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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: Mon, 12 May 2025 00:28:23 +0800
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On Sun, 2025-05-11 at 10:38 -0500, olcott wrote:
> On 5/11/2025 9:34 AM, wij wrote:
> > On Sat, 2025-05-10 at 21:19 -0500, olcott wrote:
> > > On 5/10/2025 9:09 PM, wij wrote:
> > > > On Sat, 2025-05-10 at 20:56 -0500, olcott wrote:
> > > > > On 5/10/2025 8:44 PM, wij wrote:
> > > > > > On Sat, 2025-05-10 at 20:26 -0500, olcott wrote:
> > > > > > > On 5/10/2025 8:17 PM, wij wrote:
> > > > > > > > On Sat, 2025-05-10 at 17:03 -0500, 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:
> > > > > > >=20
> > > > > > > >=20
> > > > > > > > You don't know the counter example in the HP proof, your D =
is not the case what HP
> > > > > > > > says.
> > > > > > > >=20
> > > > > > >=20
> > > > > > > Sure I do this is it! (as correctly encoded in C)
> > > > > > >=20
> > > > > > > typedef void (*ptr)();
> > > > > > > int HHH(ptr P);
> > > > > > >=20
> > > > > > > int DD()
> > > > > > > {
> > > > > > > =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 if (Halt_Status)
> > > > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 HERE: goto HERE;
> > > > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 return Halt_Status;
> > > > > > > }
> > > > > > >=20
> > > > > > > int main()
> > > > > > > {
> > > > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 HHH(DD);
> > > > > > > }
> > > > > > >=20
> > > > > > >=20
> > > > > >=20
> > > > > > Try to convert it to TM language to know you know nothing.
> > > > > >=20
> > > > >=20
> > > > > I spent 22 years on this. I started with the Linz text
> > > > >=20
> > > > > When =C4=A4 is applied to =E2=9F=A8=C4=A4=E2=9F=A9
> > > > > =C4=A4.q0 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* embedded_H =E2=9F=
=A8=C4=A4=E2=9F=A9 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* =C4=A4.qy =E2=88=9E
> > > > > =C2=A0=C2=A0=C2=A0=C2=A0 or
> > > > > =C4=A4.q0 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* embedded_H =E2=9F=
=A8=C4=A4=E2=9F=A9 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* =C4=A4.qn
> > > > >=20
> > > > > (a) =C4=A4 copies its input =E2=9F=A8=C4=A4=E2=9F=A9
> > > > > (b) =C4=A4 invokes embedded_H =E2=9F=A8=C4=A4=E2=9F=A9 =E2=9F=A8=
=C4=A4=E2=9F=A9
> > > > > (c) embedded_H simulates =E2=9F=A8=C4=A4=E2=9F=A9 =E2=9F=A8=C4=A4=
=E2=9F=A9 ...
> > > > >=20
> > > > > Thus =E2=9F=A8=C4=A4=E2=9F=A9 =E2=9F=A8=C4=A4=E2=9F=A9 correctly =
simulated by embedded_H
> > > > > cannot possibly reach its simulated final halt state
> > > > > =E2=9F=A8=C4=A4.qn=E2=9F=A9
> > > > >=20
> > > > > > To refute the HP, you need to understand what it exactly means =
in TM.
> > > > >=20
> > > > > I have known this for 22 years.
> > > >=20
> > > > A working TM.=C2=A0Build it explicitly from transition function, th=
en explain
> > > > your derivation. You know nothing.
> > > >=20
> > >=20
> > > That would be like examining how an operating system
> > > works entirely from its machine code.
> >=20
> > You are refuting a CS foundamental theorem (i.e. HP) officially.
> > So, yes, and actually MORE need to be done (beyond your imagination).
> >=20
> > Knowing a car or smart phone,... is far different from making one.
> > Knowing E=3Dmc^2 is far from knowing relativity, making A-bomb (actuall=
y, making
> > A-bomb don't need to know E=3Dmc^2, people are often fooled by popular =
saying)
> > =C2=A0=20
> > Every chapter of Linz's book, C text textbook has exercises, you need t=
o those
> > exercises AT LEAST to comment CS (and computation theory is more advanc=
ed topic
> > than TM). Saying so is because we know you can't do the exercise and bo=
ast lots
> > about TM stuff (and pretty much anything else from just reading words),=
 even
> > about theorem.
> >=20
>=20
> When =C4=A4 is applied to =E2=9F=A8=C4=A4=E2=9F=A9
> =C4=A4.q0 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* embedded_H =E2=9F=A8=C4=A4=
=E2=9F=A9 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* =C4=A4.qy =E2=88=9E
> =C2=A0=C2=A0 or
> =C4=A4.q0 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* embedded_H =E2=9F=A8=C4=A4=
=E2=9F=A9 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* =C4=A4.qn
>=20
> (a) =C4=A4 copies its input =E2=9F=A8=C4=A4=E2=9F=A9
> (b) =C4=A4 invokes embedded_H =E2=9F=A8=C4=A4=E2=9F=A9 =E2=9F=A8=C4=A4=E2=
=9F=A9
> (c) embedded_H simulates =E2=9F=A8=C4=A4=E2=9F=A9 =E2=9F=A8=C4=A4=E2=9F=
=A9
>=20
> All that I need to know is that I proved that
> embedded_H correctly recognizes the repeating
> pattern where its correctly simulated =E2=9F=A8=C4=A4=E2=9F=A9 =E2=9F=A8=
=C4=A4=E2=9F=A9
> cannot possibly reach its own simulated final
> halt state of =E2=9F=A8=C4=A4.qn=E2=9F=A9
>=20
> https://www.liarparadox.org/Linz_Proof.pdf
>=20
> > > We only have to actually know one detail:
> > > Every counter-example input encoded in any model
> > > of computation always specifies recursive simulation
> > > that never halts to its corresponding simulating
> > > termination analyzer.
> >=20
> > More example here that you don't understand nearly all CS terms.
> >=20
>=20
> Mere empty rhetoric entirely bereft of any supporting
> reasoning. The x86 language is comparable to a RASP
> machine that is equivalent to a Turing machine.

Question:
1. Do you understand that you can't do the exercises in Linz's book?
2. Do you understand your ability of C/assembly/TM is less than 1 year CS l=
evel?