 Figure 1
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Can all but one square of an n by n chessboard be
covered by L-shaped trominoes?
In general, it can not necessarily be done,
but if n is a power of 2, it can!
See the Figure for an 8x8 example.
Is there a method for finding such a tiling?
Yes, and we can find it by induction!
Presentation Suggestions:
Ask students to try to tile an 8x8 or a 16x16 board with
L-shaped tiles before showing them the proof.
The Math Behind the Fact:
A simple proof by induction shows this property.
The simplest case is a 2x2 chessboard. Clearly if one
square is removed, the remainder can be tiled by one
L-shaped tromino.
So, given a chessboard of size 2nx2n,
with one square removed, it can be divided into four
2(n-1)x2(n-1) smaller chessboards,
one of which contains the removed tile.
In the remaining three
2(n-1)x2(n-1) boards, we'll remove
a tile from each by
taking the adjacent center corners of the three
to give an L-shaped hole.
The inductive hypothesis will show how to cover the
four smaller boards, since they each have a square removed,
and are size 2(n-1)x2(n-1).
The L-shaped hole can also be covered by a tromino.
This covers the original board except for the original
missing square!
In Figure 1 above, applying the inductive step removes
the yellow tiles near the center of the 8x8 board. The
8x8 is divided into 4 quadrants, each with a square
missing. Then applying the inductive step removes the
green tiles near the center of each 4x4 quadrant. This
gives several 2x2 boards with a square removed---the
base case!
How to Cite this Page:
Su, Francis E., et al. "Inductive Tiling."
Mudd Math Fun Facts.
<http://www.math.hmc.edu/funfacts>.
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