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Diagonal Cellular Automata

Posted 5 years ago
POSTED BY: Ed Pegg
12 Replies

Yes, that's a good way too. You can't go wrong with powers of two.

POSTED BY: Todd Rowland

Thanks! I got it this way:

CellularAutomaton[{10710, {2, {{0, 8, 0}, {4, 2, 1}}}, 1, 2}, {{{1}, {1}}, 0}, 31]
POSTED BY: Ed Pegg
POSTED BY: Todd Rowland

Todd, how would CellularAutomaton be set up to do rule 214 on NKS 437 ?

POSTED BY: Ed Pegg
POSTED BY: Todd Rowland
POSTED BY: Todd Rowland

Nice!

n=10;
array=Transpose[CellularAutomaton[{588,{3,{{1,0},{1,1}}},1/2,2},PadLeft[{{1}},{2,2^n},0,{0,2^n-1}],2^n-2]];
new=Table[RotateLeft[array[[n]],n-1],{n,1,Length[array]}];
ArrayPlot[Transpose[new], PixelConstrained->1, Frame-> False]
POSTED BY: Ed Pegg
POSTED BY: Todd Rowland
POSTED BY: Ed Pegg
POSTED BY: EDITORIAL BOARD
POSTED BY: Vitaliy Kaurov

420 -- 3 colors

POSTED BY: Ed Pegg
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