A standard 8×8 chessboard can easily be covered (tiled) with non-overlapping dominoes (1×2 pieces): simply use 4 dominoes in each row.

But what if we remove two squares—one each from diagonally opposite corners of the chessboard? Can this modified chessboard be completely covered by non overlapping dominoes?

We give a simple proof by contradiction that this is not possible. Suppose it were possible to completely cover the modified chessboard with non-overlapping dominoes. Now in any covering, every domino must cover exactly one white square and one black square. Thus the modified board must have exactly the same number of black and white squares. On the other hand, notice that the two removed squares must have been the same color because they came from diagonally opposite corners. See Figure 1. Thus there cannot be the same number of white squares and black squares in the modified chessboard!

Therefore it must be impossible to cover the modified board with non-overlapping dominoes!

**Presentation Suggestions:**

If you don't have a transparency, you can draw a small checkered 4×4 board as an example. Take a poll to see who does and doesn't think it is possible before you tell them… this will force them to form an opinion!

**The Math Behind the Fact:**

Imposing more structure on a problem sometimes helps reveal the structure that is there!

If you like this problem, here's another in the same vein. Is it possible to tile an 8x8x8 *cube* with two diagonally opposite corners removed, using 1x1x3 “trominoes”? Does a similar argument work? For a hint or answer, see the reference.

**How to Cite this Page:**

Su, Francis E., et al. “Dominoes on a Chessboard.” *Math Fun Facts*. <https://www.math.hmc.edu/funfacts>.

**References:**

J. Konhauser, D. Velleman, S. Wagon, *Which way did the bicycle go?*, pp. 48-49.

**Fun Fact suggested by:**

Michael Moody