# Diagonal Cellular Automata

Posted 3 years ago
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 I was amused by the post 1-D Cellular Automata With a Twist, so I thought I'd take a look. 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] 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 1504Here's 1240 Here's 1263
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Posted 3 years ago
 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 3 years ago
 420 -- 3 colors
Posted 3 years ago
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Posted 3 years ago
 As an answer to Vitaliy's question, you can use the CellularAutomaton function, but there is subtle issue with the boundary conditions. In the Diagonal CA, you start with a single gray cell (value 1 out of 1,2, or 0) and when you need a new boundary cell it is always white (or 0). In the CellularAutomaton function, if the rule sends white cells to white (or quiescent) then it is the same, but otherwise the boundary cells are not necessarily going to be white (or 0). Because of this the evolutions are not the same in general.But when the rule is quiescent then they are the same. One other difference should be mentioned, that Ed's rule numbering is backwards from the usual conventions. So Ed's rule 420 = 0 1 2 0 1 2 0 = CellularAutomaton 588 = 0 2 1 0 2 1 0, which is quiescent. The rule type is a second order CA because it depends on the previous diagonal and the diagonal before that. Here is an implementation as a radius 1/2 rule.  hist=CellularAutomaton[{588, {3, {{1, 0}, {1, 1}}}, 1/2, 2}, PadLeft[{{1}}, {2, 32}, 0, {0, 31}], 30]; Graphics[MapIndexed[If[#2[[1]] < #2[[2]], Nothing, {GrayLevel[1 - #/2], Rectangle[{#2[[2]], 30 - #2[[1]] + #2[[2]]}]}] &, hist, {2}]] (I'd post an image, but something strange happened with the interface.)
Posted 3 years ago
 Todd, how would the 3D diagonal CA be set up? It seems obvious that we should find out if a "1D" cellular automaton can build a menger sponge.
Posted 3 years ago
 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 3 years ago
 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 3 years ago
 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}}}]] 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 2 years ago
 Todd, how would CellularAutomaton be set up to do rule 214 on NKS 437 ?
Posted 2 years ago
 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 2 years ago
 Thanks! I got it this way: CellularAutomaton[{10710, {2, {{0, 8, 0}, {4, 2, 1}}}, 1, 2}, {{{1}, {1}}, 0}, 31] 
Posted 2 years ago
 Yes, that's a good way too. You can't go wrong with powers of two.
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