Stoyan and Nikolai have two $100\times 100$ chess boards. Both of them number each cell with the numbers $1$ to $10000$ in some way. Is it possible that for every two numbers $a$ and $b$, which share a common side in Nikolai's board, these two numbers are at a knight's move distance in Stoyan's board (that is, a knight can move from one of the cells to the other one with a move)?
Nikolai Beluhov
An easy one, though I'm a bit rusty ig. The answer is no. The idea is to look at the corners. It's easy to see the set of cells at the corners of both tables are identical as they have only two distinct neighbors. Observe that any $2\times 2$ tile on Nikolai's table corresponds to some rhombus on Stoyan's. Consider a pair of corresponding corners in both tables: $(a_{1}, a_{100})$ on Nikolai's and $(b_{1}, b_{100})$ on Stoyan's table. Since $b_{1}$ and $b_{100}$ are two vertices of a parallelogram (formed by the corners of a chain of connected rhombuses) and that's clearly impossible as the side parallel to $b_{1}b_{100}$ exceeds the borders of Stoyan's table.