Shapovalov A.V.

A square grid $9 \times 9$ is formed from matches, with each cell side being one match. Petya and Vasya take turns removing one match, with Petya starting. The winner is the one after whose move there are no whole $1 \times 1$ squares left. Who can act to ensure their victory, regardless of how their opponent plays?
Note that before Vasya's move, there will always be an odd number of matches left. It is clear that the winner will be the one who ends up with a position where there is one or two adjacent (by side) squares. Therefore, Vasya just needs to avoid leaving such a position after his move, and then he will win, since a draw is impossible. Let's show how he can do this in different cases.

1) More than three squares are left. He will take the outermost match, spoiling no more than one square.
2) Three squares are left. He will take a match not from them, and if there are no such matches, then due to the odd number of matches, it is clear that two squares are adjacent, while the third is not adjacent to them; then he will spoil one of the adjacent squares.

3) Two squares are left, and they are not adjacent. Due to the odd number, there is a match not included in them, which Vasya will take.

All positions have been considered.

## Answer

Vasya.

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