Re: This function proves that only the outermost HHH examines the execution trace

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Sujet : Re: This function proves that only the outermost HHH examines the execution trace
De : acm (at) *nospam* muc.de (Alan Mackenzie)
Groupes : comp.theory
Date : 27. Jul 2024, 22:16:09
Autres entêtes
Organisation : muc.de e.V.
Message-ID : <v83o2p$2nhr$3@news.muc.de>
References : 1 2 3 4 5 6 7 8 9
User-Agent : tin/2.6.3-20231224 ("Banff") (FreeBSD/14.1-RELEASE (amd64))
olcott <polcott333@gmail.com> wrote:
On 7/27/2024 3:20 PM, Alan Mackenzie wrote:
olcott <polcott333@gmail.com> wrote:
On 7/27/2024 1:14 PM, Alan Mackenzie wrote:
olcott <polcott333@gmail.com> wrote:

Stopping running is not the same as halting.
DDD emulated by HHH stops running when its emulation has been aborted.
This is not the same as reaching its ret instruction and terminating
normally (AKA halting).

I think you're wrong, here.  All your C programs are a stand in for
turing machines.  A turing machine is either running or halted.  There is
no third state "aborted".

Until you take the conventional ideas of
(a) UTM
(b) TM Description
(c) Decider
and combine them together to become a simulating partial halt decider.

Where does the notion of "aborted", as being distinct from halted, come
from?


After all of these years and you don't get that?

"Aborted" being distinct from halted is an incoherent notion.  It isn't
consistent with turing machines.  I was hoping you could give a
justification for it.

A simulating partial halt decider can stop simulating
its input when it detects a non-halting behavior pattern.
This does not count as the input halting.

Says who?  Well, OK, it would be the machine halting, not the input, but
that's a small point.

Why do you say that this cessation of simulation doesn't count as the
machine halting?  As the machine is no longer running, it must be in the
halted state.

Maybe one of the experts in the group might like to chime in here, and
explain what, if anything, I've misunderstood here.

The key difference between a partial decider and a decider is that
the former case only needs to get at least one input correctly.

That doesn't seem to have anything to do with my point.


*This is indirectly related to your above question*
A decider must be all knowing it is not allowed to get one
input incorrectly. A partial halt decider is only required
to get at least one input correctly.

That doesn't seem to have anything to do with my point, either.  :-(

--
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).


Date Sujet#  Auteur
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