How to compute number of bit change among consecutive binary numbers?

% A = 110; 101; 011; 111; 100; 001; 010
A = dec2bin([6; 5; 3; 7; 4; 1; 2],3)
A = 7×3 char array
    '110'
    '101'
    '011'
    '111'
    '100'
    '001'
    '010'
N = size(A,1);
C = zeros(N);
for ii = 1:N
    for jj = 1:N
        C(ii,jj) = nnz(A(ii,:) ~= A(jj,:));
    end
end
disp(C);
     0     2     2     1     1     3     1
     2     0     2     1     1     1     3
     2     2     0     1     3     1     1
     1     1     1     0     2     2     2
     1     1     3     2     0     2     2
     3     1     1     2     2     0     2
     1     3     1     2     2     2     0

10 个评论
显示 8更早的评论隐藏 8更早的评论

Voss 2022-12-16

在 MATLAB Online 中打开

What is matde?

matde = [6; 5; 3; 7; 4; 1; 2]
matde = 7×1
     6
     5
     3
     7
     4
     1
     2
matbi=de2bi(matde,3,'left-msb')
matbi = 7×3
     1     1     0
     1     0     1
     0     1     1
     1     1     1
     1     0     0
     0     0     1
     0     1     0
yy=[];
for i=1:7
    aas=matbi(i,:);
    xxy=[];
    for ii=1:7 
        y=xor(aas,matbi(ii,:));
        xy=sum(y);
        xxy=[xxy xy];
    end
    yy=[yy; xxy];
end
yy
yy = 7×7
     0     2     2     1     1     3     1
     2     0     2     1     1     1     3
     2     2     0     1     3     1     1
     1     1     1     0     2     2     2
     1     1     3     2     0     2     2
     3     1     1     2     2     0     2
     1     3     1     2     2     2     0

yy seems to be correct, in that it is the C you asked for in your question.

I'm not sure what the remaining part of the code is for (constructing dxy). You're removing the zeros on the diagonal of yy, then putting the remaining elements of yy into a zeros matrix at the non-diagonal positions, so you end up with the same thing you started with (dxy = yy).

yy(yy==0)=[];
dxxy=reshape (yy,[6,7]);
S = size(dxxy)+[1,0];
dxy = zeros(S);
dxy(~eye(S)) = dxxy
dxy = 7×7
     0     2     2     1     1     3     1
     2     0     2     1     1     1     3
     2     2     0     1     3     1     1
     1     1     1     0     2     2     2
     1     1     3     2     0     2     2
     3     1     1     2     2     0     2
     1     3     1     2     2     2     0

ASHA PON 2022-12-16

Actually while forming yy matrix, my size will be 6*7. But i need 7*7, so i included 0 diagonally to proceed with further calculation.

Voss 2022-12-16

yy is 7x7 before you remove the zeros, as shown above.

请先登录，再进行评论。

Answer 2

Star Strider 2022-12-15

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1877742-how-to-compute-number-of-bit-change-among-consecutive-binary-numbers#answer_1128592

在 MATLAB Online 中打开

Use pdist with the 'hamming' distance metric then squareform to create the matrix —

b = {'110'    '101'    '011'    '111'    '100'    '001'    '010'};
d = pdist(b, 'hamming');
Warning: Converting cell data to double.
dmtx = squareform(d*3);                                             % Differences Are Reported As Percentages, So Multiply By The Length Of The Strings To Get Integers
VNRN = cellfun(@(x)sprintf('%s',x),b, 'Unif',0);
Result = array2table(dmtx, 'VariableNames',VNRN, 'RowNames',VNRN)   % Create A 'table' To Get  The Desired Result
Result = 7×7 table
           110    101    011    111    100    001    010
           ___    ___    ___    ___    ___    ___    ___

    110     0      2      2      1      1      3      1 
    101     2      0      2      1      1      1      3 
    011     2      2      0      1      3      1      1 
    111     1      1      1      0      2      2      2 
    100     1      1      3      2      0      2      2 
    001     3      1      1      2      2      0      2 
    010     1      3      1      2      2      2      0 

The variable names and row names do not have to be valid MATLAB variable names, however if you want to use them as such, change the ‘VNRN’ function to:

VNRN = cellfun(@(x)sprintf('%_s',x),b, 'Unif',0);

or something similar so that they all begin with a non-numeric character.

.