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### Tiling a simply connected figure with bars of length 2 or 3

Eric Rémila
1996 Discrete Mathematics
We first prove that if F has no peak (a peak is a cell of F which has three of its edges in the contour of F), then F can be tiled with rectangular bars formed from 2 or 3 cells.  ...  Let F be a simply connected figure formed from a finite set of cells of the planar square lattice.  ...  of tiling with 2-bars and 3-bars: the input is a simply connected fgure F and the algorithm tells whether F can be tiled with 2-bars and 3-bars or not.  ...

### Tiling with Small Tiles [article]

Anne Kenyon, Martin Tassy
2015 arXiv   pre-print
We also present a result to a more classic tiling question with dominoes and L-shape tiles.  ...  We look at sets of tiles that can tile any region of size greater than 1 on the square grid.  ...  Figure 3 . 3 Local move 1 Figure 4 . 4 Local move 2 Figure 5 . 5 Set S 2 : a domino, L-tile, 3-bar, T-tile, and plus-tile Lemma 3 . 2 . 32 The set S 2 given inFigure 5is a fountain set.Proof.  ...

### Page 5341 of Mathematical Reviews Vol. , Issue 97I [page]

1997 Mathematical Reviews
{For the entire collection see MR 97h:05003.} 971:05025 05B45 52C20 Rémila, Eric Tiling a simply connected figure with bars of length 2 or 3.  ...  We first prove that if F has no peak (a peak is a cell of F which has three of its edges in the contour of F), then F can be tiled with rectangular bars formed from 2 or 3 cells.  ...

### Tiling groups: New applications in the triangular lattice

E. Rémila
1998 Discrete & Computational Geometry
Afterward, we use a similar method to get a linear algorithm to tile polygons with m-leaning bars (parallelograms of length m formed from 2m cells of A) and equilateral triangles (whose sides have length  ...  m) and we produce a quadratic algorithm to tile polygons with m-leaning bars.  ...  Kenyon [KK] , which obtained an algorithm to tile a simply connected figure with m-bars (rectangles of length m and unit width). In Section 2 we recall the notions introduced by J. H. Conway, J. C.  ...

### A note on tiling with integer-sided rectangles [article]

Richard Kenyon
1994 arXiv   pre-print
We show how to determine if a given simple rectilinear polygon can be tiled with rectangles, each having an integer side.  ...  Suppose we want to tile a simply connected polyomino with rectangles, each of which has a side of length n.  ...  In the case of polyominoes, this lattice is just Z, and so the algorithm yields in this case a tiling with bars of length 1 × n and n × 1.  ...

### Page 6220 of Mathematical Reviews Vol. , Issue 99i [page]

1999 Mathematical Reviews
The main results of the paper are the following two algorithms: a linear al- gorithm of tiling a simply connected figure with “leaning barsof length m (i.e. parallelograms with large sides of length  ...  m and small sides of unit length) and equilateral triangles with sides of length m as well as a quadratic algorithm of tiling with leaning bars.  ...

### Tiling with bars under tomographic constraints [article]

Christoph Durr, Eric Goles, Ivan Rapaport, Eric Remila
2001 arXiv   pre-print
We wish to tile a rectangle or a torus with only vertical and horizontal bars of a given length, such that the number of bars in every column and row equals given numbers.  ...  We present results for particular instances and for a more general problem, while leaving open the initial problem.  ...  In the same way, each V j is tillable by using only vertical bars R 1×v with each column having m bars. Figure 3: A (2 · · · 2, 3 · · · 3)-tiling of T 15×10 by R 2×1 and R 1×3 .  ...

### A Note on Tiling with Integer-Sided Rectangles

Richard Kenyon
1996 Journal of combinatorial theory. Series A
We show how to determine if a given simple rectilinear polygon can be tiled with rectangles, each having an integer side.  ...  Suppose we want to tile a simply connected polyomino with rectangles, each of which has a side of length n.  ...  Figure 1 : 1 Inductive step in the proof that a tiling gives a product of lassos. Figure 2 : 2 Removing an interior maximum. Figure 3 : 3 Lowest and highest tilings of a polygon.  ...

### Phase Transitions in Random Dyadic Tilings and Rectangular Dissections

Sarah Cannon, Sarah Miracle, Dana Randall
2018 SIAM Journal on Discrete Mathematics
We consider a weighted version of these Markov chains where, given a parameter λ > 0, we would like to generate each rectangular dissection (or dyadic tiling) σ with probability proportional to λ |σ| ,  ...  A similar edge-flipping chain is also known to connect the state space when restricted to dyadic tilings, where each rectangle is required to have the form where s, t, u and v are nonnegative integers.  ...  The lengths of the vertical boundary segments of G are integer multiples of h, and G is simply connected as it is the union of a connected region with all of its holes.  ...

### Phase Transitions in Random Dyadic Tilings and Rectangular Dissections [chapter]

Sarah Cannon, Sarah Miracle, Dana Randall
2014 Proceedings of the Twenty-Sixth Annual ACM-SIAM Symposium on Discrete Algorithms
We consider a weighted version of these Markov chains where, given a parameter λ > 0, we would like to generate each rectangular dissection (or dyadic tiling) σ with probability proportional to λ |σ| ,  ...  A similar edge-flipping chain is also known to connect the state space when restricted to dyadic tilings, where each rectangle is required to have the form where s, t, u and v are nonnegative integers.  ...  The lengths of the vertical boundary segments of G are integer multiples of h, and G is simply connected as it is the union of a connected region with all of its holes.  ...

### Tiling Rectangles with Gaps by Ribbon Right Trominoes

Premalatha Junius, Viorel Nitica
2017 Open Journal of Discrete Mathematics
We show that the least number of cells (the gap number) one needs to take out from a rectangle with integer sides of length at least 2 in order to be tiled by ribbon right trominoes is less than or equal  ...  If the sides of the rectangle are of length at least 5, then the gap number is less than or equal to 3.  ...  For any simply connected region that can be tiled by ∑ or ∑ , the difference between the number of 1 T tiles and the number of 2 T tiles that appear in the tiling is an invariant.  ...

### Tiling simply connected regions with rectangles

Igor Pak, Jed Yang
2013 Journal of combinatorial theory. Series A
Although #P-completeness is known for tilings of general regions with right tromino and square tetromino [MR], nothing was known for tilings with rectangles, or for tilings of simply connected regions.  ...  Theorem 1.2 There exists a finite set R of at most 10 6 rectangular tiles, such that counting the number of tilings of simply connected regions with R is #P-complete.  ...  We make a stronger conjecture that for every tileset T of two bars [1 × k] and [ℓ × 1], where k, ℓ ≥ 2, (k, ℓ) = (2, 2), the counting of tilings by T of simply connected regions is #P-complete.  ...

### Fixed Parameter Undecidability for Wang Tilesets

Emmanuel Jeandel, Nicolas Rolin
2012 Electronic Proceedings in Theoretical Computer Science
Deciding if a given set of Wang tiles admits a tiling of the plane is decidable if the number of Wang tiles (or the number of colors) is bounded, for a trivial reason, as there are only finitely many such  ...  One of the main new tool is the concept of Wang bars, which are equivalently inflated Wang tiles or thin polyominoes.  ...  Acknowledgements The first author thanks Daniel Gonçalves and Pascal Vanier for some interesting discussions that lead to the proof of the case of 2 Wang Bars.  ...

### Tiling figures of the plane with two bars

Danièle Beauquier, Maurice Nivat, Eric Remila, Mike Robson
1995 Computational geometry
Given two "bars", a horizontal one, and a vertical one (both of length at least two), we are interested in the following decision problem: is a finite figure drawn on a plane grid tilable with these bars  ...  Given a general pair of bars, we give two results: (1) a necessary condition to have a unique tiling for finite figures without holes, (2) a linear algorithm (in the size of the figure) deciding whether  ...  There exists a polynomial algorithm which: (1) decides the existence of a tiling of a f nite figure F with \$2, or not. (2) provides a tiling of F with \$2,/f such a tiling exists.  ...

### Tiling with Bars and Satisfaction of Boolean Formulas

Eric Rémila
1996 European journal of combinatorics (Print)
In this lattice , the problem of tiling a simply connected figure with 2-bars can be solved in the linear time (see  ) , and the problem of tiling a figure with 3-bars seems to be very dif ficult ,  ...  The problem of tiling a figure with bars of length 2 or 3 can be reduced in linear time to the logic problem 2 -SAT . P ROPOSITION 3 . 1 . Let F be a finite figure .  ...
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