On an $8×8$ board, you and your friend play the following game. Initially, all the unit squares are white. First, you color any $n$ squares blue. Second, your friend chooses $4$ rows and $4$ columns, then colors all squares in those rows and columns black. What is the minimal $n$ such that, regardless of your friend's move, the board has always at least one blue square remaining?
Alright, I believe the minimal $n$ is $13$. I have proved that for $n = 12$, your friend always wins. There are always $4$ rows which have at least $8$ blue squares together. He can color those and $4$ columns for each remaining blue square (in the worst case). If you choose $8$ squares on the diagonal your friend can color first $4$ columns and last $4$ rows, if such diagonal goes from top left to bottom right.
I have tried to find a construction for $n = 13$ but, so far I have failed to do so.
Best Answer
Is this a possible case for $n=13$ that is impossible for your friend to remove?