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Diagonal Cellular Automata

Ed Pegg, Wolfram Research
Posted 6 years ago

I was amused by the post 1-D Cellular Automata With a Twist, so I thought I'd take a look.

enter image description here

Here's rule 420.

Clear[a];
tab=Table[{a[0,k]=0, a[k,0]=0},{k,0,800}]; 
a[0,0]=1;
rule=IntegerDigits[420,3,7];
a[n_,k_] := a[n,k]= rule[[1+ a[n-1,k -1]+a[n -1,k]+a[n ,k-1]]];
tab=Table[a[m,d-m],{d,2,729},{m,1,d-1}];
ArrayPlot[Table[a[x,y],{x,1,729},{y,1,729}], Frame-> False, PixelConstrained->1]

rule 420

Here are some other nice rules:

nice = {381, 382, 383, 384, 385, 386, 414, 415, 416, 417, 418, 419, 
  420, 421, 422, 424, 478, 613, 622, 623, 624, 625, 626, 657, 658, 
  663, 694, 717, 721, 864, 865, 866, 867, 868, 869, 870, 871, 872, 
  883, 903, 904, 906, 936, 937, 960, 1110, 1111, 1112, 1113, 1114, 
  1115, 1143, 1144, 1145, 1146, 1147, 1148, 1149, 1150, 1230, 1231, 
  1232, 1233, 1234, 1235, 1236, 1237, 1238, 1239, 1240, 1242, 1243, 
  1244, 1245, 1246, 1247, 1254, 1255, 1256, 1257, 1258, 1259, 1260, 
  1261, 1262, 1263, 1265, 1266, 1272, 1274, 1278, 1279, 1280, 1284, 
  1287, 1288, 1342, 1351, 1352, 1354, 1355, 1386, 1387, 1392, 1504, 
  1527, 1531, 1539, 1547, 1548, 1549, 1556, 1557, 1558, 1559, 1560, 
  1561, 1562, 1563, 1564, 1585, 1593, 1594, 1595, 1596, 1597, 1598, 
  1599, 1600, 1601, 1608, 1612, 1632, 1633, 1634, 1635, 1666, 1774, 
  1839, 1840, 1841, 1842, 1843, 1844, 1872, 1874, 1875, 1876, 1877, 
  1878, 1879, 1882, 2071, 2080, 2081, 2082, 2083, 2084, 2121, 2175}

Here's 1504

rule 1504

Here's 1240

rule 1240

Here's 1263 rule 1263
POSTED BY: Ed Pegg
12 Replies
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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

This is one of the reversible rules? 214R ?

It's a second order rule, and there are a few ways of doing that. In general the rule is a list of length four where the last number is the order of the rule.

The best implementation is to convert to a four color rule and then take it Mod[_,2] Going to four colors encodes the previous state. There are more details in my conference presentation from NKS 2007 which was about 37R. Using the general method of going from second order to first order leads to a faster implementation in CellularAutomaton.

RevTo4ColorRule[rn_] := 
 FromDigits[
  FromDigits[{Mod[#[[2]], 2], 
      Mod[Reverse[IntegerDigits[rn, 2, 8]][[
         FromDigits[Mod[#, 2], 2] + 1]] + Quotient[#[[2]], 2], 2]}, 
     2] & /@ Tuples[{3, 2, 1, 0}, 3], 4]
POSTED BY: Todd Rowland

Todd, how would CellularAutomaton be set up to do rule 214 on NKS 437 ?
POSTED BY: Ed Pegg

The 3D case is more complicated to render, and the choice of rendering determines where the offsets go with the weighting. Please forgive me for posting a guess.

    ArrayPlot[CellularAutomaton[{114, {2, 
{{{1, 0}, {0, 0}}, {{1, 0}, {1, 1}}, {{1,1}, {0, 1}}}}, 
{1/2, 1/2}, 3}, {{{{1}}, {{0}}, {{0}}}, 0}, {{{30}}}]]

enter image description here

Not a completely random guess. Look at this graphic, showing the diagonal sheets. It is clearly an order 3 rule. Each sheet is a square lattice (but at a funny angle), and the first and second step back each have three cells, being the closest 3 in the parallelogram. Some combination with one 1 and 3 1's and 3 1's for weights will be correct.

Graphics[MapIndexed[{Hue[Mod[#2[[1]]/5, 1]], Point[#]} &, 
  Select[Union[#.{{1, 1}, {-1, 0}, {0, -1}}] & /@ 
    SortBy[MaximalBy[GatherBy[Tuples[Range[5], 3], #.{1, 1, 1} &], 
      Length, 3], #[[1]].{1, 1, 1} &]
POSTED BY: Todd Rowland

Ed, let me first try amending my earlier post, before thinking about the 3D case.

The best reference for the search for simple rules with interesting behavior is Stephen Wolfram's A New Kind of Science.

The choice of display for the diagonal CA is connected to whether the neighborhood weights are {{1,0},{1,1}} or {{0,1},{1,1}}.

The display I had last time was more simple with

ArrayPlot[ CellularAutomaton[{588, {3, {{1, 0}, {1, 1}}}, 1/2, 2}, {{{1}, {0}},  0}, 30]]

Note that you can take advantage of CellularAutomaton's initial condition syntax for infinite backgrounds.

Many ordinary ECA can be realized as diagonal CA (but not totalistic like these are) including rule 30, and their evolutions can be evaluated in even more bizarre ways.
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

enter image description here -- you have earned Featured Contributor Badge enter image description here

Your exceptional post has been selected for our editorial column Staff Picks http://wolfr.am/StaffPicks and Your Profile is now distinguished by a Featured Contributor Badge and is displayed on the Featured Contributor Board. Thank you, keep it coming, and consider contributing to the The Notebook Archive!
POSTED BY: EDITORIAL BOARD

Very interesting! I wonder if CellularAutomaton function can be harnessed somehow to implement this system for an arbitrary rule. Also rule 420 for how many colors? For elementary binary case of 2 colors should not be there also 256 rules like for Elementary Cellular Automata because the neighborhood is 3 cells? Also, it is interesting to get the analog of Wolfram table for all possible rules of elementary case for simplest non-trivial initial conditions: https://www.wolframscience.com/nks/p55--more-cellular-automata
POSTED BY: Vitaliy Kaurov

420 -- 3 colors
POSTED BY: Ed Pegg
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