From: Benja Fallenstein (email@example.com)
Date: Tue Jan 06 2009 - 12:17:46 MST
On Tue, Jan 6, 2009 at 7:17 PM, Peter de Blanc <firstname.lastname@example.org> wrote:
> I think "X simulates Y, therefore K(X) > K(Y)," is pretty unambiguous in its
> meaning, and it's wrong.
Well, given the context, I did read it as equivalent to your "it
wouldn't be possible to pick out just one program of high Kolmogorov
complexity, and only pay attention to it, while you yourself have low
Kolmogorov complexity" :-)
On the other hand, as I said, the inequality needs to be non-strict
(>=), so there's no contradiction.
> Also, two AIs certainly could conduct a dialog by
> simulating each other, if we're talking about Turing machines with infinite
Yes, although afaict indeed only if K(X) = K(Y). (I don't immediately
see a proof that goes by constructing other machines that do halt, but
if X "simulates" Y, it presumably needs a precise description of Y,
which implies K(X) >= K(Y); by a symmetrical argument, K(Y) >= K(X).
Matt mentioned that argument in a different mail.)
>> Matt, as far as I can see you're wrong, though; K(X) = K(Y), no
>> contradiction. Yes, if X "did something besides simulating Y," in a
>> certain sense, then you would have K(X) > K(Y).
> "Kolmogorov complexity" does not refer to the same thing as the English word
> "complexity." In particular, doing more things can actually decrease your
> Kolmogorov complexity. A program that outputs Wikipedia would be pretty
> long, but a program that outputs every string, including Wikipedia, can be
> very short.
Yes, but that's not the particular "certain sense" I was referring to.
Which was: If X contains instructions to simulate Y, specifically, and
additionally contains other instructions which Y does not include, say
for the computing the utility of possible universes if you like pie,
then K(X) > K(Y), and Y cannot simulate X specifically and conduct a
dialog with it. (But, as was my argument to Matt, since Y by
assumption wants to simulate X, it can compute the source code of X,
which means that Y *does* contain bits that describe the function to
calculate pie-utility, even though Y does not calculate pie-utility
All the best,
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