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We prove that if a measurable domain tiles ℝ or ℝ² by translations, and if it is "close enough" to a line segment or a square respectively, then it admits a lattice tiling. We also prove a similar result for spectral sets in dimension 1, and give an example showing that there is no analogue of the tiling result in dimensions 3 and higher.

Tiling Polygons with Parallelograms.

S. Kannan, D. Soroker (1992)

Discrete & computational geometry

Tiling with L's and squares.

Chinn, Phyllism, Grimaldi, Ralph, Heubach, Silvia (2007)

Journal of Integer Sequences [electronic only]

Tiling with smooth and rotund tiles

Victor Klee, Elisabetta Maluta, Clemente Zanco (1986)

Fundamenta Mathematicae

Tilings and isoperimetrical shapes I: Square lattice

Sylvain Gravier, Charles Payan (2001)

Acta Universitatis Palackianae Olomucensis. Facultas Rerum Naturalium. Mathematica

Tilings and isoperimetrical shapes. II. Hexagonal lattice

Sylvain Gravier (2001)

Acta Universitatis Palackianae Olomucensis. Facultas Rerum Naturalium. Mathematica

Tilings associated with beta-numeration and substitutions.

Berthé, Valérie, Siegel, Anne (2005)

Integers

Tilings associated with non-Pisot matrices

Maki Furukado, Shunji Ito, E. Arthur Robinson (2006)

Annales de l’institut Fourier

Suppose $A \in G l_{d} (ℤ)$ has a 2-dimensional expanding subspace $E^{u}$ , satisfies a regularity condition, called “good star”, and has $A^{*} \geq 0$ , where $A^{*}$ is an oriented compound of $A$ . A morphism $θ$ of the free group on ${1, 2, \dots, d}$ is called a non-abelianization of $A$ if it has structure matrix $A$ . We show that there is a tiling substitution $Θ$ whose “boundary substitution” $θ = \partial Θ$ is a non-abelianization of $A$ . Such a tiling substitution $Θ$ leads to a self-affine tiling of $E^{u} \sim ℝ^{2}$ with $A_{u} {: = A |}_{E_{u}} \in G L_{2} (ℝ)$ as its expansion. In the last section we find conditions on $A$ so...

Tilings in topological spaces.

Arenas, F.G. (1999)

International Journal of Mathematics and Mathematical Sciences

Tilings of convex polygons

Richard Kenyon (1997)

Annales de l'institut Fourier

Call a polygon rational if every pair of side lengths has rational ratio. We show that a convex polygon can be tiled with rational polygons if and only if it is itself rational. Furthermore we give a necessary condition for an arbitrary polygon to be tileable with rational polygons: we associate to any polygon $P$ a quadratic form $q (P)$ , which must be positive semidefinite if $P$ is tileable with rational polygons.The above results also hold replacing the rationality condition with the following: a polygon...

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