Finding a binary matrix such that the given hamming weight is constant

Question

Given a square binary matrix. I want to get all possible binary matrices which are at d Hamming distance apart.

Suppose

A=[1 0 1; 
   0 1 1; 
   1 1 0]. 

Then a matrix which is one (d) Hamming distance apart is

[0 0 1; 
 0 1 1; 
 1 1 0].

Any help in Matlab base coding?

Answer 1

I am hoping that I got the definition of hamming weight right in the given context. Based on that hope/assumption, this might be what you were after -

combs = dec2base(0:2^9-1,2,9)-'0'; %//'# Find all combinations
combs_3d = reshape(combs',3,3,[]); %//'# Reshape into a 3D array

%// Calculate the hamming weights between A and all combinations.
%// Choose the ones with hamming weights equal to `1`
out = combs_3d(:,:,sum(sum(abs(bsxfun(@minus,A,combs_3d)),2),1)==1)

Thus, each 3D slice of out would give you such a 3 x 3 matrix with 1 hamming weight between them and A.

It looks like you have 9 such matrices -

out(:,:,1) =
     0     0     1
     0     1     1
     1     1     0
out(:,:,2) =
     1     0     1
     0     1     1
     0     1     0
out(:,:,3) =
     1     0     1
     0     0     1
     1     1     0
out(:,:,4) =
     1     0     1
     0     1     1
     1     0     0
out(:,:,5) =
     1     0     0
     0     1     1
     1     1     0
out(:,:,6) =
     1     0     1
     0     1     0
     1     1     0
out(:,:,7) =
     1     0     1
     0     1     1
     1     1     1
out(:,:,8) =
     1     1     1
     0     1     1
     1     1     0
out(:,:,9) =
     1     0     1
     1     1     1
     1     1     0

---

Edit

For big n, you need to use loops it seems -

n = size(A,1);
nsq = n^2;

A_col = A(:).';
out = zeros(n,n,nsq);
count = 1;
for k1 = 0:2^nsq-1
    match1 = dec2bin(k1,nsq)-'0';
    if sum(abs(match1-A_col))==1
        out(:,:,count) = reshape(match1,n,n);
        count = count + 1;
    end
end