tiling deficient boards using l- pentominoes
Post on 24-Feb-2016
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Tiling Deficient Boards Using L-Pentominoes
By: Akhil Upneja
L Pentomino• Polyomino consisting of five 1x1 squares• All eight reflections and rotations
Deficient Boards
PolysolverPowerful programming toolIf types of tiles and type of board specified, will
tell you if tileable or not, and give you all possible tilings
Proof relies on this program in some parts
Theorem 1All deficient nxn boards can be tiled if:
n=1,4,6, or 9 (mod 10)n≥14
Four cases here to prove:n=1 (mod 10)n=4 (mod 10)n=6 (mod 10)n=9 (mod 10)
Case 1: n=4 (mod 10)4x4 deficient board is simply too smallJump to the next case: 14x14 deficient board
Wedge Lemma
Proof of Wedge LemmaConsider: The shaded square shown
Recall: Square has eight symmetries
Vertical Symmetry
Horizontal Symmetry
Diagonal or Rotational Symmetry
Application of Wedge Lemma
Using Polysolver and the Wedge Lemma, we conclude that all of the 14x14 deficient boards can be tiled
However, need generalization for all n=4 (mod 10) cases
Lemma 2If any deficient board with side length n can be
tiled, all mxm boards such that m (mod 10)=n mod (10) and m>n can be tiled as well
Proof of Lemma 2Consider the 24x24 deficient board in
conjunction with the wedge lemmaWedge has dimensions 12 by 12
Proof of Lemma 2 (cont.)Fit the 14x14 board inside the 24x24 board
All the wedge squares are filled because 14>n/2
Proof of Lemma 2 (cont.)Ignore the L shape left behindRecall: All deficient 14x14 boards are tileableTherefore, if we ignore the L shape, then all
deficient 14x14 boards created by removing a square from the wedge are also tileable
Last Step of ProofFinal step in this example is making sure L
shape can be tiled in the general case
How do we know this works every time?
Region II = 10xn
Region III = 10xm
Last LemmaAny 10xm rectangle such that m≥4 can be tiledProof: 2x5 and 5x2 blocks created using the L
pentominoes
Tiling Exists
Case 1 proven
Case 2: n=6 (mod 10)Using Polysolver, all 16x16 deficient boards can
be tiledBy extension, all deficient boards such that n=6
(mod 10) and n≥16 can be tiled
Case 3: n=9 (mod 10)All 19x19 deficient boards can be tiled,
according to PolysolverBy extension, all deficient boards such that n=9
(mod 10) and n≥19 can be tiled
Case 4: n=1 (mod 10)
Putting the 14x14 board into the 21x21 board satisfied the wedge lemma, and the L shape was tileable
Theorem 2 If n=6, then the deficient board is tileable if and only if the
deficient square has coordinates (1,2), (2,1), (5,1), (6,2), (1,5), (2,6), (5,6) or (6,5)
If n=9, then the deficient board is tileable if and only if the deficient square has coordinates (2,3), (2,7), (1,1), (1,3), (1,5), (1,7), (1,9), (5,1), (5,3), (5,5), (5,7), (5,9), (9,1), (9,3), (9,5), (9,7), (9,9), (3,1), (3,2), (3,5), (3,8), (3,9), (7,1), (7,2), (7,5), (7,8), or (7,9)
If n=11, then the deficient board is tileable if and only if the deficient square does not have coordinates (4,11), (8,11), (1,8), (1,4), (4,1), (8,1), (11,4), or (11,8)
Proof by Polysolver
Still to ComeFormal generalization for deficient rectanglesY-pentominoes?
Thanks!
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