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

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?

Asked By : pintoch

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.

Consider this instead:

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

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.

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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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