Negative of combinatorial game

Question

I am having problem understanding what negative of a combinatorial game $G$ exactly means in combinatorial game theory. Does it mean that if I have normal game, if I create inverse, i.e., $-G = \{-G^R \mid -G^L\}$ I get a Misere game?

Answer 1

No, the negative a game is simply the game in which the player's roles are swapped, hereditarily. You can see this in the definition you provided $$-G=\{ -G^R\mid -G^L\}$$ since the left options in $-G$ are the (negations of) the right options in $G$ and vice versa.

If we take the sum of a game $G$ and its negative $-G$, we get a game in which any move by any player can be copied in the other part of the game, which implies $G+(-G)=0$.

The negative of a game is not the same as the misère game, since in $-G$ it is not that you have the same options as before and are trying to lose, but rather that you have your opponent's options and are trying to win as though playing as them.

For example, the negative of a game of chess is still very much a game of chess—you are simply playing now as Black instead of White. But misère chess looks totally different in the manner of play, since players are exposing their queens and so forth in attempt to lose.

Answer 2

Others have already pointed out to you that the answer is negative. Let me add the following take: if you imagine a game as a bicolored tree (where the root is the starting position, and the edges are colored red or blue according as they are traversable by the right or left player, and where we play following the normal “who cannot play has lost” rule):

• To form the negative of a game, you simply exchange the colors of the edges.

• To form the misère of a game, you take each node which has no red (resp. blue) child (i.e., no red, resp. blue, edge leading away from the root) and you add a new red (resp. blue) edge from that node leading to a new terminal node¹, allowing the player who would have lost in the normal version of the game to declare victory in the misère form. That new node ensures that the opposite player has lost, thus “converting” the misère convention to the normal convention.

A concrete example where the two constructions differ is a typical impartial game (“impartial” meaning both players have identical options in any position): the negative of an impartial game will be the game itself, whereas the misère will typically differ. In the game of nim, for example, while the misère's version strategy is the same for most of the game, it differs at the end: clearly if there is only one row of sticks left, in the normal version the winning move is to take them whereas in the misère version you leave one, forcing your opponent to take it.

Another thing that must be pointed out is that the negative of a game is compatible with Conway equality (Conway equality of games $G\doteq H$ means that $G + (-H)$ is a second-player win): that is, if $G\doteq G'$ then $(-G) \doteq (-G')$, so “negative” is a well-defined construction on the game values. On the other hand, “misère” depends on the specific form of a game, it is not defined on the game values: that $G\doteq G'$ does not imply $\operatorname{mis\grave ere}(G) \doteq \operatorname{mis\grave ere}(G')$, as evidenced by the fact that the nim positions $1+1$ and $2+2$ are both second-player wins in normal play (so they have nim value $0$: they are Conway-equal) whereas in misère play the first is a first-player win and the second is a second-player win.

1. Maybe that node should have a red (resp. blue) edge leading back to itself? This wouldn't change the game itself, but it changes the way it gets added to other games. The misère construction isn't entirely well-specified (as far as I know).