Shortest Game of Life



Conway's Game of Life is the classic example of cellular automation. The cells form a square grid and each has two states: alive or dead. On each turn, each cell simultaneously updates according to its state and those of its eight neighbours:

  • A live cell remains alive if it has exactly two or three live neighbours
  • A dead cell becomes alive if it has exactly three live neighbours

Your mission, should you choose to accept it, is to code the shortest Game of Life implementation in your favourite language.

The rules:

  • The grid must be at least 20x20
  • The grid must wrap around (so the grid is like the surface of a Torus)
  • Your implementation must allow the user to input their own starting patterns
  • GoL is a bit pointless if you can't see what is happening, so there must be visual output of the automaton running, with each turn's result being shown for long enough to be seen!


Posted 2011-08-13T22:37:27.493

Reputation: 4 349

Excellent question! It's a shame I got to read this so late. Well, alas, too late, but anyway, I submitted Mathematica 333 character version with interactive features. I had to submit an entry for the only language with built in CellularAutomaton function ;-)

– Vitaliy Kaurov – 2012-07-30T21:20:27.353

I can use a conway's game of life engine to win this challenge, it technically only takes the input number of cells to run. :) (I'm just kidding though) – KrystosTheOverlord – 2019-02-09T22:44:09.003


Previously on Stack Overflow: Code Golf: Conway's Game of Life, and be sure to look at the APL implementation link in the comments.

– dmckee --- ex-moderator kitten – 2011-08-14T03:07:08.393

1Ah, I did not see that. But this is slightly different no (save me deleting the work putting the challenge together? – Griffin – 2011-08-14T03:15:49.733

6It's not a problem. Many puzzles already run on Stack Overflow have been done here too, but people will tell you that I am obsessive about linking to similar challenges. – dmckee --- ex-moderator kitten – 2011-08-14T03:17:48.697

@Griffin: You can remove all those ; before }s. Also vars can be eliminated at times (if it doesn't break your code). And for one-line fors, ifs etc, you can eliminate the { } completely: for(...) for(...) dosomething(). – pimvdb – 2011-08-14T14:21:50.107

@pimvdb, cheers, I haven't fully golfed it yet, haven't had the time. just wanted to show that I had a go too, rather than idly setting a challenge. Will golf it to the max soon. – Griffin – 2011-08-14T17:53:00.327

@pimvdb, how's that? :D – Griffin – 2011-08-19T08:27:46.387

@Griffin: Wow, more than 50%! – pimvdb – 2011-08-19T09:31:49.293

@pimvdb, squeezed a bit more out, but I think that may be the limit. I take my hat off to you if you find any more. – Griffin – 2011-08-19T09:36:42.880

You can save a couple more on your inner for loops since the third argument is optional. So for(j=0;j<r;++j) becomes for(j=0;j<r;) and you move the increment operator into the statement: document.getElementById(z+"_"+j).innerHTML=b[z][j++]?"#":"." – migimaru – 2011-08-19T10:01:05.230

Also, since K(z,j) always returns an integer, you can shorten your comparison n<=3&&n>=2:n==3 to n<4&&n>1:n==3. You also use document 3 times, so you can save a bit by making a variable to refer to document instead. – migimaru – 2011-08-19T10:15:46.647

@migimaru, well spotted, thanks – Griffin – 2011-08-19T11:07:30.910

What's the generally accepted rule for generations? Does it have to go until it reaches a stable state, or can it go forever, or for an arbitrary number of generations? – Igby Largeman – 2011-08-22T00:19:56.340



HTML5 Canvas with JavaScript, 940 639 586 519 characters

<html><body onload="k=40;g=10;b=[];setInterval(function(){c=[];for(y=k*k;y--;){n=0;for(f=9;f--;)n+=b[(~~(y/k)+k+f%3-1)%k*k+(y+k+~~(f/3)-1)%k];c[y]=n==3||n-b[y]==3;r.fillStyle=b[y]?'red':'tan';r.fillRect(y%k*g,~~(y/k)*g,g-1,g-1)}if(v.nextSibling.checked)b=c},1);v=document.body.firstChild;v.width=v.height=g*k;v.addEventListener('click',function(e){b[~~((e.pageY-v.offsetTop)/g)*k+~~((e.pageX-v.offsetLeft)/g)]^=1},0);r=v.getContext('2d');for(y=k*k;y--;)b[y]=0"><canvas></canvas><input type="checkbox"/>Run</body></html>

I always wanted to do something with canvas, so here is my attempt (original version online). You can toggle cells by clicking (also possible in running mode).

You can now also try the new version here.

Unfortunately there is an issue I couldn't work around yet. The online version is 11 characters longer because jsFiddle puts a text node just before the canvas (why?) and thus the canvas is no longer the first child.

Edit 1: Lots of optimisations and restructurings.

Edit 2: Several smaller changes.

Edit 3: Inlined the complete script block plus minor changes.


Posted 2011-08-13T22:37:27.493

Reputation: 23 109

Technically, you could omit the <html> tag. – Conor O'Brien – 2015-09-28T22:33:42.170

The attached stack snippet that wizzwizz4 added doesn't actually seem to work, so I'll roll it back for now. – FlipTack – 2017-01-28T22:10:50.247

2You can remove the html and body tags. It will function the same – arodebaugh – 2017-03-08T11:52:33.413

@Griffin Why is this accepted when there is a much shorter answer?

– Adám – 2017-04-27T09:00:40.020

Nice, but change the interval delay to 1 makes it as fast as mine rather than that slow stepping. Also if you want to implement drawing (rather than clicking on each square) you can round the mouse position to the nearest blocksize and fill the rectangle at that point. More characters but more points. – Griffin – 2011-08-15T00:32:54.480

You can replace new Array('#FFF','#800') with ['#FFF','#800']. – Lowjacker – 2011-08-15T15:59:56.880

Though saying that about drawing, my super-golfed doesn't allow drawing and is ugly as sin. Haha. You can set your two colours in the s array to tan and red since they are the two colours with the shortest representations - saves you two chars. Also, if possible, put the literal version of j into the interval. I'm sure there's lots more to squeeze out too. – Griffin – 2011-08-19T14:46:40.743

@Griffin and Lowjacker: thank you very much. I am also pretty sure that you can golf this much more (and have already some ideas). Unfortunately I didn't find the time to do so. A better golfed version will follow tomorrow - I hope... – Howard – 2011-08-19T15:07:41.210

@Howard Just tried your new version in jsFiddle and it seems to have a mismatched brace somewhere, so it won't run. – Gareth – 2011-08-20T10:16:45.137

@Gareth It is not a mismatched brace but jsFiddle seems to have the canvas not as the first child element (???). Therefore my new version with document.body.firstChild does not work but you have to get the canvas by id again: jsFiddle.

– Howard – 2011-08-20T10:47:26.967

Damn, Howard, you've blown everyone out of the water. You can save 2 more chars: you don't actually have to supply a interval delay it seems, if you leave out the ,1 it defaults to 0ms. – Griffin – 2011-08-20T12:38:43.267

@Griffin I didn't get it running without the ,1. Seems that it is not optional. But I found other places to save few chars. – Howard – 2011-08-20T18:17:30.007

@Howard, I don't have the ,1 in mine, maybe it's browser differences. – Griffin – 2011-08-20T18:30:13.100


Python, 219 chars

I went for maximum golfage, with just enough interface to satisfy the question.

import time
while 1:
 for i in N:print''.join(' *'[i*20+j in P]for j in N)
 time.sleep(.1);Q=[(p+d)%400 for d in(-21,-20,-19,-1,1,19,20,21)for p in P];P=set(p for p in Q if 2-(p in P)<Q.count(p)<4)

You run it like this:

echo "[8,29,47,48,49]" | ./

The numbers in the list represent the coordinates of the starting cells. The first row is 0-19, the second row is 20-39, etc.

Run it in a terminal with 21 rows and it looks pretty snazzy.

Keith Randall

Posted 2011-08-13T22:37:27.493

Reputation: 19 865

1This totally should have won. I guess 'ease of input' was weighted fairly high. – primo – 2012-08-03T10:13:38.923

@primo I'd even go so far as to suggest mma should have a separate competition. – luser droog – 2013-01-26T04:35:43.997

2So is this Life of Py then? – Christopher Wirt – 2017-03-10T15:59:10.990

You can always save one more char... 2-(p in P) == 2-({p}<P). But then you would have to change your input to {8,29,47,48,49} :) – JBernardo – 2017-08-17T18:26:30.513


TI-BASIC, 96 bytes (87 for non-competing entry)

For your TI-84 series graphing calculator (!). This was quite a challenge, because there is no easy way to write a buffered graphics routine (definitely nothing built in), and the graph screen has only four relevant graphics commands: Pxl-On(), Pxl-Off(), Pxl-Change(), and pxl-Test().

Uses every accessible pixel on the screen, and wraps correctly. Each cell is one pixel, and the program updates line by line horizontally to the right across the screen. Because the calculators have only a 15MHz z80 processor and BASIC is a slow interpreted language, the code only gets one frame about every five minutes.

User input is easy: before running the program, use the Pen tool to draw your shape on the graph screen.

Adapted from my entry to a code golf contest at the calculator forum Omnimaga.

While 1
If 1=pxl-Test(Y,X)+int(3fPart(3cosh(fPart(6ֿ¹iPart(sum(Ans,Y+1,Y+3

Omnimaga version (87 bytes)

This code has an additional feature: it detects if it is being run for the first time, and if randomizes the screen state. In subsequent runs it automatically continues the simulation if stopped after a frame is finished. However, it is not a competing entry because it does not wrap the screen; the cells on the outer border will always be considered dead if the graph screen is cleared beforehand.

While 1
If 2rand>isClockOn=pxl-Test(Y,X)+int(3fPart(3cosh(fPart(6ֿ¹iPart(sum(Ans,Y,Y+2

This version is probably the most golfed code I have ever written, and contains some truly nasty obfuscatory optimizations:

  • I use the clock state as a flag. At the start of the program, the date/time clock is enabled, and I use the value of the global isClockOn flag to determine whether it is the first iteration. After the first frame is drawn, I turn the clock off. Saves one byte over the shortest other method and about four over the obvious method.

  • I store the states of the three columns next to the one being updated in a 63-element array of base-7 numbers. The 49's place holds the column to the right, the 7's place holds the middle column, and the units place holds the left column--1 for a live cell and 0 for a dead cell. Then I take the remainder mod 6 of the sum of the three numbers around the cell being modified to find the total number of live neighbor cells (it's just like the divisibility by 9 trick—in base 7, the remainder mod 6 equals the sum of the digits). Saves about 10 bytes by itself and gives the opportunity to use the next two optimizations. Example diagram (let's say there is a glider centered at a certain column at Y=45:

    Row # | Cell State       | Stored number | Mod 6 = cell count
    44      Live, Live, Live   49+7+1 = 57     3
    45      Dead, Dead, Live   49+0+0 = 49     1
    46      Dead, Live, Dead   0+7+0  = 7      1

    The center cell will stay dead, because it is surrounded by exactly five live cells.

  • After each row is completed, the numbers in the array are updated by dividing the existing numbers by 7, discarding the decimal part, and adding 49 times the values of the cells in the new column. Storing all three columns each time through would be much slower and less elegant, take at least 20 more bytes, and use three lists rather than one, because the values of cells in each row must be stored before cells are updated. This is by far the smallest way to store cell positions.

  • The snippet int(3fPart(3cosh( gives 1 when the input equals 3/6, 2 when it equals 4/6, and 0 when it equals 0, 1/6, 2/6, or 5/6. Saves about 6 bytes.


Posted 2011-08-13T22:37:27.493

Reputation: 20 331


Mathematica - 333


  • Interactive interface: click cells to make your patterns

  • Nice grid

  • Buttons: RUN, PAUSE, CLEAR

The code is below.


enter image description here

If you want to get a feel how this runs, 2nd example in this blog is just a more elaborate version (live Fourier analysis, better interface) of the code above. Example should run right in your browser after free plugin download.

Vitaliy Kaurov

Posted 2011-08-13T22:37:27.493

Reputation: 1 561

2+1, nice one for having a go. Yeah that's the problem with this site, there are tonnes of old questions which one tends to miss. – Griffin – 2012-07-31T13:10:35.733

@Griffin thanks for noticing it at all ;) – Vitaliy Kaurov – 2013-09-13T01:35:16.057


C 1063 characters

As a challenge, I did this in C using the golf-unfriendly Windows API for real time IO. If capslock is on, the simulation will run. It will stay still if capslock is off. Draw patterns with the mouse; left click revives cells and right click kills cells.

#include <windows.h>
#define K ][(x+80)%20+(y+80)%20*20]
#define H R.Event.MouseEvent.dwMousePosition
#define J R.Event.MouseEvent.dwButtonState
HANDLE Q,W;char*E[3],O;Y(x,y){return E[0 K;}U(x,y,l,v){E[l K=v;}I(){E[2]=E[1];E[1]=*E;*E=E[2];memset(E[1],0,400);}A(i,j,k,l,P){while(1){Sleep(16);for(i=0;i<20;++i)for(j=0;j<20;++j){COORD a={i,j};SetConsoleCursorPosition(Q,a);putchar(E[0][i+j*20]==1?'0':' ');}if(O){for(i=0;i<20;++i)for(j=0;j<20;++j){for(k=i-1,P=0;k<i+2;++k)for(l=j-1;l<j+2;++l){P+=Y(k,l);}U(i,j,1,P==3?1:Y(i,j)==1&&P==4?1:0);}I();}}}main(T,x,y,F,D){for(x=0;x<21;++x)puts("#####################");E[0]=malloc(800);E[1]=E[0]+400;I();I();W=GetStdHandle(-10);Q=GetStdHandle(-11);SetConsoleMode(W,24);INPUT_RECORD R;F=D=O=0;COORD size={80,25};SetConsoleScreenBufferSize(Q,size);_beginthread(A,99,0);while(1){ReadConsoleInput(W,&R,1,&T);switch(R.EventType){case 1:O=R.Event.KeyEvent.dwControlKeyState&128;break;case 2:switch(R.Event.MouseEvent.dwEventFlags){case 1:x=H.X;y=H.Y;case 0:F=J&1;D=J&2;}if(F)U(x,y,0,1);if(D)U(x,y,0,0);}}}

The compiled EXE can be found here

Edit: I've commented up the source. It's available Here


Posted 2011-08-13T22:37:27.493

Reputation: 641

I'd love to see a commented version of this! – luser droog – 2012-12-21T02:50:38.087

1Sure, if I can remember what I was thinking... =p – Kaslai – 2012-12-22T23:18:56.567


@luserdroog Here it is

– Kaslai – 2012-12-22T23:59:38.513

This is just awesome. – rayryeng - Reinstate Monica – 2016-12-12T22:07:03.367


J (39 characters)

l=:[:+/(3 4=/[:+/(,/,"0/~i:1)|.])*.1,:]

Based on this APL version (same algorithm, toroidal convolution).

Example usage:

   r =: (i.3 3) e. 1 2 3 5 8
0 1 1          NB. A glider!
1 0 1
0 0 1

   R =: _1 _2 |. 5 7 {. r
0 0 0 0 0 0 0  NB. Test board
0 0 0 1 1 0 0
0 0 1 0 1 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0

   l R
0 0 0 0 0 0 0  NB. Single step
0 0 0 1 1 0 0
0 0 0 0 1 1 0
0 0 0 1 0 0 0
0 0 0 0 0 0 0


Posted 2011-08-13T22:37:27.493

Reputation: 584


Mathematica, 123 characters

A very rudimentary implementation that doesn't use Mathematica's built-in CellularAutomaton function.



Posted 2011-08-13T22:37:27.493

Reputation: 201


Ruby 1.9 + SDL (380 325 314)

EDIT: 314 characters, and fixed a bug with extra cells appearing alive on the first iteration. Upped the grid size to 56 since the color routine only looks at the lowest 8 bits.

EDIT: Golfed down to 325 characters. Grid width/height is now 28 since 28*9 is the largest you can have while still using the value as the background colour. It also processes only one SDL event per iteration now, which obviates the inner loop completely. Pretty tight I think!

The simulation starts paused, with all cells dead. You can press any key to toggle pause/unpause, and click any cell to toggle it between alive and dead. Runs an iteration every tenth of a second.

