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

Op 02.mrt.2025 om 20:06 schreef olcott:

On 3/2/2025 12:38 PM, Fred. Zwarts wrote:

Op 02.mrt.2025 om 15:42 schreef olcott:

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.

>
Olcott could as well challenge everyone to draw a correct square circle.
>

>
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.
>

>
No one will attempt to draw a square circle. Does that imply that it is wrong to say that another failed attempt to draw a square circle is incorrect?
>
Similarly, HHH cannot possibly simulate itself correctly.

 This C code conclusively proves that HHH does correctly
emulate self emulating DD correctly.
https://github.com/plolcott/x86utm/blob/master/Halt7.c

I think we can agree on the following:
It is clear that this HHH cannot possibly reach the 'ret' instruction of DD in the finite string that (when given for direct execution, or to another simulator, like HHH1) has been proven to reach the 'ret' instruction without problem.
So, we can agree that HHH correctly reports that this DD is not computable for HHH. (I.e., that a correct simulation cannot be done by HHH.)
Well done! Olcott has been able to construct a simulator that reports for certain inputs that it cannot possibly do the computation.
We agree that DD is not computable for HHH. What is the next step of the proof?
Face it.
There is a finite string DD that
1) When given for direct execution reaches its 'ret' instruction. (Computable by direct execution.)
2) When given to many simulators (including HHH1) reaches its 'ret' instruction. (Computable by these simulators.)
3) When given to a yellow carrot does not even start to run. (Not computable by a yellow carrot.)
4) When given to HHH starts, but does not reach its 'ret' instruction. (Not computable by HHH.)
Since Olcott already admitted that DD reaches the 'ret' instruction in the direct execution and the simulation by HHH1 we reached complete agreement. HHH is correct to report that DD is not computable by HHH. What is the next step of the proof?

 That you lack the technical capability to understand
that code is far less than no actual rebuttal at all.
 

Irrelevant remark, showing Olcott's lack of knowledge of my technical capabilities.