Miseré Havannah

Anyone up for trying Miseré Havannah? (where you try to lose instead of win).
We did it recently with dots and boxes here http://www.littlegolem.net/jsp/forum/topic2.jsp?forum=110&topic=138

or is there some reason it wouldn't work? What boards size would you suggest?

I have a feeling that it would be very easy for the player that does NOT have to make the last move to win, but I could be wrong. Also, the games would take a very long time to finsih, since you end up filling out all the hexes before the game is over.

@Christian: It's Mis \e re, not Miser /e.

@Marius: I don't know if this passes for very easy, but in Misere Hex it is at least true that perfect play leads to a full board.

Would anyone be willing to play a trial game or two? Which size do you think is appropriate?

I've sent you an invitation for size 4.

cool

@Christian K: I've sent you one too, also in size 4.

@Tasmanian Devil: notice that this result about perfect play doesn't hold for irregular boards of size 1xN with N > 1 (-: (thus I guess the theoretical winner is player 2 if and only if the board has an odd number of cells or a side of size 1).

If you allow non-rhombic boards, then a lot of things can happen, and it’s more complicated than you suggest.

On 1xN with N>1, the player with the shortest sides wins, regardless of who makes the first move.

On 2xN with N>2, there is a simple winning strategy for the player with the shortest sides if he is second player. The idea is to label a segment of the board as follows:

... | 1 | 1 | 3 | ...
... 3 | 2 | 2 | ....

Whenever the opponent plays in the segment, our hero simply responds with a cell with the same label. He is then guaranteed not to get a connection from left to right, since such a connection must pass through two cells with the same label.

Here is an illustration for a first player strategy which works for N>5.

... | 1 | 1 | - | 4 | 4 | ...
... 5 | 2 | 2 | 3 | 3 | 5 ...

The last player strategy can easily be extended to larger values for the shortest sides, for example:

... 1 | 1 | 5 | 5 | 9 ...
... | 2 | 2 | 6 | 6 | ...
... - | 3 | 3 | 7 | 7 ...
... | 9 | 4 | 4 | 8 | 8 ...

...you are talking about hex, right?

Yes.

I meant: in Misère Hex (the board being a parallelogram M by N), the theoretical winner is

- player 'M' (of sides separated by M cells) if M > 1 and N = 1,
- player 'N' if N > 1 and M = 1,
- player 1 if otherwise MN is even,
- player 2 if otherwise MN is odd.

Is that incorrect?

Thanks for the clarification, though your answer didn't help me at all to find why I was wrong (-:

I found my mistake anyway: the proof in your link assumes a player can follow the same strategy than the opponent, but this is not always possible if their respective sides are not at the same distance...

By the way, how does the hero play if the board cannot be divided into an integral number of such segments?

Generally we can expect that the winner is the player with the sides that a separated by the highest number of cells.

I'm sorry but my "first player strategy" above is a sidetrack. The first pattern works fine for the player who is to move from left to right whether he plays first or second. He won't play a labelled cell before the opponent has taken the other one with that label unless he has to, but there is no way the opponent can force him to take both.

So with a simple generalization of the patterns we have already seen, the winner is:

-player 'M' if N is fixed and M is larger than some boundary that depends on N
-player 'N' if M is fixed and N is larger than some boundary that depends on M

Sorry, I deleted the post you responded to in order to fix something.

We don't need to divide the whole board into such segments. We only need one. It does not matter how the moves outside of it are played, since a connection from left to right must go through it.

Hmm, yeah, good remark o-:, thanks.

I just realized...when M > N, player 'M' can win by adapting the standard strategy that player 'N' can use to win normal Hex. :-P

In Misére Havannah on a board of size 2 (7 cells), the first player is doomed.

How did you find out? Bruteforce?

For such a small size, the loser is the one that takes 2 adjacent corners or takes the centre. The former is easy to avoid for both players, whereas player 2 can force player 1 to play the centre since there is an even number of corners (while player 1 doesn't play the centre player 2 can just mirror the moves).

If player 1 begins with a corner, player 2 can also win by taking the centre. Player 1 can then either (a) lose instantly by taking an adjacent corner, (b) take a corner adjacent to an adjacent corner or ( c ) take the opposite corner. In case ( c ) he will lose in the next move no matter what player 2 does. In case (b), player 2 will take the corner not adjacent to any of player 1's cells, and player 1 will again lose in the next move.

Ah, yes, that's true, thanks, I was trying to make it short but it ended up being false. Seems it was more like 'that takes the centre if the number of pieces played was even', that is, 'that is player 1' (-:

In the general case, do you think there is something to do against the mirror strategy? Playing the centre doesn't seem to help a lot for player 1 (but I haven't given it much thinking).

I think you are right. Player 1 is in big trouble from the start as he must choose between taking the undesirable centre or risk facing the mirror strategy.

To avoid the issue, the swap rule could be the one of Steven W. Meyers' Anchor: player 1 first plays three moves, among which one is of the opponent's colour; then player 2 chooses whether to swap.

Anyone up for a game of misere havannah?

I will accept an invitation, but don't expect anything but the most amateurish of moves!