Support we have an n * m table, and two players play this game. They rule out cells in turn. A player can choose a cell (i, j) and rule out all the cells from (i,j) to (n, m), and who rules out the last cell loses the game.
For example, on a 3*5 board, player 1 rules out cell (3,3) to (3,5), and player 2 rules out (2,5) to (3,5), current board is like this: (O means the cell is not ruled out while x mean it is ruled out)
3 O O x x x
2 O O O O x
1 O O O O O
1 2 3 4 5
and after player 1 rules out cells from (2,1) to (3,5), the board becomes
3 x x x x x
2 x x x x x
1 O O O O O
1 2 3 4 5
Now player 2 rules out cells from (1,2) to (3,5), which leaves only (1,1) clean:
3 x x x x x
2 x x x x x
1 O x x x x
1 2 3 4 5
So player 1 has to rules out the only (1,1) cell, since one player has to rule out at least one cell in a turn, and he loses the game.
It is clearly that in n*n, 1*n, and 2*n (n >= 2) cases, the one who plays the first wins.
My problem is that, is there any strategy for a player to win the game in all cases? Should he plays first?
P.S
I think it is related to strategies like dynamic programming or divide-and-conquer, but has not come to an idea yet. So I post it here.
The answer
Thanks to sdcwc's link. For tables bigger than 1*1, the first player will win. The proof is follow: (borrowed from the wiki page)
It turns out that for any rectangular starting position bigger than 1 × 1 the 1st player can win. This can be shown using a strategy-stealing argument: assume that the 2nd player has a winning strategy against any initial 1st player move. Suppose then, that the 1st player takes only the bottom right hand square. By our assumption, the 2nd player has a response to this which will force victory. But if such a winning response exists, the 1st player could have played it as his first move and thus forced victory. The 2nd player therefore cannot have a winning strategy.
And Zermelo's theorem ensures the existence of such a winning strategy.