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What is an estimation of the Kolmogorov Complexity for the first N integers?
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Problem
I'm aware some ints have higher or lower Kolmogorov Complexities. For example, the number
5.41126806512 has a very low complexity as it can be expressed by 17/pi. I'm also aware that, though the KC varies depending on the expression language, it is always the same up to a given constant. So, I ask: is there a way to calculate an approximation of the KC for the first N ints?Solution
No, there is no general algorithm to compute a close approximation to the Kolmogorov complexity of the sequence $1,2,\dots,n$. Any candidate algorithm you come up with will have some inputs where it gives a bad answer (a poor approximation to the correct answer).
Denote by $[n]$ the binary encoding of the natural number $n$, and let $[[n]] = [1],[2],\ldots,[n]$ denote a comma-separated encoding of the sequence $1,\ldots,n$. It is not hard to check that $|K([n]) - K([[n]])| = O(1)$. Since computing $K([n])$ is undecidable, it follows that computing a good approximation to $K([[n]])$ is also undecidable.
Furthermore, it is known that for "most" integers $n$, $K([n]) = \Theta(\log n)$. So for "most" integers $n$, $K([[n]]) = \Theta(\log n)$.
Denote by $[n]$ the binary encoding of the natural number $n$, and let $[[n]] = [1],[2],\ldots,[n]$ denote a comma-separated encoding of the sequence $1,\ldots,n$. It is not hard to check that $|K([n]) - K([[n]])| = O(1)$. Since computing $K([n])$ is undecidable, it follows that computing a good approximation to $K([[n]])$ is also undecidable.
Furthermore, it is known that for "most" integers $n$, $K([n]) = \Theta(\log n)$. So for "most" integers $n$, $K([[n]]) = \Theta(\log n)$.
Context
StackExchange Computer Science Q#19615, answer score: 6
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