0 || 1 |} Algorithm X with Knuth's suggested heuristic for selecting columns solves this problem as follows: Level 0 Step 1—The matrix is not empty, so the algorithm proceeds. Step 2—The lowest number of 1s in any column is two. Column 1 is the first column with two 1s and thus is selected (deterministically): {| 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 |} Step 3—Rows A and B each have a 1 in column 1 and thus are selected (nondeterministically). The algorithm moves to the first branch at level 1… Level 1: Select Row A Step 4—Row A is included in the partial solution. Step 5—Row A has a 1 in columns 1, 4, and 7: {| 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; column 4 has a 1 in rows A, B, and C; and column 7 has a 1 in rows A, C, E, and F. Thus, rows A, B, C, E, and F are to be removed and columns 1, 4 and 7