The Complexity of Two-Team Polymatrix Games with Independent Adversaries

TL;DR

This work proves hardness of finding any stationary point in the simplest type of non-convex-concave min-max constrained optimization problem, namely for a class of bilinear polynomial objective functions.

Abstract

Adversarial multiplayer games are an important object of study in multiagent learning. In particular, polymatrix zero-sum games are a multiplayer setting where Nash equilibria are known to be efficiently computable. Towards understanding the limits of tractability in polymatrix games, we study the computation of Nash equilibria in such games where each pair of players plays either a zero-sum or a coordination game. We are particularly interested in the setting where players can be grouped into a small number of teams of identical interest. While the three-team version of the problem is known to be PPAD-complete, the complexity for two teams has remained open. Our main contribution is to prove that the two-team version remains hard, namely it is CLS-hard. Furthermore, we show that this lower bound is tight for the setting where one of the teams consists of multiple independent adversaries. On the way to obtaining our main result, we prove hardness of finding any stationary point in the simplest type of non-convex-concave min-max constrained optimization problem, namely for a class of bilinear polynomial objective functions.

Figures (2)

• Figure 1: Classes of total search problems. Arrows are used to denote containment. For example, $\CLS$ is contained in $\PLS$ and in $\PPAD$. The class $\TFNP$ contains all total search problems, i.e., problems which are guaranteed to have efficiently checkable solutions. $\P$ contains all such problems solvable in polynomial time.
• Figure 2: The intuition behind exact ($\varepsilon = 0$) KKT points for minmax problems. The formulation features a min-variable $x$, a max-variable $y$ and the bounding box constraint $(x, y) \in [0, 1]^2$. Some feasible points together with their respective gradient $(q^x, q^y)$ are depicted. Green points are valid KKT points whereas red ones are not. For instance, $(x_1, y_1)$ is not a KKT point because $q_1^y > 0$ but $y < 1$. On the other hand, $(x_2, y_2)$ is a valid KKT point because $q^y_2 = 0$ and $q^x_2 < 0$ with $x = 1$.

Theorems & Definitions (15)

• Theorem 1.1: Informal
• Definition 1
• Definition 2: Two-team Polymatrix Zero-Sum Game
• Definition 3
• Definition 4
• Theorem 3.1: Precise formulation of \ref{['theorem:CLS-hardness']}
• Lemma 3.1
• Claim 3.1
• proof
• Claim 3.2