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

[ 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.

DDD correctly emulated by H0 DOES NOT TERMINATE NORMALLY.

There is no concept of "normal" termination in a turing machine.  The
thing is either running or it's halted.

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.

As has often been said, it is then no longer a universal turing machine.

When this UTM stops simulating this Turing machine description we
cannot correctly say that this looping input halted.

Yes, we can.  It has been designed to count to 42 then halt.  It is then
in the halted state.

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

Indeed, not.

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

--
Alan Mackenzie (Nuremberg, Germany).