Kolakoski Sequence

▲ Kolakoski Sequence(en.wikipedia.org)

60 points by surprisetalk 6 days ago | 12 comments

▲ MontyCarloHall 6 hours ago | parent | next [-]

For those not getting this immediately (I sure didn't):

              _____________
   sequence   1 2 2 1 1 2 1 2 2 1
   run lens   1 2-- 2-- 1 1 2-- 1

Read out the bottom sequence of run lengths, and be amazed that it's the same as the first 7 digits as the top sequence. Extend the bottom to continue to recapitulate the top sequence, and add terms to the top sequence accordingly to reflect the run lengths in the bottom sequence. Repeat infinitely.

▲ AnotherGoodName 5 hours ago | parent [-]

Can't you trivially force this to happen for any sequence?

1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 4, 4,

Goes to

1, 3, 3, 3, 2... Etc.

I could extend this trivially too since the bottom sequence trails the sequence we write up top. If i wanted another '2' down the bottom whatever number i choose up top i just write twice right?

So there's nothing about this particular sequence? I can just create any such sequences trivially; 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.

It seems that this particular variant is uninteresting in the broader picture right? I could write another similar one

2, 2, 1, 1, 2, 1

2, 2, 1, 1, ... etc.

I don't get the specialness here?

▲ MontyCarloHall 5 hours ago | parent [-]

>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.

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AnotherGoodName 5 hours ago | parent [-]

Pick 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).

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flufluflufluffy 4 hours ago | parent | next [-]

I 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)

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MontyCarloHall 5 hours ago | parent | prev [-]

You are still choosing random numbers here. The Kolakoski sequence has zero randomness whatsoever.

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AnotherGoodName 5 hours ago | parent [-]

Do 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.

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MontyCarloHall 5 hours ago | parent [-]

This is just the Kolakoski sequence starting from the second term.

	▲	AnotherGoodName 3 hours ago \| parent [-]
		Ahh that makes sense. Thank you!

▲ nwellnhof 5 hours ago | parent | prev | next [-]

There's a fascinating way to generate the Kolakoski sequence with bit fiddling: https://11011110.github.io/blog/2016/10/14/kolakoski-sequenc...

▲ vindex10 5 hours ago | parent | prev | next [-]

Is it a coincidence that it is number 2 in the OEIS?)

▲ globalnode 15 minutes ago | parent | prev [-]

does it repeat? or is it like pi? -- oh the wiki says its an infinite sequence. that might be handy for generating pseudo random numbers i guess. i wonder if theres a function that generates the correct digit at position n? doh, wiki says theres a recursive formula for the i'th term.