From: Rui Ferreira (email@example.com)
Date: Mon Jan 05 2009 - 22:19:23 MST
It's subtle. When X simulates Y and K(X) < K(Y), you can't ask X what
would Y output for a certain input. In your example, if you wanted X
to predict the output of Y, it would have to be coded in X that Y is
the nth program generated and, on that case, K(Y) would be equal (or
less than) K(X).
Anyway, as you said, it's perfectly possible that X "simulates" Y and
K(X) < K(Y). Was this the point you were trying to make about the
difference of simulation and "actual computation", Peter?
On Tue, Jan 6, 2009 at 3:20 AM, Peter de Blanc <firstname.lastname@example.org> wrote:
> Mike Dougherty wrote:
>> On Mon, Jan 5, 2009 at 12:56 PM, Peter de Blanc <email@example.com>
>>> Matt Mahoney wrote:
>>>> False. If X simulates Y, then K(X) > K(Y) because X has an exact model
>>>> the mental state of Y. This implies that Y cannot also simulate X
>>>> because it
>>>> would require K(Y) > K(X).
>>> Matt, please stop posting pseudomathematics.
>> Seriously? K is http://www.google.com/search?q=kolmogorov+complexity
>> You might as well say, "Matt, please stop posting rational statements"
> My objection is to the statement "If X simulates Y, then K(X) > K(Y)."
> There's no such theorem. For example, you could write a program which
> simulates every possible program. This program would have some fixed
> complexity K, but since it simulates every program, it will simulate some
> with complexity >K.
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