California State University, Northridge

TILING WITH POLYOMINOES

A graduate thesis submitted in partial fulfillment of the requirements For the degree of Master of Science in

By

Gabriel Cedillo

August 2010 The graduate thesis of Gabriel Cedillo is approved:

Dr. Joel Zeitlin Date

~ardoA.&'ijJI Date

Dr. Silvia Fernandez, Chair Date

California State University, Northridge

11 DEDICATION

I thought the day would never come when I would see the final product of what has been in process for many days and sleepless nights. I cannot express on paper the amazing love and support of my wife throughout this journey. This thesis is dedicated to my beautiful wife Brenda Cedillo; I could have not done this without her support and words of encouragement. Thank you for all you do! To my daughter Frieda Kamila, and my two sons Diego Gabriel and Mateo Nikolas, without whom I would most certainly be lost. Thank you for your love, patience, and understanding, and for waiting for me every time you wanted me to go outside to play with you. I love you all profoundly.

111 ACKNOWLEDGEMENT

First of all, I would like to thank God for giving me the strength and the wisdom to finish this thesis. I want to especially thank Dr. Silvia Fernandez for all the hard work, dedication, and endless hours she put into this thesis to make sure that the outcome was unique, and original. Thank you for the inspiration, but most importantly for the motivation to embark on this project. You have been an inspiration to me, you were there when I most needed support and positive encouragement through my Number Theory and Combinatories classes, and obviously through finishing my thesis, and you are still here. I cannot find the words to express my admiration for your strong devotion to students like me. You are an amazing advocate for many students at California State University, Northridge and a strong believer of many of us. Thank you for believing in me! I want to thank Dr. Abrego and Dr. Zeitlin for accepting to be a part of this project. Your comments and suggestions contributed to a unique and original work as well. Thank you! Finally, I want to thank my parents, family and friends for all their support!

IV TABLE OF CONTENTS

Signature Page 11 Dedication lll Acknowledgment lV Abstract Vl

Chapter I: Tiling with Polyominoes Introduction 1 Definitions and Examples of Polyominoes and Non-polyominoes 2 The Existence Problem 3 Divisibility Condition 3 The Counting Problem 3 Polyominoes with at most 3 cells 4

Chapter II: Tiling with Monominoes and Dominoes Tiling with Dominoes 5 Boards with even area at most 16 6 Number of ways to tile Bmxn with copies of D 8 Number of ways to tile B3x4 with copies of D 12 Number of ways to tile B4x4 with copies of D 14 Variations for Dominoes 20

Chapter III: Tiling with the S-Triominoes 22

Chapter IV: Tiling with the L-Triomino 25

References 33

v ABSTRACT

MATHEMATICS:

TILING WITH POL YOMINOES

By

Gabriel Cedillo

Master of Science in Mathematics

The objective of this graduate thesis is to introduce the idea of tiling rectangular boards made up of with polyominoes. It is intended to present the abstract idea and the complexities of polyominoes through a simple mathematical application that can be used in any mathematics high school classroom. Mathematics is probably one of many, if not the only, subject that is misunderstood by many people in the world. The misconception comes because people believe that mathematics is about computations and memorization of formulas, and because we as teachers present the subject as a series of complex problems. But the reality is that in the real world, mathematics plays a completely different role. Real world problems are part of a broader context and in the real world situations, individual problems are not that terribly difficult.

Mathematicians currently find ways to relate mathematical ideas into real world and suggest real life applications on textbooks to teachers of all different levels. Hence, the idea of this thesis is to minimize the misconceptions of mathematics. As a mathematics teacher, I feel it is important to join the quest to help ease the frustration of many students and teachers when learning and teaching mathematics. Therefore, this thesis aims to stimulate the student's interest and abilities to learn mathematics and to become better thinkers in broader mathematical terms.

Vl Introduction

Mathematics is seen by many as a fearful subject, yet there are many topics that are fun and exciting in mathematics, generally called . Have you ever played ? The game tetris is based on a very exciting topic of mathematics? Tetris is a creation of Polyominoes. Polyominoes have been used in popular since at least 1907. The name was invented by Solomon W. Golomb[3] in 1953 and the topic has been popularized ever since then. In particular they were popularized by beginning in 1956. Gardner was the editor of a children's magazine named "Humpty Dumpty" but in 1956, he quit his job as an editor to work on a monthly column in recreational mathematics for the magazine Scientific American which would later turn into a column in mathematical games that influenced a generation of mathematicians. Gardner never took a mathematics course beyond high school, yet he was able to bring mathematics to a great number of admiring readers, writing his columns in such a way that people could understand them[5]. After his retirement in 1981 , Gardner wrote a large collection of books related to recreational mathematics. He died at age 94 on May 22, 2010. In tribute to his influence many mathematicians have dubbed him a "mathemagician".

In this work we consider the problem of tiling rectangular boards made up of squares with polyominoes. We give necessary and sufficient conditions to tile a rectangular board with copies of a fixed polyomino with at most three cells, we give the exact number of different ways to tile for some small boards, and bounds on this number for finite families. We also consider some variations for tiling non-rectangular boards with dominoes.

All the work on this thesis is original. However, in some cases, a similar or even better result is known. For example, in 1961 Temperley and Fisher[4], and independently Kasteleyn[2], found a formula to count the exact number of ways to tile an even area in rectangular boards with dominoes. See [1] for a nice survey and related problems.

1 Chapter I: Definitions and Problems

Definitions

A polyomino is a collection of unit squares, called, cells arranged in such a way that every is connected to every other square through a sequence of shared edges.

Figures 1 and 2 present a few examples of non-polyominoes and polyominoes, respectively.

a) b) c) d)

Figure 1. Non-polyominoes. a) Cells overlap; b) Cells are connected by vertices only; c) Although most of the cells are connected by the edges, cell 4 is connected to the rest only by two of its vertices; d) Each connected part is an example of a polyomino, but the two parts together do not form a polyomino. -

I I

t Cl I I I I a) b) c)

d)

Figure 2. a), b) and c) represent different types of polyominoes. In all cases, each cell is connected to at least another cell by an edge, and it is possible to go from any cell to any other cell by a sequence of edge-neighboring cells.

2 The Existence Problem

Our main interest in this work is to determine whether or not a given rectangular board can be tiled with copies of a polyomino.

Let P be a polyomino. Determine the rectangular boards with integer sides that can be tiled by congruent copies of P. In other words, which boards can be covered by congruent copies of P such that the copies do not go out of the board and do not overlap except along the edges? When this is possible, we say that P covers the board. Think of tiling with copies of P as you would tile a kitchen floor or the bathroom of a house, with the restriction that the tiles cannot be cut. We refer to this problem as the existence problem.

Let B111x 11 denote a rectangular board with sides m and n, where m and n are positive integers. Let I P I denote the area of P, which corresponds to its number of cells. Figure 3 shows the board B4xs tiled with copies of the polyomino P shown on the left.

Polyomino P

Figure 3. Tiling the board B 4x8 with copies of P.

Divisibility Condition

In order to cover a kitchen floor, you first need to know the number of tiles needed. In our problem this corresponds to the number of copies of a polyomino P needed to tile a board Bmxn· Since the polyominoes cannot be cut, we need exactly mn/ IP I polyominoes, which means that IP I must divide mn. We refer to this as the divisibility condition. This condition is necessary but not always sufficient as we show later on.

The Counting Problem

After finding a board that can be tiled by a polyomino, we are interested in finding different ways to tile. Given a fixed polyomino P, in how many ways can the board Bmxn be tiled by copies of P? We refer to this problem as the counting problem. Here, two ways of tiling a board are considered different even if they become the same after a series of rotations and/or reflections. (See Figure 4.)

3 We denote by T(P, B111xn) the total number of ways to tile the board B 111xn with copies of the polyomino P. Note that T(P, B 111x11 ) = T(P, Bnxm ) and thus it is enough to consider only one of the boards Bmxn or Bnxm in our analysis.

a) b)

Figure 4. a) Two different tilings of B2x2; a rotation is needed to match the tilings; b) Two different tilings of B3x2; a is needed to match the tilings.

Polyominoes with at most 3 cells

In this work, we study the existence and counting problems described above when P is one of the following polyominoes. (See Figure 5.)

Monomino (M) - A monomino is a single 1x1 square.

Domino (D) - A is the unique polyomino with 2 cells.

Triomino (S & L) - Trinominoes are with 3 cells. There are two types of triominos: The straight triomino and the L-shaped triomino. These are denoted by S and L, respectively.

I I I I

M=Monomino D=Domino S=Straight-triomino L=L-triomino

Figure 5. Polyominoes with up to three cells.

4 Chapter II: Tiling with Monominoes and Dominoes

The simplest member of the polyomino family is the monomino M. When P = M, the answers to the existence and the counting problem are trivial: M can cover any board Bmxn in a unique way!

Tiling with Dominoes

Consider the domino D. For which m and n, the board Bmxn can be covered by D?

If Bmxn can be tiled by D, then by the divisibility condition mn must be even. When m and n are both odd, mn is odd and thus Bmxn cannot be tiled by D. When at least one of m or n is even, mn is even, we show that in this case it is always possible to tile Bmxn with copies of D.

Without loss of generality, suppose that m is even, then we can cover Bmxn simply by using vertical copies of D. Every copy of D covers 2 cells. Therefore, no matter how big or small m is, as long as m is even, we can cover each column of the board with m/2 vertical copies of D. Thus D covers the board Bmxn. (See Figure 6.)

The following theorem summarizes these results.

Theorem 1. The board Bmxn can be tiled by copies of D if and only if mn is even.

nodd

m even

Figure 6. Tiling Bmxn with copies of D when m is even.

Now, we find the value of T(D, Bmxn) for all pairs m, n where mn :S 16. (See Figure 7.) For the boards B Ixn with 2 :Sn :Si 6 even we use the following observation.

5 [[] B fx2 I I I I I B fx4 I I I IB lx6

I I I I I I I I I B fx8

I I I I I I I B Ix /O

I I I I I I I I I I IB/x/2

I I I I I I I I I I I IB/x /4

I I I I I I I I I I I I IB1 x/6

DJ EEE 111111 B 2x2 B 2x3 B 2x4 B 2x5

I I I I I I I I I I I I I I I B 2x6 B 2x7

I I I I I I I I I B 2x8

B3x4

Figure 7. Boards with even area at most 16.

6 Proposition 1. For every n even, T(D, B1xn) = 1.

Proof: Since n is even, and D covers exactly two cells of n, then n can be covered with n/2 horizontal copies ofD in a unique way. (See Figure 8.) Therefore T(D, Bi xn) = 1. • [I]

Bix4

Figure 8. Tiling Bixn with copies of D when n is even.

Now we look at boards of the form B 2xn, for n 2: 2.

Theorem 2. T(D, B2x1 ) = 1, T(D, B2x2) = 2, and If n 2: 3, then

T(D, B 2xn) = T(D, B 2x(n-I)) + T(D, B 2x(n-2) ) 0

Proof. By Proposition 1, T(D, B2x1) = 1 and T(D, B2x2 ) = 2. (See Figure 4a.) Let n 2: 3, and consider the board B2xn and label the upper left cell with 1 and the lower left cell with 2. (See Figure 9.)

If we cover cell 1 with a vertical copy of D (see Figure 9a), then we are left with a B 2x(n-J) board, which can be tiled in T(D, B 2x(ii-I)) ways. If we cover cell 1 with a horizontal copy of D (see Figure 9b) , then it forces cell 2 to be horizontally tiled, and we are left with a B 2x(n-2J> which can be tiled in T(D, B 2x(ii-2J ) ways.

a) b)

Figure 9. Tiling B 2xn with dominoes.

In general, for n 2: 2, the number of ways to tile B 2xn is given by the following recursive formula T(D, B2xn) = T(D, B 2x(n-I)) + T(D, B 2x(n-2J ) . •

Note that the recursive formula in Theorem 2 corresponds to the Fibonacci numbers with Fo=F1=1.

1 Corollary 1. T(D, B 2xn) = F 11 where Fn is the n h Fibonacci number.

7 We use this corollary to find the following values.

T(D, B2x1) = F1 =J T(D, B2xs) = F s =8 T(D, B2x2 ) = F 2 =2 T(D, B2x6) = F6= 13 T(D, B2xJ ) = F3=3 T(D, B2x1) = F1=21 T(D, B2x4) = F4=5 T(D, B2x8) = F8=34.

All different ways to cover these boards are shown in Figures 10-13.

T(D,B2x2 ) = 2 EE

T(D,B2x3 ) = 3

T(D,B2x4) = 5

Figure 10. Tiling the board B2xm with D for m=2, 3 and 4.

8 T(D,B2xs) = 8 •• • •• •• T(D,B2x6) = 13 ••

I I I I I I II I I I I I I I I I I I I I

I I I I I I I I I I I I I I I I I I I I I

I I I I I I I I I I •• I I I I ••

•• I I I I I I I Figure 11. ••Tiling the board B2xmfor m=5 and 7.

9 T(D,B2x1) = 21 11111111 11111111 11111 111

I I I I I I I I I I I I I I I I I I I I I I I I

111 111 11 111 11 111 11 111111

I I I I I I I I I I I I I I I I I I I I I I I I

I I I I I I I I I I I I I I I I I I I I I I I I

I I I I I I I I I I I I I I I I I I I I I I I I

I I I I I I I I I I I I I I I I I I I I I I I I

Figure 12. Tiling the board B2x7 ·

10 T(B2xs) =34

I I I I I I I I I I I

I I I I I I I I I I I

Figure 13. Tiling the board B 2x8·

11 We now find the number of ways to tile B3x4 and B4x4 with copies of D using tree­ diagrams

Proposition 2. T(D, BJx4 ) = 11

Proof: To find T(D,B3x4), we start by considering all ways to cover its upper-left cell, labeled 1, and subsequently study all other possibilities to cover cell 2, then cell 3, and so on. We construct a tree diagram that includes all these possibilities. See Figures 14, 14a and 14b. Case 1 Im Figure 14. There are two ways to cover cell 1.

In Case 1, (See Figure 14a) cell 2 must be covered with a horizontal copy ofD. We then construct a tree diagram where each branch considers all possible ways to cover the newest labeled cell. Each leaf of the tree corresponds to one way to tile the board by copies of D. So there are 4 ways of tiling the board in Case 1. We construct a second three diagram for Case 2 (see Figure 14b), which shows 7 more ways to tile B3x4. Therefore T(D, BJx4 ) = 4 + 7 = 11. •

4

4 5 6

4 4 5 6 5 6

Figure 14a. Tree diagram for tiling B3x4 with copies of D, in Case 1.

12 2

...... l;..) 4 415 4 -5 6

4 4 4 415 --5 516 5 -6 6

4 4 Figure 14b. Tree diagram for tiling -5 -516 B3x4 with copies of D in Case 2. -6 Proposition 3. T(D, B 4x4 ) = 36.

Proof: As in the previous proof, we start by covering cell 1 (the upper-left cell) and subsequently study all other possibilities to cover cell 2, then cell 3, and so on. We have two ways to cover cell 1 (a horizontal or vertical copy of D) which determine our two cases. See Figure 15. In Case 1, cell 2 can again be covered in two different ways namely, Cases la and lb. Similarly Case 2 generates Cases 2a and 2b.

Case 1 Case 2

Case 2 -~~ 2 2

la lb 2a 2b

3

Figure 15. Four cases: 1 a, 1b , 2a, 2b.

Figure 16 takes care of Case la. It shows the corresponding tree diagram with 6 leaves. Note that when 4 copies of D have been used on the right branch, a board B2x4 is left to cover. We know from above that T(D, B 2x4 ) = 5, so the rest of this branch corresponds to the 5 ways to cover B 2x4·

14 5

5 7 8 6

~------1 B 2x3 ---+T(D, B2x3 )=3

5 6 7 5 6 7 5 7

8 6

5 6 7 8 5 6 7 5 7 5 7 8

8 6 8 6

Figure 16. Sub tree diagram for Case 1a.

15 Similarly, Figure 17 takes care of the left branch for Case 1b. Note again, that after using 5 copies of Don the left branch, we are left with a B2x3 board. Since T(D, B2x3 ) = 3, each of these branches has 3 leaves.

3

4 5 4

5

4 5 4 5 7

8

Figure 17. Sub-tree diagram for Case 1b .

On the right branches we are left with a copy of B2x4 after using 4 copies of D. This corresponds to 5 more leaves. On the left branch a copy of B2x3 is left. Since T(D, B2x3 ) = 3 and T(D, B 2x4 ) = 5, then Case 1b generates 3 + 1 + 3 + 5 = 12 leaves.

16 Now we take care of Cases 2a and 2b. Cell 2 branches out to two possibilities. Figure 18 shows the tree diagram for Case 2b. Note that there is only one way to cover cell 3 (a vertical copy of D).

7

6 Continue on page 18

7 6 8

Figure 18. Tree diagram for Case 2a.

Note that the left branch in Figure 18 has a B2x2 board left to be covered. This means that there are only T(D, B2x2) = 2 ways to continue this branch which correspond to two different tilings of B4x4. Similarly, the right branch has a B2x4 board left which means that there are T(D, B2x4) = 5 ways to continue this branch. This is explicitly shown in Figure 19.

17 5

5 7 5 6 B 2x2 -+T(D, B2x2 )=2 6

5 6 7 5 6 7 5 7 8 5 7

8 6 6 8

5 6 7 8 5 6 7

8

Figure 19. Subtree diagram for Case 2a.

18 Finally, in Case 2b (See Figure 20) a board B3x4 is left to cover which can be done in T(D, B3x4) = 11 ways. Therefore, there are 6+12+7+ 11 =36 ways to tile B4x4 •

3

Figure 20. Sub-tree diagram for Case 2b.

We can use the exact results for small boards to find the lower bounds on T(D, Bmxn).

Suppose you have a board of size B111x11 .

111 4 Proposition 4. T(D, B111xn) ::'.': 2 n1 whenever m and n are even.

Proof: Consider the board B 111x11 with m and n even. We divide Bmxn into copies of B2x2 (See Figure 21). There are T(D, B2x2)=2 different ways to tile each copy of B2x2· 1111114 Since there are (n/2)(m/2) = mn/4 copies of B2x2, then there are at least T(D, B 111xn) = 2 ways of tiling Bmxn •

B, ., B ' ..

D r2

Figure 21. B,mn divided into nm/4 copies of B2x2· There are 2 ways to tile each copy of B2x2 with dominoes.

We can improve this bound simply by dividing into copies of B4x4. We can do this whenever 4 divides m and n. (See Figure 22.)

11111116 Proposition 5. T (D, B111x 11 ) ::'.': 36 •

Proof: Consider the board B 111x11 with m and n multiples of 4. As before, we divide B111x11 into copies of B4x4 (See Figure 22.). There are T(D, B4x4)=36 different ways to tile each copy of B4x4 · Since there are (n/4)(m/4) = mn/16 copies of B4x4, then there are at least 11111116 T(D, B 111x11) = 36 ways of tiling Bmxn ·•

19 F R

E JxJ

. :

Figure 22. Bmxn divided into copies of B 4x4

Note that Proposition 5 improves Proposition 4 when m and n are multiples of 4 because

mn 11111 11111 mn

Variations for Dominoes

There are several variations of our problem when the board is not necessarily rectangular. As an example, we present a popular problem often found in puzzles and challenges books.

Example 1. Suppose you have an 8 x 8 chessboard. Remove two diagonally opposite corners. (See Figure 23 .) Can you cover the remaining 62 cells with 31 dominoes? • •• • • • • • •• • • • • • • •• • • • • • •• • • • • • Figure 23. Chessboard, B8x8 with two diagonally opposite corners removed.

The answer is no, even when the divisibility condition is satisfied. (The total number of cells is even, namely 62). Note that a domino covers one black and one white cell. Hence, any set of cells that can be covered by dominoes must contain the same number of black and white squares. Since we need 31 dominoes to cover 62 cells, these 31 would

20 have to cover 31 black cells and 31 white cells. The only problem here is that the diagonally opposite corners of an 8 x 8 chessboard have the same color, and thus we have 30 white squares and 32 black cells (or vice versa) which is impossible to cover with 31 dominoes.

Inspired by the previous example, we looked at the following problem.

Example 2. Consider the following pyramid with 11 cells forming its base. Is it possible to cover the pyramid with copies of D?

Figure 24. The pyramid with base 11.

By the divisibility condition, in order for D to cover a board, the area of the board must be even. The area of the pyramid is 1+ 3+5+7+9+ 11 = 36 which satisfies the divisibility condition and implies that 18 dominoes would be needed to tile the pyramid. However, each domino covers exactly one unit square of different color but we have 21 gray unit squares and 15 white unit squares. Hence, the answer is no and once again we have proved that the divisibility condition is not always sufficient.

We can generalize this result for any pyramid with an odd number of cells in its base.

Proposition 6. Any pyramid with 2n-1 cells in its base, cannot be tiled with copies of D.

Proof: By the divisibility condition, in order for D to cover a board, the area of the board must be even. If the base of the pyramid has 2n-1 cells, n ?:.1 , then the area of the pyramid is 1+3 + 5 + ... + (2n-1) =n2 which satisfies the divisibility condition if and only if n is even. However, the sum of the cells in white is equal to 1+2+3+4+ ... +n=n(n+1)/2 and the sum of the cells in gray is equals to 1+2+3+4+ ... +(n-1) = n(n-1)12. Since each domino covers one cell of each color, the number of white cells must be equal to the number of gray cells. But n(n+ 1)12 > n(n-1)12. Thus D cannot cover the pyramid. •

21 Chapter III: Tiling with Straight Triominoes

We now consider the straight triomino denoted by S. (See Figure 5.) We start with the following question.

For which m and n, the board Bmxn can be covered by copies of S?

Given S, we at least need m 2: 3 or n 2: 3. The smallest possible board Bmxn that can be tiled by Sis a B 3xl·

The divisibility condition in this case states that the number of copies of S necessary to cover Bmxn is mn/3 . In other words, if the board Bmxn can be tiled by S, then mn must be divisible by 3 which in turn implies that 3 I m or 3 I n. It turns out that this is a necessary and sufficient condition.

Theorem 3. The board Bmxn can be tiled by S if and only if 3 I mn.

Proof: Without loss of generality, if m is divisible by 3 and n is any positive integer, then we can cover Bmxn simply by using vertical copies of S. Indeed, no matter the size of n, we can cover any column of the board with rn/3 copies of S. The divisibility condition takes care of the other direction. •

n=l , 2, 3, ...

m multiple of 3

Figure 25. Tiling with copies of S.

Similarly to the dominoes case, we look at the counting problem for B Jxn·

Theorem 4. For n 2: 3, we have the following recursive formula for the number of ways to tile B Jxn with copies of S:

T(S, B3xn)= T(S, B 3x(n - l)) + T(S, B 3x(n-3J).

22 Proof. Consider the B3xn board and label the upper left cell by 1 and labeled the next two cells down by 2 and 3. If we cover cell 11 with a vertical copy of S, then we are left with a B3x(n- I) board which can be tiled in T(S, B 3x(n- I)) ways. See Figure 26. If we cover cell 1 with a horizontal copy of S, then it forces cell 2 and cell 3 to be horizontally tiled, and we are left with a B 3x(n-3) which can be tiled in T(S, B 3x(n-3J) ways. •

I I

2 [8 3x(n- ) 2 B Jx(1 -3)

3 3

Figure 26. Tiling B 3xn with S when n 2: 3.

Figure 27 shows all different ways to cover B Jx J, B3x2, and B 3x3 with copies of S. Then T(S, B3x1)= T(S, B Jx2) =1 and T(S, B 3x3 ) = 2. The next few values of T(S, B3xn) can be obtained using Theorem 4. T(S, B3x4)= 1+ 2 = 3 T(S, BJxs)= 1+ 3 = 4 T(S, B 3x6)= 2+ 4 = 6 T(S, B 3x1)= 3+ 6 = 9 ...

B 3x2

Figure 27. All ways to cover B3xn when n=l, 2, and 3.

We can use these values to find general lower bounds for boards of size B 111x 11 , whenever, m and n are divisible by 3.

9 Proposition 7. T(D, Bmxn) 2: 2mnl whenever m and n are divisible by 3.

Proof: Consider the board Bmxn with m and n divisible by 3. We divide Bmxn into copies ofB3x3· (See Figure 28 .) There are T(D, B 3x3)=2 different ways to tile each copy ofB3x3·

23 1111119 Since there are (n/3)(m/3) = mn/9 copies of B 3x3, then there are at least T(D, Bmxn) = 2 ways of tiling Bmxn . •

~ Jx

Figure 28. Bmxn divided into copies of B3x3· There are 2 ways to tile each copy of B3x3 with the S-triomino.

We can improve this bound by simply dividing into copies of B3x6. We can do this whenever 3 Im and 6 In. (See Figure 29.)

111 Proposition 8. T (D, B111xn) 2: 6 n1 ts.

Proof: Consider the board B 111xn with m and n are divisible by 3. Partition B111xn into copies of B3x6· (See Figure 29.) There are T(D, B3x6)=6 different ways to tile each copy of B 3x6· Since there are (m/3)(n/6) = mn/18 copies of B3x6, then there are at least T(D, Bmxn) = 11111118 6 ways of tiling Bmxn. •

B x6

B x•

Figure 29. Bmxn divided into copies of B3x6· There are 6 ways to tile each copy of B3x6 with the S-triomino.

Note that Proposition 8 improves Proposition 7 when m and n are multiples of 3 and at least one is even.

11111 19 = 1111111s < 11111 11 s < T(D B ) 2 4 6 - ' mxn ·

24 Chapter IV: Tiling with L-shape Triominoes

We turn our attention to the smallest polyomino that is not a ; the L-shaped triomino denoted by L. (See Figure 5.) For which m and n the board Bmxn can be covered by copies of L?

By the divisibility condition, to cover Bmxn with copies of L, it is necessary that 3 divides mn. The first interesting fact about this triomino is that the divisibility condition is not always sufficient. For example the L-triomino does not tile the board B3,d, even though the total number of cells, namely 9, is divisible by 3.

Lemma 1. The board B 3x3 cannot be tiled by L.

Proof. Consider the board BJxJ· here are three ways to cover the upper-left corner, cell 1, with the L-triomino. (See Figure 30.) Notice that in each case the cell 2 can be tiled in a unique way but the remaining three cells cannot be covered by a copy of L.•

Case 1 Case 2 Case 3

Figure 30. BJxJ cannot be tiled by L.

Lemma 2. The smallest rectangular board that L can cover is the board B 3x2. (See Figure 31.) This can be done in 2 ways, that is, T(L, B3x2)=2.

Figure 31. Tiling B3x2 with copies of L.

This lemma can be used to tile arbitrarily large boards.

25 Lemma 3. If 3 J m and 2 I n, then the board B 111x 11 can be tiled by copies of L. Moreover 111 6 T(L, Bmxn) 2: 2 n .

Proof: Since m is divisible by 3 and n is divisible by 2, then B111x11 can be divided into (m/3)(n/2) = mn/6 copies of B3x2 (See Figure 32.) Since B3x2 can be tiled by copies of Lin 111 6 exactly two ways, then the board Bmxn can be tiled with copies of L in at least 2 n1 ways.•

BJx2

Figure 32. Tiling B 111x 11 with copies of L when 3 I m and 2 I n.

Let us consider all possible remaining boards. Namely, boards B111x 11 where 3 j m, and n is odd.

We have previously mentioned that L does not tile a board of size B 3x3. Consider the next case, that is, m = 3 and n = 5. Is it possible to cover a board of size B 3x5 with copies of L? This certainly satisfies the divisibility condition. We show that this again is not sufficient.

Lemma 4. The board BJx5 cannot be tiled with copies of L.

Proof. Consider the board B3x5 . There are three possible ways to cover the upper-left corner, cell 1. (See Figure 33.) Notice that Cases 1 and 3 have a unique way to cover cell 2 and the uncovered region is a B3x3 board, which by Lemma 1, cannot be tiled by L. In Case 2, cell 2 cannot be covered by a copy of L. •

26 Case 1 Case 2 Case 3

B 1x1

Figure 33. B3x5 cannot be covered with copies of L.

The same argument applies to any board B 1xn with n odd.

Lemma 5. Any board of the form B3xn , where n is odd, cannot be tiled with copies of L.

Proof: We prove this result by induction. By Lemma 1, we know that B3.d cannot be tiled with copies of L. Assume that B 3x(2k+JJ cannot be tiled for k?.l. We prove that B 3x(2(k+JJ+JJ = B 1x(2k+3J cannot be tiled with copies of L. There are 3 ways to cover the upper-left cell of the board B3x(2k+J), labeled cell 1. (See Figure 34.) In Cases 1 and 2, cell 2 can be covered in a unique way. In both cases, the region left to be covered corresponds to a copy of B 3x(2k+JJ, which by hypothesis cannot be tiled by L. Finally, in Case 3, cell 2 cannot be covered. •

3 3

3 B x(2k+ J B 1. (2k+I B 1. (2k+I

Case 1 Case2 Case 3

Figure 34. B Jx(2k+3J cannot be covered by copies of L.

Next, let us consider the boards B6x5 and B9x5, can these be tiled with copies of L?

Lemma 6. The boards B6x5 and B9x5 can be tiled by copies of L.

In fact the board B6x5 can be decomposed into copies of B3x2, as shown in Figure 35. By Lemma 2, the boards B3x2 can be tiled with copies of L in 2 different ways, and thus 5 T(L, B 6x5) ?. 2 .

27 5

6

Figure 35. B 5x6 decomposed into copies of B 3x2

Can the board B9xs be tiled using copies of B3x2? The answer is no, because each B3x2 requires 6 cells and although mn is divisible by 3, it is not divisible by 6. Yet, B9xs can be tiled with copies of L. (See Figure 36.)

Figure 36. Three different tilings of the board B5x9 by copies L.

We can generalize this result for the boards B5x3n with n ?:.2.

Lemma 7. Any board of the form B5x311 can be tiled with copies of L whenever n?:.2.

28 Proof. We consider two cases: n even and n odd.

Case 1. If n is even, then 3n is divisible by 6. This implies that BsxJn can be decomposed into 3n/6 = n/2 copies of Bsx6· Thus the board Bsx3n can be tiled by copies of L by 5 5 12 Lemma 6. (See Figure 37.) Moreover, since T(L, Bsx6)2: 2 , then T(L, Bsx3n) 2: 2 n . n=2, Bsx6 n=4, Bsx/2

5 5

6 12 n=6, Bsxl 8v

5

18 n=even, Bsx3n

5

3n

Figure 37. Bsx3n divided into n/2 copies of Bsx6· Each copy of Bsx6 is then tiled by copies ofL.

Case 2. If n 2: 3 and odd, then 3n = 6[(n-3)/2}+ 9(1). This implies that the board Bsx3n can be divided into (n - 3)12 copies of Bsx6 and 1 copy of B5x9. (See Figure 38.) Because Bsx6 and Bsx9 can be tiled with copies of L (Lemma 6) we can conclude that BsxJn can also be tiled with copies of L. •

29 Bsx21; n=9; 3n=27= 6(3) + 9(1)

Bsx6 Bsx6 5

6 + 6 + 6 + 9 n= odd and 3n=6a + 9b, then Bsx3n can be decomposed into copies of Bsx6 and Bsx9·

5 5 9

6 + 6 + + 9

Figure 38. Bsx3n divided into (n - 3)12 copies of Bsx6 and I copy of Bsx9·

Although the proof uses one copy of B5x9 and many copies of B5x6, there are other tilings. In fact, for any odd b such that 9b :S 3n, there is a positive integer a such that 3n= 6a + 9b which corresponds to decomposing Bsx3n into a copies of Bsx6 and b copies of Bsx9· This in turn gives a tiling of Bsx3n with copies of L.

This last result can be generalized to larger boards.

Lemma 8. Any board of the form Br2m+I)xJiv with m ?:.2 and n~2, can be tiled with copies of L.

Proof. Since 2m+ I =2(m-2)+5 , then Br2 111 +l)x3n = Bsx3n + B2rm-2)x3n, see Figure 39. Bsx3n is tiled by copies of L by Lemma 7, and B 2(m-2)x3n can be tiled with copies of L by Lemma 3 because 2(m-2) is even. Then the board Br2m+I)xJn can always be tiled with copies of L. •

30 m-5= even

m

5

n

Figure 39. B r2 111 +I)x(3n) = Bsx(3 n) + B r211 -4Jx(3n)·

Take B11x24 for example, see Figure 40. Notice that B11x24 can be decomposed into B6x24 and Bsx24 . By Lemma 7, Bsx24 can be tiled with copies of L. Similarly B6x24 can be tiled with copies of L by Lemma 3 since 3 j 6 and 2 j 24. Therefore, B 11x24 can be tiled with copies of L...... I I I I I I I I I I I I ......

Figure 40. B llx24 decomposed into Bsx24 and B6x24·

The following theorem completely settles the existence problem for rectangular boards and the L-shaped triomino.

31 Theorem 5. Let m, n be positive integers. If 3 j mn, then the board B111x11 can always be tiled by copies of L except if a) m=l or b) n=l or c) m=3 and n odd or d) n=3 and m odd.

Proof. Because L is 2 units wide and 2 units long, L cannot tile the board B111x11 if m = 1 or n=l; and by Lemma 5, L cannot tile a board satisfying (c) or (d) above. Now assume that 31 mn and none of (a) - (d) above holds. We can assume without loss of generality that

3 m. If n is even, then B 111x11 can be tiled with copies of L by Lemma 3. If n?:.5 is odd and m?:.6, then Bmxn can be tiled by copies of L by Lemma 8. Finally, if n=3 and m is even, then B111x11 can be tiled by copies of L by Lemma 3. •

Take B 7x I8 for example (See Figure 41.) Since m=7?:.3, n=l8?:.3 and 3 j mn, then by Theorem 5 the board B 7x 18can be tiled with copies of L. Indeed B 7x I8= B5x18+ B2x18 (See Figure 41a.) following the proof of Lemma 8. B5x 18 can be decomposed into 3 copies of B5xl8 (following the proof of Lemma 7) and B2xI8 into 6 copies of B3x2 (following the proof of Lemma 3). Finally, each B5x6 and B2x3 can be tiled by Las shown in Figure 35 and 31. Another tiling can be obtained by writing B 7x l8 = B4x l8 + B3xI8 and using the proof of Lemma 3. (See Figure 41 b.)

a)

b)

Figure 41. B 7x I8 tiled with copies of L using copies of B5x6 and B3x2·

32 References

[ 1] John, P. E. and Sachs, H. "On a Strange Observation in the Theory of the Dimer Problem." Disc. Math 216, 211-219, 2000.

[2] Kasteleyn, P. W. , The statistics of dimers on a lattice: I. The number of dimer arrangements on a quadratic lattice, Physica, vol. 27, issue 12, (1961) pp. 1209-1225

[3] Solomon W. Golomb "Polyominoes" 2°ct Edition. Princeton University Press.

[4] Temperley, H. N. V. and Fisher, M. Dimer problem in statistical mechanics-an exact result, Philosophical Magazine, vol. 6, issue 68, (1961) pp. 1061-1063.

[5] Tierney, John. "For Decades, Puzzling People With Mathematics" The New York Times: Science. 19 January 2010.

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