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

On 2024-06-19 14:05:29 +0000, olcott said:

On 6/19/2024 4:29 AM, Alan Mackenzie wrote:

olcott <polcott333@gmail.com> wrote:

On 6/18/2024 4:36 PM, Alan Mackenzie wrote:

[ Followup-To: set ]

 

In comp.theory olcott <polcott333@gmail.com> wrote:

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.

 

No.  You're wrong, here.  A turing machine is either running or it's
halted.  There's no third alternative.  If your C programs are not in one
of these two states, they're not equivalent to turing machines.

 

Although I agree with this there seems to be nuances of
disagreement across the experts.

 I doubt that very much.  The whole point of turing machines is to remove
ambiguity and unneeded features from the theory of computation.  A third
alternative state is unneeded.
 

 Some people say that a TM can halt in a non-final state.

People may use different words to express the same facts. What some
people call "halting in a non-final state" is called "rejecting" by
some other people. But the facts are what they are independently of
the words used to express them.
--
Mikko