Beatrix's board and holes

Question

This is the last question from UKMT's JMC 2016:

Beatrix places dominoes on a 5 × 5 board, either horizontally or vertically, so that each domino covers two small squares. She stops when she cannot place another domino, as in the example shown in the diagram.

When Beatrix stops, what is the largest possible number of squares that may be still uncovered?

The answer is 7, as shown below:

Is there a general answer for $n\times n$ board? For example, is there a way to count, covering a $19\times 19$ board, how many holes Beatrix can leave on the board?

Answer 1

This is equivalent to computing the independent domination number of a graph with a node for each of the $2n(n-1)$ dominoes and an edge for each pair of dominoes that share a cell. See Minimum number of dominoes on an $n \times n$ chessboard to prevent placement of another domino., which links to https://oeis.org/A280984.

For $n=5$, number the horizontal dominoes $1$ through $20$ (top to bottom and then left to right), and number the vertical dominoes $21$ through $40$ (left to right and then top to bottom). Then dominoes $1$ and $6$ share cell $(1,2)$, and the full set of edges is $$\{(1,6),(1,21),(1,22),(2,7),(2,21),(2,22),(2,26),(2,27),(3,8),(3,26),(3,27),(3,31),(3,32),(4,9),(4,31),(4,32),(4,36),(4,37),(5,10),(5,36),(5,37),(6,11),(6,22),(6,23),(7,12),(7,22),(7,23),(7,27),(7,28),(8,13),(8,27),(8,28),(8,32),(8,33),(9,14),(9,32),(9,33),(9,37),(9,38),(10,15),(10,37),(10,38),(11,16),(11,23),(11,24),(12,17),(12,23),(12,24),(12,28),(12,29),(13,18),(13,28),(13,29),(13,33),(13,34),(14,19),(14,33),(14,34),(14,38),(14,39),(15,20),(15,38),(15,39),(16,24),(16,25),(17,24),(17,25),(17,29),(17,30),(18,29),(18,30),(18,34),(18,35),(19,34),(19,35),(19,39),(19,40),(20,39),(20,40),(21,26),(22,27),(23,28),(24,29),(25,30),(26,31),(27,32),(28,33),(29,34),(30,35),(31,36),(32,37),(33,38),(34,39),(35,40)\}$$

The independent domination number for $n=5$ is $9$, which implies that the largest remaining number of uncovered cells is $5^2-2\cdot9=7$, as you found. For $n=19$, the independent domination number is $122$, which implies that the largest remaining number of uncovered cells is $19^2-2\cdot122=117$.