How do I create a matrix with all binary combinations?

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Christian P
Christian P 2020-5-13
评论： Pranay Agarwal 2022-12-2
Hi
I want to create a matrix with all binary combinations. If N is the length of the binary code, there would be possible combinations. Below are , and the 16 possible binary combinations are displayed.
1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0
1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0
1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
How can I create such a matrix automatically? I don't really care which order they are in, as long as every combination is present, and N is a variable.
Thanks in advance.

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采纳的回答

the cyclist
the cyclist 2020-5-13
There's an incredible obfuscated hack for this:
N = 4;
dec2bin(0:2^N-1)' - '0'
  1 个评论
Christian P
Christian P 2020-5-13
Thanks to both of you, this worked like a charm! My N is usually less than 10, so memory shouldn't be a big problem :)

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更多回答（2 个）

Fangjun Jiang
Fangjun Jiang 2020-5-13
I though it should be
ff2n(4)
James Tursa
James Tursa 2020-5-13
编辑：James Tursa 2020-5-13
dec2bin(0:2^N-1) - '0'
Note that this is only practical for relatively small values of N. Even moderatly large N can cause this to exceed your available memory.
  5 个评论
显示 3更早的评论
Voss
Voss 2022-12-1
Ran in:
@Pranay Agarwal: You can use dec2base:
N = 4; % length
base = 3;
output = dec2base(0:base^N-1,base)-'0';
disp(output);
0 0 0 0 0 0 0 1 0 0 0 2 0 0 1 0 0 0 1 1 0 0 1 2 0 0 2 0 0 0 2 1 0 0 2 2 0 1 0 0 0 1 0 1 0 1 0 2 0 1 1 0 0 1 1 1 0 1 1 2 0 1 2 0 0 1 2 1 0 1 2 2 0 2 0 0 0 2 0 1 0 2 0 2 0 2 1 0 0 2 1 1 0 2 1 2 0 2 2 0 0 2 2 1 0 2 2 2 1 0 0 0 1 0 0 1 1 0 0 2 1 0 1 0 1 0 1 1 1 0 1 2 1 0 2 0 1 0 2 1 1 0 2 2 1 1 0 0 1 1 0 1 1 1 0 2 1 1 1 0 1 1 1 1 1 1 1 2 1 1 2 0 1 1 2 1 1 1 2 2 1 2 0 0 1 2 0 1 1 2 0 2 1 2 1 0 1 2 1 1 1 2 1 2 1 2 2 0 1 2 2 1 1 2 2 2 2 0 0 0 2 0 0 1 2 0 0 2 2 0 1 0 2 0 1 1 2 0 1 2 2 0 2 0 2 0 2 1 2 0 2 2 2 1 0 0 2 1 0 1 2 1 0 2 2 1 1 0 2 1 1 1 2 1 1 2 2 1 2 0 2 1 2 1 2 1 2 2 2 2 0 0 2 2 0 1 2 2 0 2 2 2 1 0 2 2 1 1 2 2 1 2 2 2 2 0 2 2 2 1 2 2 2 2

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