Re: Failure to meet this challenge proves that all of my reviewers are wrong

On 3/3/2025 6:42 AM, Mikko wrote:

On 2025-03-02 14:42:03 +0000, olcott said:
 

HHH is an emulating termination analyzer that emulates
the x86 machine code located at the address of a function
using a fully functional x86 emulator.
>
When HHH recognizes a non-terminating pattern in the
execution trace of its emulated input it aborts this
emulation and returns 0.
>
typedef void (*ptr)();
int HHH(ptr P);
>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
_DD()
[00002133] 55         push ebp      ; housekeeping
[00002134] 8bec       mov ebp,esp   ; housekeeping
[00002136] 51         push ecx      ; make space for local
[00002137] 6833210000 push 00002133 ; push DD
[0000213c] e882f4ffff call 000015c3 ; call HHH(DD)
[00002141] 83c404     add esp,+04
[00002144] 8945fc     mov [ebp-04],eax
[00002147] 837dfc00   cmp dword [ebp-04],+00
[0000214b] 7402       jz 0000214f
[0000214d] ebfe       jmp 0000214d
[0000214f] 8b45fc     mov eax,[ebp-04]
[00002152] 8be5       mov esp,ebp
[00002154] 5d         pop ebp
[00002155] c3         ret
Size in bytes:(0035) [00002155]
>
I challenged everyone here to provide the machine address
by machine address (AKA line by line) execution trace
of DD correctly emulated by HHH that reaches its own
"ret" instruction.
>
No one made any attempt to do this because they know that
this would prove that they are stupidly wrong to say that
my trace is incorrect.

 Wrong conclusion. That sometthing is not proven only means it is unproeven.
It may be correct, it may be incorrect. Without a proof we can only say
that you claim more than you know.
 

That you do not understand this code is much less
than no rebuttal at all.
https://github.com/plolcott/x86utm/blob/master/Halt7.c
The above code proves that:
(a) HHH correctly emulates itself emulating DD.
(b) DD correctly emulated by HHH cannot possibly
reach its own "ret" instruction and terminate normally.
(c) The behavior of the input to HHH(DD) is different
than the behavior of the directly executed DD because
DD calls HHH(DD) in recursive emulation and the directly
executed DD does not call HHH(DD) in recursive emulation.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer