Consider the problem of finding the maximum number of knights that can be placed on a chessboard without two of them attacking each other. The answer is 32: it's not too difficult to find a perfect matching (the graph induced by knight moves is bipartite, and there's a perfect matching for a 4 × 4 board), which is obviously a minimum edge cover. It's also not difficult to prove that the answer is $\left\lceil \frac{mn}{2} \right\rceil$ for an $m \times n$ chessboard whenever $m,n \geq 3$: it suffices to show matchings for $3 \leq m,n \leq 6$ and do a bit of induction footwork.
On the other hand, if the chessboard were toroidal and $m, n$ were even, the proof wouldn't even require showing a matching for small boards: the map $(x, y) \rightarrow (x+1, y+2)$ has only even-length cycles so there must be a perfect matching.
Is there any equivalent for rectangular chessboards, i.e. is there any simpler way to show that for sufficiently large $m, n$ there is always a perfect matching of the chessboard? For large boards, the rectangular board and the toroidal board are almost equivalent in the sense that the fraction of missing edges goes to zero, but I'm not aware of any theoretical results that would guarantee a perfect matching in that case.
What if, instead of jumping $(1, 2)$ in either direction, a knight jumped $(2, 3)$ squares in either direction? Or, for that matter, $(p, q)$ squares, with $p+q$ odd and $p, q$ coprime? If there is a simple way of proving that the answer is $\left\lceil \frac{mn}{2} \right\rceil$ for sufficiently large $m, n$ (say, $m, n \geq C(p, q)$), what does $C(p, q)$ look like?