3 ms·
>Can't you trivially force this to happen for any sequence? No, because there's no deterministic way to infinitely extend that sequence. In your first example:
by MontyCarloHall 9mo ago
>Can't you trivially force this to happen for any sequence?
No, because there's no deterministic way to infinitely extend that sequence. In your first example:
1 3 3 3 2 2 2 1 1 1 4 4 x x y y
1 3---- 3---- 3---- 2-- 2-- 2--
What are the values of x and y?
>Whenever you start a new count, choose a random number and repeat for a many times as needed for the trailing sequence to match the top sequence.
You answered your own question. The Kolakoski sequence is special because it does not just pick a random number: the sequence is deterministically encoded by the run lengths, and vice versa.
- AnotherGoodName 9mo agoPick any number that's not 4 and repeat it twice. For the next 2 after that pick any number that's not that previous number and repeat it twice. So on. It's not like i had any difficulty coming up with that sequence i wrote to that point. 1 3 3 3 2 2 2 1 1 1 4 4 5 5 6 6 8 1 2... as an example of how trivial it is to continue to this. I get that if you limit yourself to '1' or '2' you force the choice of next number but even then there's two possibilities of this sequence (start on 1 vs start on 2).
- MontyCarloHall 9mo agoYou are still choosing random numbers here. The Kolakoski sequence has zero randomness whatsoever.
- AnotherGoodName 9mo agoDo we care similarly about the version of this that starts on 2? 2, 2, 1, 1, 2, 1, 2... 2, 2, 1, 1, Again no randomness. Just a variant of the above and trivial to continue this.
- MontyCarloHall 9mo agoThis is just the Kolakoski sequence starting from the second term.
- AnotherGoodName 9mo agoAhh that makes sense. Thank you!
- flufluflufluffy 9mo agoI guess you could say the Kolakoski sequence is special in being the “simplest” version of such a sequence (ignoring the finite trivial case {1} xD)