The wrapping is a bit wonky.

SDL.init W=56
b=[] S=W*9,S,0,0
r['yU']?$_^=1:r[?Q]?exit: r['nU']?b[e.y/9*W+e.x/9]^=1:0{|i|v=[~W,-W,-55,-1,1,55,W,57].select{|f|b[(i+f)%T]}.size;v==3||v==2&&b[i]}if$_{|i|s.fillRect i%W*9,i/W*9,9,9,[b[i]?0:S]*3}
sleep 0.1}

Looks like this:

Screenshot of the app in action

Fun challenge! I welcome any improvements anybody can see.

Paul Prestidge

Posted 2011-08-13T22:37:27.493

Reputation: 2 390

Nice try but I can see straight away that you've gone wrong. You can't have a pattern like that in GoL. Have another read of the rules:

– Griffin – 2012-08-01T08:01:20.577

@Griffin I think the screenshot was taken after pausing and toggling some cells manually - I'll check the rules again, though. Thanks! – Paul Prestidge – 2012-08-02T00:44:34.867

7@Griffin cant the seed pattern be in any possible configuration? – ardnew – 2012-08-08T04:30:52.403


Pure Bash, 244 bytes

Works on a toroidally-wrapped 36x24 universe:

mapfile a
for e in {0..863};{
for i in {0..8};{
[ "${a[(e/36+i/3-1)%24]:(e+i%3-1)%36:1}" == O ]&&((n++))
[ "$c" == O ]&&((--n==2))&&d=O
printf -vo %s\\n "${b[@]}"
echo "$o"
exec $0<<<"$o"

Since this is a shell script, the method of input is congruent with other shell commands - i.e. from stdin:

$ ./ << EOF



  O O                                                       

  O O                                                       

... etc

We can redirect input from any text source, piped through a tr filter to get interesting initial generations, e.g.

man tr | tr [:alnum:] O | ./

Digital Trauma

Posted 2011-08-13T22:37:27.493

Reputation: 64 644


Scala, 1181 1158 1128 1063 1018 1003 999 992 987 characters

import swing._
import event._
object L extends SimpleSwingApplication{import java.awt.event._
import javax.swing._
var x=Array.fill(w,h)(0)
def n(y:Int,z:Int)=for(b<-z-1 to z+1;a<-y-1 to y+1 if(!(a==y&&b==z)))d+=x((a+w)%w)((b+h)%h)
def top=new MainFrame with ActionListener{preferredSize=new Dimension(500,500)
menuBar=new MenuBar{contents+=new Menu("C"){contents+={new MenuItem("Go/Stop"){listenTo(this)
reactions+={case ButtonClicked(c)=>r= !r}}}}}
contents=new Component{listenTo(mouse.clicks)
reactions+={case e:MouseClicked=>var p=e.point
override def paint(g:Graphics2D){for(j<-0 to h-1;i<-0 to w-1){var r=new Rectangle(i*c,j*c,c,c)
x(i)(j)match{case 0=>g draw r
case 1=>g fill r}}}}
def actionPerformed(e:ActionEvent){if(r){var
for(j<-0 to h-1;i<-0 to w-1){d=0
x(i)(j)match{case 0=>if(d==3)t(i)(j)=1
case 1=>if(d<2||d>3)t(i)(j)=0}}
val t=new Timer(200,this)


import swing._
import event._

object Life extends SimpleSwingApplication
    import java.awt.event._
    import javax.swing._
    var x=Array.fill(w,h)(0)
    def n(y:Int,z:Int)=for(b<-z-1 to z+1;a<-y-1 to y+1 if(!(a==y&&b==z)))d+=x((a+w)%w)((b+h)%h)
    def top=new MainFrame with ActionListener
        preferredSize=new Dimension(500,500)
        menuBar=new MenuBar
            contents+=new Menu("Control")
                contents+={new MenuItem("Start/Stop")
                        case ButtonClicked(c)=>run= !run
        contents=new Component
                case e:MouseClicked=>
                    var p=e.point
            override def paint(g:Graphics2D)
                for(j<-0 to h-1;i<-0 to w-1)
                    var r=new Rectangle(i*c,j*c,c,c)
                    x(i)(j) match
                        case 0=>g draw r
                        case 1=>g fill r
        def actionPerformed(e:ActionEvent)
                for(j<-0 to h-1;i<-0 to w-1)
                    x(i)(j) match
                        case 0=>if(d==3)t(i)(j)=1
                        case 1=>if(d<2||d>3)t(i)(j)=0
        val timer=new Timer(200,this)

The larger part of the code here is Swing GUI stuff. The game itself is in the actionPerformed method which is triggered by the Timer, and the helper function n which counts neighbours.


Compile it with scalac filename and then run it with scala L.
Clicking a square flips it from live to dead, and the menu option starts and stops the game. If you want to change the size of the grid, change the first three values in the line: var(w,h,c,d,r)=(20,20,20,0,false) they are width, height and cell size (in pixels) respectively.


Posted 2011-08-13T22:37:27.493

Reputation: 11 678

I found 2 golfing-improvements: import java.awt.event._ and contents+=m("Go",true)+=m("Stop",false)}}, leading to 1093 characters. – user unknown – 2011-08-14T19:20:24.303

@user unknown Thanks. I found a few improvements myself - down to 1063 now. – Gareth – 2011-08-14T19:47:10.600

Damn you've been busy. Keep it up! I will be testing answers when a few more people post them. – Griffin – 2011-08-14T23:56:04.827


JavaScript, 130

Not totally responding to the challenge, but for the record, here's a Game of Life engine in 130 bytes made by Subzey and I in 2013.

/* Fill an array with 0's and 1's, and call g(array, width, height) to iterate */
g=function(f,c,g,d,e,b,h){g=[];e=[c+1,c,c-1,1];for(b=c*c;b--;g[b]=3==d||f[b]&&2==d,d=0)for(h in e)d+=f[b+e[h]]+f[b-e[h]];return g}


Posted 2011-08-13T22:37:27.493

Reputation: 5 523

This seems to have some problems with the first row. For example setting @@\n@@ (2 by 2 square in top left corner) or .@\n.@\n.@. (1 by 3 column) – AnnanFay – 2018-02-05T21:03:39.463


GW-BASIC, 1086 1035 bytes (tokenised)

In tokenised form, this is 1035 bytes. (The ASCII form is, of course, a bit longer.) You get the tokenised form by using the SAVE"life command without appending ",a in the interpreter.

10 DEFINT A-Z:DEF SEG=&HB800:KEY OFF:COLOR 7,0:CLS:DEF FNP(X,Y)=PEEK((((Y+25)MOD 25)*80+((X+80)MOD 80))*2)
20 X=0:Y=0
30 LOCATE Y+1,X+1,1
40 S$=INKEY$:IF S$=""GOTO 40
50 IF S$=CHR$(13)GOTO 150
60 IF S$=" "GOTO 130
70 IF S$=CHR$(0)+CHR$(&H48)THEN Y=(Y-1+25)MOD 25:GOTO 30
80 IF S$=CHR$(0)+CHR$(&H50)THEN Y=(Y+1)MOD 25:GOTO 30
90 IF S$=CHR$(0)+CHR$(&H4B)THEN X=(X-1+80)MOD 80:GOTO 30
100 IF S$=CHR$(0)+CHR$(&H4D)THEN X=(X+1)MOD 80:GOTO 30
110 IF S$="c"THEN CLS:GOTO 20
120 GOTO 40
130 Z=PEEK((Y*80+X)*2):IF Z=42 THEN Z=32ELSE Z=42
140 POKE(Y*80+X)*2,Z:GOTO 40
150 LOCATE 1,1,0:ON KEY(1)GOSUB 320:KEY(1) ON
160 V!=TIMER+.5:FOR Y=0 TO 24:FOR X=0 TO 79:N=0
170 Z=FNP(X-1,Y-1):IF Z=42 OR Z=46 THEN N=N+1
180 Z=FNP(X,Y-1):IF Z=42 OR Z=46 THEN N=N+1
190 Z=FNP(X+1,Y-1):IF Z=42 OR Z=46 THEN N=N+1
200 Z=FNP(X-1,Y):IF Z=42 OR Z=46 THEN N=N+1
210 Z=FNP(X+1,Y):IF Z=42 OR Z=46 THEN N=N+1
220 Z=FNP(X-1,Y+1):IF Z=42 OR Z=46 THEN N=N+1
230 Z=FNP(X,Y+1):IF Z=42 OR Z=46 THEN N=N+1
240 Z=FNP(X+1,Y+1):IF Z=42 OR Z=46 THEN N=N+1
250 Z=PEEK((Y*80+X)*2):IF Z=32 THEN IF N=3 THEN Z=43
260 IF Z=42 THEN IF N<2 OR N>3 THEN Z=46
270 POKE(Y*80+X)*2,Z:NEXT:NEXT:FOR Y=0 TO 24:FOR X=0 TO 79:Z=PEEK((Y*80+X)*2):IF Z=46 THEN Z=32
280 IF Z=43 THEN Z=42
290 POKE(Y*80+X)*2,Z:NEXT:NEXT
310 IF INKEY$=""GOTO 160

This is the maximum-golfed version, but still featureful: upon starting, you get an editor, in which you can move with the cursor keys; space toggles bacteria on/off on the current field, c clears the screen, Return starts game mode.

Here follows a less-obfuscated version, which also sets an initial game board with two structures (a circulary-rotating thing and a glider):

1000 REM Conway's Game of Life
1001 REM -
1002 REM Copyright (c) 2012 Thorsten "mirabilos" Glaser
1003 REM All rights reserved. Published under The MirOS Licence.
1004 REM -
1005 DEFINT A-Z:DEF SEG=&hB800
1007 DEF FNP(X,Y)=PEEK((((Y+25) MOD 25)*80+((X+80) MOD 80))*2)
1010 PRINT "Initial setting mode, press SPACE to toggle, RETURN to continue"
1020 PRINT "Press C to clear the board, R to reset. OK? Press a key then."
1050 CLS
1065 DATA 3,3,4,3,5,3,6,3,7,3,8,3,3,4,4,4,5,4,6,4,7,4,8,4
1066 DATA 10,3,10,4,10,5,10,6,10,7,10,8,11,3,11,4,11,5,11,6,11,7,11,8
1067 DATA 11,10,10,10,9,10,8,10,7,10,6,10,11,11,10,11,9,11,8,11,7,11,6,11
1068 DATA 4,11,4,10,4,9,4,8,4,7,4,6,3,11,3,10,3,9,3,8,3,7,3,6
1069 DATA 21,0,22,1,22,2,21,2,20,2,-1,-1
1070 RESTORE 1065
1080 READ X,Y
1090 IF X=-1 GOTO 1120
1100 POKE (Y*80+X)*2,42
1110 GOTO 1080
1120 X=0:Y=0
1125 LOCATE Y+1,X+1,1
1130 S$=INKEY$
1140 IF S$="" GOTO 1130
1150 IF S$=CHR$(13) GOTO 1804
1160 IF S$=" " GOTO 1240
1170 IF S$=CHR$(0)+CHR$(&h48) THEN Y=(Y-1+25) MOD 25:GOTO 1125
1180 IF S$=CHR$(0)+CHR$(&h50) THEN Y=(Y+1) MOD 25:GOTO 1125
1190 IF S$=CHR$(0)+CHR$(&h4B) THEN X=(X-1+80) MOD 80:GOTO 1125
1200 IF S$=CHR$(0)+CHR$(&h4D) THEN X=(X+1) MOD 80:GOTO 1125
1210 IF S$="c" THEN CLS:GOTO 1120
1220 IF S$="r" GOTO 1050
1225 IF S$=CHR$(27) THEN END
1230 GOTO 1130
1240 Z=PEEK((Y*80+X)*2)
1250 IF Z=42 THEN Z=32 ELSE Z=42
1260 POKE (Y*80+X)*2,Z
1270 GOTO 1130
1804 LOCATE 1,1,0
1900 ON KEY(1) GOSUB 2300
1910 KEY(1) ON
2000 V!=TIMER+.5
2010 FOR Y=0 TO 24
2020  FOR X=0 TO 79
2030   N=0
2040   Z=FNP(X-1,Y-1):IF Z=42 OR Z=46 THEN N=N+1
2050   Z=FNP(X  ,Y-1):IF Z=42 OR Z=46 THEN N=N+1
2060   Z=FNP(X+1,Y-1):IF Z=42 OR Z=46 THEN N=N+1
2070   Z=FNP(X-1,Y  ):IF Z=42 OR Z=46 THEN N=N+1
2080   Z=FNP(X+1,Y  ):IF Z=42 OR Z=46 THEN N=N+1
2090   Z=FNP(X-1,Y+1):IF Z=42 OR Z=46 THEN N=N+1
2100   Z=FNP(X  ,Y+1):IF Z=42 OR Z=46 THEN N=N+1
2110   Z=FNP(X+1,Y+1):IF Z=42 OR Z=46 THEN N=N+1
2120   Z=PEEK((Y*80+X)*2)
2130   IF Z=32 THEN IF N=3 THEN Z=43
2140   IF Z=42 THEN IF N<2 OR N>3 THEN Z=46
2150   POKE (Y*80+X)*2,Z
2160  NEXT X
2170 NEXT Y
2200 FOR Y=0 TO 24
2210  FOR X=0 TO 79
2220   Z=PEEK((Y*80+X)*2)
2230   IF Z=46 THEN Z=32
2240   IF Z=43 THEN Z=42
2250   POKE (Y*80+X)*2,Z
2260  NEXT X
2270 NEXT Y
2280 IF TIMER<V! GOTO 2280
2290 IF INKEY$="" GOTO 2000

I wrote this in 15 minutes while bored and waiting for a friend, who was code-golfing with his “apprentice” for Conway’s Game of Life at the same time.

It functions like this: It immediately uses the 80x25 text mode screen buffer (change the initial DEF SEG to use &hB000 if you’re on a Hercules graphics card; these settings work with Qemu and (slower) dosbox). An asterisk * is a bacterium.

It works two-pass: first, birthplaces are marked with + and death marks its targets with .. In the second pass, + and . are replaced with * and , respectively.

The TIMER thing is to make it wait for half a second after each round, in case your Qemu host is very fast ☺

I’m not hoping for a shortest-wins price here but for a coolness one, especially considering the initial board setup. I’ve also got a version where the game engine was replaced by assembly code, in case you’re interested…


Posted 2011-08-13T22:37:27.493

Reputation: 422

Considering you incremented your labels by 1 on the non-golfed version, is it possible to do the same on the golfed version? (i.e., 1, 2, 3, etc.) Or do the line numbers not count? – Zacharý – 2017-08-17T20:12:20.887

line numbers, when tokenised, count as word (16 bit), if I’m not totally mistaken – mirabilos – 2017-08-18T00:23:07.127

Okay then, guess I must've been thinking of some other BASIC dialect then. – Zacharý – 2017-08-18T10:51:40.463

@Zacharý Click on “GW-BASIC tokenised program format” then on “Program format” here to see that, indeed, line numbers are constantly two bytes, and for more in-depth details on the token format.

– mirabilos – 2017-08-22T08:48:28.937


Mathematica, 115 bytes

Here's an easy cop-out to this:



Posted 2011-08-13T22:37:27.493

Reputation: 151

1Mathematica is fine, but as the rules state the program must allow the user to input their own patterns. This rule is intentional since a few languages allow short implementations like this but with no user interaction. Sure you can put your own array in there, but it's not gonna win. – Griffin – 2011-08-14T23:58:33.903

"input" in Mathematica is mostly through the notebook interface, so I don't think "user interaction" is really possible. You just replace the RandomInteger argument to the CellularAutomaton function with whatever you want, and re-evaluate the code. – JeremyKun – 2011-08-15T00:16:18.327

3User interaction is possible. The most simplistic method I can think of right now is an array of buttons. Give it a go man. – Griffin – 2011-08-15T00:19:41.497


C# - 675 chars

I've always wanted to write a version of this program. Never knew it would only take a lazy half hour for a quick and dirty version. (Golfing it takes much longer, of course.)

using System.Windows.Forms;class G:Form{static void Main(){new G(25).ShowDialog();}
public G(int z){var g=new Panel[z,z];var n=new int [z,z];int x,y,t;for(int i=0;i<z;
i++)for(int j=0;j<z;j++){var p=new Panel{Width=9,Height=9,Left=i*9,Top=j*9,BackColor
p);}KeyUp+=(s,_)=>{for(int i=0;i<99;i++){for(x=0;x<z;x++)for(y=0;y<z;y++){t=0;for(int 
c=-1;c<2;c++)for(int d=-1;d<2;d++)if(c!=0||d!=0){int a=x+c,b=y+d;a=a<0?24:a>24?0:a;b=


  • Enter a starting pattern by clicking cells to turn them on (alive).
  • Start the game by pressing any keyboard key.
  • The game runs for 99 generations every time a key is pressed (I could have made it 9 to save a char, but that seemed too lame).

Golfing compromises

  • You can only turn cells on with the mouse, not off, so if you make a mistake you have to restart the program.
  • There are no grid lines, but that doesn't hurt playability too much.
  • Update speed is proportional to CPU speed, so on very fast computers it will probably be just a blur.
  • Living cells are red because "black" uses 2 more characters.
  • The smallness of the cells and the fact that they don't use up all the form space are also character-saving compromises.

Igby Largeman

Posted 2011-08-13T22:37:27.493

Reputation: 353


Java (OpenJDK 8) - 400 388 367 bytes

Second and (probably) Final Edit: Managed to golf an extra 21 bytes after finding these (imo) goldmines - definitely recommend new people to read these (especially if you're going to try some of these challenges using Java).

Resulting code (probably going to end up golfing even more if I find out how to shorten those double nested for loops...):

u->{int w=u.length,h=u[0].length,x,y,i,j,n;Stack<Point>r=new Stack<Point>();for(;;){for(Point c:r)u[c.x][c.y]=1;r.clear();for(x=0;x<w;++x)for(y=0;y<h;++y){boolean o=u[x][y]>0;n=o?-1:0;for(i=-2;++i<2;)for(j=-2;++j<2;)if(u[(w+x+i)%w][(h+y+j)%h]>0)++n;if(o&n>1&n<4|!o&n==3)r.add(new Point(x,y));System.out.print(u[x][y]+(y>h-2?"\n":""));}for(int[]t:u)Arrays.fill(t,0);}}

Try it online!

(Original post starts here.)

I actually thought for a moment that I would be able to at least challenge the best Python answer with my (arguably limited) knowledge of Java lol...It was a challenge I nevertheless enjoyed partaking in (despite having joined the party perhaps just a little bit late...)

There's not much to it really - basic explanation as follows (ungolfed):

 * Explanation of each variable's usage:
 * w=height* of array
 * h=width* of array
 * x=y* coord of point in array
 * y=x* coord of point in array
 * i and j are counters for calculating the neighbours around a point in the array
 * n=neighbour counter
 * r=temporary array to store the cells from the current generation
 * u=the 2d array used for all the calculations (parameter from lambda expression)
 * c=temporary variable used to help populate the 2d array
 * o=boolean variable that stores the value of whether the cell is alive or not
u-> // start of lambda statement with u as parameter (no need for brackets as it's only one parameter being passed)
    int w=u.length,h=u[0].length,x,y,i,j,n; // defines all the necessary integer variables;
    Stack<Point>r=new Stack<Point>(); // same with the only array list needed (note how I only use two data structures);
    for(;;) // notice how this is still an infinite loop but using a for loop;
        for(Point c:r)u[c.x][c.y]=1; //for every point in the "previous" generation, add that to the 2D array as a live (evil?) cell;
        r.clear(); // clears the array list to be populated later on
        for(x=0;x<w;++x) // a pair of nested for loops to iterate over every cell of the 2D array;
                // sets o to be the presence of a live cell at (x,y) then uses said value in initialising the neighbour counter;
                boolean o=u[x][y]>1;n=o?-1:0;
                for(i=-2;++i<2;) // another pair of nested for loops - this one iterates over a 3x3 grid around *each* cell of the 2D array;
                {                // this includes wrap-around (note the modulus sign in the if statement below);
                        if(u[(w+x+i)%w][(h+y+j)%h]>0)++n; // this is where the first interesting thing lies - the bit which makes wrap-around a reality;
                if(o&n>1&n<4|!o&n==3)r.add(new Point(x,y)); // this is the second interesting bit of my code - perhaps more so as I use bitwise operators to calculate the number of neighbours (x,y) has;
                                                            // (since I'm technically dealing with 0s and 1s, it's not a total misuse of them imo);
                System.out.print(u[x][y]+(y>h-2?"\n":""));  // that extra part of the print statement adds a newline if we reached the end of the current 'line';
        // since the information about the new generation is now in the array list, this array can be emptied out, ready to receive said info on the new generation;
} // end of lambda statement

(more information about lambda statements in Java 8 here)

Yes, there's a catch with my approach.

As most of you probably noticed, my golfed code as it currently stands will loop forever. To prevent this, a counter can be introduced at the top and used in the while loop to only display n (in this case, 5) iterations as follows (notice the new b variable added):

u->{int b=0,w=u.length,h=u[0].length,x,y,i,j,n;Stack<Point>r=new Stack<Point>();for(;++b<6;){for(Point c:r)u[c.x][c.y]=1;r.clear();for(x=0;x<w;++x)for(y=0;y<h;++y){boolean o=u[x][y]>0;n=o?-1:0;for(i=-2;++i<2;)for(j=-2;++j<2;)if(u[(w+x+i)%w][(h+y+j)%h]>0)++n;if(o&n>1&n<4|!o&n==3)r.add(new Point(x,y));System.out.print(u[x][y]+(y>h-2?"\n":""));}for(int[]t:u)Arrays.fill(t,0);}}

Additionally, a few points worth mentioning. This program does not check if the input is correct and will, therefore, fail with (most likely) an ArrayOutOfBoundsException; as such, make sure to check that the input is valid by completely filling in a part of an array (skewered arrays will throw the exception mentioned above). Also, the board as it currently is looks 'fluid' - that is, there is no separation between one generation and the next. If you wish to add that in to double-check that the generations being produced are indeed valid, an extra System.out.println(); needs to be added just before for(int[]t:u)Arrays.fill(t,0); (see this Try it online! for clarity). And last, but not least, given that this is my first code golf, any feedback is greatly appreciated :)

Old code from previous 388 byte answer:

u->{int w=u.length,h=u[0].length,x,y,i,j,n;ArrayList<Point>r=new ArrayList<Point>();while(true){for(Point c:r)u[c.x][c.y]=1;r.clear();for(x=0;x<w;++x){for(y=0;y<h;++y){boolean o=u[x][y]==1;n=o?-1:0;for(i=-2;++i<2;)for(j=-2;++j<2;)if(u[(w+x+i)%w][(h+y+j)%h]==1)++n;if(o&n>1&n<4|!o&n==3)r.add(new Point(x,y));System.out.print(u[x][y]);}System.out.println();}for(int[]t:u)Arrays.fill(t,0);}}

And from the initial 400 byte answer:

int w=35,h=20,x,y,i,j,n;ArrayList<Point>l=new ArrayList<Point>(),r;while(true){int[][]u=new int[w][h];for(Point c:l)u[c.x][c.y]=1;r=new ArrayList<Point>();for(x=0;x<w;++x){for(y=0;y<h;++y){boolean o=u[x][y]==1;n=o?-1:0;for(i=-2;++i<2;)for(j=-2;++j<2;)if(u[(w+x+i)%w][(h+y+j)%h]==1)++n;if(o&n>1&n<4|!o&n==3)r.add(new Point(x,y));System.out.print(u[x][y]);}System.out.println();}l.clear();l.addAll(r);}


Posted 2011-08-13T22:37:27.493

Reputation: 131

Amazing first post, welcome to PPCG! – Zacharý – 2017-08-14T01:20:32.003

Thanks, I'm definitely going to be doing more of these - they're fun :) – NotBaal – 2017-08-14T01:26:05.577

Join us, we have Dennis. Also, this is not a complete program nor a function, which it needs to be, IIRC. – Zacharý – 2017-08-14T01:28:56.267

Oh right forgot the 'program' part :P Editing that in in a bit. – NotBaal – 2017-08-14T02:04:33.507

It can just be a function too. – Zacharý – 2017-08-14T02:06:43.507

Woo hoo finally fixed it. That should do it right? – NotBaal – 2017-08-14T07:59:20.283

Yeah, again, amazing job considering this is your first golf here. – Zacharý – 2017-08-14T11:09:30.990

Thanks :) Just one final question - should I remove all the text with the strikethrough formatting or should I leave it for people who may view this in the future and can compare between my submissions (and what was wrong with the first one which can serve as a lesson for other people)? – NotBaal – 2017-08-14T11:39:10.540

You should probably remove the strucken text if you have that much of it. – Zacharý – 2017-08-14T18:07:24.327

There we go - finally got around to change and improve this :) – NotBaal – 2017-08-17T17:18:07.143


Stencil, 6 bytes

Not my favourite language, but it is short…

4 code bytes plus the nlist and Torus flags.


