Andriy and Bogdan take turns putting integers from $1$ to $n$ inclusive in the cells of an $n \times n$ board in such a way that no row or column contains two equal numbers. Andriy begins, and the player who cannot make a move loses. Depending on $n$, which player has a winning strategy? Proposed by Fedir Yudin
Problem
Source: Ukraine MO 2021 8.7
Tags: combinatorics, game
Kamran011
02.05.2021 15:45
The problem is rather well-known (I mean it's sudoku with a friend ). Answer. Bogdan wins for even $n$ and Andriy wins for odd $n$. Strategy for even n. If $n$ is even, it's enough for Bogdan just to play centrally symmetrically. Strategy for odd n. In this case it's obvious that Andriy has to "occupy" the central square, let him do so. Then he writes $(n+1)/2$ there and after that if Bogdan writes the number $i$ somewhere, Andriy writes $(n+1-i)$ in the cell symmetric to it.
nbdaaa
02.05.2021 16:04
Kamran011 wrote: The problem is rather well-known (I mean it's sudoku with a friend ). Answer. Bogdan wins for even $n$ and Andriy wins for odd $n$. Strategy for even n. If $n$ is even, it's enough for Bogdan just to play centrally symmetrically. Strategy for odd n. In this case it's obvious that Andriy has to "occupy" the central square, let him do so. Then he writes $(n+1)/2$ there and after that if Bogdan writes the number $i$ somewhere, Andriy writes $(n+1-i)$ in the cell symmetric to it. Ye it's exactly the Sudoku games