Example 4.2.5 How many $1 \times 2$ dominoes must be placed on a $6 \times 6$ chessboard to achieve a saturated cover.
Solution: As shown in Figure 4-20, 12 $1 \times 2$ dominoes are placed on a $6 \times 6$ chessboard, making it impossible to place any more dominoes. Now we prove that on a $6 \times 6$ chessboard, after placing any 11 dominoes, at least one more domino can still be placed.
Proof: There are $6 \times 6 - 2 \times 11 = 14$ empty squares on the chessboard. Dividing the chessboard into 4 $3 \times 3$ sub-boards, by the pigeonhole principle, at least one $3 \times 3$ sub-board must have at least 4 empty squares. Without loss of generality, assume the top-left $3 \times 3$ sub-board has 4 empty squares. For convenience, we label the $3 \times 3$ sub-board as shown in Figure 4-21.
If no more dominoes can be placed on the $6 \times 6$ chessboard, then no more dominoes can be placed in the $3 \times 3$ sub-board, leading to a contradiction.
If 9 is an empty square, then $2, 4, 6, 8$ are not empty squares. However, the dominoes at $2, 4, 6, 8$ must occupy additional squares, making it impossible for the $3 \times 3$ sub-board to have 4 empty squares.
If 9 is not an empty square, then this domino must occupy another square. (1) If this domino is placed horizontally at 8,9, then among $1, 2, 3, 4, 5, 6, 7$, at least 4 squares are still empty, and there must be two adjacent empty squares, allowing another domino to be placed; (2) If this domino is placed vertically at 9,6, then the sequence $7, 8, 1, 2, 3, 4, 5$ is equivalent to $1, 2, 3, 4, 5, 6, 7$, and the same proof applies.