Re: DDD simulated by HHH cannot possibly halt (Halting Problem) --- mindless robots

Liste des GroupesRevenir à cl c 
Sujet : Re: DDD simulated by HHH cannot possibly halt (Halting Problem) --- mindless robots
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theory
Date : 14. Apr 2025, 10:25:01
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <63af93cb608258cc3e12b9bab3a2efa0b7ee7eee@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Sun, 13 Apr 2025 21:11:56 -0500 schrieb olcott:
On 4/13/2025 6:11 PM, Richard Damon wrote:
On 4/13/25 5:00 PM, olcott wrote:
On 4/13/2025 3:00 PM, dbush wrote:
On 4/13/2025 3:59 PM, olcott wrote:
On 4/13/2025 3:54 AM, joes wrote:
Am Fri, 11 Apr 2025 10:56:32 -0500 schrieb olcott:
On 4/11/2025 3:24 AM, Richard Heathfield wrote:
On 11/04/2025 08:57, Mikko wrote:

Mr Olcott can have his principle if he likes, but only by EITHER
proving it (which, as you say, he has not yet done) OR by taking
it as axiomatic, leaving the world of mainstream computer science
behind him,
constructing his own computational 'geometry' so to speak, and
abandoning any claim to having overturned the Halting Problem.
Navel contemplation beckons.
Axioms are all very well, and he's free to invent as many as he
wishes, but nobody else is obliged to accept them.
>
*Simulating termination analyzer Principle*
It is always correct for any simulating termination analyzer to
stop simulating and reject any input that would otherwise prevent
its own termination.
Sure. Why doesn’t the STA simulate itself rejecting its input?
>
Because that is a STUPID idea and categorically impossible because
the outermost HHH sees its needs to stop simulating before any inner
HHH can possibly see this.
>
In other words, you agree that Linz and others are correct that no H
exists that satisfies these requirements:
Given any algorithm (i.e. a fixed immutable sequence of instructions)
X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes the
following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed
directly
>
No stupid! Those freaking requirements are wrong and*
anchored in the ignorance  of rejecting the notion of a simulating
termination analyzer OUT-OF-HAND WITHOUT REVIEW.
No, those "freeking requirement" *ARE* the requirements
AND AS STUPID AS {REQUIRING} A GEOMETRIC SQUARE CIRCLE IN THE SAME
TWO-DIMENSIONAL PLANE.
Nothing is stupid about wanting a halt decider. It’s just not obvious
that it’s impossible.

--
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.

Date Sujet#  Auteur
24 May 25 o 

Haut de la page

Les messages affichés proviennent d'usenet.

NewsPortal