1. A Tromino Simulation by Andre Berthiaume

2. Tiling with trominoes
Definition: a tromino is a group of three unit squares
arranged as follows:
Definition: a pq board is a p  q array of unit squares.
Consider: pq boards with one square removed.
Definition: Tiling a board with one square removed means
every unit square of the board is covered
No tromino covers the missing square
No two tromino overlap
No tromino extends beyond the board

3. Tiling with trominoes
Definition: a tromino is a group of three unit squares
arranged as follows:
Definition: a pq board is a p  q array of unit squares.
Consider: pq boards with one square removed.
Definition: Tiling a board with one square removed means
Every unit square of the board is covered
No tromino covers the missing square
No two tromino overlap
No tromino extends beyond the board

4. Theorem:  n  1, any 2n  2n board with one square removed can be tiled by trominoes.
Proof by mathematical induction: Let the property P(n) be the sentence“any 2n2n board with one square removed can be tiled by trominoes.”
Show that the property is true for n = 1: There are 4 possible 2  2 boards with one square removed.
Clearly, each of them can be tiled with a single tromino.

5. Thm:  n  1, any 2n  2n board with one square removed can be tiled with trominoes.
The property P(n) is the sentence“any 2n2n board with one square removed can be tiled by trominoes.”
Show that for all integers k  1, if the property is true for n = k then it is true for n = k + 1:
Let k be any integer with k  1, and suppose that any 2k  2k board with one square removed can be tiled by trominoes. [This is the inductive hypothesis.]
We must show that a 2k+1  2k+1 board with one square removed can be tiled by trominoes.

6. Tiling Problem: Induction Step
Consider an 2k+1 2k+1 board with one square removed:
Then add a tromino
in the center as follow.
First, divide
the board in
4 quadrants.
Now, separate the 4 quadrants

7. Tiling Problem:Induction Step
Consider each
blue square and
the black square
as “missing” for
each sub-board.
By the inductive
hypothesis, each
of these sub-boards
can be tiled.

8. Tiling Problem:Induction Step
These tilings are
provided by the
induction process.
Now, we join
the sub-boards
together...

9. Tiling Problem :Induction Step
… and we find a
tiling for the full
board.

10. Summary: Show that for all integers k  1, if the property is true for n = k then it is true for n = k + 1:
Let k be any integer with k  1, and suppose that any 2k  2k board with one square removed can be tiled by trominoes. [This is the inductive hypothesis.]
Suppose we have a 2k+1  2k+1 board with one square removed. Divide it into 4 quadrants, each consisting of a 2k  2k board. In one of the quadrants, one square will have been removed, and so, by inductive hypothesis, the remaining squares in that quadrant can be covered by trominos.
The other three quadrants meet at the center of the board, and the center serves as a corner of a square from each of the three quadrants. A tromino can, therefore, be placed on those three central squares. By inductive hypothesis, the remaining squares in each of those three quadrants can be completely covered by trominos.
Thus the 2k+1  2k+1 board with one square removed can be tiled with trominos.

11. Tiling Problem: Simulation
How can you tile an
8  8 board with
one square removed?

12. Tiling Problem: Simulation

13. Tiling Problem: Simulation

14. Tiling Problem: Simulation

15. Tiling Problem: Simulation

16. Tiling Problem: Algorithm
Global: Board[1..k,1..k] where k is 2n
Program Tile( Board, n, x, y )
/* we deal with the 2^n x 2^n Board
/* the missing square is at Board[x,y]
if n = 1 then \* we have a 2x2 deficient board
place the only tromino that works
else
divide Board into UL, UR, LL, LR boards
Properly place a tromino in the center
Let [x1,y1], [x2,y2], [x3,y3], [x4,y4] be the coordinates
of the missing/covered squares in each of the 4 quadrants
Tile(UL, n-1, x1, y1)
Tile(UR, n-1, x2, y2)
Tile(LL, n-1, x3, y3)
Tile(LR, n-1, x4, y4) 1 2 3 4

17. Tiling Problem: Algorithm
Global: Board[1..k,1..k] where k is 2n
Program Tile( Board, n, x, y )
/* we deal with the 2^n x 2^n Board
/* the missing square is at Board[x,y]
if n = 1 then \* we have a 2x2 deficient board
place the only tromino that works
else
divide Board into UL, UR, LL, LR boards
Properly place a tromino in the center
Let [x1,y1], [x2,y2], [x3,y3], [x4,y4] be the coordinates
of the missing/covered squares in each of the 4 quadrants
Tile(UL, n-1, x1, y1)
Tile(UR, n-1, x2, y2)
Tile(LL, n-1, x3, y3)
Tile(LR, n-1, x4, y4) 1 2 3 4

18. Copyright and Fair Use Information
All materials on these slides are Copyright 2004 by Andre Berthiaume.
Permission to make a copy of the entire set of slides or of any part thereof, including posting on personal and class web pages, for educational and scientific use is granted without fee provided that it is not made or distributed for profit or commercial advantage and that the copy include the copyright notice found on the title page. To copy otherwise or to republish requires specific permission from Andre Berthiaume.