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!