When finishing this game I realized that much of the essential to understand as a beginner to this game are (albeit very compact) contained in this game.

testing the link

Fast reference:

blue wins with odd number of chains or no chains, red with even (except 0 chains)

a loop counts for two chains.

The first thing thing a beginner should know is the concept of a chain like this:

chain in the middle

A real chain consists out of at least 3 boxes.

A loop is chain like this:

two loops in the upper right and lower left, and 4 too short chains of length two on the sides.

End Game

First consider this game with an alternative win condition: the player that fills the last box wins the game. And consider the situation linked to above. It is of enormous importance to understand how to finish of this game.

The red line in the upperleft is the move 32 made by blue. This means that red has to move. Red has to move. In essence he can choose to over two boxes or six to his opponent. He offers two and then another blue gives two back and so they exchange ' boxes each.

Ending here, where each move of red is equivalent. At this point blue can win easily by taking all 6 which I didn't do because I already decided to use this game here. By taking all 6 and giving the other 6 to the opponent I would have got 10-10 in the endgame (and due to my early 2-3 lead have won 12-13), but it is not the best I could do. I can win the endgame 8-12. And this is the important lesson to learn. By not taking all but only two boxes, my opponent was forced to give me the other 6 anyway. This way I won 10-15 and I would have even won from trailing 4-1 this way.

The point I want to make is that I controlled the endgame. Actually I did a very early offer to achieve this.

But first this: we have seen that you can keep control by giving a way 4 boxes in a loop (note that in 10 boxes loop I could have given 4 as well), the cost of giving a way a chain is only 2 boxes: dubble cross on a chain.

How did we both know at move 14 that I would get control? (This is the reason he gave away 5 boxes and I only took 3 of them)

Well here is the crux.

Independent on which moves you have played blue has control if there are zero or an odd number of chains and red if there are an even number of chains (except zero). The only remark to add is that if during the game each dubble cross (donation of two stones when filling a chain) changes this.

I find the mathematical proof of this very interesting but I want to avoid it here.

The player with control can always get the last box, and mostly but not always wins the game.

But if he offers to much in his quest for control he might loose the game. One way of achieving a win against control is using quads: a quad is a square of 2x2 and counts as a loop but if you dubbelcross you win nothing and loose 4.

I hope I help beginners with this and I hope others will clarify where my English or Game theory were off to much :-)

In addition:

move 13 is a 'losing move' (a move after which you have certainly lost the game) since after taking 3 boxes, the blue player can either take the other two and move somewhere else or give away a doublecross leaving his opponent with the same situation on the board. It is quite easy to prove this with a man-in-the-middle-argument and I'm almost sure that wccanard has proven this.

Here are some comments on the game in question.

Of course, the symmetry of this game plays a huge part in player 2's win. Games with rotational symmetry like this are definitely the exception rather than the rule. When the game has rotational symmetry, both players have to be careful. If player 2 is simply obsessively copying player 1's moves with rotational symmetry, then at some stage player 1 has to sacrifice the central box—a move that has no symmetric counterpart. But he must do it before the central box becomes part of a long chain.

Player 2's move 10 is provably a winning move. The position is symmetric and the central box is now in a long chain. But the only proof I know that move 10 is a winning move is non-constructive—it doesn't give a winning line. Player 1 probably shouldn't have ended up in this position, but, given that he has, his best move now is probably to open the chain immediately (a move which is provably a losing move). Then player 2 will probably take one and give two away, and he's now winning the chain battle but is a box down and player 2 now has to try and rig it so that the long chain battle is worth zero (which is possible to do, especially if player 1 now naively copies player 2).

The reason that opening the long chain is provably a losing move is, as Arjan Feenstra points out, a non-constructive strategy-stealing argument. I have called it “the golden rule” in the past, in my notes at wccanard.wetpaint.com, but the idea is decades old and even the application to dots and boxes is *certainly* not due to me—it's much older than that. The “golden rule” is: if you open a chain (resp. a loop) and if, after your opponent has taken all but two (resp. all but four) of the boxes, you are not winning the box count, then you have lost. And the proof, as Arjan Feenstra says, is simply that there are two lines now for your opponent: either they take the last two (resp. 4) and play first in the remaining region, or they sacrifice the last two (resp. 4) and play second. These positions are exact opposites of each other and hence one must have a non-negative value under optimal play, so your opponent chooses that line [**assuming he can work out which one it is!!**] and must win.

Note the **assuming** phrase! If P1 had opened the 3-chain with his move 11 then P2 would have to decide which option to take and it's tough to know for sure what the right one would be.

But after P1's move 11, this is now not only a provable win for P2, but it's easy in practice to find the win. P2 now just copies. When the chain is inevitably opened, P2 keeps control, because he'll be 1 box *ahead* now; he can now just copy until the end (as he basically did).

It's good if someone found this game instructive—but as I already said, it's rather a special case. In some sense it illustrates some of the mathematical principles of the game, but in a rather specialised situation. Normally you can't just win the chain battle by following an obvious algorithm, you have to think a little. And, even more crucially, winning the chain battle doesn't always guarantee winning the game!

I think move 7 was still winning for red in that game.

Generally if there's a chain through the center of the game and the board is symmetric player 2 always wins (if no sacrifaces have been made).

If player 1 moves into the 3-chain at move 11, taking them all leads to victory.

-michael

wccanard,

I never said this game was instructive in is whole, I only say that the concepts a beginner needs to understand to play reasonable are all present in this game.

Michael I believe what you say, but I agree that finding the victory is not obvious enough for me.

Hjallti: I think I was trying to say that a more instructive game for beginners would be where one person wins the chain battle because they actually *win* it (e.g. with a sacrifice that cleverly kills the (n+1)st potential chain, and leaves the position with a definite n chains). This would demonstrate that an early sacrifice might be a great move, which is a fundamental basic point.

In your game above you win the chain battle because of some application of a symmetry argument—you know you have won it even when you don't know how many chains there will be (maybe the top right and bottom left will be chains, maybe they will be loops). All my point is, is that this is not *normal*. This is a rare occurrence. Usually you know you've won the chain battle because you can see how many chains there are going to be.

I have this example at hand :

[game;id:998946;move:12;title:Aldiris making the winning early sacrifice]

… but please don't comment end of the game as it is still in progress even if we both know the winner.

Another one :

[game;id:998941;move:12;title:Carroll making the losing early sacrifice]

Here as Michael pointed after [game;id:998941;move:13] there will be a second forming chain to the left and a sure win for him.

Granted wccanard, but I wanted only to explain the start and not the second part… How to win the chain battle.

I agree Hjallti. You were able to link to various individual points in this game which very clearly demonstrate some of the concepts that a beginner needs to understand before they can get to grips with things like the chain rule.

People need to have a good understanding of these before they can move on to the more advanced things that the others are talking about.

Thanks, but I should have pointed out in the first place that that was my goal. I didn't even like playing that game because I found it stupid to win the game with that strategy.

Hjallti: here was a game I enjoyed playing. I didn't know whether my move 7 was sound but I came through in the end.

http://www.littlegolem.net/jsp/game/game.jsp?gid=875185&nmove=6

[note that I was careful not to let the middle become a 3-chain, but things were getting so crowded that I sacrificed the middle box very early because I couldn't really see anything else to do]

May I note that you could have won 14-11 but probably refrained from it because it would make the game one move longer in the end. I think I should study these kind of games to understand more, but at this moment I just want to get the feel back.