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From: wij <wyniijj5@gmail.com>
Newsgroups: comp.theory
Subject: Re: Every sufficiently competent C programmer knows --- Very Stupid
 Mistake or Liars
Date: Fri, 14 Mar 2025 23:33:27 +0800
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On Fri, 2025-03-14 at 10:13 -0500, olcott wrote:
> On 3/14/2025 9:10 AM, Richard Damon wrote:
> > On 3/13/25 11:53 PM, olcott wrote:
> > > On 3/13/2025 10:03 PM, Richard Damon wrote:
> > > > On 3/13/25 10:07 PM, olcott wrote:
> > > > > On 3/13/2025 6:09 PM, Richard Damon wrote:
> > > > > > On 3/13/25 9:41 AM, olcott wrote:
> > > > > > > On 3/13/2025 6:18 AM, Dan Cross wrote:
> > > > > > > > In article <vqud4e$36e14$3@dont-email.me>,
> > > > > > > > Fred. Zwarts <F.Zwarts@HetNet.nl> wrote:
> > > > > > > > > Op 12.mrt.2025 om 16:31 schreef olcott:
> > > > > > > > > > [snip]
> > > > > > > > > > When N steps of DDD are correctly emulated by every ele=
ment
> > > > > > > > > > of the set of C functions named HHH that do x86 emulati=
on and
> > > > > > > > > >=20
> > > > > > > > > > N is each element of the set of natural numbers
> > > > > > > > > >=20
> > > > > > > > > > then no DDD of the set of HHH/DDD pairs ever reaches it=
s
> > > > > > > > > > "return" instruction and terminates normally.
> > > > > > > > >=20
> > > > > > > > > In other words no HHH of the set of HHH/DDD pairs ever su=
cceeds to
> > > > > > > > > complete the simulation of a halting program. Failure to =
reach=20
> > > > > > > > > the end
> > > > > > > > > of a halting program is not a great success. If all HHH i=
n this set
> > > > > > > > > fail, it would be better to change your mind and start wo=
rking on
> > > > > > > > > something more useful.
> > > > > > > >=20
> > > > > > > > He seems to think that he's written a program that detects =
that
> > > > > > > > his thing hasn't 'reached its "return" instruction and
> > > > > > > > terminate[d] normally', given some number of steps, where t=
hat
> > > > > > > > number is ... the cardinality of the natural numbers.
> > > > > > > >=20
> > > > > > > > I wonder if he knows that the set of natural numbers is
> > > > > > > > infintite, though I suspect he'd say something like, "but i=
t's
> > > > > > > > countable!"=C2=A0 To which I'd surmise that he has no idea =
what that
> > > > > > > > means.
> > > > > > > >=20
> > > > > > >=20
> > > > > > > void DDD()
> > > > > > > {
> > > > > > > =C2=A0=C2=A0 HHH(DDD);
> > > > > > > =C2=A0=C2=A0 return;
> > > > > > > }
> > > > > > >=20
> > > > > > > Everyone here knows that when N steps of DDD are correctly
> > > > > > > simulated by HHH that DDD never reaches its own "return"
> > > > > > > instruction and terminates normally thus never halts.
> > > > > > > *AND THEY LIE ABOUT IT BY ENDLESSLY CHANGING THE SUBJECT*
> > > > > >=20
> > > > > >=20
> > > > > > No, the PARTIAL EMULATION done by HHH can't reach that point,
> > > > >=20
> > > > > But a complete emulation can?
> > > > >=20
> > > >=20
> > > > Yes, but an HHH that gives an answer doesn't do one, due to the=20
> > > > pathological design of the template used to build DD to the HHH it=
=20
> > > > calls (which is the only HHH that can exist, or you have violated t=
he=20
> > > > basic rules of programing and logic).
> > > >=20
> > > > We have two basic cases,
> > > >=20
> > > > 1) if HHH does the partial emulation you describe, then the complet=
e=20
> > > > emulation of DD will see that DD call HHH, and it will emulate its=
=20
> > > > input for a while, then abort and theu return 0 to DD which will th=
en=20
> > > > halt.
> > > >=20
> > >=20
> > > int main()
> > > {
> > > =C2=A0=C2=A0 HHH(DDD); // No DDD can possibly ever return.
> > > }
> > >=20
> >=20
> > Since HHH doesn;t call DDD, the statement is vacuous and shows a=20
> > fundamental ignorance of what is being talked about.
> >=20
> > Yes, No HHH can emulated DDD to the end, but since halting is DEFINED b=
y=20
> > the behavior of the program, and for every HHH that aborts and returns,=
=20
> > the program of DDD, as tested with:
> >=20
> > int main()
> > {
> > =C2=A0=C2=A0=C2=A0 DDD()
> > }
> >=20
> > will return to main, that shows that every HHH that returns 0 fails to=
=20
> > be a Halt Decider or Termination Analyzer. PERIOD.
> >=20
> > The fact that it may meet your defintion of POOP says maybe you have=
=20
> > solvec the POOP problem, but it seems you can't even properly define it=
=20
> > in the nornal field of Computation Theory.
>=20
> void DDD()
> {
> =C2=A0=C2=A0 HHH(DDD);
> =C2=A0=C2=A0 return;
> }
>=20
> The only difference between HHH and HHH1 is that they are
> at different locations in memory. DDD simulated by HHH1
> has identical behavior to DDD() executed in main().
>=20

https://en.wikipedia.org/wiki/Halting_problem
In computability theory, the halting problem is the problem of determining,=
 from a description of an
arbitrary computer program and an input, whether the program will finish ru=
nning, or continue to run
forever.

It is important to understand the meaning of the terms used in your questio=
n.
Otherwise, you are solving POO Problem. But, POOP is OK, just don't mix the=
m.

> The semantics of the finite string input DDD to HHH specifies
> that it will continue to call HHH(DDD) in recursive simulation.
>=20
> The semantics of the finite string input DDD to HHH1 specifies
> to simulate to DDD exactly once.
>=20
>=20