Re: [PO] Proving a negative is hard

From: Simon G Best (s.g.best_at_btopenworld.com)
Date: 08/24/04 

Date: Tue, 24 Aug 2004 02:30:23 +0000 (UTC)

LOOK!

Pay Attention!

Peter Olcott wrote:
> "Simon G Best" <s.g.best@btopenworld.com> wrote in message news:41296FA9.8050706@btopenworld.com...
>
>>Peter Olcott wrote:
>>
>>>"Robert Low" <mtx014@linux.services.coventry.ac.uk> wrote in message news:cgaq0l$qd0$1@sunbeam.coventry.ac.uk...

Pay attention to this part in particular:

>>>>writing a symbol on the tape (maybe the symbol already
>>>>there),
>>>
>>For example:
>>
>> q_n 'B' 'B' L q_m
>>
>>which means: 'When in internal state q_n; when the scanned symbol is
>>'B': erase the scanned symbol and write a 'B' in its place; then move
>>the head one square to the left; and move into internal state q_m.'

Writing a 'B' in place of a 'B' has the same effect as not writing
anything in place of that 'B', and just leaving that 'B' in place.

[...]
>
> It could make sense to have an implicit read operation.
> But This is not what Mitch Claimed:
>
>>>>>>Every transition in a TM writes something to the tape.

And his very next sentence was: "It may be what was there before."

So, Mitch Harris had said:

> Every transition in a TM writes something to the tape. It
> may be what was there before.

And Robert Low had said:

> A transition of a TM *by definition* comprises [...]
> writing a symbol on the tape (/maybe the symbol already
> there/)[...].

(emphasis mine).

And I then gave you some simple examples to illustrate what they were
telling you, and yet you /still/ go on to say:

> I don't think that it can make sense to have an implicit read
> operation, AND an implicit write operation. You might wipe
> out data that you need.

It is therefore unmistakeably obvious that you do not even have a basic
understanding of what a TM is.

And now I've got a sudden, severe hankering for chips by the sea-side.
Preferably on the west Welsh coast. It's a /bad/ hankering!

Simon

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