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Sep 5, 2017 at 14:11 vote accept yters
Sep 3, 2017 at 14:25 comment added Yuval Filmus No, the theorem only states that there is a limit $L$ above which you cannot $K(x) > L$. Evidently this limit is larger than 0.
Sep 3, 2017 at 14:24 comment added yters For the prefix free variant, if the conditional Kolmogorov complexity is always positive, does that result in the contradiction with Chaitin's incompleteness theorem?
Sep 3, 2017 at 7:57 history answered Yuval Filmus CC BY-SA 3.0