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# [Solved]: When does the function mapping a string to its prefix-free Kolmogorov complexity halt?

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Problem Detail:

In Algorithmic Randomness and Complexity from Downey and Hirschfeldt, it is stated on page 129 that

$\qquad \displaystyle \sum_{K(\sigma)\downarrow} 2^{-K(\sigma)} \leq 1$,

where $K(\sigma)\downarrow$ means that $K$ halts on $\sigma$, $\sigma$ being a binary string. $K$ denotes the prefix-free Kolmogorov complexity.

When does $K$ halt? I think it only halts on a finite number of inputs, since the classical proof on non-computability of the Kolmogorov complexity gives an upper bound on the domain of $K$. But then, the finite set of inputs on which $K$ halts can be chosen arbitrary (one just needs to store the finite number of complexities in the source code).

So is this sum well-defined? In other words, is the domain of $K$ well defined?

#### Answered By : Raphael

I think you are right; $K$ is a specific function which can not be computed. The author likely means to use some (arbitrary) approximative implementation; so no, this does not seem to be well-defined, if you are pedantic. You can also call it abuse of notation.

$\qquad \displaystyle \forall {M \in \mathcal{M}_K}.\ \sum_{M(\sigma)\downarrow} 2^{-K(\sigma)} \leq 1$

with $\mathcal{M}_K = \{M\ \mathrm{TM} \mid M(\sigma)\!\downarrow\ \implies\ M(\sigma)=K(\sigma) \}$ the set of all Turing machines that compute subfunctions of $K$.

In essence, this means: the bound holds no matter for which strings your implementation can compute the Kolmogorov complexity.

As Carl notes in the comments, it is plausible that the notation has nothing to do with halting or computing, as $K$ is not computable. Read $\sum_{K(\sigma)\!\downarrow}$ as sum ranging over the domain of $K$.

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