Shuffle matrix rows so that common elements are in completely different order than in a said row

Say we have a matrix row 'current_row'

1 2 3

and a matrix A

1 2 5

1 3 6

2 4 3

The task is to re-arrange matrix A so that the common elements with 'current:_row' in each row are in compelely different indices.

One solution would be:

5 1 2

6 3 1

2 3 4

So that

for i=1:size(A,1)
    find(current_row == A(i,:))
end

gives always three []'s

[Matt J]

gives always three 0s

You mean gives []'s.

[Pseudoscientist]

Oh yes true

Answer 1

[Matt J]

Here is a vectorized implementation of what John described.

[m,n]=size(A);

I=repmat((1:m).',1,n);

B=A; %copy for shuffling

pool=find(any(B==current_row,2));

while ~isempty(pool)
    
    p=numel(pool);
    
    [~,J]=sort(rand(p,n),2);
    
    B(pool,:)=B(  sub2ind([m,n], I(pool,:), J) );  %shuffle rows in 'pool'
    
    pool=find(any(B==current_row,2)); %narrow the search
    
end

result=Anew;

Answer 2

[John D'Errico]

What you are looking for is a variation of derangement - thus a permutation such that no element is in the same location as it started.

A simple solution? Just continue to shuffle each other row until it meets the goal, that no common elements are in the same location as they are in current_row. Stop when it does, and then go on to the next row. WTP?

You can probably improve on that scheme a bit, by finding the elements in common between the two rows. Then first permute them into ANY locations, as long as they are not the same as the current_row. Having done that derangement operation, then permute the other elements into any random permutation.

[Pseudoscientist]

Hmm I don't know how long this would take, there are hundreds of thousands of current_rows and a 3x3 for every current row