Re: decision theory papers

From: Marcus Hutter <>
Date: Tue, 23 Apr 2002 11:50:03 +0200

Dear Everyboy on the Everything list,

After having followed the discussions in this list for a
while I would like to make my first contribution:

The paradox between computability and free will vanishes through
careful reasoning:

That a part of the universe is computable is defined
as follows:

Assumption 1: Given a box (part of the universe) in state s at time
t we can compute the next (or some farther future) state
s' at time t'>t IF there is no interaction of the box with the
rest of the universe during time t...t'.

Without this independence assumption in time-interval [t,t'] the
possibility of correct prediction cannot be guaranteed!

Assumption 2: Assume that the brain is computable. It gets input x
at time t and computes action y at time t'. During the thinking
period [t,t'] it is completely separated from the environment.

After input x the brain B is in a state s and Assumption 1
applies, i.e. we can compute, say with algorithm A:X->Y, the
brains decision y. We can't tell the brain in the period [t,t']
this decision without violating Assumption 2.

Assume we allow this interaction, then the brain B' maps input
(x,y) to an action y', which is possibly different from y. There
is no contradiction, since A maps X to Y whereas B' maps X x Y ->
Y, so these functions have nothing to do with each other.

Assume now that a part B2 of brain B=B1 can simulate B1. Since B2
is assumed to behave identically to B1 it must itself contain a
part, say B3 which simulates B2, etc. We have an infinite
recursion as already mentioned by Brent Meeker in this thread. The
first question is NOT WHAT the output of B is and whether it is
finitely computable but whether this infinite recursion has a
value (is mathematically sound) AT ALL!

What we need is a fixed point. Insert a function A into brain B
(as a possible candidate for B2) and look whether B computes the
same function. If yes, A (and B) is a fixed point of the
recursion. If such a fixed point exists (and is unique) we may
define the value of the infinite recursion as this fixed point
value. Finally we would have to check whether this fixed point
can be found by a finite algorithm.

It is well known that not every recursion y=f(y) has a fixed
point. The paradox in our case is just that we implicitly assumed
the existence of a (unique) fixed point. The paradox resolves by
noting that this fixed point simply does not exist.

Sometimes fixed points can be found by iteration. You start with
some value y1 for y and iterate y2=f(y1), ..., yn=f(y_n-1). If the
limit y_\infty exists, then it is a fixed point.

Assume our function B as act 1 if B2 predicts 0 and vice versa.
This is the paradox discussed in this thread. In this case
y_n=1-y_n-1 oscillates and y_\infty does not exist. (In the case
of a binary decision this proofs that a fixed point does not
exist). We are talking about non-existent fixed-points. We simply
cannot construct a self-contradictory brain from Assumptions 1 and

Marcus Hutter

P.S. I maintain the Kolmogorov complexity mailing list.
Maybe you want to have a look at
Dr. Marcus Hutter, IDSIA
Istituto Dalle Molle di Studi sull'Intelligenza Artificiale
Galleria 2 CH-6928 Manno(Lugano) - Switzerland
Phone: +41-91-6108668 Fax: +41-91-6108661
Received on Tue Apr 23 2002 - 02:56:55 PDT

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