Denote each cell as (i, j) for row i, column j, so bottom left is (1,1), top right is (n, n) We say all a(i, j) build a wall when i+j=k+1, k is integer and 1<=k<=2n-1. There are 2n-1 walls. It's obvious that a path from (1,1) to (n, n) will across all walls: considering the sum of i, j of each cell on a "path", every step you move on the path, the sum +1 or -1, and the start number is 2, the end number is 2n, so each of 2~2n mush appears on the path("discrete version of IVT", not sure if it's the name in English) 1.obviously, if a>=2n-1, Alice can paint a green path from (1,1) to (n, n). Alice Win 2. if a<2n-1, there must be some "walls" whose all cells are not colored by Alice(as there are 2n-1 walls), we call them "non-green" walls. The max of their minimum length is [a/2]+1. So, 2.1. if b>= [a/2]+1, Bob can pick the non-green wall whose length is minimum, and color all of its cells to blue, and Bob Win. 2.2. if b<[a/2]+1, Alice can color the bottom row cells from(1, 1) to (1, [a/2]), and the right column cells form (n, n) to (n, [a/2]+1). Before Bob paints any cell, there are at least [a/2]+1 uncolored paths to connect these 2 green tiles, and these paths are not crossing each others. (I just don't know how to insert a pic here, so hopefully below ASCII chart (for n=5, a=6) can show the idea, I use '.' and '|' to show the paths, 'x' is placeholder to make the chart looks ok): .........G | .......G | | .....G | | |xxx| GGG..| To cut off all of these paths, Bob needs paint at least [a/2]+1 cells, one cell for each path. it means if b<[a/2]+1, Alice will Win. In summary, Bob will win only if (a<2n-1)&&(b>= [a/2]+1) is true, otherwise, Alice will win.