Re: How do simulating termination analyzers work? (in C and C++)

On 6/19/2025 11:47 AM, Bonita Montero wrote:

Am 19.06.2025 um 16:38 schrieb olcott:

On 6/19/2025 8:25 AM, Bonita Montero wrote:

Am 19.06.2025 um 01:04 schrieb olcott:
>

I don't need any more than the behavior of the posted
C functions.

>
Your question is generic to any language.
>

>
*Non-the-less I still need the answer*
>
You could just give me this answer,
(its very simple and won't take ten minutes)
or as they have done on comp.theory spend three
years and thousands of messages dodging the question.
>
void Infinite_Recursion()
{
   Infinite_Recursion();
   return;
}
>
void Infinite_Loop()
{
   HERE: goto HERE;
   return;
}
>
void DDD()
{
   HHH(DDD);
   return;
}
>
int Sipser_D()
{
   if (HHH(Sipser_D) == 1)
     return 0;
   return 1;
}
>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
My claim is that each of the above functions correctly
simulated by any termination analyzer HHH that can possibly
exist will never stop running unless aborted by HHH.
Can you affirm or correctly refute this?
>
*No one has ever been able to refute this in three years*
>
*Fully operational code*
https://github.com/plolcott/x86utm/blob/master/Halt7.c

 You deal with a simple question for decades.
That's crazy.

It is a very important simple question, yet the
reason why it is very important is outside of
the scope of this group.
You can spend hundreds of posts saying why you
won't answer the question or one post and ten
minutes answering it.
The people on comp.theory spent thousands of
posts and three years saying why they won't
answer it. That is the *only* reason why I
am here.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer