I recently got one of those 8x8 LED matrices and I was playing with some Game of Life patterns when I found this pretty repeating pattern. I found it by starting with some random patterns. If you look closely you can see the pattern becoming a mirrored version of itself halfway through. Apparently the pattern doesn't repeat like this on an infin…

game of life with 8x8 bicolor led matrix

/*
I recently got one of those 8x8 LED matrices  and I was playing with some Game of Life patterns when I found this pretty repeating pattern. I found it by starting with some random patterns. If you look closely you can see the pattern becoming a mirrored version of itself halfway through. Apparently the pattern doesn't repeat like this on an infinite grid but on this wrapping 8x8 grid it does ;-) 

FYI, the LED matrix is a bicolor one (green/red) and has an I2C interface (http://www.adafruit.com/products/902). I'm using the colors as follows:
 - newly created cells are green
 - cells that are at least 10 generations old are red
 - other living cells are yellow (simultaneously green+red)

It's hookup up to my Arduino Uno r3.

here's a video: http://www.youtube.com/watch?v=Ee2hOaQ2RDI

*/

#include <Wire.h>
#include "Adafruit_LEDBackpack.h"
#include "Adafruit_GFX.h"

boolean cells[8][8];

Adafruit_BicolorMatrix matrix = Adafruit_BicolorMatrix();

// game of life
int next[8][8];

void setup() {
  Serial.begin(9600);
  Serial.write("hello");

  randomSeed(analogRead(0));
  for (int r=0 ; r<8 ; r++) {
    for (int c=0 ; c<8 ; c++) {    
      if (random(2) >0)
        next[r][c] = 1;
    }
 } 
 
  matrix.begin(0x70);  // pass in the address
}

void loop() {

  game_of_life();
}

int current[8][8] =
{ {0,0,0,0,0,0,0,0},
  {0,0,0,0,0,0,0,0},
  {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} };
  
int mod(int a) { return (a+8)%8; }

void game_of_life() {
  
  matrix.clear();
  
  // draw
  for (int r=0 ; r<8 ; r++) {
    for (int c=0 ; c<8 ; c++) {
      int color;
      if (next[r][c] == 0)
        color = 0;
      else if (next[r][c] == 1) 
        color = LED_GREEN;
      else if (next[r][c] > 10)
        color = LED_RED;
      else
        color = LED_YELLOW;
      matrix.drawPixel(c,r,color);
    }
  }
  matrix.writeDisplay();
      
  // calc next state
  for (int r=0 ; r<8 ; r++) {
    for (int c=0 ; c<8 ; c++) {    
      // count alive neighbors
      int alive = 0;
      alive += current[mod(r+1)][mod(c)  ] != 0;
      alive += current[mod(r)  ][mod(c+1)] != 0;
      alive += current[mod(r-1)][mod(c)  ] != 0;
      alive += current[mod(r)  ][mod(c-1)] != 0;
      alive += current[mod(r+1)][mod(c+1)] != 0;
      alive += current[mod(r-1)][mod(c-1)] != 0;
      alive += current[mod(r+1)][mod(c-1)] != 0;
      alive += current[mod(r-1)][mod(c+1)] != 0;
      if (current[r][c])
        if (alive < 2 || alive > 3)
          next[r][c] = 0;
        else
          next[r][c] = current[r][c] + 1;
      else
        if (alive == 3)
          next[r][c] = 1;
    }
  }    
  
  for (int r=0 ; r<8 ; r++) {
    for (int c=0 ; c<8 ; c++) {    
      current[r][c] = next[r][c];
    }
  }  
  delay(100);
}

wonderful clean code, i use it on my 8x8 matrix just fine.
as i am no mathematician, when the life runs out after some time,
how do you check the end of all iterations, when nothing changes/live anymore?

i like the changing on the matrix while life goes on and want to start it all over again,
but it seems a litte bit over my (coding-)head to find the right code to start the game after its end again?

do you mind bothering you to give me a little hint? 😄
thanks a lot (for sharing) 👍

i tried a first approach to check if the iteration/generation of any cell is about > 100, than there was a stable pattern before we start new.
i all cells (or alternativly the sum-count of alive) is zero, than the matriy is empty and we start new.

any other ideas?