Fixed Points, Nash Equilibria, and the Existential Theory of the Reals

Abstract

We introduce the complexity class $\exists \mathbb {R}$ based on the existential theory of the reals. We show that the definition of $\exists \mathbb {R}$ is robust in the sense that even the fragment of the theory expressing solvability of systems of strict polynomial inequalities leads to the same complexity class. Several natural and well-known problems turn out to be complete for $\exists \mathbb {R}$; here we show that the complexity of decision variants of fixed-point problems, including Nash equilibria, are complete for this class, complementing work by Etessami and Yannakakis [13].

Appendix A: Proof of Corollary 3.4

Let $S = \{x \in \mathbb {R}^{n}: \varphi (x)\}$ be the semi-algebraic set defined by φ of complexity at most L. If S = ∅, there is nothing to show, so we can assume that S≠∅. By Lemma 3.1 there is a conjunction $\psi = \bigwedge _{i=1}^{\ell } s_{i} \triangle _{i} t_{i}$ with △ _i ∈{<,=} so that $S^{\prime } = \{x \in \mathbb {R}^{n}: \psi (x)\}$ is a non-empty subset of S and |ψ|≤|φ|≤L. In particular, ℓ < L. Now $S^{\prime } = \{x \in \mathbb {R}^{n}: s_{i} - t_{i} \triangle _{i} 0\}$. Let f _i : = s _i −t _i , and s = ℓ. Then s < L, each f _i has degree d at most L−2 (we need two symbols for △ _i and 0), and the bitlength of each coefficient of f _i is bounded by L−1, so τ < L (these are wildly generous bounds). Hence, we can apply Theorem 3.1 to conclude that S ^′, and therefore S contains a point at distance at most R from the origin if it is non-empty. We are left with the estimate of R. Let us first simplify the expression for R:

$$\begin{array}{@{}rcl@{}} R & \leq & ((4DN^{2}) 2^{2N(\tau^{\prime}+b(2N)+b(2DN+1))})^{1/2} \\ & \leq & 2DN 2^{N(\tau^{\prime}+b(2N)+b(2DN+1))} \\ & \leq & 2^{b(N) + b(2D) + N(\tau^{\prime}+b(2N)+b(2DN+1))}. \end{array} $$

We know that d ^′<2L (using L≥4); then D ≤ n d ^′<2n L, and 2D+1≤4n L. With this b(2D+1)≤b(4n L)≤3+ log(n L)≤n L+3 (we’re using b(x)≤ log(x)+1). Now N≤(d ^′)ⁿ≤(2L)ⁿ, so b(N) = b((2L)ⁿ)≤1 + n log(2L)≤n L+1 (for L≥4), and b(2N)≤n L+2. We can now evaluate the τ-values: τ ₀≤2L+(n+1)b(L) + b(4L)≤2L+(n+2) logL+(n+4)≤5n L (for L≥4); with that, τ ₁≤D(τ ₀+4b(2D+1) + b(N))≤2n L(5n L+4(n L+3)+(n L+1))≤27n ² L ², and τ ₂≤τ ₁+2n(b(N) + b(n))≤τ ₁+2n ² L ²≤29n ² L ², τ ^′≤N(τ ₂+(n L+1)+2(n L+3)+1)≤(2L)ⁿ(31n ² L ²+8)≤(2L)ⁿ32n ² L ²≤32n ² L ³ⁿ.This allows us to evaluate the expression τ ^′ + b(2N) + b(2D N+1)≤τ ^′+1+2b(N) + b(2D+1)≤32n ² L ³ⁿ+3n L+5≤35n ² L ³ⁿ. Finally,

$$\begin{array}{@{}rcl@{}} R & \leq & 2^{b(N) + b(2D) + N(\tau^{\prime}+b(2N)+b(2DN+1))} \\ & \leq & 2^{(nL+1) + (nL+3) + N(35n^{2}L^{3n})} \\ & \leq & 2^{(2nL+4) + (2L)^{n} (35n^{2}L^{3n})} \\ & \leq & 2^{35n^{2} L^{5n}} \\ & \leq & 2^{L^{8n}}, \end{array} $$

which is what we had to show.