Xjyuk

Alistair and Roberta are playing a game of reverse dots and boxes.

• 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). Alistair goes first.


• If a move completes a $1times1$ box, the player gets one point and has to make another move. If two boxes are completed with a single move, the player gets two points but only has to make one additional move. The player keeps making moves until they make a move which does not complete a $1times1$ 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?

First player wins

Capture two squares from three of the four wings, then cut off the last wing. Player two can no longer make any moves that don't result in a capture.

This wins, 6-7.

Example:

Second player wins.

First player's first move, assuming they don't take any squares, can only be to split an arm with two on the end or cut off an arm, taking one of the sides on the middle square.

After that,

If first player split an arm, second player cuts off that arm, taking one square. If first player cut off an arm, second player ignores it. Now the center square has one edge filled.

Then,

Second player takes an entire arm. That leaves the center with two edges marked. There are no more branches, just a chain of 2 or 3, and a chain of 7. Second player reduces the chain of 7 by taking the end of an untouched arm. Now there's a chain of 6. They split that in two.

On first player's second turn,

They face chains of length 2,3,3 or 3,3,3. Either way, every move they take completes a square, so they take all remaining squares.

In the end,

The second player has taken 4 or 5 squares, and first player has taken 8 or 9.

Alternately,

First player can force the second player to end the game on their turn. They take two entire arms and take the end of one arm, leaving a single chain of 6. They then split that chain in half, leaving the second player with chains of length 3,3. Second player must then take all the rest. First player still loses with 7 squares to second player's 6. But this strategy allows them to lose by the least amount.