GPT 6 Astra Ultra attempt on Econ-203
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Strategically centered zero-sum certificates for search-and-mix
Problem ID:C72-199. Status: unresolved.Target and outcome
The target is a fully specified polynomial-time search-and-mix algorithm with a fixed worst-case additive regret below one third for all rational bimatrix games in $[0,1]$. The survey [Survey, Section 3.4 and open-problem item 1] records the $1/3+\delta$ benchmark and the improvement question. We give a complete algorithm with a computable structural certificate that beats one third on a declared class, and a polynomial mixing optimization for fixed candidate-list sizes. Neither result yields a uniform sub-third guarantee over all games.For $x\in\Delta_m$, $y\in\Delta_n$, define \[ r(x,y)=\max\left\{\max_i e_i^TAy-x^TAy, \max_j x^TBe_j-x^TBy\right\}. \] The two terms are nonnegative because a best response is at least the current mixed payoff. All algorithms below operate on exactly encoded rational input.
Centering out strategically irrelevant terms
Let $M=A+B$. Solve the linear program \begin{equation}\label{wsref-0-63-65-6e-74-65-72-69-6e-67} \epsilon_* =\min_{c\in{\mathbb{R}}^m,d\in{\mathbb{R}}^n,\epsilon\ge0}\epsilon \quad\text{subject to}\quad |M_{ij}-c_i-d_j|\le\epsilon\quad\text{for all }i,j. \end{equation} An arbitrary constant can be transferred between $c$ and $d$, so impose $c_1=0$ as a gauge. Write $C_{ij}=M_{ij}-c_i-d_j$ for an optimal residual, and form \[ A'_{ij}=A_{ij}-d_j,\quad B'_{ij}=B_{ij}-c_i, \quad H=(A'-B')/2. \] Then $A'=H+C/2$ and $B'=-H+C/2$. Subtracting $d_j$ from the row player's payoff does not change a unilateral row deviation at a fixed opponent strategy. Subtracting $c_i$ from the column player's payoff likewise preserves all column deviations.Theorem (An exact structural regret certificate): The centering program and a Nash equilibrium $(x^H,y^H)$ of the zero-sum game $(H,-H)$ can be computed in polynomial bit complexity. Their output satisfies \[
r(x^H,y^H)\le\epsilon_*.
\] More precisely its two regrets are bounded respectively by \[
\frac12\left(\max_i e_i^TCy^H-(x^H)^TCy^H\right),\qquad
\frac12\left(\max_j(x^H)^TCe_j-(x^H)^TCy^H\right).
\] In particular, for any fixed $\eta\in(0,1/3)$, all inputs with $\epsilon_*\le1/3-\eta$ admit the requested sub-third output by this algorithm. When $M_{ij}=c_i+d_j$ exactly, the output is an exact Nash equilibrium of the original game.
Proof: The centering program has a feasible point with $\epsilon=1$: choose $c_i=0$ and $d_j=1$, since $0\le M_{ij}\le2$. Under the gauge and $\epsilon\le1$, the row $i=1$ constraints imply $d_j\in[-1,3]$. Then column $j=1$ implies $c_i\in[-4,4]$. Thus imposing these compact rational bounds loses no optimal solution. A rational optimum of polynomial bit length can be computed by a polynomial-time rational linear-programming algorithm. The zero-sum equilibrium is obtained from the primal and dual minimax linear programs for $H$, also rational and of polynomial size. Finite zero-sum minimax, equivalently linear-program duality, ensures their solutions satisfy \[
(e_i-x^H)^THy^H\le0,\qquad
(x^H)^T(-H)(e_j-y^H)\le0
\] for every pure deviation. This is the standard LP algorithm for finite zero-sum games; no real-algebraic output is required.
The use of the symmetric surrogate $H$ matters: approximating only one player's matrix by a zero-sum game loses the factor one half in the residual oscillation. Removing row and column shifts further avoids charging for payoff terms that never affect that player's deviations.Strategic invariance and the decomposition give \[ (e_i-x^H)^TAy^H =(e_i-x^H)^THy^H+\tfrac12(e_i-x^H)^TCy^H \le\tfrac12(e_i-x^H)^TCy^H. \] Take the maximum over $i$ for the first detailed certificate. The second follows identically from $B'=-H+C/2$. Every entry of $C$ is between $-\epsilon_*$ and $\epsilon_*$, so every displayed difference of its averages is at most $2\epsilon_*$. Dividing by two proves the uniform certificate. The exact and sub-third specializations follow immediately. ∎
A fully specified general search-and-mix template
The structural subroutine can be incorporated in a template that is defined on every input, not merely on the favorable class. Choose the first row $i_0$. Let $j_0$ be a pure best response of the column player to it, and let $i_1$ be a pure best response of the row player to $j_0$. Break ties by smallest index. Put \[ x^0=\tfrac12(e_{i_0}+e_{i_1}),\qquad y^0=e_{j_0}. \] Compute $(x^H,y^H)$ as above. The candidate lists are \[ X=(e_{i_0},e_{i_1},x^H),\qquad Y=(e_{j_0},y^H). \] Evaluate regrets exactly at $(x^0,y^0)$ and $(x^H,y^H)$ and return the better of the two, with a fixed tie rule. Each is a specified convex combination from these lists.Proposition (A global guarantee with a structural improvement): This deterministic template has polynomial bit complexity and, for every input, returns a profile with \[
r(x,y)\le\min\{1/2,\epsilon_*\}.
