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Is there an infinite set of strings whose Kolmogorov complexities are computable?

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    @Aidan Rocke: well, whatever the Kolmogorov complexity of $\langle 0101 \rangle$ is, that is its complexity, which we can hard-code into a program. The issue with computation is when we want a program that can take arbitrary strings as inputs, not just one particular string. Separately, in some universal prefix free codes we do know the complexity of particular strings. For example we can arrange any particular string we want, $s$, to have complexity $1$ by setting up the universal prefix free code $\phi$ so that $\phi(\langle 0\rangle) = s$. That still leaves all strings of form $1\sigma$2016-06-14

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I think you are asking this: is there an infinite r.e. set of pairs $(\sigma,n)$ where $\sigma \in 2^{<\omega}$ is a string of Kolmogorov complexity $n$. The answer to that is no.

For a contradiction, assume such a list is r.e. - then there arbitrarily long strings in it, and thus strings of arbitrarily high Kolmogorov complexity. Define a function $P$ that takes input $\tau \in 2^{<\omega}$ and does the following. First, it effectively enumerates that list until it finds a pair $(\sigma, n)$ where $n > |\tau|$. Then it prints out $\sigma$.

The assumptions we have made ensure that $P$ is a total computable function. Therefore, applying Kleene's recursion theorem to $P$ gives a program $e_0$ that, when run with no input, computes $P(e_0)$. Thus the output of program $e_0$, run with no input, is a string of Kolmogorov complexity larger than $|e_0|$, which is impossible.

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    @Ross Millikan: The problem is that the $\log(n)$ type bound is just an upper bound on the complexity. The question asks for an infinite set on which we can compute the complexity exactly, not just provide an upper bound. An exact computation like that is impossible, which is what my answer is intended to prove.2010-12-09