Re: 195 page execution trace of DDD correctly simulated by HH0

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

On 6/27/2024 4:45 AM, Fred. Zwarts wrote:

Op 26.jun.2024 om 15:30 schreef olcott:

On 6/26/2024 3:01 AM, Fred. Zwarts wrote:

Op 25.jun.2024 om 21:30 schreef olcott:

On 6/25/2024 2:17 PM, Fred. Zwarts wrote:

>
It might be true, but it is irrelevant, because the simulated H0 is aborted prematurely. The simulating H0 aborts after two cycles,

>
*I am not even talking about a simulating halt decider yet dumbo*

>
Neither am I. Why do you mention a simulating halt decider? (Who is the dumbo?)
>

>
Simulated H0 is never aborted.
Only termination analyzers do that.

>
So, you change subject again.

 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.
We are discussing an H0 that aborts after two cycles. I do not tolerate to go away from this point.
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);
        }
For which you proved that H returns after two cycles, although the simulated H did not return, because it was aborted prematurely when it had only one cycle to go.
This is exactly the same as in the DDD example. There is no infinite recursion, only a H0, the simulation of which is aborted when it had only one cycle to go.
Only an infinite simulation needs to be aborted.
There is no need to abort a simulation of an aborting simulation. So, H0 has no need to abort itself. It only needs to abort another Hx, that does not abort. Saying that it must abort is showing insufficient insight in the verified facts.