rows D and E; column 5 has a 1 in row D; and column 6 has a 1 in rows D and E. Thus, rows D and E are to be removed and columns 3, 5, and 6 are to be removed: {| 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 |} Row F remains and columns 2 and 7 remain: {| class="wikitable" ! !! 2 !! 7 |- ! F | 1 || 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 2 is the first column with one 1 and thus is selected (deterministically). Row F has a 1 in column 2 and thus is selected (nondeterministically). The algorithm moves to the first branch at level 3… Level 3: Select Row F Step 4—Row F is included in the partial solution. Row F has a 1 in columns 2 and 7: {| class="wikitable" ! !! 2 !! 7 |- ! F | 1 || 1 |} Column 2 has a 1 in row F; and column 7 has a 1 in row F. Thus, row F is to be removed and columns 2 and 7 are to be removed: {| class="wikitable" ! !! 2 !! 7 |- ! F | 1 || 1 |} Step 1—The matrix is empty, thus this branch of the algorithm terminates successfully. As rows B, D, and F are selected, the final solution is: {| class="wikitable" ! !! 1 !! 2 !! 3 !! 4 !! 5 !! 6 !! 7 |- ! B | 1 || 0 || 0 || 1 || 0 || 0 || 0 |- ! D | 0 || 0 || 1 || 0 || 1 || 1 || 0 |- ! F | 0 || 1 || 0 || 0 || 0 || 0 || 1 |} In other words, the subcollection {B, D, F} is an exact cover, since every element is contained in exactly one of the sets B = {1, 4}, D = {3, 5, 6}, or F = {2, 7}. There are no more selected