Just a couple of quick dots puzzles. They might have been solved somewhere, you are welcome to link.

1) What is the shortest game on 5x5 boxes? What about on NxN?
2) What are the longest?
3) How would the chain rule work with 3 players?

The chain rule can't work with 3 players, because one of the players will have to decide which one of his opponents wins the game by either doublecrossing or taking the boxes. Though you can find a “chain rule” for the case of exactly one long chain.

Yeah you are right, sorry. Let me rephrase, how does the number of chains dictate who will first be forced to make a looney move.

I once worked that out. Too lazy to remember exactly, but as far as I can tell you, there is a chain rule indeed and it works analogously with modulo 3.

The number of moves is equal to the number of segments plus one minus the number of captures (the plus one is because the last move is a capture, but it doesn't “gain its move”).

The number of captures is the number of boxes minus the number of double-crosses.

The number of double-crosses is somewhere between 0 and one-half of the number of boxes.

Thus, the minimum number of moves is 2N(N+1)+1 - N^{2 = N}2+2N+1 = (N+1)^2.

If N is even, the maximum is 2N(N+1)+1 - N^{2/2 = 3N}2/2 + 2N + 1.

If N is odd, the maximum is 2N(N+1)+1 - (N^{2+1)/2 = (3N}2+1)/2 + 2N.

So, for 5x5, the answers are 6^2 = 36 and 76/2 + 10 = 48.

I'm not going to touch #3.

Let pk:=p(k):=k-th player, n be the number of nodes and c be the number of upcoming chains. Counting the total number of moves z made in a game, considering that taking a box loses a move except for the last one and every double-cross gains a move with no double-cross made in the last move, we obtain:

z=edges-(boxes-1)+(c-1)=edges-surfaces+c=c+n-1 (Using Euler's polyhedron formula).

This is valid, if we assume that there is a double-cross for every chain except for the last one. But then we know that exactly two moves per chain are played during the loony endgame, which leads to:

x=z+1-2c=(c+n-1)+1-2c=n-c

,where x is the move number of the first chain sacrifice. Note that p1 plays moves 1,4,7,…≡1 mod 3, p2 plays 2,5,8,…≡2 mod 3 and p3 plays 3,6,9,…≡0 mod 3, so the player who is forced to sacrifice the first chain is p(n-c).

Sorry for the bad formatting. By those random signs I actually mean the congruence relation. :D

Note, that p(n-c) should be considered modulo 3 (kinda obvious).

Also note that loops have no impact on my argument:

Again we assume that a double-cross is made whenever possible except for the last chain/loop. Let l be the number of loops, we obtain:

z=n+c+2l-1, because taking a loop's two double-crosses or taking a chain's one double-cross costs one move (considering last chain/loop capture).
Also x=z+1-2c-2l, because no matter if chain or loop, it always “costs” two moves.
==>x=(n+c+2l-1)+1-2c-2l=n-c and the above result holds.