How to select some black color cells in a chess board such that each white cell has exactly 1 adjacent selected black cell?

Question

Given a chess board of size $n$ x $n$, where $n$ is an even number. How do i prove that for any given even number n, there exists a way to select some of the black cells of the chess board such that each of the white cells in the chess board has exactly 1 selected black square adjacent to it.
Two cells are adjacent if they share an edge.

I tried looking for some pattern in the way black squares are selected and then tried to generalize it. But i failed to do so.

I have another approach but cant fully prove why this will work.

The approach is to traverse the chess board in a row major way (i.e go to the first row and check all cells from left to right, then go to the second row and check all the cells from left to right and so on). And while doing this if we reach a black square such that all the adjacent white cells to this black cell has 0 selected black cells adjacent to them, then we select this black square.
This approach works but i cant prove why its correct?

Answer 1

Suppose w.l.o.g. that the top left square is black. Divide the board into $2\times 2$ regions. These regions will be three kinds: $A$, $B$, or unmarked. In each $A$ region, we will mark the top left black square. In each $B$ region, we will mark the bottom right black square.

First we assign the regions along the top-right/bottom-left diagonal. These are all $A$ when $n\equiv 2\pmod4$, and the diagonal immediately below it is all $B$. On the other hand, if $n$ is a multiple of $4$, then we make that diagonal all $B$, with the diagonal immediately above it all $A$.

Moving out along parallel diagonals, above each $A$ diagonal is a blank one, followed by another $A$ diagonal. Below each $B$ diagonal is a blank one, followed (if there's still room) by another $B$ diagonal.

To verify that this works, we need to check white squares in the three types regions:

1. Each $A$ region has either blanks or $B$s below it and to the right, and it either has blanks or the board's edge above it and to the left. Its white squares are adjacent to their region's own marked black square, and free-and-clear across region boundaries to the right and down.
2. Repeat the preceding argument, swapping "$A$"/"$B$", "above"/"below", "top"/"bottom", and "left"/"right".
3. Each blank region has its white squares taken care of, either by the $A$ regions below it and to its right, or by the $B$ regions above it and to its left.

Before I started looking at it diagonally, I couldn't make heads or tails of it. The "row-major" process you describe does appear to generate the same result, although I don't immediately see why it should do that.