# Simulate Rule 110

Rule 110 is a cellular automaton with some interesting properties. Your goal is to simulate a rule 110 in as few characters as possible.

For those who don't know, rule 110 is simulated line by line in a grid. Each square in a line of the grid looks at the squares above, above-left and above-right to determine what cell it should be.

current pattern  111 110 101 100 011 010 001 000
new cell          0   1   1   0   1   1   1   0


Input: numbers from 0 to 39 representing top row nth input square, in any reasonable format (comma-separated string, list, function arguments). To accommodate 1-indexed languages, numbers may also be 1-indexed and so range from 1 to 40.

Example input:

38,39


Output: A 40 x 40 grid representing the automata running including the first row. You should leave 0 as blank and 1 as any visible printing character. Trailing spaces are allowed, as long as the actual grid can reasonably be distinguished. The bottom of the grid may have a newline but there should not be blank lines between grid lines.

Example output:

                                  XX
XXX
XX X
XXXXX
XX   X
XXX  XX
XX X XXX
XXXXXXX X
XX     XXX
XXX    XX X
XX X   XXXXX
XXXXX  XX   X
XX   X XXX  XX
XXX  XXXX X XXX


etc.

Note: A similar question about 1D cellular automata has already been asked, but I hope that, by using only one rule, shorter answers can be written.

• Do the patterns wrap around (i.e. does the left-most cell check the right-most cell in the line above it)? – Ventero Jul 14 '14 at 7:15
• If it's singular, then it's a cellular automaton. – ClickRick Jul 14 '14 at 9:14
• The answers may be fractionally shorter than Simulate any 1D cellular automaton because the rule is hard-coded rather than having to be parsed from the input, but other than that the answers are going to be the same. If it were a different rule then there would be potential for savings, but how on Earth would special-casing a Turing-powerful rule save anything over a general implementation? – Peter Taylor Jul 14 '14 at 12:57
• @Ventero They do not in this version. – qwr Jul 14 '14 at 22:17
• @BMO this is an old question, but since nowadays consensus is to allow flexible input formats, I'll modify the question to allow it – qwr Dec 5 '18 at 20:46

# CJam - 47

S40*l',/{i'!t}/{N40,S3$S++f{>3<2b137Yb='!^}}39*  It uses ! for "1" cells. Try it at http://cjam.aditsu.net/ Explanation: S40* makes a string (array) of 40 spaces l',/ reads a line and splits by comma {…}/ executes the block for each item (the numbers in string form) - i'!t converts the number to integer and sets the item at that position in the previous string (initially 40 spaces) to '!' At this point we have obtained the first line. {…}39* executes the block 39 times - N adds a newline - 40, makes the array [0 1 … 39] - S3$S++ copies the previous line (position 3 on the stack) and pads it with a space on each side
- f{…} executes the block for {each number from 0 to 39} and {the padded line}
-- >3< takes a slice of 3 items from the padded line starting at the current number
-- 2b converts from base 2; the items we sliced are not base-2 digits, but characters get converted to their ASCII values and ' ' mod 8 is 0 and '!' mod 8 is 1
-- 137Yb converts 137 to base 2 (Y = 2), obtaining [1 0 0 0 1 0 0 1], which is 110 reversed and negated (on 8 bits)
-- ='!^ gets the corresponding base-2 digit (the array wraps around so the index is taken mod 8) and xor's it with the '!' character, resulting in '!' for 0 and ' ' for 1

## Ruby, 113 characters

c=[0]*41
eval"[#{gets}].map{|i|c[i]=1}"+'
c=(0..39).map{|x|putc" X"[u=c[x]]
110[4*c[x-1]+2*u+c[x+1]]}<<0;puts'*40


Takes input on stdin. To use a different rule, simply replace the 110 in the last line with whatever rule you want to try.

