This is a follow-up to my first Dots and Boxes question. It is just a little bit harder than the first one. Each one I post will be a bit harder than the previous one, and they will all be from games I have recently played.
In the following game of Dots and Boxes, it is your move.
Your opponent is a computer who will play perfectly. What is the best move, and why?
The player can't win. Analysis to follow.
Here I've marked the "safe" moves in blue. We will call them 1, 2, and 3 counter-clockwise starting from the left. The sacrifices I have marked in green. Let us call them A above and B below. Finally I have marked a move inside the lower right square in red.
Any move inside the square is bad:
Assumption - It is determinable whether the first person to make a move outside the square will win or lose.
Making a move inside the square gives the opponent the opportunity to decide who makes the first move outside the square. This is because they can either complete all the boxes in the square then move outside, or continue our line inside the 2x2 square to split it into 2 2x1 squares. In the latter case we are forced to make the first move outside the square (after completing the four boxes). Since our computer opponent is perfect, it will know whether it wants to make the first or second move outside, and will choose accordingly. We cannot give the computer that choice.
We need to note that 2 is mutually exclusive with each of 1 and 3. Also, A and B are mutually exclusive. Any non-safe, non-sacrifice move will be as bad as the red move. Finally, A has interactions with our three safe moves. If A is played: with 3 on the board, 1 and 2 become starts to long chains (bad), and without 3 on the board 1 and 2 become sacrifices.
If we make a move in one of the safe lines, the computer can play A leaving us with no good moves like so:
case 1: 1, A, complete the squares...in doing so we have played 2 and have no more valid moves (2 invalidates 3, and A invalidated B)
case 2: 2, A, complete the squares then no more valid moves (likewise 2 invalidates 3, and A invalidates B)
case 3: 3, A, complete the square then no more valid moves (A invalidates B, and 3 + A means 1 and 2 are starts to long chains)
So what if we start with a sacrifice? The computer can follow with a safe that leaves no good moves.
case A: A, complete the square then 3, no good moves (A invalidates B, and 3 + A means 1 and 2 are starts to long chains)
case B: B, complete the squares then 2, no good moves (B invalidates A, and 2 invalidates 1 and 3)
We're left with no way to win this, but the best move is probably 1 or 2 as that forces the computer to sacrifice 2 squares.
The best you can win is 10 boxes. The computer always wins by getting at least 15 boxes.
In this image, playing any of the red edges gives you 10 boxes if the computer plays optimally. If you play the green edge, you get only 9 boxes. Playing any of the blue edges gives you even less, only 8.
The computation was done using the same code I used for the previous question. The computation was made easier since there are only 30 edges instead of 32.
For a greedy strategy by both you and the opponent,
will set you up for the first chain (of five).
