Thanks for you kind words about my blog.
In the general size square Sudoku board, you have an $\kappa\times\kappa$ array of $\kappa\times\kappa$ sub-arrays. I think of the local block as a kind of neighborhood, and every location in it having its local coordinates $(\alpha,\beta)$, for $\alpha,\beta<\kappa$. But we have the $\kappa\times\kappa$ array of those local boards, and so the whole local block has its own coordinates $(\gamma,\delta)$. Thus, every location on the Sudoku board is specified by giving the coordinates $(\gamma,\delta)$ of the neighborhood block and the local coordinates $(\alpha,\beta)$ within that block. So four coordinates in all $$(\gamma,\delta,\alpha,\beta)$$ The boundaries between the local boards play no meaningful role.
On the $\mathbb{Z}$-Sudoku board, for example, which I considered on my blog, we have a $\mathbb{Z}\times\mathbb{Z}$ array of $\mathbb{Z}\times\mathbb{Z}$ local boards, and there are essentially no boundaries between these sub-boards.
Assymmetric Sudoku is somewhat more general, since it works with rectangular sub-boards. For any two cardinals $\kappa$ and $\lambda$, one may consider a $\lambda\times\kappa$ array of local boards with shape $\kappa\times\lambda$. Thus, again, every location is specified by the coordinates $(\gamma,\delta)$ of the board, and the local coordinates $(\alpha,\beta)$ within that board, where now $\gamma,\beta<\lambda$ and $\delta,\alpha<\kappa$.
A Sudoku solution is a function $f:(\lambda\times\kappa)\times(\kappa\times\lambda)\to L$, where $L$ is the set of labels (of size $\lambda\times\kappa$) that obeys the three requirements:
- Every row is a bijection with $L$. So $f(\gamma,\delta,\alpha,\beta)$ is a bijection, if you fix $\delta<\kappa$ and $\beta<\lambda$.
- Every column is a bijection with $L$. So $f(\gamma,\delta,\alpha,\beta)$ is a bijection, if you fix $\gamma<\lambda$ and $\alpha<\kappa$.
- Every local board is a bijection with $L$. So $f(\gamma,\delta,\alpha,\beta)$ is a bijection, if you fix $\gamma<\lambda$ and $\delta<\kappa$.
Note that if $G$ is a group of size $\lambda$ and $H$ is a group of size $\kappa$, then we can use these groups as location coordinates, and define $f(g,h,h',g')=(gg',hh')$, taking $L=G\oplus H$ as the set of labels. This is a solution, because if you fix $h$ and $g'$, it is a bijection; and similarly if you fix $g$ and $h'$, or if you fix $g$ and $h$. So every assymmetric Sudoku board has a solution arising in this way. This answers a question asked by Gerhard in a comment on an earlier post.