Kanel-Belov A.Y.

The game takes place on a $9 \times 9$ grid of squared paper. Two players take turns. The player who starts the game places crosses in free cells, while his partner places noughts. When all cells are filled, the number of rows and columns $K$ in which there are more crosses than noughts, and the number of rows and columns $H$ in which there are more noughts than crosses, are counted. The difference $B=K-H$ is considered the win of the player who starts. Find such a value of $B$ that

1) the first player can ensure a win of at least $B$, no matter how the second player plays;
2) the second player can always achieve that the first player gets a win of no more than $B$, no matter how the first player plays.
One possible strategy for the first player: the first move is to the center of the board; thereafter, for each move the second player makes to any cell, the first player responds by moving to the cell symmetric to the center. This is possible because the second player always breaks the symmetry.

In the final position, in each pair of cells symmetric relative to the center, there is a cross and a nought. Therefore, in the central row and the central column, there are more crosses than noughts; and for each row (column) not passing through the center, there is a corresponding symmetric row (column), where in one of them there are more noughts, and in the other fewer, than crosses. Thus, with this strategy, the first player's win is 2.

Now let's indicate a strategy for the second player. If he has the opportunity to occupy the cell symmetric (relative to the center) to the cell just occupied by the first, he does so. Otherwise, he makes any move. With this strategy, after the second player's move, the set of occupied cells is either symmetric (until the first player occupies the center) or differs from the symmetric one by exactly one cell. In each pair of occupied symmetric cells, there will be one cross and one nought. With his last move, the first player is forced to create the position described in the previous paragraph, and his win is again 2.

## Answer

$\mathrm{B}=2$.

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