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From: wij <wyniijj5@gmail.com>
Newsgroups: comp.theory
Subject: Re: Incorrect requirements --- Computing the mapping from the input
 to HHH(DD)
Date: Sun, 11 May 2025 02:09:49 +0800
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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:
> > > > > > > > > >=20
> > > > > > > > > > > Look at their replies to this post.
> > > > > > > > > > > Not a one of them will agree that
> > > > > > > > > > >=20
> > > > > > > > > > > void DDD()
> > > > > > > > > > > {
> > > > > > > > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 HHH(DDD);
> > > > > > > > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 return; // final=
 halt state
> > > > > > > > > > > }
> > > > > > > > > > >=20
> > > > > > > > > > > When 1 or more instructions of DDD are correctly
> > > > > > > > > > > simulated by HHH then the correctly simulated DDD can=
not
> > > > > > > > > > > possibly reach its "return" instruction (final halt s=
tate).
> > > > > > > > > > >=20
> > > > > > > > > > > They have consistently disagreed with this
> > > > > > > > > > > simple point for three years.
> > > > > > > > > >=20
> > > > > > > > > > I guess that not even a professor of theoretical comput=
er
> > > > > > > > > > science would spend years working on so few lines of co=
de.
> > > > > > > > > >=20
> > > > > > > > >=20
> > > > > > > > > I created a whole x86utm operating system.
> > > > > > > > > It correctly determines that the halting problem's
> > > > > > > > > otherwise "impossible" input is actually non halting.
> > > > > > > > >=20
> > > > > > > > > int DD()
> > > > > > > > > {
> > > > > > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 int Halt_Status =3D =
HHH(DD);
> > > > > > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 if (Halt_Status)
> > > > > > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 HERE: go=
to HERE;
> > > > > > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 return Halt_Status;
> > > > > > > > > }
> > > > > > > > >=20
> > > > > > > > > https://github.com/plolcott/x86utm
> > > > > > > > >=20
> > > > > > > >=20
> > > > > > > > Nope.
> > > > > > > > =C2=A0=C2=A0=C2=A0=C2=A0From I know HHH(DD) decides whether=
 the input DD is "impossible" input or not.
> > > > > > > >=20
> > > > > > >=20
> > > > > > > DD has the standard form of the "impossible" input.
> > > > > > > HHH merely rejects it as non-halting.
> > > > > > >=20
> > > > > >=20
> > > > > > You said 'merely' rejects it as non-halting.
> > > > > > So, POOH do not answer the input of any other function?
> > > > > >=20
> > > > >=20
> > > > > 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.
> > > > >=20
> > > > > The same thing applies to the Linz proof yet cannot
> > > > > be understood until after HHH(DDD) and HHH(DD) are
> > > > > fully understood.
> > > > >=20
> > > >=20
> > > > HHH(DDD) (whatever) at most says DDD is a pathological/midtaken inp=
ut.
> > > > Others of what you say are your imagine and wishes, so far so true.
> > > >=20
> > >=20
> > > 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.
> > >=20
> >=20
> > 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..
> >=20
>=20
> 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.=20
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'=20
will be an infinite recursive which cannot make any decision, which=C2=A0
you had agreed)