Re: How to write a self-referencial TM?

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.