I know several specific variations of the domino tiling problem has been determined to be decidable or undecidable, such as the seed domino problem. I have a variation which I have not been able to find, but as I am not familiar with this subject, I thought someone here might know better.


Given a Wang tile $T=(T\_N,T\_E,T\_S,T\_W)$ with colors $\mathcal{C}$, each entry corresponds to the color in the north, east, south or west sides. Given two Wang tiles $T^{(1)},T^{(2)}$ that agree on opposite sides, their fusion would be gluing them on the side on which they agree. For example, if $T^{(1)}\_N=T^{(2)}\_S$, their fusion along South-North side is a new Wang tile


$$T^{(1)} \oplus\_{SN} T^{(2)}=\Big( T^{(2)}\_N, (T^{(2)}\_E,T^{(1)}\_E), T^{(1)}\_S, (T^{(2)}\_W,T^{(1)}\_W) \Big),$$


over a different set of colors obviously. We can similarly define $T^{(1)}\oplus\_{WE} T^{(2)}$ when they satisfy $T^{(1)}\_E=T^{(2)}\_W$. If we take a Wang tile set $\mathcal{T}$ with colors $\mathcal{C}$, we can obtain new Wang tiles sets


$$ \mathcal{T\_1} = \big\{ T \oplus\_{SN} T' : T,T' \in \mathcal{T}, T\_N=T'\_S \big\} $$
and


$$ \mathcal{T\_2} = \big\{ T \oplus\_{WE} T' : T,T' \in \mathcal{T}\_1, T\_E=T'\_W \big\}. $$


$\mathcal{T}\_2$ will again be a collection of Wang tiles over the new color set $\mathcal{C}\times \mathcal{C}$. I can consider two related questions:


1. Does $\mathcal{T}\_2$ tile the plane periodically?
2. Does $\mathcal{T}\_1$ tile the plane periodically?


Is it known whether question (1) being *un\decidable* imply question (2) is *un\decidable*? I am assuming that $\mathcal{T}\_1$ tiles the plane.


This seems like a natural question to me, but I really have a bad intuition on the subject, so it may be obviously resolvable. I would be glad to find appropriate references, or even solutions.



A variation of domino tiling problem with fusions
Each tiling with tiles from $\mathcal{T}$ gives rise to exactly two tilings with tiles from $\mathcal{T}\_1$ and exactly two tilings with tiles from $\mathcal{T}\_2$, given by taking the two choices of fusing neighbouring pairs (either $2n\longleftrightarrow 2n+1$ or $2n-1 \longleftrightarrow 2n$). This process also preserves periodicity.


Up to translation, this correspondence is also reversible and again preserves periodicity. So $\mathcal{T}$ admits a periodic tiling if and only if $\mathcal{T}\_1$ admits a periodic tiling if and only if $\mathcal{T}\_2$ admits a periodic tiling.


The same argument works for non-periodic tilings of course.


