Re: How to write a self-referencial TM?

On Sun, 2025-05-18 at 19:09 -0400, Richard Damon wrote:

On 5/18/25 6:09 PM, olcott wrote:

On 5/18/2025 4:46 PM, wij wrote:

On Sun, 2025-05-18 at 15:57 -0500, olcott wrote:

On 5/18/2025 3:35 PM, wij wrote:

On Sat, 2025-05-17 at 14:39 -0500, olcott wrote:

On 5/17/2025 2:26 PM, wij wrote:

On Sat, 2025-05-17 at 15:45 +0100, Mike Terry wrote:

On 17/05/2025 04:01, wij wrote:

On Fri, 2025-05-16 at 23:51 +0100, Mike Terry wrote:

On 16/05/2025 20:35, wij wrote:

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):
 
void D() {
            D();
}
 
Easy?
 
 

 
That is not a TM.

 
It is a C program that exists. Therefore, there must
be a
equivalent
TM.
 

To make a TM that references itself the closest
thing is a UTM that simulates its own TM source-code.

 
How does a UTM simulate its own TM source-code?
 

 
You run a UTM that has its own source-code on its tape.

 
What is exactly the source-code on its tape?
 

 
Every UTM has some scheme which can be applied to a (TM
& input tape)
that
is to
be
simulated.
The
scheme says how to turn the (TM + input tape) into a
string of symbols
that
represent
that
computation.
 
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.
 
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.
 
 
Mike.

 
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?

 
Yes, a UTM can simulate any TM including itself. 
(Nothing magical changes
when
a
UTM
simulates
itself, as opposed to some other TM.)

 
Supposed UTM exists, and denoted as U(X), X denotes the
tape contents 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 definition.
But, U(U(f)) would fall into a 'self-reference' trap.

 
There is no self-reference trap.
 
In your notation:
 
-  f represents some computation.
-  U(f) represents U being run with f on its tape.
          Note this is itself a computation, distinct from
f of course
          but having the same behaviour.
-  U(U(f)) represents U simulating the previous computation.
 
There is no reason U(f) cannot be simulated by U.  U will
have no knowledge that
it
is
"simulating
itself", and will just simulate what it is given.
 
 
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 tape
would not be defined. Saying "UTM can simulate any TM" is
misleading because
no such TM (UTM as TM) exists.

 
What do you mean "the contents of the tape would not be
defined"?  A TM is
/equipped/
with
an
infinite tape, but the /contents/ of that tape are not a part
of that TM's
definition.
 
For example we could build a TM P that decides whether a
number is prime.  Given a
number n,
we
convert n into the input tape representation of n, and run P
with that tape as
input.
 
It's essentially no different for UTMs.  Such a UTM certainly
is a "complete TM",
equipped
with
its
own input tape.  Of course we don't know what's on the input
tape because nobody has
said
yet
what
computation we are asking it to simulate!  [Similarly we
don't know what's on P's
input
tape,
until
we know what n we want it to test for primeness.]  Once you
say what computation you
want
the
UTM to
simulate we can build a tape string to perform that
particular simulation.  That is
the
case
/whatever/ computation we come up with, so it is simply the
case [not misleading]
that
the
UTM
can
simulate any computation.
 
 
Mike.

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

 
Correct, and correct.
 
So... What do you mean "the contents of the tape would not be
defined"?
 
 
Mike.

 
In "UTM simulates itself", denoted as U(U(f)), the f would not
be defined.

 
Eh?  The f was something /you/ introduced!  You said it
represents some computation which
UTM U
simulates.  How can f suddenly become undefined after you defined
it?
 
Do you mean that f would not be on the input tape for (outer)U? 
That's not the case at
all.  In
U(f), the input tape for U contains a representation of f.  When
(outer)U simulates (inner)U
simulating f, (outer)U's tape contains a representation of
computation U(f), which
internally
contains the original representation of f.  The f is still there
and equally well defined in
U(U(f)).
 
I think you would benefit from being more explicit and generally
more careful in your
notation!
 
Using notation <P,I> to mean U's input tape representation of "TM
P, running with input I":
 
       Your U(f)      is U(<fp,fi>)        // fp = TM(f),
fi=InputTape(f)
       Your U(U(f))   is U((<U,<fp,fi>>)
 
f is still there!  It has not become "undefined".
 
You gloss over the details and become confused - just think it
through step by step.
 
 
Mike.

 
It seems you are addressing notational problems.
You introduced angle bracket and one more variable, and seemingly
attacking
that a variable cannot be undefined because it is there. And said
I should
not gloss over the detail and should think it through step by step.
No idea what you are talking about.
 
 

 
U(<U>)    is the most conventional notation

 
U is not a complete TM. U(<U>) is worst, also not a TM.
 

 
It is stipulated that U is a UTM that
has its own source-code as its input.

 
UTM is not a complete TM.

 
Textbooks define a UTM as a complete TM.
 

 
As I pointed out, there are two definitions of what a Turing Macine is.
 
It can be just the algorithm part, or the combination of the Algorithm
and Input.
 
In either case
 
"U <U>" isn't a valid computation, as in the TM is just tha algoritm
case, where <U> is a valid string for the representation of the TM, a
UTM needs as its input the combination of the representation of the
input program and the representation of its input (if it takes any), so,
since the program is a UTM, and that needs an input, you need to add it.

Note that TM cannot *read* input as external information, everything is fixed.
I have no problem with the 'two definition' saying, you just have to make 
it clear (but as usual in your real number, you invent your own stuff, you 
are not following your 'standard' and condemn others not following it).
However this 'other' definition of TM if exists will have no computation 
(nothing in the tape). You cannot talk about 'computation'.

This causes the problem of simulating a UTM with a copy of itself blow
up, as you keep needing to add more inputs for every representaiton of a
UTM added to the tape.
 
That is why H^ is defined to take just one piece of input, the
representation of a program (which will be of itself) and it duplicates
that to pass it to the decider, which needs a program and its input.
 
Thus H^ <H^> -> H <H^> <H^> which decides on H^ <H^> which is where we
started.