This is incorrect. The problem is that you won't be able to tell if certain small Turing machines halt to give s, or loop forever. (So, if you are dovetailing t

This is incorrect. The problem is that you won't be able to tell if certain small Turing machines halt to give s, or loop forever. (So, if you are dovetailing through every possible Turing machine, the first one to output s may not be the minimal one. If you are not dovetailing, your search procedure will not halt, as you'll become stuck enumerating a Turing machine which does not halt.) (Besides, that there is an "algorithm to compute K(s) for any particular s" directly contradicts the non-computability of K(s).)