Re: How to write a self-referencial TM?

On 5/16/2025 10:01 PM, 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.
 I was considering also expressing the idea that undecidable is caused by 'semantic
self reference'.
Ex1: The truth value of "This sentence is true" is also undecidable.
 

You have to know what a truth-bearer is to understand
that the Liar Paradox has the same truth value as this
sentence: "What time is it?" I have been working on
that for 22 years too.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer