_{1}

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Let
and let be the set of four ribbon L-shaped n-ominoes. We study tiling problems for regions in a square lattice by . Our main result shows a remarkable property of this set of tiles: any tiling of the first quadrant by , n even, reduces to a tiling by and rectangles, each rectangle being covered by two ribbon L-shaped n-ominoes. An application of our result is the characterization of all rectangles that can be tiled by , n even: a rectangle can be tiled by , n even, if and only if both of its sides are even and at least one side is divisible by n. Another application is the existence of the local move property for an infinite family of sets of tiles: , n even, has the local move property for the class of rectangular regions with respect to the local moves that interchange a tiling of an square by n/2 vertical rectangles, with a tiling by n/2 horizontal rectangles, each vertical/horizontal rectangle being covered by two ribbon L-shaped n-ominoes. We show that none of these results are valid for any odd n. The rectangular pattern of a tiling of the first quadrant persists if we add an extra tile to
, n even. A rectangle can be tiled by the larger set of tiles if and only if it has both sides even. We also show that our main result implies that a skewed L-shaped n-omino, n even, is not a replicating tile of order k^{2} for any odd k.

In this article, we study tiling problems for regions in a square lattice by certain symmetries of an L-shaped polyomino. Polyomines were introduced by Golomb [

The L-shaped polyomino we study is placed in a square lattice and is made out of

parallel to the first bisector

The inspiration for this paper is the recent publication [^{2} for any odd k. The question is equivalent to that of tiling a k-inflated copy of the straight L-tetromino using only four out of eight possible orientations of an L-tetromino, namely those orientations that are ribbon. The question is solved in [

The discussion above shows that the limitation of the orientations of the tiles used in a tiling problem can be of interest. The extension of the results in [^{2} for any odd k. This question is equivalent to that of tiling a k-inflated copy of the straight L n-omino by

set of tiles

In order to avoid repetition, we assume for the rest of the paper that, unless otherwise specified (and this will be the case only in the introduction), n is even and

Definition 1. A tiling of

Our main result is the following.

Theorem 1. Every tiling of

Theorem 1 is proved in Section 2.

It follows from Theorem 1 that:

Corollary 1. Every tiling of

Theorem 1 is optimal, as the following lemma shows.

Lemma 1. The addition of any even ´ odd or odd ´ odd rectangle to the set of tiles

Proof. As the concatenation of two odd ´ odd rectangles is an even ´ odd rectangle, we can consider only the last type. Also, a concatenation of an odd number of copies of an even ´ odd rectangles can be used to construct an even ´ odd rectangle of arbitrary large length and height. Assuming the existence of such a rectangle in the tiling, a tiling of

Some other consequences of the main result are listed below. The proofs of Corollaries 2, 3, 4 are similar to those of ( [

Corollary 2. Every tiling of a rectangle by

Definition 2. A k-copy of a polyomino is a replica of it in which all

Corollary 3. Let ^{2} for any odd k.

Corollary 4. A half-infinite strip of odd width cannot be tiled by

The following problem is open for

Problem 1. Show that a double infinite strip of odd width cannot be tiled by

It was proved by de Brujin [

Lemma 2. A rectangle with integer sides can be tiled by ribbon polyominoes made of k cells if and only if k divides one of the sides of the rectangle.

Proof. Assume that the rectangle is

and the double sum:

Each term in the last multiple sum corresponds to a cell in the rectangle. These terms can be grouped together in blocks of k terms each, combining terms corresponding to cells belonging the same ribbon polyomino with k cells. In such a group of terms, as we move from one end of the polyomino to the other, either the index ℓ_{1} decreases by 1, or the index ℓ_{2} decreases by 1 for each additional cell. It follows that the contribution of such a group to the double sum is zero. We infer that the double sum over all cells vanishes and therefore one of the factors

On the other hand,

which forces A to be divisible by k if

In conjunction with Theorem 1 this gives:

Theorem 2. Any tiling of a rectangle by

Not much is known about tiling integer sides rectangles with an odd side if we allow in the set of tiles all 8 orientations of an L-shaped n-omino, n even. It is known, and can be easily proved via a coloring argument, that if

The following result was found by Herman Chau, undergraduate student at Stanford University, during the Summer programme Research Experiences for Undergraduates, 2013, at Pennsylvania State University, Univer- sity Park.

Proposition 3. A rectangle of odd integer sides cannot be tiled by

Proof. This is obvious if n is even. If n is odd, it follows from Lemma 2 that the rectangle has a side divisible by n. Then we employ a ribbon tiling invariant introduced by Pak [

Pak showed that the function

for any tile in

It would be interesting to find an elementary proof of Proposition 3 that is independent of Pak result. The result in Proposition 3 is not valid for the set of tiles

No definitive results are known about tiling odd integer sides rectangles if the set of tiles consists of all 8 orientations of an L-shaped n-omino, n odd, despite serious computational effort invested in this question by various authors. We refer to the recent paper of Reid [

Assume that n odd.

The following notions were introduced by Thurston [

Definition 3. Given a set of tiles

It is shown in [

In conjunction with Theorem 1, the remarks above give:

Theorem 4. The set of tiles

Theorem 5. The set of tiles

The set of local moves for

_{n} and

The following two propositions are proved in Section 4.

Proposition 6. For general regions, and in particular for row-convex and column-convex regions, the set of tiles

Proposition 7. For general regions, and in particular for row-convex and column-convex regions, the set of tiles

One may wonder if a tiling by

pattern if it reduces to a tiling by

Theorem 8. 1. If m odd, n even, p odd, any tiling of

In the following cases tilings of

2. m even, n even, p even;

3. m even, n odd, p even;

4. m odd, n even, p even;

5. m even, n even, p odd;

6. m odd, n odd, p odd;

7. m odd,

8. m even,

9.

10.

Case 1) in Theorem 8 is an immediate corollary of Theorem 1 due to the fact that the tiles in

Let us summarize the main results of the paper and offer some perspective. We introduce a set of tiles that has some limitation in the orientations of the tiles. The set appears naturally from problems in recreational mathematics such as replicating properties of skewed polyominoes. For this set of tiles we are able to solve the problem of tiling of an arbitrary rectangle and are able to prove local move property.

We believe that the paper has certain heuristic value. Emerging from our work is a new method for discovering sets of tiles with interesting properties. We start by dissecting a rectangle, of base greater or equal then 2, in two irregular pieces. Then we symmetrize the pieces about the first bisector

Theorem 8 is included here in order to support the direction of research pointed above. The dissections appearing in Theorem 8 are those of rectangles of base equal to 2. When the height of the dissected rectangle is odd, the problem is completely solved by showing that always there exist tilings by the tile set that do not follow the rectangular pattern. The situation is more complex when the height of the dissected rectangle is even. If the tiles appearing from the dissection are congruent, the problem is completely solved. We identify both tiling sets that follow the rectangular pattern and tiling sets that do not follow the rectangular pattern. They appear in infinite families. If the tiles appearing from the dissection are not congruent, several cases are solved, identifying both types of tiling sets, and several cases are left open.

This suggests the following conjecture:

Conjecture 1. Fix a quadrant Q. If the height of the dissected rectangle is a multiple of the base, then both tiling sets that follow the rectangular pattern and tiling sets that do not follow the rectangular pattern are possible with respect to Q, for infinite families of rectangles and dissections.

More evidence supporting the conjecture is shown in the recent paper [

In this section we prove Theorem 1. For simplicity we refer to the

Definition 4. A

The

Definition 5. Assume that

1. The lower left corner has even coordinates;

2. The 2-squares on the left and below the upper level of the gap follow the rectangular pattern;

3. The 2-squares below the gap follow the rectangular pattern;

4. No part of the right side of an even length gap is covered by tiles;

5. The lower length 1 part of the right side of an odd length gap is not covered by tiles.

If the length of a gap is even, the gap is called even, otherwise the gap is called odd.

Pictures of gaps are shown in

For the following three lemmas we assume that a tiling of

Lemma 3. Assume that there exists an odd gap of length

Proof. Let d be the distance between the right side of the gap and the y-axis. We proceed by induction on d. For the induction step, we show that a

We consider first the case when the left side of the gap is based on the y-axis. Here we will finish with a contradiction. This includes the base case

diagonally adjacent to 1. We only need a right edge of height 1 for the odd gap. If cell 1 is tiled by a

Consider now the general case. We look at the tiling of cell 1, the lower leftmost in the gap. We cannot use a

Lemma 4. Assume that the leftmost

Proof. If

Lemma 5. Assume that a

Proof.

Lemma 6. A tiling of

Proof. Assume that a

The statement about the

Proof of Theorem 1. We show that every 2-square follows the rectangular pattern. We do this by induction on a diagonal staircase at 2k, shown in

For the induction step we prove that the 2-squares

Case 1. Let a

Case 2. Assume that a

Subcase 1. Assume that a

Subcase 2. Assume that a

In this section we show tilings of

Case 2. We place

Case 3. We place

Case 4. We place

Case 5. This case is similar to Case 3.

Case 6.

as in

Cases 7 and 8. We consider two cases. In

Cases 9 and 10.

In this section, we prove Proposition 6 and Proposition 7.

Proof of Proposition 6. For each fixed n, there exists an infinite family of row-convex (and column-convex) regions that admit only two tilings by

Proof of Proposition 7. For each fixed n, there exists an infinite family of row-convex (and column-convex)

regions, each instance of the region having exactly

family of regions is obtained by introducing more copies of the gray subregion in the middle. After factoring the

The number of tilings by

cover a cell in the top row of the region. Indeed, if this is the case, on the right side of the

If an horizontal

V. Nitica was partially supported by Simons Foundation Grant 208729. The author would like to thank several anonymous reviewers that read versions of the paper for their patience, for their generosity in sharing new ideas, and for many helpful suggestions that contributed to the improvement of this paper. The author would also like

to thank the following undergraduate students that worked with him at various tiling projects during the Summer programmes Research Experiences for Undergraduates, 2012, 2013, 2014, at Pennsylvania State University, University Park: M. Chao, D. Levenstein, R. Sharp, H. Chau, A. Calderon, S. Fairchild, S. Simon. Frequent discussions with them help the author better formalize the results in this paper.