Re: How do simulating termination analyzers work?

On 6/18/25 12:01 PM, olcott wrote:

On 6/18/2025 10:50 AM, joes wrote:

Am Wed, 18 Jun 2025 09:50:18 -0500 schrieb olcott:

On 6/18/2025 9:05 AM, joes wrote:

Am Wed, 18 Jun 2025 08:46:16 -0500 schrieb olcott:

*It is not given that any of them abort*

Huh? They contain the code to abort, even if it is not simulated.
>

*none of them ever stop running unless aborted* yes or no?

 

If HHH(DDD), where DDD() only calls HHH(DDD), is simulated by a pure
simulator such as HHH1 (not by itself, which aborts), it stops running
by aborting (the simulator also terminates). All inner invocations of
HHH would have stopped running had the outermost one (which is still
only simulated by the real HHH1) not stopped simulating.
HHH1 simulating HHH1 of course doesn't stop running, but we don't have
HHH1 involved in either role here. (For completeness, HHH1(HHH) and
HHH(HHH1) also halt.)
>

 *That is a great answer to the wrong question*
*Here is the original question again*

No. the REAL original question is:
Does the machine described by the input halt when run?
Your STRAWMAN just shows your brain is made of straw.

 void Infinite_Recursion()
{
   Infinite_Recursion();
   return;
}
 void Infinite_Loop()
{
   HERE: goto HERE;
   return;
}
 void DDD()
{
   HHH(DDD);
   return;
}
 When it is understood that HHH does simulate itself
simulating DDD then any first year CS student knows
that when each of the above are correctly simulated
by HHH that none of them ever stop running unless aborted.
 Can any of the above functions correctly simulated
by HHH ever stop running without ever being aborted?
 

Since HHH never correctly simulated DDD, your statement is just in error.