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Diagonal Cellular Automata

Posted 8 months 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.

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

12 Replies

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 8 months ago

420 -- 3 colors

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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 7 months 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 7 months 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]

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.

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} &]

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

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] 

Thanks! I got it this way:

CellularAutomaton[{10710, {2, {{0, 8, 0}, {4, 2, 1}}}, 1, 2}, {{{1}, {1}}, 0}, 31]

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

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