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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.

enter image description here

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

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2 Answers 2

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I would suggest an alternate (simpler) strategy:

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

Assured 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

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    $\begingroup$ This solution also invalidates the other two answers. If the other two solutions are attempted then Bob wins. $\endgroup$
    – LeppyR64
    Commented Jul 8, 2019 at 10:25
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    $\begingroup$ This is the intended answer. Nice job! $\endgroup$
    – Jafe
    Commented Jul 8, 2019 at 10:44
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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:

enter image description here

Enclose 2 boxes:

enter image description here

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