how to expand a Square matrix and reverse to its original form ?

if you want to expand a 2x2 to 4x4, or 8x8 to 16x16 or any square matrix, how can you do it using bitxor operation and reverse it to original matrix form? the concept operate like Hadamard code ..example a 2x2 original matrix has been converted to 4x4 matrix called new matrix as shown below

original=[1 2;3 4];

b=bitxor(1,[0 2;3 4]);
c=bitxor(2,[1 0;3 4]);
d=bitxor(3,[1 2;0 4]);
e=bitxor(4,[1 2;3 0]);
newmatrix = [b c;d e]

newmatrix = 4×4

1 3 3 2 2 5 1 6 2 1 5 6 3 7 7 4

The values at position (1,1)=1,(1,4)=2,(4,1)=3 and (4,4)=4 which are equal to the original matrix. and during the operation in each multiplier position we replace it by zero so that it remain with the same value as original after operation

Answer 1

[Voss]

Here's some code that generalizes the example for any n (size of original matrix), except this expands the matrix to size n^2-by-n^2. (It's not clear (to me) how the process would work for n > 2 in order to generate a matrix of size 2n-by-2n.)

original = [1 2; 3 4];

n = size(original,1);
newmatrix = zeros(n^2);
for ii = 1:n
    rows = n*(ii-1)+(1:n);
    for jj = 1:n
        cols = n*(jj-1)+(1:n);
        temp = original;
        temp(ii,jj) = 0;
        newmatrix(rows,cols) = bitxor(original(ii,jj),temp);
    end
end

newmatrix

newmatrix = 4×4

1 3 3 2 2 5 1 6 2 1 5 6 3 7 7 4

And to get the original matrix back:

idx = 1+(n+1)*(0:n-1);
neworiginal = newmatrix(idx,idx)

neworiginal = 2×2

1 2 3 4

[dani elias]

Thank you, I do appreciate for what you have done it.

[Voss]

You're welcome!

[dani elias]

For small matrix n^2 by n^2 works better,as the dimensions increases it takes a lot of time to computer, I stand with you on finding the solution of 2n by 2n...such as 1024by1024, computer halt,result to blue screen

[Bruno Luong]

newmatrix = zeros(2*n)

I pretend your allocation is ineffective; the size is n^2.

[Voss]

@Bruno Luong Thank you. Corrected.

Answer 2

[Bruno Luong]

Not fully check:

original=[1 2;3 4]

original = 2×2

1 2 3 4

n = size(original,1);
A = reshape(original,[n 1 n 1]);
A = repmat(A,[1 n 1 n]);
[I,J] = ndgrid(1:n,1:n);
A(sub2ind(n+zeros(1,4),I,I,J,J)) = 0;
B = reshape(original,[1 n 1 n]);
C = reshape(bitxor(A,B),n^2+zeros(1,2))

C = 4×4

1 3 3 2 2 5 1 6 2 1 5 6 3 7 7 4

[Bruno Luong]

Recover the original

% reverse
n2 = size(C,1);
n = sqrt(n2);
[I,J] = ndgrid(1:n,1:n);
K = sub2ind(n+zeros(1,4),I,I,J,J);
original = reshape(C(K),[n n])

[dani elias]

I do appreciate. Thank you