By the way, we could also consider the case where you can play one or two moves, i.e. where you can pass on the second one... Cris On Dec 30, 2014, at 9:35 PM, Cris Moore <moore@santafe.edu> wrote:
Right. Both 0 and 1 are losses for the first player. But clearly these are not equivalent, since 0+1 is a loss for the first player, while 1+1 is a win.
Clearly we need distinguish running out of moves on your first turn (like 0) from running out of moves on your second turn (like 1). These are two distinct ways to lose, which should be regarded as different values.
Cris
On Dec 30, 2014, at 9:25 PM, James Propp <jamespropp@gmail.com> wrote:
Let's begin by noting that for double-moves, 1+1+1 (a sum of three Nim heaps of size 1) is not equal to 1 (a heap of size 1): the former is a second-player win while the latter is a first-player win.
Let's say that G is equivalent to H (maybe that's not the standard word to use in this context; I forget) if G+X has the same outcome (in terms of whether it's a first-player win or a second-player win) as H+X, for all X.
My guess is that for double-move play, Nim has small equivalence classes (many of them singletons), as opposed to the usual Nim theory, which has big congruence classes (corresponding to the nimbers).
Jim Propp
On Tuesday, December 30, 2014, Cris Moore <moore@santafe.edu> wrote:
So is there an interesting theory of adding games/numbers with double moves? or k moves for general k?
On Dec 30, 2014, at 8:44 PM, James Propp <jamespropp@gmail.com <javascript:;>> wrote:
That's because the induction arguments that make the theory work are sensitively tuned to the fact that when you make a move in a disjunctive sum, you make a move in exactly ONE of the summands --- whereas, when you are faced with the game G+H and you want to make a double-move, you can make a double-move in G, a double-move in H, or a single move in each of the two sub-games.
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