Try it online!

3 3
 a member of
m the moore-neighbourhood-count with self or
e the moore-neighbourhood-count without self


Posted 2011-08-13T22:37:27.493

Reputation: 37 779


Perl, 218 216 211 202 bytes

$,=$/;$~=AX3AAAx76;$b=pack('(A79)23',<>)x6;{print unpack'(a79)23a0',$b;select$v,$v,$v,0.1;$b=pack'(A)*',unpack'((x7a/(x13)X4Ax!18)1817@0)4',pack'((a*)17xx!18)*',unpack"x1737(AA$~Ax$~AA$~@)2222",$b;redo}

(No newline at the end of this code.)

Reads the starting pattern from standard input, as a text file where live cells are represented as 1, dead cells represented as a space, lines are separated by a newline. The input shall not have characters other than these. Lines can be variable length, and will be padded or truncated to exactly 79 wide. Example input is a glider gun:

                                1 1
                      11      11            11
                     1   1    11            11
          11        1     1   11
          11        1   1 11    1 1
                    1     1       1
                     1   1


As the program runs Game of Life, every state is dumped to the standard output in a format similar to the input, then delays 0.1 seconds. The delay can be customized by changing the fourth argument of the select call.

The game board is hard-coded to size 79x23. It is wrapped in a torus: if you leave the board on the bottom, you end up at the top; if you leave on the right side, you end up at the left side but shifted one row down.

Here's an alternate version that doesn't read any input and starts from a random board:

$,=$/;$/=AX3AAAx76;$b=pack("(A)*",map{rand 3<1}0..1816)x6;{print unpack'(a79)23a0',$b;select$v,$v,$v,0.1;$b=pack'(A)*',unpack'((x7a/(x13)X4Ax!18)1817@0)4',pack'((a*)17xx!18)*',unpack"x1737(AA$/Ax$/AA$/@)2222",$b;redo}

This code is derived from an obfuscated game of life perl program I wrote years ago. I have changed it a lot to make the board toroidal and golf the code.

This is probably not the shortest method to implement Game of Life in perl, but it's one of the less comprehensible ones.

The board is stored in $b as a string of '1' and ' ', one for each cell, only the whole thing is repeated at least three times. The third unpack call extracts 17 values for each cell: there's one for the cell itself and two for each of the eight neighboring cells, in an arbitrary order, and each value is '1' or an empty string. The cell should be live in the next iteration iff the number of '1' values among these 17 values is 5, 6, or 7. The third pack call concatenates these 17 values to a 18 character wide field left aligned and padded with nul bytes on the right. The second unpack call takes such a 18 wide field, dispatches on the character at position 7, unpacks the space from position 17 if it's a '1', or unpacks the character from position 4 otherwise. This result is exactly the value the cell should have in the next generation.


Posted 2011-08-13T22:37:27.493

Reputation: 341


J, 45

I thought I'd give J a try. It's not particularly well golfed yet, but I'll give it another try soon.



   f =: 5 5 $ 0 1 0 0 0   0 0 1 0 0   1 1 1 0 0   0 0 0 0 0    0 0 0 0 0
0 1 0 0 0
0 0 1 0 0
1 1 1 0 0
0 0 0 0 0
0 0 0 0 0
   f (]+.&(3&=)+)+/((4&{.,(_4&{.))(>,{,~<i:1))&|. f
0 0 0 0 0
1 0 1 0 0
0 1 1 0 0
0 1 0 0 0
0 0 0 0 0


Posted 2011-08-13T22:37:27.493

Reputation: 6 466


Processing 536 532

int h=22,t=24,i,j;int[][]w=new int[t][t],b=new int[t][t];int[]q={1,0,-1};void draw(){if(t<9){clear();for(i=2;i<h;i++){for(j=2;j<h;j++)w[i][j]=b[i][j];w[i][1]=w[i][21];w[i][h]=w[i][2];w[1][i]=w[21][i];w[h][i]=w[2][i];}for(i=1;i<23;i++)for(j=1;j<23;j++){t=-w[i][j];for(int s:q)for(int d:q)t+=w[i+s][j+d];b[i][j]=w[i][j]>0&(t<2|t>3)?0:t==3?1:b[i][j];}a();}}void keyPressed(){t=0;}void mousePressed(){int i=mouseX/5+2,j=mouseY/5+2;w[i][j]=b[i][j]=1;a();}void a(){for(i=0;i<h-2;i++)for(j=0;j<h-2;j++)if(w[i+2][j+2]==1)rect(i*5,j*5,5,5);}

I believe this satisfies all of the requirements.


int h=22,t=24,i,j;
int[][]w=new int[t][t],b=new int[t][t];
void draw(){
      for(int s:q)
        for(int d:q)
void keyPressed(){
void mousePressed(){
  int i=mouseX/5+2,j=mouseY/5+2;
void a(){


Posted 2011-08-13T22:37:27.493

Reputation: 299


Scala - 799 chars

Run as a script. A mouse click on a square toggles it on or off and any key starts or stops generation.

import java.awt.Color._
import swing._
import event._
import actors.Actor._
new SimpleSwingApplication{var(y,r,b)=(200,false,Array.fill(20,20)(false))
lazy val u=new Panel{actor{loop{if(r){b=Array.tabulate(20,20){(i,j)=>def^(i:Int)= -19*(i min 0)+(i max 0)%20
var(c,n,r)=(0,b(i)(j),-1 to 1)
for(x<-r;y<-r;if x!=0||y!=0){if(b(^(i+x))(^(j+y)))c+=1}
if(n&&(c<2||c>3))false else if(!n&&c==3)true else n}};repaint;Thread.sleep(y)}}
preferredSize=new Dimension(y,y)
reactions+={case e:MouseClicked=>val(i,j)=(e.point.x/10,e.point.y/10);b(i)(j)= !b(i)(j)case _:KeyTyped=>r= !r}
override def paintComponent(g:Graphics2D){g.clearRect(0,0,y,y);g.setColor(red)
for(x<-0 to 19;y<-0 to 19 if b(x)(y))g.fillRect(x*10,y*10,9,9)}}
def top=new Frame{contents=u}}.main(null)


Posted 2011-08-13T22:37:27.493

Reputation: 31


Matlab (152)

s=@(m)imfilter(m,[1 1 1;1 0 1;1 1 1],'circular')
while 1

I dont have Matlab installed right now to test it, I just golfed the code that I wrote a few years ago.

%% initialize
Bsize = 20;
nsteps = 100;
board = uint8(rand(Bsize)<0.2); % fill 20% of the board
boardsum = @(im) imfilter(im,[1 1 1; 1 0 1; 1 1 1], 'circular');
step = @(im, sumim) uint8((sumim==3) | (im & (sumim==2)) );

%% run
for i = 1:nsteps
    imshow(kron(board,uint8(ones(4))), [])
    ss(p,i) = sum(board(:));
    board = step(board, boardsum(board));
  • Boardsize is hardcoded but can be anything
  • wraps around
  • for user input one can change the inital board either by hardcoding another matrix or using the variable editor. Not pretty, but it works
  • 20 chars can be saved if the graphical output is skipped, the board will still be printed as text at every iteration. One-pixel cells which change every millisecond are not very useful anyway


Posted 2011-08-13T22:37:27.493

Reputation: 2 811

works in R2014a, just tested – masterX244 – 2014-10-06T13:12:37.233


Processing 270,261 249 bytes

Grid is the 100*100 pixels of screen, input comes in the form of a png picture

void setup(){image(loadImage("g.png"),0,0);}void draw(){loadPixels();int n,i=0,j,l=10000;int[]a=new int[l],p=pixels;for(;i<l;a[i]=n==5?-1<<24:n==6?p[i]:-1,i++)for(j=n=0;j<9;j++)n+=j!=4?p[(i+l-1+j%3+100*(j/3-1))%l]&1:0;arrayCopy(a,p);updatePixels();}


void setup() {
  image(loadImage("g.png"), 0, 0);
void draw() {
  int c=100, i=0, n, l=c*c, b=color(0);
  int[]a=new int[l], p=pixels;
  for (; i<l; i++) {
  arrayCopy(a, pixels, l);



Posted 2011-08-13T22:37:27.493

Reputation: 653


Lua + LÖVE/Love2D, 653 bytes

l=love f=math.floor t={}s=25 w=20 S=1 for i=1,w do t[i]={}for j=1,w do t[i][j]=0 end end e=0 F=function(f)loadstring("for i=1,#t do for j=1,#t[i]do "..f.." end end")()end function l.update(d)if S>0 then return end e=e+d if e>.2 then e=0 F("c=0 for a=-1,1 do for b=-1,1 do if not(a==0 and b==0)then c=c+(t[((i+a-1)%w)+1][((j+b-1)%w)+1]>0 and 1 or 0)end end end g=t[i][j]t[i][j]=(c==3 or(c==2 and g==1))and(g==1 and 5 or-1)or(g==1 and 4 or 0)")F("t[i][j]=t[i][j]%2")end end function l.draw()F("[i][j]==1 and'fill'or'line',i*s,j*s,s,s)")end function l.mousepressed(x,y)S=0 o,p=f(x/s),f(y/s)if t[o]and t[o][p]then t[o][p]=1 S=1 end end

or spaced out:

for i=1,w do
    for j=1,w do
    loadstring("for i=1,#t do for j=1,#t[i] do  "..f.." end end")()
function l.update(d)
    if S>0 then
    if e>.2 then
        for a=-1,1 do
            for b=-1,1 do
                if not(a==0 and b==0)then
                    c=c+(t[((i+a-1)%w)+1][((j+b-1)%w)+1]>0 and 1 or 0)
        t[i][j]=(c==3 or(c==2 and g==1))and(g==1 and 5 or-1) or (g==1 and 4 or 0)]])
function l.draw()
    F("[i][j]==1 and'fill'or'line',i*s,j*s,s,s)") end
function l.mousepressed(x,y)
    if t[o]and t[o][p] then

Click on the field to add living cells. Click outside of the field to run it.

Try it online!

enter image description here


Posted 2011-08-13T22:37:27.493

Reputation: 161


Python, 589 bytes

Mouse buttons: left - put a cell, right - remove a cell, middle - start/stop.

from Tkinter import*
import copy
f=[F*[0]for i in'7'*F]
def p(x,y,a):f[y][x]=f[y][x]or c.create_oval(x*S,y*S,x*S+S,y*S+S)if a else c.delete(f[y][x])
def R(e):global r;r=1-r
exec("c.bind('<Button-%i>',lambda e:p(e.x/S,e.y/S,%i));"*2%(1,1,3,0))
def L():
 if r:return
 for y in z(F):
	for x in z(F):
	 for j in z(-1,2):
		for i in z(-1,2):
		 if i or j:n-=not g[(y+j)%F][(x+i)%F]
	 if 1<n<4:
		if n==3and not g[y][x]:p(x,y,1)

And here is a version where you can drag mouse to draw. Graphics are a bit more pleasant.

from Tkinter import*
import copy
f=[F*[0]for i in'7'*F]
def p(x,y,a):f[y][x]=f[y][x]or c.create_rectangle(x*S,y*S,x*S+S,y*S+S,fill='gray')if a else c.delete(f[y][x])
def R(e):global r;r=1-r
exec("c.bind('<Button-%i>',lambda e:p(e.x/S,e.y/S,%i));c.bind('<B%i-Motion>',lambda e:p(e.x/S,e.y/S,%i));"*2%(1,1,1,1,3,0,3,0))
def L():
 if r:return
 for y in z(F):
  for x in z(F):
   for j in z(-1,2):
    for i in z(-1,2):
     if i or j:n-=not g[(y+j)%F][(x+i)%F]
   if 1<n<4:
    if n==3and not g[y][x]:p(x,y,1)

Oleh Prypin

Posted 2011-08-13T22:37:27.493

Reputation: 706

This doesn't follow the game of life rules correctly. – Steven Rumbalski – 2011-08-23T13:23:37.397

1@StevenRumbalski: Oh really? – Oleh Prypin – 2011-08-23T14:00:51.697

2really. You have an indentation error in your second version. The section starting with if 1<n<4: should be indented at the same level as for j in z(-1,2): – Steven Rumbalski – 2011-08-23T18:36:13.213


Python 2, 456 bytes

While I know this is an old post, I couldn't help myself from giving it a shot. The initial board can be any size as long as you draw a border around it and have an extra space on the last line.

import time,itertools as w,sys;t,q=map(lambda x:list(x[:-1]),sys.stdin.readlines()),list(w.product(range(-1,2),range(-1,2)));del q[4];n=map(lambda x:x[:],t[:])
while time.sleep(0.1)==None:
 for j in range(1,len(t)-1):
  for i in range(1,len(t[j])-1):x=sum(map(lambda s:1 if t[j+s[0]][i+s[1]]in'@'else 0,q));f=t[j][i];n[j][i]='@'if(f=='@'and(x==3 or x==2))or(f==' 'and x==3)else' '
 t=map(lambda x:x[:],n[:]);print'\n'.join(list(map(lambda x:''.join(x),t)))

Input.txt (note the extra space on the last line)

|                    @                   |
|                     @                  |
|                   @@@                  |
|                                        |
|                                        |
|                                        |
|                                        |
|                                        |
|                                        |
|                                        |
|                                        |
|                                        |
|                                        |
|                                        |
|                                        |
|                                        |
|                                        |
|                                        |
|                                        |
|                                        |

How to Run

python < input.txt


Posted 2011-08-13T22:37:27.493

Reputation: 137

time.sleep(0.1)==None => not time.sleep(.1), (f=='@'and(x==3 or x==2))or(f==' 'and x==3) => x==3or f=='@'and x==2 – CalculatorFeline – 2017-06-22T03:35:24.467

^, You forgot one, 1 if=>1if. – Zacharý – 2017-08-17T20:05:02.090


Python 2: 334 Bytes

Only 6 years late.

import time
while 1:
 r=[[0]*k for i in z(k)]
 for i in z(k*k):
  m,g=sum([n(s[(a+c)%k][(b+d)%k])for c in l for d in l if c|d]),n(s[a][b])
  r[a][b]=n((m==2)&g or m==3)
  print'*'if r[a][b]else' ',
  if b-k+1==0:print

You can run it like:


Where the 0s and 1s represent dead and live cells, an extra newline at the end begins execution.

Grids must be square.

It's easier to run than the shortest python one, supports any sized grids, and looks pretty when run.

It's also 100 bytes more, so there's that.


Posted 2011-08-13T22:37:27.493

Reputation: 41


JavaScript 676

Sorry Griffin, I just couldn't look at your code and not re-write it slightly... had to shave off two characters but it was damn worth it!

b=[];r=c=s=20;U=document;onload=function(){for(z=E=0;z<c;++z)for(b.push(t=[]),j=0;j<r;j++)with(U.body.appendChild(U.createElement("button")))t.push(0),id=z+"_"+j,style.position="absolute",style.left=s*j+"px",*z+"px",onclick=a}; ondblclick=function(){A=E=E?clearInterval(A):setInterval(function(){Q=[];for(z=0;z<c;++z){R=[];for(j=0;j<r;)W=(c+z-1)%c,X=(c+z+1)%c,Y=(r+j-1)%r,Z=(r+j+1)%r,n=b[W][Y]+b[z][Y]+b[X][Y]+b[W][j]+b[X][j]+b[W][Z]+b[z][Z]+b[X][Z],R.push(b[z][j++]?4>n&&1<n:3==n);Q.push(R)}b=Q.slice();d()})};function a(e){E?"_");b[P[0]][P[1]]^=1;d()}function d(){for(z=0;z<c;++z)for(j=0;j<r;)U.getElementById(z+"_"+j).innerHTML=b[z][j++]-0}

But as they say, it's easier to ask for forgiveness than permission... ;)


Posted 2011-08-13T22:37:27.493

Reputation: 6 949


Octave (153)

the same as Matlab by DenDenDo at Shortest Game of Life , but had to change imshow to imagesc:

s=@(m)imfilter(m,[1 1 1;1 0 1;1 1 1],'circular')
while 1

Martin Novy

Posted 2011-08-13T22:37:27.493

Reputation: 11


Postscript 529 515

Started with the example from Rosetta Code. Invoke with a filename argument (gs -- pulsar), the file containing 20*20 binary numbers (separated by space). Infinite loop: draw board, wait for enter, calculate next generation.

