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

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.
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.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer