# Tiling with arbitrary tiles

• Published in 2015
In the collections
Let $T$ be a tile in $\mathbb{Z}^n$, meaning a finite subset of $\mathbb{Z}^n$. It may or may not tile $\mathbb{Z}^n$, in the sense of $\mathbb{Z}^n$ having a partition into copies of $T$. However, we prove that $T$ does tile $\mathbb{Z}^d$ for some $d$. This resolves a conjecture of Chalcraft.

## Other information

key
Tilingwitharbitrarytiles
type
article
2022-02-23
date_published
2015-09-14

### BibTeX entry

@article{Tilingwitharbitrarytiles,
key = {Tilingwitharbitrarytiles},
type = {article},
title = {Tiling with arbitrary tiles},
author = {Vytautas Gruslys and Imre Leader and Ta Sheng Tan},
abstract = {Let {\$}T{\$} be a tile in {\$}\mathbb{\{}Z{\}}^n{\$}, meaning a finite subset of
{\$}\mathbb{\{}Z{\}}^n{\$}. It may or may not tile {\$}\mathbb{\{}Z{\}}^n{\$}, in the sense of
{\$}\mathbb{\{}Z{\}}^n{\$} having a partition into copies of {\$}T{\$}. However, we prove that
{\$}T{\$} does tile {\$}\mathbb{\{}Z{\}}^d{\$} for some {\$}d{\$}. This resolves a conjecture of
Chalcraft.},
comment = {},
}