Path: news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail
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 12:32:03 +0800
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On Fri, 2025-05-09 at 23:18 -0500, olcott wrote:
> On 5/9/2025 10:43 PM, wij wrote:
> > 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 effe=
ctively
> > > > > > > > > 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 i=
tself 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 r=
egains
> > > > > > > control after the first recursion. To return, it must abort t=
he
> > > > > > > simulation, which means the simulation fails.
> > > > > > >=20
> > > > > > > > void DDD()
> > > > > > > > {
> > > > > > > > =C2=A0=C2=A0=C2=A0=C2=A0 HHH(DDD);
> > > > > > > > =C2=A0=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 assumin=
g it
> > > > > > > does nothing else, your code would be equivalent to this:
> > > > > > >=20
> > > > > > > =C2=A0=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=C2=A0 =
DDD();
> > > > > > > =C2=A0=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=C2=A0 }
> > > > > > >=20
> > > > > > > Then the return statement (which is unnecessary anyway) will =
never be
> > > > > > > reached.=C2=A0 In practice, the program will likely crash due=
 to a stack
> > > > > > > overflow, unless the compiler implements tail-call optimizati=
on, in
> > > > > > > which case the program might just run forever -- which also m=
eans 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 Gi=
ven
> > > > > 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 wou=
ld
> > > > > 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 di=
d not
> > > > > address any issues of partial simulation, where olcott claims tha=
t
> > > > > 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 mach=
ine
> > > > > 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 simula=
tes
> > > > > DDD, then HHH correctly simulates DDD".=C2=A0 (My understanding f=
rom
> > > > > posts by others, whom I presume to be sufficiently knowledgeable,
> > > > > is that HHH logically cannot accurately simulate DDD.)=C2=A0 I wo=
uld
> > > > > prefer that olcott refrain from using my words to support any of
> > > > > his arguments beyond the scope of what he and I directly discusse=
d.
> > > >=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=C2=A0 H(D)=3D1 if D() halt.
> > > > =C2=A0=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
> >=20
> > What is the correct mapping?
> >=20
>=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,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
> Computing the mapping of DDD emulated by HHH
> according the the rules of the x86 language
> to its behavior by HHH actually emulating DDD.

The above says you have no idea what the mapping is.

> > If POOH are not talking about the mapping:
> >=20
> > =C2=A0=C2=A0 H(D)=3D1 if D() halt.
> > =C2=A0=C2=A0 H(D)=3D0 if D() not halt.
> >=20
>=20
> The way that simulating termination analyzers process
> their input by showing all of the steps of how the mapping
> must be computed refutes the above simplistic view.

No (real) problem with that. But the HP asks:

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