Example:

$ruby 110.rb <<< 38,39 XX XXX XX X XXXXX XX X XXX XX XX X XXX XXXXXXX X XX XXX XXX XX X XX X XXXXX XXXXX XX X XX X XXX XX XXX XXXX X XXX XX X XX XXXXX X XXXXXXXX XX XXX XX XXXX XX X XXX XX X XXXXX XX X XXX XXXX X XXXXX XX XXX X XX XX X XXXXX X XX XXX XXX XX XX XXXXXXXX X XX X XXXXXX XX XXX XXXXXXX X XXX XX X XX X XXXX X XXXXX XXX XX XX XXX XX X XX X XXX XXX XX X XXX XX XXXXX XX XXX XXXXXX XX X XXX XX X XXXXX XXX XXXXXXXX X XXX XXXX XXX X XX XXX XX X XX X XX XXX XXX XX X XXXXXXXX XX XXXXX X XX X XXXXX XX XXXXXX XXXXXXXX XX X XXX XX X XX X XXX XX XX X XXX XX XXX XX XX X XXX XXXXX XX X XXXXX X XXXXXXXXXX X XX X XXXXX XX XXX XX XXX XXX XX XX XXXX XX X XXX XX X XX X XXXXXX XX X XXXXX XX X XXXXX XXXXXX X XXX XXXX XXXXXX XX X  ## Mathematica, 122 bytes f[a_]:=Riffle[CellularAutomaton[110,Array[If[MemberQ[ToExpression["{"<>a<>"}"],#-1],1,0]&,40],39]/.0->" "/.1->"X"," "]<>""  Yes, you might view this as abusing this loophole, but a) that loophole is quite disputed, b) a Cellular Automaton question needs a Mathematica answer (especially one about Rule 110) and c) Ventero's Ruby answer is shorter anyway, so I don't think any harm is done. Most of the characters are used for input parsing and output formatting. The actual automaton is simulated using CellularAutomaton[110,initialGrid,39]  This uses periodic boundary conditions (so the grid wraps around). # Python - 141 i=input() o=range(40) l=''.join(' X'[c in i]for c in o) for r in o:print l;l=''.join('X '[l[c-1:c+2]in('XXX',' ','X ','',' ')]for c in o)  Run as e.g. python 110.py <<< 38,39 • ['X',' '] could be changed to 'X ' to save 5 characters. – Calvin's Hobbies Jul 14 '14 at 8:14 • My favourite fruit is now an o=range() – kitcar2000 Jul 15 '14 at 17:19 # q, 6762 58 bytes Assumes no wrap-around: {40{not(2 sv'flip 1 0 -1 xprev\:x)in 0 4 7}\@[40#0b;x;~:]}  Old version {40{not(flip(prev;::;next)@\:x)in 3 cut 111100000b}\@[40#0b;x;not]} {40{not(flip 1 0 -1 xprev\:x)in 3 3#111100000b}\@[40#0b;x;~:]}  # Python, 186 def r(s,m=range(40)): s=[int(i in s)for i in m] for g in m:print''.join([' X'[i]for i in s]);s=[int(not''.join(map(str,s[i-1:i+2]if i else s[:2]))in'111 100 000 00'.split())for i in m]  Decent but probably not optimal. You didn't specify how input is gotten so I just made a function. Use example: r([38,39]) Output:  XX XXX XX X XXXXX XX X XXX XX XX X XXX XXXXXXX X XX XXX XXX XX X XX X XXXXX XXXXX XX X XX X XXX XX XXX XXXX X XXX XX X XX XXXXX X XXXXXXXX XX XXX XX XXXX XX X XXX XX X XXXXX XX X XXX XXXX X XXXXX XX XXX X XX XX X XXXXX X XX XXX XXX XX XX XXXXXXXX X XX X XXXXXX XX XXX XXXXXXX X XXX XX X XX X XXXX X XXXXX XXX XX XX XXX XX X XX X XXX XXX XX X XXX XX XXXXX XX XXX XXXXXX XX X XXX XX X XXXXX XXX XXXXXXXX X XXX XXXX XXX X XX XXX XX X XX X XX XXX XXX XX X XXXXXXXX XX XXXXX X XX X XXXXX XX XXXXXX XXXXXXXX XX X XXX XX X XX X XXX XX XX X XXX XX XXX XX XX X XXX XXXXX XX X XXXXX X XXXXXXXXXX X XX X XXXXX XX XXX XX XXX XXX XX XX XXXX XX X XXX XX X XX X XXXXXX XX X XXXXX XX X XXXXX XXXXXX X XXX XXXX XXXXXX XX X  • I did specify input: in your case you would have to use input() and format the input as specified in the original post. – qwr Jul 15 '14 at 19:09 # Mathematica, 113 chars Another Mathematica answer using CellularAutomaton. Print@@" "["X"][[#]]&/@CellularAutomaton[110,SparseArray[#+1->1&/@ImportString[InputString[],"CSV"][[1]],40],39];  • Interesting, how does " "["X"][[#]]& work? – Martin Ender Jul 14 '14 at 10:59 • @m.buettner " "["X"][[1]] is "X". " "["X"][[0]] returns the head of " "["X"], namely " ". – alephalpha Jul 14 '14 at 11:10 • Oh, I see. So that's just generally saving a character for lists. That's really clever. I guess you could add it to codegolf.stackexchange.com/questions/12900/… – Martin Ender Jul 14 '14 at 11:15 # C - 178 This code depends on the fact that each row in a matrix is stored in contiguous memory. Also, it does not print the first row, but it prints the next 40 ones, since the rules only specified a 40x40 grid. Indented for readability only, the byte count only includes necessary code. a[41][42],i,j,*t; main(){ while(scanf("%d,",&j)>0) a[i][j]=1; for(;i<40;i++,puts("")) for(j=0;++j<40;) t=&a[i][j], putchar((*(t+42)=1&(110>>(*(t+1)?1:0)+(*t?2:0)+(*(t-1)?4:0)))?88:32); }  # Lua - 351 Not the ideal language for golfing. s,n,t,u=arg[1],{},table.remove,table.insert for i=1,40 do u(n,i,'.') end for i in s:gmatch("%d+")do u(n,i,'x');t(n)end function a(b) c="";for i=1,40 do c=c..b[i] end;print(c);return c end for i=1,40 do z= n[40]..a(n)..n[1];for k=2,41 do y=string.sub(z,k-1,k+1);if y=="xxx"or y=="x.." or y=="..." then u(n,k-1,'.')else u(n,k-1,'x')end;t(n)end end  • do u(n,i,'x') that’s intentional, isn’t it? – Stan Strum Mar 2 '18 at 19:41 # Haskell, 175 170 169 136 127 124 bytes −9 bytes thanks to @bmo t(a:b:r:_)=mod(b+r+b*r+a*b*r)2 w%a=take 40.map w.iterate a d l=t%tail$0:l++[0]
f l=map(" #"!!)%d$(fromEnum.(eleml))%succ$0


