revised version of notes Dots and Boxes
5 replies. Last post: 2007-12-22
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5 replies. Last post: 2007-12-22
Reply to this topic Return to forumI put a new version of my notes up at wccanard.wetpaint.com. In this revised version I typed up a proof of the 5+5=6 lemma and also wrote down the only proof I know of the following result: in a simple loony endgame (i.e. just isolated loops and long chains) with no 3-chains, one optimal way to play it is to open all the loops first (smallest first) and then all the chains. I also tidied things up a bit, and put in proofs that I'd found of results which previously were only computational observations. These notes are not basic hints about how to play the game: they are typically rigorous proofs of results that experts know by intuition already. In particular I suspect that they won't be much help for someone who is trying to learn the game: for those people I would recommend Berlekamp's book.
I am pretty sure I'm missing something with the no-3-chain result though; my proof is awfully long.
In the future I plan on writing two more documents: one on how to count positions and one on how to defend when you've lost the chain battle. Both will be for beginners who know about chain counting but don't know much else, but I don't know when I'll find time to write them!
I'm confused. Isn't the value of {Lx,C4}
max { open Lx, DDX: (x-4)-4+4 = x-4 open Lx, take all: x -4 = x-4 open C4, DX: 2 -2+x = x open C4, take all: 4 -x }
e.g. x=6
+-+-+-+-+-+| | + +-+ + +-+| | +-+-+-+-+-+
Right: it's different people deciding (1) whether to open the chain or the loop and (2) whether to take all or leave some. So you need to do some kind of minimax thing; and then to work out the optimal line you have to go backwards and see at which choices the min/max was attained. I know the answer for an x-loop and a 4-chain: the fully controlled value is x-8 and the terminal bonus is 4 so the value should be x-4.