How can I create a random binary matrix with a specified number of 1's and 0's?

fi = 1/3

x = randi([0 1],1,9);

    while (mean(x)~= fi)

      x = randi([0 1],1,9); 

    end

This finds a random 3 x 3 binary matrix with 3 ones, but for a 4x4 matrix and larger this can take forever.

Basically what I'm trying to do is find all possible combinations of a m x n binary matrix with a specified number of 1's.

Answer 1

[Jan]

n = 3;
a = zeros(5, 6);
a(randperm(numel(a), n)) = 1;

Or for older Matlab versions, which do not support RANDPERM with 2 inputs:

index = randperm(numel(a));
a(index(1:n)) = 1;

For both methods you get a zero matrix with exactly n ones.

But for your goal to count the number of possible combinations this approach is extremely bad. There are 2^(5*6) different binary [5x6] matrices.

[Aryama Mandal]

what will be the code if n=1, as this code is not running in case of n=1

Answer 2

[Lotanna Ohazuruike]

Here's a neat trick I use.

r4 = rand(15,15); % generates uniformly distributed random number 15x15 array
r4(r4>0.3) = 1;     % sets every element with a value greater than 0.3 to become 1
r4(r4<=0.3) = 0;   % sets every element with a value less than or equal to 0.3 to become 0

Hope that helps!

Answer 3

[Azzi Abdelmalek]

a=zeros(5);
n=3   % number of 1
a(randi(numel(a),3,1))=1

[Jan]

There is a high chance that RANDI replied repeated values.

[Azzi Abdelmalek]

Exact

[Jan]

But this drawback remains in the method you suggest. Do you post it because the chance to get repetitions in randi(numel(a),3,1) is at least smaller than to get exactly 3 ones in randi([0 1], 4, 4)?

[Nick Suttell]

Sometimes this will return a matrix with only two 1's. Any idea why this is?

[Nick Suttell]

Oh ok it's because randi gave repeated values.

Answer 4

[Jan]

Another solution which works until matrices with 52 elements:

M = 5;
N = 6;
d = rem(floor(rand * 2^52 * pow2(-51:0)), 2);
a = reshape(d(1:M*N), M, N);

Now you get the bit representation of a random number between 0 and 2^52.