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

On 4/23/2025 12:23 PM, Fred. Zwarts wrote:

Op 23.apr.2025 om 17:53 schreef olcott:

On 4/23/2025 3:59 AM, Fred. Zwarts wrote:

Op 22.apr.2025 om 21:24 schreef olcott:

On 4/22/2025 7:42 AM, joes wrote:

Am Mon, 14 Apr 2025 17:48:36 -0500 schrieb olcott:

On 4/14/2025 4:29 AM, joes wrote:

Am Sun, 13 Apr 2025 16:00:43 -0500 schrieb olcott:

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:

>

No proof of this principle has been shown so its use is not
valid.

No proof of Peano's axioms or Euclid's fifth postulate has been
shown.
That doesn't mean we can't use them.
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.
As anyone can see HHH MUST REJECT ITS INPUT OR GET STUPIDLY STUCK IN
NON-TERMINATION. If people were not mindless robots they would have
immediately acknowledged this years ago.

But why does it not return „I know this halts, but I can’t simulate
it”?

Because it is not a liar and tells the truth for every input in its
domain.

Aha. Then why does it not simulate it and say that it halts?
>

>
I have proven that the directly executed DD and DD
emulated by HHH according to the semantics of the
x86 language have a different set of state changes
many hundreds of times for several years.
>

>
You did not prove it, you dreamed about it for many years, but you failed to show the first state change were the simulation differs from the simulation. You only showed that HHH failed to complete the simulation.

>
_DD()
[00002133] 55         push ebp      ; housekeeping
[00002134] 8bec       mov ebp,esp   ; housekeeping
[00002136] 51         push ecx      ; make space for local
[00002137] 6833210000 push 00002133 ; push DD
[0000213c] e882f4ffff call 000015c3 ; call HHH(DD)
[00002141] 83c404     add esp,+04
[00002144] 8945fc     mov [ebp-04],eax
[00002147] 837dfc00   cmp dword [ebp-04],+00
[0000214b] 7402       jz 0000214f
[0000214d] ebfe       jmp 0000214d
[0000214f] 8b45fc     mov eax,[ebp-04]
[00002152] 8be5       mov esp,ebp
[00002154] 5d         pop ebp
[00002155] c3         ret
Size in bytes:(0035) [00002155]
>
By merely knowing that HHH emulates DD until it
sees itself about to emulate DD a third time
(mathematical induction proof that DD is stuck in
recursive emulation) we can know that

 We know that the simulating HHH fails to see that the recursion is finite

 From the point of view of the HHH that is simulating DD
the recursion will never stop unless this HHH stops it.
On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
 > I don't think that is the shell game.  PO really /has/ an H
 > (it's trivial to do for this one case) that correctly determines
 > that P(P) *would* never stop running *unless* aborted.
...
 > But H determines (correctly) that D would not halt if it
 > were not halted.  That much is a truism.
That you don't know software engineering well enough to
see this is less than no rebuttal at all.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer