A standard 8x8 chessboard can easily be covered (tiled)
with non-overlapping dominoes (1x2 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!
If you don't have a transparency, you can draw a small
checkered 4x4 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.