Question

Given a binary matrix in which every row and column contains exactly only one 1, I need to rearrange the matrix columnwise so that it will become an identity matrix. For example, given a binary matrix:

Binary = [ 0     1     0     0     0
           0     0     1     0     0
           1     0     0     0     0
           0     0     0     0     1
           0     0     0     1     0 ]

To get the identity matrix we rearrange the column as 2 3 1 5 4.

How can we optimally rearrange the columns for any given arbitrary square binary matrix?

Answer 1

A very simple way to do this is to use the function FIND like so:

[index,~] = find(Binary.');  %'# Transpose the matrix and find the row indices
                              %#   of the non-zero entries

And you can test that it work as follows:

>> Binary(:,index)

ans =

     1     0     0     0     0    %# Yup, that's an identity matrix alright!
     0     1     0     0     0
     0     0     1     0     0
     0     0     0     1     0
     0     0     0     0     1

OLD APPROACH:

This isn’t as compact or efficient as the above solution, but you could also transpose the matrix and use SORTROWS to sort the columns (now transposed into the rows) and return the sort indices. This will actually sort values in ascending order, which will give you an anti-diagonal matrix, so you will want to flip the vector of indices using FLIPUD. Here’s the code:

[~,index] = sortrows(Binary.');  %'# Transpose and sort the matrix
index = flipud(index);            %# Flip the index vector