I have a function: \begin{equation*} f(a_1,\ldots,a_7,b_1,\ldots,b_4)=-\tfrac12 a_1 + a_1 b_1+\tfrac12 a_1 b_2-\tfrac12 a_1 a_2 b_2 + 2 a_1 a_2 - \tfrac32 a_1 a_2 b_1 - \tfrac12 a_4 + a_4 b_1 + \tfrac12 a_4 a_5 + \tfrac12 a_4 a_5 b_2 - a_4 a_5 b_1 + \tfrac32 a_6 - \tfrac12 a_6 b_3 + \tfrac12 a_6 b_4 + \tfrac32 a_7 - \tfrac32 a_7 b_3 - \tfrac12 a_7 b_4 \end{equation*}
$$\begin{array}{cc}
\forall i: 0 \le a_i \le 1, & \forall j: 0 \le b_j \le 1
\end{array}$$
Two players are playing a game, where player $A$ is trying to maximize the function by picking $a_i$, and player $B$ is trying to minimize it by picking $b_j$. What's the Nash equilibrium?
What's the most efficient way to compute this (e.g., in Mathematica or something else), especially when you have a lot more variables?
Revised answer:
The restriction that $\forall i: 0\le a_i,b_i \le 1$ limits things rather severely. Player $B$ will always choose $0$ for all positive-coefficient terms (this will be best for $B$ regardless of what $A$ may do). Player $A$ will always choose $0$ for all negative-coefficient terms (this will be best for $A$ regardless of what $B$ may do). So the function will always have a value of $0$ in a Nash equilibrium. There are, however, infinitely many equilibria, as it doesn't matter what $A$ chooses for positive-coefficient terms or what $B$ chooses for negative-coefficient ones.