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From: wij <wyniijj5@gmail.com>
Newsgroups: comp.theory
Subject: Re: How to write a self-referencial TM?
Date: Sat, 17 May 2025 03:35:17 +0800
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On Fri, 2025-05-16 at 16:33 +0100, Mike Terry wrote:
> On 16/05/2025 12:40, wij wrote:
> > 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 =
(which calls itself):
> > > > > > > > > > > >=20
> > > > > > > > > > > > void D() {
> > > > > > > > > > > > =C2=A0=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 equivalent 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 & inp=
ut tape) that is to be
> > > > > > > simulated.
> > > > > > > The
> > > > > > > scheme says how to turn the (TM + input tape) into a string o=
f symbols 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 i=
s to be simulated.
> > > > > > >=20
> > > > > > > If you're looking for the exact string symbols, obviously you=
 would need to specify the
> > > > > > > exact
> > > > > > > UTM
> > > > > > > being used, because every UTM will have a different answer to=
 your question.
> > > > > > >=20
> > > > > > >=20
> > > > > > > Mike.
> > > > > >=20
> > > > > > People used to say UTM can simulate all TM. I was questing such=
 a UTM.
> > > > > > 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 m=
agical 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 conten=
ts of 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 de=
finition.
> > > > 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=C2=A0 Note this is itself a computation, distinct =
from f of course
> > > =C2=A0=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 n=
o knowledge that it is
> > > "simulating
> > > itself", and will just simulate what it is given.
> > >=20
> > >=20
> > > Mike.
> >=20
> > Sorry for not being clear on the UTM issue (I wanted to mean several th=
ings in one post).
> > You are right there is no self-reference.
> > I mean 'UTM' is not a complete, qualified TM because the contents of th=
e tape
> > would not be defined. Saying "UTM can simulate any TM" is misleading be=
cause
> > no such TM (UTM as TM) exists.
>=20
> What do you mean "the contents of the tape would not be defined"?=C2=A0 A=
 TM is /equipped/ with an=20
> infinite tape, but the /contents/ of that tape are not a part of that TM'=
s definition.
>=20
> For example we could build a TM P that decides whether a number is prime.=
=C2=A0 Given a number n, we=20
> convert n into the input tape representation of n, and run P with that ta=
pe as input.
>=20
> It's essentially no different for UTMs.=C2=A0 Such a UTM certainly is a "=
complete TM", equipped with its=20
> own input tape.=C2=A0 Of course we don't know what's on the input tape be=
cause nobody has said yet what=20
> computation we are asking it to simulate!=C2=A0 [Similarly we don't know =
what's on P's input tape, until=20
> we know what n we want it to test for primeness.]=C2=A0 Once you say what=
 computation you want the UTM to
> simulate we can build a tape string to perform that particular simulation=
..=C2=A0 That is the case=20
> /whatever/ computation we come up with, so it is simply the case [not mis=
leading] that the UTM can=20
> simulate any computation.
>=20
>=20
> Mike.

TM has no I/O mechanism. 'Computation' always means the contents of the tap=
e
is defined (fixed before run).