**From:** Rui Ferreira (*ruiandreferreira@gmail.com*)

**Date:** Mon Jan 05 2009 - 22:19:23 MST

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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 <peter@spaceandgames.com> wrote:

*> Mike Dougherty wrote:
*

*>>
*

*>> On Mon, Jan 5, 2009 at 12:56 PM, Peter de Blanc <peter@spaceandgames.com>
*

*>> wrote:
*

*>>>
*

*>>> Matt Mahoney wrote:
*

*>>>
*

*>>>> False. If X simulates Y, then K(X) > K(Y) because X has an exact model
*

*>>>> of
*

*>>>> 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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