Try it online!

# Haskell, 135 131 130 bytes

-1 byte thanks to Ørjan Johansen (rearranging take 40)

Completely different approach to FrownyFrog's answer but about the same length:

(a?b)r=mod(b+r+b*r+a*b*r)2
r x=0:(zipWith3(?)x=<<tail$tail x++[0]) f y=take 40$map(" o"!!)<$>iterate r[sum[1|elem i y]|i<-[0..40]]  Uses $$\1\$$-indexing and has a leading space on each line, try it online! ## Explanation We're going to work with length-$$\41\$$ lists with values $$\\texttt{0}\$$, $$\\texttt{1}\$$, so let's start with the correct array: f y= [sum[1|elem i y]|i<-[0..40]]  Next we're going to iterate the rule $$\40\$$ times:  take 40$              iterate r


And finally map each $$\\texttt{0}\$$ and $$\\texttt{1}\$$ to some fancy character:

            map(" o"!!)<$>  The function r which applies the $$\\texttt{110}\$$-rule is pretty simple: Using zipWith3 and some padding we can outsource the actual decision for the next cell to (?): r x=0:(zipWith3(?)x=<<tail$tail x++[0])


The (?) operator is the most interesting part of the solution: Previously I used a Boolean rule generated with a Karnaugh map, but turns out there is an even more concise way:

(a?b)r=mod(b+r+b*r+a*b*r)2

• Save a byte by putting take 40$ before map(" o"!!)<$> . – Ørjan Johansen Dec 5 '18 at 10:48

# Husk, 31 28 bytes

Hah, Husk is beating Jelly!

