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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 11:43:51 +0800
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On Fri, 2025-05-09 at 22:24 -0500, olcott wrote:
> On 5/9/2025 10:13 PM, wij wrote:
> > On Fri, 2025-05-09 at 19:40 -0700, Keith Thompson wrote:
> > > olcott <polcott333@gmail.com> writes:
> > > > On 5/9/2025 4:40 PM, Richard Heathfield wrote:
> > > > > On 09/05/2025 21:15, olcott wrote:
> > > > > > On 5/9/2025 3:07 PM, Richard Heathfield wrote:
> > > > > > > On 09/05/2025 20:46, olcott wrote:
> > > > > > > > We have not begun to get into any of those points.
> > > > > > > > We are only asking can DDD correctly simulated
> > > > > > > > by any HHH that can exist ever reach its own
> > > > > > > > "return" instruction.
> > > > > > >=20
> > > > > > > DDD can't be correctly simulated by itself (which is effectiv=
ely
> > > > > > > what you're trying to do when you fire up the simulation from
> > > > > > > inside DDD).
> > > > > >=20
> > > > > > How the Hell did you twist my words to say that?
> > > > > I haven't touched your words. What I have done is to observe that
> > > > > DDD's /only/ action is to call a simulator. Since DDD isn't itsel=
f a
> > > > > simulator, there is nothing to simulate except a call to a
> > > > > simulator.
> > > > > It's recursion without a base case - a rookie error.
> > > > > HHH cannot successfully complete its task, because it never regai=
ns
> > > > > control after the first recursion. To return, it must abort the
> > > > > simulation, which means the simulation fails.
> > > > >=20
> > > > > > void DDD()
> > > > > > {
> > > > > > =C2=A0=C2=A0=C2=A0 HHH(DDD);
> > > > > > =C2=A0=C2=A0=C2=A0 return;
> > > > > > }
> > > > > >=20
> > > > > > When 1 or more statements of DDD are correctly
> > > > > > simulated by HHH then this correctly simulated
> > > > > > DDD cannot possibly reach its own =E2=80=9Creturn statement=E2=
=80=9D.
> > > > > On what grounds can you persuade an extraordinarily sceptical
> > > > > readership that HHH 'correctly simulated' DDD?
> > > >=20
> > > > Any competent C programmer can see that
> > > > the call from DDD to HHH(DDD) (its own simulator)
> > > > is equivalent to infinite recursion.
> > > >=20
> > > > On 5/8/2025 8:30 PM, Keith Thompson wrote:
> > > > > Assuming that HHH(DDD) "correctly simulates" DDD, and assuming it
> > > > > does nothing else, your code would be equivalent to this:
> > > > >=20
> > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 void DDD(void) {
> > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 DDD();
> > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 return;
> > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 }
> > > > >=20
> > > > > Then the return statement (which is unnecessary anyway) will neve=
r be
> > > > > reached.=C2=A0 In practice, the program will likely crash due to =
a stack
> > > > > overflow, unless the compiler implements tail-call optimization, =
in
> > > > > which case the program might just run forever -- which also means=
 the
> > > > > unnecessary return statement will never be reached.
> > >=20
> > > I had not intended to post again, but I feel the need to make
> > > a clarification.
> > >=20
> > > I acknowledged that the return statement would never be reached
> > > *given the assumption* that HHH correctly simulates DDD.=C2=A0 Given
> > > that assumption, a call to DDD() should be equivalent to a call
> > > to HHH(DDD).
> > >=20
> > > I did not address whether the assumption is valid.=C2=A0 I merely
> > > temporarily accepted it for the sake of discussion, just as I would
> > > accept that if I were ten feet tall I would bump my head against
> > > the ceiling in my house.
> > >=20
> > > The discussion I had with olcott did not reach the point of
> > > discussing *how* HHH could correctly simulate DDD, or whether it
> > > would even be logically possible for it to do so.=C2=A0 I also did no=
t
> > > address any issues of partial simulation, where olcott claims that
> > > HHH can "accurately simulate" only a few x86 instructions rather
> > > than simulating its entire execution.=C2=A0 I did not participate in
> > > any discussion that would require knowledge of x86 machine or
> > > assembly code.=C2=A0 (I have no doubt that I could learn x86 machine
> > > and assembly code reasonably well if motivated to do so, but I am
> > > not so motivated.)
> > >=20
> > > What I acknowledged was barely more than "if HHH correctly simulates
> > > DDD, then HHH correctly simulates DDD".=C2=A0 (My understanding from
> > > posts by others, whom I presume to be sufficiently knowledgeable,
> > > is that HHH logically cannot accurately simulate DDD.)=C2=A0 I would
> > > prefer that olcott refrain from using my words to support any of
> > > his arguments beyond the scope of what he and I directly discussed.
> >=20
> > Don't know why you people stick on the 'simulation' stuff so long.
> > The HP simply asks for such an H (in function form. POOH does not
> > resemble TM):
> > =C2=A0 H(D)=3D1 if D() halt.
> > =C2=A0 H(D)=3D0 if D() not halt.
>=20
> My invention of a simulating termination
> analyzer shows exactly how to compute the
> mapping that the input that HHH(DD) specifies
> into a correct answer for the halting problem's
> otherwise impossible input.
>=20
> All rebuttals are based on failing to compute
> this mapping correctly.
>=20

What is the correct mapping?

If POOH are not talking about the mapping:

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

POOH is likely nothing to do with HP