Re: Sequence of sequence, selection and iteration matters

Op 09.jul.2024 om 23:18 schreef olcott:

On 7/9/2024 3:31 PM, Fred. Zwarts wrote:

>
You understand x86 insufficiently, because you think that a two cycle recursion means an infinite recursion.
>

 _DDD()
[00002163] 55         push ebp      ; housekeeping
[00002164] 8bec       mov ebp,esp   ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
 I am saying that DDD correctly emulated by HHH
can't freaking make it past the above line of
code no freaking matter what.

Which proves that HHH does not interpret x86 code correctly. So, the set of HHH that correctly emulates is empty.
It performs an abort halfway the simulation, where the x86 code does not allow it.
You can repeat it hundred times, but that does not make it correct.
HHH cannot possibly simulate itself correctly.
You may want very very much that it is correct, but that does not change the situation that it is incorrect.
A correct simulator shows that HHH *does* abort and halt. So, when HHH is unable to reproduce it, it only shows that its simulation is incorrect.
A two cycle recursion is not an infinite recursion. Is that already over your head?
void Finite_Recursion (int N) {
   if (N > 0) Finite_Recursion (N - 1);
}
Maybe it is too complex for you, because you add an unneeded function DDD. Without DDD it is much simpler to understand:
        int main()
        {
          return HHH(main);
        }
It is clear that HHH returns, but that its simulation thinks it does not halt, which makes it simply incorrect.
So, Sipser was right, but he was talking about a correct simulation, not about the incorrect simulation of HHH.

 The chances of this are the same as a billion
geometric square circles hitting you in the face
and knocking you on your ass.

Irrelevant text ignored.

 [00002170] 83c404     add esp,+04
[00002173] 5d         pop ebp
[00002174] c3         ret
Size in bytes:(0018) [00002174]