# Mathematical Recreations: Tweetable Game Of Life 10

Posted by **Loren Shure**,

Today I’d like to introduce a guest blogger, Matt, who is consultant over in The MathWorks UK. Some of you may have come across
his customer projects and series of demos on using `MATLAB in Physics`. He’s going to talk about an interesting one line MATLAB program.

### Contents

### Introduction

One Saturday afternoon I saw a link on `Hacker News` to a `YouTube` video showing Conway's Game of Life in one line of APL. This made me wonder whether MATLAB® could perform as well.

The answer: maybe not quite a single line, but definitely a `tweet`.

The code is reproduced below:

s=[1 1 1]'; n=@(a)conv2(s,s',1*a,'same')-a; lf=@(a)n(a)==2&a|n(a)==3; a=rand(128)>0.8; for ii=1:500,spy(a);drawnow;a=lf(a);end

In the code below I've commented out the population initialisation and visualisation code and replaced it with various alternatives for this published version, together with an animated gif of the output.

A = accumarray([3 3; 3 4; 4 2; 4 4; 5 4],1,[7 7]); imagesc(~A); colormap(gray); for ii=1:9,imagesc(~A);drawnow;A=lf(A);end

In this post I'll show the evolution of the code from initial conception to finished version, and how thinking in matrices can lead to new algorithms.

### Conway's Game Of Life

`Conway's Game of Life` is a two-dimensional cellular automation that demonstrates complex behaviour.

It consists of a grid of cells, each of which may be in one of two possible states, 0 ('dead') or 1 ('alive'). The state of each cell in the next iteration of the game is dependent on the state of the surrounding eight cells as follows:

- a live cell with 2 neighbours remains alive
- any cell with 3 neighbours becomes or remains alive
- any other cell dies of loneliness or overcrowding

The core of the algorithm consists of two steps:

- calculate the number of live neighbours of each cell
- using this, determine the new state of each cell based on its current state

### Calculating The Number Of Neighbours

To calculate the number of neighbours for each cell one approach would be to loop over each cell in the array, add up the number of live cells in the surrounding 3x3 square, and finally subtract the state of the cell itself.

The YouTube video used the different approach of creating 8 copies of the original array, offseting each by a step horizontally and/or vertically and then summing the arrays.

The key part of development of my algorithm was to realise that this looping-with-sum-over-neighbourhood is just a 2D convolution.
In MATLAB we can do a 2D convolution using `conv2`.

```
S = [1 1 1; 1 0 1; 1 1 1]
neighbours=@(a) conv2(double(a),S,'same')
```

S = 1 1 1 1 0 1 1 1 1 neighbours = @(a)conv2(double(a),S,'same')

The 'same' argument is to restrict the output of conv2 to be the same size as the original input matrix.

For example:

b = rand(5)>0.5 neighbours(b)

b = 0 0 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1 1 0 0 0 0 1 0 1 ans = 2 3 3 3 2 2 4 5 5 4 3 6 4 4 1 2 4 3 5 2 1 3 2 3 0

### Calculating The Next State

After calculating the number of neighbours, the next step is to use it to determine the next state using the rules of Conway's Game of Life.

life = @(a) (neighbours(a)==2 & a)... % Rule for 2 neighbours | (neighbours(a)==3) % Rule for 3 neighbours b life(b)

life = @(a)(neighbours(a)==2&a)|(neighbours(a)==3) b = 0 0 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1 1 0 0 0 0 1 0 1 ans = 0 1 1 1 1 1 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 1 1 0

Let's use this as a reference implementation to test that modifications to the function do the same thing:

t = @(fOrig,f,a) assert(all(all(fOrig(a) == f(a))));

### Reducing The Size

`Twitter` allows users to post 'tweets' of up to 140 characters in length. Fitting in the Game of Life evolution function *and* visualisation in this limit required a few tricks:

Use the syntax of conv2 that uses 1D vectors for the filter. This happens to also be faster but requires an extra element-wise matrix subtraction:

```
s = [1 1 1]';
n2 = @(a) conv2(s,s',double(a),'same')-a;
t(neighbours, n2, b)
```

Drop some more characters by using an implicit cast to double by multiplying a by 1. Also remove spaces:

```
n=@(a)conv2(s,s',1*a,'same')-a;
t(neighbours, n2, b)
```

Now shorten the Life rule function by removing spaces and parentheses, instead relying on the operator precedence rules:

lf=@(a)n(a)==2&a|n(a)==3; t(life,lf,b)

Finally, initialise a random population and visualise using `spy` in a loop:

a=rand(128)>0.8;for ii=1:25,spy(a);drawnow;a=lf(a);end

### Modifications

It is possible to lose a few more characters by changing the 'for' loop to a 'while 1' loop (to go forever), changing the definition of s to be a row vector, changing the 0.8 to .8, etc. I stopped at the point where I did since it allowed me to fit the code into a tweet, with room for a #MATLAB tag.

A modification that *would* be worth making is to introduce periodic boundary conditions:

makeperiodic = @(a) [a(end,end), a(end,:),a(end,1);... a(:,end),a,a(:,1);... a(1,end),a(1,:),a(1,1)]; unperiodic = @(a) a(2:end-1,2:end-1); np = @(a) unperiodic(n(makeperiodic(a))); lfp = @(a) np(a)==2&a|np(a)==3;

(an alternative is to use the `Image Processing Toolbox` `imfilter` command which has an option to use periodic boundaries)

... and making the visualisation nicer:

A = accumarray([3 3; 3 4; 4 2; 4 4; 5 4],1,[7 7]); imagesc(~A); colormap(gray); for ii=1:25,imagesc(~A);drawnow;A=lfp(A);end

Other things to try are left as an exercise for the reader:

### Conclusion

I've shown here how thinking in terms of operations on matrices rather than looping constructs can result in compact code. In this case the motivation was just for fun, but in general it can make use of the built-in MATLAB functionality and avoid having to reimplement existing functions.

Do you have any good one line (or one tweet) MATLAB programs? I'd love to hear from you here.

Get the MATLAB code

Published with MATLAB® 7.9

**Category:**- Fun,
- Function Handles,
- Image Processing

### Note

Comments are closed.

## 10 CommentsOldest to Newest

**1**of 10

Hello

I am a big fan of your blog, and usually I spent some time trying to figure out the ways you reduce in one line what I would spend like 4.

Keep up the good work, and keep good programs coming.

Leon

**2**of 10

Is Lauren on Twitter btw? Failed to find her.

**3**of 10

Gabe-

Spell my name as my parents did and you will find me on twitter. I am not very active there. Matt was the real author of this particular post.

–loren

**4**of 10

Fun :-). Just tried a few tricks to shorten it further… This is what i got:

a=rand(128)>0.8;

for ii=1:500,spy(a);drawnow;a=abs(filter2(ones(3),a)*2-6-a)<2;end

**5**of 10

And a high-life version:

a=rand(256)>0.8; for ii=1:500,spy(a);drawnow;f=filter2(ones(3),a);a=(f-6)+2*a==0|f==3;end

Test with this initial “a”:

a=zeros(200); a(101:105,101:105)=[0 0 1 1 1;0 1 0 0 1; 1 0 0 0 1;1 0 0 1 0;1 1 1 0 0];

**6**of 10

Aslak-

Very nice! I like your abs(filter2(…)…) approach, although I think the conv2 with 1D vectors will still win in terms of speed for large arrays.

I’m working on an approach using accumarray and the integral image of ‘a’ (google Viola-Jones for reference), I’ll add it to the thread when I get it working.

Now, if we can just get HashLife into a tweet…

-Matt

**7**of 10

Here’s another fun piece of code inspired by cellular automata. This one generates what looks like a labyrinth… I made it as a music visualizer for winamp many years ago.

And now i re-implemented it in matlab.

c=-ones(5); c(2:4,2:4)=2;

c(3,3)=13; c=c/8;

A=rand(300)>.99;

close all

colormap gray

for ii=1:100; A=max(min(filter2(c,A),1),0); imagesc(A==0); drawnow; end;

**8**of 10

Hello,

here is simple code to calculate for an optional vector a[i], i=1,..,n the sum by i of products, containing multipliers (for all j!=i) such as a[j]/(a[j]-a[i]). It can be proved that for all vectors this expression equals to 1. The code does not contain any loops:

A=x'*ones(size(x)); B=A./(A-A'+diag(x)); res=sum(prod(B))

**9**of 10

Heh! The same APL video also inspired me to write a Life program in Matlab. Not quite tweetable, however!

**10**of 10

In case you’re interested, someone made a code golf challenge on Stack Overflow:

http://stackoverflow.com/questions/3499538/code-golf-conways-game-of-life

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