matlab - 找到一个二进制矩阵，使得给定的汉明权重是恒定的

Question

给定一个二元方阵。我想得到所有可能的二进制矩阵，它们d相距汉明距离。

认为

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

那么相隔一 (d) 汉明距离的矩阵是

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

Matlab基础编码有什么帮助吗？

Answer 1

我希望我hamming weight在给定的上下文中得到了正确的定义。基于这种希望/假设，这可能就是您所追求的 -

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)

因此，每个 3D 切片out都会为您提供这样一个3 x 3矩阵，1它们之间具有汉明权重和A.

看起来你有9这样的矩阵 -

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

---

编辑

对于 big n，您似乎需要使用循环 -

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