[/f ARGUMENTS 0 get(r)file/n 20>>begin[/m
n 1 sub/b[n{[n{f token pop}repeat]}repeat]/c 400
n div/F{dup 0 lt{n add}if dup n ge{n sub}if}>>begin{0
1 m{dup 0 1 m{2 copy b exch get exch get 1 xor setgray
c mul exch c mul c c rectfill dup}for pop pop}for
showpage/b[0 1 m{/x exch def[0 1 m{/y exch def 0
y 1 sub 1 y 1 add{F dup x 1 sub 1 x
1 add{F b exch get exch get 3 2 roll add exch
dup}for pop pop}for b x get y get sub b x get y get
0 eq{3 eq{1}{0}ifelse}{dup 2 eq exch 3 eq

Spaced, with a few stack comments (just the ones I needed).

/f ARGUMENTS 0 get(r)file
/n 20
/sz 400
%/r{rand 2147483647 div}
/m n 1 sub
%n{[n{r .15 le{1}{0}ifelse}repeat]}repeat
 n{[n{f token pop}repeat]}repeat
/c sz n div
/F{dup 0 lt{n add}if dup n ge{n sub}if}
    0 1 m{dup % y y
    0 1 m{ % y y x
        2 copy b exch get exch get 1 xor setgray
        c mul exch c mul c c rectfill
    }for pop pop}for
    /b[0 1 m{/x exch def
      [0 1 m{/y exch def
          y 1 sub 1 y 1 add{F dup %s y y
          x 1 sub 1 x 1 add{F %s y y x
              b exch get exch get %s y bxy
              3 2 roll add exch %s+bxy y
              dup %s y y
          }for pop pop}for
          b x get y get sub
          b x get y get
          0 eq{3 eq{1}{0}ifelse}{dup 2 eq exch 3 eq or{1}{0}ifelse}ifelse

pulsar data file:

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0
0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0
0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0
0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 0 0 0
0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0
0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0
0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

luser droog

Posted 2011-08-13T22:37:27.493

Reputation: 4 535


PHP, 201 bytes (not tested)

for($s=file(f);print"\n";$s=$t)foreach($s as$y=>$r)for($x=-print"
"<$c=$s[$y][++$x];print$t[$y][$x]=" X"[$n<4&$n>2-$a])for($n=-$a=$c>A,$i=$x-!!$x-1;$i++<=$x;)for($k=$y-2;$k++<=$y;)$n+=$s[$k][$i]>A;

Run with -nr.


for($s=file(f);                         # import input from file "f"
    print"\n";                              # infinite loop: 1. print newline
    $s=$t)                                  # 3. copy target to source, next iteration
    foreach($s as$y=>$r)                    # 2. loop through lines
        for($x=-print"\n";"\n"<$c=$s[$y][++$x]; # print newline, loop $x/$c through line characters (before line break)
            print                                   # 5. print new cell
                $t[$y][$x]=" X"[$n>2-$a&$n<4])      # 4. new cell is alive if neighbour count<4 and >2 (>1 if alive)
            for($n=-                                # 2. init neighbour count: exclude self
                $a=$c>A,                            # 1. $a=cell is alife
                $i=$x-!!$x-1;$i++<=$x;)             # 3. loop $i from one left to one right of current position
                for($k=$y-2;$k++<=$y;)                  # loop $k from one above to one below current position
                    $n+=$s[$k][$i]>A;                       # increase neighbor count if neighbour is alife


Posted 2011-08-13T22:37:27.493

Reputation: 13 814


VBA (Excel), 546* Bytes

Sub t()
For Each c In f.Cells
If g(r,0,0,l,-1,0)Then h=h+1
If g(r,0,0,l,1,21)Then h=h+1
If g(r,-1,0,l,0,0)Then h=h+1
If g(r,1,21,l,0,0)Then h=h+1
If g(r,1,21,l,-1,0)Then h=h+1
If g(r,-1,0,l,1,21)Then h=h+1
If g(r,-1,0,l,-1,0)Then h=h+1
If g(r,1,21,l,1,21)Then h=h+1
If h=3Or(h=2And c.Interior.Color=0)Then c.Value="A"
For Each c In f
If c.Value="A"Then c.Interior.Color=0
End Sub
Function g(r,e,f,c,q,z)
End Function

Create the living cells by highlighting cells in A1:T20 and coloring them black.

*When t is run, it executes a single tick. I added a button on mine to click repeatedly. If this doesn't satisfy the requirements, I'll add a do-->while[counta(f)]>0 and a DoEvents for +34 bytes.

[I'll edit this with a link when I upload a copy of my macro-enabled Excel doc]


Posted 2011-08-13T22:37:27.493

Reputation: 181

This answer appears to be non-functional for a couple of reasons, namely the declaration of f should include a set clause as otherwise f.cells will invoke an object required error as f will be instanciated as a 20 x 20 Array. Secondly, when this is corrected the given solution does not implement the rules of conways game of life properly. I suggest using a simple blinker oscillator as a future test case – Taylor Scott – 2019-02-15T19:11:11.657


HTML and Javascript, 248 characters

<script>b=[];setInterval(()=>x.innerHTML=(,y)=>,x)=>+(([0,1,2,4,6,8,9,10].reduce((s,n)=>s+((b[y-1+(n>>2)]||[])[x+n%4-1]||0),0)|c)==3)))).map(r=>`<tr>${>`<td>${'⬜⬛'[v]}`).join('')}`).join(''),999)</script><table id='x'>

Run it by opening the web developer console and setting the global variable b (for "board") to a 2D-array of 0 (dead) and 1 (alive) values, for example:

b=new Array(32).fill(0).map(() => new Array(32).fill(0).map(() => Math.floor(Math.random() * 2)))

The core of this is the following:

Javascript, 114 characters


Annotated as follows:

l=       // l="Life"
  b=>    // Function that takes a board (2D array of ints that are 1 or 0) // map each row 
      r, // the row (array of 1s and 0s)
      y  // index is the y position of the row
    )=> //map each cell
        c, // the cell value (0, or 1)
        x  // index is the x position
      )=>(+( //this whole thing will return true or false, convert to 1 or 0
        // these magic numbers represent the eight neighbors of a cell
        // they map to x, y offsets (used below):
        // value  x     y
        //      n%4-1 (n>>2)-1
          0, //  -1    -1
          1, //   0    -1
          2, //   1    -1
          4, //  -1     0
          6, //   1     0
          8, //   1    -1
          9, //   1     0
          10 //   1     1
        .reduce( //get the number of neighbors that are alive (===1)
            s, // the accumulator (sum of alive neighbors so far)
            n  // the number from the list of magic numbers above
          s+(  // add to the accumulator
              b[            // find the row containing the neighbor
                y-1+(n>>2)  // A shorter way of expressing y+Math.floor(n/4)-1
              ]||[]         // if index is -1 or b.length, empty array
            )[              // within the row, find the value
              x+n%4-1       // offset x value (see table above)
            ]||0            // coalesce undefined(x is -1 or r.length) to 0
          ),0               // initial value for neighborCount accumulator
        )|c                 // | bitwise or with the value of the cell, 1 or zero
                            // if the cell is dead (0), this has no effect
                            // if the cell is alive (1), this changes 2 to 3
      )==3)                 // 3 means either:
                            // 3 neighbors or
                            // 2 neighbors and an alive cell (the rules for life)

More info here:


Posted 2011-08-13T22:37:27.493

Reputation: 1