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From: wij <wyniijj5@gmail.com>
Newsgroups: comp.theory
Subject: Re: How to write a self-referencial TM?
Date: Fri, 16 May 2025 19:40:17 +0800
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On Fri, 2025-05-16 at 03:26 +0100, Mike Terry wrote:
> On 16/05/2025 02:47, wij wrote:
> > On Fri, 2025-05-16 at 01:40 +0100, Mike Terry wrote:
> > > On 15/05/2025 19:49, wij wrote:
> > > > On Thu, 2025-05-15 at 17:08 +0100, Mike Terry wrote:
> > > > > On 14/05/2025 18:53, wij wrote:
> > > > > > On Wed, 2025-05-14 at 12:24 -0500, olcott wrote:
> > > > > > > On 5/14/2025 11:43 AM, wij wrote:
> > > > > > > > On Wed, 2025-05-14 at 09:51 -0500, olcott wrote:
> > > > > > > > > On 5/14/2025 12:13 AM, wij wrote:
> > > > > > > > > > Q: Write a turing machine that performs D function (whi=
ch calls itself):
> > > > > > > > > >=20
> > > > > > > > > > void D() {
> > > > > > > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 D();
> > > > > > > > > > }
> > > > > > > > > >=20
> > > > > > > > > > Easy?
> > > > > > > > > >=20
> > > > > > > > > >=20
> > > > > > > > >=20
> > > > > > > > > That is not a TM.
> > > > > > > >=20
> > > > > > > > It is a C program that exists. Therefore, there must be a e=
quivalent TM.
> > > > > > > >=20
> > > > > > > > > To make a TM that references itself the closest
> > > > > > > > > thing is a UTM that simulates its own TM source-code.
> > > > > > > >=20
> > > > > > > > How does a UTM simulate its own TM source-code?
> > > > > > > >=20
> > > > > > >=20
> > > > > > > You run a UTM that has its own source-code on its tape.
> > > > > >=20
> > > > > > What is exactly the source-code on its tape?
> > > > > >=20
> > > > >=20
> > > > > Every UTM has some scheme which can be applied to a (TM & input t=
ape) that is to be
> > > > > simulated.
> > > > > The
> > > > > scheme says how to turn the (TM + input tape) into a string of sy=
mbols that represent that
> > > > > computation.
> > > > >=20
> > > > > So to answer your question, the "source-code on its tape" is the =
result of applying the
> > > > > UTM's
> > > > > particular scheme to the combination (UTM, input tape) that is to=
 be simulated.
> > > > >=20
> > > > > If you're looking for the exact string symbols, obviously you wou=
ld need to specify the
> > > > > exact
> > > > > UTM
> > > > > being used, because every UTM will have a different answer to you=
r question.
> > > > >=20
> > > > >=20
> > > > > Mike.
> > > >=20
> > > > People used to say UTM can simulate all TM. I was questing such a U=
TM.
> > > > Because you said "Every UTM ...", so what is the source of such UTM=
?
> > >=20
> > > Yes, a UTM can simulate any TM including itself.=C2=A0 (Nothing magic=
al changes when a UTM simulates
> > > itself, as opposed to some other TM.)
> >=20
> > Supposed UTM exists, and denoted as U(X), X denotes the tape contents o=
f the
> > encoding of a TM. And, U(X) should function the same like X.
> > Given instance U(U(f)), it should function like f from the above defini=
tion.
> > But, U(U(f)) would fall into a 'self-reference' trap.
>=20
> There is no self-reference trap.
>=20
> In your notation:
>=20
> -=C2=A0 f represents some computation.
> -=C2=A0 U(f) represents U being run with f on its tape.
> =C2=A0=C2=A0=C2=A0 Note this is itself a computation, distinct from f of =
course
> =C2=A0=C2=A0=C2=A0 but having the same behaviour.
> -=C2=A0 U(U(f)) represents U simulating the previous computation.
>=20
> There is no reason U(f) cannot be simulated by U.=C2=A0 U will have no kn=
owledge that it is "simulating=20
> itself", and will just simulate what it is given.
>=20
>=20
> Mike.

Sorry for not being clear on the UTM issue (I wanted to mean several things=
 in one post).
You are right there is no self-reference.
I mean 'UTM' is not a complete, qualified TM because the contents of the ta=
pe
would not be defined. Saying "UTM can simulate any TM" is misleading becaus=
e
no such TM (UTM as TM) exists.