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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 13:27:47 +0800
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On Sat, 2025-05-10 at 00:19 -0500, olcott wrote:
> On 5/10/2025 12:13 AM, wij wrote:
> > On Sat, 2025-05-10 at 00:06 -0500, olcott wrote:>>
> > > When mathematical mapping is properly understood
> > > it will be known that functions computed by models
> > > of computation must transform their input into
> > > outputs according to the specific steps of an
> > > algorithm.
> > >=20
> > > _DDD()
> > > [00002172] 55=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 push eb=
p=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 ; housekeeping
> > > [00002173] 8bec=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 mov ebp,esp=C2=A0=
=C2=A0 ; housekeeping
> > > [00002175] 6872210000 push 00002172 ; push DDD
> > > [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
> > > [0000217f] 83c404=C2=A0=C2=A0=C2=A0=C2=A0 add esp,+04
> > > [00002182] 5d=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 pop ebp
> > > [00002183] c3=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 ret
> > > Size in bytes:(0018) [00002183]
> > >=20
> > > For example HHH(DDD) only correctly map to the
> > > behavior that its input actually specifies by correctly
> > > emulating DDD according to the rules of the x86 language.
> > >=20
> > > This causes the first four instructions of DDD
> > > to be emulated followed by HHH emulating itself
> > > emulating the first three instructions of DDD.
> > >=20
> > > It is right at this recursive simulation just
> > > before HHH(DDD) is called again that HHH recognizes
> > > the repeating pattern and rejects DDD.
> >=20
> > Yes, but you still did not answer the question: Is POOH exactly about H=
P?
> >=20
>=20
> =C2=A0>>>>> H(D)=3D1 if D() halt.
> =C2=A0>>>>> H(D)=3D0 if D() not halt.
>=20
> Right now it is mostly about proving the
> above requirements are is mistaken.
>=20

Why is the requirement invalid?

H(D)=3D1 if D() halt.
H(D)=3D0 if D() not halt.