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Balti Tee 2024 · Ülesanne 8

Kombinatoorika

Let a,b,na, b, n be positive integers such that a+b≤n2a+b \leq n^{2}. Alice and Bob play a game on an (initially uncoloured) n×nn \times n grid as follows:

  • First, Alice paints aa cells green.
  • Then, Bob paints bb other (i.e. uncoloured) cells blue.

Alice wins if she can find a path of non-blue cells starting with the bottom left cell and ending with the top right cell (where a path is a sequence of cells such that any two consecutive ones have a common side), otherwise Bob wins. Determine, in terms of a,ba, b and nn, who has a winning strategy.

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Official solution diagram for Baltic Way 2024 Problem 8 (Figure 7).

Figure 7

Official solution diagram for Baltic Way 2024 Problem 8 (Figure 8).

Figure 8

If a≥2n−1a \geq 2 n-1, Alice can win, for example by painting all the cells in the leftmost column and the topmost row green, ensuring that there will be a green path (Fig. 7). If 2n−1>a≥2b2 n-1>a \geq 2 b, Alice can also win. Indeed, note that n>bn>b, so Alice can make sure to color (at least) the bottommost bb cells in the leftmost column and the rightmost bb cells in the topmost row green. There are now b+1b+1 disjoint paths from some green square in the leftmost column to some green square in the topmost row (Fig. 8), so there is no way for Bob to block all of the paths. If however a<min⁡(2b,2n−1)a<\min (2 b, 2 n-1), Bob wins. Indeed, note that min⁡(2b,2n−1)\min (2 b, 2 n-1) is the number of all descending diagonals of length at most bb. Hence after Alice has made her move, there must still be some of these diagonals with no green cells in it. Bob can color all cells in it blue and win. Hence Alice wins iff a≥min⁡(2b,2n−1)a \geq \min (2 b, 2 n-1).

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