1. Tiling Deficient Boards Using L-Pentominoes
By: Akhil Upneja

2. L Pentomino Polyomino consisting of five 1x1 squares
All eight reflections and rotations

3. Deficient Boards

4. Polysolver Powerful programming tool
If 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

5. Theorem 1 All 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)

6. Case 1: n=4 (mod 10) 4x4 deficient board is simply too small
Jump to the next case: 14x14 deficient board

7. Wedge Lemma

8. Proof of Wedge Lemma Consider: The shaded square shown
Recall: Square has eight symmetries

9. Vertical Symmetry

10. Horizontal Symmetry

11. Diagonal or Rotational Symmetry

12. 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

13. Lemma 2
If 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

14. Proof of Lemma 2
Consider the 24x24 deficient board in conjunction with the wedge lemma
Wedge has dimensions 12 by 12

15. Proof of Lemma 2 (cont.) Fit the 14x14 board inside the 24x24 board
All the wedge squares are filled because 14>n/2

16. Proof of Lemma 2 (cont.) Ignore the L shape left behind
Recall: All deficient 14x14 boards are tileable
Therefore, if we ignore the L shape, then all deficient 14x14 boards created by removing a square from the wedge are also tileable

17. Last Step of Proof
Final step in this example is making sure L shape can be tiled in the general case

18. How do we know this works every time?
Region II = 10xn
Region III = 10xm

19. Last Lemma Any 10xm rectangle such that m≥4 can be tiled
Proof: 2x5 and 5x2 blocks created using the L pentominoes

20. Tiling Exists
Case 1 proven

21. Case 2: n=6 (mod 10)
Using Polysolver, all 16x16 deficient boards can be tiled
By extension, all deficient boards such that n=6 (mod 10) and n≥16 can be tiled

22. Case 3: n=9 (mod 10)
All 19x19 deficient boards can be tiled, according to Polysolver
By extension, all deficient boards such that n=9 (mod 10) and n≥19 can be tiled

23. Case 4: n=1 (mod 10)
Putting the 14x14 board into the 21x21 board satisfied the wedge lemma, and the L shape was tileable

24. 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

25. Still to Come Formal generalization for deficient rectangles
Y-pentominoes?

26. Thanks!