A “simple loony endgame” in dots and boxes is a position comprising just of isolated long chains (of length 3 or more) and isolated loops: I won't even allow a “dipper” (a chain one of whose ends is a loop) or any short chains. These endgames come up often enough to warrant some serious analysis. The big questions are: (a) what is the value of such an endgame, and (b) what move to play? There are only loony moves left—the player not in control will have to open either a chain or a loop—and general principles tell you that it should be either the smallest chain or the smallest loop. On the other hand, it's sometimes tricky to decide which! For example in the position 3+5+Q+Q (a 3-chain, a 5-chain and two quads) one should open the 3-chain first; opening the quad is a bad move. But in 3+8+Q+Q one should open a quad first: opening the 3-chain is a bad move.

What would be great would be a simple algorithm to work out the value of these simple positions, and which move to play in them—by a “simple algorithm” I mean something simple enough to be implemented in ones head rather than on a piece of paper!

I still don't have such an algorithm. However yesterday I found a proof of the following:

Theorem: In a simple loony endgame with no 3-chains, first open all the loops (smallest first) and then all the chains (smallest first).

My proof is _very_ long! I can't believe it's the best proof. Am I missing something? Is the theorem completely obvious??

wcc

PS whether the controlling player should leave dominoes or take everything is a different matter! He needs to compute values to work this out. But computing values is easy once you know the order to play the moves.

Is it true, as I'm guessing it is, that opening a 3 may be correct, but opening a chain longer than 3 when there are quads remaining is never correct?

Gah! That's what you just said. I posted after reading only your first paragraph.

I think considering the numeric sacrifice to keep control will reveal why starting a 4 is never right. The cost of a quad is -4, while it's +4 to give up control. So control must be worth 8.

The cost of a 3 is -1, while the payoff for ceding control is +3. So both Qs and 3s present the second player with a meaningful choice.

But the cost of a 4 is 0 and the cost of a 5 or greater is positive. One cannot improve one's position by letting the second player maintain control without paying a cost.

Life's not as easy as this, I don't think, because sometimes in a complicated game player 2 is going to give up control whatever you do and then your intuition gets messed up. In particular I think I can construct a game (not a simple loony endgame but a general game) with a 4-chain and a quad, where opening the 4 is better than opening the quad. I'll get back to you if I manage to do it. The reason I think that such a game exists is that if G is a game and G+4 is worth x and G+Q is worth y, then x and y have the same parity and \|x-y\|<=4 but I think that this is about all you can say about x and y, and this isn't enough to prove that opening the quad is always at least as good as opening the chain in G+x+y.

I'll think more about this today and perhaps post an example in about 12 hours.

EDIT: “…opening the chain in G+4+Q”.

So in the game 3+4+5+Q+Q (that is, a 3-chain, a 4-chain, a 5-chain and two 4-loops) opening the 4-chain is better than opening the quad: in fact opening the 4-chain is an optimal move. The reason is that after opening the 4-chain it doesn't matter if your opponent leaves the domino or takes the lot, he's still only going to make a net gain of two boxes. But if you open the quad instead he'll grab the lot and what's left is worth zero.