3 ms·

You pick an obvious encoding (such as binary) yourself, in the same way your computer is not outputting some platonic ideal "A" but a series of electrical impul

You pick an obvious encoding (such as binary) yourself, in the same way your computer is not outputting some platonic ideal "A" but a series of electrical impulses that your monitor plus your eyes and brain interprets as "A". Sure, you can object that the encoding is "outside" the TM, but for the purposes of discussing complexity these objections are pretty trivial, again for the same reasons (whatever encoding you pick the conversion process is a program you can write down, and once you write it down it means the Kolmogorov Complexity is the same between different TMs up to the length of whatever encoding/decoding program you come up with). Put another way, a TM with alphabet is {0, 1} is technically not the same as the TM with alphabet {A, B}. But it's obvious to us that the TMs are equivalent.

rhelz 3y ago
Well, here's why it's not that simple....I can make any number N have a kolmogorov complexity number equal to 1--if the Turing machine has N+1 symbols :-) I just express the number in base N. (which will bet "10" for any base :-) It true that we typically limit ourselves to binary when we are proving theorems, etc in Kolmogorov complexity. You can prove that for any two turing machines U and V, KU(X) <= KV(X) + O(1).....but this relies on the fact that there is an O(1)-sized program which lets U emulate V, and V emulate U. And that is only true if U and V share the same symbol set. If they don't, then the kolmogorov complexity of the two machines can be made arbitrarily different from each other, just by changing the symbol set.