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

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

On 6/30/2024 5:42 AM, Fred. Zwarts wrote:

Op 29.jun.2024 om 22:03 schreef olcott:

On 6/29/2024 2:44 PM, joes wrote:

Am Fri, 28 Jun 2024 14:28:20 -0500 schrieb olcott:

On 6/28/2024 2:18 PM, joes wrote:

Am Fri, 28 Jun 2024 12:53:46 -0500 schrieb olcott:

On 6/28/2024 12:41 PM, joes wrote:

Thanks for leaving the unanswered questions in place, though I’d
rather have you answer them.
Am Fri, 28 Jun 2024 12:05:18 -0500 schrieb olcott:

On 6/28/2024 11:26 AM, joes wrote:

Am Fri, 28 Jun 2024 10:25:36 -0500 schrieb olcott:

On 6/28/2024 8:14 AM, joes wrote:

Am Thu, 27 Jun 2024 12:30:38 -0500 schrieb olcott:

>

To the caller DDD, which then returns to its own caller H0, which
returns „halting” to main… hold on.

Why doesn’t the first recursive H return?

Question still not answered.
>

HHH(DDD)
simulates DDD that calls HHH(DDD) that simulates DDD that calls HHH(DDD)
that proves to the outer directly executed HHH that it must abort and
reject.

>

How does it do that?

Over your head. I have explained it too many times
and you just can't get it.
>
Technically it is called detecting a repeating state.
I don't mean "New Jersey, New Jersey".
>

>
But a repeating state is not by definition an infinitely repeated state.
In case of HHH we know it is repeated only twice.
But it is over olcott's head that two is different from infinite.

 The call from DDD to HHH(DDD) when N steps of DDD are
correctly emulated by any pure function x86 emulator
HHH at machine address 0000217a cannot possibly return.
 _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.