Re: Halting Problem: What Constitutes Pathological Input

Am Wed, 07 May 2025 23:17:58 -0500 schrieb olcott:

On 5/7/2025 10:08 PM, Richard Damon wrote:

On 5/7/25 9:51 AM, olcott wrote:

On 5/7/2025 4:48 AM, Mikko wrote:

On 2025-05-06 18:40:16 +0000, olcott said:

On 5/6/2025 10:53 AM, joes wrote:

Am Tue, 06 May 2025 10:29:59 -0500 schrieb olcott:

On 5/6/2025 4:35 AM, Mikko wrote:

No, it does not. The input is DD specifides exactly the same
sequence of steps as DD. HHH just answers about a different
sequence of steps instead of the the seqeunce specified by its
input.

As agreed to below:
>

*input D* is the actual input *would never stop running unless
aborted*
is the hypothetical H/D pair where H does not abort.

H should simulate its actual input D that calls the aborting H, not
a hypothetical version of D that calls a pure simulator.

*would never stop running unless aborted*
refers to the same HHH that DD calls yet this hypothetical HHH does
not abort.

Yes, this hypothetical HHH would not abort, but the actual HHH called
by the input still does.

You cannot possibly show the exact execution trace where DD is
correctly emulated by HHH and this emulated DD reaches past its
own machine address [0000213c].

Duh, no simulator can simulate itself correctly. But HHH1 can
simulate DD/HHH.

HHH does simulate itself correctly yet must create

No, it cannot simulate itself to the point where it returns.

HHH(DD) does return.

>

In other words HHH caught DD trying to cause itself to halt.

And what's wrong with that? HHH should let DD halt by itself.

HHH only need simulate DD until it correctly determines *simulated D
would never stop running unless aborted*
This refers to a hypothetical HHH/DD pair where HHH never aborts.

Right, that a correct simulation of D would never stop running,

No because that would make the termination analyzer fail to terminate
and that is not allowed dumbell bee.

Obviously you can't completely simulate a nonterminating input AND halt.

That includes using the fact that H does what it does, and the input is
what it is.
And thus NOT the hypothetical HHH/DD pair, but the hypothetical HHH
looking at the actual DD which still calls the origianl HHH

It is an inherent aspect of the nature of simulating termination
analyzer is that they must always correctly predict what the behavior of
their input would be IF THEY NEVER ABORTED. If they don't always do this
they will sometimes get stuck in non-termination.

If they, the simulator, didn't abort the unchanged input, i.e. that still
calls an aborting simulator.

You just don't understand what (correct) simulation means in the field,
or what a program actually is.

Everyone here seems to think they they get to stipulate what correct it,
rather what it really is.

Your HHH really is not a simulator. I want mine to produce the same
behaviour as direct execution.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.