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El'endia Starman
El'endia Starman
00:48
Feb 15 at 0:20, by El'endia Starman
So one can generate a row of the Sierpinski triangle, chop it up into chunks of n, reverse every other chunk, sum up the columns, and look at the parity to see what cells are alive (odd if they are, even if not).
@orlp: I never did hear what you thought about this.
orlp
orlp
sec
need to think
@El'endiaStarman that won't be any faster unless you've got some special formula to get a row of the siepinski triangle
El'endia Starman
El'endia Starman
That's basically what I'd like to see. If we can figure out a formula, we can possibly work backwards from there and find a closed or iterative/recursive solution to the original problem.
orlp
orlp
@El'endiaStarman also we can only use this to test upper bound conjectures for periods
actually
if we have a conjecture for a period, and it works we only also need to check its divisors
El'endia Starman
El'endia Starman
yeah
orlp
orlp
01:09
oftenpaper.net/sierpinski.htm
@El'endiaStarman there's a formula
pascal's triangle, see if the numbers are odd
El'endia Starman
El'endia Starman
01:25
I'm not sure that'll work for the CA version.
There are gaps...
orlp
orlp
@El'endiaStarman actually, there's another problem
to get to generation n you need to generate a triangle with n elements (and then reverse and fold over itself)
so if we need to get to generation 2^100...
El'endia Starman
El'endia Starman
Yeah, that's not gonna work.
If we can calculate a row directly, that'd help (I think).
El'endia Starman
El'endia Starman
01:47
@orlp This is an amazing rabbit hole...
By the way, where he talks about Towers of Hanoi, I think that's the structure we get with the CA.

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