\]
Proof: At $y^0$, row $i_1$ is a best response, so the row regret of $x^0$ is $\tfrac12(A_{i_1j_0}-A_{i_0j_0})\le1/2$. For any column deviation $j$, \[
(x^0)^TB(e_j-e_{j_0})
=\tfrac12(B_{i_0j}-B_{i_0j_0})
+\tfrac12(B_{i_1j}-B_{i_1j_0})\le1/2,
\] since the first parenthesis is nonpositive by the choice of $j_0$ and the second is at most one. Thus the fallback has regret at most $1/2$. Theorem Theorem (An exact structural regret certificate) bounds the other profile. Exact comparison of the two computed regrets proves the bound. Best-response scans and rational linear programs have the claimed polynomial complexity. ∎
The fallback proof is the elementary best-response mixing construction, not an improvement on the survey's general $1/3+\delta$ algorithm. Its role here is to make the new certificate part of a complete, independently specified template without invoking an unexpanded subroutine.Optimizing mixing for constant-size candidate lists
The next result applies to arbitrary rational candidate lists, including the three-by-two lists above. Candidate generation and mixing optimization are separate issues in the target.Lemma (Regret continuity): For all probability vectors $x,x',y,y'$, \[
|r(x,y)-r(x',y')|\le2\left(\|x-x'\|_1+\|y-y'\|_1\right).
\]
Proof: For entries of $A$ in $[0,1]$, changing $y$ changes each pure-row payoff by at most $\|y-y'\|_1$, and hence changes their maximum by at most that amount. The current payoff changes by at most $\|x-x'\|_1+\|y-y'\|_1$, by adding and subtracting $(x')^TAy$. The row regret therefore changes by at most $\|x-x'\|_1+2\|y-y'\|_1$. The column argument interchanges the two coefficients. The absolute difference of maxima of two numbers is at most the larger absolute difference of the two entries, yielding the stated convenient bound. ∎
Theorem (Polynomial approximate mixing at fixed list size): Fix positive constants $s,t$. Given lists $X=(x^1,\ldots,x^s)$ and $Y=(y^1,\ldots,y^t)$ of rational strategies and rational $\tau>0$, one can compute mixtures with regret at most \[
\min_{\lambda\in\Delta_s,\mu\in\Delta_t}
r\left(\sum_a\lambda_ax^a,\sum_b\mu_by^b\right)+\tau
\] in time polynomial in input bit length and $1/\tau$. If specified additional mixture pairs are also evaluated, the output is no worse than any of them.
Proof: If $s=t=1$, there is only one mixture. Otherwise choose an integer $K\ge4(s+t-2)/\tau$ and enumerate probability vectors whose coordinates are integer multiples of $1/K$. For any $\lambda\in\Delta_s$, round its first $s-1$ coordinates downward and put the remaining probability in the last coordinate. The resulting $\widehat\lambda$ obeys $\|\lambda-\widehat\lambda\|_1\le2(s-1)/K$. The same procedure gives $\|\mu-\widehat\mu\|_1\le2(t-1)/K$. Taking convex combinations of probability vectors contracts this bound: for example \[
\left\|\sum_a(\lambda_a-\widehat\lambda_a)x^a\right\|_1
\le\sum_a|\lambda_a-\widehat\lambda_a|\|x^a\|_1
=\|\lambda-\widehat\lambda\|_1.
\] A minimizing pair exists by continuity on compact simplices. Lemma Lemma (Regret continuity) shows that a grid pair within the above rounding distances has regret at most its optimum plus $4(s+t-2)/K\le\tau$.
Applied to Proposition Proposition (A global guarantee with a structural improvement), including its two output candidates among the evaluated pairs retains the bound $\min\{1/2,\epsilon_*\}$ while approaching the best possible mixture from the chosen lists. The exponent in the grid size depends on $s+t$, so this is not a polynomial algorithm for lists of unrestricted length.There are $\binom{K+s-1}{s-1}\binom{K+t-1}{t-1}$ grid pairs. For fixed $s,t$ this is polynomial in $K$, hence in $1/\tau$. Their regrets are exactly evaluated by scanning the original pure actions and performing rational arithmetic of polynomial bit length. Return a minimum, including any extra specified pairs in the same comparison. This proves both assertions. ∎
Remaining universal gap
The centering residual is not uniformly below one third. For example, $A=B=\left(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right)$ has $\epsilon_*=1$: its alternating sum $M_{11}+M_{22}-M_{12}-M_{21}=4$ is unchanged by row/column shifts, while four residual entries of magnitude at most $\epsilon$ have alternating sum at most $4\epsilon$; $\epsilon=1$ is attainable. This game itself is easy, but it proves that the certificate alone cannot imply the desired universal constant.The unresolved task is a candidate-generation argument guaranteeing a sub-third mixture for every game, together with polynomial complexity. Theorem Theorem (Polynomial approximate mixing at fixed list size) addresses only fixed-list mixing once such candidates are given. The survey HTML was checked on 9 October 2026, including the $1/3+\delta$ running-time statement and the explicit improvement question. No numerical experiment or claim about the current best universal bound is needed for these restricted results.
References
- [Survey] H. Li, W. Huang, Z. Duan, D. H. Mguni, K. Shao, J. Wang, and X. Deng, A survey on algorithms for Nash equilibria in finite normal-form games, arXiv:2312.11063, 2023. Section 3.4 and "Open problems and future directions," item 1. https://arxiv.org/html/2312.11063.