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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: Sat, 10 May 2025 23:28:03 +0800
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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 HHH(DDD);
> > > =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 cannot
> > > possibly reach its "return" instruction (final halt state).
> > >=20
> > > They have consistently disagreed with this
> > > simple point for three years.
> >=20
> > I guess that not even a professor of theoretical computer
> > science would spend years working on so few lines of code.
> >=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 int Halt_Status =3D HHH(DD);
> =C2=A0=C2=A0 if (Halt_Status)
> =C2=A0=C2=A0=C2=A0=C2=A0 HERE: goto HERE;
> =C2=A0=C2=A0 return Halt_Status;
> }
>=20
> https://github.com/plolcott/x86utm
>=20

Nope.=C2=A0
From I know HHH(DD) decides whether the input DD is "impossible" input or n=
ot.