Is parquetability decidable?

Let $$T\neq \emptyset$$ be a finite subset of $$\mathbb{Z}\times\mathbb{Z}$$. We say that $$\mathbb{Z}^2 = \mathbb{Z}\times\mathbb{Z}$$ is parquettable by $$T$$ if there is a partition $$\frak P$$ of $$\mathbb{Z}^2$$ such that for every $$P\in {\frak P}$$ there is a bijective linear map $$\ell_P: \mathbb{Z}^2 \to \mathbb{Z}^2$$ and an element $$c_P\in \mathbb{Z}^2$$ such that for the affine map $$f_P:\mathbb{Z}^2\to\mathbb{Z}^2$$ defined by $$x\mapsto \ell_P(x) + c_P$$ we have $$f_P(T) = P.$$

Given $$T\subseteq \mathbb{Z}^2$$, is it decidable whether $$\mathbb{Z}^2$$ is parquettable by $$T$$?

• Just to restate: $T$ parquets $\mathbf{Z}^2$ if there's a partition of $\mathbf{Z}^2$ by affine translates of $T$.
– YCor
Dec 1, 2022 at 18:35
• The problem is at worst co-c.e., since having a tiling is equivalent to tiling arbitrarily large finite regions of the plane, and a failure of this can be discovered by a finite stage of searching. So we can answer No correctly in finite time. Dec 1, 2022 at 19:09
• Right @YCor - that's the elegant & concise formulation. Dec 2, 2022 at 9:09