Sorting a binary 2D matrix?

Question

I\'m looking for some pointers here as I don\'t quite know where to start researching this one.

I have a 2D matrix with 0 or 1 in each cell, such as:

  1         


        

Answer 1

I came up with the below algorithm, and it seems to work correctly.

Phase 1: move rows with most 1s up and columns with most 1s right.

1. First the rows. Sort the rows by counting their 1s. We don't care if 2 rows have the same number of 1s.
2. Now the columns. Sort the cols by counting their 1s. We don't care if 2 cols have the same number of 1s.

Phase 2: repeat phase 1 but with extra criterions, so that we satisfy the triangular matrix morph.
Criterion for rows: if 2 rows have the same number of 1s, we move up the row that begin with fewer 0s.

Criterion for cols: if 2 cols have the same number of 1s, we move right the col that has fewer 0s at the bottom.

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Example:

Phase 1

  1 2 3 4                     1 2 3 4                   4 1 3 2
A 0 1 1 0                   B 1 1 1 0                 B 0 1 1 1
B 1 1 1 0  - sort rows->    A 0 1 1 0  - sort cols->  A 0 0 1 1
C 0 1 0 0                   D 1 1 0 0                 D 0 1 0 1
D 1 1 0 0                   C 0 1 0 0                 C 0 0 0 1

Phase 2

  4 1 3 2                     4 1 3 2
B 0 1 1 1                   B 0 1 1 1
A 0 0 1 1  - sort rows->    D 0 1 0 1  - sort cols-> "completed"
D 0 1 0 1                   A 0 0 1 1
C 0 0 0 1                   C 0 0 0 1

Edit: it turns out that my algorithm doesn't give proper triangular matrices always.
For example:

Phase 1

   1 2 3 4                    1 2 3 4                
A  1 0 0 0                  B 0 1 1 1                
B  0 1 1 1 - sort rows->    C 0 0 1 1  - sort cols-> "completed"
C  0 0 1 1                  A 1 0 0 0                
D  0 0 0 1                  D 0 0 0 1                

Phase 2

   1 2 3 4                    1 2 3 4                   2 1 3 4
B  0 1 1 1                  B 0 1 1 1                 B 1 0 1 1
C  0 0 1 1 - sort rows->    C 0 0 1 1  - sort cols->  C 0 0 1 1
A  1 0 0 0                  A 1 0 0 0                 A 0 1 0 0
D  0 0 0 1                  D 0 0 0 1                 D 0 0 0 1
                           (no change)

(*) Perhaps a phase 3 will increase the good results. In that phase we place the rows that start with fewer 0s in the top.