# 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.

### BibTeX entry

@article{Tilingwitharbitrarytiles,
title = {Tiling with arbitrary tiles},
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.},
url = {http://arxiv.org/abs/1505.03697v2 http://arxiv.org/pdf/1505.03697v2},
year = 2015,
author = {Vytautas Gruslys and Imre Leader and Ta Sheng Tan},
comment = {},
urldate = {2022-02-23},
archivePrefix = {arXiv},
eprint = {1505.03697},
primaryClass = {math.CO},
collections = {easily-explained,fun-maths-facts,geometry}
}