Re: Incorrect requirements --- Computing the mapping from the input to HHH(DD)

On 5/9/25 11:03 AM, olcott wrote:

On 5/9/2025 1:17 AM, Keith Thompson wrote:

olcott <polcott333@gmail.com> writes:

On 5/9/2025 12:31 AM, Keith Thompson wrote:

[...]

You didn't respond to the above.  I'll ask directly.
Would I need to understand x86 code to understand your claims?  Yes
or no.

>
It makes it much easier because the state transition graph
of the control flow at the x86 level is unequivocal.

>
What a pity you couldn't have made that clear much sooner.
>

 Your help on the C part was very useful.
The x86 part is only needed to understand
the internals of the simulating termination
analyzer.
 _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]
 You don't need to understand the first two instructions.
The next two instructions simply call HHH(DDD) in
recursive emulation.
 At the x86 machine code level HHH can see that the
first four instructions of DDD repeats. It sees this
after it emulates DDD and then emulates itself
emulating DDD once.
 

As I said, I'm not very familiar with x86 code, and it's
absolutely not worth my time and effort to learn it for the sake
of understanding your claims.
>
Bye.
>

 

And, since you say that is *ALL* of the input, it is CATEGORICALLY IMPOSSIBLE for a correct emulator to emulate the instruction, as it is missing the details.
Sorry, but you are just showing that youy are so dense that you are unable to learn basic definitions in programing.