I am looking for a simple way to get all combinations of an binary matrix. I tried already the function perms() but did not get a proper result.
I have for example an matrix N x N filled up with 1 and -1. With N=2 there would be 2^4 possible combinations of 1 and -1 like
(1 1) (1 1) (-1 -1)
M(1) = (1 1) , M(2) = (1 -1) , M(3) = ( 1 1) and so on...
When I use perms() I don't get for example the first matrix.
How can I fix that?
You can represent all numbers between 0 and 2^(N^2)-1 as binary numbers, and then reshape:
N = 2;
v = (1:2^(N^2))-1;
A = dec2bin(v)' - '0'; %'// Or use: decimalToBinaryVector(v)';
A(A==0) = -1;
A = reshape(A,N,N,2^(N^2));
2^N-1 instead of 2^N²-1). Plus I'd point out that this approach doesn't force you to store all of these combinations in memory, as it gives a 1:1 correspondance between your matrices and 0:2^N²-1, meaning if you just need to get one of these matrices at random for example, you can just pick a number and then apply the equivalence
A. (Seems like an easy fix to me). In case of this answer, no idea, it does indeed deliver you all 16 possibilities for a 2x2 matrix containing 1 and -1.
1s and -1s is fixed.
decimalToBinaryVector is specific to the Data Acquisition Toolbox. I believe you should be able to do the same with A=(dec2bin(v)-'0').'; though.decimalToBinaryVector. I'll edit my answer. Thanks!A simple hack is as follows:
v = [1 -1 1 -1];
P = perms(v);
for ii = 1:size(P,1)
A = reshape(P(ii,:),2,2)
end
which results in:
A =
-1 -1
1 1
...
Still there are some identical matrices in the result which should be removed.
I think that I found an solution to my problem
L = 2;
N = L^2;
v = cell(N,1);
for k = 1:N
v{k} = linspace(-1,1,2);
end
ne=numel(v);
x=cell(ne,1);
[x{1:ne,1}]=ndgrid(v{end:-1:1});
p=reshape(cat(ne+1,x{:}),[],ne);
F = cell(length(p),1);
for k=1:length(p)
F{k} = reshape(p(k,:),L,L);
end
clear all in SO codes, as people tend to copy these codes in their own and a clear call messes up everything. Second: p is a 65536x16 double, which seems a bit large for your permutations. All desired permutations are there, but way too often. You went a bit overboard with your dimensions I think.v is just a 16x1 cell with 16 times the same matrix: [-1 1]. That entire loop is just a difficult way of writing those two numbers. (As a side note, I did not vote on this answer)clear, that was not the main point of this solution being wrong. Just run it and see for yourself that v contains 16 matrices all being [-1 1], and p is way larger than you'd need. This makes the answer unusable.
v, see a previous comment as explanation. This results in all cells of x containing duplicate matrices, making this code not very efficient. Also I'd suggest not using a cell for F, but a 3D array as such: F(:,:,k) = reshape(p(k,:),L,L); All in all it's working, but not pretty nor efficient. I removed the downvote though, as it does work.