hypothetical HHH the emulates an arbitrary number
of steps of DDD according to the semantics of the x86 language.
It is always at machine address 000015d2. This same HHH also
has the ability to emulate itself emulating DDD.

But the HHH that DDD calls isn't supposed to be that, but the original HHH that you claim gives the correct answer.
A program doesn't depend on things outside of it, unless given as its input, and DDD doesn't take HHH as an input, so it must be a fixed quantity.
And, as you have shown, HHH can NOT correctly emulate itself (at least none that give an answer) as it will ALWAY have aborted before reaching the end,
And thus, when we look at the ACTUAL correct emulation of that DDD, we see it will halt, despite the fact that HHH, with its INCORRECT logic failed to do so.
All you have done is turned one failure, into an infinite number of them, as each "problem" must have a specific HHH and its specific DDD, and ALL the HHH that abort and return 0 are shown to be wrong, as the actual correct (non-aborted) emulation of that input will halt, it just can't be done by HHH, as HHH wasn't defined to do that correct emulation.

 *Here is what the looks like*
 _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]
   machine   stack     stack     machine    assembly
  address   address   data      code       language
  ========  ========  ========  =========  =============
New slave_stack at:1038c4
Begin Local Halt Decider Simulation   Execution Trace Stored at:1138cc
[00002172][001138bc][001138c0] 55         push ebp      ; housekeeping
[00002173][001138bc][001138c0] 8bec       mov ebp,esp   ; housekeeping
[00002175][001138b8][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:14e2ec
[00002172][0015e2e4][0015e2e8] 55         push ebp      ; housekeeping
[00002173][0015e2e4][0015e2e8] 8bec       mov ebp,esp   ; housekeeping
[00002175][0015e2e0][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][0015e2dc][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
 

Sorry, but you are just painting yourself into the corner and revealing your fraud.