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From: wij <wyniijj5@gmail.com>
Newsgroups: comp.theory
Subject: Re: Why I need to cross-post to comp.lang.c --- CORRECTLY REFUTED
Date: Sat, 10 May 2025 23:39:42 +0800
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On Sat, 2025-05-10 at 10:13 -0500, olcott wrote:
> On 5/10/2025 2:15 AM, Mikko wrote:
> > On 2025-05-09 03:01:40 +0000, olcott said:
> >=20
> > > On 5/8/2025 9:23 PM, Keith Thompson wrote:
> > > > Richard Damon <richard@damon-family.org> writes:
> > > > > On 5/8/25 7:53 PM, olcott wrote:
> > > > [...]
> > > > > > void DDD()
> > > > > > {
> > > > > > =C3=82=C2=A0 HHH(DDD);
> > > > > > =C3=82=C2=A0 return;
> > > > > > }
> > > > > > We don't need to look at any of my code for me
> > > > > > to totally prove my point. For example when
> > > > > > the above DDD is correctly simulated by HHH
> > > > > > this simulated DDD cannot possibly reach its own
> > > > > > "return" instruction.
> > > > >=20
> > > > > And thus not correctly simulatd.
> > > > >=20
> > > > > Sorry, there is no "OS Exemption" to correct simulaiton;.
> > > >=20
> > > > Perhaps I've missed something.=C2=A0 I don't see anything in the ab=
ove that
> > > > implies that HHH does not correctly simulate DDD.=C2=A0 Richard, yo=
u've read
> > > > far more of olcott's posts than I have, so perhaps you can clarify.
> > > >=20
> > > > If we assume that HHH correctly simulates DDD, then the above code =
is
> > > > equivalent to:
> > > >=20
> > > > void DDD()
> > > > {
> > > > DDD();
> > > > return;
> > > > }
> > > >=20
> > > > which is a trivial case of infinite recursion.=C2=A0 As far as I ca=
n tell,
> > > > assuming that DDD() is actually called at some point, neither the
> > > > outer execution of DDD nor the nested (simulated) execution of DDD
> > > > can reach the return statement.=C2=A0 Infinite recursion might eith=
er
> > > > cause a stack overflow and a probable program crash, or an unending
> > > > loop if the compiler implements tail call optimization.
> > > >=20
> > > > I see no contradiction, just an uninteresting case of infinite
> > > > recursion, something that's well understood by anyone with a
> > > > reasonable level of programming experience.=C2=A0 (And it has nothi=
ng to
> > > > do with the halting problem as far as I can tell, though of course
> > > > olcott has discussed the halting problem elsewhere.)
> > > >=20
> > > > Richard, what am I missing?
> > > >=20
> > > *****
> > > Now you are seeing what I was talking about.
> > > Now you are seeing why I needed to cross post
> > > to comp.lang.c
> >=20
> > What were you told in comp.lang.c that you were not told in comp.theory=
?
> >=20
>=20
> void DDD()
> {
> =C2=A0=C2=A0 HHH(DDD);
> =C2=A0=C2=A0 return;
> }
>=20
> People quickly realize that when DDD is correctly
> simulated by HHH that DDD cannot possibly reach
> its "return" statement (final halt state).
>=20
> Once you know this then you can see that the
> same thing applies to DD.
>=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
> Once you know this then you know that the halting
> problem's otherwise "impossible" input is non-halting.
>=20
> Once you know this then you know that the halting
> problem proof has been correctly refuted.

Nope.
POOH is (at most) about whether the input D is "impossible" input or not.