The Dinitz conjecture states that, for each $n$ and for every collection of $n$-element sets $S_{ ij}$, an $n\times n$ partial latin square can be found with the $(i,j)$\<th entry taken from $S_{ ij}$. The analogous statement for $(n-1)\times n$ rectangles is proven here. The proof uses a recent result by Alon and Tarsi and is given in terms of even and odd orientations of graphs.

Categories: math.CO

The Dinitz problem solved for rectangles