Re: [PO] Re: Can a regular Turing Machine provide Protected Memory?

From: Simon G Best (s.g.best_at_btopenworld.com)
Date: 08/30/04 

Date: Mon, 30 Aug 2004 14:05:35 +0000 (UTC)

Peter, make sure you read *all* the way through before replying. If you
don't, you will be leaving whatever you don't bother to read unrefuted.
  If you ignore anything when replying, you will be leaving whatever
you're ignoring unrefuted.

Peter Olcott wrote:
> "Owen Jacobson" <angstrom@lionsanctuary.net> wrote in message news:pan.2004.08.30.07.41.46.286184@lionsanctuary.net...
>
>>>From that statement, we can construct a Turing machine that
>>contradicts whatever Halts reports about that machine (already proven and
>>belaboured elsewhere).
>
> This has not yet been shown.

Yes it has.

As you often don't bother to read all the way through posts you respond
to, you are not in a credible position to claim that it has "not yet
been shown". In fact, it's been shown many, many times that such a
Turing Machine can be constructed.

The key is this: it's not Halts that's part of another TM, it's the
'UTM' it runs on that's part of another TM. Halts' final states still
have no further transitions defined for them, but the 'UTM' does have
further transitions out of its own (not Halts') 'final' transitions.

Let's call Halts' state transition table T_Halts, and your 'UTM''s state
transition table T_UTM.

We are not making any changes *at all* to T_Halts. The 'final' states
in T_Halts are *still* final states. There are no further transitions
out of those states *at all*.

We are, however, making some changes to T_UTM. The 'final' states in
T_UTM are no longer final states.

Your 'UTM' has, on its tape, the state transition table T_Halts. It
does not have its own state transition table, T_UTM, on its tape. It
cannot tell whether or not T_UTM has been changed.

As T_Halts has not been changed, Halts will run *exactly the same way*.

Simon

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