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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 02:38:54 +0800
Organization: A noiseless patient Spider
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Message-ID: <07c4f2302645a7e58957b5e5bffed80397a6ddae.camel@gmail.com>
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On Sun, 2025-05-11 at 12:40 -0500, olcott wrote:
> On 5/11/2025 12:21 PM, wij wrote:
> > On Sun, 2025-05-11 at 12:00 -0500, olcott wrote:
> > > On 5/11/2025 11:28 AM, wij wrote:
> > > > 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=C2=A0=C2=A0 int Halt_S=
tatus =3D HHH(DD);
> > > > > > > > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 if (Halt_S=
tatus)
> > > > > > > > > > > =C2=A0=C2=A0=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=C2=A0=C2=A0 return Hal=
t_Status;
> > > > > > > > > > > }
> > > > > > > > > > >=20
> > > > > > > > > > > int main()
> > > > > > > > > > > {
> > > > > > > > > > > =C2=A0=C2=A0=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 nothi=
ng.
> > > > > > > > > >=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=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 co=
rrectly 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 exactl=
y means in TM.
> > > > > > > > >=20
> > > > > > > > > I have known this for 22 years.
> > > > > > > >=20
> > > > > > > > A working TM.=C2=A0Build it explicitly from transition func=
tion, then 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 imagina=
tion).
> > > > > >=20
> > > > > > Knowing a car or smart phone,... is far different from making o=
ne.
> > > > > > Knowing E=3Dmc^2 is far from knowing relativity, making A-bomb =
(actually, making
> > > > > > A-bomb don't need to know E=3Dmc^2, people are often fooled by =
popular saying)
> > > > > > =C2=A0=C2=A0=C2=A0=20
> > > > > > Every chapter of Linz's book, C text textbook has exercises, yo=
u need to those
> > > > > > exercises AT LEAST to comment CS (and computation theory is mor=
e advanced topic
> > > > > > than TM). Saying so is because we know you can't do the exercis=
e and boast 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=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.
> > > >=20
> > > > Question:
> > > > 1. Do you understand that you can't do the exercises in Linz's book=
?
> > >=20
> > > Everything is 100% irrelevant besides the fact that
> > > I have shown that =E2=9F=A8=C4=A4=E2=9F=A9 =E2=9F=A8=C4=A4=E2=9F=A9 c=
orrectly simulated by
> > > embedded_H cannot possibly reach its own simulated
> > > final halt state =E2=9F=A8=C4=A4.qn=E2=9F=A9. Thus when embedded_H re=
ports
> > > on the behavior that its input specifies it can
> > > correctly transition to =C4=A4.qn.
> > >=20
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