# Tiling with arbitrary tiles

- Published in 2015
- Added on

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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.

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### 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} }