Kolmogorov complexity

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Intuitively, the Kolmogorov complexity of an integer [math]n[/math] is the least number of bits one needs to describe [math]n[/math] in some suitable language. For instance, one could use the language of Turing machines: an integer [math]n[/math] has Kolmogorov complexity at most [math]k[/math] if there exists a Turing machine with length at most [math]k[/math] which, when run, produces [math]n[/math] as the output.

Thus, for instance, any [math]k[/math]-bit integer has Kolmogorov complexity at most [math]k+O(1)[/math] (the [math]O(1)[/math] overhead being for the trivial program that outputs the remaining [math]k[/math] 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 [math]k[/math] (we'll refer to this as the counting argument). As a consequence, most [math]k[/math]-bit integers have Kolmogorov complexity close to [math]k[/math].

On the other hand, a deterministic algorithm which takes [math]k[/math] as input and runs in time polynomial in [math]k[/math] 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 [math]O(1)[/math] 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 [math]k[/math] is).