Kolmogorov complexity: Difference between revisions
New page: Intuitively speaking, the Kolmogorov complexity of an integer n is the least number of bits one needs to describe n in some suitable language. For instance, one could use the language of ... |
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On the other hand, a deterministic algorithm which takes k as input and runs in time polynomial in k can only produce integers of Kolmogorov complexity <math>O(\log k)</math>, since any number produced can be described by a Turing machine of length <math>O(\log k) + O(1)</math> (the O(1) is to describe the algorithm, and the <math>O(\log k)</math> bits are to describe how long the program is to run and what k is). | On the other hand, a deterministic algorithm which takes k as input and runs in time polynomial in k can only produce integers of Kolmogorov complexity <math>O(\log k)</math>, since any number produced can be described by a Turing machine of length <math>O(\log k) + O(1)</math> (the O(1) is to describe the algorithm, and the <math>O(\log k)</math> bits are to describe how long the program is to run and what k is). | ||
* [[wikipedia:Kolmogorov_complexity|The Wikipedia entry for Kolmogorov complexity]] | |||
Revision as of 16:09, 19 August 2009
Intuitively speaking, the Kolmogorov complexity of an integer n is the least number of bits one needs to describe n in some suitable language. For instance, one could use the language of Turing machines: an integer n has Kolmogorov complexity at most k if there exists a Turing machine with length at most k which, when run, will halt to produce n as output.
Thus, for instance, any k-bit integer has Kolmogorov complexity at most k+O(1) (the O(1) overhead being for the trivial program that outputs the remaining k bits of the Turing machine). On the other hand, it is obvious that there are at most [math]\displaystyle{ 2^k }[/math] integers of Kolmogorov complexity at most k (we'll refer to this as the counting argument). As a consequence, most k-bit integers have Kolmogorov complexity close to k.
On the other hand, a deterministic algorithm which takes k as input and runs in time polynomial in k can only produce integers of Kolmogorov complexity [math]\displaystyle{ O(\log k) }[/math], since any number produced can be described by a Turing machine of length [math]\displaystyle{ O(\log k) + O(1) }[/math] (the O(1) is to describe the algorithm, and the [math]\displaystyle{ O(\log k) }[/math] bits are to describe how long the program is to run and what k is).