†!¨↑¨↑40¡ȯẊȯ!ḋ118ḋėΘ:0M#ŀ40


Try it online!

## Explanation & Ungolfed

Before adding an explanation, let me ungolf this a bit.. Let's first remove the various compositions, add explicit parentheses and uncompress the ¨↑¨ string. Also let's replace 40 with 4 for a more readable explanation:

†!"t "↑4¡(Ẋ(!ḋ118ḋė)Θ:0)M#ŀ4  -- example input: [3]
ŀ4  -- lower range of 4: [0,1,2,3]
M     -- map over left argument
#    -- | count in list
-- : [0,0,0,1]
¡(              )      -- iterate the following indefinitely (example with [0,1,1,1])
:0       -- | append 0: [0,1,1,1,0]
Θ          -- | prepend 0: [0,0,1,1,1,0]
Ẋ(       )           -- | map over adjacent triples (example with  1 1 0
ė            -- | | create list: [1,1,0]
ḋ             -- | | convert from base-2: 6
-- | | convert 118 to base-2: [1,1,1,0,1,1,0]
-- | | 1-based index: 1
-- | : [1,1,0,1]
-- : [[0,0,0,1],[0,0,1,1],[0,1,1,1],[1,1,0,1],[1,1,1,1],[1,0,0,1],...]
↑4                       -- take 4: [[0,0,0,1],[0,0,1,1],[0,1,1,1],[1,1,0,1]]
†                              -- deep map the following (example with [1,1,0,1])
!"t "                         -- | use each element to index into "t ": "tt t"
-- : ["   t","  tt"," ttt","tt t"]


# Java, 321 characters

Input passed as argument from command line, for example java R 38,39

I have never written more obfuscated java code :-)

class R{public static void main(String[]a) {
Integer s=40;boolean[]n,o=new boolean[s];
for(String x:a[0].split(","))o[s.valueOf(x)]=s>0;
for(Object b:o){n=o.clone();
for(int j=0;j<s;j++){
boolean l=j>1&&o[j-1],r=o[j],c=j+1<s&&o[j+1];
n[j]=!(l&c&r|l&!c&!r|!(l|c|r));
System.out.print((r?"X":" ")+(j>s-2?"\n":""));
}o=n;}}}


Update: Correct output example here (with 40 lines not 50): New output below (removed previous one for brevity):

                                      xx
xxx
xx x
xxxxx
xx   x
xxx  xx
xx x xxx
xxxxxxx x
xx     xxx
xxx    xx x
xx x   xxxxx
xxxxx  xx   x
xx   x xxx  xx
xxx  xxxx x xxx
xx x xx  xxxxx x
xxxxxxxx xx   xxx
xx      xxxx  xx x
xxx     xx  x xxxxx
xx x    xxx xxxx   x
xxxxx   xx xxx  x  xx
xx   x  xxxxx x xx xxx
xxx  xx xx   xxxxxxxx x
xx x xxxxxx  xx      xxx
xxxxxxx    x xxx     xx x
xx     x   xxxx x    xxxxx
xxx    xx  xx  xxx   xx   x
xx x   xxx xxx xx x  xxx  xx
xxxxx  xx xxx xxxxxx xx x xxx
xx   x xxxxx xxx    xxxxxxxx x
xxx  xxxx   xxx x   xx      xxx
xx x xx  x  xx xxx  xxx     xx x
xxxxxxxx xx xxxxx x xx x    xxxxx
xx      xxxxxx   xxxxxxxx   xx   x
xxx     xx    x  xx      x  xxx  xx
xx x    xxx   xx xxx     xx xx x xxx
xxxxx   xx x  xxxxx x    xxxxxxxxxx x
xx   x  xxxxx xx   xxx   xx        xxx
xxx  xx xx   xxxx  xx x  xxx       xx x
xx x xxxxxx  xx  x xxxxx xx x      xxxxx
xxxxxx    x xxx xxxx   xxxxxx     xx   x


Doing another puzzle I learned something interesting about nesting statements in for loops in php, and suddenly they are far more complex than I originally thought. When I get time I reckon I can beat this score considerably. For now though it remains unchanged at a non-competitive 408.

