Difference between revisions of "Kolmogorov complexity"

From Polymath1Wiki
Jump to: navigation, search
(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 ...)
 
(add a link to Kolmogorov complexity as explained on wikipedia)
Line 4: Line 4:
  
 
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 17:09, 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]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]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).