On Tilings of Quadrants and Rectangles and Rectangular Pattern

The problem of tiling rectangles by polyominoes generated large interest. A related one is the problem of tiling parallelograms by twisted polyominoes. Both problems are related with tilings of (skewed) quadrants by polyominoes. Indeed, if all tilings of a (skewed) quadrant by a tile set can be reduced to a tiling by congruent rectangles (parallelograms), this provides information about tilings of rectangles (parallelograms). We consider a class of tile sets in a square lattice appearing from arbitrary dissections of rectangles in two L-shaped polyominoes and from symmetries of these tiles about the first bisector. Only translations of the tiles are allowed in a tiling. If the sides of the dissected rectangle are coprime, we show the existence of tilings of all (skewed) quadrants that do not follow the rectangular (parallelogram) pattern. If one of the sides of the dissected rectangle is 2 and the other is odd, we also show tilings of rectangles by the tile set that do not follow the rectangular pattern. If one of the sides of the dissected rectangle is 2 and the other side is even, we show a new infinite family of tile sets that follows the rectangular pattern when tiling one of the quadrants. For this type of dissection, we also show a new infinite family that does not follow the rectangular pattern when tiling rectangles. Finally, we investigate more general dissections of rectangles ,3 k n n k × ≤ < , with ( ) 1 gcd , n k n < < . Here we show infinite families of tile sets that follow the rectangular pattern for a quadrant and infinite families that do not follow the rectangular pattern for any quadrant. We also show, for infinite families of tile sets of this type, tilings of rectangles that do not follow the rectangular pattern.

tiles about the first bisector.Only translations of the tiles are allowed in a tiling.If the sides of the dissected rectangle are coprime, we show the existence of tilings of all (skewed) quadrants that do not follow the rectangular (parallelogram) pattern.If one of the sides of the dissected rectangle is 2 and the other is odd, we also show tilings of rectangles by the tile set that do not follow the rectangular pattern.If one of the sides of the dissected rectangle is 2 and the other side is even, we show a new infinite family of tile sets that follows the rectangular pattern when tiling one of the quadrants.For this type of dissection, we also show a new infinite family that does not follow the rectangular pattern when tiling rectangles.Finally, we investigate more general dissections of rectangles ,3 k n n k × ≤ < , with ( ) Here we show infinite families of tile sets that follow the rectangular pattern for a quadrant and infinite families that do not follow the rectangular pattern for any quadrant.We also show, for infinite families of tile sets of this type, tilings of rectangles that do not follow the rectangular pattern.

Introduction
In this article, we study tiling problems for regions in a square lattice by polyominoes.
Polyominoes were introduced by Golomb in [1] and the standard reference about this subject is the book Polyominoes [2].The polyominoes are made out of unit squares, or cells.In a a b × rectangle, a is the height and b is the base.
Understanding tilings of rectangles by particular polyominoes, even of simple shape, is a difficult combinatorial problem with a long history.See for example the paper of Golomb [3].Among the pioneering contributions, we mention those of Klarner [4].On page 113 of [4], Klarner emphasizes the difficulty of classifying rectangles tileable by L-shaped n-ominoes for which two copies can be assembled in a rectangle: It seems impossibly difficult to characterize the rectangles which can be packed with an n-omino of order 2. A theorem of this kind restricted to the L-shaped n-ominoes of order 2 would probably still be too difficult to formulate.A partial result in this direction ap- pears in Reid [5], where it is shown that any L-shaped polyomino of order 2 for which the basic rectangle has coprime sides has odd order.We recall that the order of a polyomino is the minimal number of tiles that can be assembled into a rectangle.
A similar problem can be investigated for parallelograms in a skewed lattice by using instead of polyominoes skewed tiles that have all sides parallel to the sides of the parallelogram.The problems are independent and probably of the same level of difficulty.A first observation is that the problem of tiling a parallelogram is equivalent to that of tiling a rectangle by polyominoes (the straightened tiles), allowing only a reduced set of orientations for those polyominoes.Some progress was done in the case of L-shaped n-ominoes of order two in several recent papers of the author and collaborators [6] [7] [8] [9] [10].The results are consequences of more general tiling results for quadrants, showing that for many tiling sets there exists at least a quadrant for which all tilings can be reduced to tilings by congruent rectangles built out of two tiles from the tiling set.If this is the case, we say that the tile set and the corresponding tilings follow the rectangular pattern.Some caution is needed, as some pairs of tiles of order 2 can be assembled in two different rectangles.To eliminate the ambiguity, the class of tile sets that follow the rectangular pattern is restricted to those that have a single basic rectangle.Another motivation for the study of tile sets with a reduced set of orientations comes from the study of skewed replicating tiles [11] [12].
The result in [5] shows that, if the full set of orientations of an order 2 L-shaped tile is allowed, there exist plenty of tile sets that do not follow the rectangular pattern.We show, in Theorem 1, a related result when only a reduced set of orientations is allowed.
The tile sets appear from arbitrary dissections of rectangles in two L-shaped polyominoes and from symmetries of these tiles about the first bisector.Only translations of the tiles are allowed in a tiling.In Theorem 2, we show the existence of tilings for certain half-infinite strips.Theorems 1 and 2 leave open the question of tiling rectangles without following the rectangular pattern by our tile sets.We answer this question if the dissected rectangle has base 2 and odd height, improving in Theorem 3 some results obtained in [7].

V. Nitica
If the dissected rectangle has base 2 and even height, we show in Theorem 4 new examples of tile sets that follow the rectangular pattern for tilings of a quadrant, complementing the results in [7].They are given by an infinite subfamily of tile sets In the examples from [7] and [10], one has ( ) ( ) The paper ends with several open questions and a conclusive summary.

Main Results
Our argument can be applied to a larger class of tile sets then those generated by L-shaped n-ominoes of order 2. We start with a rectangle in the square lattice and dissect it in two L-shaped polyominoes, not necessarily congruent, by a vertical cut.Due to symmetry, the case of a horizontal cut is also covered by our argument.For a given tile sets appearing from a dissection as in Figure 2(a) of type 1, and respectively as in Figure 2(b), of type 2. A tile set of type 1 is shown in Figure 3.To establish some terminology, we refer to tile sets of type 1 or 2 as 2-dissection tile sets, or simply by dissection tile sets.We call the dissected rectangle the basic rectangle of the dissected tile set.
Theorem 1 is a corollary of Theorem 2.
Theorem 1.If , m n are coprime, all dissection tile sets have tilings of all quadrants that do not follow the rectangular pattern.We can arrange for a single tile to be out of the rectangular pattern.
Theorem 2. If , m n are coprime, all dissection tile sets have tilings of half-infinite strips that do not follow the rectangular pattern.We can arrange for a single tile to be out of the rectangular pattern.
Proof.Due to symmetries, it is enough to show the proof for a tile set of type 1.The tile set is symmetric about the first bisector, so it is enough to show the proof only for a half-strip opening to the right and for one opening to the left.The tilings are shown in   + .Then region II is a rectangle of base qwm and height tzn, region IV is a half-infinite strip of width n, region V is a half-infinite strip of width syn, and region III is a half-infinite strip of width twm.
Problem 1. Decide if all dissection tile sets have a tiling of a torus (or/and of a cylinder) that does not follow the rectangular pattern and in which not all tiles are in an irregular position.
The problem can be solved for torus if we allow all tiles to be in an irregular position.Indeed, any notched rectangle in the square lattice has a doubly periodic tiling of the plane that does not follow the rectangular pattern.An instance is shown in Figure 5.
Other natural problem is: Problem 2. Determine which of the dissection tile sets tile rectangles without following the rectangular pattern.
The case when the base of the dissected rectangle has length 2 and ( ) gcd , 1 k n = is solved in Theorem 3.This complements Theorem 1 in [7].The general problem of tiling some rectangle by a tile set is undecidable [13].
Figure 6 shows a family of dissected tile sets studied in [7]  In Cases 2. through 6. there exist tilings of rectangles by , , m n p T that do not follow the rectangular pattern.
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; Proof.First part of Case 1 and Case 6 are proved in Theorem 8, [7].For Cases 2 -5 Theorem 8, [7] shows only tilings of the first quadrant that do not follow the rectangular pattern.For Cases 7 -8 Theorem 8, [7] shows tilings of the first quadrant if 1 n = (the pictures a), b) in Figure 18 in [7] should be interchanged and assume 1 m > ) and shows two cases of irregular tilings of rectangles, for 3, We prove the second part of Case 1 and Cases 2 -6 below (correcting some misprints from [7] in Case 6).During the proof we use that multiples of ( ) ( )          Theorem 4. The tile sets 1,1, p T , p even, follow the rectangular pattern for the first quadrant and there are tilings of the other three quadrants that do not follow the rectangular pattern.
Proof.The tile set consists of three tiles: a tromino and two other tiles which we call horizontal/vertical.
We show the proof for the first quadrant.A 2 2 × square with all vertices of even coordinates is called a 2-square.We show that every 2-square is covered by a rectangle with even vertices that is covered by two tiles from 1,1, p T .If this is the case, we say that the 2-square follows the rectangular pattern.We proceed by induction on a diagonal staircase shown in Figure 13.We assume that every 2-square below the staircase satisfies the hypothesis and show that every 2-square i X above the staircase also satisfies it.
It is easily checked that the corner of the first quadrant can be tiled only following the rectangular pattern.We show now the induction step.Choose the rightmost square i X , that does not follow the rectangular pattern.The lower left cell in that square can be covered only by the tromino.The cell in i X not covered by the tromino can be covered by either one of the tiles in an irregular way.If covered by the tromino or the horizontal tile, this leads to a cell that cannot be covered further.See Figure 14.If covered by a vertical tile, we look at the 2-square 1 i X + .If already covered regularly, this leads to a cell above i X that cannot be covered.Otherwise, the lower left cell in 1 i X + can be covered only by the tromino.A repeat of the argument leads either to a contradiction as before, or to the appearance of a staircase that propagated towards to y-axis, finally leading to a cell adjacent to the y-axis that cannot be covered.See Figure 15.
For the negative results, due to symmetry, it is enough to show examples of tilings only for the second and third quadrant.See Figure 16.In the second quadrant, regions I, II, III, IV, V, VII, VIII are half infinite strips of even width, region VI is a copy of the second quadrant and region IX is a ( ) rectangle.In the third quadrant, regions I, II, III, V, VI, VII are half infinite strips of even width, region IV is a copy of the third quadrant and region VIII is a ( ) ( ) If a 2 2 × square is added to 1,1, , p T p even, the new tiling set is called 1,1, p T + .As it was the case in [6] and [7], the new tiling set preserves the rectangular pattern.The proof of next theorem is similar to that of Theorem 4.
Figure 15.The end of the induction step.
Figure 16.Tilings of the second and third quadrants by 1,1, p T + , p even.
Theorem 5.The tile sets 1,1, p T + , p even, follows the rectangular pattern for the first quadrant and there are tilings of the other three quadrants that do not follow the rectangular pattern.The results for the other type can be obtained via a symmetry about the y-axis.
The positive results in Theorem 4 and Theorem 5 cannot be found using coloring invariants.We refer to [9] for a discussion of this topic and relevant examples.We leave the formal proof as an exercise for the reader.
Theorem 6.The tile sets 1,2 1,2 do not follow the rectangular pattern for rectangles.
In particular, a rectangle ( ) ( ) Proof.The case 1 n = is shown in [7].The pattern of the general construction follows easily from Figures 17-19, showing the cases 2, 3, 4 n = . All regions labeled by roman numerals follow the rectangular pattern.
The rectangles in Theorem 6 have the feature that admit a unique irregular tiling.
They also are of minimal height with the property that allow an irregular tiling.We T + for which the number of tiles in an irregular position is bounded with respect to n.
Next theorem shows that a 2 × horizontal inflation of    The tile sets ( T C n mn T C n mn + are studied in detail in [10].We show that for each of these tile sets, with the possible exception of ( ) , for which we could not decide, there exists at least a quadrant for which any tiling has to follow the rectangular pattern.In particular, any tiling of a rectangle by the tile sets ( ) k n = then the tile sets ( ) , , i T C n k do not follow the rectangular pattern, a particular case of Theorem 1 in this paper.The case when ( ) completely open, including the particular case 4, The next theorem gives some partial results for tile sets generated by this type of dissections in the simplest case when ( ) gcd , 2 k n = .While these results are not final, they illustrate some differences between the tile sets considered here and those corresponding to dissections discussed in Theorem 1 and in [10].Heuristically, the existence of a smaller foundational square, as compared to the size of the other tiles in the tile set, allows for more freedom in the tiling, therefore to the existence of more tilings that do not follow the rectangular pattern.Nevertheless, as we see from the next theorem, the rigid behaviour is still possible, even if we include the foundational square  1) The tile set ( )

, , T C n k
+ has tilings of rectangles that do not folow the rectangular pattern, therefore does not follow the rectangular pattern for any quadrant.

, , T C n k
+ has tilings of rectangles that do not folow the rectangular pattern, therefore does not follow the rectangular pattern for any quadrant.
3) The tile set Proof.Due to symmetries, it is enough to prove 1), 3) and 5).The foundational square is a 2 2 × square.

Open Problems
We   We recall [10] that the tile set ( )

Conclusion
The goal of the paper is to study tiling problems in a square lattice by specific tile sets.
The tile sets appear from dissections of rectangles in two L-shaped polyominoes and from symmetries of these tiles about the first bisector.Only translations of the tiles are allowed in a tiling.We investigate of the quadrants and of rectangles.Our results have applications to tilings of paralelograms in a skewed lattice and to the study of replicating figures in a skewed lattice.These problems were mostly overlooked in the literature, which concentrated on tiling rectangles using all symmetries of a single polyomino.A crucial observation made in [6] [7] and [10] is that many of tile sets of type studied here follow the rectangular pattern, that is, any tiling reduces to one by rectangles, each rectangle is tiled by two pieces from the tile set.As observed in [8] and [9] these results do not follow from coloring invariants.We show in the paper that if the sides of the dissected rectangle are coprime, then the tile set allows for tilings of all quadrants that do not follow the rectangular pattern.If the sides of the dissected k n × rectangle satisfy ( ) 1 gcd , k n n < < , then both tile sets that follow the rectangular pattern and tile sets that do not follow the rectangular pattern are possible.These results complement those in [10], where we study tile sets appearing from dissection of rectangles k n × with k multiple of 3 n ≥ .We also complement the results in [7], where we study tile sets appearing from dissection of rectangles 2 k × .If one of the sides of the dissected rectangle is 2 and the other side is even, we show in the paper a new infinite family of tile sets that follows the rectangular pattern when tiling one of the quadrants.For this type of dissection, we also show a new infinite family that does not follow the rectangular pattern when tiling rectangles.Our results answer some questions left open in [7] and [10].Several new open problems are listed in the Open Problems section.

Figure 1 .
Figure 1.The tile set 1,1,2T that follows the rectangular pattern for the first quadrant.

Figure 4 .Figure 2 .
Figure 4.As m,n are coprime, there exists positive integers , , , x y z w such that 1, 1 xm yn zn wm − = − = .For the half-strip opening to the right, region I is a rectangle of base qxm qyn q = + and height tzn twm t = + , region II is a rectangle of base qyn and height sxm, region IV is a half-infinite strip of width n, region V is a half-infinite strip of width twm, and region III is a half-infinite strip of width yns.For the half-strip

Figure 3 .
Figure 3.A complete tile set of type 1.

Figure 4 .
Figure 4. Tilings of half-infinite strips that do not follow the rectangular pattern.

1 )
If m odd, n even, p odd, any tiling of the first quadrant by , , m n p T follows the rectangular pattern.In this case there exist irregular tilings of tori.

Figure 5 .
Figure 5.A double periodic tiling of a plane by notched rectangles.

Figure 6 .
Figure 6.The set of tiles , , m n p T .

Figure 8 .
Figure 8. Case 7. The tiling is shown in Figure 11.Region I is a rectangle ( ) ( )

Case 8 .
Figure 11.Case 1, second part.The tiling is shown in Figure 12.Region I is a rectangle ( ) ( ) 2 m n p m n p + + × + + .Region II is a rectangle ( ) ( ) m n p m n p + + × + + .Region III

Figure 12 .
Figure 12.An irregular tiling of a torus, second part of Case 1.

Figure 1 LR and their reflections 2 R
Figure 22.Tiling a rectangle by square a d d × square.If an extra foundational square, denoted S, is added to the tile set, we denote the tile set of a region in the plane is said to follow the rectangular pattern if it reduces to a tiling by k n × and n k × rectangles, each tiled in turn by two pieces from the tile set.A tiling by of a region in the plane is said to follow the rectangular pattern if it reduces to a tiling by d d × squares and by k n × and n k × rectangles.

+
follows the rectangular pattern.It is also shown in[10] that if ( ) gcd , 1

++ 6 )
has tilings of first, second and fourth quadrants that do not follow the rectangular pattern.has tilings of first, second and fourth quadrants that do not follow the rectangular pattern.5)Assume in addition that 2 p = .Then any tiling of the third quadrant by Assume in addition that 2 p = .Then any tiling of the third quadrant by ( )4 , , T C n k+ follows the rectangular pattern.

Figure 25 .
Figure 25.Regions I through V can be tiles by 2 2 × squares.Region I is a rectangle

3 )
Tilings of first and fourth quadrant by of quadrants of half infinite strips of even width that can be covered by 2 2 × squares.A tiling for the second quadrant follows by symmetry.

Problem 3 .
Figure 27.The induction staircase in the third quadrant.

Figure 29 .
Figure 29.Tilings of rectangles by enhanced dissection tile sets.

Figure 30 .
Figure 30.Tilings of cylinders by enhanced dissection tile sets.
n n ≥ , follows the rectangular pattern with respect to all quadrants.Problem 9. Find a dissection tile set generated by the dissection of a k n × rectangle, ( ) 1 gcd , k n n < < , which follows the rectangular pattern with respect to all four quadrants.
. It is indexed by positive tile sets that allows for irregular tilings.Proof.The tilings are shown in Figure21, if n is even, and in Figure22, if n is odd.