Dots and boxes, 2 against 1

Question

To start the Monthly Topic Challenge #3: Pencil and Paper Games : A simple question (not necessarily with a simple answer).

3 fanatic dots-and-boxes players come together.
Of course no one wants to wait while the others play, and no one likes king making.
So they decide on a team match: 2 versus 1
To compensate their numerical disadvantage, the single player team gets 2 advantages:

• They only need one of the boxes to win.
• They can set the player order

Warm-up question:
Assuming perfect play, will the single player team win on a 3*3 area?

Real question:
Assuming perfect play, will the single player team win on a 5*5 area?

Since wikipedia mentions variants: They start with an empty board. For the rules see e.g https://en.wikipedia.org/wiki/Dots_and_Boxes

Hint:

On a 3 by 3 the single player has a big advantage when choosing to play first but winning is not automatic.
For example if player1 plays orange (in picture 1), the second position will be possible at player1's next turn, and the area played can (with the help of player2) always be taken by player3, who then can play the safe spot in the other area (e.g. picture 3)

Hints/strategy suggestions

The single player can win if they can take the last square. Thus player 1 can play 'defensively': If scoring 2 boxes with 1 line can be prevented while picking the appropriate play order, player 1 will win.
This is e.g. trivially possible in 1 by 1 (by starting first).

This is not possible for 5 by 5 (and larger). The 2-player team can force the formation of a 2 by 2 or larger enclosed area and then score 2 boxes with 1 line when filling that area!

Player 1 will also win however if there are 3 or more areas in the end phase (when all moves open scoring for the next player). Thus player 1 can also play 'offensively': Create as many areas as possible. (I do not see a clear way to do this fast enough.)

For 5 by 5 a combination seems best: Choose 3rd position so that the 2-player team must force a box. Create as many areas as possible, while hindering them forming their box whenever useful. After some playing against myself, I expect that the team wins, but it is a close call, and I did not work out all possibilities

Answer 1

First, the warm-up: I brute-forced the smaller cases up to 30 edges. What I found is ...

If my program is correct, for the smaller sizes, player 0 (solo) can always win by choosing the player order and then playing optimally.

Here is the output of my program.
The players play in order 0-1-2-0-1-2-..., player 0 is team 1, player 1 and 2 are team 2.
The W or L indicates a win or loss for player 0 depending on who plays first, resp. player 0, 1 or 2. For example, with size 1x2, player 0 can win if they play 1-2-0-1-2...

Size 1x1: W L L
Size 1x2: L W L
Size 1x3: L L W
Size 1x4: W L L
Size 1x5: L W L
Size 1x6: L L W
Size 1x7: W L L
Size 1x8: L W W
Size 1x9: W L W
Size 2x1: L W L
Size 2x2: L W L
Size 2x3: L W L
Size 2x4: L W L
Size 2x5: L W L
Size 3x3: W L L

So for the 3x3 case we have:

Player 0 can win if and only if she decides to play first.

What about larger cases?

From the calculations above, it seems that player 0 always can win. It even tends to become a bit easier as the board gets larger.

My first thought was that team 2 would always win. The idea is that as soon as an opportunity arises, player 1 can prepare a box and player 2 can then capture the box leaving no opportunity for player 0 to ever capture a box.

But what happens when we reach the end game? At end game, when reaching the stage where no move is possible without completing the 3rd side of a box and providing a capture opportunity, if there are multiple corridors left then team 2 has a problem. Player 1 must close a corridor, and player 2 must capture all the boxes in the corridor. But then player 2 still has a move to do and she is forced to give a box opportunity for player 0.

But can player 0 make sure that there are multiple corridors left at the end? After all, team 2 could consume corridors as soon as they appear.

Note that what I call a corridor could be just a single box. A box with 2 opposite sides completed is already a corridor. An angle is also a corridor, one where one side is redued to a single point.
In fact, any box with 2 sides completed is a problem for team 2 in the end game.

And it is easy to create a corridor. Just add a parallel line to an existing line. Or create an angle with an existing line. Player 0 can create a new single-box corridor with every move.
She sometimes can create two at a time, for instance by adding an angle in the middle of a wall. Or, from an existing line, player 0 can draw a little "plus" sign, creating four 2-sided boxes in three moves.

Team 2 will loose if there are just 2 corridors left when they reach end game. So they must neutralize every single corridor player 0 creates as she creates it.
I believe given enough space it is easy for player 0 to create a state where there are many independent corridors which makes player 0 win.

What about 5x5?

I am pretty sure with 5x5, possibly even 4x4, the board is large enough for player 0 to win regardless of who starts. She just has to keep busy creating small corridors all over the place. Team 2 will not be able to catch up and end up being forced to provide a box opportunity for player 0.