are to be removed: {| class="wikitable" ! !! 1 !! 2 !! 3 !! 4 !! 5 !! 6 !! 7 |- ! A | 1 || 0 || 0 || 1 || 0 || 0 || 1 |- ! B | 1 || 0 || 0 || 1 || 0 || 0 || 0 |- ! C | 0 || 0 || 0 || 1 || 1 || 0 || 1 |- ! D | 0 || 0 || 1 || 0 || 1 || 1 || 0 |- ! E | 0 || 1 || 1 || 0 || 0 || 1 || 1 |- ! F | 0 || 1 || 0 || 0 || 0 || 0 || 1 |} Row D remains and columns 2, 3, 5, and 6 remain: {| class="wikitable" ! !! 2 !! 3 !! 5 !! 6 |- ! D | 0 || 1 || 1 || 1 |} Step 1—The matrix is not empty, so the algorithm proceeds. Step 2—The lowest number of 1s in any column is zero and column 2 is the first column with zero 1s: {| class="wikitable" ! !! 2 !! 3 !! 5 !! 6 |- ! D | 0 || 1 || 1 || 1 |} Thus this branch of the algorithm terminates unsuccessfully. The algorithm moves to the next branch at level 1… Level 1: Select Row B Step 4—Row B is included in the partial solution. Row B has a 1 in columns 1 and 4: {| class="wikitable" ! !! 1 !! 2 !! 3 !! 4 !! 5 !! 6 !! 7 |- ! A | 1 || 0 || 0 || 1 || 0 || 0 || 1 |- ! B | 1 || 0 || 0 || 1 || 0 || 0 || 0 |- ! C | 0 || 0 || 0 || 1 || 1 || 0 || 1 |- ! D | 0 || 0 || 1 || 0 || 1 || 1 || 0 |- ! E | 0 || 1 || 1 || 0 || 0 || 1 || 1 |- ! F | 0 || 1 || 0 || 0 || 0 || 0 || 1 |} Column 1 has a 1 in rows A and B; and column 4 has a 1 in rows A, B, and C. Thus, rows A, B,