Chapter 1, Problem 1E

To determine

To show: An $m\text{and}n$ chessboard has a perfect cover by dominoes if and only if at least one of m and n is even.

Explanation of Solution

To show this, we have to add 3 cases here.

Case 1: Assume that one of $m\text{and}n$ is even and the second one is odd. Here, cover rows with even number of slabs, that is, cover the entire chessboard.

For an example, consider a $4\text{-by-}3$ chessboard.

Case 2: Assume that both $m\text{and}n$ are even. Now cover rows with even number of slabs, That is cover the entire chessboard.

For an example, consider a $4\text{-by-4}$ chessboard.

Case 3: Suppose that both $m\text{and}n$ are odd, and the chessboard has a perfect cover.

Here, the numbers of two color slabs are different, because one domino covers exactly one of the each.

So in this case, the chessboard can’t have a perfect cover. This is a contradicts.

Thus, our assumption is wrong. When both $m\text{and}n$ are odd, then chess board does not have a perfect cover.

In other words, if the chessboard has a perfect cover, at least one of $m\text{and}n$ is even.

For an example, consider a $4\text{-by-4}$ chessboard.

Thus, by cases 1, 2, and 3, and $m\text{and}n$ chessboard has a perfect cover by dominoes if and only if at least one of $m\text{and}n$ is even.