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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 14:19:12 +0800
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On Sat, 2025-05-10 at 01:06 -0500, olcott wrote:
> On 5/10/2025 1:00 AM, wij wrote:
> > On Sat, 2025-05-10 at 00:41 -0500, olcott wrote:
> > > On 5/10/2025 12:27 AM, wij wrote:
> > > > 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 ebp=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,e=
sp=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 HP?
> > > > > >=20
> > > > >=20
> > > > > =C2=A0=C2=A0=C2=A0>>>>> H(D)=3D1 if D() halt.
> > > > > =C2=A0=C2=A0=C2=A0>>>>> H(D)=3D0 if D() not halt.
> > > > >=20
> > > > > Right now it is mostly about proving the
> > > > > above requirements are is mistaken.
> > > > >=20
> > > >=20
> > > > Why is the requirement invalid?
> > > >=20
> > > > H(D)=3D1 if D() halt.
> > > > H(D)=3D0 if D() not halt.
> > > >=20
> > >=20
> >=20
> > > The notion that the behavior specified by the finite
> > > string input to a simulating termination analyzer
> >=20
> > POOH reads(takes) its input as a function, not 'finite string'.
> > Are you talking about POOH now? There is no POOH that takes
> > 'finite string'.
> >=20
>=20
> It <is> a finite string of x86 bytes.

Disagree.
The D in Halt7.c (I just saw once) does not treat H as 'finite string',
D calls H. H also does not treat D as 'finite string'.

> > > does sometimes differ from the behavior of its direct
> > > execution. It is a provably different sequence of steps.
> >=20
>=20
> This is a verified fact.
> The pathological relationship that inputs can have
> with their simulating termination analyzer changes
> the behavior of these inputs relative to their direct
> execution.

So, you redefined the halting problem should be about the behavior of D=20
decided by POOH, not the 'direct' behavior of D?

I.e, H(D)=3D1 iff H(D)=3D1?

> > So, you are talking about the behavior of the 'simulating termination a=
nalyzer'
> > i.e. POOH? (not really about the HP)
> >=20
> >=20
> >=20
> >=20
>=20
>=20