Re: Incorrect requirements --- Computing the mapping from the input to HHH(DD)

On 5/10/2025 6:03 PM, olcott wrote:

On 5/10/2025 4:44 PM, wij wrote:

On Sat, 2025-05-10 at 14:29 -0500, olcott wrote:

On 5/10/2025 2:02 PM, wij wrote:

On Sat, 2025-05-10 at 13:47 -0500, olcott wrote:

On 5/10/2025 1:37 PM, wij wrote:

On Sat, 2025-05-10 at 13:17 -0500, olcott wrote:

On 5/10/2025 1:09 PM, wij wrote:

On Sat, 2025-05-10 at 12:17 -0500, olcott wrote:

On 5/10/2025 12:01 PM, wij wrote:

On Sat, 2025-05-10 at 11:47 -0500, olcott wrote:

On 5/10/2025 11:29 AM, wij wrote:

On Sat, 2025-05-10 at 11:19 -0500, olcott wrote:

On 5/10/2025 11:06 AM, wij wrote:

On Sat, 2025-05-10 at 10:45 -0500, olcott wrote:

On 5/10/2025 10:28 AM, wij wrote:

On Sat, 2025-05-10 at 09:33 -0500, olcott wrote:

On 5/10/2025 7:37 AM, Bonita Montero wrote:

Am 09.05.2025 um 04:22 schrieb olcott:
>

Look at their replies to this post.
Not a one of them will agree that
>
void DDD()
{
           HHH(DDD);
           return; // final halt state
}
>
When 1 or more instructions of DDD are correctly
simulated by HHH then the correctly simulated DDD cannot
possibly reach its "return" instruction (final halt state).
>
They have consistently disagreed with this
simple point for three years.

>
I guess that not even a professor of theoretical computer
science would spend years working on so few lines of code.
>

>
I created a whole x86utm operating system.
It correctly determines that the halting problem's
otherwise "impossible" input is actually non halting.
>
int DD()
{
           int Halt_Status = HHH(DD);
           if (Halt_Status)
             HERE: goto HERE;
           return Halt_Status;
}
>
https://github.com/plolcott/x86utm
>

>
Nope.
        From I know HHH(DD) decides whether the input DD is "impossible"
input
or
not.
>

>
DD has the standard form of the "impossible" input.
HHH merely rejects it as non-halting.
>

>
You said 'merely' rejects it as non-halting.
So, POOH do not answer the input of any other function?
>

>
The input that has baffled computer scientists for 90
years is merely correctly determined to be non-halting
when the behavior of this input is measured by HHH
emulating this input according to the rules of the x86
language.
>
The same thing applies to the Linz proof yet cannot
be understood until after HHH(DDD) and HHH(DD) are
fully understood.
>

>
HHH(DDD) (whatever) at most says DDD is a pathological/ midtaken input.
Others of what you say are your imagine and wishes, so far so true.
>

>
DDD emulated by HHH accor not the 'HHH' that makes the final decision

(otherwise, it will be an infinite recursive call which you agreed)
>

ding to the rules of
the x86 language specifies recursive emulation
that cannot possibly reach the final halt state
of DDD.
>

>
I have no problem with that. And, you said HHH merely rejects it as non-halting.
You had denied HHH can decide the halting property of any input, except DDD/DD/D..
>

>
As long as HHH correctly determines the halt status
of a single input that has no inputs then HHH is
a correct termination analyzer for that input.

>
Go it, that is a stronger statement that HHH ONLY decides DD.
I have no problem with that, but be noticed that the HHH inside DD
is not the 'HHH' that makes the final decision (otherwise, the 'HHH'
will be an infinite recursive which cannot make any decision, which
you had agreed)
>

>
HHH(DD) correctly determines that its input specifies
recursive emulation when this input is emulated by HHH
HHH according to the rules of the x86 language.

>
   From the about, so you are talking about 'the HHH' which does not compute the final
decision.
>

>
HHH does recognize the recursive emulation pattern
of DDD emulated by HHH according to the rules of
the x86 language.
>

*Thus exactly meets the following specification*
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
        If simulating halt decider H correctly simulates its
        input D until H correctly determines that its simulated D
        would never stop running unless aborted then
>
        H can abort its simulation of D and correctly report that D

>
This H won't be the same HHH inside the DD, otherwise an infinite recursive call happens.
>

>
It must always be the outermost HHH that does this
because it has seen one entire recursive emulation
more than the next inner HHH.

>
No problem. H is not HHH.
>

>
The H is the template that Professor Sipser agreed to.
HHH is a specific implementation of H.
>

This is also a pitty no one here understand POOH can help AI industry and mankind, even so mini.
>

>
It is the same halting problem after its mistake
has been corrected. So just like how ZFC corrected
the error in set theory so that Russell's Paradox
could be correctly decided, HHH corrects the error
in the halting problem proof so that the otherwise
impossible input is correctly decided.

>
I don't know what that part of set theory works.
(My feeling is that they are garbage, for reasons,
unless you are doing logic researches)
>

The original set theory is now called naive set
theory after its mistake has been corrected. Thus
the original halting problem proofs can now be
called the naive halting problem proofs.

>
Traditional logic (or the part mostly used) that won't
cause confusion is more reliable.
>

The halting problem itself remains the same, yet
loses its most important proof.

>
HP is based on TM. Proof of any other kind other than TM have to be cautious.

 Unless this is done as an actual simulating termination
analyzer

That assumes one exists.  Linz and others proved it doesn't, and you have *explicitly* agreed with that.
On 5/5/2025 5:39 PM, olcott wrote:
 > On 5/5/2025 4:31 PM, dbush wrote:
 >> Strawman.  The square root of a dead rabbit does not exist, but the
 >> question of whether any arbitrary algorithm X with input Y halts when
 >> executed directly has a correct answer in all cases.
 >>
 >
 > It has a correct answer that cannot ever be computed