Re: Simulating termination analyzers by dummies --- What does halting mean?

On 6/18/2024 12:57 PM, joes wrote:

Am Tue, 18 Jun 2024 12:25:44 -0500 schrieb olcott:

On 6/18/2024 12:06 PM, joes wrote:
void DDD()
{
    H0(DDD);
}
DDD correctly simulated by any H0 cannot possibly halt.

DDD halts iff H0 halts.

So H0 returns "doesn't halt" to DDD, which then stops running,
so H0 should have returned "halts".
 

This was three messages ago.
I had to make sure that you understood that halting
does not mean stopping for any reason and only includes
the equivalent of terminating normally.
DDD correctly emulated by H0 DOES NOT TERMINATE NORMALLY.

Some TM's loop and thus never stop running, this is classical
non-halting behavior. UTM's simulate Turing machine descriptions.
This is the same thing as an interpreter interpreting the source-code of
a program.

Some TMs do not loop and do not halt.
 

A UTM can be adapted so that it only simulates a fixed number of
iterations of an input that loops. When this UTM stops simulating this
Turing machine description we cannot correctly say that this looping
input halted.

Yes. We also cannot say that that input was simulated correctly.
 

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