You could say this—doing so would be like describing your own language in which things involving 12,345,346,437,682,315,436 can be expressed concisely.
So Kolmogorov complexity is somewhat language-dependent. However, given two languages in which you can describe numbers, you can compute a constant such that the complexity of any number is off by at most that constant between the two languages. (The constant is more or less the complexity of describing one language in the other). So things aren’t actually too bad.
But if we’re just talking about Turing machines, we presumably express numbers in binary, in which case writing “3” can be done very easily, and all you need to do to specify 3^^^3 is to make a Turing machine computing ^^^.
However, given two languages in which you can describe numbers, you can compute a constant such that the complexity of any number is off by at most that constant between the two languages.
But can’t this constant itself be arbitrarily large when talking about arbitrary numbers? (Of course, for any specific number, it is limited in size.)
The constant depends on the two languages, but not on the number. As army1987 points out, if you pick the number first, and then make up languages, then the difference can be arbitrarily large. (You could go in the other direction as well: if your language specifies that no number less than 3^^^3 can be entered as a constant, then it would probably take approximately log(3^^^3) bits to specify even small numbers like 1 or 2.)
But if you pick the languages first, then you can compute a constant based on the languages, such that for all numbers, the optimal description lengths in the two languages differ by at most a constant.
The context this in which this comes up here generally requires something like “there’s a way to compare the complexity of numbers which always produces the same results independent of language, except in a finite set of cases. Since that set is finite and my argument doesn’t depend on any specific number, I can always base my argument on a case that’s not in that set.”
If that’s how you’re using it, then you don’t get to pick the languages first.
You do get to pick the languages first because there is a large but finite (say no more than 10^6) set of reasonable languages-modulo-trivial-details that could form the basis for such a measurement.
You could say this—doing so would be like describing your own language in which things involving 12,345,346,437,682,315,436 can be expressed concisely.
So Kolmogorov complexity is somewhat language-dependent. However, given two languages in which you can describe numbers, you can compute a constant such that the complexity of any number is off by at most that constant between the two languages. (The constant is more or less the complexity of describing one language in the other). So things aren’t actually too bad.
But if we’re just talking about Turing machines, we presumably express numbers in binary, in which case writing “3” can be done very easily, and all you need to do to specify 3^^^3 is to make a Turing machine computing ^^^.
But can’t this constant itself be arbitrarily large when talking about arbitrary numbers? (Of course, for any specific number, it is limited in size.)
Well… Given any number N, you can in principle invent a programming language where the program
do_itoutputs N.The constant depends on the two languages, but not on the number. As army1987 points out, if you pick the number first, and then make up languages, then the difference can be arbitrarily large. (You could go in the other direction as well: if your language specifies that no number less than 3^^^3 can be entered as a constant, then it would probably take approximately log(3^^^3) bits to specify even small numbers like 1 or 2.)
But if you pick the languages first, then you can compute a constant based on the languages, such that for all numbers, the optimal description lengths in the two languages differ by at most a constant.
The context this in which this comes up here generally requires something like “there’s a way to compare the complexity of numbers which always produces the same results independent of language, except in a finite set of cases. Since that set is finite and my argument doesn’t depend on any specific number, I can always base my argument on a case that’s not in that set.”
If that’s how you’re using it, then you don’t get to pick the languages first.
You do get to pick the languages first because there is a large but finite (say no more than 10^6) set of reasonable languages-modulo-trivial-details that could form the basis for such a measurement.