and C are to be removed and columns 1 and 4 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 |} Rows D, E, and F remain and columns 2, 3, 5, 6, and 7 remain: {| class="wikitable" ! !! 2 !! 3 !! 5 !! 6 !! 7 |- ! D | 0 || 1 || 1 || 1 || 0 |- ! E | 1 || 1 || 0 || 1 || 1 |- ! F | 1 || 0 || 0 || 0 || 1 |} Step 1—The matrix is not empty, so the algorithm proceeds. Step 2—The lowest number of 1s in any column is one. Column 5 is the first column with one 1 and thus is selected (deterministically): {| class="wikitable" ! !! 2 !! 3 !! 5 !! 6 !! 7 |- ! D | 0 || 1 || 1 || 1 || 0 |- ! E | 1 || 1 || 0 || 1 || 1 |- ! F | 1 || 0 || 0 || 0 || 1 |} Step 3—Row D has a 1 in column 5 and thus is selected (nondeterministically). The algorithm moves to the first branch at level 2… Level 2: Select Row D Step 4—Row D is included in the partial solution. Step 5—Row D has a 1 in columns 3, 5, and 6: {| class="wikitable" ! !! 2 !! 3 !! 5 !! 6 !! 7 |- ! D | 0 || 1 || 1 || 1 || 0 |- ! E | 1 || 1 || 0 || 1 || 1 |- ! F | 1 || 0 || 0 || 0 || 1 |} Column 3 has a 1 in