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

Op 28.jun.2024 om 16:27 schreef olcott:

On 6/28/2024 3:23 AM, Fred. Zwarts wrote:

Op 27.jun.2024 om 19:30 schreef olcott:

>
When you prove that you are totally overwhelmed and confused
by the original issue I break it down into simpler steps.
>
If you don't have a slight clue about the C programming
language then the first step is you must learn this language
otherwise it is like trying to talk to someone about
differential calculus that does not know how to count to ten.

>
If... But since this if does not apply, the the is irrelevant.
You keep repeating irrelevant texts to hide that you cannot show any error in my reasoning.
>

>
typedef void (*ptr)();
int H0(ptr P);
>
void Infinite_Loop()
{
   HERE: goto HERE;
}
>
void Infinite_Recursion()
{
   Infinite_Recursion();
}
>
void DDD()
{
   H0(DDD);
}
>
int main()
{
   H0(Infinite_Loop);
   H0(Infinite_Recursion);
   H0(DDD);
}
>
Every C programmer that knows what an x86 emulator is knows that when H0
emulates the machine language of Infinite_Loop, Infinite_Recursion, and
DDD that it must abort these emulations so that itself can terminate
normally.
>
When this is construed as non-halting criteria then simulating
termination analyzer H0 is correct to reject these inputs as non-halting
by returning 0 to its caller.
>
Simulating termination analyzers must report on the behavior that their
finite string input specifies thus H0 must report that DDD correctly
emulated by H0 remains stuck in recursive simulation.
>

>
Another attempt to distract from the subject.You claim you are not talking about halt-deciders or termination analyzers, but now you bring them up again.
>

 https://liarparadox.org/HHH(DDD)_Full_Trace.pdf
 I only do this because you have gotten overwhelmed.
I prove my point step-by-step and because you don't
understand any of the steps you leap to the conclusion
that I am wrong.
 

We are discussing an H0 that aborts after two cycles. I do not tolerate to go away from this point.
>

 I updated all of my names in my code.
// HHH(DDD) and HHH1(DDD) are the standard names for DDD input
// DDD calls HHH(DDD). HHH1 is identical to HHH.
 // HH(DD,DD) and HH1(DD,DD) are the standard names for (DD,DD) input
// DD calls HH(DD,DD) and HH1 is identical to HH.
  You haven't shown that you even understand that Infinite_Recursion()
doesn't halt. You must understand this before you can understand
the more complex example of DDD.

We agreed to talk only about the simulator which aborts after two cycles of recursive simulation.
You are unable to show any error in what I said, so now your are trying to bring up an infinite recursion again.
Is a two cycle recursion already over your head?
Is is too difficult for you to stay at the point?
I cannot tolerate any distraction from the point, because then we will run in circles again.
So, stay at the point. We need to agree about finite recursion before we can go to the next step of infinite recursion.

 

The examples of Infinite_Loop and Infinite_Recursion do not apply. They are completely different from an H0 that aborts after two cycles.
>
The last example can be formulated much simpler:
>
        int main()
        {
          return H(main, 0);
        }
>

 That example is merely a more difficult to understand
version of HHH(DDD). If you can't even understand that
Infinite_Recursion() doesn't halt you won't be able
to understand the more complex example of DDD.
 

No, HHH(DDD) is a more difficult example. DDD is completely superfluous. But you like more complex examples to hide the simple fact you do not even understand the simple ones.