# Reverse dots and boxes

Alice and Bob are playing a game of reverse dots and boxes. The rules are simple:

• The players take turns adding one horizontal or vertical line in one free spot on the grid (marked with light gray lines in the below image). Alice goes first.
• If a move completes a $$1\times1$$ box, the player gets one point and has to make another move. The player keeps making moves until they make a move which does not complete a $$1\times1$$ box.
• The game ends when all possible lines have been drawn.
• Since this is a reverse game, the player with the most points loses.

Which of the players can win the game played in the above grid? What strategy should they use?

I would suggest an alternate (simpler) strategy:

By sacrificing five squares immediately, Alice can be assured of victory

Then by splitting the remaining 6 into 2x3 chunks, Bob has no option but to fill in the remaining 6.

Using this strategy Alice always wins.

As pointed out by LeppyR64 - if Alice does not take boxes from the outset, and only creates a 2 or 3 block section, Bob can always win by taking 3 boxes, and leave her with 3+2+3 or 2+3+3

• This solution also invalidates the other two answers. If the other two solutions are attempted then Bob wins. – LeppyR64 Jul 8 '19 at 10:25
• This is the intended answer. Nice job! – jafe Jul 8 '19 at 10:44

My solution...

The grid contains $$11$$ boxes. You can split them up into $$2+2+2+2+3$$ or $$2+3+3+3$$. So the first sum is $$4$$ moves and the second sum is $$3$$ possible moves.

The last player who can set a line without enclosing one box wins, because his opponent will enclose one box after the next. Enclosing means, that you can't set a line in this area without making a $$1×1$$ box.

The player who starts wants to make the second sum, because it is $$3$$ moves (first player, second player, first player, now the second player will enclose all boxes). The second player wants to make the first sum, because it is $$4$$ moves. So if the first player plays good he will win because he encloses $$3$$ boxes, then the second player will enclose $$2$$ boxes, because $$3$$ would mean the second sum. Then the first player will enclose $$3$$ boxes for the second sum.

Images:
Enclose 3 boxes:

Enclose 2 boxes: