Subfield: Combinatorics


Question: Let $n$ be a positive integer. Dominoes are placed on a $2 n \times 2 n$ board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each $n$, determine the largest number of dominoes that can be placed in this way.

(A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)

Solution:
Let $M$ denote the maximum number of dominoes which satisfy the condition of the problem. We claim that $M=n(n+1) / 2$. The proof naturally splits into two parts: we first prove that $n(n+1) / 2$ dominoes can be placed on the board, and then show that $M \leq n(n+1) / 2$ to complete the proof. To prove that $M \leq$ $n(n+1) / 2$, we again border the $2 n \times 2 n$ square board up to a $(2 n+2) \times(2 n+2)$ square board. In fact, we shall ignore the corner border cells as they cannot be covered anyway and consider only the $2 n$ border cells along each side. We prove that out of each four border cells next to each other at most two can be covered. Suppose three out of four cells $A, B, C, D$ are covered. Then there are two possibilities below:

<img_3563>

or

<img_3278>



The first option is that $A, B$ and $D$ are covered (marked with + in top row). Then the cells inside the starting square next to $A, B$ and $D$ are covered by the dominoes, but the cell in between them has now two adjacent cells with dominoes, contradiction. The second option is that $A, B$ and $C$ are covered. Then the cells inside the given square next to $A, B$ and $C$ are covered by the dominoes. But then the cell next to $\mathrm{B}$ has two adjacent cells with dominoes, contradiction.

Now we can split the border cells along one side in groups of 4 (leaving one group of 2 if $n$ is odd). So when $n$ is even, at most $n$ of the $2 n$ border cells along one side can be covered, and when $n$ is odd, at most $n+1$ out of the $2 n$ border cells can be covered. For all four borders together, this gives a contribution of $4 n$ when $n$ is even and $4 n+4$ when $n$ is odd. Adding $4 n^{2}$ and dividing by 8 we get the desired result.
Consider the number of pairs of adjacent cells, such that one of them is covered by a domino. Since each cell is adjacent to one covered cell, the number of such pairs is exactly $4 n^{2}$. On the other hand, let $n_{2}$ be the number of covered corner cells, $n_{3}$ the number of covered edge cells (cells with 3 neighbours), and $n_{4}$ be the number of covered interior cells (cells with 4 neighbours). Thus the number of pairs is $2 n_{2}+3 n_{3}+4 n_{4}=4 n^{2}$, whereas the number of dominoes is $m=\frac{n_{2}+n_{3}+n_{4}}{2}$.

Considering only the outer frame (of corner and edge cells), observe that every covered cell dominates two others, so at most half of the cells are ccovered. The frame has a total of $4(2 n-1)$ cells, i.e. $n_{2}+n_{3} \leq 4 n-2$. Additionally $n_{2} \leq 4$ since there are only 4 corners, thus

$8 m=4 n_{2}+4 n_{3}+4 n_{4}=\left(2 n_{2}+3 n_{3}+4 n_{4}\right)+\left(n_{2}+n_{3}\right)+n_{2} \leq 4 n^{2}+(4 n-2)+4=4 n(n+1)+2$

Thus $m \leq \frac{n(n+1)}{2}+\frac{1}{4}$, so in fact $m \leq \frac{n(n+1)}{2}$.
We prove that this is the upper bound (and also the lower bound!) by proving that any two configurations, say $A$ and $B$, must contain exactly the same number of dominoes.

Colour the board in a black and white checkboard colouring. Let $W$ be the set of white cells covered by dominoes of tiling $A$. For each cell $w \in W$ let $N_{w}$ be the set of its adjacent (necessarily black) cells. Since each black cell has exactly one neighbour (necessarily white) covered by a domino of tiling $A$, it follows that each black cell is contained in exactly one $N_{w}$, i.e. the $N_{w}$ form a partition of the black cells. Since each white cell has exactly one (necessarily black) neighbour covered by a tile of $B$, each $B_{w}$ contains exactly one black tile covered by a domino of $B$. But, since each domino covers exactly one white and one black cell, we have

$$
|A|=|W|=\left|\left\{N_{w}: w \in W\right\}\right|=|B|
$$

as claimed.

Final answer: $\frac{n(n+1)}{2}$