How can I obtain the permutation of 3 binary variables?

Hi everyone!

Can anybody tell me how to obtain the classical permutation matrix of binary variables?

for example given n=3 boolean variables A B C:

ABC

000

001

010

011

100

101

110

111

I hope there would be a command for this.

thank you!!

Answer 1

[Rik]

The easiest way to do this is with bin2dec. You can loop from 0 to 2^n-1 to get all combinations. To convert the character array back to a numeric vector you can simply use =='1'.

[Matteo Masto]

It sounds good but still I don't get it.. Can you please be clearer? Thanks a lot!!

[Rik]

I notice a typo, I meant dec2bin, not bin2dec.

Did you read the documentation?

for k=1:4
    x=dec2bin(k,3)
    x=='1'
end

x = '001'

ans = 1×3 logical array
   0   0   1

x = '010'

ans = 1×3 logical array
   0   1   0

x = '011'

ans = 1×3 logical array
   0   1   1

x = '100'

ans = 1×3 logical array
   1   0   0

[Matteo Masto]

thanks a lot! I had to fix it a bit but now it works :)

n = 3;
H = zeros(2^n,n);

for k=0:2^n-1
    x=dec2bin(k,n)
    
    for q = 1:1:n
    H(k+1,q) = str2num(x(q));
    end
end

[Rik]

str2num is not a fix, it introduces a call to eval, which is completely unnecessary. Comparing to '1' will already convert to a numeric vector. You can also pass a vector to dec2bin:

n=3;
str=dec2bin(0:2^n-1,n);
H= str=='1';
disp(H)

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