From: Tennessee Leeuwenburg (tennessee@tennessee.id.au)
Date: Tue Feb 22 2005 - 21:37:10 MST
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Ben,
I read your idea on actions based on proven theories, subject to a
blacklist against which unsafe options are blocked. You seemed to be
worried that such a system would be susceptible to improve itself to a
local minima - that is, it would be hill-climbing to a potentially
limiting endpoint.
Well, here is a possible way out. (I am attempting to express myself
formally for the first time in a long time, sorry if I make a hash of it)
Let IQ(AGI) be the intelligence of AGI.
Let OPTIONS be the set of future actions known to the AI, including
doing nothing
Let V(X) be the heuristic value of option X upon evaluation, including
safety margin
Let V(now) be the heuristic value of doing nothing
Let Max(Options) be the option within OPTIONS whose V(X) is greater
than all other options
Let us assume that V(X) = M(X) * S(X)
Let M(X) represent the motive value of X - a representation of the
distance towards AGI's goals it is able to move
Let S(X) represent the safety factor.
Now, I am assuming that *increasing intelligence* is a factor in the
motive value of X.
Proposition : A more intelligent AGI will be able to percieve more
options than a less intelligent one - in your parlance, perhaps it
will be able to evaluate more theorems, have better heuristics where
heuristics are necessary, evaluate more complex heuristics etc.
Let Mag(OPTIONS) represent the number of elements contained in the set
OPTIONS.
Mag(OPTIONS) is proportional to IQ(AGI).
Here's my argument :
X = Max(OPTIONS)
If X != now & V(X) > V(now) then IQ(AGI,X) > IQ(AGI,now) which implies
Mag(OPTIONS,AGI, X) > Mag(OPTIONS, AGI, now).
Corrolary : Regardless of the speed at which the intelligence of the
AGI grows, the options available to the AGI increase with the
intelligence of AGI. Incremental increases to IQ do not result in
local minima, because the horizon of the AGI is pushed wider with each
improvement.
Cheers,
- -T
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