https://aops.com/community/p11481346 liekkas wrote: The answer is $10$. For Alice, she can open her strategy: choose square $(1,4),(1,8),(2,1),(3,6),(4,3),(5,8),(6,1),(6,5),(8,3),(8,7)$. For the picture see the attachment below (I don't know how to draw a picture with LaTeX ).It can be found that any two squares have Descartes distance at least $4$, so Bob can't connect them to form one component. For Bob, define a "box" to be a pack of squares, such that any two squares of it have Descartes distance at most $3$. Now we can dissect the chessboard to $10$ boxes (see the attachment). It's easy to see that Bob can guarantee there is no more than $1$ component in any box (Bob shouldn't color the square that is not in the same box as Alice's) .