My php version 408 characters:

This was a great puzzle. I also spent ages playing with the inputs as these are fascinating things it must be said. Anyway, here is my PHP version (which is nowhere near as good as some of the answers posted but is complete. In 0th position only take above and above right, in 39th position only take above and above left, ie no wrapping. So here is my version:

<?php $a='38,39';$b='';$d=explode(',',$a);for($i=0;$i<40;++$i){$c=' ';
foreach($d as$k=>$v){if($v == $i){$c='x';}}$b.=$c;}echo $b."\n"; for($x=1;$x<41;++$x){$o='';for($i=1;$i<41;++$i){if(($i>1)AND(substr($b,$i-2,1)=='x')){$l=1;}else{$l=0;}if((substr($b,$i-1,1))=='x'){$v=1;}else{$v=0;}if((substr($b,$i,1))=='x'){$r=1;}else{$r=0;}if((($l+$v+$r)==2)OR(($v+$r)==1)){$o.='x';}else{$o.=' ';}}
echo $o."\n";$b=$o;}?>  You can see it and run it here: http://codepad.org/3905T8i8 Input is a input string at the start as$a='38, 39';

Output is as follows:

xx removed as was too long originally - had 50 lines, not 40 xx


Hope you like it!!!

PS I had to add a few line breaks to the code so you could see all of it and not have it stretch accross the page with a scroll bar.

• Your output has 50 lines – aditsu Jul 15 '14 at 19:20
• Ah, that was because I was playing with it after I finished and seeing what happened. Altering the rules slightly has such interesting affects. Anyway have changed it to 40 now and sorry for missing that. – Paul Drewett Jul 16 '14 at 5:36
• You may want to change the output too :p – aditsu Jul 16 '14 at 6:35
• Corrected the output and added new codepad link with correct value. Thank you again. – Paul Drewett Jul 16 '14 at 12:49

# Stax, 24 bytesCP437

╦♥µ╤u{£┬íQ<;▀ΦΣ╢╕╚äZ↕áû↑


Run and debug online!

Uses the codepoint 1 in CP437 for "1" cells.

Excellent case to show the power of this language.

## Explanation

Uses the unpacked version (29 bytes) to explain.

0]40X*,1&xDQ0]|S3B{:b^374:B@m
0]40X*                           Prepare a tape with 40 cells
,1&                        Assign 1 to the cells specified by the input
xD                      Repeat the rest of the program 40 times
Q                     Output current tape
0]|S                 Prepend and append a 0 cell to it
3B               All runs of length 3
{         m    Map each run with block
:b            Convert from binary
^           Increment (call this value n)
374:B      The binary representation of 374
[1,0,1,1,1,0,1,1,0]
which is 01101110 reversed and prepended a 1
@     Element at 0-based index n


# K (ngn/k), 44 35 bytes

{"X "39{(2\145)@2/'3'1,x,1}\^x?!40}


Try it online!

{ } function with argument x

!40 list of ints from 0 to 39

x? find their indices in x, use 0N (the "integer null") for not found

^ which of them are nulls? this gives us the input, negated

39{ }\ apply 39 times, collecting intermediate results in a list

1,x,1 surround the list with 1s (negated 0s)

3' triples of consecutive items

2/' binary decode each

@ use as indices in ...

2\145 binary encode 145 (negated bits of 110)

"X " finally, use the 40x40 matrix as indices in the string "X " (the @ here is implicit)

# Jelly, 29 bytes

‘;41ṬṖŻ;0Ḅe“¢£¤¦©‘Ɗ3ƤƊ39Ð¡Yo⁶
`

Try it online!