Vladimir_M comments on Does Solomonoff always win? - Less Wrong

11 Post author: cousin_it 23 February 2011 08:42PM

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Comment author: Vladimir_M 24 February 2011 05:21:18AM *  0 points [-]

From what I understand, Omega is defined simply as a function F : L -> {0,1}, where L is the set of all strings over {0,1}, and F((b_1,...,b_n)) is interpreted as its prediction of what the bit b_{n+1} will be given the initial bits b_1,...,b_n. Then you can always define the "malevolent" bit sequence simply as b_{n+1} = ~F((b_1,...,b_n)). F can be wildly uncomputable, of course, so it makes no sense to ask about the Kolmogorov complexity of these bit sequences.

The generalization where Omega also assigns probabilities rather than just making binary guesses seems obvious.

Comment author: ciphergoth 24 February 2011 10:32:57AM 3 points [-]

it makes no sense to ask about the Kolmogorov complexity of these bit sequences.

The bit sequences are finite length at each step, so they do have a Kolmogorov complexity.