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

I am having 7 random binary numbers of 3 bit. Now, I need to compute how many number of bits are changed among consecutive numbers.

Example:

A= 110; 101; 011; 111; 100; 001; 010

Expected output:

C=

110 101 011 111 100 001 010

110 - 2 2 1 1 3 1

101 2 - 2 1 1 1 3

011 2 2 - 1 3 2 1

111 1 1 1 - 2 2 2

100 1 1 3 2 - 2 2

001 3 1 1 2 2 - 2

010 1 3 1 2 2 2 -

Answer 1

[Voss]

% 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

[Askic V]

I would also like to propose another very similar solution:

A = [110; 101; 011; 111; 100; 001; 010]

A = 7×1

110 101 11 111 100 1 10

N = length(A);
M = zeros(N); % Initialize

for ii = 1:N
    for jj = 1:N
        cc = bitxor(A(ii), A(jj)); % use xor bitwise
        M(ii,jj) = mod(cc, 9);
    end
end

[ASHA PON]

Thank you for the reply.

[ASHA PON]

I am not getting correct answer with this procedure. Kindly share me some other method.

[Voss]

@ASHA PON: Kindly demonstrate your code with inputs and outputs.

[ASHA PON]

Thank you Voss. Actually I myself wrote the code and got the output.

[Voss]

What output did you get? What input did you use? What code did you write?

[ASHA PON]

This is the code which i have written. My in put is 3 bit binary numbers and my output is the same as I mentioned in the above example.

Code:

matbi=de2bi(matde,3,'left-msb');
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==0)=[];
dxxy=reshape (yy,[6,7]);
S = size(dxxy)+[1,0];
dxy = zeros(S);
dxy(~eye(S)) = dxxy;

[Voss]

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]

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]

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

Answer 2

[Star Strider]

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.

.

[ASHA PON]

Thank you for the reply. Actually, I have solved the problem.

[Star Strider]

My pleasure!