Two players in turns color the squares of a $4 \times 4$ grid, one square at the time. Player loses if after his move a square of $2\times2$ is colored completely. Which of the players has the winning strategy, First or Second? (4 points)
Problem
Source: Fall 2007 Tournament of Towns Junior P-Level #5
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Popescu
31.03.2015 18:18
The second one has a winning strategy.If the first player colors the square (a,b), the second one colors the square (5-a,5-b).By symmetry, the square (5-a,5-b) isn't already colored, so the second player can make this move(color this square).In conclusion, the second player has a winning strategy.
mavropnevma
31.03.2015 18:25
What is $n$?
Popescu
31.03.2015 20:02
I meant 5, sorry
mavropnevma
31.03.2015 20:31
So First player colors square $(2,2)$, and by your strategy Second player colors square $(3,3)$. Now First player colors square $(2,3)$, and by your strategy Second player colors square $(3,2)$. Only that now he has lost, since he completely colored a $2\times 2$ square. Back to the drawing board ...
Popescu
31.03.2015 21:36
Indeed, back to the drawing board.The second player wins.The first player colors the square (a,b), then the second one colors the square (a,b+2), where 5=1 and 6=2