A Tromino Simulation by Andre Berthiaume 2004 1

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 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

Theorem: n 1, any 2 n board with one square removed can be tiled by trominoes. Proof by mathematical induction: Let the property P(n) be the sentence“any 2 n 2 n 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. 4

Thm: n 1, any 2 n board with one square removed can be tiled with trominoes. The property P(n) is the sentence“any 2 n 2 n 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 2 k board with one square removed can be tiled by trominoes. [This is the inductive hypothesis. ] We must show that a 2 k+1 board with one square removed can be tiled by trominoes. 5

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

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. 7

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

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

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 2 k board with one square removed can be tiled by trominoes. [This is the inductive hypothesis. ] Suppose we have a 2 k+1 board with one square removed. Divide it into 4 quadrants, each consisting of a 2 k 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 2 k+1 board with one square removed can be tiled with trominos. 10

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

Tiling Problem: Simulation 12

Tiling Problem: Simulation 13

Tiling Problem: Simulation 14

Tiling Problem: Simulation 15

Tiling Problem: Algorithm Global: Board[1. . k, 1. . k] where k is 2 n 1 2 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 2 x 2 deficient board place the only tromino that works 3 4 else divide Board into UL, UR, LL, LR boards Properly place a tromino in the center Let [x 1, y 1], [x 2, y 2], [x 3, y 3], [x 4, y 4] be the coordinates of the missing/covered squares in each of the 4 quadrants Tile(UL, n-1, x 1, y 1) Tile(UR, n-1, x 2, y 2) Tile(LL, n-1, x 3, y 3) Tile(LR, n-1, x 4, y 4) 16

Tiling Problem: Algorithm Global: Board[1. . k, 1. . k] where k is 2 n 1 2 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 2 x 2 deficient board place the only tromino that works 3 4 else divide Board into UL, UR, LL, LR boards Properly place a tromino in the center Let [x 1, y 1], [x 2, y 2], [x 3, y 3], [x 4, y 4] be the coordinates of the missing/covered squares in each of the 4 quadrants Tile(UL, n-1, x 1, y 1) Tile(UR, n-1, x 2, y 2) Tile(LL, n-1, x 3, y 3) Tile(LR, n-1, x 4, y 4) 17

Copyright and Fair Use Information n n 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. 18