Let $x \in \left(0..n-1\right)^m$.
I want to construct a bijection $g$ : $\left(0..n-1\right)^m \to \left(0..n-1\right)^m$ such that if we know $m'$ components of $g(x)$ and $m-m'$ components of $x$, we can deduce both $x$ and $g(x)$.
The identity function doesn't work, if we take $m=2$ and know the first component of both $x$ and $g(x)$, we have no help figuring the second component of $x$ or $g(x)$.
I'm ashamed to say I have a hard time even figuring out a suitable function for $m=2$ and a given $n=8$, the modulo makes things hard.
Can you help me figure out a way to construct such a function, if not for an arbitrary $m$, at least for $m=2$?