how to expand a Square matrix and reverse to its original form ?
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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]
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
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Voss
2022년 8월 28일
편집: Voss
2022년 8월 30일
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
And to get the original matrix back:
idx = 1+(n+1)*(0:n-1);
neworiginal = newmatrix(idx,idx)
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Bruno Luong
2022년 8월 30일
newmatrix = zeros(2*n)
I pretend your allocation is ineffective; the size is n^2.
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Bruno Luong
2022년 8월 30일
Not fully check:
original=[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))
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Bruno Luong
2022년 8월 30일
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])
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