Re: Uncomputable Natural Numbers

On Sep 5, 10:09 pm, reaste...@xxxxxxxxx wrote:
Neither is the finite case.

I rebutted,

You can just DECREE that when such and such a string occurs, THE INPUT
IS JUST *OVER*.

And WE HAVE so decreed.

RE ignorantly replies,
But you can't guarantee that you will ever read the string
that says computation over.

Well, yes, there are some TMs that loop within a finite
proper subpart of the input, and never get to the end of it.
But that does not stop the input from being finite.

 THAT IS WHY WE NEED
to make the ASSUMPTION that the input is finite.

Then why read the tape at all?

Because it MATTERS a great deal WHICH finite number
the input is! THAT'S why!

Halt and return true because we already know the
input is a natural number. Certainly, we can
assume the input is a natural number.

If that's all you want to know, THEN YES, YOU CAN do EXACTLY that.
The point is, that is NEVER what anybody wants to know. It goes
WITHOUT
saying that the empty set and the WHOLE set, the TM that accepts
everything
and the TM that rejects everything, are trivially recursive.

If we don't make such an assumption,
we can prove the laguage of natural number
representations is not TM decidable.

That is ridiculous. The language of representations of TMs
is REGULAR, under reasonable assumptions.
A TM is just a list of input-direction/new-state pairs(one element of
the list
for each OLD-state). The language of representations of TMs is ( L+R
1(0+1)* )*
where the bit-string after the L/R is just the binary number for the
new state, and
state-numbers can't be lower than 1.
You can define a transition to a state not on the list as a reject-
halt.
Do people compute natural numbers?

This is a stupid question. It is not even meaningful to speak of
computing 3 or computing 69 or computing 142857 or computing
999999999999999999999999999999.

(on proving that some machine can only read finite strings)
I do prove it.

No, you don't.

No one has shown any error in my proof.

The set of unread positions will always be infinite.
The set of read positions will always be finite.

That is only under the assumption that the TM has stopped.
If the TM is still running then it cannot have failed.
But that is not the point. Obviously a machine with a lower
bound on how long it takes to read 1 character cannot read
an infinite string. But that IS NOT RELEVANT since MEANINGFUL
INPUTS in this paradigm are always going to be FINITE ANYWAY.
You were doing better back when you were talking about inability
to confirm the finite case.


Yes, the TM will halt if it reads a blank
or any other character that is not a "1".
But, as long as the TM is reading "1's"
there is nothing we can say about the
state of the TM other than "processing".

If the input is finite then there IS a blank eventually,
so the TM CAN eventually get there UNLESS it is programmed to loop
on some substring of the input. That loop does not CHANGE the fact
that the input was finite.
But you are no longer clear which side of this you are arguing.

Even if we assume the TM is "infinitely fast"
or we wait until the end of time, there will
always be an infinite number of finite positions
that haven't been read by the TM.


THAT is NOT true.
We don't even NEED to assume that the TM "infinitely" fast.
If the amount of time it takes to read the nth cell is 2^-n seconds,
then
the TM will finish reading the whole infinite tape in 1 second.
Even though it never "reaches infinite" speed.
But the point is, anything that CAN do THAT is NOT a TM.
.

Quantcast