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

On 4/14/2025 12:15 PM, Fred. Zwarts wrote:

Op 14.apr.2025 om 13:56 schreef olcott:

On 4/14/2025 4:25 AM, joes wrote:

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.
>

>
When people insist that a termination analyzer reports
on behavior other than the behavior that its finite string
input specifies this is isomorphic to requiring a perfectly
geometric square circle in the same two dimensional plane,
simply logically impossible, thus an incorrect requirement.

 It is very clear what its finite string input specifies: when exactly this same finite string input is used in direct execution or in world- class simulators,

And by this same reasoning we can say that
int sum(int x, int y) { return x + y; }
returns 7 for sum(2,3) because it returns 7 for sum(5,2).

we see that it specifies a halting program according to the unique semantics of the x86 language.
It is not clear what a geometric square circle is. So, your comparison fails.
But I think we agree that there is no algorithm that can determine for all possible inputs whether the input specifies a program that (according to the semantics of the machine language) halts when directly executed. Correct?
 

--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer