This article shines new light on the classical problem of tiling rectangles with squares efficiently with a novel method. With a twist on the traditional approach of resistor networks, we provide new and improved results on the matter using the theory of Diophantine Approximation, hence overcoming long-established difficulties, such as generalizations to higher-dimensional analogues. The universality of the method is demonstrated through its applications to different tiling problems. These include tiling rectangles with other rectangles, with their respective higher-dimensional counterparts, as well as tiling equilateral triangles, parallelograms, and trapezoids with equilateral triangles.
Let a polygon be composed of equal rectangles. We find all quadratic irrationals r for which the polygon can be tiled by similar rectangles with given side ratio r.
Let $m,k$ be fixed positive integers. Determining the generating function for the number of tilings of an $m\times n$ rectangle by $k\times 1$ rectangles is a long-standing open problem to which the answer is only known in certain special cases. We give an explicit formula for this generating function in the case where $m<2k$. This result is used to obtain the generating function for the number of tilings of an $m\times n \times k$ box with $k\times k\times 1$ bricks.
In this paper, we study 2-port networks and introduce new concepts of voltage drop and $Π$-equivalence. The main result is that each planar network is $Π$-equivalent to a network with no more than 5 edges. This implies that if an octagon in the shape of the letter $Π$ can be tiled by squares then it can be tiled by no more than 5 rectangles with rational aspect ratios. Kenyon's theorem from 1998 proves this only for 6 rectangles.
Straight polyomino tilings of rectangles and special rim-hook tableaux
University of British Columbia（不列颠哥伦比亚大学）；Indian Statistical Institute（印度统计研究所）；Indraprastha Institute of Information Technology Delhi (IIIT-Delhi)（德里因德拉普拉斯塔信息技术学院）
Signed tilings by ribbon L n-ominoes, n even, via Groebner bases
令$\mathcal{T}_n$为某个$n\ge 4$偶数的带状$L$形状的$n$ -ominoes的集合，令$\mathcal{T}_n^+$为带有额外$2\times 2$正方形的$\mathcal{T}_n$.本文利用$\mathcal{T}_n$和$\mathcal{T}_n^+$研究了矩形的符号镶嵌问题。我们证明了一个矩形有符号平铺$\mathcal{T}_n$当且仅当矩形的两边都是偶数，其中之一是可分的$n$，或如果一边是奇数，另一边是可分的$n\left (\frac{n}{2}-2\right ).$我们还证明了一个矩形有符号平铺$\mathcal{T}_n^+, n\ge 6$偶数，当且仅当矩形的两边都是偶数，或如果一侧是奇数和另一侧是可分的$n\left (\frac{n}{2}-2\right ).$我们的证明是基于展览的显式Gröbner基地的理想所产生的多项式相关联的平铺集。特别是，我们表明，\emph{ V.~Nitica, Every tiling of the first quadrant by ribbon $L$ $n$-ominoes follows the rectangular pattern. Open Journal of Discrete Mathematics, {\em 5}, (2015) 11--25,}中的一些定期平铺结果不…
Tilings are around us everywhere, and our curiosity draws us to study their properties. A tiling is a way of arranging pieces on a board, such that there is no space left uncovered, nor any space covered by more than one tile. In particular, we study fault-free tilings of boards with dominoes. To be fault-free every line that intersects the tiling must also intersect the interior of at least one of the tiles. Fault-free rectangular boards have been well studied, however we look at boards that are cylinders, tori, and Möbius strips. Using dominoes we study the various shapes of boards and see h…
We consider the tiling of an $n$-board (a board of size $n\times1$) with squares of unit width and $(1,1)$-fence tiles. A $(1,1)$-fence tile is composed of two unit-width square subtiles separated by a gap of unit width. We show that the number of ways to tile an $n$-board using unit-width squares and $(1,1)$-fence tiles is equal to a Fibonacci number squared when $n$ is even and a golden rectangle number (the product of two consecutive Fibonacci numbers) when $n$ is odd. We also show that the number of tilings of boards using $n$ such square and fence tiles is a Jacobsthal number. Using combi…
We apply Diophantine analysis to classify edge-to-edge tilings of the sphere by congruent almost equilateral quadrilaterals (i.e., edge combination a3b). Parallel to a complete classification by Cheung, Luk and Yan, the method implemented here is more systematic and applicable to other related tiling problems. We also provide detailed geometric data for the tilings.
我们考虑用T形四分块来平铺大小为4 m x 4 n的矩形。每个图块都被分配了一个权重，该权重取决于其在晶格上的方向和位置。对于一个特定的选择的权重，平铺的生成函数被示出为多变量Tutte多项式Z\_G（Q，v）（也称为物理学家的分拆函数的Q-状态Potts模型）的评价（m-1）x（n-1）矩形G，其中参数Q和边缘权重v可以采取任意值取决于瓦片的权重。
University of Nebraska – Lincoln（内布拉斯加大学林肯分校）；Faculty of Mathematics and Computer Science, University of Science, Ho Chi Minh City（胡志明市科技大学数学与计算机科学学院）；Vietnam National University, Ho Chi Minh City（胡志明市越南国立大学）
莫泽问，尺寸为1 x 1/2，1/2 x 1/3，1/3 x 1/4，...，总面积等于1的正方形的集合是否可以无重叠地装入单位正方形，边长为1/2，1/3，1/4，.可以无重叠地组合成一个面积为pi ^2/6-1的矩形。人们已经分别对将这些集合包装成边长为1+ π的正方形和面积为π ^2/6-1+ π的矩形进行了计算研究，并且可以考虑这样一个显然较弱的问题：对于每个正数π，这种包装是否可能。在本文中，我们建立了一个一般定理序列的几何包装，这意味着特别是“为每一个ε”的版本，这两个问题实际上是等价的原始平铺问题。
