Avoider-Enforcer game

The winning condition of an Avoider-Enforcer game is exactly the opposite of the winning condition of the Maker-Breaker game on the same ${\displaystyle {\mathcal {F}}}$. Thus, the Avoider-Enforcer game is the Misère game variant of the Maker-Breaker game. However, there are counter-intuitive differences between these game-types.

For example, consider the biased version of the games, in which the first player takes p elements each turn and the second player takes q elements each turn (in the standard version p=1 and q=1). Maker-Breaker games are bias-monotonic: taking more elements is always an advantage. Formally, if Maker wins the (p:q) Maker-Breaker game, then he also wins the (p+1:q) game and the (p:q-1) game. Avoider-Enforcer games are not bias-monotonic: taking more elements is not always a disadvantage. For example, consider a very simple Avoider-Enforcer game where the losing sets are {w,x} and {y,z}. Then, Avoider wins the (1:1) game, Enforcer wins the (1:2) game and Avoider wins the (2:2) game.

There is a monotone variant of the (p:q) Avoider-Enforcer game-rules, in which Avoider has to pick at least p elements each turn and Enforcer has to pick at least q elements each turn; this variant is bias-monotonic.[1]^: 45–46

Similarly to Maker-Breaker games, Avoider-Enforcer games too have fractional generalizations.

Suppose Avoider needs to avoid taking at least a fraction t of the elements in any winning-set (i.e., take at most 1-t of the elements in any set), and Enforcer needs to prevent this, i.e., Enforcer needs to take less than a fraction t of the elements in some winning-set. Define the constant: ${\displaystyle c_{t}:=(2t)^{t}\cdot (2-2t)^{1-t}=2\cdot t^{t}\cdot (1-t)^{1-t}}$(in the standard variant, ${\displaystyle t=1,c_{t}\to 2}$).

• If ${\displaystyle \sum _{E\in {\mathcal {F}}}{c_{t}}^{-|E|}<1}$, and the total number of elements is even, then Avoider has a winning strategy when playing first.[3]^{: thm A5}

Biased positional game#A winning condition for Avoider